Penney Perspectives in Mathematics 1972

PERSPECTIVESIN

MATHEMATICS

DAVID E. PENNEY, The University of Georgia

!~~t~A~~~ ~ W. A. BENJAMIN, INC.~ ~

~{IS't\~ Menlo Park, California

Copyright © 1972 by W. A. Benjamin, Inc. Philippines copyright 1972 by W. A. Benjamin,Inc.

All rights reserved. No part of this publication may be reproduced, stored in a retrievalsystem, or transmitted, in any form or by any means, electronic, mechanical, photocopying,recording, or otherwise, without the prior written permission of the publisher. Printed inthe United States of America. Published simultaneously in Canada. Library of CongressCatalog Card No. 70-166540.

For B. J. Pettis

Harry H. CorsonA. C. Woods

Paul S. Mostert

G. O. SabidussiPaul F. Conrad

Fred B. Wright

A. D. WallaceA. H. Clifford

G. S. Young, Jr.

Frank D. Quigleyand especially L. B. Treybig

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PREFACE

This book was planned primarily as a text for a one year course in mathematics for students not intending to take calculus. However, it could beused as a sourcebook by teachers and mathematicians or as outside readingby undergraduate mathematics majors. The usual topics from collegiateprecalculus courses have specifically been excluded, as has calculus itself.Each chapter contains a rather detailed study of a topic chosen from one ofthe major branches of modern mathematics. Although some of this materialis available elsewhere, it has not been gathered before into a single volumewritten for the reader who does not have an extensive background inmathematics. I quite frankly admit to choosing topics I found particularlyinteresting, and I hope that the reader will be pleased with some of thesechoices. The problems at the end of each chapter are in many cases meantto open up avenues of deeper exploration of the subject of the chapter. Thechapters themselves are almost wholly independent of one another, thusenabling the student to learn about those topics most appropriate for him,in any order, and in the necessary degree of mathematical rigor. This shouldprovide the flexibility desirable for so diverse an audience-even undergraduate mathematics majors are not commonly exposed to the majority ofthe material included, and I hope that they too might enjoy this book.

Although the formal prerequisites for one who wishes to profit from thisbook are minimal-some high-school algebra, a little geometry, and aptitudeto do college-level work-it is not an easy book. It contains only incidentalreferences to the history of mathematics. It should acquaint the reader withthe various branches of modern mathematics, and each student is expectedto find out what mathematics is by doing mathematics: guessing patterns,making conjectures, and (most important) proving theorems.

ix

x Preface

Most technical terms are defined upon the occasion of their first appearance, and in this case the term appears in boldface. The statements oftheorems appear in italics. The reader is advised to have paper and pencilalways available, and should not hesitate to consult the index. Generally,the last problems ofeach chapter are the most difficult, but there are numerousexceptions; the difficult problems have not been so indicated in order toincrease the similarity to mathematical research.

I wish to thank for their patience those undergraduates who were exposedto this book in its preliminary form, the Department of Mathematics of theUniversity of Georgia for making it possible for me to write this book, andmy former teachers of mathematics, to whom it is dedicated. Special thanksare due Professor Gail S. Young of the University of Rochester and ProfessorFrank D. Quigley of Tulane University for their help with the secondchapter, Professor F. A. Roach of the University of Houston for his helpwith the third, ProfessorH. S. M. Coxeter of the University of Toronto forhis help with Chapter 5, and Professor D. B. Hinton of the University ofTennessee for his help with Chapter 8. Finally, lowe a great deal to my wifeCarol, who read every word, suggested innumerable improvements, andinvented a number of the better problems.

Athens, GeorgiaNovember 1971

D.E.P.

CONTENTS

Chapter 1 The Bolyai-Gerwin Theorem 1

1.1 The fundamental definitions 21.2 The Bolyai-Gerwin theorem 61.3 How to cut a polygon into triangles 81.4 How to cut a triangle to make a parallelogram 111.5 How to cut a parallelogram to make a rectangle 111.6 How to cut a rectangle to make a square 121.7 How to cut several squares and reassemble the pieces to form a

single square . 141.8 How to cut a square into pieces which can be reassembled to

form a given polygon of the same area 161.9 Some concluding remarks 17

Chapter 2 Brunnian Links 25

2.1 The simple closed curve 272.2 Links and their properties 302.3 A simple algebra. 332.4 Brunnian 4-links. 392.5 Notation and examples. 432.6 Commutators 442.7 The final generalization. 46

Chapter 3 The Well-Tempered Clavichord

3.1 Properties of logarithms3.2 A peculiar manipulation3.3 Continued fractions.

xi

52

535759

xii Contents

3.4 The value of a continued fraction .3.5 Applications to baseball and grade distributions3.6 Harmony.3.7 Tuning a piano, old style3.8 Tuning a piano, new style3.9 Improving the octave

Chapter 4 Group Theory

4.1 Some examples of groups4.2 Subgroups4.3 Cyclic groups and Abelian groups.

626670737678

82

8295

102

Chapter 5 Polyhedra 108

5.1 The definition of polyhedron 1085.2 Euler's formula 1185.3 Regular solids 1235.4 A converse of Euler's formula . 1275.5 Map coloring 134

Chapter 6 Infinite Sets 145

6.1 Sets 1466.2 Functions. 1546.3 More on one-to-one correspondences. 162

Contents xiii

6.4 The Cantor-Schroeder-Bernstein theorem 1646.5 Properties of finite and infinite sets 1706.6 Nondenumerable infinite sets 177

Chapter 7 Number Theory 184

7.1 Divisibility 1847.2 Well-ordering 1897.3 The fundamental theorem of arithmetic 1967.4 The greatest common divisor 2057.5 Applications . 210

Chapter 8 Animal Populations 220

8.1 Unrestricted growth of a single species 2218.2 Growth of a single species under limiting conditions 2298.3 The case of two competing species 2338.4 The predator-prey case . 244

Chapter 9 The Art Gallery Theorem 254

9.1 Convex sets 2549.2 Intersections of convex sets 2589.3 Hulls and kernels 2609.4 Helly's theorem . 2669.5 Krasnoselskii's theorem. 2719.6 L-convexity 276

xiv Contents

Chapter 10 The Real Number System 280

10.1 The rational numbers 28010.2 Nested intervals of rational numbers 28810.3 Construction of the real numbers . 29510.4 The order relation on R 30210.5 Are there more numbers? 30710.6 An unusual set of real numbers 311

Epilogue 316

Answers and Hints 320

Index 345

CHAPTER 1

THEBOLYAI-GERWIN

THEOREM

You have probably seen some of the numerous puzzles and games involvinga small number of plastic pieces of various shapes. In one variety of these,an accompanying booklet gives outlines of various figures, and one attemptsto form the pieces into these figures. Also enjoying some popularity areboard games in which one uses the varied shapes of his pieces to force orprevent certain moves by his opponent. It should not surprise you to findthat many mathematicians like such puzzles, and moreover that some of themathematical ideas behind such puzzles have been studied by mathematiciansfor many years.

The puzzles usually feature one extra problem when one tires of constructing figures; for example, the plastic pieces come in a flat rectangularbox, and we find it surprisingly difficult to return the pieces properly to thebox. This chapter will reveal, among other things, a method by which youcan construct such puzzles of your own, even ones which will fit into arectangular box, if you wish. You will be able to cut several plastic pieceswhich will fit into the box, but which you can reassemble to form a triangle,a hexagon, a symmetrical five-pointed star-or all three.

Of course, one restriction immediately becomes apparent. If one suchfigure can be cut up and the pieces reassembled to form another figure, thetwo must have the same area. On the other hand, it is hard to believe that asquare could be cut into a finite number of pieces which could be reassembledto form a circular disk of the same area, but we cannot answer whether ornot this is possible. This question presently enjoys the status of an UnsolvedProblem in Mathematics. (Do not confuse this with the famous problem ofSquaring the Circle-the distinction between the two will be discussed in the

1

2 The Bolyai-Gerwin Theorem 1.1

exercises.) Because circles, curved cuts, and curved figures in general leadus beyond the frontiers of current mathematical knowledge, we shall restrictour attention to polygons and straight-line cuts. In these cases the mathematical problems involved have been solved, and the solution is remarkablysimple. All that is required is that the two figures have the same area. Ifso, then either may be cut into pieces and reassembled to form the other.This is the essence of the Bolyai-Gerwin Theorem.

1.1 THE FUNDAMENTAL DEFINITIONS

We plan to lead you through a proof of the Bolyai-Gerwin Theorem for tworeasons. First, it is in some sense a very "rich" proof in that the proof itselfanswers numerous questions which may already have occurred to you.Second, the proof is a good model of mathematical thinking, and you shoulddevelop the habit of mathematical thinking in order to profit from this book.Have no fear, though, that we are trying to turn you into a mathematicianwe hope to convince you that mathematical thinking is not at all mysterious,but merely ordinary careful reasoning. A little geometry will be helpfulto you but not essential. Your intuitive idea of the area of a plane figure willserve, and we need but two definitions. Before the first definition, a smallbit of notation will be introduced for clarity. If L is a straight-line segmentin the two-dimensional plane, we will write L = [a, b] to indicate that a isone endpoint of L, and b is the other. The interval notation is used to remindus of what is in fact the case, that [a, b] consists of the two points a and btogether with all the points between them on the straight line through a andb. And we call a the first point of Land b the last point of L.

A polygon is a plane figure of finite area bounded by a finite number ofstraight line segments L b L 2 , L 3 , .•• , L", such that the last point of L 1 isthe first point of L 2 , the last point of L 2 is the first point of L 3 , and so on,and the last point of L" is the first point of L 1 ; and moreover, other than asindicated above, no two of these line segments have any points in common.These line segments are called the edges of the polygon, and the endpointsof these line segments are called the vertices of the polygon. The pointsbelonging to the edges of a polygon are to be considered part of the polygon;that is, the polygon consists of all its interior points together with all itsboundary points. Consequently, when we speak of a point x belonging to thepolygon P, it is permissible that x be either an interior point or a boundarypoint of P.

We require that the figure have finite area to avoid certain paradoxes(to be discussed in the exercises) and also so that we may attach the commonmeanings to many terms such as "square." We certainly want to mean by"square" a square boundary together with what's inside, and the requirement

1.1 The fundamental definitions 3

that a polygon have finite area prevents us from the necessity of dealing witha square boundary together with what's outside.

Two polygons are said to be equidecomposable if it is possible to cut oneup, using straight cuts, into a finite number of pieces which may be reassembled, omitting none, to form the other.

You may object that this definition is somewhat vague. But an attemptto make it much more precise would also make it several times as long, andrequire the introduction of several other terms which would also have to bedefined precisely. Again, as with the idea of area, we ask that you acceptyour natural understanding of the concept of equidecomposability as thecorrect one. At this point you may test your concept of area and yourunderstanding of equidecomposability by means of the following theorem.

Theorem 1.1 If two polygons are equidecomposable, then they have the samearea.

Ifyou feel that this proposition is obvious, then you surely understand enoughto proceed. After all, whatever we may mean by "area," the area of a planefigure should not be altered by rigid (congruence) motions-so all that wouldbe necessary for a proof of the above theorem is to cut the first polygoninto several pieces and measure the area of each such piece. Then observethat as the pieces are assembled into each of the two given polygons theresulting total areas must be equal, for each is the sum of the same set ofnumbers. But on a deeper level there are always a few "obvious" thingswhich become less obvious when more closely examined. If we were to startwith certain axioms about the geometry of the two-dimensional plane andattempt to define "area," the development of this definition would be lengthy,and the above theorem would have a rather complicated proof. For example,it is conceivable that some of the pieces into which a polygon might be cutsimply cannot have area. It would be necessary to establish that if onlystraight cuts are used to form such pieces then each piece must in fact havearea, and then that the total area of the polygon is equal to the sum of theareas of the individual pieces. Further consideration of these topics will befound in the exercises.

Exercises

1.1 Which of the figures shown in Fig. 1.1 is a polygon? Exactly in whatway does each that is not a polygon fail to satisfy the definition?1.2 What is the smallest number of vertices that a polygon can have? Provethat your answer is correct.1.3 Construct a polygon having one and only one vertex in each of thefour quadrants and not containing the origin-or prove that no such polygoncan exist.


(a) (b)

(e)

Fig.1.1 Which of the above figures is a polygon 7

1.1 The fundamental definitions 5

1.4 Need a polygon have positive, rather than zero, area? Why?

1.5 Show that a polygon must have at least one interior angle less than 1800•

1.6 The relation of equality between numbers enjoys the following threeproperties.

a) The Reflexive Property: If x is a number, then x = x.

b) The Symmetric Property: If x and yare numbers such that x = y, theny = x.

c) The Transitive Property: If x, y, and z are numbers such that x = yand y = z, then x = z.

Such a relation is said to be an equivalence relation.

If P and Q are polygons, let us write P '" Q provided that P and Q areequidecomposable; that is, if it is possible to cut P up, using straight cuts,into a finite number of pieces which can be reassembled, omitting none, toform Q. Show that equidecomposability is an equivalence relation:specifically, if P, Q, and R are polygons, then

a) P '" P.

b) If P '" Q, then Q '" P.

c) If P '" Q and Q '" R, then P '" R.

1.7 Two polygons are said to be nonoverlapping if they have, at most,boundary segments in common. In particular, polygons with only one point,or nothing, in common are nonoverlapping. We take for granted in ourdefinition and subsequent discussion of equidecomposability that the piecesone assembles to form a given polygon are to be nonoverlapping, but thatthey may have common edges.

Show that a triangle is not the union of nonoverlapping parallelogramseach pair of which are congruent. Is this still true if we drop the restrictionthat each two parallelograms are congruent, and require only that finitelymany parallelograms be used?

1.8 Plane figures not of finite area can exhibit surprising behavior, even ifthe boundaries are required to be straight-line segments meeting in commonendpoints, much as in the definition of a polygon. We might as well allowthe use of half-infinite lines as well, with only one endpoint, that is also anendpoint of another such half-infinite line or line segment. Here, unfortunately, the notion of geometric congruence breaks down rather badly.It turns out that there is an example of such a figure which is congruent toa proper part of itself. Thus there can be no way to define "area" for suchfigures so that congruence motions would preserve area, or so that proportional figures with the same area would be congruent.


Find an example of such a plane figure with polygonal boundary congruent to a proper part of itself. (The figure must be unbounded-that is,contained in no circle, no matter how large the radius of the circle.)

1.9 In the terminology of the previous exercise, find an example of an unbounded plane figure with polygonal boundary such that the figure has finitearea. It is permissible to use infinitely many segments to form the boundary.

1.10 It is also very easy to find an example of a plane figure that has a finiteperimeter but infinite area. Please do so.

1.2 THE BOlYAI-GERWIN THEOREM

The Bolyai-Gerwin Theorem is just the converse of Theorem 1.1. The beautyof this theorem lies in the extremely simple condition for equidecomposabilityof two polygons and the curious fact that the theorem is unexpectedly true.

Theorem 1.2 (Bolyai-Gerwin) If two polygons have the same area then theyare equidecomposable.

The proof has its own beauty: it is constructive. The statement of the theoremwould be of little use to you should you wish to construct such puzzles aswere mentioned earlier, but the proof provides a "how to do it" recipe. Ifthe proof were not broken into a sequence of short steps, it would be easy tolose the thread of the argument, so we shall present the proof in the followingsequence. We shall show, in order, how to cut

1) a polygon into triangles,

2) a triangle to make a parallelogram,

3) a parallelogram to make a rectangle,

4) a rectangle to make a square,

5) several squares to form a single square when the pieces are reassembled,and

6) a square into pieces which can be reassembled to form a given polygon ofthe same area.

Given two polygons of the same area, we can cut one into pieces whichcan be assembled into a square, using the first five steps above. We can dothe same with the other polygon and assemble both into squares. The twosquares will have the same area, and so can be superimposed. If all cuts nowvisible are made, then either square can be made into either polygon. Hencethe first polygon can be made into the second, with one of the squares as ahalfway point.

Finally, before we begin our sequence of six short proofs, nute that ineach proof, each construction can be performed by so-called "pure" geometric methods-that is, with unmarked straightedge and compass. This

1.2 The Bolyai-Gerwin Theorem 7

Fig.1.2 The Texan Rectangle.

is not necessary for the truth of the Bolyai-Gerwin Theorem, but it doesserve as an added attraction and makes the construction of puzzles considerably simpler. Also note that in each proof all cuts are straight-linecuts, as required in the definition of equidecomposability, and finally that itis never necessary to lift a piece and turn it over.

Exercises

1.11 Let R be a rectangle with short side of length one unit and long side oflength two units, and let S be a square of the same area as R. Show how to cutR into a finite number of pieces, using straight cuts, so that the pieces can bereassembled to form S. Can you do this with only three cuts?

1.12 Show how to cut up the Texan Rectangle shown in Fig. 1.2, boundedby two parallel rays and a segment perpendicular to each, into infinitelymany squares that can be reassembled to form a strip twice as wide. Thisexample is one reason why we restrict our attention to figures of finite areawhen dealing with equidecomposability.

1.13 Show how to cut the Texan Rectangle into squares that can bereassembled to form the entire two-dimensional plane.

1.14 Show how a given triangle can be divided into two right triangles with asingle cut.

1.15 Let T be a right triangle. Show how you can cut T into two pieces whichcan be reassembled to form the mirror image of T, without turning any pieceover.

1.16 In the next section we will show how to cut any polygon into triangles.Thus the two previous exercises show how turning pieces over in our study ofequidecomposability could be avoided, even if thought necessary in someconstruction. Explain how.


1.17 What is a theorem? A corollary? A lemma? Look up the answers.What is the difference between a mathematical definition and the sort ofdefinition found in a dictionary?

1.18 How would you phrase the Bolyai-Gerwin Theorem for the onedimensional case-that is, for sets which are subsets of the real number line?In order to answer this question, you will have to decide on a one-dimensionalanalogue of "polygon," and define equidecomposability for a figure likethis.

1.19 Do you believe that the theorem you phrased in the previous exerciseis true?

1.20 See Exercise 1.6. Given real numbers a and b, let us say that a isequivalent to b, and write a ~ b, provided that a - b is a whole number.Is the relation ~ an equivalence relation? If so, prove it; if not, show whynot.

1.3 HOW TO CUT A POLYGON INTO TRIANGLES

It will be much more convenient if the mathematical objects with which wework have short and easily remembered names, so we begin by letting P bea plane polygon. Let its vertices be called Vi' V2' V3' ..• ,Vn• We shall alsosuppose that we have so named these vertices that the edges of P are the linesegments [Vh V2], [V2' v3], ... , [Vm Vi]. If P has only three vertices, it isalready a triangle and no further construction is necessary. If P has four ormore vertices, there must be at least one at which the interior angle is lessthan 1800 (why?). We may suppose that we have named this vertex Vi' andthus by the definition of polygon, Vi lies on the two edges [Vi' V2] and[vm Vi] of P.

Consider the line segment K joining V2 with Vn• Now K will not be anedge of P itself, but it may happen that except for the endpoints of K, K liesentirely within the interior of P. If so, we cut P along this new line segment,and thus obtain two new polygons. One is, of course, the triangle T whosevertices are Vi' V2' and Vm and the other is a new polygon P' with one vertexfewer than P. On the other hand, what if K does not lie entirely within P?In this case we can produce another new line segment L, also joining twovertices of P, and lying entirely within P, as follows.

Imagine a straight line M, parallel to the segment K, and passing throughthe vertex Vi' Then imagine this line M moving slowly toward K while remaining always parallel to it. Since K does not lie entirely within the polygonP, as M moves toward K it must eventually meet some part of the boundaryof P that lies between [v h V2] and [Vm Vi]. To be precise, M must intersectsome part of the boundary of P that lies within the triangle T whose verticesare Vi' V2' and Vn, as shown in Fig. 1.3. But when M first meets part of the

1.3 How to cut a polygon into triangles 9

Fig. 1.3 Triangulationof a polygon.

boundary in this fashion, during its journey from V l to K, then M must simultaneously meet at least one vertex Vi of P. Let L be the line segment fromVl to Vi. Then L lies, except for its endpoints, wholly within the interior of P.

Thus we can always produce some new line segment, either K or L,joining two vertices of P not previously joined by an edge of P, and lyingexcept for its endpoints wholly within the interior of P. If we cut P alongthis new line, we produce two new polygons with the following importantproperty: Each has fewer vertices than did P. We repeat this process on eachresulting polygon that has more than three vertices until the process is forcedto come to a halt by virtue of the fact that P has been cut entirely intotriangles. All the cuts are straight-line cuts, and each construction can beperformed with straightedge alone. The method is the "most efficientpossible" in that it does not require the creation of new vertices. Thus wehave shown how it is possible to cut a given polygon into triangles.

Exercises

1.21 Suppose that P is a polygon with vertices V h V2, V3, ... ,Vn• In theconstruction of this section, in which a polygon such as P is decomposed intotriangles, how many cuts must be made, and how many triangles will beobtained? The answer of course depends on the value of n, and should givethe correct value in particular for n = 3 and n = 4.

1.22 Suppose that f/ is a statement meaningful for natural numbers; that is,f/ might be the statement "Every natural number is composite." In order tobe meaningful it is not necessary that f/ be true, merely that for each naturalnumber f/ is either true or false. Let us consider the statement f/, as anexample, that "The sum of the first n natural numbers is n(n + 1)/2." Inthis case f/ happens to be always true, and for such statements there isfrequently the possibility of proving them true by the method of induction.

To prove by the method of induction that such a statement is true, wemust first establish that the statement is true for n = 1; and then, assumingthe truth of the statement for n = k, show that it follows that the statementis also true for n = k + 1. Consequently, since the statement is true for 1,it is then true for 2; then, since it is true for 2, it is also true for 3, and so on.Thus the statement must be true for each natural number. Use the method ofinduction to prove that your answer to the previous exercise is correct.

1.23 Continuing the previous exercise, let f/ be the statement

n(n + 1)1+2+3+"'+n= .

2

Prove f/ by the method of induction.

1.24 The method used in this section to prove that each polygon can betriangulated is actually a proof by induction in disguise. Explain how torephrase this proof so that the use of the method of induction becomesmore apparent.

1.25 In the proof of the last section, we showed how to cut up a polygoninto triangles without the introduction of new vertices. Do you think ananalogous process will work for three-dimensional polyhedral solids? Inother words, can each polyhedral solid be cut up into tetrahedra (not necessarily regular) without the introduction of new vertices?

1.26 See Exercise 1.22. Prove by induction: If0 < a < 1 and n is a naturalnumber, then 0 < an < 1. Of course, you may use anything you knowabout inequalities.

1.27 Prove by induction: If n is a natural number, then

(-It = 1

(_l)n+t = -1

if n IS even,

if n is odd.

1.5 How to cut a parallelogram to make a rectangle 11

1.28 Prove by induction: If n is a natural number, then 4n- 1 is evenly

divisible by 3. Note that if m is a natural number evenly divisible by 3, thenthere exists a natural number k such that m = 3k.

1.29 Give an example of a pair of rectangles of the same area such that it isnot possible to cut one into finitely many squares that can be reassembledto form the other.

1.30 Give an example of a rectangle that cannot be cut into finitely manysquares each two of which have the same area.

1.4 HOW TO CUT A TRIANGLE TO MAKE APARALLELOG RAM

This is much easier. In fact, you have probably already guessed how to do it.If not, one hint is that it can be done with only one cut.

Let T be a triangle and let a and b be midpoints of two of its sides. Joina with b by the straight-line segment [a, b], and cut T along [a, b]. This willproduce a small triangle and a trapezoid. Holding the trapezoid fixed,rotate the triangle half a turn about either a or b. The resulting figure will bea parallelogram. That it is indeed a parallelogram is the only fact notimmediately obvious. Establishing this fact is left for the next exercise.

Exercises

1.31 Establish, as indicated in the proof of this section, that the reassembledtriangle actually does form a parallelogram.

1.32 With what type of triangle will the method of this section produce arectangle rather than just a parallelogram?

1.5 HOW TO CUT A PARALLELOGRAM TOMAKE A RECTANGLE

This is almost as easy as the previous proof. It is possible to construct in theparallelogram (which we suppose is not already a rectangle) a line from onevertex where the interior angle exceeds 90° to the opposite side, such that thisline lies within the parallelogram and is perpendicular to the opposite side.Cut along this line; you will obtain a trapezoid and a triangle. If the triangleis moved without rotation to the opposite side of the trapezoid, the two willfit together to form a rectangle. Again, the only problem is to show that thenew figure is indeed a rectangle, and we leave this for the exercises.

Exercises

1.33 Show how to construct the perpendicular needed for the cut in theproof of this section.

12 The Bolyai-Gerwin Theorem

q

1.6

Fig.1.4 Finding thesolution of x2 = abo

1.34 Prove that the reassembled parallelogram in the proof of this sectionactually does form a rectangle.

1.35 Under what circumstances will the reassembled parallelogram in thelast proof form a square, rather than just a rectangle?

1.6 HOW TO CUT A RECTANGLE TO MAKE A SQUARE

There is a trick to this construction. We must first determine the size of thesquare, draw it, and then use it to determine where to make the cuts on therectangle. To draw the desired square, we use a method known to the ancientGreek geometers. Let the rectangle be called R, let its short side have lengtha, and let its long side have length b. (If R is already a square we are finished,and no construction is necessary.)

Draw a straight line of length a + b, bisect it, and draw a circle with thecenter at the point of bisection so that the segment of length a + b forms adiameter of that circle. Since this diameter has length a + b, there is a pointp on it which divides the diameter into two segments, one of length a and oneoflength b. Draw a perpendicular to the diameter at the pointp. See Fig. 1.4.

This perpendicular must meet the circle at some point q. It turns out thatthe length x of the segment [p, q] has the property that x 2 = ab (to be

1.6 How to cut a rectangle to make a square

s

13

Fig. 1.5 How to turna rectangle into a square

of the same area. R

established in an exercise), and hence a square of side length x will have thesame area as the rectangle R.

There will now be a small technical problem if the rectangle R is morethan four times as long as it is wide, but in an exercise we will ask you to showhow such a rectangle can be converted into one which is less than four timesas long as it is wide. Hence we may assume that R itself is in fact less thanfour times as long as it is wide.

Draw R and the square, which we shall call S, in such fashion that theyoverlap as shown in Fig. 1.5. Draw the extra line L also shown in the samefigure, and cut R along that portion of L which lies in R. Then cut R alongthe line segment M, that part of the right-hand side of S which lies below L.

Now R consists of several regions, numbered from 1 to 4 in the figure.S consists of the regions numbered 1, 3, 5, and 6. We reassemble the piecesof R to form S as follows:

• Leave piece 1 alone.

• Move piece 2 to the position of piece 5.

• Move the triangle composed of pieces 3 and 4 to the position of the trianglecomposed of pieces 3 and 6.


No rotations are used in these motions. All we need to do is show thattriangle 2 is congruent to triangle 5 and that the triangle composed of pieces3 and 4 is congruent to the triangle composed of pieces 3 and 6. Again, theproof is left for the exercises.

We wanted the rectangle R to be less than four times as long as wide,to ensure that the line L will angle down steeply enough to guarantee theexistence of triangle 3. In any case, we have shown that with two straightcuts a rectangle can be reassembled into a square.

Exercises

1.36 In Section 1.6, an important preliminary to the construction is to showhow to construct a line segment of length x such that x 2 = ab (see Fig. 1.4).It is shown in many courses in plane geometry that if the lines from q to theendpoints of the diameter are drawn, then the angle formed at q is a rightangle. Use this fact to show that x 2 = abo

1.37 In the proof of this section, it was necessary to use a rectangle less thanfour times as long as wide. Suppose that R is a rectangle more than fourtimes as long as wide. Show how to convert R by straight line cuts into arectangle of equal area less than four times as long as wide. What if R isexactly four times as long as wide?

1.38 The construction in this section states that the rectangle R has to beless than four times as long as it is wide in order to ensure the existence oftriangle 3 (see Fig. 1.5). Establish this by showing the following to be true:If the rectangle R is less than four times as long as it is wide, then the line Langles down steeply enough to guarantee the existence of triangle 3, and nototherwise.

1.39 The last step in the proof of this section is to show that triangle 2 iscongruent to triangle 5, and that the triangle composed of pieces 3 and 4 iscongruent to the triangle composed of pieces 3 and 6. Please do so.

1.7 HOW TO CUT SEVERAL SQUARES ANDREASSEMBLE THE PIECES TO FORM A SINGLE SQUARE

Actually all we will show is how to assemble two squares 8 1 and 82 into asingle square 8. If there should happen to be a third square 8 3 then we wouldrepeat the process with 8 and 8 3 to obtain a single square, and so on.

Let 8 1 and 82 be two squares. If they should happen to have the samearea, then there is an extremely easy way to cut and reassemble them into asingle square, a way left for you to discover. So let us suppose that the areaof 8 1 exceeds the area of 82 . As in Section 1.6, a preliminary construction isneeded to determine where the cuts should be made in 8 1 and 82 in order that

1.7 Cutting squares and reassembling as one square 15

A 'Y

/ ..........,9 T/ " S

5 S210I ...................

/ "/ " b/

.........................../ 8/

I "/ 4 ' ..../ " ....

[- f-/ 8~/

3 I /I /

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Fig. 1.6 How toturn two squares

into a single square.

the resulting pieces can be reassembled to form the square 8. Draw 8 1 and82 as shown in Fig. 1.6, as well as the auxiliary square A, and let 8 1 and 8 2

have side lengths a and b, respectively.Note in the figure that each of the two points y and {) is located at distance

b from the next vertex of A counterclockwise around A. Locate the twopoints rL and psimilarly. Draw the segments [rL, P], [P, y], [y, {)], and [{), rL],shown as dashed lines in the figure. These segments form the boundaryof the desired square 8. Cut 8 1 and 82 along the parts of the boundary of 8which lie in each. Now 8 1 consists of three pieces numbered 1, 2, and 3;82 consists of two pieces marked 4 and 5; and 8 itself must be formed byrearranging these pieces so as to cover the regions numbered 1, 4, 6, and 8.We may ignore the pieces numbered 7, 9, and 10, since they do not enter theconstruction-they are merely artifacts introduced to help discover whereto make the cuts, and are not part of either 8 1, 82 , or 8. Here is the recipefor reassembling the pieces of 8 1 and 82 to form 8:

• Leave piece 1 alone.

• Remove piece 4 and leave it on the side for a while.


• Pick piece 2 up and move it, without rotation, so that it covers region 4and part of region 8; the point marked y goes to position ~.

• Rotate piece 5 counterclockwise 90° about the point ~, thus covering thetop half of region 6.

• Use piece 3 to cover the rest of region 6; the point marked {3 on piece 3goes to position y.

• Find piece 4 (left on the side for a while) and use it to cover the rest ofregion 8; the point marked ~ on piece 4 goes to position IX.

A little easy geometry involving lengths and right angles can be used toshow that the pieces actually fit correctly and do form a square. We chosethis particular construction because Fig. 1.6 can also be used to prove thePythagorean Theorem. There are other proofs of the Pythagorean Theorem,just as there are other proofs, using different figures, that two squares can beassembled into a single square, but we prefer this one because it does twojobs instead of one.

Exercises

1.40 Show that the polygon 8 bounded by the segments [IX, {3], [{3, y],[y, b], and [~, IX] is in fact a square. See Fig. 1.6.

1.41 Show that the reassembly in which the squares 8 1 and 82 are cut intopieces which form 8 actually works-that is, that the implied congruencesare actual.

1.42 The Pythagorean Theorem states that if a right triangle has legs oflength a and b, respectively, and hypotenuse of length c, then a2 + b2 = c2

•

Show how ideas in the proof in this section and use of Fig. 1.6 can be usedto establish this theorem.

1.8 HOW TO CUT A SQUARE INTO PIECES WHICH CANBE REASSEMBLED TO FORM A GIVEN POLYGON OFTHE SAME AREA

We discussed this procedure immediately after we listed in Section 1.2 thesix steps in the proof of the Bolyai-Gerwin Theorem. What we do, of course,is make the polygon into a square, using the first five steps. Clearly, then,the actual process of assembling the pieces of the polygon to form a squaremay be reversed. Or if you prefer, Exercise 1.6 has some bearing on thisproblem, and can be used to show the existence of a method for cutting up asquare into pieces which can be reassembled to form a given polygon of thesame area.

1.9 Some concluding remarks 17

To summarize, suppose that P and Q are two polygons of the same area.Chop each into triangles. Make each triangle into a parallelogram, eachparallelogram into a rectangle, and each rectangle into a square. Assembleall the squares obtained from P into one giant square S, and all the squaresobtained from Q into one giant square T. Then Sand T will have the samearea, since each has the area common to P and Q. If you draw T togetherwith its cuts as a transparent overlay and place this overlay upon S, you willsee where to make additional cuts on S. Using these cuts, we can reversethe construction of T from Q, using these new pieces of S and so reassembleS to form Q. This maneuver shows that P and Q are equidecomposable, andconcludes the proof of the Bolyai-Gerwin Theorem.

Exercise

1.43 Exactly where in the proof of the Bolyai-Gerwin Theorem did we usethe hypothesis that the polygons P and Q have the same area?

1.9 SOME CONCLUDING REMARKS

We have derived a great deal of pleasure from the actual construction ofpuzzles, using the techniques of the proof of the Bolyai-Gerwin Theorem.If you think you too would enjoy such constructions, here are two pieces ofadvice. One is aesthetic; one pertains to mechanical details.

For the aesthetic advice, an important point to remember is that thepuzzles will be unsatisfactory if some of the pieces are too small or if there isa great disparity in their relative sizes. Constructions proposed for puzzlesshould be drawn carefully first, perhaps using cardboard, to verify that thereare no tiny pieces. In addition, there is a code of honor among puzzlemakers to the effect that no unnecessary cuts should be made merely for thepurpose of confusing the puzzle-worker. Since the only practical way toapply the proof of the theorem to the invention of a puzzle is to turn somepleasing polygon into a square, rather than the reverse, this is indeed the rightapproach; but variations in where the cuts are made should be tried notonly to avoid very tiny pieces, but also to use the smallest possible numberof pieces in the puzzle. It is quite difficult to work such a puzzle with morethan fifteen pieces, and tremendously difficult to work one with more thantwenty-five.

Mechanically, a very satisfactory method ofconstructing the puzzle piecesthemselves is as follows: Cut a close-grained light hardwood about onequarter inch thick into the shape of some polygon, such as a regular hexagon.Follow the steps of the construction, lightly drawing lines on the wood with apencil. Make the necessary cuts after each step using a fine-bladed jigsaw,and the reassembled pieces may be glued in the new shapes to a piece of stiffcardboard to hold them fast for subsequent drawing and cutting. The glue


and the pencil marks may be sanded off after the final cuts are made and thecardboard removed. The grain of the wood may give a few clues to the puzzleworker, but he will appreciate this. Keep a record of your construction incase your victim challenges you to work the puzzle.

A natural question that would occur at this point to a mathematician isthe following: Does a theorem analogous to the Bolyai-Gerwin Theoremhold for a three-dimensional figure? That is, is it true that if we have twosolid polyhedra with the same volume, then we can show that they areequidecomposable, still of course using only straight-line cuts? The answeris no. Ofcourse, there are some pairs of solid polyhedra with the same volumewhich are equidecomposable-thickening up any two equidecomposablepolygons, as you would in effect do if you constructed a puzzle, wouldprovide such an example. That the proposed theorem is not true does notmean that it is always false, merely that there do exist pairs of polyhedrawith the same volume which are not equidecomposable. In fact, what maybe the simplest possible example-a cube and a regular tetrahedron of thesame volume-provides an example of a pair of nonequidecomposablepolyhedra with the same volume. This example was discovered by the Germanmathematician M. Dehn and first published in 1902.

One might next ask if, given two polyhedra with the same volume, itis possible to tell by a simple test whether or not they are equidecomposable.The Swiss mathematician H. Hadwiger found the answer to this questionin 1949, but to provide the necessary definitions here to express the answerwould take us too far afield. Let it suffice to say that, as one might expect,the question is resolved on the basis of the dihedral angles of the twopolyhedra.

Exercises

1.44 A famous problem of antiquity, studied by early geometers, was theproblem known as Squaring the Circle. Given a circle, the problem was toconstruct a square of equal area, using only an unmarked straightedge andcompass. This problem is usually mentioned along with two companions,that of trisection of a given angle and the duplication of the cube (given acube, to construct with straightedge and compass the edge length of a cubewith double the volume of the first). These problems remained unsolved forthousands of years, but the solutions are now known. The answer is that ineach of the three cases cited above no such construction can exist. Please donot misinterpret this statement. We do not mean that nobody can squarethe circle with straightedge and compass because nobody knows how.What we do mean is that in each case it has been proved that no such construction exists, nor ever can exist. Of course, mathematics departments ofvarious famous universities receive from time to time proposed constructions.The tendency is to ignore them, or return them to the proposer with a note


to the effect that the construction is not correct, and that if the proposerwants to know where the error is, he should be willing to pay a fee for professional services. This has given mathematicians quite a reputation fordogmatism and narrow-mindedness, but most would prefer to work onthe many problems to which the solution is yet unknown rather than searchfor an error in a construction they know in advance to be incorrect.

The reason that these constructions are impossible has to do with thefact that, given a line segment of length 1, unmarked straightedge, and compass, certain lengths cannot be constructed. However, if a number can beconstructed, such as 2 or 1/4 (how?), its square root can also be constructed.The technique is concealed in the proof of the Bolyai-Gerwin Theorem,in Section 1.6 of this chapter. You are invited to discover the technique anduse it to construct a line segment whose length is the square root of 5, givenstraightedge, compass, and a line segment of length 1.

Curiously enough, the impossibility of squaring the circle with straightedge and compass has little bearing, as far as we know, on the UnsolvedProblem in Mathematics mentioned early in this chapter-the problem ofwhether a square and a circle of the same area are equidecomposable in thegeneral sense that, naturally, curved cuts are to be allowed, but still that onlyfinitely many pieces should be used. We do not require that only straightedgeand compass be used. The problem here is one of finding where to make thecuts, or alternately, to show that there can be no way to make the cuts thatwill work. This problem was first stated by A. Tarski in FundamentaMathematicae in 1925.

1.45 The first three stages in the construction of the Snowflake Curveare shown in Fig. 1.7. Some techniques of topology may be used to show thatit makes sense to talk about "the figure obtained by continuing this process,over and over, once for each natural number." That is, of each point in theplane it can be determined whether that point is eventually (and thus permanently) within the curve, or never within. The boundary of the figure,which is what we mean by the snowflake curve itself, consists of those pointswhich eventually get on the boundary at some stage of the construction andthen stay on the boundary in each successive stage.

Since the figure is bounded its area is finite. However, the perimeter isinfinite; that is, given any whole number n no matter how large, at somestage in the construction of the snowflake curve its perimeter exceeds n,and the perimeter increases at each stage. In fact, the perimeter is multipliedby 4/3 each time we pass to the next stage. So if the initial perimeter is 1,we obtain the following sequence of perimeters for the successive stages ofthe construction:

1, 4/3, (4/3)2, (4/3)3, (4/3)4, (4/3)5, ... ,

and this sequence of numbers increases without bound. Why?


2

Fig. 1.7 First three stagesin the constructionof the Snowflake Curve.

3

If you know a little about infinite geometric series, you can use thisknowledge to calculate the area of the figure bounded by the snowflakecurve. Such a series is one in which each term is a fixed multiple, say r, ofthe previous one; that is, the series has the form

where a2 = rat, a3 = ra2' a4 = ra3, and so on. The sum of such a series isthe first term divided by 1 - r. This formula works only if - 1 < r < 1,but such will be the case if you find the right series for computing the area ofthe snowflake curve. See whether you can do this.

1.46 It is possible that a plane figure has no area. We do not mean "zeroarea," nor do we mean "infinite area." A much stranger phenomenon cantake place.

What ought "area" to mean? Suppose we list the properties it shouldhave, and then ask first if any such thing exists. That is, does there exist afunction A which assigns to each bounded subset of the two-dimensionalplane a nonnegative real number, called the area of the subset, with thefollowing properties:

a) The area of a line segment, or a point, or the "empty set," is zero.

b) If Rand S are congruent bounded plane sets, then they have the samearea.

c) If Rand S are bounded plane sets which overlap in a set of zero area, thenthe area of R u S is the sum of the area of R and the area of S.

d) If ReS then the area of R does not exceed the area of S.

e) The area of a rectangle is the product of its length and its width.

The Polish mathematician S. Banach published in 1923 a proof that nosuch function A can exist. The only way, then, to get an effective areameasuring function such as we would like A to be is to drop one or more ofthe above stipulations as to its behavior. If property (e) is dropped, then itturns out that there is only one such function, and it is a rather dull one;it assigns' area zero to every set. So we might as well keep property (e). Butwe surely do not want to give up any of the first four properties listed, and sothe only solution is to drop the stipulation that the area function A measuresevery bounded subset of the plane. If we do this, it is then possible to construct a function A with all the desirable properties listed above; in fact, Awill coincide with our intuitive notion of "area" for geometric figures. Forexample, A will assign the "correct" value for the area of a circle. Whatgoes wrong, as shown by Banach, is that there will exist bounded planefigures to which A cannot assign any area value at all while still satisfyingall the listed properties. Such a set is said to be nonmeasurable.

No one has ever constructed an example of such a set; we merely have aproof of its existence. This is understandably not very satisfying, but it isalso unavoidable. Incidentally, nonmeasurable sets exist in every dimension.

Can you use the techniques of this chapter to construct an area functionA with the properties listed above so that A does assign an "area" to everybounded subset of the plane with polygonal boundary?


1.47 The idea of volume in dimension three is a natural one, although wehave already mentioned that there are polyhedral figures of equal volumewhich are not equidecomposable. On the other hand, S. Banach and A.Tarski found that the situation is far worse than one might expect. It turnsout that the solid ball of radius 1 and the solid ball of radius 2 are equidecomposable-though here we use the term in a much more general sensethan previously, for the cuts are not straight, and most of the pieces intowhich the balls are cut are nonmeasurable (see the previous exercise). Onedramatic way of putting it is this: If matter were homogeneous, it would bepossible to cut the sun into pieces and reassemble these into a solid ball thesize of a pea.

This sort of thing cannot happen for bounded plane sets, but it is possibleto solve the following problem: Show how to cut into infinitely many piecesa square of side 1 so that these pieces may be reassembled into a rectangleof twice the area of the square.1.48 One method of cutting into pieces a polygon P and reassembling theminto another polygon Q may be called more efficient than another suchmethod if fewer pieces are involved. Very little is known about most efficientprocedures in the general case other than the mere fact of their existence.Our methods, applied to turning an equilateral triangle into a square,involve cuts to form a total of seven pieces. Find a more efficient way,using shortcuts.1.49 Suppose that P and Q are rectangles with short sides a and c, respectively, and long sides band d, respectively; and suppose in addition thatP and Q have different areas and that b is larger than d. Show how to cut Pup into finitely many polygonal pieces which can be reassembled into arectangle which has one side equal to d. This shows that two rectangles canbe assembled into a single rectangle. How?

1.50 Can you use Exercise 1.49 and other information in this chapter toprovide a shorter proof of the Bolyai-Gerwin Theorem? How?

1.51 It was mentioned in the last section that there do exist polyhedralsolids of the same volume which are not equidecomposable. However, theBolyai-Gerwin Theorem may well hold for certain types of polyhedra. Forexample, is it true that two rectangular parallelepipeds of the same volumeare equidecomposable? Explain.1.52 At the beginning of this chapter, it was stated that one can make apuzzle of a number of pieces that fit into a rectangular box, and which canbe reassembled to form a triangle, a hexagon, a symmetrical five-pointedstar-or all three. Of course, the Bolyai-Gerwin Theorem shows how to cuta rectangle into pieces that can be reassembled to make a hexagon, but howdoes it follow that a rectangle can be cut into pieces that can be reassembledto form all three of the above figures?

Notes and references 23

1.53 Is it possible to cut up two cubes of equal volume into pieces whichcan be reassembled to form a single cube? Explain.

1.54 Continuing the previous exercise, is it possible to cut up two cubes ofequal volume into a number of small cubes of equal volume which may bereassembled to form a single cube? Reduce this problem to one of solvinga certain simple algebraic equation.

1.55 Is it possible to assemble finitely many squares, no two of which havethe same area, into a rectangle? Explain.

NOTES AND REFERENCES

The most convenient text covering the material of this chapter, as well asassociated topics, is V. G. Boltyanskii's Equivalent and EquidecomposableFigures (Heath, 1963; translated by A. K. Henn and C. E. Watts from thefirst Russian edition, Moscow, 1956). In particular, the results of Dehn andHadwiger about equidecomposable polyhedra are given in detail.

For those who wish to pursue these matters in the one-dimensionalcase, W. Sierpinski's long article "On the Congruence of Sets and TheirEquivalence by Finite Decomposition" is excellent. This was originallypublished in Vol. 20 of Lucknow University Studies (1954), and is availablein English in Monographs, published by Chelsea, containing the article bySierpinski as well as one by Klein, one by Runge, and one by Dickson onother topics. One might also see Sierpinski's article, "Sur quelques problemes concernant la congruence des ensembles de points," in Elemente derMathematik, Vol. 5, pages 1-4 (1950).

With respect to Exercise 1.46, the paper of Banach's referred to is "SurIe Probleme de Mesure," in Fundamenta Mathematicae, Vol. 4, pages 7-33(1923). A general reference covering a wide variety of problems about planesets is the excellent text by Hadwiger, Debrunner, and Klee, CombinatorialGeometry in the Plane, published by Holt, Rinehart, and Winston (1964).A mention of equidecomposability is made on page 52, and several additionalreferences are given.

Additional topics on triangulation, construction of an area-measuringfunction for the plane, and an alternate proof of the Bolyai-Gerwin Theoremmay be found in E. E. Moise's Elementary Geometry from an AdvancedStandpoint (Addison-Wesley, 1963), in Chapters 14 and 24. Chapter 19 alsocontains some interesting material on straightedge and compass constructions.

In mathematics, as well as in the physical sciences, it frequently happensthat discoveries are made almost simultaneously by men working independently. In the case of the Bolyai family, we have a double coincidence.Bolyai Farkas (whose name is frequently rendered as Wolfgang Bolyai deBolya) was a Hungarian mathematician born in 1775. He studied at theUniversity at Gottingen, where he and Gauss, then also a student there,


became close friends. The elder Bolyai returned to teach at Maros-Vasarhely,and during this time his son Bolyai Janos (or Johann Bolyai de Bolya)acquired an interest in mathematics. In 1832 Farkas published his majormathematical work, the Tentamen, covering a variety of geometrical ideas,and in which he stated the theorem we have called the Bolyai-GerwinTheorem. At almost the same time-perhaps within a year-the Germanofficer and amateur mathematician Gerwin also published the same result.

Meanwhile, Janos had been working on the problem of the independenceof Euclid's famous parallel postulate, and in the process discovered that theparallel postulate could be replaced by a nonequivalent alternative. Theresulting geometry, known as non-Euclidean geometry, was at that time amajor breakthrough in mathematical thought; indeed, so important that thereader is urged to consult a history of mathematics for a full appreciation ofits implications. The work of the younger Bolyai was published as a 26-pageappendix to his father's Tentamen. Again, almost simultaneously, theRussian mathematician Lobachevsky came up with much the same result;meanwhile, when he heard of this work of the younger Bolyai, Gauss revealedto Bolyai Farkas that he too had obtained such results, but had hesitatedto publish them because he thought they might not be well received.

This was quite disappointing to Bolyai Janos, who never again publishedany mathematical results; curiously enough, he is by far the better known ofthe two Bolyais. This is probably as it should be, inasmuch as his discoveryof non-Euclidean geometry has had major implications in both the foundations of mathematics itself as well as in Einstein's Theory of Relativity.

Bolyai's work went almost unnoticed for thirty-five years, until it wasnoted after his death by Richard Baltzer in 1867. Finally, in 1894, a memorialstone was placed on Bolyai Janos' grave in Maros-Vasarhely.

CHAPTER 2

BRUNNIANLINKS

Examine the three rings shown in Fig. 2.1, close the book, and try to reproducethe drawing.

Many people have some difficulty in correctly drawing the three rings ofFig. 2.1, known as the Borromean Rings, and will draw instead three ringslinked much as the three shown in Fig. 2.2. As soon as the difference betweenthe two is seen, though, it becomes easy to draw the Borromean Rings. Thedistinguishing property of the Borromean Rings is that each of the rings liescompletely over one of the other two, and completely under the other. Thethree rings shown in Fig. 2.2, on the other hand, have the property thateach actually links each of the other two. We shall examine in this chapter theimplications of this difference, as well as properties of such figures made ofother numbers of rings.

For the rest of this chapter, we shall assume that all such figures lie inordinary three-dimensional space, so that if you wish you may constructexamples made of string or wire. The circular shapes into which your materialis formed will not have to be perfectly circular, nor is the material itselfimportant with regard to the mathematical properties that will concern us.All we care about is the manner in which the various curves link one another.Hence we immediately pass to a convenient mathematical abstractionthat of the simple closed curve. Just as Euclid saw in the real world variousrough approximations to a perfectly straight line (such as the line where thewall meets the ceiling), and passed to the abstraction of the geometric straightline without width or end, so also do mathematicians who wish to examineproperties of knots and linking curves pass to the abstract idea of a simpleclosed curve.

25

26 Brunnian links 2.1

Fig.2.1 The Borromean rings.

Fig. 2.2 A (3, 1)-Brunnian link.

2.1

Fig.2.3 A wildsimple closed curve.

The simple closed curve 27

2.1 THE SIMPLE CLOSED CURVE

Let us consider the reality from which the concept of the simple closed curveis abstracted. Imagine an ordinary piece of string with the ends woven together so as to form a homogeneous and continuous curve; then think of themidline of this curve. This abstraction, the midline, is an example of asimple closed curve. The important properties are these:

a) No point separates a simple closed curve. That is, no single scissors-cutcan separate it into two pieces.

b) Each set of two points does separate a simple closed curve. That is,two scissors-cuts will separate the curve into two (why not more?)pIeces.

c) The simple closed curve is one-dimensional-every sufficiently smallconnected piece of it has the same dimensional properties as a smallpiece of Euclid's perfect geometric straight line.

d) The simple closed curve is bounded-it is contained in some sphere ofsufficiently large radius.

Mathematicians were surprised, as you might be, to discover that theabove four properties leave something to be desired if they are supposed to bedefining properties of the common notion of a simple closed curve; at least,something to be desired if this definition is to be an accurate abstraction fromreality. For this definition does not exclude such an object as is shown inFig. 2.3.


Fig. 2.4Tame simpleclosed curves.

The object shown in Fig. 2.3 does have the four properties listed above,and so ought to be considered a simple closed curve, but is not an abstractionfrom reality since it certainly cannot be tied with a piece of string. There areinfinitely many loops, decreasing in size; of course, the entire figure cannotbe drawn, and so most of it is concealed in the box. Such curves are said tobe wild, as opposed to the tame ones we shall study. We may prevent theoccurrence of any wild simple closed curves henceforth by the followingagreement: We consider a curve only if it can be continuously deformedwithout self-intersection onto a polygonal curve (one formed of a finitecollection of straight-line segments). Mostly for aesthetic reasons we shallcontinue to draw our tame simple closed curves without corner points;just imagine if you wish that all the corners have been rounded off slightly.Each of the simple closed curves shown in Fig. 2.4 is tame-the one on theleft is an approximation by a polygonal simple closed curve to the one on theright.

Since the curves shown in Figs. 2.1 and 2.2 can be continuously deformedwithout self-intersections onto polygonal curves, they too are tame; but thecurve shown in Fig. 2.3 does not have this property. (This is not obvious.)

Note that the Borromean Rings of Fig. 2.1 have the property that eachring lies completely over the one immediately clockwise to it. This meansthat if anyone of these three simple closed curves is removed from theBorromean Rings, the remaining two come apart. However, you will have

2.1 The simple closed curve 29

little trouble in convincing yourself that the three do not come apart. Becauseof this interesting property, the Borromean Rings have been used for centuriesin the Christian religions as a symbol of the Holy Trinity. On a less reverentnote, they have also been used as the trademark of a well-known manufacturer of beer and ale. The mathematician H. Brunn may have been thefirst to study generalizations of the Borromean Rings, about 1892, and thesegeneralizations are known to some mathematicians as Brunnian links forthis reason.

What generalizations? Why, one would naturally ask if it is possible toconstruct four simple closed curves with the Borromean Property: Thefour are linked together but removal of anyone causes the remaining threeto fall apart. And why stop with four? Do there exist ten, or twenty, or evena hundred simple closed curves, the totality linked together, but such thatthe removal of anyone causes the rest to fall apart? We shall examinethese generalizations as well as others in this chapter.

Exercises

2.1 We listed earlier the four properties of a simple closed curve:

a) No point separates a simple closed curve.b) Each two-point set separates a simple closed curve.c) A simple closed curve is one-dimensional.d) A simple closed curve is bounded.

If there exists a figure in three-dimensional space which is not a simple closedcurve and has three of these properties but not the fourth, it shows that thefourth property is essential for an abstract definition of simple closed curve.Can you construct a figure having properties (a), (c), and (d), but not property(b)? What about other combinations?

2.2 Which of the defining properties of a simple closed curve does a straightline segment have?

2.3 Which of the defining properties of a simple closed curve does an infinitestraight line have?

2.4 Which of the defining properties of a simple closed curve does a thetacurve (a figure shaped like the Greek letter 8) have?

2.5 Which of the defining properties of a simple closed curve does a flattwo-dimensional circular disk have?

2.6 Construct an example of a wild simple closed curve essentially differentfrom that shown in Fig. 2.3.

2.7 Construct an example of four simple closed curves in three-dimensionalspace with the property that the removal of a certain one of these causesthe other three to come apart, but removal of anyone of the other threeleaves the remaining three linked together.


2.8 Each of the curves shown in Fig. 2.4 is said to be knotted because neithercan be continuously deformed into a circle in three-dimensional spacewithout self-intersection at some stage. On the other hand, a circle or asquare is an unknotted simple closed curve. Show that a square is unknotted.

2.9 Your intuition should tell you that a tame simple closed curve is unknotted (see Exercise 2.8) if and only if it can be continuously deformed inthree-dimensional space until it lies in a flat plane without self-intersection.Show that this last condition is equivalent to the following: The tame simpleclosed curve can be continuously deformed in three-dimensional space untilit lies on the surface of a round two-dimensional sphere without self-intersection.

2.10 It is difficult to show that each of the curves of Fig. 2.4 is knotted, butyou can show that either-and thus both-can be deformed so as to lieon the surface of a two-dimensional torus without self-intersection. Pleasedo so. (A torus is the mathematician's abstraction of the surface of a onehole doughnut.)

2.2 LINKS AND THEIR PROPERTIES

We have already introduced some technical terms and concepts withoutdefinition. We need these definitions to ensure that the writer and the readeragree on precisely what is meant, but we will define these terms-"linking,""falling apart"-in such a way as to be as near as possible to the commonand natural interpretations of these terms. This is how we have been able sofar to use these terms in a very nonmathematical fashion-without previousdefinition-with some assurance that no confusion as to precise meaning hasyet arisen. However, after the definitions, we will begin to use the mathematical terms now current to remind you that these terms have been definedin a certain way, and to prevent the connotations associated with the commonterms from creeping in and clouding the issues.

An n-Iink is a collection of n simple closed curves in three-dimensionalspace. (Remember that we shall deal only with tame simple closed curves; nis merely a positive whole number.) See Fig. 2.5.

An n-link is splittable if it is possible to deform it continuously in threedimensional space in such a way that part of the link lies within Bland therest of the link lies within B2 , where B 1 and B2 are mutually exclusive solidballs in three-dimensional space. Figure 2.6 shows an example of a splittable3-link.

Your intuition is in good working order if it tells you that if an n-linkcomposed of curves C1 , C2 , ... , Cn is splittable, then no one of the curvesC i can lie partly within B 1 and partly within B2 • What does happen is this:Some (but not all) of the curves lie entirely within B 1 , some (but not all) of

2.2 links and their properties 31

Fig. 2.5 A 4-link.

Fig. 2.6 A splittable 3-link.

Fig.2.7 A 4-link andone of its 3-sublinks.

the curves lie entirely within B2 , and each curve lies entirely within exactlyone of the two balls. For example, if a 3-link is splittable, then it can becontinuously deformed so that two of its curves lie entirely within B 1 andthe third entirely within B2 , where (as before) B1 and B2 are disjoint solidballs.

An n-link composed of the curves C1 , C2 , ••• , Cnis said to be completelysplittable provided that there exist mutually exclusive balls B1, B2 , ••. , Bn

in three-dimensional space such that the link can be deformed continuouslyso that each Ci lies entirely within the ball Bi•

The example of Fig. 2.6 is a splittable 3-link which is not completelysplittable, because C2 and C3 cannot be split apart. So all that splittabilitymeans is that the link comes at least partly apart; completely splittable linkscome entirely apart. If we say a link is nonsplittable, we mean that not evenone of the curves involved, or any pair, or any combination, can be separatedfrom the rest without cutting.

Finally, given any n-link composed of the curves C1 , C2 , .•. , Cn' asublink of this link is simply some subcollection of the curves C1 , C2 , ••• , Cn-obtained, if you like, by erasing the ones not in the collection. A ksublink of the link L is just a k-link which is a sublink of L. This of coursemakes sense only for 1 :::: k < n, where n is the number of curves used toform L.

Figure 2.7 shows on the left a 4-link, and on the right one of its 3-sublinks.Please note that a collection ofcurves in a sublink retains the linking propertiesit enjoyed in the whole link, although removal of some curves to form a

2.3 A simple algebra 33

sublink out ofthe remainder may cause splittability where none existed before.For example, consider the 2-sublinks of the Borromean 3-link.

We may use our new terminology to describe the property of interestthat distinguishes the Borromean Ring example from the example shown inFig. 2.2: The Borromean Rings form an example of a nonsplittable 3-linkevery 2-sublink of which is completely splittable. And the generalizations weseek may be described in the following way:

a) Does there exist a nonsplittable 4-link such that each of its 3-sublinksis completely splittable?

b) Given a whole number n > 5, does there exist a nonsplittable n-linkwith each of its (n - 1)-sublinks completely splittable?

c) Given whole numbers nand k with 1 ~ k < n, does there exist a nonsplittable n-link such that each of its k-sublinks is completely splittablebut each of its (k + l)-sublinks is nonsplittable? In particular, doesthere exist an example of a nonsplittable 4-link with every 3-sublinkalso nonsplittable, but with every 2-sublink completely splittable?

Exercises

2.11 Is every splittable 2-link also completely splittable? Why?

2.12 Is every splittable 3-link also completely splittable? Explain youranswer.

2.13 Show how to construct, for each whole number n > 2, an example of anonsplittable n-link each of whose sublinks is also nonsplittable.

2.14 Show how to construct, for each whole number n > 3, an exampleof a nonsplittable n-link one of whose (n - 1)-sublinks is completely splittable. Can this be done in such a way that one and only one of the (n - 1)sublinks is completely splittable, and all the other (n - l)-sublinks arenonsplittable?

2.15 Construct infinitely many circles C1 , Cz, . .. in three-dimensionalspace forming a nonsplittable link L such that removal of anyone of thecurves C j from L produces a splittable link. For the purposes of this exercise,how is the word "link" being used? What is a reasonable definition of"splittable"?

2.3 A SIMPLE ALGEBRA

We shall answer the questions raised at the end of the preceding section byuse of an extremely simplified form of analytic geometry. Analytic geometrymay be described as the process of assigning algebraic equations or expressions to geometric objects so that we can make deductions which may notbe geometrically apparent. The simplification we make lies in our new simple


algebra. We shall use the customary symbols a, b, c, ... of ordinary algebra,but the rules of our algebra are very few in number. Here they are:

1. The only operation is "multiplication." We shall denote the product ofa and b byab.

2. There is a multiplicative identity, which we shall denote by 1. It has theproperty that for each a, la = a = a1.

3. The associative law for multiplication holds-that is, we may ignoreparentheses. It will always be true that (ab)c = a(bc), so that we maysimplify an expression such as a(b[cd(b)]) to abcdb.

4. Finally, for each symbol a in our algebra, there exists an element denotedby a-I (and pronounced "a-inverse") in our algebra such that aa- 1 =1 = a- 1a.

To save space, we shall abbreviate such expressions as aaaa by a4, andthe usual laws of exponents can be shown valid:

d"d' = d"+n and (d"t = d"n

for all whole numbers m and n and each element a of our algebra. Inparticular,

( -l)n -na = a and aa- 1 = aO = 1,

thus justifying our use of the notation a-I for the inverse of the element a.What we cannot use is the so-called commutative law; in general it will

not be true that ab = ba. The only complicated thing is to determine whatthe symbols a, b, c, . .. represent. They are certainly not whole numbers,nor are they real numbers. Moreover, the operation which we have soblithely called "multiplication" certainly cannot be ordinary multiplication,since the objects we are "multiplying" are not numbers. We need to explainwhat the symbols a, b, c, and so on represent, and what the meaning of the"product" ab is.

Let A be a circle in three-dimensional space, such as is shown in Fig. 2.8.Warning: So far, "A" is just the name of a circle, and the symbol A willnot be an element of our algebra. The object a of our algebra will, however,be closely associated with the circle A, and the association is so natural as tocause us to use almost the same name-differing only in upper as opposed tolower case-for two different things.

Imagine a curve in three-dimensional space that starts at the tip of yournose, passes through the circle A once going away from you, curves around,and returns to the tip of your nose. (The tip of your nose is denoted by p inFig. 2.8.) This new curve is almost the object a of our algebra. But all wecare about is the fact that this new curve passes through the circle A a netof one time away from you, and it is the collection of all such curves that


Fig. 2.8 A and B.

p

deserves the name a. Thus what the element a of our algebra represents is thenet effect ofpassing through the circle A one time going away, or alternatively,the set of all such curves that start and end at the tip of your nose and passthrough the circle A once going away.

Your intuition should tell you that if x is any other curve also passingthrough A a net of one time going away, then the curve x can be continuouslydeformed to the curve representing a (as shown in Fig. 2.8) without cutting xor the circle A, and in such a way that x never touches the circle A, and theends of the curve x are never removed from the tip of your nose. Conversely,any curve x which can be so continuously deformed must pass through thecircle A a net of one time going away. So if you imagine all possible curvesthat start at your nose, pass through the circle A a net of one time goingaway from you, and return to your nose, you may consider each of these asrepresenting the element a of our algebra, since anyone can be deformed toany other such in the manner described above.

Now that you know what the symbols a, b, C, • •. of our algebra represent, it is time to define the operation of multiplying two of them together.Let A and B be circles in three-dimensional space as shown in Fig. 2.9. Theelement a of our algebra can be represented by a curve that starts at p,passes through the circle A once going away from you (and not passingthrough B at all), and finally returns to p. Similarly, the element b can be


p

Fig. 2.9 A curverepresenting the"product" abo

represented by a curve that behaves the same way with respect to the circleB. We define the product ab to be represented by each and any curve thathas the net effect of "first doing a, then doing b." That is, a typical representative of ab is a curve that starts at p, passes away from you through A, thenpasses away from you through B, and finally returns to p. It is permissibleto remove the end of a and the beginning of b from the point p and thenjoin these ends together close by, as we have shown in Fig. 2.9, in order toclarify the picture.

A little experimentation with wire circles and two pieces of string shouldconvince you that, at least in this example, ab =1= ba; that is, the curve abcannot be deformed to the curve ba without cutting it or cutting the circlesA and B or removing its ends from p.

What sort of curve represents a- 1 ? Why, each curve that passes oncethrough the circle A toward you. What is the meaning of a2 ? This stands forthose curves that have the net effect of passing through the circle A twiceaway from you. What about the multiplicative identity 1 of our algebra?That is represented by any curve that passes through no circle at all. Thesephenomena are illustrated in Fig. 2.10.


Fig.2.10 Somealgebraic illustrations.

A

A

A


It is easy to see that la = a = al for each element a of our algebra,and that aa- l = I = a-lao We are, of course, interpreting equality as"has the same effect as," or "can be continuously deformed to," or "passesthrough the same curves in the same direction in the same order as." Theassociativity, which allows us to write abc for either (ab)e or a(be), will bediscussed in the exercises. You should now spend some time convincingyourself that the laws of our simple little algebra in fact hold, when youinterpret the objects a, b, e, . .. and the operation of "multiplication" aswe have done.

From our point of view, one of the most useful things about this algebrais that we can tell when two "objects" or expressions in the algebra areactually equal by a very simple test: First, perform the only algebraicsimplifications allowable in the algebra:

Replace aa - 1 (or a- 1a) by 1.

Replace 1a (or a1) by a.

Simplify expressions such as aaaa to a4•

Perform all these simplifications on each expression. If the resulting expressions are identical, then the original expressions must have been equal.For example, we may ask whether

aba2a- 1bbeae - lee and

are equal. Each simplifies to abab2 eae. Since they have identical simplifications, they are equal.

Exercises

2.16 The definition of "group" is given in Section 4.1. In our simplealgebra developed for the study of Brunnian links, we started with a set ofobjects a, b, e, ... , each representing a way of sending a simple closed curvethrough some fixed circle in three-dimensional space, and a "multiplication"of these objects, by which the product ab was interpreted to mean the effectof first doing a, then doing b. Verify informally that this set of objectstogether with this multiplication does indeed satisfy the definition of a group.

2.17 We saw that our simple algebra has the following property: Twostrings of symbols are equal if they have identical simplifications. However,there may be equal expressions without identical simplifications. If thecircles A and B are linked together as shown in Fig. 2.11, then ab = ba, butthis relation cannot be derived from the simplifications listed in the precedingsection. Why is ab = ba in this example?

2.18 Let us study just a little more the algebra associated with the 2-linkshown in Fig. 2.11. Show that every expression in this algebra is equal to oneof the form d'bm

, where nand m are whole numbers (possibly zero; we

2.4 Brunnian 4-links 39

Fig.2.11 ab = ba.

interpret XO as the identity, 1). For example, b- 1a2b3a5 = a7b2. This showsthat each expression in this algebra has a "standard form" that looks liked'bm

• Are two expressions in this algebra equal if and only if their standardforms are identical? Explain your answer.

2.19 We return in this exercise to the general case of an algebra known onlyto satisfy the group axioms (see Exercise 2.16). Show that (xy)-l = y-1X- 1.Your proof should work not only in the case that x and yare single symbolsin the algebra, but also in the case that x and y stand for more complicatedexpressIOns.

2.20 Show that the two expressions

aab2b-lca2b3b-la-lb3

and

are equal.

2.4 BRUNNIAN 4-LlNKS

Although we are going to use a very simple algebra, it will turn out to be apowerful tool in our examination of Brunnian links. You will now see howa fairly complex geometric problem can be transformed into a remarkablysimple algebraic one.

Suppose you had the task of drawing four simple closed curves to form anonsplittable link each 3-sublink of which is to be completely splittable.


Fig.2.12 Threefourths of a{4, 3)-Brunnian link.

Since each 3-sublink is to be completely splittable, it should be possible todraw the desired 4-link by first imagining one of the curves not present.Since the remaining three split completely, they could be drawn as separatedcircles much as in Fig. 2.12. The only problem would be to draw the fourthcurve in such a way as to guarantee that the resulting 4-link would be nonsplittable, but such that all four possible 3-sublinks would be completelysplittable. One-fourth of the latter task has already been completed: Clearly,if the fourth curve, not yet drawn, is removed, then the remaining threecome apart.

You will not be convinced of the difficulty of finding a geometric solutionto this problem unless you give it a try-so spend some time trying to drawthat fourth curve. Remember, in order to obtain the desired example, itmust be true that if anyone of the curves A, B, or C is removed from yourfigure, then the remaining three, two circles and your curve, must come

2.4 Brunnian 4-links 41

Fig.2.13 TheBorromean rings again.

p

completely apart. Do not try to draw a circle for the fourth curve; so far aswe know, it cannot be done that way. All you need is any tame simple closedcurve for the fourth curve.

Let us see if the miniature algebra we have developed will serve to tellus why the Borromean Rings form a nonsplittable 3-link each 2-sublink ofwhich is completely splittable. Perhaps this will provide some insight thatwill help in the construction of other examples. If you carefully draw apartthe upper two rings of the Borromean example shown in Fig. 2.1, you willobtain a figure much like the one shown in Fig. 2.13.

We have two separated circles A and B, which is what we might wellwant to start with if we wanted to invent the example of the Borromean rings.The tip of your nose is shown as the point p on the curve C, for we wish tocalculate the algebraic formula of this curve. It first passes away from youthrough A, then toward you through B, then again toward you through A,and finally away from you through B. The algebraic formula we havelearned to associate with such a curve is ab-1a-1b. This simple formulatells us a great deal about the Borromean rings.

First, since no simplification of the formula ab-1a-1b is possible, it doesnot reduce to I. (This does not contradict Exercise 2.17, since the curvesA and B are not linked.) Hence the curve representing this formula actuallylinks the union of the curves A and B and cannot be removed without cutting.


Second, observe the effect of cutting and removing the circle A. Theeffect on the formula ab- 1a- 1b is both simple and striking. When the circleA is taken away, the effect on the formula is to delete all occurrences of thesymbol a (and a- 1 as well). The formula becomes b- 1b, which simplifies to 1.This shows that if A is removed, then Band C are completely splittable.Similarly, if B is removed, then we obtain aa- 1 = 1, and so A and Carecompletely splittable. By our construction, since we drew A and B alreadyseparated, A and B split completely if C is removed. This shows that every2-sublink of the Borromean rings is completely splittable.

If the 3-link itself were splittable it would then be completely splittableby the above discussion. But this is not the case since ab- 1a- 1b #= 1.Hence the 3-link is nonsplittable. This demonstrates that the Borromeanrings actually do provide an example of a nonsplittable 3-link each 2-sublinkof which is completely splittable.

The one item of crucial importance about the formula ab- 1a- 1b, theone phenomenon that makes the example work, is this: The deletion of alloccurrences of anyone symbol in the formula causes the formula to collapseto I. What this means is that if anyone circle is removed from the exampleof the Borromean rings, then the remaining two come apart. To applysimilar algebraic methods to the construction of a nonsplittable 4-link each3-sublink of which is completely splittable, let us return to Fig. 2.12 andsimply ask what sort of formula the fourth curve we must draw should have.

First, the formula should involve all three of the symbols a, b, and e,so that the four curves together will form a nonsplittable link.

Second, the formula must not collapse to I. This too is needed to ensurethat the 4-link is nonsplittable.

Finally, the formula must have the property that the deletion of alloccurrences of anyone symbol-either a, b, or e-will cause the formula tocollapse to I. This will, as in the example of the Borromean rings discussedabove, guarantee that each 3-sublink is completely splittable.

Exercise

2.21 Can you write down a formula involving all three symbols a, b, and e,such that the formula does not collapse to I, but the deletion of all a's, orall b's, or all e's, does cause the resulting formula to collapse to I? Or,alternatively, can you prove this impossible, thus demonstrating there is noanalogy to the Borromean rings with four simple closed curves?

We are sure you will be able to discover the answer for yourself, and inorder for you to enjoy fully the thrill of mathematical discovery, we ask younot to read beyond this paragraph until you have worked on this problemfor a while. Go back and look at the formula associated with the third curvein the Borromean rings; it will probably help.

2.5 Notation and examples 43

2.5 NOTATION AND EXAMPLES

Let us introduce at this point a small bit of notation that will serve as a usefulabbreviation.

By an (n, k)-Brunnian link we mean a link of n simple closed curves inthree-dimensional space such that each k-sublink is completely splittable, buteach (k + 1)-sublink, and each (k + 2)-sublink, and so on up to the n-linkitself, is nonsplittable. Thus an (8, 5)-Brunnian link would be a collection ofeight simple closed curves such that the set of all eight, or any seven, or anysix, would be nonsplittable, but any 5-sublink would be completely splittable.Clearly, if each 5-sublink is completely splittable, then so is any 4-sublink,any 3-sublink, and any 2-sublink. Of course, it makes sense to talk about an(n, k)-Brunnian link only if 1 ~ k < n; if k = 1 the definition still makessense-Fig. 2.2 shows a (3, 1)-Brunnian link. And, of course, the Borromeanrings form a (3, 2)-Brunnian link.

We have previously asked you to construct, first geometrically and thenalgebraically, an example of a (4, 3)-Brunnian link. If you have succeededin doing this, we invite you to try a much harder problem-the constructionof a (4, 2)-Brunnian link. Here is the hint or observation that may simplifythe task: If a (4, 2)-Brunnian link exists, then the removal of anyone curvefrom this link should produce a (3, 2)-Brunnian link. Thus one should perhaps begin by drawing a (3, 2)-Brunnian link and then attempt to find whereto put the fourth curve so as to produce a (4, 2)-Brunnian link. Make use ofthe algebra we developed specifically to solve problems of this sort.

Let us return to the problem of producing a (4, 3)-Brunnian link, startingwith the three circles A, B, and C as shown in Fig. 2.12. As we hope youhave discovered, there does indeed exist a formula using the symbols a, b, andc such that the deletion of all occurrences of anyone of these symbols causesthe resulting formula to collapse to 1. One such formula is

aba - 1b- 1cbab - 1a-Ic- 1 •

The presence of the "subformula" aba- 1b- 1, so like the one associatedwith the Borromean rings, is no coincidence. As soon as this is noted, thepattern becomes clear for construction of even larger links, such as the(5, 4)-Brunnian link.

To construct a (5, 4)-Brunnian link, first draw four separated circlesA, B, C, and D. Find a formula, such as

aba-lb-lcbab-la-lc-ldcaba-lb-lc-lbab-la-ld-l ,

with the property that all four symbols a, b, c, and d appear in the formula,and such that the deletion of all occurrences of anyone of these symbolscauses the resulting formula to collapse to 1. We have not chosen the shortestsuch formula above for the (5, 4)-Brunnian link, but it is one which continues


the pattern already established. If we then draw in a fifth curve with thisformula, we can argue (as we did in the case of the Borromean rings) thatwhat is produced is indeed a (5, 4)-Brunnian link. You will also have seenthat the number of curves in these constructions is not material, except thatlarge numbers of such curves produce rather long formulas. But since theformulas have the desired properties it follows that for each whole numbern > 2, an (n, n - 1)-Brunnian link exists. Thus it is possible to construct,say, one hundred simple closed curves in three-dimensional space forming anonsplittable link, but such that the removal of anyone of these curves causesthe remaining ninety-nine to split completely.

Exercises

2.22 Draw a (2, 1)-Brunnian link.



2.25 Find a shorter formula than that given in Section 2.5 for the fourthcurve in a (4, 3)-Brunnian link.

2.26 Draw a (3, 2)-Brunnian link essentially different from the example ofthe Borromean rings shown in Fig. 2.1.

2.27 Show that an (n, 1)-Brunnian link can be constructed for each wholenumber n > 2. Compare this with Exercise 2.13.

2.28 If you have not already done so, construct an example of a (4, 2)Brunnian link.

2.29 For appropriate choice of x and y, the formula for the third curve in a(3,2)-Brunnian link was seen to have the form xyx- 1y-1. Is this also trueof the formula given in Section 2.5 for the fourth curve in a (4, 3)-Brunnianlink? See also Exercise 2.19.

2.30 Construct four simple closed curves in three-dimensional space, sayA, B, C, and D, with the following rather unsymmetrical linking property:The removal of A causes the remaining three to split completely, but of B, C,and D, exactly two must be removed to cause A and the third to split, and ifonly one of B, C, or D is removed, then the remaining three form a nonsplittable link.

2.6 COMMUTATORS

The frequent repetition of a mathematical concept justifies calling attentionto that concept and supplying it with a name. You have undoubtedlynoticed by now the frequent occurrence of a string of symbols of the formxyx - 1Y-1. In addition, you may have noticed that in our simple algebra,if we wish to construct the inverse ofa string of symbols such as ab2a- 1cab- 3

,

2.6 Commutators 45

the proper way to do it is to write the string of symbols in the reverse order,changing the sign of each exponent. (The exponent of a is 1.) Thus, in theexample just mentioned, the inverse would be b3a- 1c- 1ab- 2a-t, for theproduct of the two will clearly produce 1. In general, we may state as auseful law of exponents in our algebra that

(xy)-l = y-1x- 1

even if one or the other or both of x and yare strings of symbols rather thana single symbol. Let us abbreviate by (x, y) the string of symbols xyx-1y-l.This object is a victim of frequent study in modern mathematics, and is calledthe commutator of x and y.

Suppose we are given three symbols a, b, and c, and form first the commutator of a and b, and then form the commutator of that string of symbolswith the element c. Symbolically, we would express this as «a, b), c), whichexpands to the expression

aba-lb-lcbab-la-lc-l.

It is no coincidence that this is the formula by which we earlier produced a(4, 3)-Brunnian link, and indeed if you expand the multiple commutator«(a, b), c), d), you will obtain the formula for the fifth curve in a (5, 4)Brunnian link. The reason that commutators produce this effect is this:Any of the multiple commutators we have been writing down has the propertythat if all occurrences of anyone symbol are deleted, then the entire commutator collapses to 1. This is precisely the property needed to produce(n, n - l)-Brunnian links.

Let us turn our attention to the problem of producing a (4, 2)-Brunnianlink. As we have already observed, a reasonable starting point would beto draw a (3, 2)-Brunnian link; that is, the Borromean rings. We do thisbecause if anyone curve is deleted from a (4, 2)-Brunnian link, what is left isa (3, 2)-Brunnian link. So we draw the three, and puzzle over exactly how todraw the fourth curve. In terms of our algebra, if the three we draw arecalled A, B, and C, what is required is a formula using all the symbols a, b,and c, such that the deletion of all occurrences of anyone of these symbolscauses the formula to collapse, not to 1, but to the formula for a (3, 2)Brunnian link. In other words, can we use the symbols a, b, and c to writea formula in which the deletion of all occurrences of anyone symbol does notcause the formula to collapse to 1, but one in which the deletion of alloccurrences of any two symbols does cause the formula to collapse to I?If we could find such a formula, and draw in a fourth curve representing thisformula, we would have a (4, 2)-Brunnian link as desired.

Here is where the concept of the commutator of two symbols provesuseful. If the formula involves the commutator (a, b), then at least this muchwill remain if all occurrences of c in the formula are deleted, and some care is


taken in the construction of the formula. If in addition this were all thatremained, then after removing curve C, we would have exactly what we wanta (3, 2)-Brunnian link. So let us first write down the commutator of a and b,and be sure that c is involved in any other commutators we write down soas to guarantee the disappearance of all these other commutators if alloccurrences of c are deleted. At this point you may have already guessedwhat will happen next. About the only other sorts of things we can constructthat are commutators involving the symbol c are the commutators (a, c)and (b, c). But then we notice that the product of all three has the desiredproperty

(a, b)(a, c)(b, c) = aba-lb-laca-lc-lbcb-lc-t,

which will collapse to one of these three commutators if all occurrences of anyone symbol are deleted. Whether or not we remove the curve A, or B, or C,what is left will be a (3, 2)-Brunnian link. Hence drawing a fourth curverepresenting the above formula will produce a (4, 2)-Brunnian link.

Exercises

2.31 Show that the inverse of (x, y) is (y, x).

2.32 Prove that (x, y) = 1 if and only if xy = yx.

2.33 Show that the deletion of anyone symbol in all its occurrences from themultiple commutator «(a, b), c), d) causes the resulting expression tocollapse to 1.

2.34 Use the ideas of this section to construct a (5, 2)-Brunnian link. Note:you should start by drawing a (4, 2)-Brunnian link.

2.35 Use the methods of this section and your experience with Exercise2.34 to show that a (n, 2)-Brunnian link can be constructed for any wholenumber n > 3.

2.7 THE FINAL GENERALIZATION

At this point we have exhausted all possibilities for 4-links. As we haveshown, it is not too hard to produce a (4, 3)-Brunnian link; it is easy toproduce a (4, l)-Brunnian link; and we finally succeeded in the last sectionin producing a (4, 2)-Brunnian link. It would be reasonable at this stage toguess that for any whole numbers nand k for which the notation makessense, an (n, k)-Brunnian link can be constructed. But let us first examine thenext case still open, that of a (5, 3)-Brunnian link.

As before, let us start with all but one of the curves already drawn. Ifwe imagine a (5, 3)-Brunnian link with one of the curves removed, what wemust have is a (4, 3)-Brunnian link. However, we know that a (4, 3)Brunnian link can be constructed, so let us draw one. Let the four curves

2.7 The final generalization 47

Fig. 2.14A (4, 3)-Brunnian link.

involved be called A, B, C, and D. Then, as before, the problem is to drawa fifth curve E with certain properties-but this amounts only to finding aformula for E, a formula involving all the symbols a, b, c, and d, such that thedeletion of all occurrences of anyone of these symbols from the formulaproduces the formula for a (4, 3)-Brunnian link.

The formula for producing a (4, 3)-Brunnian link is itself a commutator,but a slightly more complicated commutator than we used in the (n, 2)Brunnian examples (see Exercise 2.35). Recall that if separated circles A,B, and C are drawn, then drawing a fourth simple closed curve D accordingto the formula (a, b), c) will produce a (4, 3)-Brunnian link, as in Fig. 2.14.

To produce a (5, 3)-Brunnian link, we must provide a formula usingthe four symbols a, b, c, and d, so that-if possible-the deletion of alloccurrences of anyone of these symbols produces a commutator in threeof these symbols, such as our original commutator (a, b), c). For example,if a were deleted, it would be sufficient to have the formula collapse to(b, c), d); if b were deleted, we would like to have the formula collapse to


«a, c), d). But observe that even these more complicated commutators,such as «a, b), c), do collapse to I upon the deletion of any symbol presentin them. Therefore, as we did for the construction of the (n,2)-Brunnianlinks, we write the product of all possible commutators involving three of thesymbols a, b, c, and d, and obtain

«a, b), c)«a, b), d)«a, c), d)«b, c), d).

If all occurrences of anyone symbol, such as c, are deleted, all the commutators involving that symbol collapse to 1. What we have left is the singlecommutator «a, b), d), not involving the symbol c. But this shows that if wedraw the fifth curve with the above formula, we have indeed produced a(5, 3)-Brunnian link.

We need but a single definition here, for the purpose of making our ideaseasier to express. If a, b, c, ... , are symbols in our algebra, and k is a wholenumber at least 2, then a k-commutator of these symbols is simply one of themultiple commutators we have been considering, one that involves exactly kdistinct symbols drawn from our algebra. For instance, using the abovesymbols, one example of a 4-commutator is «(a, b), c), d). It is importantto note that a k-commutator has the property that if all occurrences of anyone symbol are deleted from it, then the formula that remains will collapseto 1.

Let us see how the concept of a k-commutator will serve us in guessingthe general procedure necessary to produce an (n, k)-Brunnian link for anymeaningful choice of nand k. Suppose we first wished to draw a (3, 2)Brunnian link. We draw two separated circles, label them A and B, andconsider the associated symbols a and b in our algebra. Since we wish toform a (something, 2)-Brunnian link, we write down all possible 2-commutators using the symbols a and b. There is no distinction between aba- 1b- 1

and (for example) b - 1aba -1 for our purposes. All we need to do is to writedown only one such, say aba - 1b -1. We then draw in a third curve with thisformula and obtain a (3, 2)-Brunnian link.

Similarly, to produce a (4, 2)-Brunnian link, we first produce a (3, 2)Brunnian link, label the three curves A, B, and C, and consider the threeassociated symbols a, b, and c in our algebra. We form all possible 2commutators using the symbols a, b, and c, and write their product

(a, b)(a, c)(b, c).

If a fourth curve is drawn according to this formula, we obtain a (4, 2)Brunnian link.

To produce a (4, 3)-Brunnian link, we draw three separated circles,call them A, B, and C, and form the product of all possible 3-commutatorson the three symbols a, b, and c. There is only one, «a, b), c). If a fourthcurve is drawn with this formula, then a (4, 3)-Brunnian link is formed.

2.7 The final generalization 49

To produce a (5, 3)-Brunnian link, we first construct a (4, 3)-Brunnianlink, call the four curves of this link, A, B, C, and D, and then write theproduct of all possible 3-commutators using the four symbols a, b, c, and d.When a curve is drawn with this forty-symbol formula, a (5, 3)-Brunnianlink is produced.

The following is the general procedure, and you may verify for yourselfthat it works. Given an (n, k)-Brunnian link, here is how to produce an(n + 1, k)-Brunnian link. Label the n simple closed curves of the (n, k)Brunnian link C1 , C2 , C3 , ..• , Cn. Let the algebraic symbols associatedwith these curves be called ClJ C2, C3' ... , Cn. Form the product of all possiblek-commutators using these n symbols. Draw a curve with this formula. Theresulting link will be an (n + 1, k)-Brunnian link.

For example, to produce a (7, 4)-Brunnian link, first construct a (5, 4)Brunnian link. Label the five curves A, B, C, D, and E. Form the product ofall possible 4-commutators using the five symbols a, b, c, d, and e. (Of thefive possible, one such 4-commutator would be «(a, b), c), d).) Draw asixth curve with this formula; label it F. This will give you a (6, 4)-Brunnianlink. Form all possible 4-commutators of the six symbols a, b, c, d, e, andhand multiply these commutators together. Draw a seventh curve with thisformula. You will then have a (7, 4)-Brunnian link, as desired.

In general, the method for constructing an (n, k)-Brunnian link is this:Start with a (k + 1, k)-Brunnian link. By using k-commutators at eachstage, form next a (k + 2, k)-Brunnian link, form from that a (k + 3, k)Brunnian link, and continue the process until the desired (n, k)-Brunnianlink is obtained. This establishes the only theorem of this chapter, which inconclusion we state below.

Theorem 2.1 For any whole numbers nand k for which the notation makessense (I ~ k < n), there does exist an (n, k)-Brunnian link.

Exercises

2.36 Let S be an expression in our algebra; that is, S is a string of symbolswith various exponents all multiplied together. Let a be an element of thealgebra. Show that even though aSa - 1 need not be the same as S, the two doin some sense represent the same curve.

2.37 Find a shorter formula than that given in Section 2.5 for a (5, 4)Brunnian link.

2.38 Prove that the deletion of all occurrences of a single symbol from a kcommutator causes the resulting formula to collapse to 1. (You may useproof by induction on the integer k; see Exercise 1.22.)

2.39 In what sense is our proof of Theorem 2.1 a proof by induction?

2.40 Sometimes it is difficult or impossible to draw certain examples ofBrunnian links without having some of the curves pass over or under themselves. This may cause such curves to be knotted. Does this matter? Canit be prevented? How?

2.41 At certain points in the construction of Brunnian links one must writeall possible essentially distinct k-commutators using n symbols, where2 < k < n. This number of different k-commutators is the same as thenumber of different k-element subsets of a set with n elements. For sets, wemay even allow k to take on the values °and I, and as an example, let usconsider a set with four elements. The number of subsets with 0, 1, 2, 3, and 4eletp.ents is, respectively, 1, 4, 6, 4, and 1. In what other context have you seenthis sequence? Can you obtain a general formula for calculating the numberof k-element subsets of a set with n elements? See Exercise 1.22 for onemethod of attack.

2.42 Make up another problem like Exercise 2.30, but harder.

2.43 Make a Mobius strip by cutting a fairly long and narrow strip ofpaper, and then attaching the ends with glue or tape after giving the stripa half-twist. Show that the strip has only one side and only one edge.

2.44 Continuing the previous exercise, start drawing a line on the Mobiusstrip, starting at a point one-third of the way from one edge and continuingaround the strip, always staying one-third of the way from that edge, untilyou reach the starting point. Predict what will happen if the Mobius strip isnow cut in two down its center line. How many edges will the new object orobjects have? How many strips will result? Verify your predictions by theexperimental method.

2.45 Continuing the previous two exercises, what do you think will happenif a Mobius strip is cut in two by keeping the scissors always one third of theway from one edge, until the starting point is reached?

2.46 Repeat the previous three exercises for a strip with two half-twists.

2.47 Show that a strip with three half-twists has but one edge and one side,and that the edge is knotted.

2.48 Since a Mobius strip has but one edge, and that edge is a simple closedcurve, one could imagine sewing two such strips together along the edges.One would obtain a surface with no boundary edge-like a sphere, but alsovery unlike a sphere in one way. How would the resulting surface differfrom a sphere?

2.49 Can a surface have one side and two edges? Two sides and one edge?What are the impossible combinations?

2.50 A currently unsolved problem in mathematics is to find the answer to thefollowing question: Does every simple closed curve in the plane contain the

vertices of a square? One method of attack might be to show that everysimple closed curve in the plane contains the vertices of a parallelogram;another approach might be to restrict the sorts of simple closed curves underconsideration. For an example of the latter approach, can you show thatevery triangular simple closed curve contains the vertices of a square? Thatis, given a triangle, is it always possible to construct a square whose verticeslie on the boundary of the triangle?


The first examples of (n, n - 1)-Brunnian links were given by H. Brunn inhis paper "Uber Verkettung", published in Sitzungberichte der BayerischenAkademie der Wissenschaften, Mathematische-Physikalische Klasse, Vol. 22(1892), pages 77-99. Hans Debrunner published in Vol. 28 (1961) of theDuke Mathematical Journal, pages 17-23, a paper entitled "Links of BrunnianType," in which he showed that Brunn's examples had the properties claimedfor them by Brunn, and further generalized these examples to show theexistence of (n, k)-Brunnian links for arbitrary nand k, I ~ k < n. D. E.Penney's paper "Generalized Brunnian Links," published in Vol. 36 (1969)of the Duke Mathematical Journal, provided an alternative construction of(n, k)-Brunnian links.

This chapter is in essence providing sufficient material on homotopytheory for the elementary study oflinks. Additional material for the advancedstudent may be found, for example, in I. M. Singer and John A. Thorpe'sLecture Notes on Elementary Topology and Geometry, published by ScottForesman in 1967. Other material on links and knots is found in Crowelland Fox's Introduction to Knot Theory, published in 1963 by Ginn andCompany, and in Topology of 3-Manifolds (edited by M. K. Fort, Jr.),published by Prentice-Hall in 1962.

The existence of wild knots, such as the one shown in Fig. 2.3, is aconsequence of the thesis of L. Antoine, published in 1921. The study ofknots, links, and wild curves and surfaces in general, has become in recentyears an exciting and rapidly advancing branch of modern mathematics.In particular, although the structure of surfaces is now well understood, theproblem of describing high-dimensional figures is much more difficult, andto date only very partial answers to most questions have been obtained.

CHAPTER 3

THEWELL-TEMPEREDCLAVICHORD

As you probably know, the "octave" on a piano has this name becausecounting both ends-it contains eight keys; these and the black keys betweensome of them are named as shown in Fig. 3.1. If we count black keys aswell, and only the essentially different notes, there are actually twelve differentkeys in an octave, and most music of the Western Hemisphere is writtenusing these and only these twelve different notes. For this reason, such musicis referred to as twelve-tone music; but you are probably aware that musiccan be written in other systems.

We shall try to answer three questions in this chapter. First, why arethere twelve different notes in an octave on the piano? Second, what wasthe proposed modification in the twelve-tone system used in the time ofJohann Sebastian Bach, a modification that he sought to call attention to bythe publication of his Woh/temperiertes K/avier? Finally, are furtherimprovements in the scale system we now use possible, and if so, how can suchimprovements be effected?

Curiously enough, some information pertaining to the answers to allthree of these questions is supplied by the methods of continued fractions,not an especially new branch of mathematics, but one which has begun toenjoy valuable applications in fields where accurate approximations toirrational numbers are needed, such as in the use of high-speed computers.Moreover, the use of continued fractions to help answer the above questionsis particularly simple, and the only prerequisite for this particular application is a degree of familiarity with logarithms.

52

3.1 Properties of logarithms 53

C# D# F# G# A# C#' D#' F#'

Fig. 3.1The names ofthe notes on

the piano.C D E F GAB C' D' E' F'

Partly to remind you of the properties of logarithms, and partly so thatthe necessary properties will be listed in this book for your convenience, weshall first take up some topics in logarithms, leaving some proofs for theexerCIses.

3.1 PROPERTIES OF LOGARITH MS

First, if b is any positive number other than 1, and a is any positive numberwhatsoever, the equation

always has a unique solution x, which may be positive, negative, or zero.This number x is called the logarithm of a to the base b, and we write

x = 10gb a.

Thus, 10gb a is the power to which the base b must be raised in order toobtain the number a. For example,

loglo 1000 = 3 and log232 = 5

54 The well-tempered clavichord

y-axis

3.1

--+----+----------- x-axisFig.3.2 The graphof y = log,o x.

because103 = 1000 and 25 = 32.

The graph of y = loglo X is shown in Fig. 3.2. Note that the graphof y = loglo x passes over only the positive x-axis-this indicates thatonly positive numbers have logarithms. Also, the graph passes through thepoint (1, 0), since 100 = 1; the graph of y = 10gb x will have this propertyfor any acceptable value of b (namely, b positive but not equal to 1). Also,the graph of y = loglo X is always increasing-that is,

This is also the case for any other value of b so long as b exceeds 1.

3.1 Properties of logarithms 55

One useful application of logarithms is in quickly finding approximateanswers to problems involving the products and quotients of many, orlarge, numbers; another especially useful application is found in obtaininggood approximations to such numbers as (17)35.61. Because of this application, the base b = lOis used for the tables of logarithms so frequentlyfound in the backs of textbooks in algebra, trigonometry, and engineeringmathematics. By choice of base 10, it becomes especially easy to locate thedecimal point in the final answer. However, the laws of logarithms whichmake such shortcut computations possible are independent of the choice ofbase, and theoretically any other positive number (except 1) will do as well.The answers obtained by such computations in any case will usually be onlyapproximations, because although you may see from a table of logarithmsthat IOg10 2 is given as 0.30103, actually this number has been rounded offto fit in the table; 10glo 2 is in fact an infinite nonrepeating decimal. Hence10°·30103 is not equal to 2, but will be very close to 2.

The mathematical properties of logarithms which make such computations possible follow from the ordinary laws of exponentiation for realnumbers, given below. For values for which the following expressions aredefined, we have always:

aO = 1 (for a =1= 0),

-x 1a =x 'a

From these laws may be derived the various laws of logarithms. We needonly two; the others are given in the exercises. Here are the two:

Let b be a positive number other than 1. Then, for all positive numbersx and y,

10gbxy = (10gbx) + (10gb y).

Also, for every positive number x and every number y,

It happens that these are exactly the two properties which also make theapproximate calculations mentioned earlier possible. For example, supposewe wish to find, using logarithms, the approximate value of the product of82,425 and 46,037. Using a table of logarithms to the base 10, we find that

10glo 82,425 = 4.91606 and 10glo 46,037 = 4.66311.

56

Thus

The well-tempered clavichord

IOg10 (82,425)(46,037) = IOg10 82,425 + IOg10 46,037

= 4.91606 + 4.66311

= 9.57917.

3.1

Using our table of logarithms in reverse to find what number has itslogarithm equal to 9.57917, we find that number to be 3,794,600,000 (thefive zeroes at the end of this number mean that our table is not sufficientlyaccurate for us to be sure of the last five digits). The correct answer to theproblem is 3,794,599,725. This problem takes almost as long to do withlogarithms as actually multiplying the original two numbers together, but in aproblem involving several products and quotients, logarithms can immenselyshorten the time needed for the computation. The price paid is, of course,the sacrifice in accuracy.

For another example, suppose we wish to find an approximate value for250. This would take quite a while to multiply out by hand, but we find thelogarithm of 2 in our table to be 0.30103, and thus

= 50· 0.30103 = 15.05150.

Again using the logarithm table in reverse, we find that the number whoselogarithm is 15.05150 is 1,125,900,000,000,000 (where, again, the elevenzeroes indicate that only the first five digits are reliable). Here the correctanswer is 1,125,920,387,354,624.

Exercises

3.1 Why can we not use the positive number 1 as a base for takinglogarithms?

3.2 Sketch the graph of y = log2 x.

3.3 Sketch the graph of y = IOg(1/2) x.

3.4 Show that if 0 < b < 1, then the graph of y = 10gb X is always decreasing. You may assume that the graph of y = 10gb X is increasing if1 < b.

3.5 Show that if 0 < band b i: 1, then the graph of y = 10gb x passesthrough the point (1, 0).

3.6 Use logarithms to the base 2 to compute the value of 4·8.

3.7 Use logarithms to the base 2 to compute the value of 43 •

3.2 A peculiar manipulation 57

3.8 Use the laws of exponents stated in this section to prove that

10gb xy = 10& x + 10gby.

3.9 Use the laws of exponents stated in this section to prove that

log" xY = Y 10gb x.

3.10 We defined x = 10gb a to mean that bX = a. Can you use this definitionin reverse, as a definition of the meaning ofb\ to prove the laws ofexponents?Of course, you may use the properties of logarithms given in the previoustwo exercises.

3.11 Let a, b, and c be numbers for which the expressions below are meaningful. Prove that

(lo&, b) . (10gb c) = lo&, c.

Hint: Let x = lo&, b, y = 10gb c, and z = lo&, c. Then ~ = b, and so on.

3.12 Let b be a positive number not equal to I, c a positive number, and aa positive number not equal to 1. Simplify the following expressions:

10gb bc, 10gb (ljb), (10gb a) . (lo&, b).

3.13 Let band c be positive numbers, neither equal to 1. Show that

1loge b = .

10gb c

3.14 Let b be a number larger than 1; if you want to be specific, you may evenassume for the purpose of this exercise that b = 10. One law of exponentsis that if x and yare any two numbers such that x < y, then bX < bY. Usethis fact to show that also, if x < y and both x and yare positive, then

10gb x < 10gb y.

3.15 You will need a table of logarithms for this exercise. Solve for x:

2x = 3.

3.2 A PECULIAR MANIPULATION

Let us assume that all logarithms to be used from now on are to the baseb = 10, since tables of such logarithms are the most readily available, andwe may then abbreviate IOg10 x by log x.

Later, one of the most important computations we shall want to performwill involve expressing a number such as (log 3)j(log 2), which exceeds 1,as the sum ofa whole number and a number between 0 and 1. We know that(log 3)j(log 2) exceeds 1 by virtue of Exercise 3.14: since 2 < 3, also log 2 <log 3, and hence (log 3)j(log 2) > 1.

58 The well-tempered clavichord 3.2

Here is the computation by which we can express (log 3)/(log 2) as thesum of a whole number and a number between 0 and I. Note how the lawsof logarithms mentioned in Section 3.1 are used.

log 3 log [(2) . (3/2)]--=----"'-------=log 2 log 2

= log 2 + log (3/2)log 2

= log 2 + log (3/2)log 2 log 2

= 1 + log (3/2) .log 2

Now 3/2 < 2, and hence log (3/2) < log 2. Thus we have obtained thedesired result provided both log (3/2) and log 2 are positive, so that theirquotient does lie between 0 and I. You will see in the exercises at the end ofthis section that it is very easy to determine that both log (3/2) and log 2are positive, and so we have achieved the desired result. Given the number(log 3)/(log 2), we have expressed it as the sum of a whole number and anumber between 0 and I, in the form

log 3 = 1 + log (3/2) .log 2 log 2

It may seem strange to you at this stage that these peculiar manipulations oflogarithms can provide us with information about problems involvingmusical scale systems, but they will.

Exercises

3.16 Show that log (3/2) and log 2 are both larger than 0; that is, that eachis positive. Hint: Use Exercise 3.14 and the fact that log I = O.

3.17 Express the number 12/7 as the sum of a whole number and a numberbetween 0 and I.

3.18 Express the number (log 5)/(log 2) as the sum of a whole number anda number between 0 and I.

3.19 Express (log 4)/(log 2) as the sum of a whole number and a numberbetween 0 and I.

3.20 Since we now know that 3/2 < 2 and log (3/2) < log 2, and thatthe latter numbers are positive, we can write

log 2 1---> .log (3/2)

3.3

Express

log 2

log (3/2)

Continued fractions 59

as the sum of a whole number and a number between 0 and 1.

3.3 CONTINUED FRACTIONS

Consider a fraction such as 12/7. We may convert this into a compoundfraction of a certain standard form as follows.

121 + ~- -

7 7

1- 1 +-

7-

5

1 + 1-

21 + -

5

1 +1

-1

1 +-5-2

1 +1

-

1 + 11

2+-2

The "standard form" into which 12/7 has been converted is this: It is acompound fraction in which each numerator is I, all signs are "+" ratherthan "-", and all numbers are positive whole numbers. If we had begunwith a negative number, we could still obtain this standard representationfor it, except that the first number on the left would be a negative number.It should be clear that every rational number can be so expressed, and that theresulting compound fraction must be finite because the denominators decreaseat each stage. But if we allow nonterminating denominators, just as we allownonterminating decimal expansions, each real number can be expressed as


such a compound fraction-these are called continued fractions. For example,

the continued fraction for .J2 is1

1 + ---------12 + ---------

12+ -------1

2 + -----2 + 1

2+

Since we have agreed that all numerators are to be 1 and all signs positive,we can adopt a much more convenient abbreviation for a continued fraction.We merely list in order the numbers to the left of each fraction, setting thefirst such number off by a semicolon, since it, unlike the others, may be zeroor negative. The above continued fraction would then be abbreviated by

(1 ; 2, 2, 2, 2, 2, ... ),

and the continued fraction we previously obtained for 12/7 would be expressed as (1; 1, 2, 2). In the latter case, the case of a terminating continuedfraction, we need to give the last denominator.

Of course, there is no difficulty in evaluating a terminating continuedfraction, such as (1; 2, 3). Only the following simple arithmetic is needed.

(1; 2, 3) = 1 + 11

2+3

1-1 +

7

3

3 10-1+-=-.

7 7

Here is a method for "evaluating" nonterminating continued fractions inwhich the numbers repeat periodically. This method will always produce thecorrect value for the sort of continued fractions we will be dealing with. Weapply the method to the continued fraction (1; 2, 2, 2, ... ). Let x =(1; 2, 2, 2, ... ). Then x = (1; 1 + x), and so

x=I+_I_1 + x

Thusx· (1 + x) = (1 + x) + 1,

so thatx + x 2 = X + 2 or

3.3 Continued fractions 61

We discard the negative solution of this last equation for reasons to be discussed later, and find that x = .J2.

If you wish to work in the opposite direction and find the continuedfraction expansion of x = .J2, first write

(x + 1)(x -- 1) = 1.or

so that x 2-- 1 -- 1,

Since we know that x '::F --1, we may divide both sides of this last equationby x + 1, and thus obtain

x -- 1 = 11 +x

Thus

x = 1 + 1

1 +x

Since the entire right-hand side of this last equation is equal to x, we substitutethe entire right-hand side for the x in the last denominator, and obtain

x = 1 +1

1 + 1 +1

1 +xor

1 +1x=

2+1

1 +x

Ifwe continue this process of substitution for the x in the denominator of theright-hand side, we obtain x = (I; 2, 2, 2, ... ), as desired.

Exercises3.21 Evaluate the cQll,tinued fraction (I; 2, 1, 3, 2).

3.22 Evaluate the continued fraction (1; 3, 3, 3, ... ).

3.23 Find the continued fraction for .J"S.3.24 Evaluate the continued fraction (1 ; 3, 4, 3, 4, 3, 4, 3, 4, ... ).

3.25 Since.J3 is between 1 and 2, the continued fraction expansion for .J3must be of the form

.J3 = (1; a1' a2, a3, a4' ... ),

where the numbers a1, a2' a3, a4' ... are all positive whole numbers. Showthat the continued fraction expansion of .J3 cannot be of the form

(I ; a, a, a, a, ... ),

where a is a positive whole number.


3.26 Find the continued fraction expansion of .Ji

3.27 Evaluate (1; 1, 2, 2) and (1; 1, 2, 1, 1).

3.28 Evaluate (1; 1, 1, 1, 1, ... ). Call the resulting number Q. Show that ifa rectangle with sides of length 1 and Q is constructed, then divided into asquare and a rectangle by a line perpendicular to the long side, the new smallrectangle also has its sides in the proportion 1 : Q.

3.29 Show that any rectangle with the property of the one constructed in theprevious exercise must have its sides in the proportion 1 : Q, where

Q = (1; 1, 1, 1, 1, ... ).

3.30 Look up The Golden Mean in a book on the history of mathematics.

3.4 THE VALUE OF A CONTINUED FRACTION

We went through a process in the previous section by which we concluded

that the value of the continued fraction (1 ; 2, 2, 2, 2, ... ) was.Ji But howcan we say that a number is "equal" to one of these nonterminating continuedfractions when such a continued fraction, because it is nonterminating, clearlycannot be evaluated directly? It is impossible to actually perform the infinitenumber of additions and divisions necessary to "evaluate" a nonterminatingcontinued fraction, so what we must do is give a definition of the value of acontinued fraction in terms of processes which actually can be carried out.

This is how it is done. Consider again our continued fraction representa-

tion of .J2; that is, .J2 = (1; 2, 2, 2, 2, ... ). Let us imagine that we wereactually trying to evaluate this "infinite" fraction, and write the sequence ofso-called partial quotients that we obtain by evaluating it out to a certainpoint and then forgetting about the rest. In this instance, we would firstevaluate (1; ), then (1 ; 2), then (1 ; 2, 2), and so on. The partial quotientsthus obtained form the following sequence:

1,3/2,7/5,17/12,41/29,99/70,239/169, ... , or

Sl = 1.000 000 000 .S2 = 1.500 000 000 .S3 = 1.400 000 000 .S4 = 1.416 666 666 .Ss = 1.413 793 103 .S6 = 1.414 285 714 .S7 = 1.414201 183 .

If you know that .J2 = 1.414213 562 ... , you notice an interestingphenomenon. The sequence of partial quotients has bracketed the value of

3.4

Fig. 3.3 .J2 as the limit of a sequence.

The value of a continued fraction

1.4 1.5

63

.J2 much as in artillery fire; the values are bouncing back and forth oneither side of the value of .J2 and are getting closer and closer to it, asindicated in Fig. 3.3. The technical term used here is that .J2 is the limitof the above sequence of numbers, and it is in this sense that we say that thecontinued fraction (1 ; 2, 2, 2, 2, ) has the value.Ji The method we usedfor "evaluating" (1; 2, 2, 2, 2, ) in the previous section, the one in whichwe substituted the x on the left-hand side of an equation for the x on theright-hand side, is merely a device for avoiding examination of the sequenceof partial quotients. This device is very useful when it works, but as you cansee, it will work only on a continued fraction that repeats its terms periodically.

You may well wonder whether or not all continued fractions have valuesin the sense of the limit of the sequence of partial quotients. After all, it isconceivable that the sequence increases without bound. Fortunately for ourpurposes, there is a theorem in the theory of continued fractions whichguarantees that if ao is any real number whatsoever and ai' a2' a3' a4' ...are all positive whole numbers, then the sequence of partial quotients of thecontinued fraction (ao; ai' a2' a3' a4' ... ) does indeed have a limit, and it isthis limit we mean when we speak of the value of the continued fraction(ao; a h a2' a3' a4' ... ). Further topics on limits can be found in the nextset of exercises.

But the above-mentioned theorem is by no means the most interestingfact about continued fractions. We shall use the following result to answersome of our questions about musical scale systems: A continued fraction


provides us with a "best possible" sequence of rational approximations to itslimit. In the last example, you can see that 17/12 is a fairly good approxima-tion to,Ji The next fraction in the sequence is 41/29. The continuedfraction for ,J2, namely (1 ; 2, 2, 2, 2, ... ), will yield up to us the informationas to which, if any, fractions with denominators between 12 and 29 arebetter approximations to ,J2 than 17/12. (Sometimes the method will produce a few which are not so good as 17/12, but these can be eliminated byinspection.) The method will be easier to see if we first try a more complicatedexample; say,

ex = (1; 3, 1, 4, 1, 3, 1, 4, 1, ... ).

The value of ex is approximately 1.261 28. The sequence of partialquotients associated with the continued fraction expansion of ex is

1/1, 4/3, 5/4, 24/19, 29/23, ....

Now 5/4 was obtained by calculating the value of (1; 3, 1) and 24/19 wasobtained by calculating the value of (1; 3, 1, 4). To find the possibly betterapproximations to ex with denominators between 4 and 19, one simplycalculates the values of the so-called intermediate fractions. One may thinkof these as being obtained in the following way. We already have 5/4 =(1; 3, 1). The next approximation given in the above sequence, 24/19, isobtained by calculating the value of (1; 3, 1, 4). To get the intermediatefractions we instead successively calculate the values of the fractions

(1; 3, 1, 1)(1; 3,1,2)(1; 3,1,3)

or, in more mathematical notation, we calculate the values of the fractions(1; 3, 1, n), where n takes on all positive whole number values between 1 and4 (since 4 is the last number in the fraction (1; 3, 1, 4) = 24/19); the firstn - 1 of these are the intermediate fractions, and the last is just 24/19 itself.If the last term of 24/19 = (1; 3, 1, 4) had been some much larger number,such as 30, then n would have to be allowed to take on twenty-nine valuesrather than three, but the principle is the same.

In any case, in the above example the three intermediate fractions turnout to be

9/7, 14/11, 19/15,

and each can be tested to see whether or not it is a better approximation toex = 1.261 28 than 5/4. We insert these three new fractions into the sequenceof partial quotients, and obtain

... , 5/4, 9/7, 14/11, 19/15, 24/19, ...

3.4 The value of a continued fraction 65

These five fractions have the following decimal values:

1.200 00 .1.285 71 .1.272 72 .1.26666 .1.263 18 .

It then is clear that each is a better approximation to 1.261 28 than its predecessor. What the theorem guarantees is this: These are the only betterapproximations with such denominators-of all fractions with denominatorsbetween 4 and 19, only 9/7, 14/11, and 19/15 are better approximations to a;

than 5/4. Similarly, there may be such intermediate fractions between othersuccessive partial quotients. It will be necessary to examine the intermediatefractions in order to answer completely the question raised at the beginningof this chapter, on how improvements in the twelve-tone system might beaccomplished.

Exercises

3.31 Recall that we evaluated the continued fraction x = (1; 2, 2, 2, 2, ... )by algebraic methods in which we obtained the equation x 2 = 2. We thenstated that the value of x must be )2, since the negative root can be ignored.If you know the definition of the value of a continued fraction in terms of itssequence of partial quotients, you will now be able to explain why the negativeroot can indeed be discarded. Please supply this explanation.

3.32 The sequence Sh S2' S3' S4' . .. of real numbers is said to have thenumber L as its limit if the following is true: Given the positive number f,

no matter how small, there exists a whole number N (probably dependenton f) such that for all values of n > N, also ISn - LI < f.

This is the precise meaning of the word "limit" as used in the precedingsection. With the aid of this definition, you can prove such theorems as this:If a sequence of real numbers has a limit, then it has only one limit. Thereare many others you can think of and prove. The concept of limit is fundamental to calculus, perhaps the best-known and most important branchof higher mathematics.

3.33 You may have noticed that the sequence of partial quotients Sl' S2'

S3' S4' ... of the continued fraction expansion of )2 had the property that

Sl < S3 < Ss < ... < )2and

)2 < ... < S6 < S4 < S2·

Try to give an informal proof why this should always be true, restricting yourattention to continued fractions with only positive whole number entries.


3.34 Insert the intermediate fractions in the appropriate places in thesequence of partial quotients of the continued fraction (1; 2, 2, 2, 2, ... ).

Indicate which of these are better approximations to .J"2 than their predecessors in the new sequence.

3.35 A desk calculator would be very helpful in working this problem.Suppose you wanted to find the continued fraction expansion of the squareroot of 10, and you alrtiady know that the decimal expansion of the squareroot of 10 looks like 3.162 277 660 . . . . You could then find the continuedfraction expansion of 3.162 277 660 instead, since this is a good approximation to the square root of 10. You would expect that the answer would besimilar to the continued fraction expansion of the square root of 10 itselfthe entries should be the same until round-off error catches up with you.Knowing the decimal expansion, you could proceed as follows:

3.162277 660 = 3 + 0.162277 660

= 3 + I__I

0.162 277 660

= 3 + 16.162 277 666

= 3 + 1 ...6 + 0.162 277 666

At each stage, you extract the largest whole number you can from the lastdenominator, invert the remainder twice (to keep it equal to what it wasbefore while producing a number larger than 1), and continue the process.After a few stages you will have a good guess as to the actual continuedfraction expansion of the square root of 10 itself. Continue the abovecomputations, and make that guess.

3.5 APPLICATIONS TO BASEBALL AND GRADEDISTRIBUTIONS

First, here is an example of how methods of continued fractions might beused to answer a question about baseball. When a player's batting averageis given as, for example, 0.263, this number is computed by dividing the totalnumber of the player's official hits by the total number of official at-bats atthat point in the baseball season. The quotient is rounded off to three-placeaccuracy. Suppose that we are given that a player's batting average is 0.263.Can we draw any conclusions about his total number of hits and total numberof at-bats?

The number 0.263 is obtained by rounding off some longer decimal,obtained perhaps by dividing 76 by 289. But it would be far too much

3.5 Applications to baseball and grade distributions 67

trouble to try all possible fractions with denominators less than one thousand(a rather generous upper bound for the total number of at-bats availableto one player in a single season) to find out which ones might round off to0.263. However, 0.263 is itself a reasonably good approximation to theplayer's "true" or unrounded average, hence we can expand 0.263 into acontinued fraction in order to find the good rational approximations to it,and thereby find the good rational approximations to the player's trueaverage. Such rational numbers are the candidates we examine, the numerators providing us with possible numbers of hits, and the denominatorsbeing the corresponding numbers of at-bats. The continued fraction expansion of 0.263 is given below:

2630.263 = - = (0; 3, 1,4, 17,3).

1000

The corresponding sequence of partial quotients is

1/3, 1/4, 5/19, 86/327, 263/1000.

We insert the intermediate fractions and obtain

1/1, 1/2, 1/3, 1/4, 2/7, 3/11, 4/15, 5/19, 6/23, 11/42, 16/61, 21/80,

26/99, 31/118, 36/137,41/156,46/175,51/194, 56/213, 61/232,

66/251, 71/270, 76/289, 81/308, 86/327, 91/346, 177/673, 263/1000.

Of these, only 5/19 and those from 21/80 on give quotients which roundoff to 0.263. However, since 5/19 appears in the list, this prevents the occurrence of other forms of the same fraction, and so we should also include10/38, 15/57, and so on. In any case, we now have all the possible combinations of hits and at-bats that give a percentage of0.263. Ifyou have additionalinformation-such as the fact that the player is a little-used pinch-hitter, orthat it is still early in the baseball season-this might be enough to concludethat the player must have had five hits out of nineteen times at-bat. Ofcourse,with such advance knowledge we wouldn't have bothered to carry the listingof the fractions out so far, but would have terminated our list when thedenominators became sufficiently large to take care of the maximum possiblenumber of at-bats.

As a second example, suppose that you know that in a certain class of nomore than sixty students, an instructor's final grade distribution was given as

A's: 10.8%B's: 24.3%C's: 37.8%D's: 8.1%F's: 18.9%


Suppose that you wish to find exactly how many students were in the class.Consider first only the percentage of students receiving "A" grades. Wefind the continued fraction expansion of this number as follows:

108 = 0 + _1_1000 1000

108

= 0 + 19 + 28

108

= 0 + __I_I

9+108

28

=0+ __1__

9 + 13 + 24

28

= 0 + __1__

9 + 1

3 + ~7

= 0 + 1 _1

9 + ----3 + 1

11 +

6

or(0; 9, 3, 1, 6).

The associated sequence of partial quotients, together with their decimalexpansions, is

1/9 = 0.111 111 .

3/28 = 0.107 142 .

4/37 = 0.108 108 .

27/250 = 0.108000 .

3.5 Applications to baseball and grade distributions 69

Now 28 is too small a denominator, since the corresponding decimal does notround off to 0.108, and 250 is too large, since the denominator represents theclass size and there are no more than 60 students. So any other possibleapproximations that round off to 0.108 can only be found among thoseintermediate fractions immediately after 3/28 and those immediately after4/37.

But there are no intermediate fractions between 3/28 and 4/37, and thefirst one after 4/37 is (0; 9, 3, 1, 1) = 7/65, which we reject since its denominator is too large. The following intermediate fractions can have only largerdenominators, hence we are already finished. In effect, the only fractionwith denominator between 1 and 60 which does round off to 0.108 is 4/37.Thus we know that there must have been 37 students in the class, and thatfour of these students received "A" grades.

Exercises

3.36 Repeat the preceding example, using the "B" grades instead. Youshould obtain a continued fraction in the form (0; 4, 8, 1, 2, 9) for 0.243,and 9/37 the only admissible ratio.

3.37 Repeat the preceding example, using the "C" grades. You shouldobtain two admissible fractions, 14/37 and 17/45.

3.38 Do this example again, this time using the "D" grades. You shouldobtain only one admissible fraction.

3.39 Repeat the example using the "F" grades. You should obtain more thanone admissible fraction.

3.40 If you repeat Exercise 3.36, altering it only in that as many as 150students may have been in the class, what results do you obtain for thenumber of students that mayor must have been in the class?

3.41 If a baseball player's batting average at the end of a season is 0.338,what is the minimum number of times he could have batted (that is, what isthe minimum number of official at-bats)?

3.42 Use the definition of limit given in Exercise 3.32 to prove that yourguess as to the limit of the sequence

1, 1/2, 1/3, 1/4, 1/5, 1/6, ...

is correct.

3.43 The value of a continued fraction has been defined as the limit of itssequence of partial quotients. How could one analogously define the "sum"of an infinite series such as

1 + 1/2 + 1/4 + 1/8 + 1/16 + ... ?


3.44 What is your guess as to the sum of the infinite series given in theprevious exercise?

3.45 How would you prove that your answer to the previous exercise iscorrect?

3.6 HARMONY

We are finally ready to turn our attention to the musical scale. A vibratingstring sets up corresponding vibrations in the air about it, vibrations whichare perceived as sound if they are sufficiently, but not excessively, rapid.The human ear is usually capable of perceiving vibrations between thirtyand seventeen thousand hertz (the currently approved term for "cycles persecond"), although at the extremes only fairly loud sounds can be heard.

It was known to the ancient Greeks that the frequency of such a vibratingstring is inversely proportional to its length, all other pertinent factors(tension, density, diameter, ... ) remaining the same. That is, doubling thelength of a string produced a frequency half that of the original string; totriple the frequency it would be necessary to use a string one-third the originallength. In addition, it was probably known to the Greeks that a vibratingstring of length (say) twelve inches also vibrated to a certain extent as if itwere two six-inch strings joined at the middle, as well as three four-inchstrings, and so on. This phenomenon, which occurs in different proportionwith different musical instruments, is known as the production of the higherharmonics of the so-called fundamental frequency of the vibrating string,and it follows that the frequencies of these harmonics are the whole numbermultiples of the fundamental frequency. See Fig. 3.4 for an indication of theway in which the higher harmonics are produced. It is the production ofthese higher harmonics in different proportions by the various orchestralinstruments that enables a composer to create music of such wide ranges ofsound. The particular combination of harmonics is one of the main reasonswhy a violin sounds different from a saxophone.

It also follows that if two strings are set in simultaneous vibration andthe length of one is twice the length of the other, then the various harmonicsmatch up, producing a pleasant (or at least a nondissonant) effect. For in thiscase the first harmonic of the longer string would be the fundamentalfrequency of the shorter, and all higher harmonics would have frequenciesequal in pairs. However, though not unpleasant, the sound of two suchtones is not a very rich sound. You can produce such a harmony by strikingany two notes one octave apart on a properly tuned piano.

It must have been discovered early in the history of music that if twostrings were simultaneously plucked, and the length of the second was twothirds the length of the first, a pleasing harmony resulted. The shorter stringproduces a frequency three-halves that of the longer one, and many of the

3.6 Harmony 71

Fundamental

Second harmonic

~---~Third harmonic

oc;;:::::: ~"""""== :..7'Fourth harmonic

............... :;7_

Fifth harmonic

Fig. 3.4 Production of harmonics by a vibrating string.


higher harmonics are equal in frequency. The mixture of two such tonesgives a richer sound than can be obtained from the harmonics of a singlestring, as you can hear by comparing the sound of middle C on the pianotogether with the G immediately above it with the sound produced by twonotes an octave apart. (The pitch of this G is, however, not precisely threehalves that of middle C, but only very close to this ratio-later in this chapterwe shall see why this is so.)

Finally, the Greeks also discovered a principle which has come to beknown as the Law of Small Whole Numbers: If the ratios of the lengths oftwo plucked strings could be expressed using small whole numbers, such as2 : 1 in the case of an octave or 3 : 2 in the case of a fifth, then the soundswere harmonic rather than dissonant. But as the numbers needed to expresssuch ratios become larger, the sounds become more dissonant; the ratios of4 : 3 and 5: 4 produce harmonies which most people find quite pleasant,whereas only the most avant-garde composers might consider that a 37 : 29ratio produces a harmonious sound.

We may now speculate about the origin of our present twelve-tone scalesystem so widely used in our Western culture. If you will refer again to Fig.3.1 at the beginning of this chapter, you will see that X # would be theblack key of the piano immediately above the white key X. There is no blackkey between E and F, nor is there one between Band C, and so a musicianwould interpret E # to mean F and B# to mean C.

If we have information about the frequencies of the notes in a singleoctave, we may use this information to find the frequencies of every othernote on the piano, for each other note is some whole number of octavesabove or below one of these, and its frequency may be found by the appropriate amount of doubling or halving the right one in our first octave.Finally, it is common to tune middle C to about 256 hertz, but this fact needconcern us no further.

After a brief excursion into the art of tuning a piano, we will be ableto continue our study of the origins of the twelve-tone system.

Exercise

3.46 Strike the key middle C on a piano, while holding down the key C'(the C one octave above middle C), and release middle C immediately. Youwill hear C' sounding its natural frequency, showing that this frequency isone of the harmonics of C. Can you find which other keys show the samebehavior, and make a table of the higher harmonics of middle C in terms ofthe other notes on the piano. Hint: Of course, C' has twice the frequency ofC, and C" (the C immediately above C') has four times the frequency of C,and so on. So, for example, you would look for the third harmonic of middleC somewhere between C' and C".

3.7 Tuning a piano, old style 73

3.7 TUNING A PIANO, OLD STYLE

We have already mentioned that the harmony produced by two strings in a3: 2 length ratio can almost, but not exactly, be produced by striking middleC on the piano together with the G immediately above it. Why is not this Gtuned to exactly three-halves the frequency of middle C?

Suppose it were, and in fact that all such pairs of fifths-two notes sevenhalf steps apart on the piano-were so tuned. Let the frequency of middleC be denoted by v. The argument that follows turns out to be independent ofthe value of v, since it cancels out, but if you wish you may assume thatv = 256 hertz. Remember that if we move down one octave, or twelve halfsteps, on the piano, the frequency will be halved. We proceed to fill in thefrequencies of all the other notes in our octave, using the fact (which we haveassumed in order to reach an eventual contradiction) that each note a fifthabove another has frequency exactly three-halves the frequency of the other.

Cv

G

(3j2)v

D'

(9j4)v

Since the note D' above has escaped our octave, we return it by dividing itsfrequency by 2:

C

v

D

(9j8)v

G(3j2)v

From the frequency of D we can calculate the frequency of A, then E'-wethen return the latter to our octave as before.

C

v

D

(9j8)v

E

(8Ij64)v

G(3j2)v

A(27j16)v

We continue this process for the other notes in the scale, and we finally endup with the following frequencies for each:

C vC# {lj2)4(3j2fvD {lj2)(3j2)2 V

D # {lj2)5(3j2)9V

E {lj2)2(3j2)4V

F {lj2)6(3j2)11 V

F# {lj2)3(3j2)6 V

G (3j2)vG# {lj2)4(3j2)8 V

A (lj2)(3j2)3vA # {lj2)5(3j2)10V

B (lj2)2(3j2)5V

C' {lj2)6(3j2)12 V


The frequency of C', the last to be obtained and the last in the above list,should give us the value v if divided by two, since C' is one octave abovemiddle C, our starting point. Thus

(1/2)7(3/2)12v == v.

We may cancel out v, and obtain next

{1/2)7(3/2)12 - 1.

When we simplify this equation, we get

312 == 219.

But this last equality is impossible, since 312 is odd but 219 is even. Thisshows that it is impossible to have all the fifths on a piano tuned exactlyso that the frequency ratio in each is exactly 3 : 2. Of course, one could tunemiddle C to the correct frequency v and then tune all the other C's on thepiano to the correct multiples of v. Then one could tune G above middle Cto the frequency (3/2)v (this can be done quite easily by ear alone with theaid of a little experience), and then set all the other G's by that one. Proceeding in this manner, one would finally obtain the frequency {1/2)6(3/2)11 V

for the F above middle C, and then all the other F's could be tuned accordingly. All the fifths would have exactly the correct ratio except for one-thefifth from F to C. Here the ratio would be a little less than 1.48 rather than1.5, quite enough difference to sound very peculiar. Pianos were tuned inthis fashion for many years (actually, in Europe, this method of tuningdisappeared about the same time as the harpsichord did), and the intervalbetween F and C was known as the Wolf Interval, because wolves howl;this was an allusion to the howling higher harmonics of the two notes, fewof which came anywhere near matching up, thus producing a number ofdissonances.

Of course, if one were to playa composition in which the notes F and Cwere used as little as possible, as in a piece written in the key of B major,the Wolf Interval would be avoided and the music would sound quite good.(If you wished to compose in C, though, you would have to have yourpiano retuned so as to place the Wolf Interval on another, little-used, fifth.)However, transposition, changing the key of a piece by moving all notes upor down a fixed amount, was impossible in the modern sense, for variousharmonic ratios would be changed by transposition. The key of compositionwas sufficiently important to composers of the seventeenth and eighteenthcenturies so that they commonly included it in the titles of their works.

Exercises

3.47 Repeat the argument of this section, using a seven-tone scale system,and thus show that it is also impossible to have all fifths in exactly a 3 : 2

3.7 Tuning a piano, old style 75

or

ratio in such a system. Note: In a seven-tone scale, a "fifth" would be aninterval of four, rather than seven, half steps.

3.48 Repeat the argument of this section, using an n-tone scale system(where n is a positive whole number larger than 1), and thus show that itis also impossible to have all "fifths" in exactly a 3 : 2 ratio.

3.49 Suppose that middle C on the piano is tuned to the frequency v, andall other C's are set accordingly. Suppose also that each note on the pianois then tuned a fixed multiple of the frequency of the note one half step belowit; that is, calling this multiple k, then the frequency of C# would be kv,the frequency of D would be k 2v, and so on. What is the value of k?

3.50 Since 2 < .J7 < 3, we know the continued fraction expansion of J7starts off like (2; ... ? ... ). So we start by writing

.J7 = 2 + (.J7 - 2),

because .J7 - 2 is a number between 0 and 1. We invert twice to obtain anumber larger than 1, and thus

.J7 = 2 + _I_I

.J7 - 2

The formula (a + b)(a - b) = a2- b2 can be used to eliminate .J7

from the last denominator, as follows:

1 .J7+2----

.J7 - 2 (.J7 - 2)(.J7 + 2)

.)7 + 2-

7 - 2 2

.)7 + 2-

3Thus

r 1-v7 = 2 + /

-v7 + 23

Since 2 < .J7 < 3, it follows that 4 < .J7 + 2 < 5, and so

4 .J7 + 2 5-< <-333

O .J7+2 1< < .3

76

Hence

The well-tempered clavichord

,fi=2+ J .7 - 1

1 +3

3.8

If you continue this process, eventually a repetition will set in; you willfind that

../7 = 2 + 1 _1

1 +------1

1 +-----1 + 1

2 + ../7

Hence ../7 = (2; 1, 1, 1,4, 1, 1, 1,4, 1, 1, 1, 4, ... ). Repeat this process in

order to find the continued fraction expansion of ../i

3.8 TUNING A PIANO, NEW STYLE

Now 312 is not equal to 219, but the two numbers are relatively close, and

careless tuning might convince someone that it would be possible, usingtwelve notes in an octave, to return to exactly the original frequency aftertuning twelve successive "perfect" fifths. This may well have something todo with the genesis of the twelve-tone scale used almost exclusively inWestern music. Other cultures have developed scale systems with othernumbers of half steps in the octave, seven being one of the most common.But the same argument as used before, in which we obtained the contradictoryresult 312 = 219

, can be repeated for such scale systems to show that anattempt to produce perfect fifths will always result in a Wolf Interval.Perhaps the first question of this chapter has been partially answered, as towhy there should be twelve notes in an octave. We say "partially," for whatwe have done so far is to give a possible evolutionary justification for thetwelve-tone system. Later we shall see that there is a mathematical reasonas well.

Johann Sebastian Bach wrote his Well-Tempered Clavichord, a collection of forty-eight piano compositions, to draw attention to an alternatetuning method which was then being considered in European musical circles.This new method, known as well-tempering, involved tuning all the C's onthe piano as before, but subsequently effecting a compromise. Rather thanhave most of the fifths in perfect harmony and one very bad Wolf Interval, itwas proposed that all the fifths be tuned very slightly flat-with a ratio ofapproximately 1.498 307 rather than exactly 1.5-the number 1.498 307chosen because it is that number which will cause all the fifths, including the

3.8 Tuning a piano, new style 77

last one tuned, to be in the same ratio. Rather than having a piano with onehowling interval, we would have a piano with twelve, but each of these twelvewolves would howl so quietly and only at such high harmonics that theywould be undetectable except to electronic instruments or to an uncommonlywell-trained ear.

Avoiding the Wolf Interval was one of the reasons brought forth tosupport well-tempering, but another must surely have had wide appeal:Because all the intervals (not just the fifths) would have fixed ratios on a welltempered piano, transposition becomes possible. A piece written in C #could be played in D or in G without changing any of the harmonic ratios,and only a person with a well-developed sense ofpitch could tell the difference.The effect of transposition would be the same as if the piece were recordedat 33t rpm and played back slightly faster or slower. And with the possibilityof transposition, vocal music becomes much easier to write and to sing.This advantage of well-tempering must have been an important factor contributing to its eventual adoption; for adopted it was, and is now used onvirtually all pianos. And perhaps we have answered the second questionraised at the beginning of this chapter, by showing one reason for the composition of the Wohltemperiertes Klavier.

The drawback to well-tempering is, of course, that none of the harmoniesother than octaves are "perfect." As we have seen, a fifth is tuned with thehigher note about 1.498 307 times the frequency of the lower, rather than1.5 exactly. We now turn to the third question mentioned at the beginning ofthis chapter. Would further improvements in the scale system be possible,and how might they be accomplished? Should it be possible, by changing thenumber of notes in an octave from twelve to some other number, to welltemper a piano, but have the ratios of the fifths much closer to 1.5 than theyare at present? It seems likely. Theoretically we could have 1200 notes in anoctave rather than 12, and the fifth might be better approximated by strikingtwo notes 702 half steps apart rather than the 700 which would correspondto the sound of the twelve-tone fifth. Of course, the technical difficultiesinvolved in the construction of such an instrument, to say nothing of theproblem of playing the monstrosity, appear intimidating; but perhaps wedo not need so many notes in order to improve the ratios of the fifths. Well,we seem to be asking for a "next best approximation," and this brings usback to the methods of continued fractions.

Exercises

3.51 It was mentioned in this section that in well-tempering, the ratio of thefifths is 1.498 307 rather than 1.5 exactly. How was this number obtained?Hint: See Exercise 3.49.

3.52 Any scale system that improves fifths will also improve fourths-pairsof notes five half steps apart in the twelve-tone system, with an ideal (but not


actual) frequency ratio of 4: 3. Why will improving fifths also improvefourths?

3.53 In music, thirds are used very extensively in producing harmonies.In the twelve-tone piano, a third can be produced by striking middle C andthe E immediately above it, or any two notes four half steps apart. Theratio of frequencies in an ideal third would be 5 : 4, or 1.25, but such cannotbe obtained for every third in the twelve-tone system for reasons similar tothose which show that not all fifths can have the perfect ratio of 1.5. Showwhy this is so. In the well-tempered twelve-tone piano, what is the actualratio of a third?

3.54 In what keys are the forty-eight compositions in the Well-TemperedClavichord written?

3.55 You may need both a table of logarithms and a desk calculator for thisproblem. Suppose that a piano were constructed with 1200 notes in eachoctave. Would the note 702 half steps above middle C produce a harmonywith middle C closer to a perfect fifth than the note 700 half steps abovemiddle C? Would the former note be better than any other on such a pianofor this purpose?

3.9 IMPROVING THE OCTAVE

Now 2(7/12) is very nearly equal to 1.5; in fact, 2(7/12) is approximately equalto 1.498 307. What we seek is some possibly larger number, say n, of notesin an octave, in which a "better" fifth would be (say) m notes up, so that2(m/n) would be a better approximation to 1.5 than 2(7/12). As in our baseballexample, it would be both difficult and time-consuming to test all possiblefractions with denominators between 12 and 1000 to find such a betterapproximation. Instead we seek rational approximations to the solution xof the equation 2x = 1.5 by finding the continued fraction expansion of x.Note that it will be unnecessary to find explicitly the decimal expansion of thesolution x in order to do this.

Using the laws of logarithms, we first transform the equation 2X = 3/2into a more tractable one, as follows:

2x = 3/2,

log 2X = log (3/2),

x log 2 = log (3/2),

log (3/2)x---- log 2 .

3.9 Improving the octave 79

We find the continued fraction expansion of x, much as we found the fraction

for ..j7 in Exercise 3.50. Since 3/2 < 2, then also log (3/2) < log 2, soinitially we invert to obtain a fraction larger than 1 to work with. We thenapply to the denominator the methods of Section 3.2, as follows

log (3/2) 1----

log 2 log 2 '

log (3/2)

1x=----

log (3/2)(4/3)

log (3/2)

1------

1 + log (4/3)log (3/2)

The fraction in the last denominator is less than 1, so again we doublyinvert it to obtain a number exceeding 1, and proceed as before. By replacinglog (3/2) with log (4/3)(9/8), we obtain as our second stage the fraction below:

1x=------

1 + 11 + log (9/8)

log (4/3)

Again, we doubly invert the fraction in the last denominator to obtain anumber larger than 1, but this time we obtain a number between 2 and 3, andslightly trickier methods are now needed:

log (4/3) log (9/8)2(256/243)-

log (9/8) log (9/8)

= 2 + log (256/243) .log (9/8)

Thus we obtain for the third stage of our continued fraction expansion of xthe following:

1x = ----------

11 + ---------

1 + 12 + log (256/243)

log (9/8)


The numbers involved get sufficiently large to cause some practicaldifficulties even with the aid of a desk calculator in about two more stages,so we carry the above no further but provide you with the results of ourlabor.

x = (0; 1, 1,2,2,3, 1, 5, 2, 26, ... ).

The corresponding sequence of partial quotients is

1/1, 1/2, 3/5, 7/12, 24/41,31/53, ....

As you recall, the reason that the fraction 7/12 appears in this sequenceis that 7/12 is a fairly good approximation to the solution x of the equation2x = 3/2. Actually, the value of x is 0.584 962 501 ... , and the differencebetween 7/12 and x is 0.001 629 .... This is quite small, and provides us witha good mathematical reason for using twelve notes in an octave, but we willhave to insert the intermediate fractions in the above sequence to find theother possibly better approximations. Between 7/12 and 24/41 there are twointermediate fractions, 10/17 and 17/29. Some arithmetic calculations provideus with the information that the error between 10/17 and x is 0.003 272 ... ,so that 10/17 is a poorer approximation to x than 7/12. Thus we cannotimprove the harmonies of fifths by using a seventeen-tone scale. For 17/29the error is 0.001 244 ... , so that a twenty-nine-tone scale would be slightlybetter than the twelve-tone scale, but not by much. It would seem hard tojustify using more than twice as many notes in an octave in order to reducethe error by less than twenty-five percent.

However, the error between 24/41 and x is much smaller-only0.000 403 ... , and we leave it to musicians to decide whether reducing theerror in the fifths to less than a quarter of its value in the twelve-tone scalewould justify building instruments with forty-one steps in each octave andwriting music for such instruments. The important idea is this: The theoryof continued fractions guarantees to us that the "next best" number of notesin an octave, after twelve, is twenty-nine, next after that is forty-one, and so on.

Note finally that there is one intermediate fraction, namely 4/7, betweenthe two partial quotients 3/5 and 7/12. This suggests that a seven-tonescale might be reasonably harmonious, and in fact some oriental music iswritten in such a system.

Exercises

3.56 It was stated in this section that the continued fraction expansion of thesolution x of 2X = 3/2 is

x = (0; 1, 1,2, 2, 3, 1, 5, 2, 26, ... ).

But only the first four numbers were found by computations shown in thetext. Verify that the next two are correct.


3.57 This is a very long problem. If you find the continued fraction expansion of the solution y of 2Y = 5/4, you will be able to find what number ofnotes in an octave would produce better approximations to the "perfectthird" ratio of 1.25 (see Exercise 3.53) than are possible in the usual twelvetone scale system. With luck, one of these solutions might also producebetter fifths. Find the next larger number of notes after 12 which will improvethirds. If possible, find the next larger number of notes after 12 which willimprove both fifths and thirds.

3.58 Think of an argument against the well-tempering system of tuning aplano.

3.59 This is a highly speculative problem. All you have to do is think aboutit, or perhaps discuss it. The human eye perceives just about an "octave"of the spectrum of radiation frequencies, and it has been customary forcenturies to think of this "octave" as made of seven "notes": red, orange,yellow, green, blue, indigo, and violet. Why seven? Is there a theory of colorharmony lurking about somewhere? Is the division of the spectrum intoseven colors a cultural phenomenon, or is there a good physical or mathematical reason for it? Note that 4/7 is a good approximation to the solutionof 2X = 3/2. Do any cultures have twelve color names, or five, or forty-one?

3.60 Can a violinist play perfect fifths? Explain your answer.


Continued Fractions, A. Ya. Khinchine's little book published in 1964 bythe University of Chicago Press in English translation, is still a good referencework on the theory of continued fractions whose entries are real or positivewhole numbers.

Science and Music, by Sir James Jeans (Dover Publications 1968 edition),contains many topics on harmony, a thorough discussion of the history ofscale systems, acoustics, and several other topics related to the material ofthis chapter. There is a minor error on page 188 of this edition, in which SirJames has neglected the possibility of a twenty-nine-tone scale system.

Analytic Theory of Continued Fractions, by H. S. Wall (van Nostrand,1948) contains a wealth of information of a much more abstract sort, mostof which is of use only to students of advanced mathematics.

Horns, Strings, and Harmony, by Arthur H. Benade (Doubleday, 1960)is another good book, one which should serve to show that this chapter dealswith only one very simple consideration in the mathematical and physicaltheory of music.

CHAPTER 4

GROUPTHEORY

Group theory is remarkable in that from such simple beginnings so muchcan be deduced, and in the wide variety of applications it has found-notonly in almost every other branch of modern mathematics, but also in suchdiverse fields as crystallography and quantum mechanics. We shall beconcerned with only a very few major theorems in this chapter, but evenin the introductory stages of group theory you will be able to appreciate itselegance, as well as learn enough to prove a wide variety of theorems onyour own.

4.1 SOME EXAMPLES OF GROUPS

We first present some examples of mathematical systems which will turn outto satisfy our subsequent definition of "group." You should ask yourselfwhat these examples have in common, for it is just these common propertiesthat willleaQ us to the abstract definition.

Example 4.1 Consider first the set W ofall whole numbers, positive, negative,and zero, together with the single operation of ordinary addition. Thisoperation is said to be closed with respect to the set W, for if m and n aretwo whole numbers then so is m + n. In other words, we cannot get outsidethe set W using the operation of addition.

Moreover, this operation also obeys the associative law: If m, n, and pare all elements of the set W, then

(m + n) + p = m + (n + p).

82

4.1 Some examples of groups 83

Since we are dealing only with the operation of addition, this law is a greatconvenience-it permits us to neglect parentheses in many cases, and we shalldo so. To put it another way, the value of an expression such as

m+n+p

is independent of either choice of placement of parentheses, and hence isunambiguous.

Third, there is one element of W, namely 0, with the property that if mis any element of W whatsoever, then

o + m = 0 = m + O.

For this reason the number 0 is called an identity element of the set W withrespect to the operation of addition. We must say "an" identity element, forit is conceivable that there might be others. Also, if we were using a differentoperation (such as multiplication) the identity element might well be someother element of W.

Finally, with the existence of an identity element assured, we may speakof an inverse of an element of W. In this example, each element of W doeshave an inverse; that is, given an element m in W, there does exist at least oneelement m' of W such that

m' + m = 0 = m + m'.

The element m' is called an inverse for m; again, it is conceivable that anelement has more than one inverse. In this particular case, m' is of coursejust the so-called additive inverse - m of the whole number m.

To summarize, then, as Example 4.1 we have a pair (W, +) consistingof a set W, a binary operation "+" on W, such that the operation is closedand associative, W contains an identity element 0 with respect to this operation, and each element of W has an inverse with respect to this operation.

Example 4.2 For our second example, we begin with an equilateral trianglein the plane. The elements of the set G that will form our group are going tobe the congruence motions of this triangle. There are a number of differentways in which an equilateral triangle is congruent to itself, and the motionsthat produce such congruences will be the elements of the group-not thetriangle itself; the triangle has been introduced for accounting purposesonly, to keep track of the effect of each congruence motion.

Three such motions are shown and labeled in Fig. 4.1. The motion whichhas the effect of rotating the triangle one-third revolution clockwise will bedenoted by S. The motion of rotating the triangle one-third revolutioncounterclockwise will be denoted by R, and the motion (if we may be permitted use of such a term in this last case) of no movement at all will be

84 Group theory

a a

4.1

b~---~c

b

c~---~a

I

b~---~c

c

a~---~b

Fig.4.1 Three congruencemotions of an equilateraltriangle.

denoted by I. If we are prohibited from moving the triangle out of the plane,these are all the possible motions. Now we have a set containing three ratherususual elements, the motions I, R, and S; in order to have a group we needan operation on this set, and so before we consider other possible motionsof the triangle, let us see how we are going to define the necessary operation.

We shall always refer to our group operation as multiplication, exceptwhen we happen to be using ordinary addition as in Example 4.1. To continue with this analogy, we shall write the product of the two group elementsx and y as x· y or even simply as xy. In the present example, we define themultiplication of congruence motions in the following manner. Givenmotions x and y, the product xy is that motion which has the net effect offirst doing x, then doing y. In Fig. 4.2, we have indicated the product of Rwith itself, and we see that RR = S, since the net effect of first doing onerotation one-third revolution counterclockwise and then another, has thesame effect on the triangle as a single rotation one-third revolution clockwise.You can also see with the aid of figures similar to Fig. 4.2 that IR = R,RS = I, and so on.


a

c~----~a

c

----~b

b----~c

Fig.4.2 Productsof congruence

motions: R . R = S.

If we are allowed to lift the triangle out of the plane in order to performcongruence motions, three more motions become possible, as shown inFig. 4.3. We have called one of these motions by the name A because thevertex labeled a of the triangle does not move as a consequence of thismotion, one which amounts to rotation of the triangle about the bisectorof the angle with vertex at a. For similar reasons Band C have received theirnames.

The important thing to remember about this operation of "multiplication" of congruence motions is that the motion A does not mean to put thetriangle into the configuration shown as a consequence of the motion A inFig. 4.3, but means to rotate the triangle about its vertical axis, regardless ofthe names pasted on its vertices. Similarly, the motion R does not mean toput the triangle with the vertex labeled c at the top, a at the lower left, and bat the lower right, but simply means to rotate the triangle one-third revolution counterclockwise.

86 Group theory 4.1

a a

..A

b c c b

!c~b c

Fig. 4.3 Congruencemotions of an equilateraltriangle by rotation about

a c b a an angle bisector.

If you wish, you might now cut an equilateral triangle from cardboardand label its vertices a, b, and c (on both sides of the cardboard), and thenyou can discover, for example, that AR = C, BC = S, and so on. We cansummarize such information about this mathematical system by saying thatthe set of objects which will form the group consists of the six motions I, R,S, A, B, and C; the multiplication is defined as above; and a "multiplicationtable" for this system is shown below.

I R S A B C

I I R S A B CR R S I B C AS S I R C A BA A C B I S RB B A C R I SC C B A S R I

This multiplication table is interpreted in the following way: The firstelement of the product XY appears in the leftmost column, the second inthe topmost row, and the product XY itself is in the "obvious" place in thebody of the table. With the aid of this table we can examine this mathematical


system for those same four properties we abstracted from Example 4.1, theexample of whole numbers with the operation of ordinary addition.

First, since each entry in the body of the table is in fact an element of theset G = {I, R, S, A, B, C}, we have the desired closure property. That is,if X and Y both belong to the set G, then so does their product XY.

Next, the operation is associative for a simple, though perhaps subtle,reason: Both (XY)Z and X( YZ) mean, in effect, first do X, then do Y, andthen do Z. This is the interpretation no matter which way the parenthesesare placed, and the net effect on the triangle will be the same in either case.Hence for all elements X, Y, and Z of the set G, (XY)Z = X( YZ). Thusthe operation is associative.

The appearance of a row identical to the topmost or guide row-thatis, the appearance of the rowwith the element I at its left-and the appearanceof a column identical to the left-most column show us that

IX = X = XI

for all elements X of G, and hence the element I of G is an identity for thisoperation.

Finally, the appearance of the identity element I at least once in eachrow and column assures us of the existence of inverses for each element of G;to find, for example, the inverse R' of the element R, locate I in the row to theright of R in the body of the table. Then find the element at the head of thatcolumn; in this case, the element is S, and so we know that RS = 1. Weverify also that SR = I, and hence S = R', the desired inverse of R, andsimilarly R = S' as well. If the other elements of G are also checked, youwill find that for each congruence motion X, there is a congruence motionX' such that

X'X = I = XX'.

Thus each element of G has an inverse in G.

There are thus four properties common to Examples 4.1 and 4.2. Theseare just the properties that will lead us to the definition of group. If we denotethe multiplication in G by ".", then we have seen that in both the system(W, +) and the system (G, .), the operation is closed and associative, anidentity exists with respect to this operation, and each element in the systemhas an inverse in the system. However, there is very little else shared bythese two examples. For instance, the operation in (W, +) is commutativem + n = n + m is true for all values of m and n in W; but in the system(G, .), RA = B whereas AR = C. Thus the order in which group elementsare multiplied is important, and we cannot assume without prior knowledgethat a given operation is commutative.

Also, the set W is infinite while the set G is finite. The system (W, +)has an algebraic or numerical origin, but (G, .) has a geometric origin. Other

88 Group theory 4.1

than the four properties we have noted that they do have in common, thereis very little that appears to suggest common properties of the two systems(W, +) and (G, .). However, it is the aim of group theory to discover thoseproperties which must be held in common by all (or certain classes of)groups.

Example 4.3 H = {O, 1, 2}. Now you know what set is to be used in thisexample. It remains only to define an operation, which amounts to filling inthe following table.

012

!iWe are using whole numbers for group elements here simply for con

venience, since there may not be enough letters in our alphabet to cover allthe examples we might wish to consider. But how do we fill in this table?We certainly cannot use ordinary addition or multiplication, for 2 + 2 = 4,and 2·2 = 4, and 4 is not an element of H. Under these circumstancesour operation could not be closed. One way we might fill in the table isshown below.

012

o 0 1 21 I 2 02 2 0 1

At this point you may be a little disturbed, if you are asking yourselfwhat this operation "is." But if you reflect a moment, you will see that youalready know everything there is to know about this operation on H; thepreceding table gives the answer to every conceivable question about theoperation, except possibly the one question that bothers many people atthis point. When they ask what this operation "is," they are really askinghow it came into being. One possible response is that this question belongsto the realm of philosophy rather than mathematics. But mathematicsshould be able to answer reasonable questions dealing with the origin andexistence of mathematical objects. So if you return to Example 4.2, andconsider only the three congruence motions in the set {I, R, S}, you can seethat the multiplication table for this subsystem is as follows.

IRS

I IRSR R S IS SIR


So our multiplication table for H = {a, 1, 2} is really the preceding tabletraveling under an alias; we have simply replaced I by 0, R by 1, and S by 2.So the above question about the origin of Example 4.3 may be answered inthis case by demonstrating a geometric origin for the elements of H and theirmultiplication. It is appropriate to mention here that the other examples ofthis section, too, as well as all groups, can be thought of as having a geometricorigin in much the above sense.

There does remain the question of checking Example 4.3 to see that thefour properties we are so interested in hold true. Closure, the existence of anidentity, and the existence of an inverse for each element of H are easilyverified by an examination of the multiplication table. And because we havedemonstrated that H is actually of geometric origin, so that the operationin H may be thought of as combination of congruence motions as in Example4.2, the operation in H is automatically associative. This is perhaps one ofthe easiest ways to demonstrate associativity; without previous knowledgeof the origin of the operation in H, one would otherwise have to check forassociativity by verifying that a rather large number of equations hold truein H. For example, one would test 0· (1 ·2) and (0· 1) ·2 for equality, andcontinue with twenty-six other cases. There is, in fact, a test for associativityof such a multiplication table that involves "looking at the table" in much thesame way that a table can be visually examined for the other three properties,but it is quite complicated and generally takes longer than almost anyalternative.

We finally present, very briefly, a few more examples of groups. Asyou have seen, it is really sufficient to list the elements of the set and give themultiplication table. We will usually do this and no more in our examples.Later, when you are making conjectures about properties of groups, you mayfind it useful to refer to these examples to test your conjectures.

Example 4.4 J = {a, I, 2, 3}.

° 1 2 3

° ° 1 2 31 1 2 3 °2 2 3 ° 13 3 ° 1 2

Example 4.5 K = {e, a, b, c}.

e a b c

e e a b ca a e c bb b c e ac c b a e

90 Group theory 4.1

Example 4.6 Zn = {O, 1, 2, 3, ... , n - I}. xy is the remainder upon division of x + y by n.

Example 4.7 E = {... , - 4, - 2, 0, 2, 4, ... }. The operation is ordinaryaddition.

Example 4.8 R+ is the set of all positive real numbers. The operation isordinary multiplication.

Example 4.9 Let D be a circular disk of radius 1 in the plane. The elementsof the group L are to be all possible congruence motions of this disk, including infinitely many different rotations and infinitely many different waysof turning the disk over about a straight line through its center. We couldeasily name these congruence motions; for example, we could denote byR 30 the motion of rotating the disk 30 degrees counterclockwise, and byT90 the motion of flipping the disk over about a line making a 90 degreeangle with the x-axis. The operation on the elements of L will be the sameas in the case of the congruence motions of the equilateral triangle, Example4.2; namely, the product (R30) • (T90) will be that motion which has the neteffect of first doing R 30, then T90 , to the disk. In this case the product is T75 ,

as you can easily verify. Again, for the same reasons as in Example 4.2,this operation is associative. It is only closure which is not so obvious.

Example 4.10 Let a regular tetrahedron in three-dimensional space have itsvertices labeled as in Fig. 4.4. The set T is to consist of all the congruencemotions of this tetrahedron. We will name these motions in such fashionas to make the effect of each easy to remember, and it will turn out that ourlabeling method also makes computation of the product of two such motionsvery simple.

First, consider the motion of rotation of the tetrahedron one-thirdrevolution counterclockwise about the altitude passing through the vertexlabeled O. This has the effect of moving the vertex in position 1 to position 2,the vertex in position 2 to position 3, and the vertex in position 3 to position 1.The vertex labeled 0 remains fixed. Hence we abbreviate this motion by(1 2 3), where it is the order of the symbols that tells us what is going on:By the motion (1 2 3) we mean that "1 goes to 2, 2 goes to 3, and (becausehere the parentheses close up) 3 goes to 1." The omission of 0 from thesymbol (1 2 3) means that the vertex at 0 is left fixed. The symbol (0 1)(2 3)represents another possible congruence motion, the one in which the verticesat 0 and 1 change places while the vertices at 2 and 3 also change places.Caution: There are some symbols that seem meaningful, such as (1 2), whichdo not represent possible congruence motions of the tetrahedron.

As in Examples 4.2 and 4.9, the multiplication we use on this set T of allpossible congruence motions of the tetrahedron goes as follows: The productxy of two such is that congruence motion which has the net effect of first doing

4.1

2

oL.---------~

Some examples of groups

Fig. 4.4 A regular tetrahedronwith vertex positions labeled.

91

x, then doing y to the tetrahedron. As we mentioned before, our method oflabeling these congruence motions makes computation of products particularly simple. If, for example, you wish to find the product

(1 2 3)(1 4 3),

all you have to do is follow each vertex through each of these motions to findout where it finally ends up. Take the vertex in position 1. After applying(1 2 3) to the tetrahedron, this vertex lies in position 2. Then, after applying(1 4 3), the vertex in position 2 does not move. Hence the net effect of theproduct

(1 2 3)(1 4 3)

on the vertex in position 1 is to move that vertex to position 2. So we canbegin writing down the answer; the product must look like the expressionbelow:

(1 2 3)(1 4 3) = (1 2

In the incomplete symbol on the right-hand side above, we next want towrite down the symbol following the 2; since the product is to tell us wherethe vertex in position 2 ends up, we follow that vertex through the two motions(1 2 3) and (1 4 3), and find that the vertex in position 2 first moves to

92 Group theory 4.1

position 3, and then, under the action of (1 4 3), is moved to position 1.We indicate this by closing up the parentheses:

(1 2 3)(1 4 3) = (1 2).

But our work is not yet complete, for it is quite possible that the verticesin positions 3 and 4 are also moved by the product (1 2 3)(1 4 3), so wefollow each of them through the two motions (1 2 3) and (1 4 3), and finallydiscover that the desired answer is

(l 2 3)(1 4 3) = (1 2)(3 4).

It turns out that the tetrahedron will admit twelve congruence motions,which we list below; since each product can be computed directly from thesymbols involved, it is as unnecessary to give the multiplication table for Tas it would be to give the complete multiplication table for the set R+ ofExample 4.8. All one needs to know is that

T = {I, (1 2 3), (1 3 2), (1 2 4), (1 4 2), (1 3 4), (1 4 3), (2 3 4),

(2 4 3), (1 2)(3 4), (1 3)(2 4), (l 4)(2 3)},

where I stands for the "motion" in which no vertex is moved. Like theprevious example, only the closure property is not so obvious.

To summarize, in all ten examples we have a set, together with an operation, such that the operation is both closed and associative, the set containsan identity with respect to this operation, and each element of the set has aninverse with respect to this operation. Any such mathematical system isknown as a group. And you now have at your disposal an infinite number ofdifferent examples, because Example 4.6 is really infinitely many examples.On the other hand, Examples 4.4 and 4.3 are really special cases of Example4.6, and in some sense Example 4.1 is really the same as Example 4.7.

Exercises

4.1 Verify that Example 4.4 is an example of a group. Note: The associativity provides the only difficulty. This can be established by "realization"of the example as a group of congruence motions, by direct computation,or by use of the Euclidean algorithm: If m and n are two natural numbers,then there exist unique integers q and r such that n = qm + rand 0 ~ r < m.Some careful use of this theorem will produce a proof that the group givenas Example 4.4 is associative, and in fact can also be used for Example 4.6.

4.2 You saw that Example 4.3 was really in some sense the "same" as thegroup {I, R, S} of rotations of an equilateral triangle. The sameness comesfrom the fact that it is possible to rename the elements of one group andthus obtain the other group (the multiplication table, when renamed, must


also correspond to the multiplication table of the other group). Show that,as stated at the end of this section, the groups of Examples 4.1 and 4.7 arereally the "same" in the above sense.

4.3 Even though the groups of Examples 4.4 and 4.5 have the same numberof elements, they are not the same in the sense of the previous exercise.Why not?

4.4 Let a square be given in the two-dimensional plane. Name each of theeight congruence motions of this square according to some reasonablescheme, and exhibit the multiplication table for this system. (As usual, theproduct of two motions is to be the motion that has the net effect of firstdoing one, then the other.) Show that this system is a group.

4.5 Let (G, .) be an arbitrary group. Show that G cannot contain twodifferent identity elements; that is, if e andfare two elements of G such that

ex = x = xe,

and

fx = x = xl,

for all elements x of G, then e = f4.6 Let (G, .) be an arbitrary group. Show that each element of G has onlyone inverse; that is, show that if x is an element of G, and y and z are elementsof G each of which acts like an inverse for x, then y = z. Your first line ofthe proof might well be:

"Let x be an element of G, and suppose that y and z are elements of Gsuch that

yx = e = xy,

and

zx = e = xz,

where e denotes the identity of G."

4.7 Show that if x is an element of the group G such that xx = x, thenx = e, the identity of G.

4.8 By virtue of Exercise 4.6, we may refer to "the" (rather than "an")inverse of a group element. Let us denote the inverse of x by X-l. Provethat if x is an element of the group G, then (X-l)-l = x.

4.9 Let G be a group and w an element of G such that

wx = x

for some (not necessarily all) x in G. Prove that w must be the identity of G.

94 Group theory 4.1

4.10 Let M = {O, I}. Find all possible ways to fill in the multiplicationtable shown below so as to make M into a group.

o I

4.11 Let H = {O, I, 2}. Find all possible ways to fill in the multiplicationtable shown below so as to make H into a group. For simplicity, you shouldassume that in any case, 0 is to be the identity for the resulting group.

012

4.12 Verify that the system shown in Example 4.5 actually is a group.

4.13 Of the examples considered in this section, some have a commutativeoperation (xy = yx is always true) and some do not (xy = yx is sometimesfalse). List those groups for which the operation is commutative and thosefor which it is not.

4.14 See Exercise 4.13. Is there a way that one can decide if a group operation is commutative by simply looking at the group's multiplication table?

4.15 Write down the complete multiplication table for the group T givenin Example 4.10.

4.16 Show that, in the multiplication table for a group, each element of thegroup appears exactly once in each row and column (excluding, of course,the guide row and guide column).

4.17 See Exercise 4.16. Fill in the multiplication table shown below in sucha way that 0 is an identity element for the operation, each element appearsexactly once in each row and column, but the resulting system is not a group.What must go wrong?

o I 234

oI234

04.18 Besides ordinary addition and multiplication, there are numerous otherassociative operations on the set R of all real numbers. Verify that theoperation # defined by

x#y=x+y+1

4.2 Subgroups 95

for all real numbers x and y is associative. See if you can find three otherassociative operations on R. We ask that these operations be closed, butnot necessarily so chosen as to make R into a group.

4.19 An operation on a set S is said to be cancellative provided that both thefollowing are true:

a) Ifax = ay,

b) If xa = ya,

then x = y.

then x = y.

Show that every group is cancellative.

4.20 An operation on a set S is said to be cross-cancellative provided that,whenever xa = ay, then x = y. Give an example of a group in which thecross-cancellative law does not hold.

4.2 SUBGROUPS

Consider the subset

U = {I, (l 2)(3 4), (l 3)(2 4), (1 4)(2 3)}

of the group T given in Example 4.10. One would generally expect a subsetof a group not to be a group in its own right if the same operation were used;one thing that might go wrong would be that the subset does not contain anidentity (which would have to be the same as the identity of the originalgroup). However, the subset U shown above does contain the identity of T.Another difficulty could be that the subset is not closed under the operationused in the group. We test U for closure by writing down its multiplicationtable as follows:

I(l 2)(3 4)(l 3)(24)(1 4)(2 3)

I

I(l 2)(3 4)(l 3)(24)(l 4)(2 3)

(l 2)(3 4)

(l 2)(3 4)I

(l 4)(2 3)(1 3)(24)

(l 3)(24)

(l 3)(24)(l 4)(2 3)

I(l 2)(3 4)

(l 4)(2 3)

(l 4)(2 3)(l 3)(24)(l 2)(3 4)

I

We see immediately from this table that not only is the operation closedwith respect to U, but also each element of U has an inverse in U. Since theoperation is automatically associative on U, U together with the operationpreviously defined for T becomes a group in its own right. Under suchcircumstances U is said to be a subgroup of the group T, and the formaldefinition is given next.

H is said to be a subgroup of the group G provided that H is a subsetof G and (H, .) is a group, where the operation is the same as that in thegroup G.

96 Group theory 4.2

Since the operation of G, when used on a subset H of G, is automaticallyassociative, in order to show that H is a subgroup of the group G it is onlynecessary to show that

a) H is a subset of G,

b) the identity e of G belongs to H, and

c) if x and yare elements of H, then so are xy and X-i.

As mentioned in Exercise 4.8, since each group element x has one andonly one inverse, we may refer to that inverse with the notation X-i. Notethat a group G is always a subgroup of itself, and that the subset {e} of Gconsisting of the identity e of G alone is alse always a subgroup of G. Thesetwo subgroups are called improper subgroups of G; all others, if any, arecalled proper subgroups of G.

The order o(G) of a group G is just the number of elements in the set G.If this set should be infinite, we say that the group G has infinite order. Forexample, oCT) = 12, where T is the group of Example 4.10; the order of itssubgroup U discussed previously is 4. The group (W, +) of whole numberswith the operation of addition provides an example of a group of infiniteorder. A group consisting of a single element (which must necessarily act asits identity) is the only possible way one can have a group of order I. Foreach natural number n, Example 4.6 provides an example of a group of ordern, namely Zn. So there is at least one group of given order. There may bemore than one; see Exercise 4.3.

If you now check some examples of groups, such as the ones previouslygiven in this chapter, you will discover no exception to the following rule:If G is a finite group and H is a subgroup of G, then o(H) divides o(G)evenly. For example, the group T of Example 4.10 has subgroups of orderI, 2, 3, 4, and 12, and all of these numbers are divisors of 12, the order of Titself. This important principle is known as LaGrange's Theorem, and because it has so many useful consequences we now provide a proof.

Theorem 4.1 (LaGrange's Theorem) If G is a finite group of order nand His a subgroup of G of order m, then min.

Proof. The notation min is just a convenient abbreviation for the phrase"m divides evenly into n." Note that min is true if and only if there existsa whole number k such that n = mk; that is, if n is an integral multiple of m.

So let G be the group of order n, and let H be a subgroup of G of order m.Now if x is any element of G whatsoever, we can form what is called aleft coset of H in G, obtained by multiplying each element of H on the leftby x, and thus obtaining a subset of G. Our notation for the left coset ofH by x in G will be xH, and an alternate way of stating the above definition is

xH = {xh I h E H}.

4.2 Subgroups 97

We now establish the following things about the left cosets of H in G:

a) H is a left coset of H in G.

b) Any two left cosets of H in G contain the same number of elements.

c) Any two left cosets of H in G are either the same, or else have no elementsof G in common.

d) Each element of G belongs to some left coset of H in G.

Consequently, the left cosets of H in G actually have the effect ofchoppingG up into a number, say k, of nonoverlapping subsets, each of which has thesame number of elements as H itself-namely m. Hence n = mk, and hencemin, as we wish to show. In the next set ofexercises, outlines of the methodsof establishing the four propositions above will be given, with the detailsleft for you to provide. Contingent upon these four exercises, the proof ofLaGrange's Theorem will be completed.

To illustrate how this proof works in a special case, let us examine whathappens if we use this coset procedure on the group G of Example 4.2.Now o(G) = 6, so we want to show that any subgroup of G has order 1,2,3,or 6. Let us use for the subgroup H, the set H = {I, R, S}. Pretend you donot immediately see that o(H) = 3, one of the divisors of o(G).

First we list the left cosets of H in G as follows:

IH = {II, IR, IS} = {I, R, S}.RH = {RI, RR, RS} = {R, S, I}.SH = {SI, SR, SS} = {S, I, R}.AH = {AI, AR, AS} = {A, C, B}.BH = {BI, BR, BS} = {B, A, C}.CH = {CI, CR, CS} = {C, B, A}.

Note how each of the four propositions we need to establish for the proofof LaGrange's Theorem is illustrated. First, H = {I, R, S} is indeed oneof the left cosets of H in G. Second, any two of the left cosets of H in Gindeed contain the same number of elements. Third, any two left cosets ofH in G are either the same or else have nothing in common. Finally, everyelement of G appears in some left coset of H in G. Since there are exactlytwo left cosets of H in G in this example, the number of elements in G mustbe double the number in each coset. In particular, the number of elements ofG must be double the number in its subgroup H, and so since o(G) = n isan integral multiple of oCR) = m, we see that min is true.

Note that LaGrange's Theorem tells us that a group of order 30 (forexample) can have subgroups only of order 1, 2, 3, 5, 6, 10, 15, and 30-butnot that such subgroups must exist. The group T of Example 4.10 has order12 but no subgroup of order 6, even though 6 I 12. So when you useLaGrange's Theorem, be sure to use it with care.

98 Group theory 4.2

Suppose that G is a group and 9 is one of its elements. By g1 wemean g. Also, by g2 we mean gg, by g3 we mean ggg, and so on; so if n is anatural number, g" is just an abbreviation for the product of g"-1 and g.In analogy to calling our group operation multiplication, and speaking of ggas the product of 9 and g, we call n an exponent in the expression g", andcall g" the !,lth power of g.

We can also define negative powers of g. If n is a natural number, thenby g-" we mean (g-1)". Finally, we convene that gO = e, the identity of G.With these definitions of exponentiation, certain laws of exponents hold;namely,a) gmg" = gm+",

b) (gm)" = gm".

However, as a general rule, only those laws of exponents hold which involvebut one element of G; it is not in general true that if 9 and h are elements ofG, then (gh)" = g"h". There is one very useful law of exponents, though,that does hold when two elements of G are involved:

c) (gh)-1 = h- 1g- 1.

One item worth remembering is that the two laws (a) and (b) above hold forall integral values of m and n, positive, negative, and zero; for example, from(b) one can derive that

for all integral values of n.Suppose we were to pick an element 9 from a group G, and examine all

positive powers of g. For example, in our group G of Example 4.2, the setof all positive powers of R is

{Rt, R 2, R 3

, R4, R 5, R 6

, R 7, .•. } = {R, S, I, R, S, I, R, ... }.

As another example, in the group R+ of Example 4.8, the set of all positivepowers of the element 3 is

{31, 32

, 33, 34, 35

, 36, 37

, ..• } = {3, 9, 27, 81, 243, 729,2187, ... }.

It may happen that when we consider the set of all positive powers of theelement 9 of an arbitrary group G, the set

{ 1 234567 }g,g,g,g,g,g,g, ...

mayor may not contain e, the identity of G. In our first example above, theset does contain the identity; R 3 = I, the identity of the group of congruencemotions of an equilateral triangle. In the second example, in the group R+,no positive power of 3 is equal to 1, the identity of R+.

4.2 Subgroups 99

If no positive power of the element 9 in G is the identity, then 9 is saidto have infinite order. If some positive power of9 is the identity, then there isa least such positive power, and this exponent is called the order of g. So inthe above two examples, the order of R is 3, and the element 3 of R+ hasinfinite order. It should be clear that in a finite group, each element hasfinite order; an infinite group may contain only one element of finite order(the identity), or may consist entirely of elements of finite order, or may evenbe mixed.

However, in a finite group, the set consisting of all positive powers of anelement turns out to be a subgroup; for instance, in the first example above,the set of all positive powers of R is {I, R, S}, which we have already seen isa subgroup of the group G of Example 4.2. Moreover, this subgroup hasorder 3, the same as the order of the element R which generates it. This isno coincidence; if a group element 9 has finite order m, then the subset

{ 1 23m}g,g,g, ... ,g

always turns out to be a subgroup of order m; and if the group involved isfinite of order n, then min by LaGrange's Theorem. Consequently, we haveanother theorem, with details of the proof left for the exercises.

Theorem 4.2 IfG is a finite group oforder nand 9 is an element of G oforderm, then min.

In summary, the order of each subgroup of a group of order n, and theorder of each element of a group of order n, must always be a divisor of n.These two facts will be found very useful in subsequent exercises.

Exercises

4.21 Find three different subgroups of the group (W, +) of Example 4.1.Does this group contain any subgroups of finite order?

4.22 Find all the subgroups of the group G of Example 4.2. Give the orderof each subgroup of G and the order of each element of G.

4.23 Does the group G of Example 4.2 consist of all positive powers of someone of its elements? Why?

4.24 Does the group H of Example 4.3 consist of all positive powers of someone of its elements? Explain.

4.25 Give the order of each element of the group J of Example 4.4, and theorder of each element of the group K of Example 4.5. Does this informationprovide a possible method of working Exercise 4.3?

4.26 What is the order of the identity e of a group G? Do any other elementsof G have the same order as e?

100 Group theory 4.2

4.27 Does the group L of Example 4.9 contain elements of finite order otherthan its identity? How many? Does the group L of Example 4.9 contain anyelements of infinite order? How many?

4.28 Does the group L of Example 4.9 contain any subgroups of order two?How many? Does L contain a subgroup of order three? Four? Any naturalnumber? Does L contain any subgroups of infinite order in addition to Litself?

4.29 Give the order of each element of the group T of Example 4.10. Isevery divisor of o(T) = 12 represented among these numbers?

4.30 Let G be a group and g an element of G of order n. Show that, for eachelement x of G, x-lgx also has order n. Warning: It is not generally truethat (x-lgx)n = (x-Itgn~. Another warning: Just because y" = e, it doesnot immediately follow that y has order n.

4.31 Find an example of a group G in which the equation (gh)2 = g2h2

does not hold for all elements g and h of G.

4.32 Let G be a group in which, for each two elements g and h of G, (gh)2 =g2h2. Prove that G must be commutative.

4.33 Let G be a group with identity e such that x 2 = e for all elements x inG. Prove that G must be commutative.

4.34 Let G be a group with subgroups Hand K. Prove that H (") K is alsoa subgroup of G.

4.35 Can a group G contain exactly one element of order 3?

4.36 Let G be a group and a and b elements of G. Suppose that the order ofab is finite. Prove that ba must then have the same order as abo

4.37 Let G be a group and a and b elements of G, neither the identity of G.Suppose that as = e and aba- l = b2. What is the order of b?

4.38 Let G be a commutative group and let H be the set of all elements x ofG such that x 2 = e. Prove that H is a subgroup of G.

4.39 Prove that if a, b, and c are elements of a finite group, then the elementsabc, bca, and cab have the same order.

4.40 Find all the subgroups of the group T of Example 4.10. What is theorder of each of these subgroups? Is every divisor of o(T) = 12 representedamong these orders?

4.41 This is how to show part (a) of the proof of LaGrange's Theorem:If G is a group and H is a subgroup of G, then H itself is a left coset of H inG. It suffices to find an element x of G such that xH = H. What is a reasonable candidate for x? Show that your candidate works.

4.42 This is how to show part (b) of the proof of LaGrange's Theorem:Any two cosets of the subgroup H in the finite group G contain the samenumber of elements. Let xH and yH be two left cosets of H in G (of course,

4.2 Subgroups 101

x and yare elements of G). Then a typical element of xH has the form xhwhere h is some element of H. Let p = yx- 1. Simplify pxh. Show thatpxh is an element of yH. Show that the set yH can be obtained by applyingp as a left multiplier to each element of xH. Show that this transformationof xH into yH by application of p as a left multiplier actually sets up a oneto-one correspondence between the elements of xH and yH. It follows thatxH and yH have the same number of elements.

4.43 This is how to show part (c) of the proof of LaGrange's Theorem:Any two left cosets of the subgroup H in the group G are either the same,or else have no elements of G in common. Let xH and yH be two such cosets.If xH and yH have no elements in common, we have no more to prove.Suppose then that xH and yH have some element of G, say g, in common.Then 9 must have the form xh1 for some element h1 of H; but 9 must alsohave the form yh2 for some element h2 of H. Hence xh1 = yh2. From thisequation, derive the equation y - 1X = h2h11

• Then show that y -1 X mustbe an element of H. Table this result for a while.

Now, to show that xH = yH, choose a typical element, say xh(h E H)ofxH, and prove that xh must belong also to yH. Use the fact that (y-1 x)H =H, which follows from the fact that y-1x E H (why?). Of course, a similarproof can be used to show that each element of y H is an element of xH,and it follows that xH = yH. Hence if two left cosets of H in G overlap,they must be equal.

4.44 This is how to show part (d) of the proof of LaGrange's Theorem:Each element of G belongs to some left coset of the subgroup H in G. Let9 be an arbitrary element of G. What is a reasonable candidate for the leftcoset xH to which 9 might belong? Show that, for your choice of x, 9actually does belong to xH.

4.45 Show that, in a group G with elements 9 and h, (gh) - 1 = h - 19 -1 .

4.46 Show that each element of a finite group has finite order.

4.47 Let G be a finite group, and 9 an element of G of order m. Let Hconsist of the first m powers of g; that is,

H = {g1, g2, g3, ... , gm}.

Show that H must be a subgroup of G. Hint: You may use the fact that foreach natural number n, the group Zn of Example 4.6 actually is a group.

4.48 With respect to the previous exercise, show in addition that o(H) = m.Hint: Clearly, o(H) ~ m. What would happen if o(H) < m?

4.49 Let G be a finite group. Show that there exists a fixed natural numberk such that, for each element 9 of G, gk = e.

4.50 Let G be a finite group of even order. Prove that G must contain anelement of order 2.


4.3 CYCLIC GROUPS AND ABELIAN GROUPS

The group G is said to be cyclic provided that G contains an element g suchthat each element of G is a power of g.

With respect to this definition, one is allowed to use all integral powers ofg, positive, negative, and zero. For example, the group (W, +) of Example4.1 is cyclic, because every element of W is a "power" of 1. Here, because ofthe additive rather than multiplication notation, we interpret 1" as n· 1, forg" means iteration of the group operation, using the element g, n times. Soin the case of the element 1 of the group (W, +), 1" would mean the sum ofn 1's:

1+1+1+···+I=n·l.

If G is a cyclic group, then an element g of G whose powers produce allelements of G is called a generator of G, and G is said to be generated by g.It is usually the case that a cyclic group has more than one generator. Thegroup (W, +) contains another generator in addition to I-what is it?

The group G of Example 4.2 is not cyclic, as you were asked to verifyin Exercise 4.23. Of the two groups J and K of Examples 4.4 and 4.5, one iscyclic and one is not-which is which? See also Exercise 4.25.

The group G is said to be Abelian if the operation in G is commutative;that is, if gh = hg for all elements g and h of G.

Numerous examples of Abelian groups have already been examined. SeeExercises 4.13, 4.32, and 4.33. Our next theorem is sometimes quite useful.

Theorem 4.3 Every cyclic group is Abelian.

Proof Let x and y be elements of the cyclic group G, and let g be a generatorof G. Then, by definition, there exist integers m and n such that x = gmand y = g". Hence

xy = (gm)(g") = gm+" = g"+m = (g")(gm) = yx.

Therefore G is Abelian.

As is frequently the case in group theory, the converse of this theoremdoes not hold. That is, there do exist Abelian groups which are not cyclic.However, there is one very important class of groups which must be cyclicmerely as a consequence of their order. Recall that a positive whole numberp is said to be prime ifp has exactly two natural number divisors-itself and 1.For example, the first ten prime numbers are

2,3,5, 7, 11, 13, 17, 19, 23, 29.

Theorem 4.4 Every group ofprime order is cyclic.

Proof Let G be a group of prime order p. Since p > 2, G contains at leastone element g not equal to the identity e of G.

4.3 Cyclic groups and Abelian groups 103

Let H be the subgroup of G generated by g. That is,

H = {gl, g2, g3, ... , gn},

where, as we have seen, n is the order ofg. Since 9 :I: e, 9 has order at least 2.But the order n of 9 must be a divisor of p = o(G), by Theorem 4.2. Sincep is prime, its only divisors are I and p, and hence n = p.

But H also has order n = p, and hence H is a subset of G containing asmany elements as G itself does. Hence H = G. So, since each element of His a power ofg, so is each element of G. Therefore by definition, G is cyclic.

As usual, the converse of this theorem does not hold. It does not followthat a cyclic group must be of prime order.

Exercises

4.51 Is every subgroup of (W, +) cyclic?

4.52 Show that the group Zn of Example 4.6 is cyclic for each natural number n.

4.53 Show that if a group G has prime order, then G is generated by eachof its elements other than the identity. Hint: Look at the proof of Theorem4.4.

4.54 Let G be an Abelian group, n an integer, and 9 and h elements of G.Show that

(gh)n = gnhn.

4.55 Let G be an Abelian group and H the subset of G consisting of allelements of G of finite order. Show that H must be a subgroup of G. Hint:Use the previous exercise.

4.56 Let G be a finite Abelian group and let k be a fixed natural number.Let H be the set of all kth powers of all elements of G. Prove that H mustbe a subgroup of G.

4.57 Let G be a group and n a natural number such that, for all elementsx and y of G,

(xyt = ~yn,

(xy)n+l = ~+1~+t,

(xy)n+2 = ~+2yn+2.

Prove that G must be Abelian.

4.58 Prove that every subgroup of a cyclic group is cyclic. Hint: Let G be acyclic group and H a subgroup of G. Let 9 be a generator of G. Then eachelement of H is a power of g. Some positive power of G must appear in H(unless H = {e}: but then H is cyclic); let k be the least positive integer suchthat gk E H. Prove that gk generates H.


4.59 Let G be a group. The center Z of G is the set of all elements z of Gsuch that zg = gz for all 9 E G. Prove that Z is a subgroup of G.

4.60 Let G be a finite group and 9 a fixed element of G. Let N(g) be the setof all elements x of G such that gx = xg. Prove that N(g) is a subgroupofG.

4.61 This is difficult. A semigroup consists of a set together with a closedand associative binary operation; the set must be nonempty. Prove that afinite cancellative semigroup is a group. See Exercise 4.19.

4.62 Suppose that G is a finite group with no proper subgroups. Prove thatG must have prime order. Warning: The converse of LaGrange's Theoremis false.

4.63 Let G be a group, H a subgroup of G, and 9 a fixed element of G.Let g-IHg consist of all elements of G of the form g-lhg, where h is anelement of H. That is,

g-IHg = {g-lhg I hE H}.

Prove that 9 - I Hg must also be a subgroup of G.

4.64 Let G be a group and 9 and h elements of G. Prove that the equation

xgx = h

has a solution x in G ifand only ifgh = a2for some element a in G. Warning:Group G is not necessarily Abelian.

4.65 Suppose that 9 is the only element of order two in the group G. Provethat 9 must belong to Z, the center of G. See Exercise 4.59.

4.66 Let G be a group containing elements x and y such that xy2 = y3xand yx2 = x 3y. Prove that x = e = y.

4.67 The subgroup H of the group G is said to be normal in G provided thatxH = Hx for all x in G. (Of course, Hx = {hx I h E H}.) Prove that thecenter Z of a group G is a normal subgroup of G. See Exercise 4.59.

4.68 Let G be a finite group of order 2n and suppose that G contains a subgroup H of order n, where n is a natural number. Prove that H is a normalsubgroup of G. See Exercise 4.67.

4.69 Let G be a finite group of order 2n and suppose that G contains asubgroup H of order n, where n is a natural number. Prove that H containsall the elements of G of odd order.

4.70 Prove that the equation x2ax = a-I has a solution x in the group Gif and only if a = g3 for some element 9 of G.

4.71 Let A be a subset of the finite group G and suppose that A containsmore than half the elements of G. Prove that each element of G is the productof two elements of A.

4.3 Cyclic groups and Abelian groups 105

4.72 Prove that if Hand K are two normal subgroups of the group G andhave only the identity of G in common, then every element of H commuteswith every element of K. That is, if hE Hand k E K, then hk = kh. SeeExercise 4.67.

4.73 Let G be a finite group of order n, where n is not 1 and n is not prime.Prove that G must contain at least one proper subgroup.

4.74 (Azriel Rosenfeld) Prove that a group with only finitely many subgroups must be finite.

4.75 (Mira Bhargava) Prove that the union of two subgroups of a group isitself a subgroup if and only if one of the two original subgroups containsthe other.

4.76 (F. M. Sioson) Prove that any semigroup in which x2y = y = yx2

for all its elements x and y must be a group. See Exercise 4.61.

4.77 (W. A. McWorter) Let G be a group of order n2 and H a subgroup of Gof order n. Prove that for each element x of G, H () x- 1Hx contains morethan one element. See Exercise 4.63.

4.78 (Michael Gemignani) Let G be a group with identity e and A a subgroupof G such that (G - A) u {e} is also a subgroup of G. Prove that eitherA = {e} or A = G. Note: G - A is the subset of G consisting of thoseelements of G not belonging to A.

4.79 (Alan Schwartz) Let G be a finite Abelian group of order n. Then n isodd if and only if each element of G is a square. (The element 9 of G is asquare if 9 = h2 for some h E G.)

4.80 (Erwin Just and Mary R. Embry) Let G be a group in which no elementhas order 2. Prove that if (xy)2 = (yX)2 for all elements x and y of G, thenG must be Abelian.

4.81 (W. A. Donnell) Let S be a semigroup satisfying the cross-cancellationlaw. Prove that S is both Abelian and cancellative. Need S be a group?See Exercises 4.20 and 4.61.

4.82 This and the next three exercises deal with one of the major theoremsabout homomorphisms. A homomorphism cp: G --+ H is just a function cpfrom the group G to the group H that "preserves the group operations"that is, for each two elements g1 and g2 of G,

CP(g1g2) = [CP(g1)] [CP(g2)]'

Of course, the operation between the square brackets is the group operationin H. The image I of such a homomorphism cp: G --+ H is the set of elementsh of H such that

h = cp(g)

for some 9 E G. Show that, in the above notation, I is a subgroup of H.

4.83 Continuing the previous exercise, we define the kernel K of a homomorphism <p from the group G to the group H as that subset of G consistingof those elements g such that

cp(g) = e,

where e denotes the identity of H. Show that, in the above notation, K is anormal subgroup of G.

4.84 Continuing the two previous exercises, let cp: G -+ H be a grouphomomorphism with kernel K. Let GJK denote the collection of left cosetsof K in G. For two such left cosets xK and yK, we define an operation on theset GJKas follows:

(xK)(yK) = (xy)K.

Show that GJK together with this operation becomes a group.

4.85 Finally, in the notation of the previous exercises, let (): GJK -+ Iaccording to the rule

()(xK) = cp(x),

where I denotes the image of cp. Show that () is then also a homomorphism,and that in fact () is both one-to-one and onto (see Chapter 7 for definitionsof the last two terms).


There are so many excellent texts and references on group theory that itwould be almost a hopeless task to list them all. A few are given below forthe convenience of the reader; however, most of those listed are fairlyadvanced.

Birkhoff, G., and S. MacLane, A Survey of Modern Algebra (Macmillan,1963).

Hall, M., The Theory of Groups (Macmillan, 1959).

Herstein, I. N., Topics in Algebra (Blaisdell, 1964).

Kurosh, A. G., The Theory of Groups (two volumes, translated by K. A.Hirsch; Chelsea, 1955).

Jacobson, N., Lectures in Abstract Algebra, volume I (van Nostrand, 1951).

MacLane, S., and G. Birkhoff, Algebra (Macmillan, 1967).

Passman, D. S., Permutation Groups (Benjamin, 1968).


I wish to thank the editors of the American Mathematical Monthly forpermission to use several problems submitted as part of the regular problemssection of that journal. These are Exercises 4.74 (E 1522, Vol. 69, 1962),4.75 (E 1592, Vol. 70, 1963),4.76 (E 1629, Vol. 70, 1963),4.77 (E 1874, Vol.73, 1966), 4.78 (E 1764, Vol. 72, 1965), 4.79 (E 1794, Vol. 72, 1965), 4.80'(E 1996, Vol. 74, 1967, and solution in Vol. 75, 1968), and 4.81 (E 2007,Vol. 74, 1967).

Joseph Louis LaGrange, who first formulated and proved the theoremthat bears his name, was born in Turin in 1736. He was recognized as anexceptionally able mathematician while still in his teens, and many considerhim second only to Euler of the mathematicians of the eighteenth century.He solved a large number of significant and difficult problems, was one ofthe first to insist on accuracy in mathematical proofs, and late in his life hebecame known as a great teacher of mathematics. His major researches,curiously enough, were not in group theory, but in mechanics, the calculusof variations, and number theory.

CHAPTER 5

POLYHEDRA

In this chapter we shall be concerned mostly with the numerical relationsthat must exist between the numbers of faces, edges, and vertices of a solidpolyhedron in three-dimensional space, and with the connection betweensuch relations and the mapmaker's problem of coloring each face of such afigure so that faces with a common boundary are colored differently. Althoughthe material of this chapter is much simpler than much of that in the previousfour, it is unusual and surprising that under such circumstances we shallfind ourselves on the very edge of the mathematically unknown. For example,suppose you are given a map of connected countries on a large island, andyou wish to color different countries in such a way that countries with acommon boundary have different colors. It has been proved that five colorsare always sufficient, but no one has ever been able to construct such a mapin which all five colors are actually required. Since it is easy to construct amap that requires four colors, one sees that the answer to the question of howmany colors are needed is either "four" or "five"; but although amateur andprofessional mathematicians have worked on this problem for over a century,no one has yet discovered which is the correct answer.

Apparently the answer has something to do with Euler's Formula, whichgives one numerical relation that must exist between the numbers of vertices,edges, and faces of a polyhedral solid; "apparently," because the proof thatfive colors are always sufficient in the case mentioned above does requireEuler's Formula. Hence we begin with some introductory material aboutpolyhedra.

5.1 THE DEFINITION OF POLYHEDRON

A peculiarity of much of mathematics is its instability under slight changes.A theorem that is true may become false with only the slightest alterationin its wording-it is even possible that a change in the order of two clauses

108

5.1 The definition of polyhedron 109

may alter the truth of the statement of the theorem. Hence it is quite importantthat the things about which these theorems are proved are defined precisely,not out of any innate desire for precision, but from the need for accuracy.Although you may be quite sure that you know what is meant by a "solidpolyhedron in three-dimensional space," someone else may well have adefinition that encompasses a wholly different set of objects. So the problemarises of formulating a definition of such objects, merely for purposes ofcommunication alone. Of course, it would be preferable if the resultingdefinition were exactly in accord with every man's preconceived notion of theterm defined, but this is not essential and is frequently impossible. Forexample, you may well feel that a polyhedron ought to be connected-that is,come in a single lump rather than several-but you could easily imaginehow one could defend an alternative definition in which polyhedra were notrequired to be connected.

If there were in the literature of Qlathematics a single and well-establisheddefinition o(the term "polyhedron," it would be the one used here; but sincethere is not, we may define this term as we choose so long as, for practicalreasons, our definition is reasonably close to the commonly accepted meaningof the term. So we choose a definition close to the common meaning, butsuited to the purposes of this chapter.

What we seek is a way of defining a solid three-dimensional object, asubset of three-dimensional space bounded by "flats" with straight edges,and connected so as to assure that the object "comes in a single piece."Since we must begin with some primitive or undefined terms, we assume asgiven and understood the terms "point," "line segment," and "plane" ofordinary euclidean geometry. Part of the definition of polyhedron willinvolve the notion of a polygon, so first we rephrase the definition of"polygon" as given in Chapter I.

A polygon is a plane figure of finite area bounded by a finite number ofstraight-line segments with the property that each endpoint of each linesegment is in fact the endpoint of exactly two of the line segments, and suchthat each two such line segments meet, if at all, in a single common endpointof each. Finally, each polygon is to be connected, and so is the boundaryof each polygon-that is, the condition that the polygon be connected meansthat

a) For each two points x andy within the polygon, there is a polygonalline L with endpoints x and y lying entirely within the polygon.

And the condition that the boundary of each polygon be connectedmeans that

b) For each two points x and y on the boundary of the polygon, there is apolygonal line L with endpoints x and y lying entirely within the boundaryof the polygon.

110 Polyhedra 5.1

Fig. 5.1 The exterior of apolygonal boundaryis not a polygon.

To understand this definition fully, you should determine exactly whatsorts of polygon-like objects are ruled out by this definition. First, thecondition that a polygon have finite area prohibits such an object as thatshown in Fig. 5.1 from being a polygon. Next, the condition that each endpoint of each line segment be, in fact, the endpoint of exactly two line segments is really a condition on the boundary of the polygon-by which wemean the union of that collection of boundary segments-and will preventan object such as that shown in Fig. 5.2 from being a polygon.

That each two line segments meet, if at all, in a common endpoint ofeach eliminates the set shown in Fig. 5.3 as a possible polygon. In this figurethe two crossing sides are understood to cross at a point not an endpointof either side. The condition that the polygon be connected will preventthe polygonal-like object shown in Fig. 5.4 from being a single polygon:this example is the union of three polygons.

The last condition, that the boundary be connected, will prevent apolygon from having a hole in it, as shown in Fig. 5.5. This consequence is aspecial case of the Schoenflies Theorem, in which it is proved that a simpleclosed curve in the two-dimensional plane is the boundary of a topologicaldisk (a "topological disk" is one that can be continuously deformed intoa circular disk). It turns out to be surprisingly difficult to prove this "obvious"theorem, and it is beyond the scope of this text even to attempt to prove it forthe special case of a polygonal simple closed curve.


Fig. 5.2 The figure is not a polygonbecause one point lies on more than two

boundary segments.

Fig. 5.3 The figure is not apolygon because two sides

intersect in a point not acommon endpoint.

112 Polyhedra 5.1

Fig. 5.4 The figureis not a polygonbecause it is notconnected.

Fig. 5.5 The figureis not a polygon becausethe boundary is notconnected.


Fig. 5.6 A slightalteration to make

the boundaryconnected.

•

These polygons, as defined above, are to be the faces of our polyhedralsolids. We could actually alter the definition and allow polygons to havepolygonal holes in them, but over and over again in the subsequent proofswe would have to convert each such polygon into one fitting the abovedefinition by slight alterations in its structure, of the sort shown in Fig. 5.6.

Now by a polygon we will mean a plane figure as defined above togetherwith its boundary, so in future use of the term "polygon" it will be permissiblefor a subset of a polygon to lie partly or wholly on the boundary of thepolygon.

Using the term "polygon" as defined above, we can now define exactlywhat is meant by a polyhedron, or polyhedral solid.

A polyhedron K is a subset of three-dimensional space such that

a) K has positive but finite volume;

b) each two points of K can be joined by a polygonal line lying entirelywithin (or partly on the boundary of) K; and

c) the boundary of K is the union of finitely many polygons, such that eachtwo of these polygons meet, if at all, in a single common vertex or asingle common edge; and such that each three such polygons meet, ifat all, in a single common vertex.

Condition (c) of this definition is meant to prevent such an object as thatshown in Fig. 5.7 from being a polyhedron; however, it will be permissiblefor a polyhedron to have a hole in it, as shown in Fig. 5.8.

114 Polyhedra 5.1

I I II I I

I I I II I I I)--~ I ~

/ I II/ I I I

/ I II/ 1 .....""1/ _----..Lc--~--

Fig. 5.7 The figure is not a polyhedronbecause more than two faces have an edgein common.

Fig. 5.8 A polyhedron with a holerunning all the way through.


I! II I 1I I 1I (

I I \ II

I \ II \ 1

J----1 " 1--/ I I I/ I I I

I : I1 I I II 1/\1

j(t \1.....-- .... --_~ .....--,.-

/

I/

I/

//

/ -----Fig. 5.9 The polygonal closedcurve in the boundary does not

separate the boundary.

The vertices of the polygons forming the boundary of the polyhedronK are called the vertices of K; the edges of the polygons are called the edgesof K; the polygons themselves are called the faces of K.

In some of the theorems which follow we shall need an additionalhypothesis that will assure us that our polyhedra have no holes in them.Imagine a large cube with a small cube removed from its interior. We couldprevent this phenomenon by requiring that the boundary of each polygonbe connected; that is, each two points of the boundary could be joined by apolygonal line lying entirely within the boundary. But this condition wouldnot prevent a hole running all the way through the polyhedron, as in Fig.5.8. If we need to exclude such a possibility, so as to consider only suchpolyhedra as could be molded from a cube of clay without cutting or pokingholes, we could add the following condition: that each polygonal closedcurve in the boundary of a polygon separates the boundary into two parts.The polyhedron shown in Fig. 5.8 does not have this last property; Fig. 5.9shows a polygonal closed curve in its boundary that does not separate theboundary.

Since we may need both the conditions we have been discussing, inorder to prevent an internal hole or a hole all the way through, we combinethem into a single definition.

116 Polyhedra 5.1

The connected polyhedron K is said to be 2-connected provided that Khas connected boundary and each closed polygonal curve on the boundaryof K separates the boundary of K into two parts.

By virtue of condition (b) of our definition of polyhedron, each timewe use the term "polyhedron" we mean one which is connected; but werefer to a connected polyhedron in the above definition to emphasize thatthis definition is unambiguous only for connected polyhedra. In any case, a2-connected polyhedron is what most people think of as a "polyhedron"the five regular solids, for example, are each 2-connected. A large piece ofSwiss cheese, even with polyhedral holes, is not.

Exercises

5.1 Draw a map of connected countries requiring four colors for a "proper"coloring-one in which two countries that have a boundary segment incommon must be colored differently.

5.2 Is there a solution to the previous exercise in which only four countriesare drawn? Why?

5.3 Can each of the countries drawn in Exercise 5.1 be shaped like arectangle? Explain your answer.

5.4 Can each of the countries drawn in Exercise 5.1 be square? Give areason for your answer.

5.5 Suppose that a map of countries is drawn in which each vertex lies onthe boundary of exactly four countries. How many colors are needed for aproper coloring of such a map? Can you prove this?5.6 Draw a map of connected countries on the surface of a sphere-use atennis ball if you wish. Consider each country to be a "face," each boundarysegment between two vertices to be an "edge," and each point where threeor more boundary segments meet to be a "vertex." Let F be the number offaces, E the number of edges, and V the number of vertices. Evaluate thenumber V - E + F. Repeat this experiment a number of times.5.7 Repeat the previous exercise, but use a torus-a figure shaped like adoughnut-instead of a sphere. Use only countries that have connectedboundaries.5.8 What happens in the previous exercise if some countries do not haveconnected boundaries?5.9 Why was it required in the definition of "polyhedron" that a polyhedronhave finite volume?5.10 A subset of three-dimensional space is said to be convex provided thateach two points of the set can be joined by a straight-line segment lyingentirely within (or on the boundary of) the subset. Must a convex polyhedronbe 2-connected? Must a 2-connected polyhedron be convex? Explain.


Fig.5.10 A Mobius strip.

5.11 If countries are not required to be connected (such as Pakistan) butthe mapmaker still wishes to have all parts of a given country colored thesame color, the solution to the coloring problem becomes more complicated.Suppose that you know that each country comes in no more than two connected parts-can you construct a map of such countries in the plane requiring eight colors for a proper coloring? Ten? More?

5.12 Find the minimum number of colors necessary to properly color eachof the regular solids, considering each face as a different country.

5.13 A Mobius strip is formed by cutting a long rectangular piece of paper,and then joining the short sides not in the expected fashion, but after givingthe strip a half twist. An example is shown in Fig. 5.10. Can you draw amap of connected countries on a Mobius strip such that you need five colorsto color the map properly?

5.14 What happens to the mapmaker's problem in three dimensions?Consider the "countries" to be three-dimensional solids, such as rectangularpolyhedra. We require that two "countries" be colored differently if theyhave part of a boundary face in common. Can you construct a "map" ofsuch countries requiring six colors for a proper coloring? Eight? Ten?More?

5.15 Repeat Exercise 5.11, but allow the countries to come in arbitrarilylarge numbers of pieces. For example, you may have one country in twopieces, a second in three, a third in four, and so on. Is there any upper limitto the number of colors needed to color all such maps?

118 Polyhedra 5.2

5.2 EULER'S FORMULA

Let K be a 2-connected polyhedron and let V, E, and F denote respectivelythe number of vertices, edges, and faces of K. The Austrian mathematicianL. Euler discovered a simple relation that must always hold between thenumbers V, E, and F:

V-E+F=2.

This is what you should have discovered in working Exercise 5.6. It isthe aim of this section to provide a proof of this relation, known as Euler'sFormula.

Since this formula has nothing to do with the interior of the polyhedronK, we immediately forget its existence, and consider only the boundary of K,as composed of a number of vertices, edges, and faces. Choose a face of Kthat you particularly like (or dislike), and remove this face from K, leavingonly the edges that it had in common with other faces of K. Since K is 2connected, it is now possible to deform the cup-shaped figure composedof the remaining boundary polygons in such a way that it lies in a plane,without changing the number of vertices or edges or the way they are connected. The remaining faces of K thus become polygons in the plane, boundedby the same edges as before; of course, the new faces cannot remain congruent to the old ones, nor can the edges retain their original lengths in thisdeformation. Figure 5.11 shows a cube before such a deformation, andFig. 5.12 shows the resulting plane figure, which is sometimes known as thenet of K.

The net of K may be thought of as a number of polygons, some of whichmay be triangles, but some of which may not. We need next to convert allthese polygons into triangles. But first we count the number of vertices,edges, and faces in the net, denoting these numbers by V, E, and Frespectively,and form the sum V - E + F. Note that V and E are the same for Kand the net of K, while K has one more face than its net since the unboundedexterior domain of the net is not counted as a face.

In Chapter 1 we provided a proof that any plane polygon can be triangulated without introducing additional vertices. In this proof, straight-linesegments were drawn in the polygon, joining various pairs of its vertices,until the polygon had been completely triangulated. If such a line segmentis drawn in one of the polygons in the net of K, thus joining two vertices ofsome polygon of the net, the value of E will be increased by one. But thevalue of F is simultaneously increased by one as well, and consequently thevalue of V - E + F will be unchanged by this process. Thus we proceed totriangulate each polygon in the net of K, and when we are done, though thevalues of E and Fwill generally change, the value of the expression V - E + Fis not altered.

5.2 Euler's Formula 119

Fig.5.11 The boundary of a cube inthree-dimensional space.

IIIIIr------

///

Fig.5.12 After one face of the cube isremoved, its boundary can be deformed

so as to lie in a plane.

120 Polyhedra 5.2

Fig. 5.13 The net of the cube isnow completely triangulated.

In Fig. 5.13 we show the net of K completely triangulated; we are continuing with the cube as the model for this proof. We now attack thiscollection of triangles with an eraser. Our aim is to erase all the trianglesbut one, one at a time, without changing the value of V - E + F (thoughwe are certainly going to change the values of the individual terms). In theerasing process, we also want to have a connected net of triangles at eachstage. Two types of erasing will be needed. Figure 5.14 shows that one canerase a single edge, thus removing one edge and one face. The line to beerased is shown as a dashed line in that figure. Since this erasure decreasesthe value of each of E and F by 1, the value of V - E + F is unchanged.Figure 5.15 shows an example of the other type of erasing needed; one canalso erase a vertex and two edges, shown as dashed lines in that figure. Thissecond type of erasing decreases each of V and F by 1 and simultaneouslydecreases E by 2. Again, the value of V - E + F is unchanged.

At each stage of the erasing process, at least one of these two types oferasures can be performed. We take care only to erase vertices and edgeslying on the "outside" of the net, so that the net remains connected at eachstage. Since at each erasure the value of F is decreased by 1, the process canbe continued until after a finite number of steps the value of F becomes Iand a single triangle is all that remains.

5.2 Euler's Formula 121

Fig.5.14 Removing one triangle byerasing an edge.

Fig.5.15 Removing one triangle by erasingtwo edges and their common vertex.

/'I \I \

I \I \

I \I \

I \

122 Polyhedra 5.2

The value of V - E + F is the same for this triangle as it was for the netof K, but since a triangle has V = 3, E = 3, and F = I, the value of V E + F is 3 - 3 + 1 = 1. Thus the value of V - E + F must also be 1for the net of K. Now K itself has the same number of vertices and edges,and one more face, than its net, so the value of V - E + F for K must be 2.This establishes Euler's Formula: for any 2-connected polyhedron,

V - E + F = 2.

Exercises

5.16 Calculate the value of V - E + F for a polyhedron shaped roughlylike a cube with a single hole running all the way through, such as the oneshown in Fig. 5.8. Repeat this for a polyhedron with two holes, three holes,and four holes. Generalize. How might you go about proving your guessis correct?

5.17 You can see from the proof of Euler's Formula that it is not actuallynecessary for the edges involved to be straight. Hence for any map ofconnected countries on the surface ofa sphere, if F is the number of countries,E the number of boundary segments, and V the number of vertices, thenit is still true that V - E + F = 2. We can also use the formula VE + F = 2 for a map of connected countries in the two-dimensional plane,provided that we interpret the unbounded outside region as a large country.You can use this somewhat generalized form of Euler's Formula to work anumber of problems, including this one: Suppose you are given five pointsmarked in the plane and lines are drawn from each to the other four, detouring around vertices. This will make ten lines in all. Prove that twoof these lines must intersect.

5.18 Prove that three houses cannot each be connected to each of threewells by nonintersecting paths. Hint: See Exercise 5.17.

5.19 Prove that if each country of a map on a sphere has exactly three edges,then the number of countries is even.

5.20 If a map of connected countries on a sphere has 60 vertices and eachcountry has exactly three edges, how many countries are there?

5.21 Suppose that a map of connected countries on a sphere has the propertythat each vertex lies on an even number of edges. What is the minimumnumber of colors required to properly color such a map?

5.22 Prove that there is no map covering the surface of a sphere such thateach vertex lies on exactly four edges and each country has exa~tly six sides.

5.23 Suppose a map is given in the plane such that each vertex lies on exactlythree edges and no two countries have in common more than a single vertexor single edge. Show that at least one country in the map must have five or

5.3 Regular solids 123

fewer edges. Hint: For each natural number n, let Fn be the number ofcountries that have exactly n edges. Suppose that a map such as the onementioned above has no country with five or fewer edges. Then

F1 = F2 = F3 = F4 = Fs = 0,

and

F6 + F7 + Fs + ... = F.

Moreover, since each vertex lies on exactly three edges, 3V = 2E. Finally,the number of edges in the map can be counted as follows: For each countrywith six edges, we count all the edges and get 6F6 • Similarly, we count theedges using countries with exactly seven edges, and get 7F7 • The totalnumber of edges counted in this fashion is then

6F6 + 7F7 + 8Fa + ... ,and this process counts each edge exactly twice, so

2E = 6F6 + 7F7 + 8Fa + ,> 6F6 + 6F7 + 6Fa + .= 6F.

Now you have the following three relationships:

V - E + F = 2,

3V = 2E,

2E > 6F.

Does this lead to a contradiction?

5.24 Figure 5.16 shows a map of countries on an island. Prove that thismap cannot be colored with only three colors in such a way that adjacentcountries are colored differently.

5.25 Suppose the Martians have divided their planet into two countries,one occupying the Northern hemisphere and called Northia, and oneoccupying the Southern hemisphere and called Southia~There is then a singleboundary edge, the Martian equator. Does Euler's Formula hold for thissort of map? If not, what convention about boundary edges is necessaryto make the formula valid?

5.3 REGULAR SOLIDS

A regular solid is a 2-connected polyhedron each of whose faces is congruentto a given regular polygon and each of whose vertices lies on the samenumber ofedges. It was known to the ancient Greeks that there could be onlyfive regular solids, but a careful use of Euler's Formula will show a much

124 Polyhedra 5.3

Fig. 5.16 A map that cannotbe properly colored withonly three colors.

more general result. It turns out that the metric or geometric properties arenot in themselves what prevent the existence of a sixth regular solid, butmerely the numerical relationships between the numbers V, E, and F.

Theorem 5.1 There can be no more thanfive regular solids.

Proof. Suppose we have a regular solid; that is, a 2-connected polyhedronwith the property that the same number of edges meet at each vertex, andsuch that each face is bounded by the same number of edges. The lattercondition means that we are actually considering figures that are much moregeneral than regular solids, for in a regular solid not only is each face boundedby the same number of edges, but in fact each face is congruent to a fixedregular polygon. But we shall show, nevertheless, that there can be but fivesuch figures, and it then follows that only five regular solids can exist.

Since the number of edges incident at each vertex is the same, let usdenote this number by p; similarly, let q denote the fixed number of edgesbounding each face. Let us first count the number of edges by multiplyingthe number p incident at each vertex by the number V of vertices; we obtainp V. But since this process counts each edge twice, we find that

pV = 2E.

Let us also count the number of edges by multiplying the number qbounding each face by the number F of faces; we obtain qF. This processalso counts each edge twice, and hence

qF = 2E.

5.3 Regular solids 125

Using the above two equations, we express each of E and Fin terms of V,and we obtain

E = pV2

and F = pV.q

We substitute the above in Euler's Formula, V - E + F = 2, which holdsfor such a figure as we are considering, and we find that

pV pVV--+-=22 q ,

or

2Vq - pqV + 2pV = 4q.

We solve for V, and we obtain the equation

v = 4q2p + 2q - pq

Now V is a positive whole number, and so is 4q, so that the last denominatormust be positive. That is,

2p + 2q - pq > 0,

or

pq - 2p - 2q < o.

We next add 4 to both sides of this last inequality in order to make itpossible to factor the left-hand side, and we get

pq - 2p - 2q + 4 < 4,

which, when factored, becomes

(p - 2)(q - 2) < 4.

Our definition ofpolyhedron implies that each face has at least three boundingedges and that there are at least three edges incident at each vertex. Thuseach of p and q is no less than 3. Hence both p - 2 and q - 2 are positivewhole numbers, and their product is less than 4. The only possibilities arethese:

p = 3, q = 3: The tetrahedronp = 4, q = 3: The octahedronp = 3, q = 4: The cubep = 5, q = 3: The icosahedronp = 3, q = 5: The dodecahedron

126 Polyhedra 5.3

The values of V, E, and F may be found by substitution in the equations

v = 4q ,2p + 2q - pq

E = pV2 '

F = pV.q

For example, for p = 5 and q = 3, we obtain

V = 12, E = 30, F = 20,

so that the figure in question must have 20 faces, each a triangle (since q = 3);30 edges; and 20 vertices, each the meeting place of exactly five edges (sincep = 5). We have referred to such a figure above as an icosahedron, whereasthis is more properly the name of the regular solid with such a number ofvertices, edges, and faces. What we have in fact shown is this: If one wishesto glue together a number of not necessarily equilateral triangles to form a2-connected polyhedron with five edges incident at each vertex, then 20triangles must be used, and no other construction is possible. In particular,there can be at most five regular solids in the classical sense.

Exercise

5.26 We have shown in Theorem 5.1 that there can be at most five regularsolids, but the theorem does not show that the five solids actually exist.Some very careful work with blocks of wood, solid geometry, and a bandsawmay make it seem very likely that all five do exist, but there is a difficulty.No matter how carefully the bandsaw is used, and no matter how carefullyyou measure lengths and angles, there is always the possibility that the figureyou have constructed is almost, but not quite, regular, since errors in lengthsand angles may be beyond the limits of physical measurement.

Suppose you were given eight equilateral and congruent triangles, andwere asked to prove that these can be assembled to form the boundary of aregular octahedron. It would not be difficult to show that the first sevencould be matched up, each edge coinciding with another edge in most cases,but then you would have the problem of showing that the last triangle wouldexactly fit the triangular hole left in the constructed figure. This wouldprobably involve some very difficult solid analytic geometry and all sorts oflengthy equations. Can you think of a way to prove the last triangle wouldfit without all this agony? Then proceed to prove the existence of the otherfour regular solids.

5.4 A converse of Euler's Formula 127

5.4 A CONVERSE OF EULER'S FORMULA

Suppose you are given three positive whole numbers, say a, b, and c. Supposealso that these numbers satisfy the relationship of Euler's Formula: a - b +c = 2. Need there exist a 2-connected polyhedron with V = a, E = b,and F = c? Obviously not, for it seems clear that in any polyhedron, thenum1?ers V and F must be at least 4. But even if this condition is met, doessuch a polyhedron have to exist? If not, can we find conditions on thenumbers a, b, and c that will assure its existence? What we are asking for isthis: Given positive whole numbers V, E, and F, such that V - E + F = 2,what conditions on V, E, and F will assure the existence of a 2-connectedpolyhedron with V vertices, E edges, and F faces?

E. Steinitz answered this question in a paper published in 1906. Itturns out that the answer to the above question is the double inequality

V + 4 ~ 2F < 4V - 8.

This inequality says, roughly, that the number of faces must be a little morethan half the number of vertices, but not quite so much as twice the numberof vertices. If you wish, you may convert the above inequality to the alternative form

F + 4 ~ 2V ~ 4F - 8.- -

This shows that there is some symmetry hidden in the original inequality.Later we will show a more natural geometric interpretation of this inequality.We state Steinitz's result as a theorem and proceed with its proof.

Theorem 5.2 Let V, E, and F be three positive whole numbers. Then thereexists a 2-connected polyhedron with V vertices, E edges, and F faces if andonly if both of the following relations hold:

V - E + F = 2,

V + 4 ~ 2F :s 4V - 8.

This is called an "if and only if" theorem, for we must really supplytwo almost independent proofs for the two theorems that Theorem 5.2 reallyis. First we have to show that if K is a 2-connected polyhedron, then both theabove relations hold. Then we must show that if the numbers V, E, and Fsatisfy both the above relations, then there exists a 2-connected polyhedronK with the corresponding numbers of vertices, edges, and faces. The secondpart will be a constructive proof; that is, we will show how one actually wouldgo about building the needed polyhedron. But the first part is much easier,and so that is where we begin the proof.

128 Polyhedra 5.4

Proof Suppose that K is a 2-connected polyhedron with V vertices, E edges,and F faces. We have already shown in Section 5.2 that V - E + F = 2,and so it remains only to show that

V + 4 < 2F ~ 4V - 8.

Now there are at least three edges incident at each vertex. If we were tocount all the edges at each vertex and there happened to be exactly threeedges incident at each vertex, we would obtain the number 3V, and we couldthen say that 3V = 2E since in this process each edge would be countedtwice. However, since there may be more than three edges at some vertices,let us only count three and ignore any others. The number we obtain by suchcareless counting will still be 3V, but since we may have ignored some edges,we can say only that

3V ~ 2E.

Similarly, we count only three edges in the boundary of each face, eventhough some faces may have more. If each face had exactly three edges,we would find that 3F = 2E; since we ignore some edges in this processwe can say only that 3F ~ 2E. We know that Euler's Formula holds for K,and so our information to this point may be summarized as follows:

V - E + F = 2,

3V < 2E,

3F < 2E.

We transform the first equation into

4 + 2E = 2V + 2F,

and replace the quantity 2E which appears in the resulting equation withquantities known to be smaller (or at least no larger): first 3V, then 3F.We thus obtain the two inequalities

4 + 3V ~ 2V + 2F,

4 + 3F < 2V + 2F.

We simplify each; the first becomes

4+V~2F,

and the second becomes

4 + F < 2V,or

F ~ 2V - 4,or, finally,

2F ~ 4V - 8.

Fig. 5.17A cut of type A.

We now have both 4 + V ~ 2F and 2F < 4V - 8. Combining these twointo a single relationship, we get

V + 4 ~ 2F < 4V - 8,

and the first part of the proof is complete.To finish the proof of the theorem, we need to show that if V, E, and F

are three positive whole numbers satisfying both the relations

V - E + F = 2,and

V + 4 :::;; 2F ~ 4V - 8,

then there exists a 2-connected polyhedron K with V vertices, E edges, and Ffaces. We shall construct Kby starting with a regular tetrahedron and literallysawing off various chunks until we obtain the desired polyhedron. You canimagine these saw cuts as actually being done with a bandsaw. There arethree types of cuts that will be needed. Each is to be applied to a vertexwhere exactly three edges of the polyhedron meet. Since we are beginningwith a tetrahedron, it will always be possible to make the first cut. We willneed to show that in the construction of K, each cut we perform leaves us withat least one vertex where exactly three edges meet, so that the process can becontinued until K is obtained.

Cut of Type A.' See Fig. 5.17. Let P be a 2-connected polyhedron and v avertex of P where exactly three edges meet. Choose three points a, b, and c,one on each of these three edges, and much closer to v than to the other

130 Polyhedra

------ ba I\ v I\ I\ I\ I\ I\ I\ I\ I\ I\ I\ I\ I\ I

c

" \\

Fig. 5.18A cut of type B.

5.4

vertices of P. Cut along the plane determined by a, b, and c, and discard thesmall tetrahedron with vertices v, a, b, and c.

Cut of Type B: See Fig. 5.18. Let P be a 2-connected polyhedron and v avertex of P where exactly three edges meet. Choose two points a and b,one on each of two of these edges, and much closer to v than to the othervertices of P. Let c be the vertex other than v on the third edge incident at v.Cut along the plane determined by a, b, and c, and discard the small tetrahedron with vertices v, a, b, and c.

Cut of Type C: Let P be a 2-connected polyhedron and v a vertex of Pwhere exactly three edges meet. Let a be a point on one of these edges muchcloser to v than to the other vertices of P, and let band c be, respectively, thevertices of P other than v on the other two edges incident at v. As indicatedin Fig. 5.19, cut along the plane determined by a, b, and c, and discard thesmall tetrahedron with vertices v, a, b, and c.

In each case, a new vertex will be formed at the point a, and it is easy tosee that exactly three edges must be incident at this vertex. Hence if we useonly cuts of the three types described above, and begin with a tetrahedron,we shall always have available at least one vertex where exactly three edgesmeet. So this cutting process can be continued as long as we please. Moreover, it should be at least intuitively clear (though it is not difficult to prove)

5.4

Fig. 5.19A cut of type C.

A converse of Euler's Formula

~-- I_~~ I

~-- I....- I

- Ia \ I\ v I\ I\ I\ I\ I\ I\ I

\ /\ I\ I\ /

c

'.,, ,,

131

that since we begin with a tetrahedron-which is 2-connected-we producea 2-connected polyhedron at each stage as a result of each of these cuts.

It is easy to show the following important facts about the results of thesetypes of cuts. First, a cut of type A increases the number of vertices by twoand the number of faces by one. Second, a cut of type B increases the numberof vertices by one and the number of faces also by one. However, some caremust be used in considering what happens as a result of a cut of type C.We must avoid using a cut of type C at a vertex v where three edges meet,but where also the three faces meeting at v are all triangles. Figure 5.20 showswhat would happen if a cut of type C were applied in such a case. Thenumber of vertices, edges, and faces would not change, so a cut of type Chere would do us no good in constructing our desired polyhedron K. Wemust be sure, when applying a cut of type C, that not only is this cut appliedat a vertex v where exactly three edges are incident, but in addition one of thethree faces incident at v must have four or more edges. Moreover, we mustchoose the points band c indicated in Fig. 5.19 both in the boundary of theface with four or more edges. Then when a cut of type C is applied, the facewith four or more edges is divided by the line from b to c into two faces.

132 Polyhedra 5.4

\

Fig. 5.20 The case when a cutof type C should not be used.

Hence, under the right conditions, a cut of type C will increase the number offaces by one without changing the number of vertices.

Note that we do not care what happens to the number of edges as aresult of any of these cuts, for the following reason: When we: finally succeedin constructing a polyhedron K with the desired number of vertices andfaces, the number of edges will take care of itself because Euler's Formulamust hold for K.

Now we are ready to proceed with our construction of the polyhedronK, which is to have Vvertices, E edges, and Ffaces. We begin with a tetrahedron, which has four vertices, six edges, and four faces. Suppose first thatV = F. In this case, simply apply F - 4 cuts of type B. The number ofvertices and faces of the tetrahedron will then each be increased by F - 4,so the resulting polyhedron will have 4 + (F - 4) vertices, or F vertices.But in this case, V = F, so that the resulting number of vertices will be V,as desired. Moreover, the resulting polyhedron will have 4 + (F - 4) = Ffaces, also as desired. As we mentioned earlier, the value of E takes care ofitself, as in any case E = V + F - 2.

The second possibility is that F < V. If so, first apply to the tetrahedron2F - V - 4 cuts of type B and then V - F cuts of type A. Again, it is nothard to verify that the resulting polyhedron does indeed have V vertices andFfaces.

Finally, the only complicated case: that in which V < F. Here we mustuse cuts of type C, but we must make sure that such a cut is applied only to a

12 • • • • • • •11 • • • • • •10 • • • • • •9 • • • • •

:::::. 8 • • • • •..... 7 • • • •0CIl

6Q.l • • • • • •::Jco 5 • • •> • • • •

Fig. 5.21 The graphical 4 • • • • • • • •interpretation of the formula 3 • • • • • • • •V + 4 ~ 2F ~ 4V - 8.

2 • • • • • • • • •1 • • • • • • • • • •

2 3 4 5 6 7 8 9 10 11

(Values of F)

vertex where three edges meet such that, of the three faces also incident there,at least one has four or more edges. We do so as follows: We apply one cutof type B followed by one of type C, using for the second cut one of the newvertices formed by the cut of type B. It is not hard to see that a cut of type Bwill always produce at least one new face with four or more edges, and thatone of the new vertices created on the edges of this face will be a vertexwhere exactly three edges meet. So we use this vertex for the next cut oftype C.

We perform this process of first applying a cut of type B, then a cut oftype C, exactly F - V times. This is possible since V < F. After this processis completed, we next apply exactly 2V - F - 4 cuts of type B. Again, it iseasy to verify that the resulting polyhedron does have exactly V vertices andF faces, as desired. This concludes the proof of Steinitz's Theorem.

Exercises

5.27 The graph shown in Fig. 5.21 shows plotted dots, each representinga possible value pair for V and F. The two lines are the graphs of

V + 4 = 2F,and

2F = 4V - 8.

134 Polyhedra 5.5

The shaded region between the lines represents possible value pairs for Vand F for which a polyhedron can exist; outside this region none can exist.Why?

5.28 The proof of Steinitz's Theorem shows how one may begin with atetrahedron and construct a polyhedron with given values of V and F.Note that the tetrahedron itself is represented by the point of intersectionof the two lines shown in Fig. 5.21. Give a geometric interpretation of theconstruction in Steinitz's Theorem with the aid of Fig. 5.21.

5.29 Show that the application of a cut of anyone of the three types used inthe proof of Steinitz's Theorem to a convex polyhedron produces a convexpolyhedron. See Exercise 5.10. As a consequence of this exercise and Exercise5.10, the construction used in the proof of the second part of Steinitz'sTheorem works because each polyhedron used is convex. Why is this enoughto make the construction work?

5.30 Ifa 2-connected polyhedron has 20 edges, what is the maximum numberof vertices it can have? What is the minimum number?

5.31 Verify that a cut of type A does indeed increase the number of verticesof a polyhedron by two and the number of faces by one.

5.32 Verify that a cut of type B does indeed increase the number of verticesof a polyhedron by one and the number of faces by one as well.

5.33 Verify that application of a cut of type C in the proper place on apolyhedron does increase the number of faces by one without changing thenumber of vertices.

5.34 In the proof of Steinitz's Theorem, where the polyhedron K is constructed, it is shown only for the case V = F that the constructed polyhedronactually has the desired number of vertices and faces. Verify that the valuesof V and F also come out right in the other two cases, the cases of F < Vand V < F.

5.35 To continue the previous exercise, how many cuts are actually neededto produce the polyhedron K from a tetrahedron? Your answer should be interms of the number F of faces that K is supposed to have.

5.5 MAP COLORING

As we mentioned at the beginning of this chapter, it is an unsolved problemwhether or not four colors are sufficient to color every map of connectedcountries on the surface of a sphere (or, for that matter, on the surface of any2-connected polyhedron). With the aid of Exercise 5.23, which dependsitself on Euler's Formula, it is possible to show that each such map can becolored using no more than five colors; the answer to Exercise 5.1 shows thatfour colors are necessary. But no one has ever been able to construct such a

5.5 Map coloring 135

Fig.5.22 The dashedline shows which

countries and edges toremove in order to

make the torus intoa cylinder.

map requiring five colors, nor has anyone ever been able to prove that fourcolors are sufficient for every possible map. This appears to be a very difficultproblem.

However, in what should be a more complicated case, the problem hasbeen solved. We refer to the case ofa map on the surface of a torus. Actually,the problem has even been solved for a number of other surfaces as well,including the Mobius strip and the surface of a two-hole torus. It seems quitestrange that the problem has not been solved in what really ought to be thesimplest case of all-the case of the plane or sphere (the two are equivalentproblems). We present next the proof that the answer to the coloring problemfor the torus is seven; that is, that every map on the surface of a torus can becolored with seven or fewer colors, and that there do exist maps that requireseven colors. Of course, we always refer to a proper coloring, in whichadjacent countries have different colors; moreover, in the case of maps onthe torus, we require not only that each of the countries be connected, butalso that each has connected boundary and that no country encircle the torusin either of the two possible fashions. We could really eliminate these lastconditions by a simple device, but we will make these assumptions forsimplicity.

As you might expect, it will first be necessary to derive Euler's Formulafor a torus. Let a map of countries be given on a torus, subject to the conditions mentioned above. As in Fig. 5.22, draw a circle around the torusthe short way, avoiding vertices, and passing through a country no more thanonce. Remove the countries and boundaries crossed by this curve. Thisoperation will decrease the values ofF and E for this map by the same amount,

136 Polyhedra 5.5

Fig. 5.23 The plane map resultingafter the cylinder is deformed.

so that the value of V - E + F remains unchanged. If there should be anyfree vertices in the resulting map, where only two edges meet, remove theseas well. Each such operation decreases the value of V and E by one each,so that after this is done, the value of V - E + F is still unchanged.

What we now have is a map on a cylinder, which can then be deformedinto a map in the plane in the shape of an annulus, as in Fig. 5.23. We adda dummy country in the hole in the middle of the annulus, and note that wehave here a map of the sort examined in the proof of Euler's Formula inSection 5.2. There is no need to triangulate this map and remove triangles,for we already know that for this map,

V-E+F=1.

We remove the dummy country in the hole, thus decreasing Fby 1, and findthat

V - E + F = O.

But the value of V - E + F is the same for the annular map as for theoriginal map on the torus, and so we discover that Euler's Formula for atorus is

V - E + F = O.

This is a result you should have obtained in Exercise 5.7.We are now ready to prove that seven is the "map-coloring number"

for the torus.


Fig. 5.24 Adjusting amap so that only three

edges meet at eachvertex.

•

Theorem 5.3 Each map of connected countries each with connected boundaryon the surface of a torus can be colored with seven or fewer colors.

Proof The theorem is clearly true for a map consisting of seven or fewercountries. Suppose it is.false for some value of F larger than seven: Thenthere is a least such value of F for which the theorem is false; and so, lettingthat value be itself denoted by F, there is a map on a torus with F countriesthat cannot be colored with seven colors. Remember that F is the smallestnumber for which we are supposing the theorem to be false, so that forexample, any map with F - 1 countries can be colored with seven colors.

We have a map of F countries on the surface of a torus, and this maprequires eight or more colors for a proper coloring. We next adjust this mapso that only three edges meet at each vertex. One way to do this is shown inFig. 5.24, in which a few edges are moved slightly to one side or the other.This will increase the value of V and E, but F will not be changed.

Since three edges meet at each vertex, for this adjusted map we canmultiply the number of edges at each vertex by the number of vertices, andobtain the number 3V. This counts each edge twice, hence

3V = 2E,or

6V = 4E.Also, for this map

V - E + F = 0,and so

F= E- V,so that

6F = 6E - 6V.

138 Polyhedra 5.5

Since we also know that 6V = 4£, then

6F = 6£ - 4£,

or

6F = 2£.

We now claim that one of these countries has fewer than seven boundarysegments, and hence is bounded by fewer than seven other countries. Forif not, we let Fn denote the number of countries of this map that have exactlyn boundary segments, for all natural number values of n. If no country hasfewer than seven boundary segments, then

F 1 = F2 = F3 = F4 = Fs = F6 = 0,

and

F = F7 + Fa + F9 + .. '.We count the number of edges by counting the number bounding each

country. Since seven edges bound F7 countries, eight edges bound Facountries, and so on, and because this process counts each edge exactlytwice, we find that

2£ = 7F7 + 8Fa + 9F9 + .> 7F7 + 7Fa + 7F9 + .= 7· (F7 + Fa + F9 + )= 7F.

Hence 2£ > 6F. But we had previously established that, on the contrary,2£ = 6F. This contradiction shows that at least one country must have six orfewer boundary segments.

Remove such a country temporarily from the map, and allow the countries which formerly bounded it to annex the now unclaimed territory, as inFig. 5.25. These six or fewer countries bound exactly the same countries asbefore the annexation, provided that the annexation is carried out properly.This new map has F - 1 countries, too, and hence can be colored using nomore than seven colors. Color it.

We now ask the countries that took part in the annexation to cede theirnew territory back to the country that was temporarily removed, and wereplace that country. Since it is bounded by no more than six countries, ittouches only six other colors at most, and the seventh color is available tocolor it properly. Color it that color.

At the beginning of the proof, we made an adjustment of the edges sothat only three ~dges met at each vertex. We now reverse that adjustment,restoring the original map. No new boundaries between countries are set upin this process; indeed, some countries may no longer bound ones they did


Fig. 5.25Annexation of a

country by itsneighbors.

previously. Hence our coloring with seven colors will also work for theoriginal map. This contradicts the assumption that the map could not becolored using only seven colors, and hence our supposition that the theoremwas false is itself a false assumption. Thus the theorem is"true, and we havecompleted the proof of Theorem 5.3.

All that is left is to establish the existence of a map on the torus thatactually does require seven colors for a proper coloring. One such, usingin fact only seven countries, is shown in Fig. 5.26. This figure requires someexplanation. It is too confusing to draw a view of a semitransparent toruson a two-dimensional piece of paper and try to show the various countries.What appears in Fig. 5.26 is instead a recipe for building such a torus; or,if you prefer, a set of directions for painting such a map on the surface of anold inner tube. If you make a copy of Fig. 5.26 on a flat but flexible rectangleof paper and glue together the two long sides of the rectangle, you will have along cylinder. Then glue together the ends of the cylinder; you will obtaina torus. Hence, in Fig. 5.26, opposite sides of the rectangle are to be thoughtof as attached, and so a country such as number 2 (for example) has a common boundary with country number 7. You can verify that each of the sevencountries actually shares a boundary with each of the other six. We havedrawn the figure, for simplicity, so that the sides of the rectangle are alsoboundaries of countries, except for the four corners, all of which belong tocountry number 1.

140 Polyhedra 5.5

Fig.5.26 A map on thetorus requiring sevencolors.

This example, together with Theorem 5.3, shows that the map-coloringnumber for the torus is indeed seven. The example shows that seven colorsare necessary; the theorem shows that more are unnecessary.

Exercises

5.36 Each face of a certain regular solid is a pentagon, and exactly threeedges meet at each of its vertices. Use the techniques of this chapter, but notthe results of Theorem 5.1, to find the possible number of faces such a figurecan have.

5.37 Suppose that a map on the surface of a torus has II vertices and eachcountry has three edges. How many countries can there be? Why?

5.38 What is the value of V - E + F for the Mobius strip? (See Exercise5.13.) Prove that your answer is correct.

5.39 What is the map-coloring number for the Mobius strip? Hint: UseExercise 5.38.

5.40 A map on the surface of a sphere consists of pentagons and hexagonsattached as shown in Fig. 5.27. Note that:

a) five hexagons surround each pentagon;b) three pentagons and three hexagons alternate, surrounding each hexagon;c) each vertex necessarily lies on exactly three edges; andd) not all of the map is shown in Fig. 5.27.

5.5

Fig. 5.27 The patternof hexagons and

pentagons for Exercise5.40.

III

,,\

Map coloring 141

142 Polyhedra 5.5

Find how many pentagons and hexagons there can be. Hint: Let Pdenote the number of pentagons and H the number of hexagons. ThenP + H = F, where F is the number of faces; and, as usual,

V-E+F=2.

Incidentally, the figure described one possible pattern for a geodesicdome, such as has been used for large structural work, and also gives thepattern of polypeptides in the beet virus molecule.

5.41 The author's wife has called to his attention the following generalization of the previous exercise: Suppose that a map on a sphere consists ofcountries each having either five or six sides, and such that each vertex lieson exactly three edges. Then exactly 12 of the countries have five sides.See whether you can prove this.

5.42 Continuing the previous exercise, what are the possible numbers ofcountries with six sides, proceeding under the following "regularity" assumption: Each pentagonal country bounds the same number of hexagonalcountries?

5.43 If you look over the proof of Steinitz's Theorem, you will see a numberof instances in which the next step is to perform (for example) 2V - F - 4cuts of one type, and so on. Check each such statement in the proof to makesure the number of cuts given is never negative-for if it were, this wouldinvalidate the proof. Doing this exercise will help show exactly why thehypothesis

v + 4 ~ 2F ~ 4V - 8is needed.

5.44 Suppose that a map on the surface of a sphere consists of a number Tof triangular countries and a number Q of four-sided countries, such thateach vertex lies on exactly three edges, each triangular country is boundedby three four-sided countries, and each four-sided country is bounded byexactly two triangular countries. How many triangular and how manyfour-sided countries are there?

5.45 Can a map such as the one described in the previous exercise exist onthe surface of a torus?

5.46 Suppose that a map on the surface of a sphere consists of a number Tof triangular countries and a number Q of four-sided countries, such thateach vertex lies on exactly four edges, each triangular country is boundedby exactly three four-sided countries, and each four-sided country is boundedby exactly two triangular countries meeting it in opposite sides. How manytriangular countries and how many four-sided countries are there?

5.47 Can a map such as the one described in the previous exercise exist onthe surface of a torus? Explain your answer.


5.48 The situation involved in describing "regular solid tori" is in one waysimpler, in another way more complicated, than in the case of the sphere.Assume that a map on the surface of a torus consists of a number of countriessuch that

a) each country has connected boundary,b) each vertex lies on the same number d > 3 of edges, andc) each country has the same number n > 3 of sides.

(This is in analogy to the case for the ordinary regular solids.) Use the torusformula

V-E+F=O

to show that there are only three possibilities for the shapes of the countries:triangles, quadrilaterals, or hexagons.

5.49 Show that at least one map for each of the three possibilities of theprevious exercise exists.

5.50 Show that-in continuation of the previous two exercises-more thanone map of each of the sorts discovered can exist. In fact, infinitely manyexist, of each type, and thus there are infinitely many "regular tori"; infinitely many are composed of triangular countries, infinitely many ofrectangular countries, and infinitely many of hexagonal countries.


The paper of Steinitz mentioned in Section 5.4 is his "Uber die EulerschePolyederrelationen," Arch. Math. Phys. (3), 11 (1906), pp. 86-88. The proofgiven in this text is new and possibly simpler. Some other general referencesto polyhedra and map coloring are listed below:

Coxeter, H. S. M., Introduction to Geometry (Wiley, 1961).

Coxeter, H. S. M., Regular Polytopes, second edition (Macmillan, 1963).

Dynkin, E. B. and V. A. Uspenskii, Multicolor Problems, translated byN. D. Whaland, Jr., and R. B. Brown (Heath, 1963).

Griinbaum, B., Convex Polytopes (Interscience, 1967).

Hilbert, D. and S. Cohn-Vossen, Geometry and the Imagination, translatedby P. Nemenyi (Chelsea, 1952).

Lyusternik, L. A., Convex Figures and Polyhedra, translated by T. JeffersonSmith (Dover, 1963).

Ore, 0., The Four-Color Problem (Academic Press, 1967).

Tietze, H., Famous Problems of Mathematics (Graylock, 1965).

144 Polyhedra

A complete proof of the Five-Color Theorem for the sphere and planemay be found in Sherman Stein's Mathematics: The Man-Made Universe,second edition (Freeman, 1969), in the chapter on map-coloring.

The great Swiss mathematician Leonhard Euler was born in Basle in1707. He was one of the most productive of all mathematicians, continuinghis research even after he became totally blind in his sixtieth year until hisdeath seventeen years later. He spent most of his life either at St. Petersburg(Leningrad) or at Berlin, under the sponsorship of royalty, and contributedto mechanics, the calculus of variations, the three-body problem, numbertheory-indeed, to virtually every branch of mathematics. His interestswere not restricted to mathematics. In his stay in Russia he developed amathematical theory of investment (out of which our present theory ofannuities grew), wrote most of the mathematics textbooks for the Russianschool system (he was a superb textbook writer), and reformed the Russiansystem of weights and measures.

The four-color problem is not so old as most people think. The bestevidence we have is that Francis Guthrie (later a professor of mathematics)asked one of his teachers at University College, London, for a proof thatfour colors were sufficient to color any map in the plane. The teacher, whowas the well-known mathematician Augustus de Morgan, communicatedthis problem to his colleague, the famous Sir William Rowan Hamilton, in aletter written in 1852. Apparently the problem did not formally appear inprint until about 1878; at that time incorrect proofs were published by Kempeand Tait. P. J. Heawood found the flaw in Kempe's proof, and published in1890 a paper showing how Kempe's proof could be modified to show thatfive colors are sufficient.

Last-minute note: The author has just received a copy of the paper"The Four Color Theorem," by Professor Emeritus (of Duke University)Joseph Miller Thomas. In this paper Dr. Thomas states that each map on thesphere can be properly colored using no more than four colors, and givesa proof of some twelve printed pages. Perhaps this century-old problemhas finally been solved.

CHAPTER 6

INFINITESETS

Imagine a hotel with so many rooms that it takes all the positive wholenumbers to number the rooms. That is, the hotel has its rooms numbered1, 2, 3, 4, 5, ... , without any largest room number. If the hotel is filled withguests, one in each room, it might seem difficult to provide space for anadditional guest. But should such an extra guest arrive at the hotel, a cleverroom clerk can arrange for this guest to have a room to himself with onlyminor inconvenience to the other guests: The room clerk can request theguest in room 1 to move to room 2, the guest in room 2 to move to room 3,and in general request the guest in room n to move to room n + 1. He canthen assign the new guest to room 1. The hotel is still full, all the guests stillhave private rooms, and the new guest has been accommodated.

In order to repay all the other guests for their courtesy, the new guestdevises a scheme for improving the financial status of each. He posts a largesign in the hotel lobby directing each guest in rooms numbered 1 through 10to deliver one dollar to the guest in room 1, each guest in rooms numbered11 through 20 to deliver one dollar to the guest in room 2, each guest in roomsnumbered 21 through 30 to deliver one dollar to the guest in room 3, and ingeneral, each guest in rooms numbered IOn + 1 through IOn + 10 todeliver one dollar to the guest in the room numbered n + 1. (The number nis supposed to take on all positive whole number values.) Now this is nogreat inconvenience, for each guest need make no more than one trip, and thenet result is that each guest receives ten dollars while paying out one, thusmaking a profit of nine dollars. Actually, our clever friend in room 1 doesbest of all by this scheme, for he receives nine dollars just as everyone elsedoes, but he does not have to do any walking.

Of course, this plan will not work properly unless each guest has at leastone dollar to start with-but if some do not, there is still a way to handle the

145

146 Infinite sets 6.1

problem. Suppose we have the extreme case in which the guests in the oddnumbered rooms have no money, but all the others have at least one dollar.The guests in the even-numbered rooms could then be directed thus: Each isto deliver one dollar to the guest in the room with the number half his roomnumber. Then the guest in room 2 would deliver a dollar to the guest in room1, the guest in room 4 would deliver a dollar to the guest in room 2, and ingeneral the guest in room numbered 2n would deliver a dollar to the guestin the room numbered n. You can easily verify for yourself that each guestin an odd-numbered room receives one dollar, and each guest in an evennumbered room pays out and receives one dollar. Thus there would be nochange in the financial status of the guests in the even-numbered rooms, whileeach guest in an odd-numbered room would now have one dollar. Then theoriginal scheme proposed above could take place without any difficulty.

The reason that such peculiar manipulations are possible is of coursedue to the fact that we are dealing with infinite sets-the hotel has infinitelymany rooms as well as infinitely many guests, and infinitely many dollarsare involved in the financial transactions. However, in spite of the apparentcontradictions involved here, we hope to show you that such phenomenaare natural-even common-when one deals with infinite sets. Throughoutmuch of the following material, the central idea is the concept of a one-toone correspondence between two sets. With this concept we will be able tocompare the numbers ofelements of two infinite sets because it turns out to bequite possible that one infinite set actually contains "more" elements thananother. The basic tools are some ideas about sets and functions.

6.1 SETS

By a set we mean a collection of objects, thought of as a whole. The objectswhich make up the set are called the elements of the set.

We do not pretend that the above comprises a definition, at least in theformal sense that previous definitions have been given in this book. Anyattempt to define a term requires the introduction of other more primitiveterms, and so it is easy to see that some terms must be taken as absolutelyprimitive-their meaning assumed clear. However, it is helpful in such casesto attempt to give synonyms and examples. For the term "set," which wetake as such a primitive term, we can supply a large list of synonyms. Someof these could be "class," "collection," "aggregate," "conglomerate," orjust "bunch." We will soon give several examples, and ask you to use yourpowers of abstraction. One final note: In the next section we shall encounterthe same problem with the definition of the term "function."

It is not necessary that the elements of a set have any particular propertyin common. For example, it makes sense to talk of the set whose four elements

6.1 Sets 147

are the first two words on this page, the number 6, and the moon. But we doask two things:

First, that of any object potentially the member of a set under consideration, it is at least theoretically possible to determine whether or not thatobject belongs to the given set. In the above example, it is possible for you todetermine whether or not the word "however," or the positive square root of36, or the eighth largest body in the Solar System, belongs to the set mentioned.

Second, we also require that we can at least theoretically tell apart anytwo objects which do belong to a given set. One reason for this is that we willfollow the convention that objects of a given set are not to be listed in thatset more than once; otherwise, in the material which follows, we would runinto difficulties in counting the number of elements of a set. For example,if W is to be the set whose elements are the first president of the UnitedStates and George Washington, we want the list of elements of W to containonly one entry, so that we can say that W contains one element.

These two considerations-that a set should be sufficiently well definedto tell what its objects are and to tell them apart-are probably quite inaccord with your intuitive notion of what a set is, but they are also importantconsiderations for those mathematicians who study logic and the foundationsof mathematics.

There are basically two ways to specify a set. One method might be calledthe listing method. One simply writes down the elements of the set. Thismethod is quite useful for sets with only a small number ofobjects or elements,but if the number of elements is large (or if the set is infinite) we must resortto the ellipsis:

A = {I, 2, 3, 4},

B = {l, 2, 3, 4, , lOO},

C = {I, 2, 3,4, }.

When the elements of a set are listed, the list is enclosed in braces for clarity.The set A above consists of the first four positive whole numbers; B consistsof the first one hundred positive whole numbers; C consists of all the positivewhole numbers. It is certainly possible to tell whether or not any given objectis a positive whole number, and given two positive whole numbers it can bedecided whether or not they are equal. Thus these three examples satisfythe two criteria previously mentioned.

As an alternative to the listing method for specifying a set, we can use thedescriptive method.

D = {x I x is a positive whole number}.

We read this notation as follows. As soon as you see the first brace on theleft, you realize that you are about to encounter a set, and you should thinkto yourself "The set of," or "The set consisting of ...." The next word is


quite important, and it is surprising that it is omitted in the above notation;the word is "all." The letter x is simply a dummy variable, to be used in thesentence that will follow the vertical bar, so up to the vertical bar you cantranslate as follows:

"The set of all objects ... "or

"The set consisting of all elements x ...."

The vertical bar is merely translated "such that" or "with the property that."Thus what follows that vertical bar must be a declarative sentence. It gives theexact condition an object must satisfy in order that it belong to the set beingdescribed. The brace at the right-hand side tells that the description has beencompleted. Thus, the entire sentence

D = {x I x is a positive whole number}

may be correctly translated as

"D is the set consisting of all objects x such that x is a positive wholenumber",

or

"D is the set of all x such that x is a positive whole number,"

or

"D is the set of all positive whole numbers,"

or, finally,

"An object is an element of the set D if and only if the object is apositive whole number."

Of course, you see that the set D above contains exactly the same objectsas the set

C = {I, 2, 3, 4, ... }.

Since it is the aggregate ofobjects itselfwhich makes up a set, not the particularmethod of describing the set, we have here an example of what it means fortwo sets to be the same. We say that C and D are equal sets, and we writesimply C = D.

Using the symbol E we can also supply a convenient shorthand for theidea of an object's belonging to a set. If x is an element of the set C, then wewrite x E C; if not, we write x ¢ C. These two examples may be translatedas follows:

"x E C" translates as "The object x is an element of the set C" or "x isan element of C."

6.1 Sets 149

"x ¢ C" translates as "The object x is not an element of the set C" or"x is not an element of C."

In the example of the set C given above, two true statements using thisnotation are

2EC and -3¢ C.

Two false but meaningful statements are

DEC and 17 ¢ C.

Finally, using this symbol E and the set C, we can use the descriptive notationfor the sets

A = {I, 2, 3, 4},

and

B = {I, 2, 3, 4, ... , loo},

by writing the equivalent statements

A = {x E C I 1 < x < 4},

and

B = {x E C I 1 < x < loo}.

This quantification of the dummy variable x has the property of removingambiguity, for without knowing what sorts of objects the values of x werelimited to, we could not tell whether the set A, as described above, containedbut four elements, or infinitely many (in case x were allowed to take on realnumber values).

There are important relations between sets, and one of the most importantis the idea of inclusion. The set A is said to be a subset of the set B providedthat each element of A is also an element of B. Symbolically, we writeA c B as a shorthand for "A is a subset of B," and it is customary to say,if A c B, that A is contained in B, that B contains A, that A is a subset ofB, or that B is a superset of A. For A and B as in the examples above, it istrue that A c B and false that B c A; in this case we write B ¢ A.

For a formal definition, we can state the following: Let A and B be sets.Then A c B if, and only if, for each x E A also x E B. Moreover, usingthe relation of set-inclusion, we can also define set-equality: Let A and Bbe sets. Then A = B if and only if both A c Band B cA.

New sets can also be manufactured from old ones. The union of the twosets Sand T is the set consisting of all elements which belong either to Sor to T (or both), and is denoted by S u T. In symbols, then,

S u T = {x I XES or XE T}


We can also form the intersection of Sand T, consisting of all elements common to Sand T; that is,

S fl T = {x I XES and XE T}.

Since it may happen that Sand T have no elements in common, it turns outto be convenient to accept the notion of the so-called empty set, denoted by0, which contains no elements. Thus, for example, if

S = {x I x is an even integer}

and

T = {x I x is an odd integer},

then

S fl T = 0.

In the examples below, let

A = {I, 2, 3, 4},

B = {I, 2, 3, 4, , loo},

C = {I, 2, 3, 4, },

S = {... , - 4, - 2, 0, 2, 4, 6, ... },

and

T = {... , - 3, -I, I, 3, 5, 7, ... }.

Example 6.1 The following are true statements:

5 ¢ A,

5 E B,

5 E C,

5 ¢ S,

5 E T,

5¢0·


A c B,

Be C,

C ¢ S,

S ¢ T,

o c S,

T¢ C.


Au B = B,

A u C = C,

SuT=W,

where W is the set of all whole numbers.

6.1 Sets 151

u u

A ns

Fig.6.1 Venn diagrams illustrating A n B and A u B.


A (\ B = A,

A (\ S = {2, 4},

C (\ T = {l, 3, 5, 7, ... }.

A US

For some people, Venn diagrams as shown in Fig. 6.1 are very useful invisualizing the unions and intersections of sets. The rectangle U symbolizessome fixed universe of elements under consideration, and the area within thecircle A is meant to symbolize the elements of the set A. The set A (\ B isshown on the left, shaded; A u B is shaded in the right figure.

Exercises

6.1 Why are the sets {I, 2, 4} and {2, 4, I} equal?

6.2 Let A = {I, 2, 3}, B = {2, 3, 5}, and C = {4, 5, 6}. Express the setsbelow by listing their elements between braces, or by using 0 if necessary.

a) Au B

c) Au C

e) (A u B) u C

g) (A (\ B) (\ C

b) A (\ B

d) A (\ C

f) A u (B u C)

h) A (\ (B (\ C)

6.3 Using A, B, and C as in the previous exercise, compare (A u B) (\ Cand A u (B (\ C) to see if they are equal.

152 Infinite sets

c

u

6.1

Fig. 6.2 The correctway to draw threesets in a Venndiagram.

6.4 In drawing Venn diagrams involving three sets A, B, and C, they shouldoverlap as shown in Fig. 6.2 so as to allow for all possibilities of the variousintersections of these sets. Of course, it is possible that no elements lie insome region, such as B n C, but it does no harm to let Band C overlap.Shade in the following sets in two copies of Fig. 6.2:

A n (B u C),and

(A n B) u (A n C).

6.5 Your result on the previous exercise should suggest that the formula

A n (B u C) = (A n B) u (A n C)

holds for all sets A, B, and C. The picture is, of course, no substitute for aformal proof, but it can serve as a guide for construction of such a proof.

One way to invent such a proof is to choose an arbitrary element, say x,from the set A n (B u C) and show that x must necessarily belong to theset (A n B) u (A n C). Then one would choose an arbitrary element x(there is no harm in using the same symbol as before) from the set (A n B) u

6.1 Sets 153

(A n C) and show that x must also necessarily belong to the set A n (B u C).These two proofs would show that, in order,

A n (B u C) c (A n B) u (A n C),and

(A n B) u (A n C) cAn (B u C).

Hence by the definition of equality of sets,

A n (B u C) = (A n B) u (A n C).

Provide the necessary details of this proof.

6.6 The previous exercise should remind you of the distributive law ofmultiplication over addition, which holds in the real number system. Ifa, b, and c are real numbers, then

a' (b + c) = (a' b) + (a ·c).

Addition does not distribute over multiplication in the real number system;that is, it is generally false that

a + (b' c) = (a + b)' (a + c).

However, union of sets does distribute over intersection, for given setsA, B, and C, it is true that

A u (B n C) = (A u B) n (A u C).

Please prove this.

6.7 The previous two exercises suggest that there is a formal "algebra" ofsets much as there is an algebra involving addition and multiplication of realnumbers. You may verify as many of the set-algebraic properties listedbelow as you wish. Capital letters stand for sets.

A = C.A c C.

then

then

B = A.

B = C,Be C,

thenand

and

A = A.

If A = B,

If A = BIf A c B

A cA.

o cA.

A u A = A = AnA.

A u B = B u A, and A n B = B n A.

(A u B) u C = A u (B u C).

(A n B) n C = A n (B n C).

o u A = A and 0 n A = 0.A u (B n C) = (A u B) n (A u C).

A n (B u C) = (A n B) u (A n C).


6.8 In the above list of properties of the algebra of sets, one can make ananalogy with the algebra of real numbers, interpreting A, B, and C as realnumbers, intersection as multiplication, and union as addition. What roledoes 0 play in this analogy? How good is the analogy? How do you interpret the relation of set-inclusion?

6.9 We define A - Bas

A - B = {x E A I x ¢ B}.

Draw a Venn diagram, similar to the ones shown in Fig. 6.1, to illustratethe set A-B.

6.10 Let A, B, and C be sets. Prove that

A - (B n C) = (A - B) u (A - C).

See Exercise 6.9.

6.11 Continuing the ideas in the previous exercise, discover and prove valida formula resembling the one above for A - (B u C).

6.12 How would you define

Ai u A 2 U A 3 U A 4 U "',

andAi n A 2 n A 3 n A4 n "',

where, for each natural number n, An is a set?

6.13 Prove the formula

A - (Bl n B2 n B 3 n B4 n ... )

= (A - B l ) u (A - B2 ) u (A - B3) u (A - B4 ) u ....

Is a similar formula with unions and intersections interchanged also valid?

6.14 Let m and n be natural numbers, and let A be a set containing melements and B be a set containing n elements. What can be said about thenumber of elements of A u B? What can be said about the number ofelements of A n B?

6.15 In the previous exercise, if you know that the number of elements ofA n B is k, what can be said about the number of elements in A u B?

6.2 FUNCTIONS

Even if the concept of "set" is of first magnitude in the galaxy of mathematics, so is the concept of a "function."

Let A and B be sets. A functionjfrom A to B is a rule which assigns toeach element of A one and only one element of B. We writej: A -+ B, callA the domain off, and B the range off. If x is an element of A, and y is that

6.2 Functions 155

element of B assigned by f to x, we call y the value of f at x, and writey = f(x).

In this definition we encounter the same problem as in our definitionof set. We have simply substituted for the word "function" another primitiveterm, "rule," as a sort of synonym. In Exercise 6.16 we shall discuss anothermethod of defining the term "function," using a somewhat clearer and morenatural primitive term.

If f: A ~ B is a function, the phrase ''I assigns to each element of Aone and only one element of B" is not meant to be interpreted as meaningthat each element of B is used exactly once as a value off It is permissibleto use some elements of B more than once, and others not at all. For example,let R denote the set of all real numbers, and let f: R ~ R according to therulef(x) = x 2

• Thenfisjust the function which assigns to each real numberits square. Not every number in the range is used--4 is not a value offOn the other hand, some elements in the range are used more than oncef(2) = 4 and also f( - 2) = 4, so that the number 4 in the range is usedtwice.

For another example, suppose you are given a large collection ofgummedlabels, say at least a hundred thousand with the phrase "five feet, two inches"printed on them, another hundred thousand with the phrase "six feet, eleveninches" printed on them, and so on, a hundred thousand for each of thepossible heights of students you might encounter at your school. Supposenext that you walk around campus until you have met each student; uponfirst meeting each, you paste on his forehead that label most nearly indicatinghis height. (If a person claims to be exactly five feet, eleven and one-halfinches tall, just round this off to six feet.)

In this example the domain is the set of all students on campus, therange is a set of heights, and you are the rule, assigning to each element(student) in the domain one and only one element (his height) in the range.The "one and only one" part of the definition comes in because no studenthas two different labels pasted on him and because we presume you continuethis campaign sufficiently long so that each student receives a label. You,as the paster, are operating as the rule part of this function. The functionitself consists of three things, a domain, a range, and a rule; in this case, thefunction consists of the set of students on campus, the set of heights, andyourself. Note again that it is permissible for some heights to be used morethan once-you will likely encounter at least two students almost exactlythe same height-and some labels, such as the "eight feet, ten inches" label,may never be used at all, and thus the possible height "eight feet, ten inches"would be an unused element of the range.

It is sometimes convenient to consider the set of all values in its rangethat a function does use; this is called the image of the function, and we candefine it using set-theoretic notation as follows.


Let f: A ~ B be a function. Then the image off is the set

[m(f) = {y E B I y = f(x) for some x E A}.

In the study of infinite sets, there are two properties of functions whichwill be of particular interest to us. First, it may happen that a function doesindeed use each value in its range no more than once. Such a function is saidto be one-to-one. It may also happen that a function has its image equal toits range; such a function is said to be onto. Again, we provide a formaldefinition.

Let f: A ~ B be a function. The function f is said to be one-to-oneprovided that, if Xl and X2 are elements of A and Xl =F X2' then f(Xl) =Ff(X2)' If [m(f) = B, thenfis said to be onto. Moreover, iffis both one-toone and onto, then f is said to be a one-to-one correspondence from A to B.

Example 6.5 Letf: R ~ R by f(x) = x 2• Thenfis a function, because

a) each real number has a square, so that R is the domain andfcan operateon R;

b) the square of each real number is again a real number, so that the"output" offdoes lie in R, and hence R is the range off; and

c) if a E R, b E R, and a = b, then a2 = b2, and hence f(a) = feb).Hence even if a real number is known by two different names a and b,the output of the function is fixed for that number. So f has the "oneand only one" property.

However, f is not one-to-one, and neither is it onto, as we have alreadyobserved.

Example 6.6 Let f: R ~ R according to the rule f(x) = 2x • Then f is afunction, for reasons similar to those above. In addition, f is one-to-one,since if a and b are real numbers and a =F b, then 2Q =F 2b

, and hencef(a) =F feb). Butfis not onto, because for each real number x, 2x is positive.Although - 1 E R, the range off, - 1 ¢ [m(f). The graph of the function fof this example is shown in Fig. 6.3.

Example 6.7 Let f: R ~ R by f(x) = x 3- x. The graph off is shown in

Fig. 6.4. Again, f is a function, but f is not one-to-one, since 1 and - 1 areboth numbers in the domain off, but although 1 =F - 1, f(1) = 0 = f( - 1).However,fis onto, since given a number y in the range off, there does existat least one value of X in the domain off for which y = x 3

- x; that is,y = f(x).

Example 6.8 This is an example of a non-function. Let the domain andrange each be the set R of all real numbers, as in the previous example, andlet f assign to each number its reciprocal. We might write f(x) = l/x.

6.2 Functions 157

y-axis

Fig.6.3 The graph of(x) = 2)( I a one-to-onefunction fromRtoR.

=======-------i------------ x-axis

y-axis

--------,,e----"k--------,,e--------- x-axis

Fig. 6.4 The graph of(x) = x 3 - x, an onto

function from R to R.


Then I is not a function, for there is at least one number-namely O-in thedomain ofI to which no value in the range is assigned by f

Example 6.9 This is another example of a non-function. Again, let thedomain and range each be R. Let I assign to each irrational number thevalue 17. Given a rational number, express it as a fraction a/b. Then fassigns to this rational number the value a + b.

This is not a function, because there are equal numbers in the domainto which are assigned unequal numbers in the range. For example, 1/2 =2/4, but/O/2) = 3 ¥: 6 = f(2/4).

Example 6.10 This is our last example of a non-function. Let the domainbe the set R of all real numbers, and the range be the set [0, 1] of all realnumbers between °and 1 (including °and 1). Let f have the rule f(x) =2x + 3. Then/is not a function, for although the number 12 is in its domain,f(12) is technically undefined-but in any case, the only "value" f(12) couldhave, according to the rule off, is 27, and 27 does not belong to the range off. Hence f does not assign to the number 12 in its domain any value in itsrange.

Example 6.11 Let I: R --+ R according to the rule f(x) = 2x + 3. Thennot only is I a function, but in fact f is a one-to-one correspondence from Rto itself. To show the latter, we need only show that I is both one-to-oneand onto. First, suppose that Xl and X2 are two numbers in the domain offsuch thatf(x l ) = I(X2). Then

so that

and hence

Consequently, if Xl ¥: X2' then I(Xl) ¥: I(X2). Hence I is one-to-one.Now suppose that y belongs to the range R off Let

y - 3X=--.

2

Then

f(x) = 2x + 3

=2. y-

3 +32

= (y - 3) + 3 = y.

6.2 Functions 159

Hence given the number y in the range off, there does exist a value of xnamely, x = (y - 3)/2-such that f(x) = y and x does lie in the domainR off Hence f is also onto, and thus by definition is a one-to-one correspondence from R to itself.

Iff: A ~ B is a one-to-one correspondence from the set A to the set B,there is then naturally associated with this function another functiong: B ~ A according to the rule

g(y) is that element x of A such thatf(x) = y.

The function 9 is called the inverse of the function f, and is sometimesdenoted byf - 1.

Example 6.12 Letf: R ~ R according to the rulef(x) = 2x + 3. We sawin Example 6.11 that f is a one-to-one correspondence from R to R. So fmust have an inverse g. In the proofthatfis onto, we saw that the element xof the domain off that was assigned to the element y in the range off byf had the value x = (y - 3)/2. Hence the rule of 9 is given by

y - 3g(y) = 2 '

or, if you prefer,

x - 3g(x) = .

2

Note that we can use any reasonable symbol we please to specify how the ruleof a function acts, for in the above example, 9 is simply the function whichassigns to each real number that number obtained by halving the givennumber diminished by three. Note also how much clearer it is, at least in thiscase, to use symbols rather than words to describe the rule part of thefunction.

Finally, as you may have noticed, there is no reason why we shouldrestrict our attention to functions which have domains and ranges subsets ofR, and rules that look like algebraic formulas. The example with the stickersgiving a student's height shows that other sets and rules may be used. Andthe rule does not have to have any special regularity about it; as a finalexample, you could simply paste the stickers on the students' foreheads moreor less at random, so long as each student received one and only one sticker,and this would be a different function from the height function previouslymentioned.

Exercises

6.16 We give here another primitive term, that of ordered pair, by which anequivalent definition of "function" can be given. The ordered pair (a, b)


consists of two objects a and b together with the idea that a is the first, andb the second, in the pairing.

The Cartesian product A x B of two sets is defined as follows:

A x B = {(a, b) I a E A and bE B}.

A familiar example of an application of the above concept is the ordinarycoordinate system for the two-dimensional plane used in analytic geometry.One forms the Cartesian product R x R, and then a is the so-called xcoordinate, and b the y-coordinate, of the point (a, b) E R x R.

We can provide an alternate but equivalent definition of the term"function." Let A and B be sets. A function f from A to B is a subset ofA x B such that

a) for each x E A, there exists y E B such that (x, y) Ef; and

b) if (x, y) Efand (x, z) Ef, then y = z.

Iff is a function in the sense of this definition, and (a, b) Ef, does ourold notationf(a) make sense? What is the value of f(a)?

6.17 Give an example unlike Example 6.6 of a function f: R -+ R that isone-to-one but not onto.

6.18 Give an example unlike Example 6.7 of a function f: R -+ R that isonto but not one-to-one.

6.19 Give an example unlike Example 6.11 of a functionf: R -+ R that is aone-to-one correspondence from R to R.

6.20 Findf- 1 for your example above.

6.21 The open interval (- (nI2), n12) of all real numbers between - (nI2)and nl2 can be thought of as a set of angles in radian measure, and with thisdomain and with range R, the rule f(x) = tan x gives a function that is aone-to-one correspondence from ( - (nI2, n12) to R. The graph off is shownin Fig. 6.5. Sketch the graph off-I.

6.22 Let A and B be sets, and f a one-to-one correspondence from A to B.Prove that not only is f- 1 a function with domain B and range A, but alsothatf- 1 is a one-to-one correspondence from B to A.

6.23 Let A be a set. Construct a one-to-one correspondence from A toitself. Hint: This problem is very easy.

6.24 Let A, B, and C be sets, and letf: A -+ Band g: B -+ C be functions.As indicated in Fig. 6.6, we can go directly from A to C by a new function,called the composition offand g, which we construct as follows. We denotethis function by {g(f)}, and define {g(f)}: A -+ C according to the rule

{g(f)}(x) = g(f(x)).

Verify that {g(f)} : A -+ C is indeed a function.

6.2

Fig. 6.5 The graph of(x) = tan x,

- (1t/2) < x < 1t/2.

-(1T/2)

y-axis

Functions 161

oFig. 6.6

The composition{g( f) } of the

functions ( and g. {g(f)}


6.25 Using the notation and definitions of the previous exercise, verify thatif each off and g is one-to-one, then so is {g(f)}.

6.26 Using the notation and definitions of Exercise 6.24, prove that if eachoff and g is a one-to-one correspondence from its domain to its range, then{g(f)} is a one-to-one correspondence from A to C.

6.27 Continuing the previous exercise, find a formula for {g(f)} -1 in termsoff- 1 and g-l.

6.28 See Exercise 6.24. Let f: R --+ R by f(x) = x 2, and let g: R --+ R

by g(x) = x + 1. To be equal, two functions must have the same domain,the same range, and the effects of their rules must be the same, in that if xis in their common domain, then f(x) = g(x). Are the functions f and ggiven in this exercise equal functions? Why?

6.29 Using the functions f and g of the previous exercise, find {g(f)} and{f(g)}. Are the latter two equal functions? Explain your answer.

6.30 Let f: A --+ B be a one-to-one correspondence from the set A to theset B, and let g = f-l. Show that {g(f)}(x) = x for all x E A and that{f(g)}(x) = x for all x E B.

6.3 MORE ON ONE-TO-ONE CORRESPONDENCES

A,...., C.then

then

and

Recall that we say there is a one-to-one correspondence from the set A to theset B provided that there is a one-to-one and. onto function f: A --+ B.If so, we shall use the notation A ,...., B, and say that the sets A and B can beput into one-to-one correspondence. By using the notion of one-to-onecorrespondence between two sets, we shall be able to define what we mean by afinite set, an infinite set, and what it means for two sets to have the samenumber of elements even if this number should be infinite. First, however,we examine the properties of one-to-one correspondences; specifically, wewant first to show that this relation is an equivalence relation (see Exercise1.6). That is, we want to show that if A and Band C are any sets, then

a) A ,...., A.

b) If A,...., B,

c) If A,...., B

But part (a) has been taken care of in Exercise 6.23, part (b) has beentaken care of in Exercise 6.22, and part (c) has been proved in your work forExercise 6.26. So from this point on, we can eliminate a large number oftedious constructions from many proofs and exercises. For example, if youknow that each of the two sets Sand T can be put into one-to-one correspondence with a third set U, then you know that Sand T can also be putinto one-to-one correspondence with each other; knowing this, you know in

6.3 More on one-to-one correspondences 163

addition that there exists some one-to-one and onto function !: S -+ T,so you can simply "let" ! be such a function, and use! where necessary inyour proofs.

Now we can give a precise definition of what it means for a set to beinfinite-we are going to say that a set is infinite provided that it is notfinite, so first we define what we mean by a finite set.

The set S is said to be finite provided that either S = 0 or, for somenatural number n, S ,..., {I, 2, 3, ... , n}.

The set S is said to be infinite provided that S is not finite.The example that comes easily to mind for an example of an infinite set

is the setN = {I, 2, 3,4, ... },

whose elements are all the positive whole numbers. But in order to show thatN is in fact an infinite set, it is necessary to show that N =1= 0 (which is easy)and that there can exist no one-to-one correspondence between N and anyset of the form {I, 2, 3, ... ,n}. This is likely to be a formidable task, forone must show for infinitely many different values of n that no such one-toone correspondence can exist. Moreover, it is also quite difficult to prove atthis point the obvious, necessary fact that the two sets {I, 2, 3, ... , n} and{I, 2, 3, ... , m} can be put into one-to-one correspondence if and only ifm = n. In the next section we shall provide the Cantor-Schroeder-BernsteinTheorem, an almost indispensable tool in dealing with problems of this sort.

Exercises

6.31 Supply, in addition to N, two more examples of infinite sets.

6.32 Show that the set N of positive whole numbers can be put into one-toone correspondence with its proper subset E of even positive whole numbers.Note: In order to do this, you must show the existence, presumably byconstruction, of a one-to-one correspondence between Nand E; that is,you must construct a one-to-one and onto function!: N -+ E (or!: E -+ N).

6.33 Show that the set N of positive whole numbers can be put into one-toone correspondence with its proper subset M of all squares of positive wholenumbers.

6.34 Show that the set N of positive whole numbers can be put into one-toone correspondence with its subset T of all such numbers with two or moredigits.

6.35 Show that the set N of positive whole numbers can be put into one-toone correspondence with the set X = {- 1, - 2, - 3, - 4, ... }.

6.36 Give three different functions each of which is a one-to-one correspondence from the set N to itself.


6.37 Give two different functions each of which is a one-to-one correspondence from the set N to its proper subset L = {I, 3, 5, 7, 9, ... }.

6.38 Show that the set N of positive whole numbers can be put into one-toone correspondence with the set W of all whole numbers.

6.39 Show that the set of points on a line one unit long can be put into oneto-one correspondence with the set of points on a line two units long.

6.40 Show that the set (0, I) of real numbers between°and I (not including°or I) can be put into one-to-one correspondence with the set R of all realnumbers. Hint: In Exercise 6.21, an example was given of a one-to-onecorrespondence between the set ( - (nI2), n12) and the set R. If you can showthat (-(nI2), n12) and (0, I) can also be put into one-to-one correspondence,then it will follow (why?) that (0, I) and R can also be put into one-to-onecorrespondence.

Assuming that you have established a one-to-one correspondence, say f,from (0, I) to R, can you now show the existence of a one-to-one correspondence from [0, I] to R? ([0, I] = {x E RIO ~ x ~ I}.) How?

6.4 THE CANTOR-SCHROEDER-BERNSTEIN THEOREM

There are many cases in which one desires to show that two sets can be putinto one-to-one correspondence, but all that can easily be accomplished inpractice is something such as this: Sand T are two given sets, and it turns outto be possible to show that S can be put into one-to-one correspondence withsome subset B of T, and that T can be put into one-to-one correspondencewith some subset A of S. In Fig. 6.7 this situation is diagrammed, withcI> the one-to-one function from S onto Band 'P the one-to-one functionfrom Tonto A.

This situation seems to suggest that since T has at least as "many"elements as S, and S as many as T, then Sand T themselves could be putinto one-to-one correspondence. However, though this is true, finding such acorrespondence in actual practice can sometimes be rather complicated.The Cantor-Schroeder-Bernstein Theorem guarantees under these conditionsthe existence of a one-to-one correspondence between Sand T.

Theorem 6.1 (Cantor-Schroeder-Bernstein) Let Sand T be two sets, andsuppose that S '" B, where BeT, and that T '" A, where A c S. ThenS '" T.

Proof Let cI>: S -+ Band 'P: T -+ A each be one-to-one and onto functions.Both cI> and 'P exist because of our hypotheses that S '" Band T '" A.We shall produce a function E> that is a one-to-one correspondence fromS to T. The proof of the existence of E> will actually be constructive, so thata formula for E> could actually be written in a specific case; however, in such

6.4 The Cantor-Schroeder-Bernstein Theorem 165

Fig. 6.7The hypotheses of the

Cantor-SchroederBernstein Theorem: The

functions are one-to-one.

cases the formula could be so complicated that we will generally be contentwith the knowledge of the existence of 0.

Consider an element XES. If there exists an element yET such that'¥(y) = x, then we will call y a parent of x. If such an element yET inaddition has a parent Z E S, such that (z) = y, we will also call z a parentof x, and in this case {'¥(<I»}(z) = x. Note that since the subset A of S is theimage of '¥, then each element of A has at least one parent; if XES - Athen x has no parents. Similar remarks hold for Band T.

Given that XES, there are exactly three possibilities:

a) x has infinitely many parents.

b) x has but finitely many parents, and the ancestry of x begins with aparentless ancestor in S.

c) x has but finitely many parents, and the ancestry of x begins with aparentless ancestor in T.

If x has no parent at all, which happens when XES - A, then x belongsin case (b) above. So we can divide the set S up into three mutually exclusivesubsets:

Soo = {x E S I x has infinitely many parents}.

SfT = {x E S I the ancestry of x begins in S}.

Sf = {x E S I the ancestry of x begins in T}.

Note that we have chosen the subscripts so as to help us remember whichelements of S belong to each of these subsets of S. Of course, S - A C SfT'but in fact SCI contains those elements of S with an even number of parents

166

•••

Infinite sets

•••

b~-'

•••

•••

6.4

Fig.6.8 The elementa E S has infinitelymany parents; theelement b E S hasancestry beginningin S; the elementC E S has ancestrybeginning in T.

c

s T

(recall that 0 is an even number) and Sr: contains those elements of S with anodd number of parents. See Fig. 6.8.

Similarly, we divide Tup into three mutually exclusive subsets as follows:

Too = {y E T I y has infinitely many parents}.

Tu = {y E T I the ancestry of y begins in S}.

~ = {y E T I the ancestry of y begins in T}.

We emphasize that Too, Tu' and Tr; are mutually exclusive-the intersectionof any two is the empty set-and that T = Too u Tu u~. Hence eachelement of T belongs to exactly one of the sets described above. Similarremarks hold for S.

6.4 The Cantor-Schroeder-Bernstein Theorem 167

We will establish that is a one-to-one correspondence from S 00 toToo, that is also a one-to-one correspondence from Sq to Tq, and that'llis a one-to-one correspondence from ~ to St' This will prove that Soo ,..., Too,Sq ,..., Tq, and St ,..., Tt • Then we can "glue together" the functions onSoo u Sq and '1'-1 on St to obtain the one-to-one correspondence E> fromS to T, as desired.

Now Soo c S, so is defined on Soo' Moreover, if x E Soo, then x hasinfinitely many parents, so that (x) is an element of T with infinitely manyparents as well. Hence is a function with domain S00 and range Too.(Actually, this is an improper use of terminology, since by restricting thedomain of from S to Soo we are actually considering a new function;perhaps we should indicate this by calling this new function <1>00 instead ofof just <1>, but the additional notation hardly seems justified in this case.)

We need to show that <1>: Soo --+ Too is both one-to-one and onto. But is clearly one-to-one, since it is one-to-one on all of S. And if y E Too,then y has infinitely many parents, so in particular it has an immediate parentXES, for which (x) = y. But this element x must actually belong to Soo'for if x had but finitely many parents then so would y. Hence there doesindeed exist an element x E Soo such that (x) = y. Thus <1>: Soo --+ Toois both one-to-one and onto, and thus is indeed a one-to-one correspondence from S00 to Too"

By the symmetry of the remaining two cases, it is sufficient to prove onlyone of them, for example that 'II is a one-to-one correspondence from~ to St' This is left for you, in the exercises at the end of this section. Sowe may assume that we know that each of the following is a one-to-onecorrespondence:

<1>: S 00 --+ Too ,

<1>: Sq --+ Tq,

'1': ~ --+ St'

See Fig. 6.9. By Exercise 6.22, '1'-1 is also a one-to-one correspondencefrom St to ~, so that we have the following situation. The three functionsshown below are each one-to-one and onto:

<1>: S 00 --+ Too ,

<1>: Sq --+ Tq,

'1'-1: St --+ ~.

The three domains shown above are disjoint and their union is all of S;the three ranges shown above are also disjoint and their union is all of T.

168

s~

Infinite sets

•

•

6.4

T~

Fig.6.9 Each functionis a one-to-onecorrespondence on theindicated subset.

So we define 0: S --+ T as follows:

If x E SOC!,

If x E S(1'

If x E S-r'

then

then

then

0(x) = (x).

0(x) = (x).

0(x) = \}1-1(X).

It should be clear not only that 0 is a function, but in fact is a one-to-onecorrespondence from S to T. Thus S '" T, and this concludes the proof ofthe Cantor-Schroeder-Bernstein Theorem.

6.4 The Cantor-Schraeder-Bernstein Theorem 169

As an application, we show how this theorem can be used to show theexistence of a one-to-one correspondence between the set

w = {... , - 2, - 1, 0, 1, 2, 3, ... }

of all whole numbers, and the set X = W x W, or

X = {em, n) I mEW and nEW},

the Cartesian product of W with itself. (This may appear to be a surprisingresult, since X appears to be much "larger" than W.)

Let $: W --+ X by the rule $(n) = (n, 0). Then $ is clearly a one-to-onefunction from W to X. Now all we need is a one-to-one function 'P fromX to W. Among other things, if (m, n) is an ordered pair of whole numbers(a typical element of X), we must so design 'P that 'P(m, n) is a whole number,and thus an element of W.

One approach that almost works is to let 'P(m, n) = mn. Then'P is afunction from X to W, but unfortunately is not one-to-one, since

'P(1, 6) = 6 = 'P(2, 3)and

(1, 6) i= (2, 3).

Another approach that almost works is to let 'P(m, n) = 2m• 3n• Then

'P would be one-to-one, but unfortunately not a function from X to W,since (1, -1) E X but 'P(1, -1) = 2/3 and 2/3 ¢ W.

But a modification of the last approach does work. In the last approachit is only the fact that m or n might be negative that produces fractions, ratherthan whole numbers, in the output of 'P, so we define 'P as follows:

If m > 0 and n > 0, then 'P(m, n) = 2m• 3n

•

If m > 0 and n < 0, then 'P(m, n) = 2m • 5- n•

If m < 0 and n > 0, then 'P(m, n) = 7-m • 3n•

If m < 0 and n < 0, then 'P(m, n) = 7- m • 5- n•

In each case, the output of'P is a positive whole number, so that 'P: X --+ Wis a function. The fact that each positive whole number has a unique primefactorization into the product of primes means that the output of'P uniquelydetermines its input; or, in other words, 'P is one-to-one.

Neither $ nor'P is onto, but this does not matter. We have exactly thesituation given in the hypotheses of the Cantor-Schroeder-Bernstein Theorem.Wand X are sets, $: W --+ X is one-to-one, so that W '" B = Im($) c X,and 'P: X --+ W is one-to-one, so that X '" A = Im('P) c W. Hence bythe Cantor-Schroeder-Bernstein Theorem, W '" X.

Exercises

6.41 Show, in the proof of Theorem 6.1, that'P actually is a one-to-onecorrespondence from Tt to St'

6.42 In the example following Theorem 6.1, in which it is shown that W '" X,show that the function <D constructed therein is one-to-one and that neitherof the functions <D or 'P is onto.

6.43 Use the Cantor-Schroeder-Bernstein Theorem to prove that if A '" Cand A c B c C, then B '" C.

6.44 At the conclusion of the proof of Theorem 6.1, it was stated that thefunction 0 constructed therein was actually a one-to-one correspondencefrom S to T. Verify this.

6.45 In Exercise 6.38, we asked you to prove that the set N of positive wholenumbers can be put into one-to-one correspondence with the set W of allwhole numbers. Prove this by use of the Cantor-Schroeder-BernsteinTheorem. Hint: Use techniques similar to those in the example followingthe proof of Theorem 6.1.

6.5 PROPERTIES OF FINITE AND INFINITE SETS

First we should show that both finite sets and infinite sets exist. It is clearthat the sets of the form {I, 2, 3, ... , n} are finite for each positive wholenumber value of n. To show the existence of an infinite set,we prove that theset N of positive whole numbers is in fact infinite.

Theorem 6.2 The set N = {I, 2, 3, ... } is infinite.

Proof The proof is by contradiction; we suppose by way of contradictionthat the set N is finite. Then, by definition, either N = 0 or N can be putinto one-to-one correspondence with a set of the form {I, 2, 3, ... , n}, forsome positive whole number n. Since 17 E N, N =F 0, and hence there is aone-to-one and onto function f from {I, 2, 3, ... , n} to N for some positivewhole number n.

We can think of f as a one-to-one correspondence from the set {I, 2,3, ... , n} to the set {fO), f(2), f(3), ... ,f(n)}, and it would not be difficultto think of a method of selecting from the latter set its largest element, amethod which would require no more than n steps. So the set {fO), f(2),f(3), ... ,fen)} contains a largest element, say m, and m = f(j) for some jsuch that 1 <j < n.

But m + 1 is an element of the set N, and hence since f is onto, theremust exist a positive whole number k E {I, 2, 3, , n} such that f(k) =

m + 1. Sof(k) is an element of {f(1),f(2),f(3), ,fen)} larger than m.This contradicts the fact that m is the largest element of {fO), f(2), f(3), ... ,f(n)} , as m + 1 > m. Hence our original supposition, that N is finite,

6.5 Properties of finite and infinite sets 171

leads to a contradiction, and consequently N is infinite. This completes theproof of the theorem.

Note that it was not necessary to use the fact that f is one-to-one.Our next result depends on Exercise 6.46, at the end of this section:

That if S is any infinite set whatsoever and XES, then the set S - {X} isalso infinite, and therefore nonempty. Moreover, our next theorem alsoimplies that the set N is in some sense a "smallest" infinite set-the exactmeaning of this statement will be discussed in Exercise 6.47.

Theorem 6.3 If S is an infinite set, then S contains a subset M such thatM ,.., N.

Proof. Choose Xl E S. Then by Exercise 6.46, S - {xd is infinite, and thusnonempty. So we may choose X2 E S - {xd. Since S - {Xl} is infinite,it follows again by Exercise 6.46 that S - {x h x2 } is infinite and nonempty.So we next choose X3 E S - {Xl' X2}' We continue this process. It will no1.terminate, for if Xl' X2' X3' ... , Xk have been chosen, then still

S - {Xl' X2' X 3, ... , xk}

is infinite, thus nonempty, and we may choose Xk+ 1 from it. Then

is still infinite and nonempty, and the process can be continued. Thus foreach positive whole number m we can produce an element X m E S, and byour construction, if j :I: m then x j :I: X m• So the set M = {x h X2' x 3, ••. }

is a set of distinct elements of S, one for each positive whole number.Let f: N -+ M according to the rule f(n) = X n• It is clear that f is a

one-to-one correspondence, and hence that M ,.., N. This establishes thetheorem.

The next theorem seems "obvious," but it is not easy to prove, at leastwithout using the Cantor-Schroeder-Bernstein Theorem.

Theorem 6.4 Every subset of a finite set isfinite.

Proof. Suppose that F is a finite set. The theorem is clearly true if F = 0,so we consider only the case in which F ,.., {I, 2, 3, ... , n} for some positivewhole number n.

Suppose by way of contradiction that F contains an infinite subset S.By Theorem 6.3, S contains a subset M such that N ,.., M. Let f: N -+ Mbe a one-to-one correspondence from N to M.

Now {I, 2, 3, ... ,n} c N, so that, lettingf(j) = mj for eachj EN,

is a subset of M such that K ,.., {I, 2, 3, ... , n}. So K ,.., F. Also, sinceK c M and M c S, then K c S.


So we have the following situation: K eSc F and K '" F. ByExercise 6.43, which is an easy consequence of the Cantor-SchroederBernstein Theorem, it follows that S '" F. But F '" {I, 2, 3, ... , n}, andhence S '" {I, 2, 3, ... , n} as well. This is impossible because S was supposed to be infinite; if S '" {I, 2, 3, ... , n} were true, then S would befinite by definition, and again by definition "infinite" means "not finite."This contradiction shows that the finite set F can contain no infinite subsets,and establishes the theorem.

We now prove a theorem sometimes known as the "Dedekind BoxPrinciple," which together with the succeeding theorem will enable us to givean alternate, equivalent, and sometimes more useful definition of what itmeans for a set to be infinite.

Theorem 6.5 No finite set can be put into one-to-one correspondence with oneof its proper subsets.

Proof Let F be a finite set and G a proper subset of F. Suppose by way ofcontradiction that G '" F; then there exists a one-to-one correspondenceqJ from F to G. Since G c F, then G must be finite by Theorem 6.4. Henceeither G = 0 or else G = {I, 2, 3, ... , n} for some positive whole numbern. If G = 0, then F =1= 0 since G is a proper subset of F; however, forF =1= 0 and G = 0, qJ: F --+ G cannot be a function. So we can eliminatethe possibility that G = 0.

In several previous exercises, it was seen that when a one-to-one correspondence exists between two sets such as F and G, there are generallyseveral such correspondences. If we have one, such as cp, we can let

Mlp = {x E F I cp(x) =1= x}

be that subset of F of points of F moved under the action of cp. By Theorem6.4, for each such correspondence cp the set Mlp must be finite because it is asubset of the finite set F. So the number of elements in the set M lp is a positivewhole number or zero.

Choose a one-to-one correspondence cp from F to G so that M lp containsthe smallest possible number of elements. With this choice of cp we willseek to establish a contradiction.

Now Mlp =1= 0. For if Mlp = 0, then cp(x) = x for all x E F. ThenG = Im(cp) = F, so that G would not be a proper subset of F. Therefore,since Mlp =1= 0, we may choose an element x E Mlp (") G (how?).

Now cp(x) =1= x, so cp(x) = y for some y E F such that y =1= x. Moreover, cp is onto, so there exists WE F such that cp(w) = x. And if w = x,then cp(w) = cp(x) since cp is a function; but since cp(w) = x, this wouldimply that x = cp(x), which is false by the choice of x E Mlp. Hence w =1= x.

6.5 Properties of finite and infinite sets 173

Fig. 6.10(fJ moves w to x and x to y.

F

Next, y e M([). For if not, then qJ(Y) = y. But also qJ(x) = y, and cp isone-to-one. This implies that x = y, which we have shown false. HenceyeM([).

Also, we M([). For if not, then qJ(w) = w. But cp(w) = x by choice ofw, so that W = x since cp is a function. But then cp(x) = x, again contraryto our choice of x. So we M([) as well.

Hence we have the situation shown in Fig. 6.10. All three ofw, x, andybelong to M([). Also w =1= x and y =1= x (although it is possible that w = y;however, for clarity we have shown in Fig. 6.10 the more general case in whichw =1= y). Finally, qJ(w) = x and cp(x) = y.

We modify the function qJ and obtain a slightly different function I/J asfollows:

Let

Let

Let

I/J(w) = y.

I/J(x) = x.

I/J(z) = qJ(z) if Z =1= w and Z =1= x.

Now I/J is almost the same function as qJ; all that has been done to cp inorder to obtain I/J is to switch the values of qJ at wand x, as indicated in Fig.6.11. Hence it is easy to see that I/J is also a one-to-one correspondence fromF to G. But let us now consider the set M "', given by

M", = {x e F I I/J(x) =1= x}.


F

Fig. 6.11 tp is modified to becometfI, which moves w to y and leavesx fixed.

. The particular element x, previously chosen from M tp' has the propertythat q>(x) =1= x; but for this element x, t/J(x) = x. So X E Mtp' but x ¢ M",.Moreover, if v ¢ Mtp' then q>(v) = v and, in addition, v =1= wand v =1= xsince x and ware elements of Mtp. So, by definition of t/J, also t/J(v) = v.Thus if v ¢ Mtp' then v ¢ M",.

Hence M", can contain no more elements of F than Mtp' and in fact Mtpactually contains more elements of F than M", because x E Mtp but x ¢ M",.But q> was chosen so that its corresponding set of points Mtp would containthe minimum possible number ofelements ofF, and here we have constructedt/J, another one-to-one correspondence from F to G whose correspondingset M", contains fewer elements than MfP. This is in contradiction to thechoice of q>, and this contradiction establishes that no such function as q> canexist. Therefore no finite set can be put into one-to-one correspondence withone of its proper subsets, and the theorem is proved.

The mathematician Richard Dedekind suggested an alternative, butequivalent, definition of what it means for a set to be infinite; the term usedis that the set is Dedekind infinite, and this means that the set can be putinto one-to-one correspondence with one of its proper subsets. Theorem6.6 shows that this is the same property as being infinite.

Theorem 6.6 The set S is infinite if and only if Sis Dedekind infinite.

Proof. Suppose first that S is Dedekind infinite. Then, by definition, Scanbe put into one-to-one correspondence with one of its proper subsets. By

6.5 Properties of finite and infinite sets

s

175

Fig. 6.12 The function f is a one-to-onecorrespondence from S to S - {x}.

• • • • • • • •••1234567···

the previous theorem, S cannot be finite; hence by definition, S must beinfinite.

Next, suppose that S is an infinite set. By Theorem 6.3, S contains asubset M such that M ,..., N. Let () be a one-to-one correspondence fromN to M, and let x = ()(1). Then x EM c S, so that x is also an element ofS; let T = S - {x}. Tis certainly a proper subset of S. We will show that Scan be put into one-to-one correspondence with T.

We define a function f from S to T as follows. If S E S - M, just letf(s) = s. If S E M, then s = ()(n) for some positive whole number n since(): N --+ M is onto; in fact, n is uniquely determined by s since () is alsoone-to-one. Letf(s) = ()(n + 1).

Of course, what we are doing here is shoving each element of M up onenotch, sending x = ()(1) to ()(2), sending ()(2) to ()(3), and so on, whileleaving the elements of S not in M alone. This action off is indicated inFig. 6.12. You may easily verify for yourself that f is indeed a one-to-onecorrespondence from S to T. So we have shown that the infinite set Scanbe put into one-to-one correspondence with one of its proper subsets, andthus that S is Dedekind infinite. This completes the proof of Theorem 6.6.

The Dedekind Box Principle (Theorem 6.5) can be phrased very informally as follows: If a postman has n letters to put into m mailboxes andm < n, then at least one box must get at least two letters.


Exercises

6.46 Prove that if 8 is an infinite set and x E 8, then S - {x} is infinite.Note: This exercise is used to establish Theorem 6.3, so that theorem and itssuccessors may not be used in working this exercise. Hint: First suppose that8 - {x} is finite, in order to reach an eventual contradiction. Set up a oneto-one correspondence f from 8 - {x} to the set

{I, 2, 3, ... , n}.

Use f to produce a one-to-one correspondence g from (8 - {x}) u {x} tothe set

{I, 2, 3, ... , n, n + I}.

Explain why this is a contradiction. Draw the desired conclusion.

6.47 Show that if K is a subset of N, then either K is finite or K '" N. Usethis fact to show that if S is a set such that 8 '" N, and T is a subset of 8,then either T is finite or T '" N. This is the sense in which N can be considered a "smallest" infinite set, bearing in mind the next exercise as well.

6.48 Prove that if 8 is a set containing the set N of positive whole numbersas a subset, then S must be infinite. Use this fact to show that if 8 is a setcontaining a subset M such that M '" N, then 8 must be infinite.

6.49 A set 8 is said to be denumerable provided that 8 '" N. Prove thatN x N is denumerable. Hint: Use the Cantor-Schroeder-Bernstein Theorem(Theorem 6.1) and the techniques used in the application immediately afterits proof. See also Exercise 6.45.

6.50 Use Exercise 6.49 to prove that if A and B are denumerable sets, then sois A x B.

6.51 Show that the set Q of all rational numbers is denumerable. Hint:Use Exercises 6.47 and 6.50.

6.52 Prove that if A and B are denumerable sets, then so is A u B.

6.53 For each prime pEN, let A p be the set of all positive integral powersofp; that is,

A p = {p, p2, p3, p4, ... }.

Show that for each prime p, Ap is a denumerable set.Prove that the collection

{A p Ip is prime} = {A 2 , A 3 , As, A 7 , All' ... }

is denumerable. Note: There are infinitely many primes.

6.54 Use the previous exercise to show that the set N of positive wholenumbers contains infinitely many infinite sets no two of which have anyelement in common.

6.6 Nondenumerable infinite sets 177

6.55 Use the ideas developed in the previous two exercises to show that ifeach of B1, B2 , B3 , ••• is a denumerable set, then so is the set

B = B 1 U B2 U B3 U .••.

6.56 An equilateral triangle of side length 2 is drawn in the plane, and fivepoints are selected within this triangle. Prove that some two of these pointsmust lie within distance I of each other. Hint: Use the Dedekind BoxPrinciple.

6.57 Prove that at least one pair of people in Atlanta, Georgia have the samenumber of hairs on their heads.

6.58 Show that the sets

{I, 2, 3, ... , n}and

{I, 2, 3, ... , m}

can be put into one-to-one correspondence if and only if m = n.

6.59 Prove that if F is a finite set and S is an infinite set, then S '" S u F.

6.60 So far, every infinite set has turned out to be denumerable. Do youbelieve that every infinite set is denumerable? For example, let P be the setof all polynomials with whole number coefficients. Is P denumerable? Hint:See Exercise 6.55.

6.6 NONDENUMERABLE INFINITE SETS

We have shown N infinite, and N is denumerable by definition. In Exercise6.51, you were asked to show that the set Q of all rational numbers is alsodenumerable, and in Exercise 6.60 that the set P of all polynomials withcoefficients in Wis also denumerable. At this point you might wonder, quitejustifiably, if there might be only two kinds of sets-finite and denumerableand thus that all infinite sets could be put into one-to-one correspondence.Our next theorem shows that this is not the case; the familiar set R of allreal numbers is certainly infinite, but it cannot be put into one-to-onecorrespondence with the set N, and is thus nondenumerable. Hence, in avery precise sense, the set R contains "more" elements than the set N. Thiswas proved by the mathematician Georg Cantor in 1873; he later discovereda simpler proof which we next present. This proof depends on the fact thateach real number has a decimal expansion, and that two real numbers areequal if and only if their decimal expansions are identical. (There is a minorpoint here to be discussed in the exercises; 0.99999 . .. and 1.00000. .. aredifferent decimal expansions for the same real number; we assume in theproof to follow that if there should be such ambiguity, the latter form is tobe used.)


1 Al • all al2 a l3 a)4 a l5 •••

2 A 2 • a21 a22 a 23 a 24 a25 •••3 A 3 • a 31 a32 a 33 a 34 a 35 ••• Fig. 6.13 The hypothetical4 A 4 • a 4 ) a42 a 43 a 44 a45 ••• one-to-one correspondence5 As • as) a52 a53 a54 ass ••• from Nto R.

• •• •• •

Theorem 6.7 The set R ofall real numbers is nondenumerable.

Proof. We suppose, by way of contradiction, that R is denumerable. Thenthere must exist a one-to-one correspondenceffrom N to R. We arrange a"diagram" of the action off as shown in Fig. 6.13.

The left column lists the positive whole numbers, the domain off Tothe right ofeach is its value under the action off Now we have supposed thatthe one-to-one correspondence f exists, but we do not know what it is;we do not know whether fO) is 1/4, -7, or n. So we must represent thevalues offaccording to their decimal expansion. Since it turns out that we willnot have to consider the whole number part of fen), we have called thisAn; thusf(n) is shown in the form

so that anj is the jth digit in the decimal expansion of fen).Becausef is onto, each real number must appear somewhere in the right

hand column. We proceed to reach a contradiction by producing a realnumber that does not appear anywhere in the right-hand column.

We construct this real number ex by going down the diagonal all' a22 ,a33' ... , and changing each of these digits. Specifically, we construct exas follows: The decimal expansion of ex is to be

ex = 0 . blb2b3b4bs ... ,

where bi is obtained from au as follows. If au is less than 8, let b i = a ii + 1.If au is 8 or 9, let bi = O.

Now ex does not appear anywhere in the right-hand column, for thedecimal expansion of ex differs in at least one place from any of the numberslisted in the right-hand column. But ex has a decimal expansion, and thus isindeed a real number. This contradicts the fact that! is onto. Hence R isnondenumerable.

As you have seen, we count by means of one-to-one correspondence.For example, if the set A can be put into one-to-one correspondence with theset {I, 2, 3, ... , n}, then we say that the set A contains n elements. If A ,..., 0,then we say that A contains no, or zero, elements. The numbers 0, 1, 2, ... ,that we use in counting the finite sets are called finite cardinal numbers. Infact, if we consider the collection of all sets that can be put into one-to-onecorrespondence with the set {I, 2}, there is but one significant propertycommon to all sets in this collection, "twoness" or the property of containingtwo elements, and to be precise it is exactly this common property we meanwhen we speak of the cardinal number 2.

There is nothing to prevent us from giving names to infinite cardinalnumbers as well. By tradition, the name ~o has been used for the cardinalnumber of the set N of positive whole numbers, and thus is the cardinalnumber of all denumerable sets. (~is the first letter of the Hebrew alphabet,and ~o is usually pronounced "aleph-null.") The German letter c iscustomarily used for the cardinal number of the set R of all real numbers,because c is the first letter of the German word for continuum, used sometimes as a synonym for the real number line.

We shall not go into the arithmetic ofcardinal numbers, but only mentionthat it is possible to consider a natural order relation between them. Hereis how this is done.

First, if A is a set, then it has a cardinal number-finite or infinite-andwe denote this number sometimes by [A]. Next, given two sets A and B,it may be possible to find a one-to-one function!: A ~ B. If so, then we saythat [A] < [B]. If we can also show that there can be no one-to-one andonto function from A to B, then we can say in fact that [A] < [B]. Thisorder relation obeys the expected properties (with one exception: If m andn are cardinal numbers, it does not follow without an additional powerfulaxiom of set theory that either m < n, m = n, or n < m). In particular,the Cantor-Schroeder-Bernstein Theorem may be translated into the languageof cardinal numbers as follows:

If m and n are cardinal numbers and both m ~ nand n < m are true,then m = n.

Finally, Theorem 6.7 may be translated very simply:

~o < c.

But so far, the only two infinite cardinal numbers we have seen are just~o and c. Our final theorem says, in effect, that there are infinitely many.

s

9

9• Fig. 6.14 Either

Z E Kz or z ¢ Kz-but eitherpossibility leads toa contradiction.

Theorem 6.8 Let S be any set and let f/ be the collection of all subsets of S.Then there exists a one-to-one function from S to f/ but no such function canbe onto. Hence, in the terminology of cardinal numbers, [S] < [f/J.

Proof There clearly does exist a one-to-one function f from S to f/; justlet f(x) = {x} for each XES. Suppose, by way of contradiction, thatthere exists a function 9 that is a one-to-one correspondence from S to f/.What 9 must then do is assign to each element s E S a subset of S, which wewill call Ks ; thus Ks E f/, for Ks c S, and Ks is just another name for g(s).

Consider s E S. Since Ks c S there are just two possibilities: eithers E Ks or s f Ks• We are particularly interested in the latter case; in fact,we let

Now :K c S, so that :K E f/. Since g: S -+ f/ is onto, there mustexist an element Z E S such that g(z) = :K; in fact, since 9 is one-to-one,this element z is in fact uniquely determined by :K.

Sinceg(z) = :K, then:K = Kz • But where is z? Since z E Sand Kz c S,either z E Kz or z ¢ Kz • Let us consider each of the two possibilities. Fig.6.14 may be helpful.

If z E Kz, then z E :K since :K = Kzo But:K is the set of all s E S suchthat s f Ks• Since Z E Kz , it follows by definition of :K that Z f :K. This iscontrary to the fact that Z E :K.

On the other hand, if Z f K z , then Z f :K since:K = K z • But by definitionof :K, :K consists of all those elements s of S such that s ¢ Ks• Since Z ¢ Kz '

then Z E :K by definition of :K. This is contrary to the fact that Z ¢ :K.

Either way, the assumption that there exists a one-to-one correspondence9 from S to !/ leads to a contradiction. Thus there can be no such function,and this establishes the theorem.

Exercises

6.61 Prove that irrational real numbers exist, using results of this chapter.

6.62 It was mentioned just before the proof of Theorem 6.7 that the samereal number may have two different decimal expansions; the example givenwas 0.99999 . .. and 1.00000... . You can prove these two are equal byusing some of the techniques of infinite geometric series (see Exercise 1.45).Of course, you use the fact that the correct interpretation of the decimalexpansion 0.99999 ... is the sum of the series

9/10 + 9/100 + 9/1000 + ....

6.63 Modify the proof of Theorem 6.7 to show that the set of real numbersbetween 0 and 1 is nondenumerable.

6.64 Use the previous exercise and other results to show that the cardinalnumber of the set of real numbers between 0 and 1 is c.

6.65 Show that the cardinal number of the set of points in the unit square Sin the plane is also c by using the previous exercise and setting up a one-toone function from [0, 1] into S and a one-to-one function from S into [0, 1],then applying the Cantor-Schroeder-Bernstein Theorem. Note:

S = {(a, b) E R x RIO < a, b ~ I}.

Hint: If (a, b) E S, then a and b have decimal expansions of the form

andbo . b1b2b3b4 · ..

where ao and bo are each either 0 or 1. Where is the number

o . aObOalbla2b2a3b4 . .. ?

6.66 Use the definition of "Dedekind Infinite" to prove that the set R of allreal numbers is infinite.

6.67 Let E be the set of all even positive whole numbers. Translate thestatement "E is denumerable" into the language of cardinal numbers.

6.68 If % is the collection of all subsets of Rand [%] = n, what is the relation between c and n?

6.69 See Exercise 6.38; translate its statement into the language of cardinalnumbers.

6.70 Translate the statement of Theorem 6.3 into the language of cardinalnumbers.

6.71 Translate the statement of Theorem 6.4 into the language of cardinalnumbers.

6.72 Let~ denote the collection of all finite subsets of N. Prove that [~] =No. Hint: See Exercise 6.55.

6.73 If the collection of all sets were a set f/, then every subset of f/ wouldbe an element of f/. This is a contradiction-but to what? Hint: SeeTheorem 6.8.

6.74 What is the cardinal number of the collection of all (unbounded)straight lines in the plane?

6.75 Suppose that an urn is filled with No balls, numbered 1, 2, 3, .... Letus call a "stage" in the following experiment, the act of removing three ballsfrom the urn and then replacing two of the balls then outside the urn.Imagine performing an experiment, in which each "stage" is performedNo times. It is clear that after stage 1, one ball is outside the urn; after stage 2,two balls are outside the urn; and, in general, after stage n, n balls are outsidethe urn. The question is this: After No stages-that is, after stage 1, stage 2,stage 3, ... ,-how many balls are in the urn?

Be careful. This is a trick question.


Two good references are

Set Theory and Logic, by Robert R. Stoll (Freeman, 1963).

Set Theory, by Felix Hausdorff, translated by John R. Aumann (Chelsea,1957).

Recall that we have scarcely discussed the following two questions:Does there exist a cardinal number n such that No < n < c?Given two cardinal numbers m and n, must one of the three relations

m < n m = n n<mbe true?

See Paul J. Cohen's excellent book Set Theory and the ContinuumHypothesis (Benjamin, 1966) for a discussion of these matters.

Georg Cantor was born in 1845 in Russia; his father was a Dane; hegrew up in Germany, so he is somewhat international. He studied at Zurichand Berlin, and gave promise of becoming a talented and conventionalmathematician. But in 1874 his first revolutionary paper was published-apaper in which he was the first to attack the previously avoided problem of


the "infinite." In fact, most of the material of this chapter can be found inCantor's published works. These innovations shocked mathematicians ofthe day, and stimulated violent attacks on Cantor by his colleagues. Cantorwas sensitive and unable to weather the criticisms thrown his way; he hadattacks of irrational anger or overwhelming depression, beginning when hewas about forty, and he died in a mental institution in 1918. By this time hehad been belatedly recognized as the genius he was.

The German mathematician Richard Dedekind was one of Cantor'sfew allies during the above troubles. Dedekind had a long life-he lived from1831 until 1916-and a very productive one; he suffered a mild form of thesame sort of attack directed against Cantor because of his own researches.He is well known for putting the concept of an irrational number into alogically sound structure, and his brilliant ideas were not at first universallywell received. However, he did live long enough to become recognized as oneof Germany's greatest mathematicians.

CHAPTER 7

NUMBERTHEORY

The theory of numbers is meant principally to answer questions about the set

N = {I, 2, 3, 4, ... }

of positive whole numbers, or natural numbers, and sometimes about the set

Z = {... , - 2, -I, 0, I, 2, 3, ... }

of integers, or whole numbers. However, the techniques used to answer suchquestions frequently involve the rational numbers, complex numbers, or evencalculus. The questions themselves have fascinated people for centuriesindeed, number theory is one of the oldest branches of mathematics-perhapsbecause the questions themselves are easy to pose, and the privilege ofsearching for the answers is available to almost everyone. Indeed, for hisresearch in number theory Pierre de Fermat is known as the "Prince ofAmateurs"; he was a French jurist of the seventeenth century who mademany of the most important contributions to number theory.

We assume that you are familiar with the arithmetic properties of thesets Nand Z; that is, that addition and multiplication are both commutativeand associative, that multiplication distributes over addition, and so on.With this background you may plunge immediately into number theory.One of the cornerstone concepts is that of divisibility, for upon this conceptare built many of the other important definitions and theorems of numbertheory-and so it is with divisibility that we begin.

7.1 DIVISIBILITY

The integer m is said to be divisible by the integer d provided that thereexists an integer k such that m = dk. If so, then d is said to be a divisor of

184

7.1 Divisibility 185

m and m is said to be a multiple of d. If d is a divisor of m, we write theshorthand expression dim and read this expression as "d divides m."

For example, the number 6 has exactly four natural number divisors;namely, I, 2, 3, and 6 itself. If our context is the set Z of integers, then thenumber 6 would have twice as many integral divisors-the ones listed abovetogether with their negatives. Because there is such a close relationshipbetween the natural number divisors of an integer m and its integral divisors,we will in the future always mean (unless otherwise stipulated) by a divisor dof the integer m a natural number divisor.

The integers 0, - I, and I are rather special so far as divisibility isconcerned. First, 0 is divisible by all whole numbers, for given a number dit is always true that d I O. To see this, just take m = 0 and k = 0 in thedefinition of divisibility; then, since 0 = o· d, it follows by definition that dis a divisor of O. Moreover, the only integer of which 0 is a divisor is 0 itself,for the equation m = k· 0 has a solution k only for m = o. On the otherhand, I and -I are divisors of every integer, while they themselves are theironly divisors. Note that each natural number other than I has at least twodivisors, itselfand I. Some natural numbers, such as 5, have no other divisors;others, such as 6, do.

Divisibility may be studied for its own sake. It should be easy to see thatif d and m are natural numbers and dim, then I ~ d ~ m. With this inmind we can establish our first theorem.

Theorem 7.1 If a, b, and c are natural numbers, then:

a) a I a.

b) If a I b

c) If a I b

and

and

bla

blc

then

then

a = b.

a I c.

To prove this theorem it is necessary only to look at the definition ofdivisibility. The details of the proof are outlined in the exercises for you tocomplete. Note the close similarity between the behavior of the symbolsI and ~; the above theorem is true if the former symbol is replaced by thelatter.

The natural number I has but one divisor (remember, by "divisor" wemean only natural number divisors); 1 is the only such natural number, so wedivide the others into two classes, as indicated in the next definition.

The natural number p is said to be prime provided that p has exactlytwo divisors. If the natural number m has three or more divisors, then m issaid to be composite.

Note that I is neither prime nor composite. Ifyou need a name for such asituation, you may refer to I as a unit, as it is called in some branches ofmathematics. It is easy to see that there are infinitely many composite

186 Number theory 7.1

numbers (why?). On the other hand, prime numbers do not seem so plentiful.The first twenty, as you may easily verify, are as follows:

2 3 5 7 11 13 17 19 23 29

31 37 41 43 47 53 59 61 67 71

There are twenty-five primes between 1 and 100, sixteen between 1000 and1100, eleven between 10,000 and 10,100, and only six between 100,000 and100,100. Since primes tend to become less plentiful among the larger numbers,you might suspect that there are only finitely many primes. However, ithas been known for several thousand years that the opposite is true-thereare in fact infinitely many primes, just as there are infinitely many compositenumbers. An outline of the simple proof of this fact appears as one of theexercises at the end of this section; in order to supply the details of this proofall that is needed is our next result.

Theorem 7.2 Ifm is a composite natural number, then m has a prime factor;that is, m = kp where k and p are natural numbers and p is prime.

Proof Suppose that m is a composite natural number. Then m does havedivisors between 1 and m, so that the set

D = {d E Nil < d < m and dl m}

is a nonempty set of natural numbers. Moreover, this set is finite, since it cancontain no more than m - 2 numbers. Hence it is possible to select from theset D its least element, which we will call p. It now suffices to show that pmust be prime.

Suppose by way of contradiction that p is not prime. Then, since p E D,also p > 1; hence p must be composite. Thus p = ab, where a and barenatural numbers such that

l<a<p and 1 < b < p.

We need consider only the number a in order to reach a contradiction.Since a I p and p I m, it follows from our previous theorem that also a I m.Hence a E D. But p was chosen as the least element of D, and a < p. Thisis a contradiction, and hence p cannot be composite. Thus p is prime. Thisestablishes the theorem, for p is thus a prime factor of m.

As we have already mentioned, our next theorem follows easily now thatTheorem 7.2 is established, and the proof is outlined in the exercises.

Theorem 7.3 There are infinitely many primes.

It is quite true that the primes do tend to become more and more sparselydistributed among the larger numbers. This tendency is well illustrated bythe next theorem, whose proof is also outlined in the exercises.

7.1 Divisibility 187

Theorem 7.4 Given a natural number n, it is possible to find a sequence of nconsecutive natural numbers each of which is composite.

In other words, given a natural number n, there do exist two consecutiveprimes whose difference is at least n.

Exercises

7.1 List the divisors of 60, 100, 117, and 119. Circle the least such thatexceeds 1 in each case, and go on to the next exercise.

7.2 Is the least divisor of 60 (other than 1) prime? Does this also hold for100, 117, and 119?

7.3 The proof of Theorem 7.2 actually "proves" a little more than is needed.Rephrase Theorem 7.2 with the stronger conclusion that follows from theproof given.

7.4 If the natural number n has d divisors, how many integral divisors has n?

7.5 Suppose that a and b are integers such that both a I band b I a are true.What can you say about the relationship between a and b? In particular,must it be true that a = b? Why?

7.6 Let p and q be primes such that p I q. What can you say about therelationship between p and q?

7.7 List all the even primes. How many odd primes are there?

7.8 The text stated that if d and m are natural numbers such that dim,then 1 ~ d ~ m. Fill in the details of the following possible proof of thisfact.

First, it is clear that 1 ~ d (why?).Let us suppose by way of contradiction that m < d. Then the difference

d - m is a positive whole number; say,

d - m = a.

Since dim, m = dk for some natural number k (why?). Moreover,k =1= 1 (why?). Hence k > 2. But then

mk =

d '

m+a-a------

d

d a= - --d d

= 1a

d

(Justify each equality.) But aid is positive, and hence

ak=l--<1.

d

(Why?) This is a contradiction (why?), and hence d ~ m (why?). Therefore1 ~ d < m.

7.9 Suppose that a, b, and e are natural numbers such that both a I e andb I e are true. Does it follow that ab I e?

7.10 Suppose that a, b, and e are natural numbers such that a I b and a I e.Does it follow that a I be?

7.11 Suppose that a, b, and e are natural numbers such that a I be. Doesit follow that either a I b or a I e must be true?

7.12 Is it possible to have two composite natural numbers a and b such thatneither a I b nor b I a is true?

7.13 Here is the outline of the proof of Theorem 7.1. First, to show in part(a) that a I a is true, it suffices to find an integer k such that a = ak. Whatvalue do you choose for k?

Next, suppose that a I band b I a. Apply Exercises 7.5 and 7.8 in orderto conclude that a = b; or, if you prefer, use the following approach: Sincea I b, there exists an integer k such that b = ak. Since b I a, there exists aninteger j such that a = bj. Substitute b = ak into the latter equation. Whatconclusion can be drawn about the number kj? What must be the values ofk and j? Why does this imply that a = b?

Finally, for part (c), suppose that both a I band b I e are true. Applythe definition of divisibility to obtain two equations similar to the two above.As above, substitute one into the other. Does the desired conclusion a I efollow?

7.14 The text asked why there are infinitely many composite numbers.Supply the reason.

7.15 Show that there are infinitely many odd composite natural numbers.

7.16 Here is the outline of the proof of Theorem 7.3, which states that thereare infinitely many primes. Supply the details.

First, suppose by way of contradiction that there are only finitely manyprimes. If so, the primes could be listed in increasing order, as follows:

where Pi stands for the ith prime in the complete list of all n primes above.Form the number

q = PtP2P3 .. ·Pn + 1.

7.2 Well-ordering 189

Now q > 1 (why?), so q must be either prime or composite. But qcannot be prime (why?). So q must be composite. Hence q has a prime factorP (why?).

None of the primes in the complete list

is a divisor of q (why?). Hence since P I q, P cannot appear in this list.This is a contradiction (why?). Hence there are indeed infinitely manyprimes, and this establishes Theorem 7.3.

7.17 Give the reasons, where needed, in the following outline of a proof ofTheorem 7.4: Given a natural number n, it is possible to find a sequence of nconsecutive natural numbers each of which is composite.

Let the natural number n be given. By n! we mean the product of all thenatural numbers from 1 to n; that is,

n! = n' (n - 1)· (n - 2)· (n - 3)' . ·3 ·2· 1.

Show that there are n consecutive composite natural numbers in the sequence

(n + I)!, (n + I)! + 1, (n + I)! + 2, (n + I)! + 3, ... ,

(n + I)! + (n - 1), (n + I)! + n, (n + I)! + (n + 1).

7.18 In the proof of Theorem 7.3 outlined in Exercise 7.16, a finite numberof primes are multiplied together, the number 1 is added to the product,and the fact that this product is composite leads to a contradiction. It wouldseem to follow that if the first n primes were multiplied together and thenumber 1 added to the product, thus obtaining

1 + PIPZP3 ••• Pn'

then this number would be prime. Is this always so? If not, does this meanthat the proof given for Theorem 7.3 is invalid? Explain your answer.

7.19 It follows from Exercise 7.8 that no natural number can have infinitelymany divisors. Why not?

7.20 A surprising formula is that if a natural number n has k divisors, thentheir product is

~nk.

Verify this formula for three different two-digit values of n.

7.2 WELL-ORDERING

One property of the set N of natural numbers is quite important in establishing many results in number theory. It is strange that this property has nothingto do with the algebraic structure of the natural number system; at least on the


surface, it contains no mention of either addition or multiplication. Westate it next.

Well-Ordering Axiom Each nonempty subset of the set N ofnatural numberscontains a least element.

For example, there is a least prime, a least composite natural number, aleast number which is the sum of two squares (of natural numbers), a leastnumber which can be expressed as the sum of two squares in two differentways, a least common multiple of 11 and 13, and a least natural numberwhose decimal representation uses all the odd digits. The existence of eachof these numbers is shown by establishing that the set in question is nonemptyin each case; finding the least element of that set may be more difficult inpractice. In the last case, the number 971,513 is a natural number whosedecimal representation uses all the odd digits; hence the set of all suchnatural numbers is nonempty. The Well-Ordering Axiom guarantees thatthis set contains a least element, and thus there does exist a least naturalnumber whose decimal representation uses all the odd digits.

In contrast, the Well-Ordering Axiom does not hold for some othernumber systems. For example, you should have no difficulty in finding anonempty set of integers which does not contain a least element.

The Well-Ordering Axiom is logically equivalent to the very usefulInduction Principle for the set N. Induction was first discussed in Exercise1.17; we next state the Induction Principle, and then prove its equivalenceto the Well-Ordering Axiom in our following two theorems.

Induction Principle for N Suppose that f/ is a statement meaningful foreach natural number, and suppose moreover that both

a) f/ is true of the number 1, and

b) whenever f/ is true of the number n, then f/ is also true of the numbern + 1. Then f/ is true for each natural number.

It is amusing and sometimes helpful to visualize the Induction Principleby the device shown in Fig. 7.1. A row of dominoes is depicted, one for eachnatural number, and the dominoes are so arranged that whenever the dominonumbered n falls over, it knocks over the domino numbered n + 1. This isin analogy to the second hypothesis of the Induction Principle. We knockover the domino numbered I-this is in analogy to the first hypothesis of theInduction Principle-and we obtain the analogous conclusion: All thedominoes fall over. Before showing the equivalence of the Induction Principleand the Well-Ordering Axiom, we provide an example of how the InductionPrinciple may be used to prove a theorem in number theory.

7.2

Fig. 7.1 Visualizing the inductionprinciple as falling dominoes.

Well-ordering

,n

9I 8

I 7

I 6

I 5

I 4

I 3

I 2

1 -"""

~

~

I--

f--

191

•••

Example 7.1 In Exercise 7.8 we provided an outline of a proof that if d andm are natural numbers such that dim, then 1 < d < m. Here is a moreelegant proof, using the Induction Principle for N.

Let d and m be natural numbers such that dim. Since d is a naturalnumber, d > 1. Moreover, since dim, then by definition m = dk for somenatural number k. Suppose by way of contradiction that d > m.

Then let f/ be the statement, "For each natural number n, m < dn."Then f/ is certainly a statement meaningful for each natural number

value of n, for given n, it can be determined whether it is true or false thatm < dn. Moreover, part (a) of the hypotheses of the Induction Principleis satisfied, for the statement f/ is true for the value n = 1. (Recall that wehave supposed, by way of contradiction, that m < d = d· 1.)

Suppose that f/ is true for the natural number n. Then m < dn. Butn < n + 1, and so dn < d· (n + 1). Consequently, we have also thatm < d· (n + 1). So part (b) of the hypotheses of the Induction Principle isalso satisfied, for if f/ is true for n then f/ is true for n + 1.

Therefore by the conclusion of the Induction Principle, it must be thecase that f/ is true for all values of n. But recall that m = dk for some

natural number k. The fact that !/ is true for all values of n and the fact thatk is a natural number together imply that m < dk. But we know that m = dk.This is a contradiction, and hence the assumption that m < d must be false.Hence if dim, then 1 < d ~ m.

Other examples ofapplications of this principle to prove various theoremsmay be found elsewhere in this book, in the next set of exercises, and inGeorge Polya's book Induction and Analogy in Mathematics (PrincetonUniversity Press, 1954). We proceed to show the logical equivalence of theWell-Ordering Axiom and the Induction Principle for N.

Theorem 7.5 The Well-Ordering Axiom implies the Induction Principle for N.

Proof We assume that the Well-Ordering Axiom is true; that is, everynonempty set of natural numbers contains a least element. We assume alsothe hypotheses of the Induction Principle: that !/ is a statement meaningfulfor each natural number, that !/ is true of the number 1, and that whenever!/ is true of the natural number n, it follows that !/ is also true for the naturalnumber n + 1. We wish to show the conclusion of the Induction Principle,that !/ must be true for every natural number.

Suppose, by way of contradiction, that !/ is not true for every naturalnumber. Let F be the set of natural numbers for which !/ is false. Then Fis nonempty because of the above assumption, and hence contains a leastelement k because of the Well-Ordering Axiom. Moreover, k =F 1 since !/ istrue of the number 1 by hypothesis. So k > 1.

Hence k - 1 is also a natural number, and moreover, !/ is true of thenumber k - 1 because k is the least natural number for which !/ is false.

But we have one more hypothesis to use: That whenever !/ is true of agiven natural number n, then also !/ must be true for the number n + 1.In particular, since k - 1 is a natural number for which !/ is true, then also!/ must be true for (k - 1) + 1 = k. But!/ is false for k, and so we havereached a contradiction. Consequently,!/ must be true for each naturalnumber. This establishes the Induction Principle for N, and completes theproof of Theorem 7.5.

Now for the converse.

Theorem 7.6 The Induction Principle for N implies the Well-Ordering Axiom.

Proof To establish the Well-Ordering Axiom, we must begin with a nonempty subset of N and use the Induction Principle to show the existence ofa least element of that set.

So let S be a nonempty set of natural numbers. We suppose by way ofcontradiction that S contains no least element. Then, in particular, thenumber 1 is not an element of S, for if it were then it would be the leastelement of S.


Now let f/ be the statement, "If 1 ~ k ~ n, then k is not an element ofS." Then f/ is clearly a statement meaningful for each natural number valueof n, for given a natural number n we can certainly establish whether or notit is true that each natural number between 1 and n fails to belong to S. Wenow apply the Induction Principle to the statement f/.

First, f/ is certainly true for the value n = 1, for in that case the onlyvalue of k such that 1 ~ k ~ n is k = 1 as well, and we have already seenthat 1 is not an element of S.

We suppose that f/ is true of the natural number n; in order to obtainthe conclusion of the Induction Principle we need only show that it followsthat f/ is necessarily also true of n + 1. Having supposed that f/ is true ofn, we know that if 1 ~ k :::; n, then k is not an element of S; in particular,n itself is not an element of S.

If f/ were false for the number n + 1, then for some natural number kbetween 1 and n + 1, k would be an element of f/. But we have seen thatk cannot lie between 1 and n, so such a value of k could only be n + 1itself. Thus n + 1 would have to be an element of S. But then n + 1 wouldin fact be the least element of S, for no number between 1 and n belongs to S.But since we have supposed that S contains no least element, this situationis impossible, and hence the statement f/ must be true for n + 1.

So we have shown that f/ is true for 1, and that whenever f/ is true forn it must also be true for n + 1. Hence by the Induction Principle, f/ mustbe true for each natural number n. Now f/ was chosen to be the statement,"If 1 ~ k ~ n, then k is not an element of S." In particular, since f/ isknown to be true for all values of n, it follows that no natural number is anelement of S. Since S is a set of natural numbers and natural numbers only,S must be the empty set. This contradicts our assumption that Sis nonempty,and this contradiction establishes that S must contain a least element after all.This concludes the proof of Theorem 7.6.

Exercises

7.21 Use the Induction Principle to prove that the sum of the first n naturalnumbers is n(n + 1)/2; that is, that

n· (n + 1)1+2+3+···+n= .

2

7.22 Use the Induction Principle to prove that the following formula holdsfor all natural numbers n:

1 1 1 1-+-+- +... +---1·2 2·3 3·4 n·(n + 1)

n---

n + 1


7.23 In the proof of Theorem 7.6, what goes wrong if we replace the statement f/ used there by the far simpler statement, "If n is a natural numberthen n is not an element of S."?

7.24 It is frequently easier to apply the Induction Principle when it is statedin the following form:

Suppose that f/ is a statement meaningful for natural numbers, that f/is true of the natural number 1, and that whenever f/ is true of every naturalnumber less than the number n, then f/ is true for n as well. Then f/ is truefor every natural number.

Show that this form of the Induction Principle follows from the formpreviously stated. Note: Since the original form of the Induction Principlehas been shown equivalent to the Well-Ordering Axiom, it may be easier(and is certainly sufficient) to show that the above form of the InductionPrinciple follows from the Well-Ordering Axiom.

7.25 Let S be the set of all positive real numbers. Does S contain a leastelement? Does the Well-Ordering Axiom hold for the real number system?

7.26 Suppose that % is a statement meaningful for natural numbers, that% is true of the natural number 5, and that whenever % is true for the naturalnumber n, then also % is true for the natural number n + 1. For whatnatural numbers need % be true? Can you prove this?

7.27 Compute the values of the successive sums

1, 1 + 3, 1 + 3 + 5, 1 + 3 + 5 + 7, ....

Guess a formula and prove it by induction.

7.28 Does the Well-Ordering Axiom hold for the set

E = {2, 4, 6, 8, 10, ... }

of all even natural numbers? Can you prove this?

7.29 Does the Well-Ordering Axiom hold for the set of all rational numbers?Give a reason for your answers.

7.30 Compute the values of the successive sums

1,1

1 + 2 '

1 11 + - +

2 4'111

1+-+-+-2 4 8 '

1 1 1 11+-+-+-+-

2 4 8 16 '

Guess a formula and prove it by induction.

7.31 The number 125 can be expressed as the sum of two squares in twodifferent ways:

Knowing this, could you prove that there is a least natural number which canbe expressed as the sum of two squares in two different ways? Ifso, what is it?

7.32 Prove by induction: IfS is a finite set containing n elements, then S has2" subsets. Hint: Having assumed this proposition true for n, suppose that Sis a set containing n + 1 elements. Choose a E S and let T = S - {a}.Then T contains n elements, and hence has 2" subsets. Note that every subsetof S is either a subset of T or a subset of T with the element a adjoined.

7.33 Let the sequence

of real numbers be defined as follows:

151 = - = 1,

1

1 252 = 1 + - = -,

51 1

1 353 = 1 + - = -,

52 2

1 554 = 1 + - = -,

53 3

and, in general, for each natural number n,

15,,+ 1 = 1 + - .

5"

If you write a few more terms of this sequence you will notice that

Prove this by induction. Note: It suffices to prove that

if n is odd;

if n is even; and

if m is odd and n IS even.

It may help if you first est~blish that if s" = alb, then s,,+ 1 = (a + b)la.7.34 Prove by induction: If x and yare positive real numbers such thatx < y, then x" < y" for each natural number n.

7.35 Prove by induction: For each natural number n, 3 I (n 3- n).

7.3 THE FUNDAMENTAL THEOREM OF ARITHMETIC

If you factor a given composite number as much as possible, you will findthat it can be factored into the product of primes. Moreover, no matter howyou go about this factorization, you will always obtain, for a given compositenumber, the same answer; that is, the prime factorization you obtain will beunique except possibly for the order in which you write the primes down.For example, you could factor 144 as follows:

144 = 2·72

= 2·2·36

= 2·2·6·6

= 2·2·2·3·2·3

= 24• 32

•

We establish this result as follows: We first show that each compositenumber can be factored into the product of primes. Then we establish theso-called Euclidean Algorithm. Finally, using the Euclidean Algorithm,we show that the factorization that exists must be unique. Note the use ofthe Well-Ordering Axiom in each of these three theorems.

Theorem 7.7 If m is a composite natural number, then m is the product ofprimes,. that is,

m = PtP2P3 ... Pm

where Pi is prime for 1 ::;; i < n.

Proof Suppose by way of contradiction that the theorem is false. Then, bythe Well-Ordering Axiom, there exists a least composite natural number notexpressible as the product of primes; we denote that number by m.

Since m is composite, m = ab, where 1 < a < m and 1 < b < m.Since each of a and b exceeds 1, each is either prime or composite. We shallconsider only one case, the one in which a is composite and b is prime; theother cases are handled similarly.

Since m is the least composite number not expressible as the product ofprimes and a is a composite number less than m, then a can be expressed asthe product of primes; thus

a = PtP2P3 ... Pk'

where Pi is prime for 1 ::;; i ~ k. Hence

m = ab,

= PtP2P3 ... Pkb,

and, since each number in the last expression is prime, we have expressed mas the product of primes. This contradiction establishes the theorem.

7.3 The fundamental theorem of arithmetic 197

Fig.7.2 23 = (4)·5 + 3,and m = dq + r.

4

5')2320

3

q

d)7;qd

r

Perhaps you have begun to notice a pattern in most of our proofs thatuse the Well-Ordering Axiom. We first suppose that the theorem is false;then the set of natural numbers for which the theorem does not hold is nonempty, and thus contains a least element. Using this smallest exception tothe theorem, we produce smaller numbers for which the theorem must thenbe true. We use the truth of the theorem about the smaller numbers-inthe above theorem, factors of the original number-to establish the truthof the theorem about the original number, the number about which thetheorem was supposed to be false. This contradiction establishes the theorem.

We next use the Well-Ordering Axiom to establish the EuclideanAlgorithm-which says, quite simply, that it is possible to divide one integerinto another so as to obtain not only a quotient but also a nonnegative"small" remainder. We shall treat only the case in which the divisor ispositive; the case in which it is negative can be handled similarly.

Theorem 7.8 Let m be an integer and d a natural number. Then there existintegers q and r such that m = dq + rand 0 < r < d.

Proof In spite of the fact that the statement of this theorem contains nomention of divisibility, it really is a theorem about dividing one number intoanother. See Fig. 7.2.

We take as an example the case where d = 5 and m = 23. You recallfrom the chronological vicinity of the third grade the process by which onedivides 5 into 23, as shown in Fig. 7.2. The quotient is 4 and the remainderis 3. Moreover, we hope you also recall the method by which your thirdgrade teacher asked you to check your work. You were to multiply thedivisor by the quotient and add the remainder to this product; if you obtainedthe dividend, your arithmetic was correct. Finally, you should also recallthat your answer was "wrong" unless your remainder was a nonnegativeinteger less than the divisor; in this example,

o ~. 3 < 5

is true, and the arithmetic check looks like

(5 . 4) + 3 = 23.

2

SJ231013

23 = (2)' S + 13

6s}23

30-7

23 = (6)'S + (-7)Fig. 7.3 Arithmetically correctways of dividing 5 into 23.

The general case is also shown in Fig. 7.2. The divisor d is divided intothe dividend m, obtaining the quotient q and the remainder r. The equivalentcheck would be to verify that

O~r<d,

and thatdq + r = m.

But this is precisely the statement of this theorem. So the theorem reallydoes say that it is possible to divide one whole number by another and obtaina quotient and a remainder, the remainder nonnegative and less than thedivisor.

Without the requirement that the remainder r satisfy the inequalityo ::;; r < d, there would be infinitely many ways to divide d into m in whichthe arithmetic would check. For example, we could divide 5 into 23 with aquotient of 2 and a remainder of 13, as shown in Fig. 7.3; in this case wewould be informed by the third-grade teacher that our quotient is not largeenough because "5 goes into 23 more than twice." Or if you prefer, youcould obtain a quotient of 6 and a "remainder" of - 7; again, the arithmeticchecks, as indicated in Fig. 7.3. Here the third-grade teacher would informus that the remainder is supposed to be nonnegative. The restriction that theremainder be nonnegative and less than the divisor provides not only theanswer referred to as "correct" by the third-grade teacher, but also providesa unique choice of q and r, as we shall prove.

Our method of proof will be to look at all possible such divisions of dinto m, without the restriction that 0 ~ r < d, and by use of the WellOrdering Axiom select the division that gives the least nonnegative value of r.We will then show that 0 ~ r < d; the quotient q will automatically "work"in that m = dq + r. Finally, we show that this choice of q and r is indeedunique. Now for the proof of Theorem 7.8.

•••

'i m - di

•••

'3 m - d·3

'2 m - d·2

'1 m - d·1

'0 = m d·OFig. 7.4 Possible remainders

upon division of m by d. '-1 m - d·(-1l

'-2 m - d·(-2l

•••

Recall that we are given the divisor d, which is a natural number, and thedividend m, which is an integer. For each integer i, we let

'i = m - di.

In other words, as indicated in Fig. 7.4, we look at all possible ways of"dividing" d into m, and associate with each possible quotient i the arithmetically correct remainder; since'i = m - di, it follows that our arithmeticcheck m = di + 'i will be automatically true. We have denoted the remainder 'i with the subscript i because the value of the remainder dependson the choice of the quotient i.

We first show the existence of a remainder , i such that 0 < , i < d.We do this by considering the set of all nonnegative remainders appearingas in Fig. 7.4; if we can show this set is nonempty, then we can choose itsleast element, which is certainly a reasonable candidate for the inequality'i < d.

If m > 0, we let i = - 1. Then the corresponding remainder r - 1 =m + d is also nonnegative. On the other hand, if m is negative, we leti = m. Then

rm = m - dm

= m(l - d)

= (-m)(d - 1),

and since - m > 0 and d - 1 > 0, r m > O. In either case, there is at leastone nonnegative remainder. Hence the set R of all nonnegative remaindersis a nonempty subset of the nonnegative whole numbers. It should be clearthat the Well-Ordering Axiom holds for the set N u {O} as well as N, sowe can select from R its least element, which we denote simply by r, and welet the corresponding quotient be denoted by q.

We know that

r = m - dq,

and that

o ~ r;hence

m = dq + r,

and

o ~ r.

If we can show that also r < d, this will complete the proof of the existenceof the desired numbers q and r.

Keep in mind that r is the least possible nonnegative remainder upondividing d into m. Suppose by way of contradiction that d ~ r. Then thenumber s = r - d is nonnegative, and s < r because 0 < d. Now

m = dq + r

= dq + d + r - d

= d(q + 1) + r - d

= d(q + 1) + s.

But since m = d(q + 1) + s, we have expressed the division of d intom with a new quotient q + 1 and a remainder s such that 0 ~ s < r. Thisis in contradiction to the fact that r is the least possible nonnegative remainder.Hence our supposition that d :::;; r leads to a contradiction, and thereforethe desired inequality r < d must be true. Thus we have shown the existenceof integers q and r such that both

are true.

m = dq + r and O~r<d

Technically, this completes the proof of Theorem 7.8 as stated, but wewant to show one more useful fact while all this machinery is set up: that thequotient q and the remainder r chosen above are unique as well; that is, ifx and yare two numbers such that both

m=dx+yand

O~y<d

are true, then x = q and y = r. Suppose that x and yare integers satisfyingthe above two relations.

Now r was chosen to be the least nonnegative integer such that m =dq + r for some integer q; since y is such an integer, it must be true thatr ~ y. Suppose that y = r. Then

r = m - dq,and

y = m - dx,hence

m - dq = m - dx,and thus

dq = dx,

so that x = q. In this case, y = r and x = q, as we wish to show. On theother hand, suppose that r < y.

Then we have the inequality

o ~ r < y < d.Now

m = dq + r,and

m = dx + y,hence

dq + r = dx + y.Thus

dq - dx = y - r,or

d(q - x) = y - r.

Therefore d is a divisor of y - r. But since

o ~ r < y < d,it follows that

o < y - r < d - r < d,so that

o < y - r < d.

By Exercise 7.8, no natural number (such as d) can divide evenly into asmaller natural number (such as y - r). So the fact that d I (y - r) leadsto a contradiction, and shows that the inequality r < y is impossible. Onlythe previous case considered, in which y = r, can occur; and we have seenthat in this case also x = q. This establishes the uniqueness of q and r,as desired.

For example, while there are many ways to "divide" 5 into 23 to obtaina "quotient" q and a "remainder" r such that 23 = 5q + r, there is one andonly one choice of q and r-namely, q = 4 and r = 3-such that both23 = 5q + rand 0 ~ r < 5.

Theorem 7.9 (The Fundamental Theorem of Arithmetic) If m is a naturalnumber other than 1, then m can be factored into the product ofprimes, andthis factorization is unique (apart from the order of the prime factors).

Proof If m is already prime, then we understand the theorem to mean thatm is its own unique factorization; certainly m cannot be equal to two differentprimes, nor can m be factored any further. So we suppose that m is composite; moreover, we suppose by way of contradiction that the theorem isfalse. Since Theorem 7.7 guarantees that each composite natural number hasat least one prime factorization, it must then be the case that there existsa composite natural number with a nonunique prime factorization. Bythe Well-Ordering Axiom, we select the least composite natural number witha nonunique prime factorization, and denote this number by m.

Hence

and

where all the p's and q's are primes, and these two factorizations are differentin that the p's and q's differ in number or in kind, or both.

IfPI = q l' then

and since n < m, these two prime factorizations of n must be the same.Hence by properly rearranging subscripts, we have P2 = q 2, P3 = q 3' ... ,

Pi = q j' and i = j. But then the two factorizations of m shown above arethe same, for also PI = q 1. This contradicts the fact that the two givenfactorizations of m are different. And by repeating this argument for otherp's and q's, we see that none of the p's can equal any of the q's.

Since P1 '1= q l' we can suppose that the p's and q's are so named thatP1 < q 1· By Theorem 7.8, there exist integers wand r such that

q1 =P1W + r,

and

We substitute P1 w + r for q 1 in the equation

and obtain

m = (P1 W + r)q2q3··· qj

=P1wq2q3···qj + rq2q3···qj·

But m is also equal to P1P2P3 ...Pi' so we have the equation

Hence

rq2q3··· qj = P1P2P3·· ·Pi - P1 wq2q3··· qj

= P1(P2P3 ... Pi - wq2q3··· q).

So P1 is a divisor of rq 2q 3 ••• q j.

Now r < P1 andp1 < q1' so r < q1. Hence the natural number

is less than m, since m = q 1q 2q 3 ••• q j. Because m is the least naturalnumber with a nonunique prime factorization, rq 2q 3 ••• q j has a unique primefactorization. Moreover, since rq 2q 3 ••• q j is divisible by P1' P1 appears insome factorization of rq 2q 3 ••• q j. Since P1 cannot equal any of the q's,P1 must appear in the prime factorization of r, and hence P1 I r. But 0 :::;r < P1' so the only way thatp1 I r can be true is if r = O. But if so, then theprevious equation

becomes

and hence P1 I q 1. This, too, is a contradiction, as we have shown that noneof the p's can equal any of the q's; but since P1 and q 1are primes andp1 I q l'

the only way this can be true is if P1 = q 1. Thus our supposition that thereis a composite number with a nonunique prime factorization leads to acontradiction, and this establishes the theorem.

An alternate version of stating the Fundamental Theorem of Arithmeticis this: If m is a natural number other than 1, then there is one and only oneway of expressing m as the product

where the exponents nt, n2' ... , nk are natural numbers, and Pt, P2' ... ,Pkare primes arranged so that Pt < P2 < ... < Pk'

Exercises

7.36 In the proof of Theorem 7.7, only one of four cases was consideredthe one in which m = ab, where a was composite and b was prime. Showhow to handle the other three cases.

7.37 Let m = -23 and d = -5. For what values of q and r is it true that

m = dq + rand

o ~ r < -d?

(One handles the statement of Theorem 7.8 for the case of a negative divisorby changing the sign of d to obtain a pusitive number.)

7.38 Factor 6300 into the product of primes. In how many essentiallydifferent ways can this be done?

7.39 If we interpret the product of no numbers to be the number 1, and theproduct of one number to be itself, can we then state that "Every naturalnumber has a prime factorization?" Can we state that "Every naturalnumber has a unique prime factorization?"

7.40 Let a, b, and c be integers. Prove that if a I b and a I c, then a I (b + c)and a I (b - c).

7.41 Suppose that a and b are natural numbers and P is a prime such thatP I abo Prove that either P I a or P I b.

7.42 Is it true that if n is a natural number such that 12 I n2 then 12 I n?

7.43 Prove that every divisor of both m and n is also a divisor of both3m + nand m + 2n.

7.44 Suppose that p is a prime and m and n are natural numbers such thatP I (m + n). Does it follow that one of the relations P I m or pin is true?

7.45 Wilson's Theorem states that the natural number n is prime if and only if

n I (n - I)! + 1.

Use Wilson's Theorem to prove that 7 is prime.

7.46 See Exercise 7.45. Use Wilson's Theorem to prove that 20 is not prime.

7.4 The greatest common divisor 205

7.47 See Exercise 7.45. Use Wilson's Theorem to prove that l00! + 1 isnot prime.

7.48 Suppose that P and q are two different primes and n is a natural numbersuch that both pin and q I n are true. Prove that pq I n.

7.49 How could you use the prime factorization of 144 in order to writedown all the divisors of 144?

7.50 Suppose that n is a natural number with the "canonical" primefactorization

n - p(Zlp(Z2 • •• p(Zk- I 2 k'

By the "canonical" factorization we mean the one in which the exponents(Xl' (X2' ••• ,(Xk are natural numbers and the numbers PI' P2' ... ,Pk areprimes so arranged that

PI < P2 < ... < Pk'

Use this factorization to find a formula giving the number of divisors of n.Hint: See the previous exercise.

7.4 THE GREATEST COMMON DIVISOR

If m and n are two integers not both zero, then by their greatest commondivisor g we mean the largest integer g such that both g I m and gin aretrue. The greatest common divisor of m and n will be denoted by (m, n).

It should be clear that if m and n are not both zero, then (m, n) exists,for m and n do have common divisors (such as 1), and only finitely manyintegers can divide evenly into both m and n. So there must be a largest such;in fact, since I is a common divisor of m and n, then 1 ~ (m, n), and hence(m, n) is a positive integer. For example, (24, 28) = 4 and (8, 9) = 1.

The Euclidean Algorithm provides a method for computing the greatestcommon divisor of two integers. The method depends on the next theorem.

Theorem 7.10 Let m and n be natural numbers, and let q and r be integerssuch that m = qn + rand 0 ~ r < n. Then (n, m) = (r, n).

The proof of this theorem is outlined in the next set of exercises. Weillustrate with an example the method of using this theorem to find thegreatest common divisor of 21 and 78.

As indicated in Fig. 7.5, we first divide 21 into 78, obtaining a quotientof 3 and a remainder of 15. We ignore the quotient; Theorem 7.10 tells usthat (21,78) = (15,21). We continue successive divisions, and obtain thechain of equalities

(21, 78) = (15, 21) = (6, 15) = (3, 6) = (0, 3).

206

321}78

63

15

267) 193

134

59

Number theory

1

15}2115

6

1

59}6759

8

26}15

12

3

7

8)5956

3

2

3T66

aFig.7.5 Successivedivisions to find (21, 78).

7.4

23)8

6

2

12)"3

2

1

Fig.7.6 Finding (67, 193)-Notethat each remainder is next dividedinto the corresponding divisor.

Since the remainders must decrease with each division, we must eventuallyreach the remainder zero after a finite number of steps. So in any problemof this sort, our last entry in a chain such as the one above will have the form(0, r), where r is the next-to-Iast remainder. But r =1= 0; in fact, r > 0, andit is easy to see that (0, r) = r. So the greatest common divisor of m and nwill be the last nonzero remainder in the sequence of successive divisions ofprevious remainder into previous divisor.

For another example, we show the calculations to find (67, 193) in Fig.7.6. We obtain the chain of equalities

(67, 193) = (59, 67) = (8, 59) = (3, 8)

= (2, 3) = (1, 2) = (0, 1) = 1.

Hence the greatest common divisor (67, 193) of 67 and 193 is 1.


Moreover, a sort of reverse of this process can be carried out once theabove computations are made. We take the example (21, 78) = 3. Thedivisions shown in Fig. 7.5 can be expressed in the form below:

78 = (3)' 21 + 15,

21 = (1)'15 + 6,

15=(2)'6+3,

6 = (2)' 3 + O.

Recall that (21, 78) is 3, the last nonzero remainder above. We ignore thelast equation, and solve the one in which the remainder 3 appears for 3 itself,thus obtaining

3 = (1)·15 + (-2)·6.

We solve the second equation in the above list for its remainder, 6, andsubstitute

6 = (1)·21 + (-1)' 15

in the previous equation to obtain

3 = (1)' 15 + (- 2) . [(1) ·21 + (-1)' 15]

= (3)·15 + (-2)·21.

The remainder previous to 6 is 15; we solve the equation 78 = (21)' 3 + 15for 15, and substitute the result for the 15 in the above equation:

15 = (1)·78 + (-3)·21,

so that

3 = (3)'[(1)'78 + (-3)·21] + (-2)·21

= (3)·78 + (-11)·21.

What we have accomplished is the expression of 3, the greatest commondivisor of 78 and 21, in terms of 78 and 21 themselves. It should be clear thatthis process of back-substitution of remainders can always be carried out soas to express (m, n) in terms of m and n. We give one more example, using67 and 193. From the data shown in Fig. 7.6, we obtain the followingequations:

193 = (2)·67 + 59,

67 = (1). 59 + 8,

59 = (7)· 8 + 3,

8 = (2)' 3 + 2,

3 = (1)·2 + 1,

2 = (2)' 1 + O.


The last equation is unnecessary; we discard it. Our sequence of solutionsand substitutions goes as follows:

1 = (1). 3 + (-1)·2

2 = (1). 8 + (-2)·3

1 = (1). 3 + (-1). [(1) . 8 + (- 2) . 3]

= (3)' 3 + (-1)·8

3 = (1). 59 + (-7)·8

I = (3)' [(1) . 59 + (-7)· 8] + (-1)' 8

= (3)' 59 + (- 22) . 8

8 = (1)' 67 + (-1)' 59

1 = (3)'59 + (-22)'[(1)'67 + (-1)'59]

= (25)' 59 + (- 22) . 67

59 = (1). 193 + (- 2) . 67

1 = (25)' [(1). 193 + (-2)· 67] + (-22)·67

= (25)' 193 + (- 72) . 67

Hence we have expressed the greatest common divisor of 67 and 193 interms of 67 and 193, as follows:

(67, 193) = (-72)·67 + (25)' 193

If you see that this process will always work, you have in effect seen theproof of our next theorem.

Theorem 7.11 Let g = (m, n). Then there exist integers x and y such that

xm + yn = g.

Exercises

7.51 Evaluate (8334, 9612).

7.52 Find integers x and y such that

43x + 91y = 1.

7.53 One can use the Well-Ordering Axiom to prove Theorem 7.11; however, this proof will merely show the existence of integers x and y such that

xm + yn = (m, n).

One proceeds as follows: Let

S = {xm + yn I x and yare integers and xm + yn > O}.

First show that S is nonempty; then let 9 be the least element of S. Provethat 9 = (m, n). Please fill in the details of this proof.

7.54 Fill in the details of the following proof of Theorem 7.10: If m and nare natural numbers and q and r are integers such that m = qn + rando ~ r < n, then (n, m) = (r, n).

Let d be any common divisor of nand m. Show that d must be a divisorof rand n. Then show that every common divisor of rand n is also a commondivisor of nand m. Since the common divisors of nand m are thus the sameset of natural numbers as the common divisors of rand m, it follows immediately that (n, m) = (r, n).

7.55 Show that (36, 64) = 4, and find integers x and y such that 36x +64y = 4.

7.56 Show that (11, 27) = 1, and find integers x and y such that llx +27y = 1.

7.57 Using the Fundamental Theorem of Arithmetic, we can factor 4200into the prime product 23 . 3 . 52 . 7 and 4500 into the prime product 22 . 32 . 53.Show how these factorizations can be used to find the greatest commondivisor (4200, 4500) of 4200 and 4500.

7.58 Is it possible to find integers m and n such that 12m + 16n = I?Explain.

7.59 Is it possible to find integers m and n such that 12m + 13n = 6?

7.60 Find integers m and n such that

31m + 231n = 1.

7.61 Let n be a natural number. Prove that

(n, n + 3) =1= 2.

7.62 Let n be an integer. Prove that

(n, n + 2) ~ 2.

7.63 Prove that 8 I (n 2- 1) if n is an odd integer.

7.64 Let m and n be natural numbers. Prove that (m, m + n) I n.

7.65 We have seen that the set W of all whole numbers, together with thetwo operations of ordinary multiplication and addition, satisfies the Fundamental Theorem of Arithmetic, and it seems that Well-Ordering is essentialfor establishing this theorem. This is indeed the case, for consider themathematical system K defined as follows:

K = {a + b~ I a and b are integers, and e = - 5}.

210

and

Number theory

We define addition and multiplication on the set K as follows:

(a + b~) + (c + d~) = (a + c) + (b + d)~,

7.5

K',

(cxfJ)y = cx(fJy);

(a + b~)' (c + d~) = ac + ad~ + bc~ + bde

= (ac - 5bd) + (ad + bc)~.

With these two operations K becomes an algebraic system satisfying thesame axioms as the system of integers; specifically,

If ex, fJ, and yare elements of K, then

ex + fJ and exfJ belong toex + fJ = fJ + ex and exfJ = fJex;(ex + fJ) + Y = ex + (fJ + y) andex(fJ + y) = exfJ + exy;fJ = 0 + O~ has the property that fJ + ex = ex;z = 1 + O~ has the property that zcx = cx; andif cx = a + b~, then -cx = (-a) +. (-b)~ has the prop-

erty that -cx belongs to K and (-cx) + cx = fJ.

However, no ordering of K compatible with addition and multiplication canbe a well-ordering; thus it is not surprising that the Fundamental Theoremof Arithmetic does not hold for K. To see this, one defines an element of Kto be "prime" if its only factorizations are the obvious ones; that is, theelement ex E K is prime if it can only be expressed as a product of two elementsof K in the two forms

cx = zcx = (- z)( -cx).

It is possible to find a "composite" element of K with two different primefactorizations. Please do so.

7.5 APPLICATIONS

The equation ax + by = c, where a, b, and c are whole numbers, has ofcourse infinitely many solutions in the real number system (unless for examplea = 0 = band c i= 0). But in number theory we are interested in onlythose whole numbers rand s which are solutions to the equation ax +by = c. The ability to find such solutions opens the way to solve a widevariety of fascinating problems.

Let us consider the equation ax + by = c, where a, b, and c are givenfixed whole numbers. The only case giving difficulty occurs when none ofa, b, and c is zero. We begin by letting 9 = (a, b).

If9 is not a divisor of c, then the equation ax + by = c has no solutionin whole numbers, so we further suppose that 9 I c.

7.5 Applications 211

We use the methods of Theorem 7.11 to solve the equation ax + by = g.Let m and n be a pair of whole numbers that solve the equation, so that

am + bn = g.

Since g I c, there is a whole number k such that c = gk. Let r = km ands = kn. Then rand s are solutions of the original equation ax + by = c,for

ar + bs = a(km) + b(kn)

= k(am + bn)

= kg = c.

So there is no difficulty in finding one whole number solution to the equationax + by = c. However, we seek all possible solutions. Using the particularsolution ar + bs = c that we have found, it turns out to be possible toexpress all possible solutions in terms of the numbers rand s. For supposethat the whole numbers p and q give another solution, so that

ap + bq = c.

Then

ap + bq = ar + bs,

so that

a(p - r) = b(s - q).

Now 9 = (a, b), so 9 I a and 9 I b. We divide our last equation by g, andobtain

(alg)(p - r) = (blg)(s - q).

Note that all quantities in this last equation are whole numbers. Moreover,the greatest common divisor of alg and big is 1, so that

(alg) I (s - q) and (big) I (p - r).

Let t be a whole number such that (alg)' t = s - q. If we substitute(alg) . t for s - q in our previous equation, we obtain

(alg)(p - r) = t· (alg)(blg).

We just cancel alg from both sides, and thus

p - r = (big)' t.

From the relations (alg)' t = s - q and (big)' t = p - r we have obtained,we can by solving for p and q also derive that

p = r + (big)' t,

q = s - (alg)' t.

Now r, s, a, b, and g are all known, so that we have expressed the othersolution pair p, q in terms of known quantities and the "variable" t-which isan integer. Since it is not hard to verify also that each integer choice of tdoes indeed produce a solution pair p, q to the equation ax + by = c, wehave in effect established our next theorem.

Theorem 7.12 Let a, b, and c be nonzero whole numbers and let g = (a, b).Consider the equation ax + by = c. Ifg does not divide c, there is no solutionto the equation. If g I c, then all possible integral solutions are given byx = p and y = q, where

p = r + (big)· t,

q = s - (alg)· t,

where r, s is one solution pair for the original equation and t is allowed to rangethrough all whole number values.

Example 7.1 Find all solutions in whole numbers of the equation 8x +9y = 10.

Now (8, 9) = 1 and 1 I 10, so solutions do exist. We first solve 8x +9y = 1. In this simple case one can see a solution immediately: x = - 1and y = 1. Hence we multiply this solution pair by 10 to obtain a solutionto the original equation:

8· (-10) + 9· (10) = 10.

So we have a = 8, b = 9, c = 10, r = - 10, s = 10, and g = 1. ByTheorem 7.12, all solutions are given by the formulas

x = -10 + 9t,

y = 10 - 8t,

where t is a whole number.It frequently happens in such problems that we are interested only in the

positive solutions. If so, we must have both

-10 + 9t > 0,and

10 - 8t > 0,so that

8t < 10 < 9t,and thus that

t < 1 and 2 < t.

But there is no integer t satisfying both these inequalities, and so we canconclude that the equation 8x + 9y = 10 has no positive whole numbersolutions.

Example 7.2 Find all integral solutions of the equation 3x + 12y = 100.

There is no solution, since (3, 12) = 3 but 3 is not a divisor of 100.

Example 7.3 Find all positive integral solutions of the equation 5x +15y = 100.

A simplifying procedure is to divide each term by (5, 15) = 5, and theresulting equations will have the same solutions as the original equation.Thus we consider instead the equation

x + 3y = 20.

Now (1, 3) = .1, so we first solve instead

x + 3y = 1.

The obvious solution x = 1, y = 0 will of course do. We multiply each ofx and y by 20 to obtain the solution r = 20, s = 0 to

x + 3y = 20.

By Theorem 7.12, we can obtain all possible solutions from the equations

x = 20 + 31,

Y = 0 - 1,

where 1 is an integer. For positive solutions, it is further necessary that

20 + 31 > 0,and

o - 1 > O.

These relations lead to the inequalities

and thus-20 < 31 and 1 < 0,

-6 ~ 1 ~ -1.

So only the values 1 = - 6, - 5, - 4, - 3, - 2, and - 1 can produce positivesolutions. In this case, those solutions are as follows:

x = 2, y = 6,

x = 5, y = 5,

x = 8, y = 4,

x = 11, y = 3,

x = 14, y = 2,

x=17,y=1.

Example 7.4 Ifa man has ninety-five cents in dimes and quarters, how manyof each type of coin might he have?

If we let d denote the number of dimes and q the number of quarters hehas, then we need to find the nonnegative solutions of the equation

10d + 25q = 95.

As in the previous example, we solve instead the simpler equation

2d + 5q = 19.

Now (2, 5) = 1 and 1 I 19, so solutions do exist. We first solve

2d + 5q = 1

One solution, by inspection, is d = 3 and q = -1. We multiply by 19 toobtain the particular solution d = 57, q = -19 of the original equation.Then all solutions have the form

d = 57 + 51,

q = -19 - 21

for 1 an integer. For q and d both to be nonnegative, we must also have

-57 < 51 and 21 < -19,

which lead to only the two values -11 and -10 for 1. These lead to the twopossibilities

or

d=2

d=7

and

and

q = 3,

q = 1.

So the solution to the original problem is this: The man has either two dimesand three quarters, or else seven dimes and one quarter.

Exercises

7.66 Find all positive whole number solutions of

4x + 6y = 100.

7.67 Find three different solutions of

lOx - 7y = 23.

7.68 A man cashed a check at a bank for two hundred forty-five dollars,and asked the teller for some ones, ten times as many twos, and the balancein fives. In how many different ways can the teller oblige him? What arethese ways?


7.69 How would you go about solving the equation

ax + by = 0,

where a and b are whole numbers? Are the formulas given in Theorem 7.12also valid in this case?

7.70 What can be said about the whole-number solutions of the equation

ax + by = c,

where a, b, and c are integers, in case one of a and b is zero? What if botha and b are zero?

7.71 Let a, b, and c be whole numbers with neither a nor b equal to zero,and let 9 = (a, b). Prove that if the equation ax + by = c has a solutionin integers, then 9 I c.

7.72 To prove Theorem 7.12, it is necessary to show that if a and barenonzero whole numbers and 9 = (a, b), then the greatest common divisorof alg and bIg is 1. Please prove this.

7.73 Here is the method for solving the equation

ax + by + cz = d

in whole numbers, where a, b, c, and d are integers, none of a, b, and c iszero, and all solutions are desired. The method is much more complicatedthan the case of only two unknowns, so we give only the method for findingthe solutions and omit the proof.

First, there is no solution unless g, the greatest common divisor of a, b,and c, also divides d, so we suppose that 9 I d.

Let

and

-bb=-.

(b, c)

Then ([3, b) = 1, hence we can find integers ex and y such that

exb - [3y = 1.

Let

, = (a, bex + cy),

and find integers Jl and v such that

Jla + v(bex + cy) = ,.

216 Number theory

Then all possible solutions of

ax + by + cz = d

7.5

are given by the three equations below, where t and u are allowed to assumeall possible whole number values (and every solution can be obtained fromthe equations below):

dJl (bex + cy)tx = --, "

exdv exatY = T + T + fJu,

ydv yatz = - + - + bu., ,

Use this result to find all positive solutions of the equation

x + 2y + 3z = 12.

7.74 If four marks are worth one dollar, five zlotys are worth one dollar,and eight pesos are worth one dollar, how maya dollar be exchanged fairlyfor marks, zlotys, and pesos so that at least one unit in each of the foreigncurrencies is obtained?

7.75 A man paid two dollars for 100 eggs, including some new-laid eggs atten cents each, some fresh eggs at two cents each, and some old eggs at onecent each. He found that he had the same number of two kinds of these eggs.How many of each did he buy?

7.76 A man bought pomegranates at 16 cents each, oranges at two centseach, and tangerines at one cent each. If he bought twenty pieces of fruit,including at least one of each kind, and spent 80 cents in all, how many ofeach did he buy?

7.77 A man cashed a check for less than one hundred dollars at a bank.The teller confused the number of dollars on the check for the number ofcents, and paid the man forty-three dollars and fifty-six cents more than hedeserved. In how many different amounts could the check have been written?

7.78 Find three different solutions of

8x + 9y + 10z = 12.

7.79 Verify that the solutions x andy given in Example 7.1 actually work forall integral values of t.

7.80 Verify that the solutions p and q given in Theorem 7.12 actually workfor all integral values of t.


7.81 If 11 brass balls (or equal weight) weigh exactly 15 pounds, 11 copperballs weigh 16 pounds, and 11 silver balls weigh 17 pounds, how many ofeach are required to weigh exactly 11 pounds?

7.82 Prove that1 1 1 1-+-+-+ ... +-2 3 4 n

is never a whole number.

7.83 Each odd prime has the form 4n + 1 or else the form 4n + 3, wheren is a nonnegative integer. Prove that no prime of the latter form is the sumof two squares of whole numbers.

7.84 Prove that if n is a cube of a natural number, then the product of thethree consecutive integers n - 1, n, and n + 1 is divisible by 504.

7.85 Find a natural number half of which is a square (of a natural number),one-third of which is a cube, and one-fifth of which is a fifth power.

7.86 Show that if n is a natural number, then

10 I (n 5- n).

7.87 What is the last digit of 7355 ?

7.88 We consider the equation x2 + y2 = Z2, and seek natural numbersolutions. It turns out that any solution has the form

x = m 2- n2

,

y = 2mn,

Z = m 2 + n2,

(provided that, since one of x and y must be even, we let that be y) where mand n are natural numbers with m > n. Verify that the formulas given aboveactually do provide a solution for the equation x 2 + y2 = Z2.

7.89 Find three different solutions to the equation

x 2 + y2 = Z2.

7.90 In connection with Exercise 7.88, we are actually finding Pythagoreanright triangles-those right triangles with whole number sides. Find twoPythagorean right triangles with the same hypotenuse.

7.91 Find two different Pythagorean right triangles with the same perimeter.

7.92 How many Pythagorean right triangles can have a hypotenuse oflength 10?

7.93 How many Pythagorean right triangles can have one side of length 10?

7.94 Find three different Pythagorean right triangles whose legs differ by thenumber 1.


7.95 Prove that one leg of a Pythagorean right triangle always has lengthdivisible by 3 and that one side always has length divisible by 5.

7.96 Find three different Pythagorean right triangles such that the lengthof the hypotenuse and the length of one leg differ by the number 1.

7.97 Prove that every natural number n can be expressed in the form

n = a2 + b2- c2

,

where a, b, and c are integers.

7.98 Is 2100- I prime? Why or why not?

7.99 Let n be a natural number and p a prime not a factor of n. Fermat'sTheorem states that p I (nP -

1- I). Use Fermat's Theorem to prove that

7 I 999,999.

7.100 Show that every prime other than 2 and 5 divides evenly into somenumber of the form 999 ... 99 (k digits, all nines).


The following books on number theory may be of interest:

Beiler, A., Recreations in the Theory of Numbers (Dover, 1964).

Dudley, U., Elementary Number Theory (Freeman, 1969).

Gelfond, A., The Solution ofEquations in Integers, translated by J. B. Roberts(Freeman, 1961).

Griffin, H., Elementary Theory ofNumbers (McGraw-Hill, 1954).

Mordell, L., Diophantine Equations (Academic Press, 1969).

Niven, I. and H. Zuckerman, An Introduction to the Theory of Numbers(Wiley, 1960).

Rademacher, H., Lectures on Elementary Number Theory (Blaisdell, 1964).

Please do not consider this chapter as containing more than a minutefraction of what is known about the theory of numbers. We have not touchedon Farey series, the distribution of primes, twin primes, Fermat's "LastTheorem," perfect numbers, or even Gauss' Law of Quadratic Reciprocity.Since the latter is considered by many to be one of the most beautiful resultsin the field, we state it below, and invite you to verify it for yourself in anumber of special cases.

~,

Suppose that p and q are two different odd primes. The problem is infinding a natural number n such that

p I (n2- q).


The Law of Quadratic Reciprocity states that such a natural number n can befound if and only if there exists a natural number m such that

q I (m2- p),

with an exception: If p and q are both of the form 4k + 3, then the firstrelation above has a solution if and only if the second does not.

The Euclidean Algorithm of Section 7.3 is, ofcourse, named for the Greekmathematician Euclid-about whom very little is known other than that helived in the third century B.c. His Elements formed such a complete systemization ofgeometry that his work has been used as a geometry text even in thiscentury.

On the other hand, a great deal is known about the life of Carl FriedrichGauss. He was born in 1777 ofvery poor parents in Braunschweig, Germany,and only a succession of fortunate coincidences made it possible for him tobecome a mathematician. He lived seventy-eight years, and though not soprolific a writer as Euler, unquestionably made far greater contributions tomathematics: His work in number theory is particularly impressive, but healso laid the foundations for differential geometry, complex analysis, andmodern topology. He desired perfection in his publications, and thus it isthat he anticipated results ofmany later mathematicians-his journal containsa large number of valuable results which he never published, and for thisreason many of these were credited to others. In any case, the consensusseems to be that the world has produced three truly outstanding men ofgenius-Archimedes, Newton, and Gauss.

CHAPTER 8

ANIMALPOPULATIONS

We shall consider an important branch of ecology, the branch which dealswith the growth of animal populations, but we feel very strongly that twoimportant cautions should be issued at once.

First, we shall deal with only the simplest possible cases; we shall beassuming, for example, that no more than two or three species are involved,that the reproduction rate ofeach species is constant, that there are no changesin the populations in question caused by migration, and numerous othersimplifying conditions that will become apparent as we proceed.

Second, there is the philosophical consideration that mathematics reallycannot prove anything about the real world, but only about mathematicsitself. We shall be passing from a physical reality to a mathematical abstraction; such a procedure always merits cautious interpretation because ofsimplifying assumptions such as those mentioned above. In addition, we shallbe making essentially unprovable (in the mathematical sense) assumptionsabout the way in which animals behave; here the usefulness of the mathematical model lies in the fact that it can offer predictions about the way inwhich animal populations ought to fluctuate; and if such fluctuations areindeed observed by biologists and ecologists in the field and laboratory, thiscan be considered evidence in favor of the validity of such assumptions.

To make the last idea clear, mathematicians have been accused in thepast of "proving" things which are patently absurd; for example, there is arumor that a mathematician once "proved" that a bumblebee cannot fly.It was likely the case that the prover in question made the assumption thatinsect flight muscle could not metabolize its fuel any faster than mammalianmuscle, as well as a host of other such assumptions. Of course, some of theseassumptions must then have been false, provided that the "proof" itself wasmathematically valid. In this case, rather than being a futile exercise, perhaps

220

8.1 Unrestricted growth of a single species 221

such a proof could give to biologists clues as to which such assumptionsshould be experimentally verified.

As another example, it is said that a mathematician once "proved"that it was impossible for a drag racer to turn a quarter mile in less than 9.0seconds. But dragsters commonly beat this time these days; in this case, theerroneous assumption may well have been that the coefficient of frictionbetween rubber and strip did not increase as the rubber temperature increased,an assumption now known to be false.

In summary, then, one cannot do ecology with paper and pencil. Whatmathematics can contribute to ecology, as it has contributed to the othersciences, is the prediction of the behavior of a system after certain assumptions-usually simplifying ones-have been made. If the predictions correlatewith field and laboratory research, this is evidence in favor of the validity ofthe assumptions. If not, the assumptions should be examined; again, suchexamination can be done only experimentally, not mathematically. Withthese cautions in mind, we proceed to the simplest case (with the simplestassumptions).

8.1 UNRESTRICTED GROWTH OF A SINGLE SPECIES

One simple assumption about the growth of an animal population is that thegrowth rate is proportional to the number of individuals present. Thisassumption is contingent upon the further assumption that there is nothingto impede growth of the population; for example, there must be effectivelyunlimited living space and food supply. It would seem reasonable that suchshould be the case if the population were small relative to the amount ofliving space and food supply available. For example, with a reasonably smallpopulation of bacteria in a large culture medium, one would expect that ifthe bacteria population were 1000 individuals increasing at the rate of 3individuals per second, then a population of 2000 individuals would increaseat the rate of 6 individuals per second and a population of 10,000 individualswould increase at the rate of 30 individuals per second. Here is the reasonwhy this should be the case.

Suppose we have such a situation, say of a small number of bacteriain a large culture, and we denote the number of bacteria at a given time t byN(t), or more simply merely by N. It seems very reasonable that the rate ofincrease of this population, which we will denote by N'(t) or simply N',depends on the value of N itself at the time t in question. This is why wehave chosen a notation N' for the rate of increase, in order to suggest thatthe value of N itself has something to do with the rate N'. After all, if allextraneous factors could be removed, it should be the case that the value ofN' depends only on the value of N, whether or not they happen to beproportional.

222 Animal populations 8.1

1\

ji1

IWe can try to make plausible the assumption that N' is proportional to N

in this simple case. We can by virtue of the above discussion at least writethe equation

N' = f(N)

to indicate that there is some function f that gives the manner in whichN' behaves with respect to the value of N. For example, it might be that

N' = N 3,

or

N' = 2N + N 2 + NN + 1

Suppose our bacteria culture were divided into two equal populations,each thus containing N /2 individuals. This division could be done eitherphysically, by removing half the bacteria to another culture medium-or itcould be done by drawing an imaginary line down the middle of the culture.Since the actual rate of increase of each half of the population should be thesame in either case, but in the former case it should be f(N /2) and in the lattercase just N' /2, we could therefore write the equation

N' /2 = f(N /2).

Similar considerations with respect to dividing the population intothirds, or multiplying it by four, lead us to the equation

a- N' = f(aN)

for all meaningful values of a. But since

N' = f(N),

we can substitute this into the former equation, and obtain

a .f(N) = f(aN).

This means that the function f has the property that f(ax) = af(x) forall values of a and x. It turns out that the only such function with a smoothgraph must be of the form f(x) = kx, where k is a constant, and hence

N' = f(N) = k - N.

In other words, it would seem reasonable that the form of the function f isextremely simple, and that N' is indeed proportional to N. Note that theconstant k must be positive in the case of bacterial growth, since we areassuming N to be positive, and N' is also positive because the population isincreasing rather than decreasing.


Let us rewrite our equation as an equation about functions, by reinsertingthe time variable t. Then

N'(t) = k· N(t).

By the end of a course in differential calculus, most students will havelearned how to "solve" this equation in order to find explicitly the form ofthe function N(t), the real unknown in the above equation. Calculus entersthe picture here because differential calculus is itself the study of rates ofchange of functions; although we will next present the formal manipulationsused to solve the above so-called differential equation, it is not necessary thatyou understand these procedures. The reasons are these: First, many of oursubsequent equations representing growth rates of animal populations aretoo complicated to be solved with pencil and paper alone-they are usuallysolved by approximation methods involving use of high-speed electroniccomputers. Second, in our later studies we shall not be interested so much inthe actual form of the function N(t), but rather in the steady-state or limitingbehavior of the animal population, and we can discover this steady-statebehavior without the necessity of solving any differential equations. Allthat will be needed is some elementary work with inequalities.

However, given the differential equation

N'(t) = k· N(t),

the calculus student first divides by N(t), since we may assume that N(t)IS never zero.

N'(t) = k.N(t)

Then

f N'(t) dt = f k dtN(t) ,

and hencelog N(t) = kt + C,

where C is a constant and the logarithm function is to the base e (the approximate value of e is 2.71828) rather than to the base 10. From this follows

N(t) = ekt+C = eC' ~t.

Since C is constant, so is eC, and we evaluate it by introducing the further

experimental assumption that the population N(t) is known when t = 0;say, N(O) = No. Then

and henceN(t) = No' ~t.

224

N-axis

Animal populations

(k>O)

8.1

Fig.8.1 The graph ofN(t} = No ·ekt, which isincreasing at an increasingrate.

0....- t-axis

The graph of the function N(t) is shown in Fig. 8.1. In this graph, wehave assumed that the constant of proportionaHty k is positive, as we havealready mentioned. The graph increases more and more rapidly with increasing values of t, indicating that the rate of population increase is itselfincreasing. Of course, the value of k itself as well as the value of No mustbe determined individually for each experimental case; k, for example, shoulddepend not only on the species of animal involved, but also on the concentration of the food supply, the effectiveness of the food in promoting reproduction, the temperature of the environment, and numerous other experimentalfactors.

The information that we want to derive from our initial populationequation

N' = k·N


is available to us without use of the manipulations of calculus shown earlier.In this chapter we shall be mostly concerned with the eventual or long-termbehavior of the population in question. In this example, we can reason verysimply from the preceding equation in the following manner. At the beginningof the experiment, we assume N to be positive, and N' positive as well sincethe value of N is assumed to be increasing, at least initially. As we have seen,k too must then be positive. But then, so long as N remains positive, N'itself must be positive; that is, the population will continue to increase.Thus N will indeed remain positive, and the population will in fact always beincreasing. In addition, as N increases, the value of N' must also increase, sothe population will be increasing at a faster rate as time goes on. Thus thepopulation will not tend toward any steady state, but behave as the graphin Fig. 8.1 indicates.

This is no proof that E. coli, or any other living species, will eventuallytake over the whole world, for we have made one very important simplifyingassumption: that the increase in the population in no way impedes futuregrowth of the population. In actual practice-say, in a culture of E. coli ina test tube-available space and food supply are both strictly limited, and weshall see in the next section how certain very simple additional assumptionswill lead us to a better model of growth of a single species.

On the other hand, it is important that we mention that the form of thefunction N(t) that we derived using calculus,

N(t) = No· tit

has been experimentally verified in a large number of cases. That is, for asmall population with a large amount of available food and space, the graphof the actual population compares well with the graph of N(t) in Fig. 8.1,so long as the value of N itself is small. In other words, so long as there arenot too many individuals, the rate of growth of the population does behaveas if it were proportional to the population. This will be an importantassumption in much of our later work. The graphs shown in Fig. 8.2 showa typical population curve, shown as a dashed line, compared with the graphof N(t) = No· ekt

• The two curves match quite well for small values of N.This should be considered as experimental justification for our assumptionthat the rate ofgrowth of animal population is proportional to the populationitself, so long as other factors are kept equal and so long as there is no inhibition of the growth rate by a too-large population.

Exercises

8.1 Suppose for some reason we had been led to the equation

N' = ~N'

226

N-axis

Animal populations 8.1

------------"'....~';1'

~';1'

~~

'----------------- t-axis

Fig. 8.2 The population curveis well approximated byN(t) = No· ekt for smallpopulations.

where k is a constant. Assuming that the population N and its rate of increase N' are both positive at some initial time t = 0, sketch the graphshowing-very roughly-the behavior of the function N(t).

8.2 If a population N is initially positive, and its rate of increase N' isconstant, what will be the behavior of the function N(t)? Treat three cases:when the constant is positive, when it is zero, and when it is negative.

8.3 For reasons given in Section 8.1, it would be plausible to suppose thatan animal population of N individuals would have a constant birth rate b,resulting in a rate of increase in the population B = bN proportional to thevalue of N ; it is equally reasonable that the population would have a constantdeath rate d, resulting in a rate of decrease D = dN in the population alsoproportional to the value of N. The net rate of increase in the population,which we have denoted by N', should then have the form

N' = B - D.

Show how the equation

N' = kN

.can be derived from the above assumptions. What is the value of k? Whatinterpretation can be given to the constant k?

8.1

N-axis

Unrestricted growth of a single species 227

N ( t) = No· ekt (k < 0)

L....- t-axis

Fig.8.3 For negative k, the graph of N(t) = No ·ekt is decreasing.

8.4 A quantity of a radioactive substance can be thought of as a population,say in terms of the mass present or the number of atoms present. It followsfrom the assumption that radioactive decay is equally likely for any twoatoms of one substance that the decrease of the amount of the radioactivesubstance is proportional to the number of atoms present. We again obtainthe differential equation

N' = k· N,

where N is the amount present and N' is the rate of increase of the amount.Here, though, k is a negative constant-the rate of increase must be negativesince the amount of radioactive substance is actually decreasing. Nevertheless, we obtain the same solution

N(t) = No· t!'t,

and the graph of this function, for negative k, is shown in Fig. 8.3.The half-life of a radioactive substance is the time it takes for one-half

of the substance to decay; for example, in the case of Iodine-131, half will begone after eight days. Show that the "half-life" is a meaningful concept;that is, that for a given radioactive substance, the half-life is independentof the initial amount present. Hint: Let T be the half-life. Find some wayof "solving" the equation N(t) = No· t!'t for T, and show that the solutionis independent of the initial quantity No.

228

N-axis

Animal populations 8.1

8.5 Here is one method by which the graph of a function N(t) satisfying adifferential equation such as

N'(t) = k· N(t)

can be found approximately, without the necessity of actually obtaining anexplicit formula for the function N(t). We shall illustrate how the procedureworks with the equation above, assuming k > 0, and leave it to you to fillin the details.

Imagine a point in the first quadrant, as shown emphasized in Fig. 8.4.Here both Nand t are nonnegative, and since N' = k· N, N' is also nonnegative. Also, the larger the value ofN, the larger the corresponding value ofN'. So the point indicated on the graph not only represents a certain possiblepopulation at a certain time, but also can be thought of as lying on the graphof a solution to the above equation. As t increases the value of N must alsoincrease, as indicated by the arrow. For larger values of N the rate of increasewill be proportionately greater, as indicated by the equation N' = k· N.Thus arrows of steeper slope are indicated for the larger values of N. Notethat smaller values of N have associated with them arrows of smaller slope

8.2 Growth of a species under limiting conditions 229

as well, but the slope is always positive because N' is positive in the firstquadrant (except when N = 0). If you select an initial population valueNo, and sketch in a smooth curve following the trend of the arrows, you willobtain a rough idea of the shape of the graph of N(t).

You can repeat this process with the first two exercises, as well as withother differential equations such as

N'= NN + l'

and other such examples of your own invention.

8.2 GROWTH OF A SINGLE SPECIES UNDERLIMITING CONDITIONS

Since the graph of N(t) = No· ekt does not give an accurate picture of thegrowth of an animal population-at least, for large values of N-we shouldtry to improve our assumptions which led to the original differential equation

N' = k·N.

One very simple way to do this is to introduce into the equation a term whichwill, in effect, decrease the value of the rate constant k as the value of Nincreases. A very natural way of doing this is to assume that the populationis living in an environment that will support a certain maximum populationM of individuals, and the closer the value of N gets to the number M, thesmaller the value of k (and thus the smaller the value of N'). However, thesolution of the previous section turns out to be quite accurate for smallvalues of N, and hence we do not wish to modify the value of k when Nis small. What we need is a term that has the value 1 for N very small, andwhich decreases to 0 as N increases; in fact, it should become 0 when N = Mand become negative for N > M.

The reason for the latter consideration is that we can imagine an animalpopulation in an environment of insufficient resources to support thatpopulation. In such a case, it seems reasonable that the number of deathswould exceed the number of births, and thus that the value of N' would benegative, indicating a net decrease in the population.

Hence we need to multiply the constant k by a term that is nearly 1 forN close to zero, a term that decreases to zero as N gets closer and closer tothe value M, and that becomes negative if N exceeds M. One of the simplestways of inventing a formula for such a term is to use

M-NM


For N = 0, the value of the above term is 1; for N between 0 and M,its value is between 0 and 1; it becomes 0 when N = M and negative forN > M. Now the above term can be regarded as a sort of degree to whichthe potential increase of population is realized; indeed, when these conceptswere first introduced into ecology, the word equation

{act~al rate} = {pote.ntial rate} . {degree of real~zatiOn}of Increase of Increase of potentIal

was used. This translates very naturally into the differential equation

N'=k·N·M

-N

M '

and this is the equation that we shall examine. There is only moderatedifficulty in actually finding the form of the function N(t), but we shall spareyou the details, and use instead the sort of analysis that will be appropriatein later sections. Remember that, as before, k is a positive constant, and Mtoo is a positive constant, indicating a maximum population that can besupported by the environment.

We can use the same sort of analysis as was used in Exercise 8.5. Wesketch in Fig. 8.5 the arrows indicating the direction of movement of thevalue of the population. For small values of N (small, that is, relative to M)the value of(M - N)/M is very close to the number 1, so that its effect on theequation can be neglected. Hence, for small N, N' is approximately proportional to N, and so for larger values of N the arrows become steeper.

However, somewhere along the line the effect of the term (M - N)/Mbegins to make itself felt, and the arrows become no steeper; indeed, as Nincreases, the term (M - N)/M becomes quite small, effectively reducing thevalue of k and thus the value of N'. Hence the arrows begin to flatten out.For N = M, in fact, the arrows must be horizontal, indicating no change inthe population, for when N = M, the equation

N'=kN· M - N

M

becomes

N' = O.

That is, there is no change in the population.Finally, for N > M the value of (M - N)/M is negative, and the arrows

must slope downward; as N continues to increase, the downward slope of thearrows must increase since (M - N)/M is taking on values such as -1,-2, -3, ....

In Fig. 8.6 we show three different population curves, each dependingon the initial choice No of the population at time t = O. These curves are

8.2

N-axis

Growth of a species under limiting conditions 231

""","""" " """ ........... " .........,N =M ------------------------------------------------

.....,. .....,. .." .." ~ .....,.

./ / / / / /I I I I I I Fig.8.5 Approximating

the graph of a solution to

I I I I I IN' = kN·(M - N)/M.

I I I I I I

" ./ ./ / / /~ "" ~ ~.....,. ,;#'"

t-axis

N-axis

Fig. 8.6 The graphs ofthree typical solutions

to the equationN' = kN·(M - N)/M.

-----------------------------

~--------------t-axis


obtained by choosing a value of No and then sketching in the graph of N(t)using the arrows as a guide.

The hardest experimental test of this solution is in the case of the lowestcurve, which exhibits the most complicated behavior. But ultimately thisso-called sigmoid curve (because it is shaped like the letter "S") gives asurprisingly good fit to actual experimental evidence. Thus the hypothesisthat the potential rate of increase of a population is appropriately modifiedby multiplication by the degree of realization of that potential is justified,as well as the form

M-NM

for the degree of realization.Actually, some researchers in the field believe that various modifications

of the equation

N' = kN. M - NM

would give a solution N(t) that would more accurately fit the experimentaldata in more cases; for example, one might wish to consider instead theequation

or even ask, in general, for the best possible exponent rx in the equation

However, for reasonable choices of rx, the steady-state behavior of the systemwill generally be unchanged. We shall consider at present only the case of ouroriginal equation, where rx = 1. Suppose, then, we inquire into the behaviorof the function N(t) that solves

N' = kN. M - NM '

as t increases without bound. The answer is already before us. The fact thatall the arrows in Fig. 8.6 are directed toward the horizontal line where N = Mindicates that, regardless of the initial population, its size will tend toward thevalue Mas t increases; unless, of course, No = O. This simple analysis willnot be much complicated in the more complex systems we take up in thenext sections.

8.3 The case of two competing species 233

Exercises

8.6 Repeat the analysis of this section for the differential equation

N' = kN. (M ;:; N) 1/2

and show that the population tends toward one of the two values 0 or M.For what initial value of the population will it tend toward the value O?

8.7 Examine the behavior of the population equation

N' = k. (N)1/2 . M - N .M

This equation has been observed to give accurate fits to experimental dataunder certain conditions.8.8 Construct, as an alternative to (M - N)jM, a formula for the so-calleddegree of realization of potential population increase. Examine the corresponding differential equation and see if it has the right sort of generalbehavior as t increases without bound.8.9 Suppose that a cylindrical tank with vertical axis has a small hole in itsbottom, is filled with water, and the water leaves the tank at a rate proportionalto the water pressure. Make the necessary assumptions about constants(height of tank, density of water, and so on) and write down the differentialequation whose solution V(t) is the volume of water in the tank at time t.

8.10 Suppose that the increase in a certain animal population due to birthsis proportional to the number of individuals present, but that the decrease inthe population due to deaths is constant. Write down the differential equationdescribing the behavior of the population N as a function of time t, andinclude the term (M - N)jM for degree of realization of potential. What isthe behavior of the solution of this differential equation as t increases without bound?

8.3 THE CASE OF TWO COMPETING SPECIES

We now turn our attention to the case of two different species of animals witha coexistence problem. We assume that they live in the same space and arecompeting for much the same food supply. However, please remember thatwe are still making a large number of simplifying assumptions; for example,we assume that there is under such circumstances a maximum populationwhich the environment will support with respect to each species, and thatthis maximum is constant; in practice, of course, this maximum is likelyto undergo variations because of the available food supply suffering seasonalvariations, and the like.

In order to make the ideas of this section more concrete, we shall consider the case of two reasonably similar species of fish-bluegill and redear-


living in a pond free from predators and with a constant but limited foodsupply. Since adults of these species are not predatory upon one another,we shall ignore the effects of predation on immature fish, so that the mostimportant aspect of the interspecies competition is the sharing of food andliving space. The important thing here is that bluegill and redear have similar,though not identical, food preferences. To make the significance of thenotation easy to remember, we shall use the following symbols:

B(t) or B will denote the number of bluegill present at time t.R(t) or R will denote the number of redear present at time t.B'(t) or B' will denote the rate of increase of the bluegill population

as usual, if B' is negative this means that the population is actually decreasing.R'(t) or R' will denote the rate of increase of the redear population.

With redear absent, we suppose that the pond will support a certainmaximum population of bluegill, and we denote this maximum by C (sincethe letters Band C are alphabetically adjacent). Similarly, we let S denotethe maximum population of redear the pond would support in the absenceof bluegill. With redear absent, we have seen in Section 8.2 that a reasonabledifferential equation describing the population B(t) of bluegill would be

B' = kB. C - BC '

where k is a positive constant having to do with the birth and death rate ofbluegill. Similarly, with bluegill absent, we can describe the behavior of theredear population by

R' R S - R= v . .S

Again, v is a positive constant.With both species present, neither of the above equations will still be

appropriate, for the very existence of redear in the pond impinges uponthe available food and space for bluegill, and in effect decreases the value of C,the maximum possible bluegill population. In fact, it seems reasonable tosuppose that the value of C would be decreased in direct proportion to thenumber of redear present, and hence our degree of realization of potentialterm would become

C-B-exR

C

where ex is a constant that can be thought of as representing the degree towhich redear interfere with the bluegills in the latter's quest for food andspace. If a wide variety of foods were available to the two species and theirfood preferences overlapped only partially, one would expect the constantex to be somewhere between 0 and I; however, if there were only one type of

food available and the redear were more efficient in obtaining it than bluegill,then it would be reasonable to suppose that ex > 1. We shall treat allpossibilities, but remember that there are also numerous other such factorsof comparison between the two species that are concealed in the littleconstant ex.

To continue our analysis, we can write a reasonable (though oversimplified) equation for the population of bluegill as

B' = kB. C - B - exRC

and, similarly, the redear equation becomes

R' = vR. S - R - pB .S

Of course, pplays a role for redear analogous to the role of ex with respect tobluegill.

What we now ask is this: Given initial populations of the two speciesof fish, and values of the various constants, what will be the eventual population of the pond? Will one species inevitably dominate the other, so that thelatter becomes extinct and the former reaches its maximum population?Or can the two species coexist, each at a certain percentage of its maximumpopulation?

In Fig. 8.7, rather than plotting either of the functions B(t) and R(t)against the time variable t, we plot instead values of B(t) on the x-axis andvalues of R(t) on the y-axis. The reason for this will shortly become clear.We now ask for what values of Band R will B be increasing; that is, wesolve the inequality B' > O. That is done by recalling that

B' = kB. C - B - exR •C

Now if B' > 0, we have

kB. C - B - exR > o.C

Since we may assume each of k, B, and C positive (they are certainly notnegative, and if B = 0 there is no problem) each may be canceled from theabove inequality, and we see that

C - B - exR > 0,or

exR < C - B,or

C-BR < .

ex

R-axis

Fig.8.7 B' = 0 on thestraight line R = (C - B)/a.8'<0

..

.,8'>0

R =Cia

L-.---------------'""'-----8-axis8=C

Thus the population of bluegill will be increasing when R < (C - B)fa.;this is the condition that will make B' > O. If we plot the graph of R =(C - B)fa., as we have done in Fig. 8.7, this will be a straight line; it mustcross the vertical axis where B = o-that is, where R = Cfa.-and it mustcross the horizontal axis where R = 0; that is, when

C-B = 0,a.

which occurs exactly when B = C.Below this line, we have

C-BR < ,

a.

and hence in the triangular region below this line, where each point representsa possible population of bluegill and a possible population of redear, thebluegill population will be increasing. We have indicated this in Fig. 8.7by an arrow directed to the right, which indicates an increase in the valueof B-we are temporarily silent as to the behavior of R. Similarly, in theregion above the line R = (C - B)fa., we must have B' < 0, so that the


Case 1

,\

\ ,\,\\

Case 3

Case 2

Case 4

Fig. 8.8 The four important cases for the positions of the lines onwhich B' = 0 and R' = O.

bluegill population will be decreasing in that region. This is indicated by anarrow pointing to the left.

If we perform a similar analysis on the equation

R' = vR. S - R - pBS '

we can expect to obtain another straight line, with redear population increasing on one side and decreasing on the other; moreover, this line mustcross the vertical axis at S and the horizontal axis at SIP. There are, however,four possibilities, as shown in Fig. 8.8, where the "redear line" is shown as adashed line. The redear line can lie entirely over the bluegill line, entirelybeneath it, or cross it in either of two ways. (We neglect the unlikely possibilities that the two lines coincide or cross at a point exactly on one of thetwo axes, for reasons to be discussed in the exercises to come.)

R-axis

LL

8'<0

R'<O

C S/{3

Fig. 8.9 The case inwhich the redear Iine liesentirely above the bluegillline.

Which of the actual cases shown in Fig. 8.8 actually occurs depends onthe experimentally obtained values of four of the constants we have introduced; the first case shown in Fig. 8.8 corresponds to the case when

C-<8(X

and8

C < -.f3

We begin our analysis with this case. Under the solid line (or bluegillline) we have indicated with arrows pointing to the right the fact that B isincreasing, and above that line the arrows pointing to the left mean that inthat region B is decreasing. At the beginning of each of these arrows we haveattached another arrow, vertically upward if R is increasing and verticallydownward if R is decreasing. All this is shown in Fig. 8.9, and is just aconvenient way of indicating that in the small triangle, both populations areincreasing; in the middle region, the redear population is increasing while thebluegill population is decreasing, and above both lines both populations aredecreasing. Moreover, note that on the bluegill line the values of Band Rare such that B' = 0; that is, the bluegill population is steady. Similarly,the redear population is steady on the redear line.

8.3

R-axis

The case of two competing species 239

Fig.8.10 Three typicalcurves showing

population trends.

Each point on the graph where Band R are nonnegative represents apossible initial population of bluegill and redear. We have selected threesuch points in Fig. 8.10, one in each of the three essentially different regionsdetermined by the two lines, and then imagined the value of t increasing fromits initial value of 0. The curves drawn represent, as one moves along eachcurve from its initial point, the manner in which each of the two populationsmust change, as indicated by the arrows in Fig. 8.9. It is easy to see whatmust happen in each of the three cases; for any initially positive populationsof bluegill and redear, the curves lead inexorably with increasing values of tto the point (0, S), where the redear line crosses the vertical axis. When thecurves reach that point, there they stay, for both B' and R' are zero at thatpoint. At this point, the population of bluegill is zero and the population ofredear is the maximum the pond can support, the number S. Hence, in thiscase, the redear will eventually take over the pond.


One concept we have just encountered will be subsequently quite useful;that is the concept ofa critical point. The point (0, S) is a critical point for thesystem of differential equations

B' = kB. C - B - IXRC '

R' = vR. S - R - fJBS '

because when B = 0 and R = S, both B' and R' are zero, and the populationsshould not be expected to change without outside influence. In general, werefer to any such situation in which all rates of change involved are zeroas a critical point, and the location and nature of these critical points willusually give us some idea as to the eventual or limiting value toward whichthe populations are tending.

In the case above, the point (0, S) will be called a stable critical pointbecause, if the values of Band R are changed slightly from the values B = 0and R = S, the tendency will be for the values of Band R to return to thevalues 0 and S, respectively. However, the above system has two othercritical points as well, which can be found by inspection of the diagram shownin Fig. 8.9, or simply by solving the differential equations so as to find whenboth B' and R' are zero. In the latter choice of procedures, we have

kB. C - B - IXR = 0C '

andS - R - fJB

vR· = o.S

The former equation holds when k = O-a solution which we ignore-orwhen B = 0, or when

C - B - IXR = o.That is, when

C-BR = .

IX

Similarly, R' = 0 when (ignoring v = 0) R = 0 or when

R = S - fJB.

To find when both B' and R' are zero, we have the four following cases:

1) B = 0 and R = O. In this case the pond holds no fish. This critical pointis not stable because a slight change in the "population" will result in themovement of the population away from this critical point, rather thanback to it.

2) B = 0 and R = S - pB. Since B = 0, the latter simplifies to R = S.We have already seen that this is a stable critical point; the pond containsonly redear.

3) R = (C - B)ja. and R = O. Since R = 0, the former equation simplifiesto just B = C. The pond contains only bluegill; again, this critical pointis unstable, since introduction of a small number of redear will upset thebalance.

4) R = (C - B)ja. and R = S - pB. This solution is a point where the"bluegill line" and the "redear line" do not cross in the first quadrant,and thus this case gives us no critical point.

To summarize, the population of the pond will be constant if it containsno fish, only bluegill, or only redear, and only the latter case is stable.

Recall our earlier statement that some physical interpretation could beattached to the numbers a. and fJ. Let us examine that interpretation in thesituation just considered. We can see from Fig. 8.9 that since the redearline lies entirely above the bluegill line, we must have the inequalities

C-<8a.

and8

C < -.fJ

For simplicity, and because the two species are fairly similar in this example,we suppose that C = S. Then

or

8- < 8a.

1 < a.

and

and

88<-

fJ'

fJ<1.

Remember that a. can be thought of, very roughly, as measuring the degree towhich the redears interfere with the success of bluegills, and fJ the degree towhich bluegills interfere with redears. That a. > 1 and fJ < 1 thus could beinterpreted, very simply, as meaning that in this situation, redear is a "moresuccessful" species.

If we pass to the second case shown in Fig. 8.8, then the situation wouldbe exactly reversed. If it turned out, by experimental measurement, that Cand S were approximately equal and that a. 1, then the stablecritical point would be located at (C, 0), indicating that one would expectthat the population ofthe pond would tend toward the maximum of bluegilland no redear.

The third case shown in Fig. 8.8 is the most complicated. Ifyou fill in thearrows indicating the direction of population trend as in Fig. 8.9, you willobtain a diagram much like that shown in Fig. 8.11. Here, for C and Sapproximately equal, it turns out that both a. and fJ can be expected to exceed


R-axis

S

Cia

LL-----------~---~--B-axis

SI(3 C

Fig.8.11 Directions ofpopulation trends in thecase of an unstable criticalpoint in the first quadrant.

the number 1, which should indicate that each species competes more successfully with the other than with itself. This sounds like a peculiar situation,and some typical lines of population flow are shown in Fig. 8.12.

We have here two stable critical points-one where B = 0 and R = S(all redear) and one where B = C and R = 0 (all bluegill). The expectedcritical point at (0, 0) is unstable and also rather uninteresting. But thebluegill line and redear line cross at a point in the first quadrant, specifically at

rxS - CB= ,

rxf3 - 1

and

R _ ,--f3C_-_S- rxf3 - 1 .

At this point, both B' and R' are zero, indicating a steady-state population, but this state is certainly unstable since a small change will generallyforce a population movement toward one of the two stable critical points.

8.3

R-axis

The case of two competing species 243

Fig.8.12 Typical curvesof population trend in the

case of an unstable criticalpoint in the first quadrant.

L...----------~---........."'----B-axis

The curves in Fig. 8.12 also tell us how to predict the eventual fish population.If the pond starts with relatively few redear, the curves flow toward the pointof eventual extinction of redear; with relatively few bluegill, the redear willeventually dominate the pond.

The most interesting case of all, shown as the fourth case in Fig. 8.8,has been reserved for your enjoyment as the very next exercise.

Exercises

8.11 Consider the bluegill-redear competition in the fourth case shown inFig. 8.8; that is, when

cS<

exand

sc <-.p

Draw figures analogous to Figs. 8.11 and 8.12. Find the stable andunstable critical points and discuss the physical interpretation of yoursolution.

8.12 Give a reason why we can ignore such cases as those in which theredear and bluegill lines coincide, or intersect on one of the coordinate axes.

8.13 What sort of critical points are obtained in case the redear and bluegilllines do coincide?

8.14 See Exercise 8.11. Does it make sense to draw the arrows on the coordinate axes? Does it make sense to draw curves of population movementthat intersect the axes? What interpretation would you give to such a curve?

8.15 Let us consider the case examined in Exercise 8.11, in which a criticalpoint is found where both Band R are positive. Assume it is plausible that afisherman fishing for bluegill and only bluegill has the effect of decreasing thevalue of C, the maximum population of bluegill supportable by the pond.What effect will his fishing have on the relative populations of bluegill andredear in case the populations have previously stabilized at the critical pointmentioned above? What if the fisherman is so expert that he reduces the valueof C so much that Clr.x < 8?

8.16 In analogy to the case of two competing species considered in thissection, write down reasonable differential equations describing the case ofthree competing species.

8.17 Why must the "bluegill line" and the "redear line" each cross thecoordinate axes at a positive value?

8.18 In Section 8.2, the differential equation

N'=kN· M - N

M

was discussed. Suppose N represents a population of one species of fish in apond, and M the maximum population that the pond will support. Find thestable and unstable critical points for this system.

8.19 Continuing the previous exercise, what will be the effect on the stablecritical point if a fisherman catches fish from the pond in proportion to theirpopulation?

8.20 Continuing the previous two exercises, what will be the effect on thestable critical point if a fisherman catches fish from the pond at a constantrate? In particular, what is a reasonable differential equation that describesthis system?

8.4 THE PREDATOR-PREY CASE

Let us now examine the interesting case in which there are again two speciesinvolved, one of which preys on the other. Again, we shall suppose that thissituation exists in a pond containing two species of fish-bass and redearthe former the predator, the latter the prey, for while redear account for avery substantial portion of the food supply for bass in such a situation,

8.4 The predator-prey case 245

redear will not prey upon bass above a certain minimum size. We will make alarge number of simplifying assumptions here, including the following:

• We assume that neither population is so great as to necessitate the introduction of the degree of realization factor. Consequently, the rate ofincrease of redear population will be the increase due to births, minus thoseredear consumed by bass; here we further assume that such consumptioncompletely accounts for attrition of redear.

• Moreover, we assume that the increase in the bass population due tobirths is wholly dependent on the available food supply-redear and nothingelse. Thus we must introduce a term to account for the decrease in basspopulation due to deaths.

• Finally, we assume that redear are consumed at a rate proportional to thenumber of encounters between bass and redear. Since the number of suchencounters will thus be proportional to the number of bass as well as to thenumber of redear, it is not hard to see that the number of encounters will beproportional to the product of the number of bass and the number of redear.For if the number of bass were doubled, the number of encounters would bedoubled. If then, in addition, the number of redear were doubled, the numberof encounters would again be doubled, resulting in four times as manyencounters in all.

Hence we assume that the following very simple differential equationsdescribe the predator-prey situation:

B' = kBR - dB,

R' = vR - wBR.

Here, B is the number of bass present, B' is as usual the rate of increase of thebass population, R is the number of redear, and R' their rate of increase.As we have said, we assume that the rate of increase of the bass populationis proportional to the number of encounters between bass and redear, so thatthis rate of increase is kBR, where k is a positive constant. The positiveconstant d represents the death rate of bass, presumably from old age; sincethe number of deaths of bass is proportional to the bass population, theattrition of the bass population is thus dB. Since this term represents adecrease in the bass population, we subtract the number of deaths from thenumber of births to obtain the net population increase of bass, and thus weobtain the first equation

B' = kBR - dB.

We have also assumed that the increase of the redear population due tobirths is proportional to the population of redear, so that this term becomesvR, where v is a positive constant. Redear vanish from the redear population

246 Animal populations

R-axis

B'>O

8.4

Fig.8.13 Directions oftrends of bass populationin the predator-preyexample.

R = d/kt-------------------

..B'<O

.......--------------------B-axis

d/kl------------~I__--------

R-axis

LR' > 0, B' > 0

R' > 0, B' < 0

rR' <O,B' >0

R' <O,B' <0

Fig.8.14 Trends ofbass and redearpopulation.

L...- -L- B-axisw/v

at a rate proportional to the number of bass-redear encounters, and thus bythe amount wBR, where w is another positive constant. Again subtractingdeaths from births, we find that

R' = vR - wBR.

We now solve the inequality B' > 0, in order to find for what values of BandR the bass population is increasing. We have

kBR - dB> 0.

Since we may assume that B > 0, the term B may be canceled from thisinequality, and thus we see that

kR > d,or

dR>

k'

since k, too, is positive.We plot the straight line R = djk in the graph shown in Fig. 8.13,

where values of B lie on the x-axis and values of R lie on the y-axis. Thusthe line R = djk is a horizontal line passing through the point djk on they-axis, and for values of R in excess of djk-that is, above this straight linethe bass population is increasing since B' > 0, when R > djk. This makesgood sense; when there are many redear, the bass population should be onthe increase, and when there are few redear the number of bass should bedecreasing. This information is indicated by the usual arrows in Fig. 8.13.

We next ask for what values of Band R is the redear population increasing; that is, we solve R' > 0. We have

vR - wBR > 0,so that

v - wB > 0,and hence

vB < -.

w

We plot the vertical line B = vjw in the graph shown in Fig. 8.14; forvalues of B less than vjw (that is, to the left of this line) R' > 0, and so theredear population is increasing. And if B > vjw, the redear population isdecreasing. The usual arrows have been drawn in Fig. 8.14, summarizingour findings about the population changes of the two species.

Now on the line R = djk, the horizontal line shown in Fig. 8.14,B' = 0. This can be seen by substitution of R = djk in the bass equation

B' = kBR - dB.


R-axis

d1k --------------- -----

Fig. 8.15 Curves ofpopulation trend in thepredator-prey case.

L...- ----'- 8-axis

v/w

Similarly, R' = °on the vertical line B = v/w. Hence the point (v/w, d/k)where the two lines intersect is a critical point. Its stability will be discussedshortly. Another critical point is (0, 0), the case in which the two species areabsent from the pond. In practice this should probably be considered a stablecritical point, since if a few of each species are introduced the bass could beexpected to eliminate the redear and then die of starvation themselvesremember, we are assuming that the bass have no other food supply.

There is no other critical point where R = 0, for again, in the absence ofredear, the bass population will decrease to zero. One can expect a criticalpoint where B = 0, somewhere high up on the vertical axis, where thepopulation of redear will stabilize at a point much higher than if bass werepresent. This is not indicated by our differential equations, since for simplicity we omitted the degree of realization term from each equation.

In Fig. 8.15 we have indicated the apparent behavior of the curves ofpopulation trend. The behavior of the arrows in the previous figure verystrongly suggests that the curves are closed, and thus that the population


of each species will undergo periodic variations. First bass and redear increase, then the larger number of bass cause the redear population to decrease;next, the diminished food supply causes a decrease in the bass population,and finally this last decrease permits an increase in redear until there aresufficiently many to permit an increase in the number of bass.

It might actually be that the curves are slowly spiraling in to the point(v/w, d/k), in which case the latter would be a stable critical point; or,possibly, the curves could be spiraling slowly outward, and the point(v/w, d/k) would then be an unstable critical point. For these particularlysimple differential equations, it turns out that the point (v/w, d/k) is neither.The curves actually are closed, and hence a perturbation in a populationinitially with values B = v/w, R = d/k will result in a new population whichcan be thought of as having values circling about the point (v/w, d/k), andthus not really moving either away from or closer to the point (v/w, d/k).Perhaps some term could be invented here, such as calling this the case of aquasi-stable critical point.

In addition, each closed curve of the sort shown in Fig. 8.15 can also bethought of as a critical path ("path" seems a better term than "point"here), since deviations from one curve simply place the population on anearby curve. It also is sensible to ask about the duration of such a paththat is, how long does it take the population to go through one completecycle? For some species of fish and many mammals, the duration of a cycleis usually measured in terms of a few years; curiously enough, even the much"longer" paths do not seem to have much greater duration that the veryshort ones. Of course, these very simple differential equations cannot pretendto be an accurate representation of what actually takes place in nature, butsuch cycles have been observed with sufficient frequency so that they deservesome explanation, even a weak one.

In summary, the mathematics predicts the existence of periodic cyclesin populations of predators and their prey, such cycles have been observed,and there is apparently a good reason why such cycles should occur.

Exercises

8.21 In the predator-prey case involving bass and redear, it would seem tothe fisherman's advantage to fish so as to move the populations of eachspecies close to the quasi-stable critical point (v/w, d/k). Why?

8.22 Continuing the previous exercise, during which parts of the populationcycle should the fisherman fish for bass? For redear? Can he ever fish forboth? Should he ever fish for neither?

8.23 Suppose that there are two species of fish in a pond, say A and B, andthe members of species A eat the eggs of species B at a rate proportional tothe population of species A. Suppose also that the members of species B


eat the eggs of species A at a rate proportional to the population of species B.Suppose, finally, that the two species are not in competition for food or space,so that other than egg-eating the presence of each species will not inhibitthe growth of the other. Under such circumstances the degree of realizationterm (M - N)/M should probably be introduced into each resulting differential equation.

Write down a plausible pair of differential equations describing thissituation. It should be the case that your equations are sufficiently simpleto make the next exercise feasible.

8.24 Use the equations invented in the previous exercise to go through aprocess like that in the last two sections, finding stable and unstable criticalpoints and sketching curves representing population trends.

8.25 List at least five deficiencies in the oversimplified treatment of thepredator-prey situation given in the previous section. For example, do bassreproduce continuously?

8.26 Suppose that species A is parasitic on species B, which it has no difficultyin finding, and this takes place in a region of ample space and food supply.It would then be reasonable to write

A' = etA. KB - AKB '

and

B' = f3B' M - B - kA .M

Explain the origin of these equations.

8.27 Find critical points and sketch curves of population trends for theequations in the previous exercise.

8.28 One can try to improve the predator-prey equations given in the lastsection by introducing a degree of realization term for the redear. We write

B' = kBR - dB,

S-RR' = vR· - wBR

S '

where S is the maximum population of redear the pond can support, and theother constants and terms have the same meaning as in the last section.Making reasonable assumptions about the relative sizes of these constantswhere necessary, find critical points and sketch curves of population trendfor this system of differential equations.


8.29 Suppose the two sPecies Band R are symbiotic, though neither iscompletely dependent on the other, and there is no competition for food orspace. Explain the origin of the descriptive equations

B' = kB' C - B + IXRC '

R' = vR. S - R + pB ,S

where IX and p are positive constants. Hint: See Section 8.3.

8.30 Assuming that the constants IX and pof the previous exercise are relatively small, sketch curves of population trend for the equations of the previousexercise and find any critical points.

8.31 If the human race were considered as two species-male and femalecompeting for food and space, what sort of differential equations woulddescribe this system? You should assume that the birth rate of males andfemales is the same. Analyze the consequences of your differential equations,and see if the results seem to match up with reality.

8.32 In the predator-prey equations

B' = kBR - dB,

R' = vR - wBR

for bass and redear, suppose a poison is introduced into the pond whichkills members of each species at a rate proportional to the population of thatspecies. We could then write the modified equations

B' = kBR - dB - pB,

R' = vR - wBR - qR,

where p and q are the positive "poison" coefficients. Without the poisonterms, the most interesting stable critical point is (v/w, d/k), as we have seen.Where is the corresponding critical point for the new system of equations?In particular, will the number of prey rise or fall?

8.33 Ladybugs prey on aphids, and aphids are hard on many vegetables andornamentals. If you observe a stable population of ladybugs and aphids inyour garden, and you have an insecticide as effective in killing ladybugs as inkilling aphids, should you spray?

8.34 Suppose that a city of fixed area contains N(t) automobiles at time t.Suppose also that new cars are being introduced into the city at a constantrate, and that cars are being permanently removed by total destruction due totwo types of accidents-single-vehicle accidents and two-vehicle accidents.


Should the rate of attrition of cars due to single-vehicle accidents be proportional to N? To what should the rate of attrition due to two-vehicleaccidents be proportional? What differential equation would describe thebehavior of N' under these circumstances?

8.35 In Section 8.1, we discussed the case of unrestricted growth of a singleorganism. Under such circumstances, will there be a fixed time to suchthat the population will double in every interval of duration to?

8.36 Suppose that a cylindrical tank with vertical axis has a small hole in thebottom of such size that if the tank were full of water, then water would pourout of the hole at the rate of 100 gallons per minute. Suppose also that in anycase, water pours out of the hole at a rate proportional to the water pressure,and that the tank is very large compared to the other amounts used in thisproblem. If water is running into the tank from an overhead pipe at the rateof 10 gallons per minute, toward what limiting volume is the amount of waterin the tank tending? (Use appropriate constants where necessary.)

8.37 How would one test the growth of the population of the United Statesin the last hundred years to see if it fits the case of the unrestricted growth of asingle species treated in Section 8.1 ?

8.38 How would one test the growth of the population of the United Statesin the last hundred years to see if it fits the case of the growth of a singlespecies under limiting conditions, as treated in Section 8.2?

8.39 Return to the predator-prey case of Section 8.4, and assume that thedeath rate d of bass is so small that we might as well suppose that d = O.What happens to the bass-redear system in this case?

8.40 Continuing the previous exercise, a more realistic assumption mightbe the following: If the death rate of bass is ignored, then a degree of realization term should be introduced into the equation giving the rate of population increase of the bass. In this case we might have for the bass equation

B' = kBR. M - RM '

where M is some theoretical maximum population of bass the pond cansupport. Should M be proportional to R, the number of redear present?If so, how does the bass-redear system behave? What if M is assumed to beconstant?


Three interesting books concerned with animal populations are:

Gause, G. F., The Struggle for Existence (Williams and Wilkins, 1934).


Lack, D. L., The Natural Regulation of Animal Numbers (Oxford ClarendonPress, 1954).

Slobodkin, L. B., Growth and Regulation of Animal Populations (Holt,Rinehart, and Winston, 1961).

/'

Eugene Odum's well-known textbook EcOlogy (Holt, Rinehart, andWinston, 1963) is an excellent general introduction to the whole field ofecology.

If you have worked a number of the exercises, do not feel as if you havebeen doing ecology. Field work and laboratory work are essential; ecologycannot be done solely at the desk. The most one can hope for is the formulation of new hypotheses susceptible to experimental verification-orinvalidation.

CHAPTER 9

THEARTGALLERYTHEOREM

The main purpose of this chapter is to develop enough of the theory ofconvex sets to prove Krasnoselskii's Theorem, also known as the "ArtGallery" Theorem. However, there will be considerable development of sidetopics, particularly in the exercises.

Although the theory of convex sets has applications in game theory,linear programming, and other branches of mathematics much used in thesocial and managerial sciences, convex sets are quite interesting in their ownright. Moreover, numerous pictures can be drawn illustratin~ definitions andtheorems in this chapter; we recommend that you draw such pictures frequently, for they will help greatly in constructing proofs and will not usuallymislead you.

A small amount of notation from the elementary theory of sets will beused throughout this chapter, so if this notation is unfamiliar to you, it wouldbe very helpful to read over the first section of Chapter 6. On the other hand,though the study of convex sets will be presented from a very geometricpoint of view, plane geometry itself is not a prerequisite for this chapter.We will usually restrict our attention to subsets of the ordinary two-dimensional plane; although generalizations to higher dimensions are frequentlypossible, such material will usually be reserved for the exercises.

9.1 CONVEX SETS

We assume that you are familiar with the properties of straight lines andstraight-line segments in the plane. We will denote the ordinary two-

254

9.1

y-axis

b

Convex sets 255

Fig.9.1 The segment[a, b] from the point a

to the point b.

----+---.......----------- x-axisa

dimensional plane by £2, just an abbreviation for Euclidean space of twodimensions.

If a and b are points of £2, the straight-line segment joining a and bwill be denoted by [a, b], and consists of a and b together with all pointsbetween a and b on the straight line in £2 through a and b. By [a, a] wemean the set {a}, consisting of the point a alone.

Example 9.1 Let a = (0, 1) and b = (2, 2). Then [a, b] consists of allpoints (x, y) of £2 such that 1 ~ x :$; 2 and y = 2x - 2. The set [a, b]of this example is shown in Fig. 9.1.

It should be clear from the definition that [a, b] = [b, a]. You may useanything you know about plane geometry to prove our first theorem.

Theorem 9.1 If a and b are points of£2 and p E [a, b], then

[a, p] u [p, b] = [a, b].

Ofcourse, to prove that the above two sets are equal, one shows that eachpoint of [a, p] u [p, b] is a point of [a, b], and conversely.

The following is our most important definition. Let C be a subset of£2. We say that C is convex if [a, b] is a subset of C whenever a and barepoints of C. That is, for every possible pair of points a and b of C, it must

256 The art gallery theorem 9.1

be true that the segment [a, b] belongs wholly to C. And thus, to prove a setis convex, a usually fruitful approach is to select two arbitrary points a and bof the set, and then prove that [a, b] is a subset of the given set.

For some examples, the plane sets shown in Fig. 9.2 are convex; thoseshown in Fig. 9.3 are not. Some special cases of convex sets are these:

E 2 itself is convex.Each segment [a, b] is convex.A set consisting of just one point in E 2 is convex.A circular disk in E 2

, containing all, part, or none of its boundary, isconvex.

If C consists of a triangle together with all points within the triangle, thenC is convex. (We will call such a set a triangular region.)

The empty set 0 is convex.

The last assertion above is a consequence of the definition of convexity.For if a set S is not convex, it must contain at least one pair of points a andb such that [a, b] is not wholly contained in S. The empty set contains nosuch pair of points, for it contains no points at all.

On the other hand, while it is at least intuitively clear that a triangularregion is convex, a formal proof of this is difficult and depends on a carefuldefinition of "triangle." You should assume whenever necessary that atriangular region is convex, even though we will not supply a proof of thisfact.

Exercises

9.1 Professor Aardvark told his class that a convex set was one "eachtwo points of which could see each other." What did he mean?

9.2 Let a and b be points of E 2• Under what circumstances is the set {a, b}

convex?

9.3 How would you define convexity for subsets of three-dimensionalspace E 3 ?

9.4 How would you define convexity for subsets of one-dimensionalspace E 1 ?

9.5 Prove Theorem 9.1; that is, show that if a and b are points of E 2 andp E [a, b], then

[~p]u[p,~ = [~~.

9.6 Draw three convex sets in E 2 such that any two of them have at leastone point in common, but such that there is no point common to all three ofthe sets.

9.7 Prove that each segment [a, b] is convex.

9.1 Convex sets 257

Fig.9.2 Twoconvex sets in £2.

Fig.9.3 Twononconvex sets in £2.


Fig. 9.4 Proving thatC rt D is convex if

~

C and Dare.

9.8 Draw two overlapping convex sets C and D. In your example, is theirintersection C (J D also convex?

9.9 Prove that if C and D are any two convex sets whatsoever in £2, thenC (J D must be convex.

9.10 Must the union of two convex sets be convex? Explain the reason foryour answers.

9.2 INTERSECTIONS OF CONVEX SETS

One might solve Exercise 9.9-to show that the intersection of two convexsets is convex-as follows. Let C and D be two convex sets in £2. We candispose of the simplest cases first. If C (J D is empty, then it is convex; ifC (J D contains only one point, it is convex. So we might as well suppose thatC (J D contains at least two points, say p and q. To help us proceed with theargument, we draw a picture like that shown in Fig. 9.4.

What do we know about the two points p and q? Only that each belongsto C (J D. Put another way, this means that p and q belong to C, and alsothat p and q belong to D.

But since p and q belong to C, and C is given a convex set-here is wherethat hypothesis is used-it follows by the definition of convex set that[p, q] is a subset of C. For exactly the same sorts of reasons, [p, q] is alsoa subset of D. But since [p, q] is a subset of both C and D, it follows that[p, q] is a subset of C (J D.

What has happened? Two arbitrary points p and q were chosen in C (J D,and it turned out that the segment [p, q] was consequently a subset of C (J D.This means that C (J D is convex, by definition. The above argument thusestablishes our next theorem.

9.2 Intersections of convex sets 259

Theorem 9.2 The intersection of two convex sets is convex.

By mimicking this proof, you can show without difficulty that the intersection of three convex sets must be convex. For a start, let A, B, and C beconvex sets, and let D = A n B n C. Stop here and try to prove that Dmust be convex.

If you worked Exercise 9.10, you probably saw that the union of twoconvex sets need not be convex. In fact, if p and q are two points of £2,then each of the sets {p} and {q} is convex, but their union {p, q} is not,since it does not contain all the points of the segment [p, q]. Even if twoconvex sets have points in common, their union need not be convex, a factthat you can establish by means of innumerable simple examples. However,there is one situation in which the union of convex sets is convex-thissituation will be discussed in the next group of exercises.

Though you may have thought of one proof that the intersection ofthree convex sets must be convex, there are two, one proof involving segments (probably the one you thought of), and one not; the latter is the onegiven for our next result.

Theorem 9.3 The intersection of three convex sets is convex.

Proof. Let A, B, and C be convex sets, and let D = A n B n C. Then

D = (A n B) n C,

since set-intersection obeys an associative law.But since A and B are convex, so is A n B, by the previous theorem.

Hence we have written D as the intersection of two convex sets, namelyA n Band C. Again using the last theorem, it follows that D is convex.Hence the intersection of any three convex sets must be convex.

Exercises

9.11 Use the technique of the proof of Theorem 9.3 to show that the intersection of four convex sets must be convex.

9.12 Suppose that n is a natural number at least 2, and it is known that theintersection of any collection of n convex sets is convex. Use this fact andthe technique of the proof of Theorem 9.3 to show that the intersection of anycollection of n + 1 convex sets is a convex set.

9.13 Does it follow from Theorem 9.2 and the previous exercise that theintersection of any finite collection of convex sets is convex? Explaincarefully.

9.14 Does it follow from Theorem 9.2 and Exercise 9.12 that the intersectionof infinitely many convex sets is convex? Explain your answer.


9.15 In the previous section it was mentioned that there is one specialsituation in which the union ofconvex sets must be convex. This occurs whenthe collection of convex sets forms what is called a tower. That is, thecollection ~ of sets-eonvex or not-is said to be a tower if, given any twosets A and B in the collection ~, it is true that either

AcB or B cA.

Prove that if~ is a tower of convex sets, then u ~ (the union of all setsin the collection ~) is convex. Hint: Ifp and q are two points of u~, thenpEA and q E B for some sets A and B in~. How can you use the fact that~ is a tower?

9.16 Does Theorem 9.2 hold for convex sets in E 3 ? (The notation E 3 ofcourse stands for three-dimensional Euclidean space.) Explain.

9.17 Do Theorem 9.3 and the technique of its proof remain valid forconvex sets in E 3 ? Why?

9.18 Does Exercise 9.15 hold for convex sets in E 3 ? Give your reasons.

9.19 Let ~ be a collection of sets each finite subcollection of which is atower. Does it follow that ~ itself must be a tower? Explain your answer.

9.20 If~ is a tower of sets in E 2, need there be a "largest" element of ~

that is, need ~ include a set L such that L contains all the other sets in ~?

Explain carefully.

9.3 HULLS AND KERNELS

Although we can use our method of proving Theorem 9.3 to show that theintersection of any finite collection of convex sets is convex, an alternateproof can be given that shows that the intersection of any collection of convexsets is convex. This is one of the more important results about convex sets,and we present it as our next theorem.

Theorem 9.4 If ~ is any collection of convex sets, then the intersection of allthe sets in ~-denoted by r. ~-is convex.

Proof If p and q are two points of r.~, then both p and q belong to everyset in the collection~. Since each such set is convex, the segment [p, q]also belongs to every set in~. Hence [p, q] is a subset of r.~. Thereforeby definition, r.~ is convex.

It is fortunate that such an important theorem has such an easy proof.Note also that Theorems 9.2 and 9.3 are superseded by this theorem, for theyare special cases of it. The proof seems to work even in the special caseswhere CC contains only one convex set, or no sets at all; the latter somewhatpuzzling situation will be discussed in the next set of exercises.

9.3 Hulls and kernels 261

Fig. 9.5 Formingthe convex hullof the two sets

Sand T.

s

Hul (S) IHull TI

In addition, if A is any subset of £2 whatsoever, then it is meaningfulto consider the collection ~ of all convex subsets of £2 containing A. Theintersection of all these sets is called the convex hull of A, and is abbreviatedby Hul (A). It has the following properties, which we list in the form of atheorem.

Theorem 9.5 Let A be a subset of£2. Then

a) A c Hul (A).

b) Hul (A) is convex.

c) If C is any convex set containing A, then C also contains Hul (A).

d) A is convex if and only if A = Hul (A).

The proofs of each of these facts are quite easy, and are left for theexercises. We should remark at this point that, as a consequence of thistheorem, there is a unique "smallest" convex set, namely Hul (A), containingeach subset A of £2. Some examples of hull formation are shown in Fig. 9.5.

Associated with each subset A of £2, in addition to its convex hull, isanother set called its convex kernel. The kernel can be defined as follows:

Ker (A) = {x E A I if YEA, then [x,y]cA}.


T

Ker (T)

Fig.9.6 Forming the convexkernels of the twosets Sand T.

Some examples of the formation of Ker (A) from A are shown in Fig. 9.6.One convenient way to remember the definition of Ker (A) is to say thatKer (A) is the set of all points of A that can "see" all the other points of A.And, in analogy to our theorem about the convex hull, we have our nextresult.

Theorem 9.6 Let A be a subset of E 2• Then

a) Ker (A) c A.

b) Ker (A) is convex.

c) A is convex if and only if A = Ker (A).


Again, the easy proof is left for the exercises. The analogy betweenkernels and hulls is not perfect; while there is a unique "smallest" convexset containing the given set A, the kernel need be neither the largest nor thesmallest convex subset of A. However, there is a connection between Ker (A)and the "largest" convex subsets of A, and we will make this connectionexplicit in our next theorem. But, before we even state that theorem, somepreliminaries are needed.

Let ~ be a collection of sets, and let II be a property meaningful for eachset A E~. (That is, for each set A in ~ it is either true or false that A hasproperty II.) The set M in ~ is said to be maximal with respect to propertyII if

a) M has property II, and

b) M is not a proper subset of any other set in ~ having property II.

It is important that you do not confuse the property of being maximalwith the property of being maximum. A maximum set having property IIwould be a set having property II and also containing every other set havingproperty n. To illustrate the distinction, we give an example.

Let~ be the collection of all subsets of £2. Let us say that a subset of theplane is nonlinear if no three of its points lie on a straight line. Let II be theproperty of "being nonlinear." Then II is certainly meaningful for sets in thecollection~. And ~ contains sets maximal with respect to property II; forexample, let M be a circle.

Then M is maximal with respect to being nonlinear, because

a) M itself is nonlinear, and

b) if M is a proper subset of L (where L E~) then L cannot have propertyII-for since M is a proper subset of L, L contains at least one pointx ¢ M. A straight line through x and the center of the circle M containsx and two points of M, thus three points of L, since MeL.

This example shows that a set that is maximal with respect to property IIneed not be unique, for there are numerous different circles in the plane,and each is maximal with respect to being nonlinear. However, there is nosubset of £2 that is a maximum with respect to being nonlinear, for the onlysubset of £2 containing all circles is £2 itself, and £2 is not nonlinear.

The following axiom is one form of the so-called Zermelo Axiom, whichis usually assumed true by most mathematicians. We will need this axiomto prove our next theorem.

Axiom Let ~ be a collection of sets and let II be the property of "being atower." (Then II is meaningful for each subcollection d c ~, since eachsuch subcollection either is or is not a tower.) Then ~ contains a subcollection.,I( maximal with respect to property n.


As an example of an application of this axiom, let ~ be the collection ofall circular disks in E 2 each of which contains its boundary. The aboveaxiom guarantees the existence of a maximal tower vii of such disks; thatis, vii is a collection of circular disks, each disk in vii either contains or iscontained in each other disk in vii, and if some circular disk D is eithercontained in or contains each disk in vii then DEvil.

(In this example it is not actually necessary to use the axiom to find themaximal tower vii of circular disks, since it is not difficult to show that thecollection of all disks centered at the origin is a maximal tower. Note alsothat this latter collection is maximal, but not a maximum.)

We mentioned that there was a connection between Ker (A) and theconvex subsets of A. We now proceed to demonstrate that connection, firstby proving the following lemma-which shows the existence of maximalconvex subsets of a given set A-and then immediately proceeding to thetheorem in question, which says that Ker (A) is the intersection of the maximalconvex subsets of A. The axiom just given is necessary to establish the lemma.

Lemma Let A be a subset of E 2 and let p be a point of A. Then there existsa subset C of E 2 ,such that

a) p E C;

b) C c A;

c) C is convex; and

d) C is maximal with respect to the above three properties.

Proof Let ~ be the collection of all convex subsets of A containing p.Then ~ is not the empty collection, since {p} E~. By the axiom, ~ contains amaximal tower J(. By Exercise 9.15, u vii is convex. Let C = u vii. Itis not difficult to show that C has the four desired properties listed in thestatement of the lemma.

Theorem 9.7 Let A be a subset of E 2• Let f!A be the collection ofall maximal

convex subsets of A. Then Ker (A) = n f!A.

The proof is outlined in one of the next exercises.

Exercises

9.21 This and the next three exercises provide a proof of Theorem 9.5.Let A be a subset of E 2

• Show that A c Hul (A).

9.22 Let A be a subset of E 2• Show that Hul (A) is convex.

9.23 Let A be a subset of E 2 and let C be a convex subset of E 2 such thatA c C. Prove that Hul (A) c C.

9.24 Let A be a subset of E 2• Prove that A is convex if and only if A =

Hul (A).


9.25 Let A be a subset of £2, and let A(A) be the set obtained by adjoiningto A all points belonging to segments [p, q], where p and q are points of A.For example, if A is a circle, this process produces a circular disk for A(A).Need the set A(A) formed in this way be convex? Explain.9.26 Continuing the previous exercise, show that A C A(A) and thatA(A) C Hul (A).

9.27 See the previous two exercises. Is it true that, given A c £2, theapplication of Aa finite number of times will produce Hul (A)? That is, is ittrue that one of the sets

A, A(A), A(A(A»), A(A(A(A»), . ..

must be the convex hull of A? Is there a maximum number of applications ofAthat will always suffice for the production ofHul (A)? What is this number?

9.28 Repeat the previous three exercises for subsets of £3 rather than £2.

9.29 Repeat Exercises 9.25, 9.26, and 9.27 for subsets of £1 rather than £2.

9.30 Let A be a subset of £2 and let 9" be the collection of all triangularregions whose vertices lie in A. Need Hul (A) = u 9"? Why?

9.31 Let A be a subset of £3 and let 9" be the collection of all triangularregions whose vertices lie in A. Need Hul (A) = u 9"? Explain.

9.32 Give an example of a tower CC of convex subsets of £2 such that (") CCis nonempty. Then give an example such that (") CC is empty.

9.33 In the previous section we provided an example-being nonlinearas a property meaningful for subsets of £2, and a set maximal with respectto this property. Provide a different example of such a property and, ifpossible, find a subset maximal with respect to that property..9.34 Let CC consist of all convex subsets of £2 no one of which contains theorigin (0, 0). Show that there exists a maximal convex subset of £2 notcontaining the origin. Is it possible to use the axiom of the last section?Is it necessary?

9.35 If CC is the empty collection of subsets of £2, what is (") CC? Hint: Ifp ¢ (") CC, then p must fail to belong to some set in the collection CC. Whatpoints p have this property?

9.36 IfCC is the collection of subsets of £2 consisting of just one set, say Athat is, CC = {A}-then what is (") CC?

9.37 Let A be a subset of £2. Show that the point p belongs to Hul (A)if and only ifp belongs to the convex hull of a set consisting of three or fewerpoints of A. Hint: Use Exercise 9.30 if you wish.

9.38 The last line of the proof of the Lemma of the preceding section leavesverification of the four properties of the Lemma for the reader. Please verifythem.


9.39 Prove Theorem 9.6.

9.40 Here is an outline of the proof of Theorem 9.7; please fill in the details.We have given that A is a subset of £2 and that f!A is the collection of all

maximal convex subsets of A. Let B = n f!A; we need to show thatKer (A) = B.

First, we show that Ker (A) c B. There is no problem if Ker (A) = 0(why?), so let p E Ker (A). It suffices to show that if M is a maximal convexsubset of A, then p E M (why is this sufficient?). So let M be a maximalconvex subset of A.

Since p E Ker (A), [p, q] c A for each point q E M (why?). Let Cconsist of all points belonging to all such segments [p, q], where q E M.Then C is convex (why?), C is a subset of A (why?), and M c C. Since Mis a maximal convex subset of A, M = C (why?). Since p E C (why?), itfollows that p E M. As we mentioned previously, this is sufficient to showthat p E B. Hence, since p is an arbitrary point of Ker (A), it follows thatKer (A) c B, and we are half done.

Next, there remains only the problem of showing that B c Ker (A).If B = 0, there is no problem, so let x be a point of B. To show thatx E Ker (A), it is sufficient (why?) to show that [x, y] c A for each YEA.So let y be an arbitrary point of A.

By the Lemma of the last section, y is contained in a maximal convexsubset M of A (exactly how is the Lemma applied ?). But since x E B, itfollows (why?) that x E M too. Thus [x, y] c M (why?). Hence [x, y] c A(why?). It follows that x E Ker (A), and thus that B c Ker (A).

We have shown that both

Ker (A) c B and B c Ker (A)

are true, and consequently Ker (A) = B. This establishes Theorem 9.7.

9.4 HELLV'S THEOREM

Helly's Theorem will be the principal tool we use to prove the Art GalleryTheorem. First, suppose that CC is a nonempty collection of convex subsetsof £2 such that each two sets in CC have nonempty intersection. Does itfollow that nee must be nonempty? Can we even conclude that the intersection of each three sets in CC is nonempty? Try to answer these questionsbefore proceeding.

We can answer the first question very easily. In the coordinatized plane,let the set Cn consist of all points on, or to the right of, the vertical linethrough the point (n, 0) on the x-axis, where n is allowed to assume all wholenumber values. Let CC consist of all the sets Cn' Then CC is a collection of planeconvex sets, and the intersection of each two sets in CC is clearly nonempty.However, no point belongs to n~, for in order that pEn CC, it would be

9.4

Fig.9.7 The line passingthrough (n, 0) has slope n.

Helly's theorem

(5,0)

267

necessary that the x-coordinate of p be greater than every integer, which isimpossible. So the answer to the first question above is a very emphatic"No"; indeed, any finite subcollection of CC has nonempty intersection, butstill (J CC = 0. However, this example sheds little light on the secondquestion, for in this case the intersection of each three sets from CC is indeednonempty.

However, the unbounded straight lines drawn as shown in Fig. 9.7 doform a collection of convex subsets of £2, and each pair of these linesintersect since no two of the lines are parallel. However, with the help of alittle analytic geometry one can show that no three of these lines have acommon point. So the answer to the second question raised above is alsoin the negative.

But ReIly's Theorem tells us that the answer must be affirmative if weincrease the numbers of sets mentioned in both the hypothesis and theconclusion. The above examples serve to show why Helly's Theorem may be asomewhat unexpected result.


Fig. 9.8 One case in theproof of Helly's Theorem.

ABO\\\\\\\\\\\

BCD--------------------------

ABC,I,I,,,,,,,,,,,,

ACO

Theorem 9.8 (Helly's Theorem) Suppose that Cft/ is a collection of convexsubsets of£2 such that any three sets in Cft/ have a point in common. Then foreach n > 4, each n sets in Cft/ have a point in common.

This is just one version of Helly's Theorem; there are forms of Helly'sTheorem for three- and higher-dimensional Euclidean space, and forms thatguarantee even that n Cft/ itself is nonempty. However, the version statedabove will be quite sufficient for the proof of the Art Gallery Theorem, andit is the easiest version to prove.

Proof We attack the simplest case first, the case in which the number n ofthe theorem is 4. That is, suppose that Cft/ is a collection of plane convex setseach three of which have a point in common; we desire to show that eachfour of the sets in Cft/ must also have a common point. So let A, B, C, and Dbe four sets in Cft/.

Since each three sets in Cft/ must have a point in common, there must inparticular be a point common to the three sets A, B, and C. For convenience,we will denote this point by ABC, for this notation serves to remind us,among other things, that the point ABC belongs to each of the three setsA, B, and C.

Similarly, there are points ABD, ACD, and BCD common to the otherpossible combinations of three of the four sets in question. In the mostgeneral case, these points are four distinct points in £2, and one possibilityis that they form the vertices of a convex quadrilateral, as shown in Fig. 9.8.

9.4 Helly's theorem 269

In this case, the diagonals of the quadrilateral must intersect in a pointwe have called p. In Fig. 9.8, the diagonal from ABC to BCD has its endpoints in the convex sets Band C, so this diagonal must be a subset of B n C.Similarly, the other diagonal must be a subset of AnD. Hence the point pmust belong to all four of the sets A, B, C, and D. This shows that any foursets in re must have nonempty intersection in this case, the one in which thefour points form the vertices of a convex quadrilateral. Fortunately, thereare not many other cases, and all of the others are even simpler, so that wehave left their discussion for the exercises at the end of this section. Assuming, then, that the other cases can similarly be disposed of, we have shownthat if each three sets in re have a point in common, then also each foursets in re must have a point in common.

You may well guess what comes next. Knowing of re more than we didbefore-that each four sets in re must have a common point-we proceed toshow that each five sets in re must have a point in common. By continuingthis process, we may conclude that each finite subcollection of sets from remust have nonempty intersection, since each finite value of n > 4 musteventually be reached by this method. However, the "obvious" way to showthat each five sets in re have a point in common is not the best way.

For suppose we are given five sets A, B, C, D, and E in re. What weshould not do is label the point known to lie in the four sets A, B, C, and Dby the symbol ABCD, and consider the possibilities for the five points wewould thus obtain. There are too many possibilities, and for larger values ofn the situation becomes far more complicated. Instead, we use a trick likethat in the proof of Theorem 9.3 and Exercises 9.11 and 9.12.

We consider the collection fJI consisting of the five sets A, B, C, D, and E.Since these sets come from re, we know that any three of them have a pointin common, and thus, by what we have already proved, also that any fourof them must have a point in common. Consider next the new collection

d = {A, B, C, D n E}.

In d, we have four convex sets, and each three of them have a point incommon. For the only possible combinations of intersections of three ofthem are

A n B n C,

A n B n (D n E),

A n C n (D n E),

B n C n (D n E).

Because of what we have observed about fJI, each of the above sets isnonempty. Hence each three sets in d have nonempty intersection. Sinceeach set in d is convex, it follows by what we have already proved that anyfour sets in d have nonempty intersection. There are only four sets in d,so we now know that

A n B n C n (D n E)


is nonempty. Thus the five sets A, B, C, D, and E have nonempty intersection.Since these are five sets arbitrarily chosen from Cfj, this shows that each fivesets from C(j have nonempty intersection.

You can see how we continue application of this method. For example,now that we know each five sets in Cfj have nonempty intersection, we letA, B, C, D, E, and F be six sets arbitrarily chosen from Cfj, and let

f!} = {A, B, C, D, E, F},

and

if = {A, B, C, D, E n F}.

Now f!} is a subcollection ofCfj, and so by what we have already shown,each five sets in f!} have nonempty intersection. Moreover, if is a collectionof convex sets in E 2 and each four sets in if have nonempty intersectionsome of the possibilities for intersections of four sets from if are listed below:

An B n C n D,

A n B n D n (E n F),

B n C n D n (E n F).

In every case, any intersection of four sets from if can be thought of as theintersection of five or fewer sets from fi), and hence any intersection of foursets from if is nonempty. Hence, by what we have already shown, the intersection of any five sets from if must also be nonempty. The only possibilityfor such an intersection is

A n B n C n D n (E n F),

but this is the same as the intersection of all six of the sets in f!}. So theintersection of any six sets from Cfj itself is nonempty.

Thus, given n > 4, we will eventually reach the value of the integer nafter a number of repetitions of this idea, and thus we can conclude that theintersection of any finite subcollection of sets from Cfj is nonempty. Thisestablishes our version of Helly's Theorem.

Exercises

9.41 Does the following statement describe what was actually shown in ourproof of Helly's Theorem?

Suppose that Cfj is a collection of convex sets in the plane, n is a wholenumber at least 3, and any n sets in Cfj have nonempty intersection. Then anyn + 1 sets in Cfj have nonempty intersection.

9.42 Suppose that Cfj is a collection of 5000 unbounded straight lines in theplane, any three ofwhich have one point in common. What can you conclude?

9.5 Krasnoselskii's theorem 271

9.43 Suppose that rc is a collection of 10,000 circles in the plane (here, by a"circle" we mean the boundary of a circular disk, rather than the diskitself), and any three circles in rc have at least one point in common. Whatcan you conclude?

9.44 Suppose that rc is a collection of 64 solid balls in three-dimensionalspace, each three of which have a point in common. What can you conclude?

9.45 Suppose that n is an integer at least 4, and rc is a collection of n pointsin £2 such that each three of these points lie within a circle of radius 1. Canyou show that all n points must lie within a circle of radius 1?

9.46 What do you think is the correct version of Helly's Theorem for £3?

9.47 What do you think is the correct version of Helly's Theorem for £1?

9.48 In the proof of Helly's Theorem, we began by letting rc be a collectionof convex sets in the plane each three of which are known to have a commonpoint, and we sought first to prove that each four sets in rc had a commonpoint. We chose four sets A, B, C, and D from rc, and denoted the pointcommon to A, B, and C by ABC, and so on. There were several cases for thelocation of the four point ABC, ABD, ACD, and BCD in £2, and we considered only the case in which these four points lay on the vertices of a convexquadrilateral. List the other possibilities.

9.49 Continuing the previous exercise, show how in each of the cases listedone can conclude that there is a point common to all four of the sets A, B, C,andD.

9.50 Continue one step further the argument used in the proof of Helly'sTheorem; that is, knowing that rc is a collection of convex sets in £2 eachsix of which have a common point, show that each seven sets from rc mustalso have a common point.

9.5 KRASNOSELSKII'S THEOREM

We will now prove the Art Gallery Theorem itself. Watch for the point inthe proof at which Helly's Theorem is used.

Theorem 9.9 (Krasnoselskii's Theorem) If, for each three paintings hung inan art gallery, there is a spot from which those three can be viewed simultaneously, then there is a spot in the gallery from which all the paintings canbe viewed.

Of course, this statement is a little imprecise; we can phrase the theoremas M. A. Krasnoselskii did, in the language of plane convex sets.

Theorem 9.9 Rephrased Let P be a plane polygon. Suppose that, given anythree points a, b, and c on the boundary of P, there exists a point x E P suchthat [x, a] u [x, b] u [x, c] c P. Then Ker (P) =1= 0.


p

I,,,I,

I

/'11.I

I

f" "',' ... ...... ...... ...... ......-......-..... ...... ...... ...... ...... ...... ...,

II

II

II,

I,,I

II

I,I,

II,

/.............. II .............. I

I -......... II........ II -.......... I, ...... , "

I -~_

, ......... I

............ "I ..... - I/ ~'''''I

Fig. 9.9 Construction ofthe squares in the ArtGallery proof.

From the second form of the theorem, you can see that the art gallerymust be polygonal in shape and with vertical walls. There are certain otherdifferences in the statements given above for the Art Gallery Theorem, butthese differences can be resolved by a careful study of the following proof.We prove the second form, of course.

Proof As shown in Fig. 9.9, we first give the sides of P a counterclockwiseorientation. For each side (J of P thus oriented, there is an unboundedstraight line Acontaining (J, and the orientation of (J induces a like orientationon A. This orientation of A makes it possible to distinguish the points on the"left side" of A from those on the right; just imagine yourself standing onA, like a tightrope walker, facing in the direction of the orientation; the"left side" of A consists of those points of £2 to your left.


The plane set consisting of Atogether with all points to the left of Aiscalled the closed left half-plane determined by A. In this half-plane use asegment of Ato construct a square 8 with the following properties:

a) 8 has one side on the line A;

b) 8 is a subset of the closed left half-plane determined by A; and

c) 8 is so large as to contain all the points of P lying to the left of A.

We construct such a square 8 for each side (J of P in exactly the abovefashion. We want to show that each three of these squares have a commonpoint. Let 8 1, 8 2 , and 8 3 be three such squares, and let a, b, and c be pointslying, respectively, on the corresponding sides (J1' (J2' and (J3 of the polygonP. By our hypotheses, there then exists a point x E P such that

[x, a] u [x, b] u [x, c] c P.

But x must lie on the left side of each of the corresponding three linesA 1, A2' and A3' For if x were to lie to the right of (for example) A 1, then[x, a] would also lie to the right of A1, and thus there would be points of Parbitrarily close to (J 1 but to the right of (J l' This cannot happen, for thecounterclockwise orientation of the sides of P guarantees that no points of Plie immediately to the right of any side of P.

Hence x must lie to the left of each of the lines A 1 , A 2 , and A 3 , and x E Ptoo. Since each of the three squares 8 1 , 8 2 , and 8 3 is constructed so as tocontain all points of P to the left of the lines used to construct these squares,it follows that x belongs to all three of the squares 8 b 8 2 , and 8 3 , So eachof the possible triples of squares we have constructed has nonempty intersection, and in fact, given three such squares, there is a point of P belongingto all three of them.

Each square is a convex subset of E 2 (here we understand that a "square"consists of the boundary together with its interior). So we have the followingsituation: We have finitely many convex sets in E 2-namely, the squaresand each three of these sets has nonempty intersection. By Helly's Theorem,there must be a point common to all the squares. We call this point q, andnow we wish to show that q E P.

If q were not a point of P, we could draw a straight line segment from qto some interior point r of P, and so arrange matters that [q, r] does notintersect a vertex of P. Then, as one moves along [q, r] in the direction fromq to r, one must first encounter a point of some side (J of P, as shown in Fig.9.10. Call that point t. Since [q, t] meets P only at the point t, then eachpoint of [q, t] must lie to the right of the side (J and thus to the right of theline Adrawn through (J. In particular, q itself would have to lie to the rightof A, and thus q could not be a point of the square 8 built on the line A.


\\\\

q

Fig. 9.10The situation if q ¢ P.

This is impossible, since q belongs to every square, and hence we may conclude that q is, after all, a point of P.

If we can show that q E Ker (P), this will establish the theorem, for itwill show that Ker (P) =I 0. But suppose that q ¢= Ker (P). Then thereexists some point Z E P that q cannot "see"-that is, such that the segment[q, z] does not lie entirely within P. Let y be a point of [q, z] not containedin P, as shown in Fig. 9.11.

Then, as one travels along the segment [y, z] from y to z, one first meetsthe boundary of P at some point w on a side (j of P. Hence [q, w] lies to theright of the side (j, as there are points of [q, w] arbitrarily close to (j but not inP. But then, q must lie to the right of the straight line Athrough (j, and henceq cannot belong to the square S built on A. This is in contradiction to thefact that q lies in each of the squares, and this contradiction establishesthat q E Ker (P).

Therefore Ker (P) =I 0, and this establishes Krasnoselskii's Theorem.Actually, Krasnoselskii's Theorem is true for any plane figure bounded

by a closed curve, and the method of proof is quite similar. However, thisversion of the theorem requires the use of a form of Helly's Theorem for thecase of the intersection of infinitely many convex sets.


-----------~~~~

Fig. 9.11 Showingthat q E Ker(P).

-----x--------

Exercises

9.51 The proof of the Art Gallery Theorem is so long that it is difficult tounderstand without a summary. Try summarizing the proof of the theorem;make your summary as condensed as possible, omitting most reasons. Forexample, you might begin like this:

"First orient the sides of P in a counterclockwise direction, then drawan unbounded straight line through each side. On each such line, constructa square such that ... "

9.52 How could you phrase the Art Gallery Theorem for a three-dimensionalgallery, with pictures hung in all sorts of directions from an observer?Should the word "three' in the statement of the theorem then be replaced bythe word "four"?

9.53 Is it possible that although 8 is a nonempty subset of£2, Ker (8) = 0?Give your reasons.

9.54 An unbounded straight line A in £2 divides £2 into three sets: A itself,the points of £2 on one side of A, and the points of £2 on the other side of A(the latter two sets are to contain no points of A itself). The last two sets arecalled the open half-planes determined by A.


Prove that if 8 is a subset of E 2, then Hul (8) is the intersection of all

open half-planes containing 8. (Remember that the intersection of an emptycollection of sets is all of E 2

.)

9.55 Is it possible to divide E 2 into two disjoint convex sets whose unionis E 2 ?

9.56 Let f(j be a finite collection of rectangles (including boundary andinterior) in E 2 each of which has sides parallel to the coordinate axes. Theredoes exist a natural number k such that, in such a situation, if each krectangles from f(j have a point in common, then there must be a pointcommon to all the rectangles in f(j. What is the least value of k for whichthis is so? Hint: It is clear that the answer is either k = 2 or k = 3-butwhy is this clear?

9.57 In Section 9.4, there was shown in Fig. 9.7 a collection of unboundedstraight lines in E 2 such that any two intersected, and no three had a point incommon. The figure shows lines passing through the points (n, 0) on thex-axis-where n is a natural number-and the slope of the line through thepoint (n, 0) was to be n itself. If you know some analytic geometry, thiswill help in constructing a proof that no three of the lines have a point incommon. If not, see if you can find an alternative proof of this fact.

9.58 See Exercise 9.46. Give an example of four convex sets in E 3 such thateach three have a point in common but such that there is no point common toall four.

9.59 This is a version of Helly's Theorem for E 2 in which the word "three"can be replaced by the word "two," but proving the following version is noteasy. Try it anyway.

Let f(j be a finite collection of convex sets in E 2 such that each two sets inf(j have a point in common. Then, for each point p E E 2

, there exists anunbounded straight line A. passing through p and through all the sets in C(/.

9.60 Here is another difficult problem, in which Helly's Theorem can be usedto give the solution.

Suppose that the set 8 consists of n points in E 2, where n is a natural

number. Then there exists a point p in E 2 such that, for each unboundedstraight line A. passing through p, at least n/3 of the points of 8 lie in each ofthe closed half-planes determined by A..

9.6 i-CONVEXITY

One frequently fruitful approach to mathematics is the generalization ofprevious results. Sometimes such generalizations actually simplify the theoryby removing extraneous details, and they sometimes show connections between branches of mathematics that were thought to be unrelated. One

9.6 L -convexity 277

possible generalization of the idea of convexity is L-convexity. We stillrestrict our attention to subsets of £2.

Let us say that the subset K of £2 is L-convex provided that, for eachtwo points x and y of K, there exists a point z of K such that [x, z] u[z, y] c K.

The name "L-convex" comes, of course, from the idea that each twopoints of K can be joined by a vaguely L-shaped figure lying entirely within K.Every convex set is L-convex, but the converse is easily shown to be false.So we do have here a generalization of the idea of convexity, and a numberof our previous theorems still hold after appropriate modifications havebeen made. For example, even without modification, one can prove thefollowing two theorems.

Theorem 9.10 The union ofa tower of L-convex sets is L-convex.

Theorem 9.11 Let K be a subset of£2 and let p be a point of K. Then thereis an L-convex subset of K maximal with respect to containing the point p.

The L-convex kernel of a set has a very natural definition: IfK is a subsetof £2, then the point x of K is said to belong to the L-convex kernel of Kprovided that, for each y E K, there exists Z E K such that [x, z] u [z, y] c

K. We abbreviate the L-convex kernel of K by L-Ker (K).At this point, the professional mathematician would raise questions

such as the following:

Need Ker (K) c L-Ker (K)?

Need L-Ker (K) be L-convex?

Need L-Ker (K) equal the intersection of the maximal L-convex subsetsof K?

Unfortunately, it is not so easy to define an L-convex hull for the set K.We would want the L-hull to be L-convex, by analogy with properties of theconvex hull. But the "obvious" approach of forming the L-hull by intersecting all L-convex sets containing K does not work-as you will see if youlet K consist of three sides of a square in the plane. The real difficulty seemsto be that the intersection of L-convex sets need not be L-convex.

On the other hand, there are some constructions with L-convex setsthat do not work for convex sets. Given a subset S of £2 and a point p E £2,we can form the join of p and S, denoted by p # S, as follows:

p # S = u rEp, s] I s E S}.

It would seem plausible that if S is any subset of £2, then p # S wouldbe convex. Unfortunately, this is not the case. However, it is true that p # Sis L-convex. We leave further development of these ideas to the exercises.


Exercises

9.61 Give an example of an L-convex subset of £2 which is not convex.

9.62 Prove that every convex set in £2 is L-convex.

9.63 Prove Theorem 9.10: That the union of a tower of L-convex sets isL-convex.

9.64 Prove Theorem 9.11: That if K is a subset of £2 and p is a point of K,then there exists an L-convex subset of K maximal with respect to containing p.

9.65 Is it true that for every subset K of £2, Ker (K) c: L-Ker (K)?

9.66 Is it true that for every subset K of £2, Ker (K) = L-Ker (K)?

9.67 Suppose that K is a polygonal region in £2. Can you show thatL-Ker (K) must be L-convex? (So far as the author knows, this problem isunsolved.)

9.68 Suppose that K is an arbitrary subset of £2. Need it be true thatL-Ker (K) is L-convex? (Hint: See the previous exercise.) Explain.

9.69 Suppose that K is a subset of £2. Need L-Ker (K) be equal to theintersection of the maximal L-convex subsets of K? Give your reasons.

9.70 Need L-Ker (K) contain the intersection of the maximal L-convexsubsets of K for every subset K of £2? Why?

9.71 Let K consist of three sides of a square in £2. Show that K is notL-convex, and that the intersection of all L-convex sets containing K isin fact equal to K.

9.72 Give an example of two L-convex sets in £2 whose intersection is notL-convex.

9.73 Give an example of a subset S of £2 and a point P E £2 such thatP # S is not convex.

9.74 Prove that if S is a subset of £2 and p is a point of £2, then p # Sis L-convex.

9.75 Show that if C is a circle in the plane, then the closed circular diskwith boundary C is not a minimal L-convex set containing C.

9.76 Formulate an alternative generalization of the idea of convexity in theplane.

9.77 Is the complement of a circular disk (containing its boundary) in £2L-convex?

9.78 Let P be a polygonal region in £2. Suppose that each two pointsx and y of the boundary of P can be joined by the segment [x, y] with[x, y] c: P. Does it follow that P must be convex?


9.79 Let P be a polygonal region in E 2• Suppose that for each two points

x and y on the boundary of P, there exists a point z of P such that [x, z] u[z, y] c P. Does it follow that P must be L-convex?

9.80 Is it true that the subset S ofE 2 is L-convex if and only ifKer (S) :I: 0?


The books by Hadwiger, Debrunner, and Klee and by Yaglom andBoltyanskii listed below are collections of problems on convexity andrelated topics, with discussion material liberally interspersed, and with thelevel of the material not excessively high (with certain exceptions). Thebooks by Valentine and Griinbaum are rather advanced. The first studyof L-convexity known to the author can be found in the paper, "Someproperties of L sets in the plane," by Alfred Horn and F. A. Valentine,published in Volume 16 (1949) of the Duke Mathematical Journal.

Benson, R., Euclidean Geometry and Convexity (McGraw-Hill, 1966).

Griinbaum, B., Convex Polytopes (Interscience, 1967).

Hadwiger, H., Dubrunner, H., and Klee, V., Combinatorial Geometry inthe Plane (Holt, Rinehart, and Winston, 1964).

Lyusternik, L., Convex Figures and Polyhedra, translated by T. JeffersonSmith (Dover Publications, 1963).

Valentine, F., Convex Sets (McGraw-Hill, 1964).

Yaglom, I. and Boltyanskii, V., Convex Figures, translated by Kelly andWalton (Holt, Rinehart, and Winston, 1961).

CHAPTER 10

THEREALNUMBERSYSTEM

We shall begin with the assumption that you already clearly understand thesystem (Q, " +, <) of rational numbers with ordinary multiplication andaddition and the usual order relation. We shall examine four main topics:

1) the inadequacies of the rational number system,

2) remedying such inadequacies by construction of the real number system,

3) the significance of decimal expansions for real numbers, and

4) some unusual and important properties of the real number system.

10.1 THE RATIONAL NUMBERS

The system of rational numbers is quite adequate, of course, for solvingmany kinds of equations, such as

2x + 3 = 7 - 5x,

or even systems of simultaneous equations in more than one unknown;for example,

5x + 3y - 2z = 0,

2x = 3z,

yJ2 + 9x = 1001.

Moreover, it does appear that every "quantity" encountered can, in manycircumstances, be approximated numerically to within any desired degree

280

10.1 The rational numbers 281

Fig.10.1 The number.J2 measures a length.

of accuracy by rational numbers; we have in mind "quantities" such aslength, weight, volume, velocity, and even the famous rational approximation 22/7 for 1t.

However, 1t is not equal to 22/7, and such a simple equation as

x 2 = 2

cannot be solved using only rational numbers. But this equation certainlyought to have an exact solution, for the positive "number" x such that x 2 = 2can easily be visualized as the exact length of the hypotenuse of an isoscelesright triangle of leg length 1, as in Fig. 10.1. We see as a result of our firsttheorem that this length cannot be a rational number.

Theorem 10.1 No rational number is a solution of the equation

x 2 = 2.

Proof We will use Theorem 7.9, the Fundamental Theorem of Arithmetic,to give the simplest (but by no means the only) proof.

Suppose by way of contradiction that there were a rational number rsuch that r 2 = 2. Whether or not r is positive, since it is rational there must(by definition of "rational number") exist integers m and n with n :1= 0 suchthat

mr =-.

n

282

Hence

The real number system 10.1

But,2 = 2, and so

and hence

m2

2=2 'n

Now m and n are integers, but here we are concerned only with theirsquares, so we may suppose that both m and n are positive. By the Fundamental Theorem of Arithmetic, each of m and n has a unique primefactorization, and so do m2 and n2

• The prime factorization of m has somenumber-possibly zero-of 2's in it, and hence m2 has one factorization inwhich twice as many 2's appear. That is, m2 has a factorization with an evennumber of 2's. Since this factorization is unique, we see that the primefactorization of m 2 must contain an even number of 2's.

Similarly, the prime factorization of n2 contains an even number of 2's.So one factorization of 2n2 contains an odd number of 2's; again, this primefacto~ization is unique, so the prime factorization of 2n2 must contain anodd number of 2's.

But we have the equation

2n2 = m 2•

We have seen that if the term 2n2 is factored into primes an odd number of2's must appear in that factorization, while if m2 is factored into primesthis factorization must contain an even number of 2's. Hence the naturalnumber with the two names above-the two names 2n2 and m2-has twodifferent prime factorizations, one with an even number of 2's and one withan odd number of 2's. This is in contradiction to the Fundamental Theoremof Arithmetic, since that theorem guarantees a unique prime factorizationof each natural number. This contradiction means that our original supposition-that the equation

has a rational solution-must be false, and hence no rational number cansolve the above equation. In other words, the square root of 2 is irrational,our term for a real number (whatever that is) that is not rational.

Since there then exist lengths which are not rational numbers-or, ifyou prefer, there are polynomials such as p(x) = x 2

- 2 which have norational zeroes-it thus seems reasonable that there do exist numbers inaddition to the rational numbers, numbers yet to be discovered. (If you prefer,


1

you may say that such numbers have yet to be constructed. It depends onwhether you think a mathematician is an explorer or an inventor.)

An alternative formulation of the above situation is this: If S is the setof all positive rational numbers with squares no less than 2; that is,

S = {r E Q+ I r 2 > 2},

(where Q+ denotes the set of all positive rationals) then the set S containsno smallest element. The method of showing this will be outlined in one ofthe exercises at the end of this section.

But you have undoubtedly approximated .J2 by a sequence of numberssuch as

1.4, 1.41, 1.414, 1.4142, 1.41421, ....

And you have likely carried out the above decimal approximations sufficientlyfar to obtain the necessary accuracy in the context of the problem you areworking. It is not difficult to find what such a sequence is tending to, ifthere is some regularity or periodicity in the terms of the sequence. Forexample, the sequence

0.3, 0.33, 0.333, 0.3333, 0.33333, ...

can easily be seen to be tending toward the number whose decimal expansionIS

0.333 333 333 ... ,

and the "value" of this decimal can be calculated using some elementaryknowledge about geometric series. The above nonterminating decimal isactually just an abbreviation for

3 3 310 + 100 + 1000 +"',

which is a geometric series with ratio 1/10 (since each term is 1/10 theprevious) and first term 3/10. If the ratio is between -1, and 1, the sum ofsuch a series is given by the formula

a

1 - r

where a is the first term of the series and r is its ratio; in the case of the seriesfor the decimal 0.333 333 333 ... , we find its value to be

3

10 1

1 310

284 The real number system

Since no periodicity in the decimal expansion

1.414213 562 ...

10.1

is apparent, such techniqlles cannot be applied, so the question naturallyarises at this point as to what the sequence

1.4, 1.41, 1.414, 1.4142, 1.41421, ...

has as its "value" toward which it is tending; that is, what is the "limit"of the above sequence of rational numbers. Thus, the construction of thereal numbers can be thought of as giving a meaning to every possible decimalexpansion, and interpreting such a decimal expansion as measuring somelength.

Our next theorem may give some insight into the structure of the rationalnumber system; it almost says that the rational numbers are sufficient forapproximation of any quantity to any desired degree of accuracy.

Theorem 10.2 Let rand s be rational numbers with r < s. Then there areinfinitely many rational numbers between rand s.

Proof Choose integers m and n such that 0 < m < n. Then

m0<-<1.

n

Since r < s, s - r is positive. Also s - r is rational, as you will be askedto establish in the exercises, and so we have

mo < - (s - r) < s - r,n

and hencem

r < r + - (s - r) < s.n

Since min and s - r are rational, so is their product, and so is their sum

mr+-(s-r).

n

Again, the details are left for the exercises. But there are infinitely manychoices of integers m and n such that 0 < m < n, and m and n can be chosenso as to give infinitely many different values of min, and thus infinitely manydifferent values of

mr + - (s - r),

n

all of which lie between rand s. This establishes the theorem.

10.1

-(3/2)

/i\/' I \

.f7'f /' : "VL./, I \

" " \" I'" I \

,," I'

7/4

The rational numbers

3

285

By virtue of this theorem, it would appear that if the rational numberswere indicated as points on an unbounded straight line, located according totheir values as indicated in Fig. 10.2, there could be no gaps in this line.

But we have seen that .J2 is not a rational number, but could be located onthis line by a method such as that shown in Fig. 10.2. So the rational numbersystem does, after all, contain gaps, which we intend to fill with numbers we.will eventually call the irrational real numbers.

Exercises

10.1 Let a, b, C, and d be rational numbers. Show that the equation

ax + b = cx + d

has either no solution or a rational solution.

10.2 Let a and b be rational numbers. Show that the numbers

a + b a - b a·b

are rational, and that alb is rational if b i= O.

10.3 The number 1t represents a length. What length? It also represents anarea. What area?

10.4 The value of 1t accurate to 30 decimal places is

1t = 3.1415926535 89793 2384626433 83279.

Can this information be used to show that 1t is not a rational number? How?

286 The real number system 10.1

10.5 Can the decimal expansion of n be used to show that

n =1= 22 ?7

10.6 Give a better rational approximation to n than 2217.

10.7 For those who have studied Chapter 3. Use continued fractions and theinformation in Exercise 10.4 to find the "next better" rational approximationto n.

10.8 Show that ,J3 is irrational; that is, that the equation x 2 = 3 has norational solution x. Hint: Use the technique of Theorem 10.1.

10.9 Give the reasons for each step in the following alternative proof that

.J2 is not rational.

Suppose that ,J2 is rational. Then 1 < ,J2 < 2 (why?). So there existnatural numbers m and n, with n =1= 1, such that

(why?) and the fraction min is in lowest terms. Hence

m2

2 =-2 'n

and so

Hence m2 is even (why?), so m is even (why?). So m = 2" where' is anatural number (why?). So

2n2 = 4,2

(why?), thus

(why?). So n2 is even (why?), and so n is even (why?). This is a contradiction

(to what?). Therefore,J2 is not rational.

10.10 For those who have studied Chapter 3. The continued fraction

expansion of ,J2 is given by

.J2 = (1; 2, 2, 2, 2, . . .),

as seen in Section 3.3. How can this fact be used to prove that,J2 is irrational?

10.11 If the techniques ofTheorem 10.1 are used in a (vain) attempt to prove

.J4 is not rational, no contradiction is reached. Why not?

10.12 Prove that if p is prime, then ,Jp is not rational.

10.13 Exactly which natural numbers n have the property that ,J~ is notrational?

10.14 Here is an outline of how to show that the set

S = {r E Q+ I r 2 > 2}

contains no smallest element. Fill in the details.First, if rES then r 2 > 2 (why?). Suppose, then, that r is the smallest

element of S. Let

r2- 2

8=r----2r

Then 0 < s (why?), and S2 > 2 (why?). Hence S E S, since s is rational(why?). But s < r (why?). This is a contradiction (to what?), and hence Scontains no smallest element (why?).

10.15 Express

0.888 888 888 ...

as a rational number; that is, in the form min, where m and n are integersand n =1= O.

10.16 Express

0.327 327 327 ...

as a rational number.

10.17 Express

0.999 999 999 ...

as a rational number.

10.18 Prove that ,J6 is not a rational number.

10.19 Prove that,J2 + ,J3 is not a rational number. Hint: Begin much asin the proof of Theorem 10.1. Then use Exercise 10.18.

10.20 Prove that,J2 + ,J3 is not a solution of any equation of the form

ax2 + bx + c = 0,

where a, b, and c are rational numbers. Hint: There is no loss of generalityin supposing that a = I-why not?

This exercise together with the previous one shows that there are irrationalnumbers not solutions to any quadratic equation with rational coefficients.

10.21 Prove that no rational number is a solution of the equation

x3 = 2.

10.22 Prove that the equation

has no rational solution x.

10.23 Prove that if ex is not a rational number and r is a rational number,then ex + r is not a rational number.

10.24 Prove that if ex is not a rational number and r is a rational numberother than 0, then ex . r is not a rational number.

10.25 If a, b, and c are integers with a =1= 0, and the equation

ax2 + bx + c = 0

has two rational solutions, what can be said about the relationship betweenthe numbers a, b, and c?

10.2 NESTED INTERVALS OF RATIONAL NUMBERS

Let a and b be rational numbers with a < b. By the interval [a, b] we meanthe set

[a, b] = {x E Q I a ~ x ~ b}.

That is, [a, b] consists of all rational numbers between a and b, includinga and b. As we have seen in Theorem 10.2, each such set contains infinitelymany rational numbers.

If [a, b] is given, then we denote its length by A([a, b]), which is given bythe formula

A([a, b]) = b - a.

Thus if I is an interval of rational numbers, then A(I) is a positive rationalnumber, and if J c I and J is an interval of rational numbers, then

A(J) ~ A(I).

In fact, if J is a proper subset of I, then A(J) < A(I), a fact whose proofis outlined in the next set of exercises.

We shall be concerned with sequences of such intervals of the form

where each interval contains the one immediately after it In the abovesequence, and where the sequence of numbers

is approaching zero. We need a precise definition of this last concept,preceded only by a definition for convenience.

10.2 Nested intervals of rational numbers 289

If r is a rational number, then we denote by Ir I the absolute value of r,and Ir I has the value

\rl = r

Irl = -r

if

if

r > 0,

r < 0.

Thus if r is rational, Ir I is nonnegative, and just measures the distancefrom r to 0, when r is thought of as located in its natural position on anunbounded straight line.

Let

be a sequence of rational numbers, one for each natural number n. Thesequence {sn} is said to approach 0, or have limit 0, provided that given anypositive number s, no matter how small, there exists a natural number ksuch that, for all natural numbers n > k,

ISnl < s.

This definition does not mean merely that as n increases, the numbers Sn

get closer and closer to zero.

Example 10.1 For each n, let Sn = 1 + (l/n). Thus we have

2, 3/2, 4/3, 5/4, 6/5, 7/6, ...

as indicated in Fig. 10.3. As n increases, the numbers Sn are indeed gettingcloser and closer to zero. But the sequence {sn} is not approaching zero.To see this, let s = 1/4. No matter what value of k is chosen, if n > k then

1ISnl - 1 + n

> 1,

and hence ISnl is not less than 1/4 = s. Hence it does not have zero for alimit, although this sequence is getting closer and closer to zero.

The definition of the limit of a sequence also does not mean that the termsof the sequence must get steadily closer and closer to the limit, but onlythat they tend to get closer "on the average." For consider the next example:

Example 10.2 For each n, let

1S =n

n

1S =-

n 2n

if n is odd,

if n IS even.


o

3/2 2

Fig. 10.3 The sequence gets closer to 0 but does not have limit O.

1/16

/I I I I I

0

/1/5 1/4 1/3 1

1/7

Fig. 10.4 The sequence has limit 0 though it does not steadily get closer to O.

10.2

Thus we have the sequence

1 ! !, 4' 3'

Nested intervals of rational numbers

111 116' 5' 64' ;;'

291

as indicated in Fig. 10.4. The terms of this sequence do not steadily getcloser and closer to zero; however, the sequence does have limit zero. Hereis one proof of this fact; note how we follow exactly the pattern of thedefinition.

Let e > 0 be given. Since to this point, only rational numbers "exist,"e itself is rational, and thus has the form

ae = -,

b

where a and b are natural numbers. Let k = b + 1. Suppose that n > k.For n odd,

1 1ISnl = - --n n

and for n even,1 1= - <-.2n - n

Hence, if n > k, then

That is, given e > 0, we have shown the existence of a natural number ksuch that for each n > k,

ISnl < e.

Therefore, by definition, the given sequence {sn} has limit zero.An alternative way of putting the definition of the sequence {sn} having

limit zero is this: No matter how small an interval of the form (- e, e) ischosen around zero, from some point on all the terms of the sequence {sn}lie within that interval. This phenomenon is illustrated in Fig. 10.5.

We shall, in essence, construct the real numbers by locating their positionusing sequences of closed intervals of rational numbers. A typical sequencefor locating .J2 would be

11 = [1, 2],12 = [1.4, 1.5],

13 = [1.41, 1.42],

14 = [1.414, 1.415],

Is = [1.4141, 1.4142],


( I I I ""11111 I I ) I

o

Fig.10.5 All terms of the sequence from 524 on lie in the interval ( -e, e).

21.4 1.5

1.41

1.414 1.415

1.42

Fig. 10.6 Locating.j2 with a sequence of closed intervals of rational numbers.

10.2 Nested intervals of rational numbers 293

These intervals are shown in Fig. 10.6. The behavior of these intervals issuch that only the number ,J"2 can belong to all of them. However, since

,J"2 has not yet been "constructed," we cannot speak of the one numberbelonging to all of these intervals; indeed, in our context of the rationalnumbers and only the rational numbers, there is no number which belongsto all the above intervals. Hence what we will do is simply say that the number

,J"2 is the sequence {In}' This is a subtle and ingenious idea, and the way itsucceeds is aesthetically very pleasing.

What is necessary to make this idea succeed is the behavior of the abovesequence of intervals. The significant characteristics we need are these givenbelow, which you can check in the above example.

a) For each n, In+ 1 is a proper subset of In.

b) The sequence {A(In)} has limit zero.

We abstract such behavior to obtain our next definition.A nested sequence of closed intervals of rational numbers is a sequence

of closed intervals of rational numbers such that

a) for each n, In+ 1 is a proper subset of In; and

b) the sequence {A(In)} has limit zero.

Here is an outline of the construction of the real number system; thedetails will be forthcoming.

First, we define what it means for two sequences of nested interv.als to bethe same. (This should not be surprising, that two different-looking suchsequences could be considered the same; after all, the rational number 1/2 hasmany different symbols, such as 2/4, 3/6, 4/8, .... In the same way, differentlooking sequences of nested intervals will be considered as different namesfor the same real number.) We then define a real number to be a sequence ofnested intervals of rational numbers. (Technically, here, a real number willbe the set of all "same" sequences of nested intervals of rational numbers.)We show that every rational number is, in this sense, a real number, so thatQ c R, where R denotes the set of all real numbers.

Next, we define addition, multiplication, and the order relation for R,and show that these are the same as those already extant in Q. We alsoshow that all the familiar axioms concerning these operations hold true in R.

Finally, we show that when the elements of R are thought of as pointson an unbounded straight line, located according to their value, then thereare no gaps in this line-in particular,,J"2 E R. And we show that if we repeatthe above process, using sequences of nested intervals of real numbers, nonew numbers are obtained.

Exercises

10.26 If r is a rational number and {rn} is a sequence of rational numbers,how would you define the statement "The sequence {rn} has limit r."?

10.27 Let {sn} be the sequence

1, - (1/2), 1/3, - (1/4), 1/5, ....

Prove that {sn} has limit zero.

10.28 Suppose that I and J are closed intervals of rational numbers, and Jis a proper subset of I. Fill in the details of the following outline of a proofthat A(J) < A(I).

Let I = [a, b] and J = [c, dJ. Since J c I, a ::::: c and d ~ b (why?).Suppose that a = c. Then we must have d A(I), weobtain a contradiction (how, and to what?). Therefore A(J) < A(I).

10.29 Remove as many absolute value signs as possible from each of theexpressions below, without changing the value of each expression.

a) 1191c) 101e) I-xl

10.30 Prove that the sequence

b) 1-71d) Ixlf) Ilxll

1, 1/4, 1/9, 1/16, 1/25, 1/36, ...has limit zero.

10.31 What is the limit of the sequence in Example 10.1? Use your answerto Exercise 10.26 to prove that your answer to this exercise is correct.

10.32 What is the limit of the sequence

1, 0, 1, 0, 1, 0, 1, 0, ... ?

Prove that your answer is correct.

10.33 Let r be a rational number such that Ir I < 1. Prove that the sequence

234r, r ,r ,r , ...

has limit zero.

10.34 Let a and r be rational numbers and n a natural number. Show that

a(1 - r n+ 1)a + ar + ar 2 + ar 3 + ... + ar n = .

1 - r

10.3 Construction of the real numbers 295

10.35 If the sum of the infinite geometric series

a + ar + ar 2 + ar 3 + ...is defined to be the limit of the sequence

a, a + ar, a + ar + ar 2, a + ar + ar 2 + ar 3,

a + ar + ar 2 + ar 3 + ar 4, ... ,

prove that the sum of the infinite series above is then given by the formula

a ,1 - r

if Ir I < 1. Hint: Use the two previous exercises.Now evaluate the sum of the infinite series

1/2 + 1/4 + 1/8 + 1/16 + 1/32 + .. '.

10.3 CONSTRUCTION OF THE REAL NUMBERS

Let' and" each be a sequence of nested intervals of rational numbers. Then

, = {II' 12 , 13 , ••• },

and" = {JI , J2 , J3 , •.. },

where each In and each I n is a closed interval of rational numbers. We saythat' and" are equivalent, and write' = ", provided that, for every combination of natural numbers m and n, there is a point common to 1m and I n•

In other words, each interval in , overlaps each interval in ". This isjust a way of saying that the two sequences' and" are "zeroing in" on thesame point. This point will be the eventual location of the real numberwhich, by virtue of the above definition, , and" are two names for. For ifsome interval 1m is disjoint from some interval Jm then all the intervals after1m in , will be disjoint from all intervals after I n in ", and there will be apositive distance between the point "defined" by , and the point "defined"by".

As an illustration of the above definition, suppose that

In = [1 - ~, 1+ ~}and

I n = [1 - ~, 1]for each natural number n. If' = {In} and" = {In}, then' = ,,; the reasonis that every interval In as well as every interval I n contains the number 1,

r s

~[1m ] [ Kn }

r -,~------------------~L Jr .J

Fig. 10.7 A contradiction is reached because all the intervals J t have length atleast s - t.

and hence for every combination of natural numbers m and n, 1m overlapsI n in at least the number 1.

Incidentally, we also have in the above example an illustration of the wayin which a rational number can be thought of as a real number; namely,the number 1 is the only number common to all the intervals In and the onlynumber common to all the intervals I n• So in this case both' and 1] are newnames for the familiar rational number 1.

We should now show that this "equality" (which we have temporarilycalled an equivalence in the above definition) has the properties that equalityought to have.

Theorem 10.3 Let " 1], and e be sequences of nested intervals of rationalnumbers. Then

a) ,="b) if ,= 1]

c) if ,= 1]

then

and

1] = ,1] = e then

Proof Parts (a) and (b) are obvious. To establish part (c), suppose by way ofcontradiction that' i= e. Then some interval 1m in , is disjoint from someinterval K n in e. Since 1m and K n are intervals, we can suppose that eachnumber in 1m is less than each number in K m as indicated in Fig. 10.7. Inparticular, the right-hand endpoint r or 1m is less than the left-hand endpoints of Kn•

Since , = 1] and 1] = (), it is not hard to see that each interval J t inthe sequence 1] must contain both rand s, since each interval J t must intersectboth 1m and Kn• Since r < s, s - r is a positive number, and hence

J...(Jt) > s - r

for all natural numbers t, as the interval [r, sJ is a subset of J t for each naturalnumber t. Hence the sequence {J...(Jtn cannot have limit zero. This contradicts the fact that 1] is a sequence of nested intervals of rational numbers,

as by definition the corresponding sequence of lengths must have limit zero.This contradiction shows that' = (), and establishes the theorem.

We define a real number to be the collection of all equivalent sequencesof nested intervals of rational numbers. This justifies the use of the symbolof equality in the definition of equivalent sequences, for then two realnumbers are equal if and only if they are represented by equivalent sequencesof nested intervals. We will use the symbol R to stand for the set of all realnumbers.

In order to define addition on the set R, we need first to define additionfor sequences of nested intervals of rational numbers. Let' and '1 be twosuch sequences. Then

and

'1 = {J1 , J2 , J 3 , ••• }.

We define, + '1 to be the sequence

{I1 + J 1 , 12 + J2 , 13 + J3 , ••• },

where, for each natural number n,

and

Thus in order to "add" two real numbers, we choose any sequences ofnested intervals representing those two real numbers; we add these sequencesby adding corresponding intervals, and intervals are added by adding eachnumber in one to each number in the other. This makes sense, since in thelatter case we are just adding rational numbers together, an operation alreadytaken for granted as defined.

There is no difficulty in calculating the result when two closed intervalsof rational numbers are added in the above way. For example, if

and

are two such intervals, then

I = [2,3J

J = [4,7J

I + J = {a + b I a E 1 and bE J},

and this must be the interval [6, 10]. For if a E I and b E J, then

2 ~ a < 3,and

4 ~ b ~ 7,and hence

6 ~ a + b ~ 10.

Moreover, if C E [6, 10], then

6 =:;; C < 10.

In order to show that there exist numbers a E I and b E J such that c =a + b, we must deal with cases. If 6 < c ~ 7, we let

a=c-4and

b = c - a.

Then a and b are rational numbers, a + b = c, and moreover, since

6 < c < 7,then

2<c-4=a<3so that a E I. And

b = c - a = c - (c - 4) = 4

so that b E J. The other cases are handled similarly. Thus, indeed, [2, 3] +[4, 7] = [6, 10]. In general, one can prove the following theorem:

Theorem 10.4 If I and J are closed intervals of rational numbers, then so is1+ J.

But curiously enough, our definition of the sum , + 1'/ of two realnumbers needs a justification theorem before it becomes a valid definition.The reason is that in order to add the real numbers' and 1'/, one selects justone of many possible sequences of nested intervals to represent , and justone of many possible sequences of nested intervals to represent 1'/. There areother choices; we may have' represented by both

{Ii' 12 , 13 , ... }

and{Xl' X 2 , X 3 , ••• },

and 1'/ too may be represented by both

{Jh J2 , J3 , ..• }

and{Yl , Y2 , Y3,· •• }.

Although {In} and {Xn} are equivalent, they need not be identical;similar remarks hold for {In } and {Yn}. Thus you should not expect the twosequences

and


to be identical; the problem is, however, that they may not even be equivalent.Since they are both supposed to determine the same sum' + 1'/, the twosequence sums above should be equivalent, or else there is an unacceptableambiguity in our definition of real number addition.

This problem can best be illustrated by an attempt to define a methodof combining rational numbers other than addition or multiplication.Suppose, for each two rational numbers rand s, we define r # s as follows:

Represent each of rand s as quotients of whole numbers (with nonzerodenominator)-thus

mr =-

nand

as =

b '

where m, n, a, and b are integers, and neither n nor b is zero. Then r # sis to have the value

m + a

n + b

This fails to be a "valid" operation, in that the definition of r # s isambiguous: If r = 1/2 and s = 2/5, then

r # s = 0/2) # (2/5) = 3/7.

But there are alternative representations of rand s as fractions-forexample, we could write r as 2/4 and s as 6/15. Then

r # s = (2/4) # (6/15) = 8/19.

Since 3/7 =I 8/19, we see that the result of combining rational numberswith the operation # gives a result dependent on the numeral used to representeach rational number, rather than on the actual number itself. This ambiguityis just what we need to show cannot happen in the case of our definition ofaddition for real numbers. To do so, it suffices to establish the next theorem.

Theorem 10.5 Let' and 1'/ be real numbers, and let' be represented by the twosequences of nested intervals {In} and {Xn} and let 1'/ be represented by the twosequences of nested intervals {In } and {Yn}. Then the two sequences of nestedintervals {In + I n} and {Xn + Yn} are equivalent, and hence give rise to thesame real number' + 1'/.

Proof. Suppose that we have the conditions as given in the hypotheses tothis theorem. And also suppose, by way of contradiction, that the sequence{In + I n} represents the real number y, the sequence {Xn + Yn} representsthe real number (j, and that y =I (j.

Because of the last condition, the two sequences {In + I n} and {Xn + Yn}are not equivalent, so that there must be some interval of the form I k + Jk


disjoint from some interval of the form Xm + Ym• We suppose without lossof generality that m > k, and then, since 1m + 1m C Ik + I k, it followsthat 1m + 1m and X m + Ym are also disjoint.

But there does exist at least one rational number a belonging to both1m and Xm, since {In} and {Xn} are equivalent sequences of nested intervals.Similarly, there does exist a rational number b belonging to both 1m and Ym•

Hence the rational number a + bbelongs to both 1m + 1m and to X m + Ym•

This contradiction establishes the theorem.Of course, we have not yet established that if each of {In} and {In} is a

nested sequence ofclosed intervals of rational numbers, then so is the sequence{In + In}; but the proof of this is very straightforward and is outlined in thenext set of exercises.

For convenience, we next establish that each rational number "is" areal number; that is, that each rational number can be thought of as acollection of equivalent sequences of nested intervals of rational numbers.

Theorem 10.6 Let r be a rational number. Then r is a real number.

Proof For each natural number n, let

In = [r - (lIn), r + (lin)].

Then, clearly, {In} is a nested sequence ofclosed intervals of rational numbers.It is easy to see that if {In} is another such sequence, then the latter isequivalent to the sequence {In} if and only if every interval In contains thenumber r. Moreover, if so, then r is the only number common to all theintervals in the sequence {In} because of the condition that the sequence{A(In)} has limit zero. Hence the rational number r is represented by asequence of nested intervals of the necessary sort, and consequently is a realnumber.

We have two ways of adding rational numbers. If rand s are rational,we can add them with our already given method of rational number addition,or we can think of rand s as real numbers, and add them using the sequencesof nested intervals. The two results turn out to be the same, and so our newdefinition of real number addition turns out to be the same as the old rationalnumber addition when both methods apply to a pair of numbers. Also,addition is commutative and associative, the rational number 0 (when thoughtof as a real number) is the identity for this operation, and each real number'has an additive inverse, which we of course denote by -,. The details of theseassertions can be found in the exercises.

An exactly similar procedure can be used to develop the idea of amultiplication for real numbers, and exactly analogous results follow:multiplication is commutative and associative and distributes over addition;the rational number 1 is the multiplicative identity; each nonzero real numberhas a multiplicative inverse. Since the multiplication defined for R is the

same as that given on Q when both can be applied, we have thus constructeda natural extension of the rational number system (Q, ., +) to the realnumber system (R, ., +); and we can think of the former as a subsystemof the latter, for not only is it true that Q c R, but the operations are thesame on Q in either case.

There remains only the question of how the order relation on Q is to beextended to R; we take up this topic in the next section.

Exercises

10.36 To prove Theorem 10.3, it is necessary to know that if {sn} is asequence of positive rational numbers each larger than the fixed positiverational number a, then {sn} cannot have limit zero. Please prove this.Hint: Let 8 = a/2.

10.37 Supply the details of the proof that if I and J each are closed intervalsof rational numbers, then so is I + J. Hint: If I = [a, b] and J = [c, d],what are the end points of the interval 1+ J? What if some of the numbersinvolved are negative?

10.38 Suppose that each of {In} and {In} is a sequence of nested intervals ofrational numbers. By the previous exercise, the sequence {In + I n} is indeeda sequence ofclosed intervals of rational numbers-but is it a nested sequence?Fill in the details of the following outline of a proof that {In + I n} is nested.

First, for each natural number n,

In + 1 + I n+ 1 C In + I n

Second, for each natural number n,

(Why?).

(Why?).

It follows that for each natural number n, not only is I n + 1 + I n + 1 aproper subset of In + Jm but also that the sequence {A(In + I n)} has limitzero. (Why, to both.)

Therefore {In + I n} is a nested sequence of closed intervals of rationalnumbers.

10.39 Given that 1 < .J2 < 2, how would you construct a sequence ofnested intervals of rational numbers representing the real (and irrational)

number .J2-without using decimals?

10.40 Prove that if ( and Yf are real numbers, then ( + Yf = Yf + (.Hint: Let ( be represented by the nested sequence {In} and Yf by the

nested sequence {In}. Show that In + I n = I n + In for each natural numbern. It follows (why?) that ( + Yf = Yf + (.10.41 Prove that if " Yf, and 0 are real numbers, then ( + (Yf + 0) =(( + Yf) + O. Hint: Do this much like the previous exercise.

10.42 Prove that the real number 0 is an additive identity; that is, thato + ( = ( for each real number (.

10.43 Suppose that the real number ( is represented by the sequence {In} ofnested intervals of rational numbers. In terms of this sequence, what is asequence representing - (? Show that - ( + ( = 0 for all real numbers (.

10.44 Show that if ( is a real number such that some representation of (by a sequence {In} contains an interval Ik of positive rational numbers only,then there exists a representation of ( by a sequence {In} in which everyinterval contains only positive rational numbers. Show that a similar resultholds in the "negative" case.

10.45 Show how to define the product of two real numbers ( and 11 interms of products of sequences of nested closed intervals of rational numbers.

10.46 Prove that if ( is a real number, then o· ( = 0 and 1 . ( = (.

10.47 Show that multiplication of real numbers is both commutative andassociative.

10.48 Show that multiplication of real numbers distributes over addition;that is, for any three real numbers (, 11, and 0,

( . (11 + 0) = (. 11 + (. o.10.49 Show that each real number other than zero has a multiplicativeinverse. Hint: Use Exercise 10.44.

10.50 Show that the multiplication for real numbers defined by yourselfin Exercise 10.45 coincides with ordinary rational number multiplicationwhen both can be applied to two numbers.

10.4 THE ORDER RELATION ON R

Let ( and 11 be real numbers, with representations {In} and {In} respectivelyby sequences of nested closed intervals of rational numbers. If there existsa rational number r and a natural number n such that each number in Inis less than r and each number in In exceeds r, then we say that ( < 11.

Theorem 10.7 If a and b are rational numbers and a < b in the order relationalready given on Q, then a < b in the order relation on R defined above, andconversely.

Theorem 10.8 If ( and 11 are real numbers, then exactly one of the threerelations

is true.( < 11 11 < (

Theorem 10.9 If (, 11, and 0 are real numbers such that ( < 11 and 11 < 0,then ( < o.

10.4 The order relation on R 303

Note: The notation , < '7 means that either' < '7 or , = '7 is true. (Byvirtue of Theorem 10.8, at most one can be true.)

Theorem 10.10 If' is a real number and 0 < " then there exists a naturalnumber n such that

1- < ,.n

Theorem 10.11 (Archimedean Property of R) If e is a real number witho < e, and y is a real number with 0 < y, then no matter how small e is andno matter how large y is, there exists a natural number n such that

y < n·e.

Let S be a nonempty set of real numbers. The number b is said to be anupper bound for S provided that x ~ b for all XES. If c is an upper boundfor S such that c ~ b whenever b is also an upper bound for S, then c iscalled a least upper bound for S.

The next theorem really says that the real number line contains no"gaps."

Theorem 10.12 (Least Upper Bound Theorem) If S is a nonempty set ofreal numbers with an upper bound, then S has a least upper bound.

The proofs of the above six theorems are outlined in the next set ofexercises. Using Theorem 10.12, we can prove that ,J2 is a real number;that is, that there exists a real number y such that y2 = 2. Here is analternative method.

The method is to construct two sequences of positive rational numbersof the form

and

such that, for every natural number n, (an)2 < 2, (bn)2 > 2, an < an+ h

bn+ I < bn, and such that the sequence {Ibn - anI} is approaching zero.This sequence looks much like that illustrated in Fig. 10.8, arranged so that

a l < a2 < a3 < ... < ,J2 < ... < b3 < b2 < bi

in actuality, although we have not yet shown the existence of,Ji Since{Ibn - anI} has limit zero, the sequence of intervals {In} = {[am bn]} will bea nested sequence of closed intervals of rational numbers, thus representingsome real number y. It is then natural to expect that y2 = 2, and this willbe proved.


Fig.10.8 Illustrating the proof that.j"2 is a real number.

The first problem is the construction of the two sequences {an} and{bn }. We begin by letting a i = 1 and bi = 2. Then we average:

This average is a positive rational number whose square is either less than 2,or greater than 2. If less, we call it a2; if greater, we call it b2. In this case,it turns out to be b2 • Now that we have ai' bi , and b2 , we average the last anconstructed with the last bn constructed, and obtain

Again this average is a positive rational number; if its square is less than2, we call it a2; if greater, we call it b3.

We continue this process. In general, if an and bm are the last numbersconstructed in each sequence, we form the average

and call this number an+ 1 if its square is less than 2; we call it bm + 1 if itssquare is greater than 2.

10.4 The order relation on R 305

It is not hard to see that {an} and {bn} are sequences of positive rationalnumbers so arranged that

and so that, for each natural number n,

The only real problem is in showing that there are actually infinitelymany a's and infinitely many b's. But suppose, for example, that there werebut finitely many b's. Then there must be infinitely many a's, and we wouldhave

where bk is the last bn constructible by this process.At each stage of the construction, the distance between the last two

a's and b's constructed is half what it was at the previous stage; for example,we have

Ib l - all - 1,

Ibz - all - 1/2,

Ibz - azl - 1/4,

and so on. Hence the sequence {bk - an} is approaching zero (n is the variablein this sequence; k is fixed). But then, the only rational number belongingto all the intervals of the form [am bkJ is bk itself. However, by Exercise10.14, bk cannot be the smallest rational number whose square exceeds 2.So there is a smaller rational number, say c, whose square exceeds 2.

By construction, the square of each an is less than 2, and the square ofbk exceeds 2. Hence the number c must belong to each interval of the form[an, bkJ. This is a contradiction, since the only number in all such intervalsis bk itself and c "# bk • The case in which there are but finitely many a's ishandled similarly, and again, with this contradiction, we can conclude thatthere must be infinitely many terms in both the sequences {an} and {bn}.

Consequently, {[an, bnJ} is a sequence of nested closed intervals ofrational numbers, and thus represents some real number which we denoteby y. The object now is to prove that yZ = 2.

But this is quite easy, for the sequence of nested intervals

not only represents the real number l, but also has the property-again byconstruction-that it represents the number 2; for (an)Z < 2 < (bn)Z for

each natural number n. Therefore yZ = 2, and hence .J"2 exists.

Exercises

10.51 Part of the proof of Theorem 10.7 is presented below; please supplythe details.

Suppose that a and b are rational numbers and a < b in the orderrelation on Q. First construct a rational number r such that a < r < b.Represent a by the sequence {An} and b by the sequence {Bn} of nestedintervals of rational numbers. Show that, for some sufficiently large valueof n, each number in An is less than r and each number in Bnexceeds r.

It may be helpful to do this last step by contradiction. Remember that{A(A n)} and {A(Bn)} are sequences approaching zero.

The conclusion is then that a < b in the order relation on R.

10.52 The other part of the proof of Theorem 10.7 is presented below;please supply the details.

Suppose that a and b are rational numbers such that a < b in the orderrelation on R. Show why there must exist a rational number r such thata < rand r < b in the order on Q. Conclude that a < b in the order on Q.10.53 An outline of the proof of Theorem 10.8 is presented below; pleasesupply the details.

Suppose first that , < '1 and , = '1 are both true. Represent' by asequence of nested intervals {In} and '1 by a sequence of nested intervals{In}. Since' = '1, each In must interse~t each Jm.

But since' < '1, there exists a natural number n and a rational number rsuch that each number in In is less than r and each number in I n is greaterthan r. Why does this lead to a contradiction?

By similar treatment of other cases, conclude that at most one of thethree relations

can be true.Suppose that neither of the last two relations above is true, and show

that the first one must be true; since' =1= '1, there must be a natural number nso large that In and I nare disjoint. Show how to construct a rational number rsuch that each number in In is less than r and each number in I n exceeds r.This shows that' < '1, and thus establishes Theorem 10.8.

10.54 Prove Theorem 10.9. Hint: Use techniques similar to those in theabove exercise, and use the fact that Theorem 10.9 does hold for rationalnumbers.

10.55 Prove Theorem 10.10. Hint: Use Exercise 10.44.

10.56 Prove Theorem 10.11. Hint: Use Theorem 10.10.

10.57 Prove Theorem 10.12. Hint: Let S be a nonempty set of real numberswith an upper bound. Let

A = {x E R I x > s for all S E S}

10.5

and

Are there more numbers 7

B = {x E R I x <s for some S E S}.

307

Then A and Bare nonempty, A u B = R, and S c B. Moreover,each number in B is less than each number in A. Follow the technique usedafter the statement of Theorem 10.12, in proving that .J2 exists, to constructtwo sequences, one drawn from A and one from B. Construct a sequence ofnested intervals using the terms of these sequences for endpoints. If yourconstruction is much like that in showing that .J2 exists, you should obtain asequence of nested intervals representing a real number' which can be shownto be the least upper bound of B. Then, with a little care, , can also be shownto be the least upper bound of S.

10.58 From this point on, we consider intervals to be intervals of realnumbers. That is,

[a, b] = {x E R I a < x < b},

[a, b) = {x E R I a < x < b},

(a, b] = {x E R I a < x < b},and

(a, b) = {x E R I a < x < b}.

Give three different upper bounds for the set [1, 2).

10.59 Prove that if a set of real numbers has a least upper bound, it hasonly one.

10.60 What is the least upper bound of the set [1, 2)1 What is the leastupper bound of the set (1, 2] 1

10.5 ARE THERE MORE NUMBERS?

First let us try to answer the question above by showing how every lengthcan be represented by a real number. Such a length can be represented as alength measured from 0 to some point on the unbounded straight line onwhich the rational numbers can be thought of as already located accordingto their values. If for some reason the length should be thought of as negative(perhaps as representing a velocity, charge, or other signed physical quantity)we shall nevertheless suppose that it has been measured off in the positivedirection-for if we can show that some real number' measures the positivelength, then -, will measure the negative length.

So, in essence, the problem is to show that each point to the right of 0on the line is the location of some real number' already constructed. Let P

be such a point. There is certainly at least one rational number r to the rightof P and -1 is a rational number to the left of P. Hence the set

s = {x E Q I x is to the left of P}

is a nonempty set of real numbers with an upper bound. Let' be the leastupper bound of S.If' is to the left of P, then by the construction of' there must exist someclosed interval I of rational numbers with , E I and P to the right of eachnumber in I. In particular, the right-hand endpoint b of I is a rational numberto the left of P. So b E S. But' < b; this is a contradiction, since' is theleast upper bound of S.

Similarly, if' is to the right of P, a contradiction is obtained. Hence thepoint P is the location of the real number'. This establishes that each pointon the line is the location of some real number'; moreover, this is the naturallocation of' because this point is to the right of every rational number lessthan , and to the left of every rational number greater than'.

We now show how each real number can be located in a natural positionon the unbounded straight line of rational numbers. This will be easiest tosee with an example; we show how to locate the number

TC = 3.14159 26535 ....

We will not actually use this decimal expansion in order to locate the positionof TC, but only to make it clear what intervals of rational numbers are to beconstructed.

Given TC, let n be the greatest integer such that n ~ TC, and let 11

En, n + 1]. In the case of TC, we obtain the interval II = [3, 4].Next, let n1 be the greatest integer such that

nn + _1 < TC,

10 -

and let 12 be the interval

[n +!!..! n + n 1 + 1].

10' 10

In this case, using TC, we obtain 12 = [3.1, 3.2]'We continue this process, obtaining the sequence

[3, 4], [3.1, 3.2], [3.14, 3.15], [3.141, 3.142], ....

The above process will work without the necessity of knowing the decimalexpansion of TC in advance; in fact, this process is just what we use to definethe decimal expansion of each real number; the decimal expansion is just the

10.5 Are there more numbers? 309

limit of the sequence of left-hand endpoints of the above intervals (with aminor exception to be dealt with in the exercises).

In any case, we think of the closed intervals we have just obtained asclosed intervals of rational numbers. Since each has length 1/10 that of theprevious interval and each is contained in the previous interval, we have asequence of nested intervals of rational numbers, which "is" the real numbern. The location of n is then the one point on the number line lying in all theabove intervals; it is clear that there can be at most one such point, and itcan be shown using some techniques of topology that at least one such pointmust exist. This is a natural location for n, since we locate n to the right ofevery rational number less than n and to the left of every rational numbergreater than n.

This construction provides us with a bonus, as we have noted. We havedeveloped a decimal expansion for each real number, and the rules of arithmetic by which these decimals can be added, multiplied, and so on, can bedeveloped as well.

This construction partly answers the question of the title of this section.There are not more real numbers; that is, since we have established a one-toone correspondence between R and all the points on an unbounded straightline, any physical quantity which can be interpreted as a length can bemeasured exactly by one and only one real number. On the other hand,even using real numbers the simple equation

x 2 + 1 = 0

has no solution x E R; this problem is discussed in Exercise 10.65.

Exercises

10.61 Suppose that Cis a real number in the interval [n, n + I] where n isan integer. Prove that there exists a nonnegative integer m such that

[m m + 1]CE n+-,n+ .10 10

What is the maximum possible value of m?

10.62 Suppose in the construction of a sequence of nested intervals ofrational numbers, as we did for the number n, the number Cfor which thesequence is constructed lies at the right-hand end of each interval. Forexample, suppose that Cis the number 1. Then

1 E [0, 1],1 E [0.9, 1],1 E [0.99, 1],1 E [0.999, 1],

Has the above construction been carried out in the same way as was indicatedfor the number n?

Since 1 lies in each of the above intervals, it would seem reasonable that adecimal expansion for 1 might be given by 0.999 999. . .. Is this correct?See Exercise 10.17.

10.63 What real numbers have two different decimal expansions? Hint:See Exercise 10.62.

10.64 A question of some theoretical interest is this: If we were to repeatthe construction of this chapter, beginning with R rather than with Q,would any new' numbers be obtained? The answer is that none would, andthe reason is the truth of Theorem 10.12, the Least Upper Bound Theorem.

Let {In} be a nested sequence of closed intervals of real numbers. Eachinterval In is of the form [am bn], where an and bn are real numbers and an <bn. Let' be the least upper bound of the set {an I n EN}, and show that'is the only real number that belongs to all the intervals In. Since in thisalternative development, , is represented by the sequence {In}, this sequenceproduces only a number that is already a real number. Fill in the details ofthis argument.

10.65 In order to construct a solution to the equation

x2 + 1 = 0,

a procedure can be used much simpler than the construction of R. Let

bE R}.andc = {(a, b) I a E R

For (a, b) and (e, d) in C, let

(a, b) + (e, d) = (a + e, b + d)and

(a, b) . (e, d) = (ae - bd, ad + be).

Show that addition and multiplication are closed, commutative, andassociative; that (0, 0) is an additive identity and that (1, 0) is a multiplicativeidentity; and that (a, b) has an additive inverse and, if (a, b) :f;: (0, 0), then(a, b) has a multiplicative inverse.

Then show that if the real number a is identified with the number(a, 0) E C, the operations of addition and multiplication with respect to Care the same as in R.

This shows that C, the complex number system, is a natural extensionof the real number system, and that R can be thought of as a subsystem of C.

Finally, show that the equation

x2 + 1 = 0has a solution in C.

10.6 An unusual set of real numbers 311

10.6 AN UNUSUAL SET OF REAL NUMBERS

Ifwe used only the digits 0, 1, and 2 for counting, we would be counting in theso-called ternary system, as shown below:

Ternary Numeral

o12

101112202122

100101102110

Decimal Numeral

o123456789

101112

A fraction written with the numeral 1/4 in the decimal system would thenbe written 1/11 in the ternary system, and so on. The development of thereal number system from the rationals could be carried out exactly as before,and only the "decimal" expansions would look any different. The ternary"decimal" for the above fraction could be computed by division:

0·02020202···11)1·00000000· ..

22100

22100

22100

221

As in the case of ordinary decimal representations of real numbers, someternary "decimals" may differ yet represent the same real number. Forexample,

0.022222 ...

can be "evaluated" using the formula for the sum of a geometric series(Exercise 10.35), as follows:

0.022222 ... = 0 + 0/3 + 2/9 + 2/27 + 2/81 + ....This is a geometric series with "first" term 2/9 and ratio 1/3, hence its sum is

2-

9 1

11 - 3'

- -3

or, in ternary notation, 0.022222 . . . can be written as

1

10or 0.100000 ...

Here is an example of a very unusual set of real numbers, known as theCantor Ternary Set. Let K be those real numbers in the interval [0, 1] notrequiring use of the digit 1 in their ternary expansion.

Thus the number 1/3 belongs to K, since although 1/3 does have a ternaryexpanSIOn

0.100000 ...

in which the digit 1 is used, it also has a ternary expansion

0.022222 ...

in which the digit 1 is not required.If 1/3 < x < 2/3, x does require the use of the digit 1 in its ternary

expansion, since the ternary expansion of such a number must use the digit 1in the first place after the decimal. So K c [0, 1/3] u [2/3, 1].

Also, if 1/9 < x < 2/9 or if 7/9 < x < 8/9, it is necessary to use thedigit 1 in the second place after the decimal point in representing x as aternary decimal, and hence

K c [0, 1/9] u [2/9, 1/3] u [2/3, 7/9] u [8/9, 1].

If this process of elimination is continued, it can be seen that K is that subsetof [0, 1] that remains after the middle third (except for endpoints) of [0, 1]is deleted, then the middle third of each of the two resulting intervals isdeleted, then the middle third of each of the resulting four intervals is deleted,and so on. This process is shown in Fig. 10.9.

In this deletion process, once a point becomes an "endpoint" of K, itmust remain in K in spite of all subsequent deletions; thus, for example, Kcontains the infinite set

{I, 1/3, 1/9, 1/27, 1/81, ... }.

10.6 An unusual set of real numbers 313

0[-------------]1

0[-----]1/3 2/3[-----]1

[ ] [ ] [ ] [ ]

E-3E-3•••

E-3B•••

Fig. 10.9 First four stages in the construction of the Cantor Ternary Set.

Clearly, each endpoint of K has denominator a power of 3; however, Kcontains other points as well, such as 1/4, since 1/4 has the ternary decimal

0.0202020 ....

Now let us calculate the "length" of the set K. This should be 1 - Jl,where Jl is the total length of all the deleted intervals. But then

1 2 4 8Jl=-+-+-+-+ ....3 9 27 81

This series is geometric, with first term 1/3 and ratio 2/3, so its sum givesus the value of Jl:

1-3

1.Jl= -1

2- -

3

Since the length of K is I - fl, K has length zero.


Let f be a function defined on K and real-valued, operating accordingto the following rule:

Given x in K, express x in ternary decimal form without using the digit 1.Replace each 2 in this ternary expansion by the digit 1. Interpret the resultingnumeral as the binary decimal numeral for a real number r. Thenf(x) = r.

For example, given 1/4 E K, we proceed as follows tofindf(1/4).The ternary decimal 0.020202... represents 1/4. Convert this to

0.010101 .... The latter is the binary decimal expansion for

o 1 0 1-+-+-+-+ ....2 4 8 16

This series is geometric, and its sum is 1/3. Hence f(1/4) = 1/3.For another example, given 1 E K, we findf(1) as follows.The ternary decimal 0.22222... represents 1. Convert this to

0.11111 .... Sum the series 1/2 + 1/4 + 1/8 + 1/16 + .. '. The sum is 1.Hence f(1) = 1.

Now each number in [0, 1] is a value of the function f For, givenZ E [0, 1], z has a binary decimal representation; each digit in this decimalcan be doubled, obtaining a numeral which can be thought of as the ternarydecimal of some real number. This ternary decimal contains no 1's, and hasthe form O. ????? ... , hence it represents a number x E K. It should be clearthat f(x) = z. Thus every number in [0, 1] is a value off

Since f is a function, it cannot have more values than the number ofelements of K; but since K c [0, 1], K cannot contain more numbers thanare in the set [0, 1]. Hence K and [0, 1] contain the same number of points.But K has length zero. This is what is unusual about the set K.

Exercises

10.66 How do you count in the binary system, using only the digits 0 and 1?

10.67 Give the binary and ternary decimals for the numbers 1/2,2/9, and 3/7.

10.68 Show that the Cantor Ternary Set contains infinitely many pointsnot endpoints.

10.69 Evaluate f(1/3) and f(2/3) for the function f constructed in thissection.

10.70 Suppose instead of the middle third, the middle fifth is deleted from[0, 1], the middle fifth next deleted from each of the two resulting intervals,and so on. What is the length of the resulting set?


W. Rudin's Principles of Mathematical Analysis (second edition, McGrawHill, 1964) and M. J. Mansfield's Intermediate Real Analysis (Prindle, Weber,


and Schmidt, 1969) develop the real numbers from the rationals usmgDedekind cuts.

R. L. Wilder's Introduction to the Foundations of Mathematics (Wiley,1952) gives, in addition, the development of the integers from the naturalnumbers, then the development of the rational numbers from the integers.B. Kripke's Introduction to Analysis (Freeman, 1968) gives some furthertopics in the study of the real number system, and contains a valuable firstchapter on the approach to the study of mathematics.

c. Goffman's Real Functions (Rinehart, 1953), D. A. Sprecher's Elementsof Real Analysis (Academic Press, 1970), and R. R. Stoll's Set Theory andLogic (Freeman, 1963) develop the real numbers from the rationals usingequivalent convergent sequences of rational numbers.

EPILOGUE

Mathematics can be thought of as being divided into several branches. Thebranches are listed below, and we have indicated which chapters of this bookfall into each branch. In addition, the listing gives supplementary readingson related topics. Some references duplicate those previously given at theends of the chapters. The level of difficulty of these books is quite variable,but most of them would be appropriate for students who have the equivalentof an undergraduate major in mathematics, while some of the books aresuitable for college freshmen.

Algebra. A very special kind of modern algebra is commonly taught inhigh school. Chapter 4 (on group theory) and some of the material inChapter 2 belong in this category. Birkhoff and MacLane's A Survey ofModern Algebra (revised edition, Macmillan, 1953) covers many of the topicscommonly thought of as "modern algebra," and has been used as a juniorlevel textbook.

Number Theory. Perhaps only geometry antedates this very old branchof mathematics. Of course, it is closely allied to algebra, but since about 1900techniques of analysis have been very fruitful in producing advances innumber theory. Niven and Zuckerman's An Introduction to the Theory ofNumbers (Wiley, 1960) is frequently used as a junior- or senior-level textbook. Beiler's Recreations in the Theory ofNumbers (Dover, 1964) is writtenfor the layman with some familiarity with elementary mathematics, and is adelightful book. Chapter 7 belongs in this category.

Analysis. Freshman calculus, calculus of several variables, and differentialequations form the backbone of the mathematical education of studentsmajoring in the physical sciences. These topics form the beginnings ofanalysis, which together with its daughter, applied mathematics, haveproduced most of the visible effects of mathematics in our culture. (Forexample, almost all the mathematical problems involved in the flight plansof space explorations belong in this category.) The use of continued fractionsin Chapter 3 is an example of an application of a topic in analysis; of course,the differential equations of Chapter 8 are solved using techniques of analysis.The material on convex sets in Chapter 9 is really geometry, but convex setshave found their widest applications in analysis; and of course, Chapter 10might best be described as an introduction to the foundations of analysis.

316

Epilogue 317

If you wish to study more mathematics of this sort, H. S. Wall's CreativeMathematics (Texas, 1963) is an unusual book-a bright freshman with agreat deal of persistence can learn a great deal of calculus on his own withthe aid of this book. Bers' Calculus (Holt, Rinehart, and Winston, 1969)is one excellent recent text, as is Spivak's Calculus (Benjamin, 1967).

Geometry. Chapter 1 on the Bolyai-Gerwin Theorem, Chapter 5 onpolyhedra, and Chapter 9 on convex sets are all highly geometric in content.It is apparent that there is a great deal more to modern geometry than Euclid'sElements. Some interesting references are Hadwiger, Debrunner, and Klee'sCombinatorial Geometry in the Plane (Holt, Rinehart, and Winston, 1964),Coxeter's Introduction to Geometry (Wiley, 1961), and Hilbert and CohnVossen's Geometry and the Imagination (Chelsea, 1952).

Logic and Foundations. Chapter 6, about infinite sets, deals with partof set theory and the latter is usually included as a part of foundations.Many people would classify the material in Chapter 10 as belonging infoundations rather than in analysis. Stoll's Set Theory and Logic (Freeman,1961) can be used as a senior-level textbook.

Topology. Some of the material in Chapter 5 belongs in this branch ofmathematics, but only the second chapter, on Brunnian links, is reallymostly topological. Hocking and Young's Topology (Addison-Wesley,1961) and Alexandroff's Elementary Concepts of Topology (Dover, 1960)are introductory but not elementary.

Applied Mathematics. Perhaps probability and statistics belong in thiscategory; perhaps they should receive separate listings. In any case, Chapter 8on animal populations is an example of an application of mathematics. Soare the topics treated in most physics books. Somewhere in the categoryof applied mathematics belong such new branches of mathematics as gametheory, queueing theory, and others, each of which is quite likely to havea profound effect on our lives and cultures, perhaps in the very near future.With respect to game theory, Williams' The Compleat Strategyst (McGrawHill, 1954) is written for the layman, and is a very entertaining book.

Two references dealing with problems in mathematics, mostly unsolved,are given below:

Dorrie, H., 100 Great Problems of Elementary Mathematics (Dover,1965, translated by David Antin).

Tietze, H., Famous Problems of Mathematics (Graylock, 1965).

That mathematics has far-reaching and surprisingly diverse applicationscan be seen merely by examination of the two titles below:

Jakobson, R., Structure of Language and its Mathematical Aspects(Proceedings of the Twelfth Symposium in Applied Mathematics, AmericanMathematical Society, 1961).

Lomont, J. S., Applications of Finite Groups (Academic Press, 1959).

318 Epilogue

Finally, here is a list of more general books, some with intent similarto this one, some not even intended as textbooks, but all appropriate for theeducated layman:

Aleksandrov, A. D., Kolmogorov, A. N., and Lavrent'ev, M. A.,Mathematics: Its Content, Methods, and Meaning (M.I.T. Press, 1963,translated by Gould, Bartha, and Hirsch).

Beck, A., Bleicher, M., and Crowe, D., Excursions into Mathematics(Worth, 1969).

Crowdis, D. G. and Wheeler, B. W., Introduction to MathematicalIdeas (McGraw-Hill, 1969).

Fraleigh, J.B., Mainstreams of Mathematics (Addison-Wesley, 1969).Polya, G., Mathematics and Plausible Reasoning (Princeton, 1954).Stein, S., Mathematics: The Man-Made Universe (second edition,

Freeman, 1969).Wilder, R. L., Evolution of Mathematical Concepts (Wiley, 1968).Wilder, R. L., Introduction to the Foundations of Mathematics (Wiley,

1952).Of course, many fine books have been omitted from the above listing;

but the bibliographies that appear in many of those listed will serve as aguide to additional reading.

Epilogue 319

GREEK ALPHABET

A ex Alpha al'f~ patB p Beta ba't~, be't~ ~ aboutr ')' Gamma gam'~ payA b Delta del't~ petE 8 Epsilon ep's~-lon' potZ , Zeta za't~, ze't~ thinH " Eta a't~ e't~ pie,e () Theta tha't~, the't~ toeI I Iota i-6't~ bootK K Kappa kap'~ pitA A. Lambda lam'd~ outM f.l Mu myoo, moo pawN Nu - - cutv noo, nyoo..... e Xi zi, si.......0 0 Omicron om'~-kron' o'm~-kron',n 1t Pi piP P Rho ro1: q Sigma sig'm~

T 't Tau tou, toy v Upsilon up's~-lon'

<f) 4J Phi fiX X Chi, Khi ki'II t/J Psi psi, sin OJ Omega o-meg'~, o-me'g~, 6-ma'g~

ANSWERSANDHINTS

CHAPTER 1

1.1 The only polygon is (d).

1.2 Suppose that a polygon had fewer than three vertices, and reach acontradiction.

1.3 It can be done.

1.4 Yes. Choose a point p on one edge but not a vertex. A sufficientlysmall circular disk centered at p will be bisected by this edge, and part of thisdisk must lie within the polygon. But a semicircular region has positive area.

1.5 See a plane geometry textbook for a formula concerning the sum of theinterior angles of a polygon.

1.6 See Section 1.8 for one approach.

1.7 In either case, since only finitely many parallelograms may be used,there must be one with a vertex coinciding with a vertex of the triangle andwith one of the two sides incident at that vertex lying on one side of thetriangle.

1.8 See Fig. 1.2.

1.9 For each integer n > 0, let Rn be the rectangle with vertices at the planecoordinates

(n,O) (n + 1, 0)

and let P be the union of all these rectangles. Since Rn has area 2-n, the totalarea of Pis

1 + 1/2 + 1/4 + 1/8 + ... = 2.

1.10 Try a figure with a square boundary.

320

Answers and hints 321

1.11 Make R into two squares with one cut. Cut each resulting squarealong a diagonal. Reassemble.

1.12 Cut the strip into squares each of which has side length the same as thewidth of the strip. Reassemble the strip

I 2 3 4 5 6 ...

in the pattern

1 3 5 7 .2 4 6 8 .

1.13 Use the same squares as in Exercise 1.12. Start at the origin and work"circularly" outward.

1.14 Drop a perpendicular from the largest angle to the opposite side.

1.15 It can be done; you should really try to find the solution on your own.

1.18 One reasonable analogue to "polygon" in the one-dimensional casemight be a figure which can be expressed as the union of finitely many linesegments, each of which contains both its end points.

1.19 Test your theorem on the two line segments Sand T, both of length 1,where S consists of all numbers x such that 0 ~ x < 1 and T consists of allnumbers y such that 2 ~ y < 3.

1.20 The relation is an equivalence relation.

1.21 There should be n - 3 cuts, resulting in n - 2 triangles.

1.22 First rephrase the induction principle as follows:If a statement meaningful for natural numbers is true for n = 3, and

whenever it is true for all integers k with 3 ~ k < n then it is also true of n,then the statement is true for all natural numbers n > 3.

1.23 If you have assumed that

1 + 2 + 3 + ... + k _ k(k + 1)2

then

1 + 2 + 3 + ... + k + (k + 1) = k(k + 1) + (k + 1).2

1.24 Suppose that if a polygon has k vertices and 3 =:;; k < n, then thepolygon can be triangulated. Let P be a polygon with n vertices. Use theproof in Section 1.3 to divide P into two polygons each of which has fewerthan n vertices. Apply the above assumption, and then the answer to Exercise1.22.

322 Answers and hints

1.25 It cannot always be done. Try polyhedra of the sort that have a holerunning all the way through. This is not an easy problem.

1.26 You may wish to use the fact that if b < c and a > 0, then ab < ac.

1.27 This is easy; you may wish to split the proof into two cases.

1.28 If 4n- 1 is divisible by 3, then there exists a natural number k such

that 4n- 1 = 3k. Hence 4n is always of the form 3k + 1. What do you

do to 4n to turn it into 4n + I?

1.29 This is not only a difficult question, it is also a trick question. In hisbook Mathematics: The Man-Made Universe (second edition, Freeman, 1969),in Theorem 3 of Chapter 8, Sherman Stein shows that-for example-the

rectangle with sides of length 1 and ,J2 can not be cut up into finitely manysquares in any way whatsoever!

1.30 Try the rectangle of sides of length 1 and ,J2, and suppose that, by wayof contradiction, it can be cut into squares of side length a > O.

1.31 A 1800 rotation of one line does not change the fact that it is parallelto another line.

1.32 A right triangle, if cut the proper way.

1.33 There are really only two choices for which perpendicular to use.Show that at least one choice must produce a segment that lies within theparallelogram.

1.34 A motion without rotation will not change the fact that two lines areparallel.

1.35 The answer is best couched in terms of the ratio of the length of thealtitude of the parallelogram to the length of its base.

1.36 In Fig. 1.4, the lines from q to the end points of the diameter are thehypotenuses of two similar right triangles, whose sides are thus in proportion.

1.37 A little experience with inequalities of real numbers will be helpfulhere.

1.38 This is a complicated but straightforward problem involvinginequalities.

1.39 Note that no rotations are used, and that x 2 = abo

1.40 First show that the four right triangles in the "corners" of square Aare c?ng(uent to each other.

1.41 Again, this is a long but not difficult problem.


1.42 In Fig. 1.6, let the square S have side length c. Then show that a2 +b2 = c2

•

1.43 Read the summary immediately preceding this exercise.

1.44 Take a = 1 and b = 5 in the proof given in Section 1.6.

1.45 First, (4/3)3 > 2, so the numbers

(4/3)3, (4/3)6, (4/3)9, (4/3)12, ...

are, respectively, larger than

2, 4, 8, 16, ....

For the question about area, note that at each stage after the first, fourtimes as many triangles are added as in the previous stage, and each newtriangle has one-ninth the area of each triangle added in the previous stage.So the ratio of the resulting geometric series will be 4/9.

1.46 Use, axiomatically, that equidecomposable figures have the same"area." A detailed formal development of the area function A is quite long.

1.47 Cut the square into vertical line segments. Hold the one at the leftfixed, and move each other segment to the right so that it ends up twice asfar from the leftmost segment.

1.48 Go directly to a rectangle by using just one cut.

1.49 Try a construction similar to the one shown in Fig. 1.5.

1.50 Yes; one stops with congruent rectangles rather than going all the wayto two equal squares. Details are given in the book of Boltyanskii mentionedat the end of Chapter 1.

1.51 If the ratio of the length of one side of one of the parallelepipeds to thelength of one side of the other is a rational number, the construction is easy.Otherwise, the answer is still in the affirmative, but the author knows of noeasy proof. That it is possible follows from a theorem mentioned in theanswer to Exercise 1.29.

1.52 All four can be cut up and reassembled into congruent squares. Allthe cuts could then be superimposed onto a single such square.

1.53 It is possible; this follows from the answer to Exercise 1.51, but againthe author knows of no easy proof. Try Exercise 1.54 instead.

1.54 If there are n cubes along each edge of the smaller cubes, and m alongeach edge of the larger cube, one must solve

2n3 = m3


It follows from the Fundamental Theorem of Arithmetic (Section 7.3)that this equation has no solution in integers.

1.55 Yes; the ingenious method of solving this problem is given in Chapter7 of the text by Sherman Stein mentioned in the answer to Exercise 1.29.

CHAPTER 2

2.1 A circle together with one of its diameters is a candidate for an answerto the first question. For other combinations, try finding figures that satisfytwo of the properties but not the other two.

2.2 Certainly not the first.

2.3 Certainly not the last.

2.4 See the answer to Exercise 2.1.

2.5 A disk certainly does have property (a).

2.6 There are infinitely many different correct answers to this problem.

2.7 This is easy.

2.8 Circumscribe a circle about the square, and then move the points of thesquare outward along radii of the circle until the result is that the square hasbeen deformed onto the circle.

2.9 Deform the curve until it lies in a plane. It then bounds a circular disk(after some possible additional deformation in the plane). The disk canbe deformed to a hemisphere, and the other hemisphere then supplied. Thecurve has thus been deformed so as to become the "equator" of the sphere.

2.10 Procure an old inner tube, and draw a curve on it that goes aroundtwice the long way while going around three times the short way. Do youthink that every knotted simple closed curve can be deformed so as to lieon the surface of a torus?

2.11 Yes; but why?

2.12 No; but why not? (Give an example.)

2.13 Can you link each curve with every other curve?

2.14 In the first construction, first draw a completely splittable (n - 1)link.

2.15 One solution might have the curves so arranged so as to look like aninfinite chain.


2.17 Since one, and thus any, curve representing ab can be deformedcontinuously to a curve representing ba. The right convincing drawing canbe found after a little effort.

2.18 Yes; since ab = ba, then also xy = yx if x and yare any two expressions whatsoever in the algebra.

2.19 Both (y-l X-l)(xy) and (xy)-l(xy) are equal to 1.

2.20 These two expressions do reduce to identical ones.

2.21 The answer is found in Section 2.5.

2.22 See Fig. 2.11.

2.23 Begin by copying Fig. 2.12.

2.24 Draw four separated circles, and then follow the formula found inSection 2.5.

2.25 No shorter formula is known to the author.

2.28 See Section 2.6.

2.29 Yes: Let x = aba- 1b- 1 and y = c.

2.30 Here is one solution. Draw three circles A, B, and C forming theBorromean Rings, and then so arrange matters that D links A and B in theBorromean manner as well as linking A and C in the same way. One possibleformula for D is thus

aba - 1b - 1aca - 1C - 1 .

2.31 Expand and simplify (x, y)(y, x).

2.33 Use a case argument.

2.35 Show that an (n + 1, 2)-Brunnian link can be constructed from an(n, 2)-Brunnian link by the methods of Section 2.6.

2.36 The two expressions represent the same curve if the point p is allowedto move. This is not permissible in the algebra, but is allowable for thepurposes of constructing the various links of this chapter. This is the one wayin which the geometry and the algebra are not in perfect correspondence.

2.37 Try (a, b)(c, d)).

2.38 Use the form of the induction principle gIven In the answer toExercise 1.22.

2.39 Again, see Exercise 1.22.


2.40 Knottedness may be removed by reversing some of the crossings of agiven curve over itself-how?

2.41 Look up "Pascal's Triangle" in any college algebra textbook.

2.42 There are many correct answers to this problem; however, certainarrangements are impossible.

2.43 To show the strip has only one side, start coloring at one spot with acrayon, and keep expanding the colored region. The whole strip willeventually be colored on "both" sides.

2.44 Note that the cut down the middle has the drawn line on both sides atall times, and never crosses it.

2.45 Try the experimental approach.

2.48 The surface would be one-sided; it is called a Klein bottle. If you findthe surface difficult to visualize, this is probably because this figure cannotbe placed in ordinary three-dimensional space without an artificial selfintersection-the same sort of artificial self-intersection you see when youtry to draw a knotted simple closed curve on a (two-dimensional) piece ofpaper.

2.49 You can add any desired number of edges to a preexisting figure byremoving the interior of as many small circular disks as you need.

CHAPTER 3

3.1 Try solving the equation 1x = 2.

3.2 The graph looks much like the one in Fig. 3.2.

3.3 Note that the number 1/2 must be taken to a negative power to give avalue larger than 1.

3.4 First establish this for b = l/e, where e > 1.

3.5 What does y = 10gb x mean in terms of exponents?

3.7 Note that log2 (43) = 3 log2 (4).

3.8 Let 10gb xy = p, 10gb X = q, and 10gby = r. Then

bP = xy, bq = x, and

3.9 Use the techniques of the previous exercise.

3.10 For b > 0, b i= 1, and x any real number, define bX to be that realnumber a such that a > 0 and x = 10gb a. Show first that such a value of aexists and is unique.


3.12 The given expressions may be simplified, in order, to

11 + 10gb C

C

-11

3.13 Use Exercise 3.11.

3.14 Let 10gb x = p and 10gb Y = q.

3.15 If 2x = 3, then x log 2 = log 3; the logarithms may be to any fixedbase, such as 10. Why is this so? In any case, the answer correct to nineplaces is 1.584 962 501.

3.17 1 + 5/7.

3.18 The correct answer is not

1 + log (5/2) ,log 2

since the fraction there exceeds 1. Why?

3.19 Note that log 4 = 2 log 2.

3.20 Attack the numerator:

log 2 = log (3/2)(4/3)

= (log 3/2) + (log 4/3).3.21 34/25.

3.22 You should obtain the equation

x = 1 + 12 + x

3.23 If x2 = 5, then x 2- 4 = 1.

3.24 The answer is

-1 + 4../3 .3

If you got this unlikely-looking monster-which we have simplified, by theway-you almost certainly worked the problem correctly.

3.25 If../3 = (l; a, a, a, ... ), then you can obtain the equation

- 1../3 = 1 + ../ '

a-I + 3

and this should lead to a contradiction.

3.26 If you are in utter despair, see Exercise 3.50.

3.30 There is a connection between this exercise and the previous two, andyou should have discovered it.

3.31 Can a sequence of positive numbers have a negative limit?

3.33 Increasing the denominator of a fraction with positive entries decreasesits value, and conversely.

3.35 You should obtain (3; 6, 6, 6, ... )-and this is the correct answer.

3.42 First, of course, you must (correctly) guess that the limit is zero. For aproof:

Let e > 0 be given. Then l/e is a positive number, so we may choose awhole number N such that N > l/e. Suppose that n is an arbitrary wholenumber such that n > N. Then

Il/n - 01 = l/n < l/N < e

and hence the sequence has limit O.

3.43 The most natural definition of the sum, and the one used in mathematics, is the limit of the sequence

1, 1 + 1/2, 1 + 1/2 + 1/4, ...

obtained by "adding up more and more terms of the series."

3.44 See Exercise 10.35 if you wish.

3.45 By use of the definition in Exercise 3.32.

3.46 The results may surprise you.

3.48 This problem can be reduced to showing that the equation

3m = 2n

has no solution if m and n are positive whole numbers.

3.49 Note that k 12 = 2. Why is this so?

3.52 A fourth is an "inverted" fifth.

3.60 Well, theoretically, yes.

CHAPTER 4

4.3 No matter how the elements of (say) the group of Example 4.4 arerenamed, one cannot obtain the multiplication of Example 4.5, since in thelatter example the square of each element is the identity, and the formerexample does not have this property. This observation enables one to avoid


consideration of the twenty-four cases corresponding to the twenty-fourdifferent ways of renaming the four elements of the first group with the namesof the four elements of the second group.

4.4 Be sure to establish associativity the "easy way."

4.5 Consider the "value" of the product ef

4.6 Examine the element yxz.

4.7 Use the fact that x has an inverse yin G.

4.8 What is the value of (x- 1) -1 . x- 1 ?

4.10 Note that either 0 or 1 can stand for the identity of M.

4.11 There is only one way to fill in the table, if 0 is to be the identity. Why?

4.12 One can establish associativity by "realizing" the group of Example4.5 as a group of (not necessarily all) motions of some geometric figure.One figure that will work is a rectangular parallelepiped. What are theappropriate motions to consider?

4.14 Yes. How?

4.16 If the element 9 appears twice in the row to the right of x, and in thecolumns headed by y and z, then xy = 9 = xz.

4.17 If the table were associative then it would have to be a group table.

4.18 Some other associative operations are these:

x # y = x + Y + 17,

x # y = x + y - xy,

x # y = y.

4.20 Try a group that is not commutative.

4.21 (W, +) contains just one subgroup of finite order-which one?

4.24 Yes; which one? Or are there more than one?

4.26 The identity has order 1. To answer the second question, what ifXl = e?

4.27 The group L contains infinitely many elements of finite order andinfinitely many elements of infinite order.

4.28 A rotation of the disk one-seventh of the way around generates asubgroup of order 7. Generalize. The group L does contain many subgroupsof infinite order, but this may not be so obvious.

4.30 First show that (x- 1gx)n = x- 1gnx.


4.31 See the next exercise.

4.32 Note that (gh)2 = ghgh.

4.33 To show that xy = yx, consider (xy) 2•

4.35 If 9 has order 3, what is the order of g2? Can 9 and g2 be the same?

4.36 Note that (ab)" = a(ba)n- 1b.

4.37 Note that b = a- 1b2a. Substitute the b on the left-hand side for eachb on the right-hand side. Continue this process, and eventually use thefact that as = e.

4.38 Since G is commutative, (xy)2 = X2y 2 for all x and y in G (why?).



4.46 Suppose by way of contradiction that 9 were an element of infiniteorder in the finite group G.

4.49 Is there a fixed value of k such that k is a multiple of each element'sorder?

4.50 One method is to show that the identity must appear at least twice onthe main diagonal of the group table for G. Suppose that it does not.

4.51 Yes; but why?


4.57 Try this first for n = 3. Generalize your proof to the case of arbitrarynatural number values of n.

4.59 Here is one way to show that Z is closed:Suppose that y and z are elements of Z. Then yg = gy and zg = gz

for all elements 9 of G. Hence (yz)g = y(zg) = y(gz) = (yg)z = (gy)z =

g(yz). Hence (yz)g = g(yz) for all elements 9 of G. Hence yz is an elementof Z by definition of Z. Hence the operation is closed in Z.

4.60 This proof is similar to the previous exercise. Do you need to knowthat G is finite?

4.62 Let n be the order of G. Among the orders of subgroups of G areonly the two natural numbers I and n. However, this does not show thatamong all natural numbers, n has only the two divisors I and n. How wouldyou show that a group of order 12 has a proper subgroup? Generalize.

4.63 If x and yare elements of g-1 Hg, then for some hand k in H,

x = g-1hg and y = g-1kg.


4.64 Note that this is an "if and only if" proof.

4.66 If this is too easy, why don't you try a really tough problem:Let G be a group containing elements x and y such that, for some fixed

natural number n,

and

Prove that x = e = y.

4.71 Even though A is not a sllbgroup, it has "cosets" such as gA, where 9is an element of G. See the proof of LaGrange's Theorem to see whatproperties gA must have even though A is not a subgroup of G.

CHAPTER 5

5.2 Yes.

5.3 Yes.

5.4 Yes.

5.5 A man walking around a vertex passes through an even number ofcountries. Will this fact help in showing that a two-color "checkerboard"coloring pattern will work?

5.9 To prevent the boundary and exterior of (say) a cube from being apolyhedron.

5.10 The answer to the second question is "no."

5.11 The maximum possible is 12. Finding a map that requires all 12 colorsis difficult; proving that 12 is sufficient for all such maps is very difficult.

5.13 It can be done. Curiously enough, it does not matter whether countries"go all the way through" the strip or whether one has different countriesand non-coincident boundaries on the two "sides" of the strip. Can you seewhy not?

5.14 Yes; you can construct such "maps" requiring any given number ofcolors for a proper coloring.

5.15 There is no upper limit.

5.16 The proof might involve consideration of what happens to the value ofV - E + F when one hole is "plugged up."

5.17 If the lines do not intersect, you have a net in the plane for whichV - E + F = 1. What is the value of F? Reach a contradiction byconsidering the possible number of boundary edges of each country.


5.19 First show that 3F = 2E.

5.21 See Exercise 5.5.

5.22 Use the fact that each vertex lies on four edges to show first that 4V =2E. Then, since each country is to have six sides, you can show that 6F = 2E.Since you also know that V - E + F = 2, you may be able to reach acontradiction using these three formulas.

5.24 Name the countries, and then proceed with a coloring scheme chosenso as to avoid cases.

5.25 A boundary edge must be a segment rather than a closed curve. Soone should just introduce two (artificial) vertices onto the equator.

5.27 The points in the shaded region are exactly those satisfying the inequality of Steinitz's Theorem.

5.28 Try a few simple examples with small values of V and F.

5.30 Since E = 20, V + F = 22. Solve for F and use the inequality inSteinitz's Theorem to find the desired conditions on V.

5.35 Curiously enough, the answer is always F - 4. Why?

5.36 First establish that 3V = 2E, and that 5F = 2E. Then use Euler'sTheorem.

5.42 Use a case argument.

5.44 Use the techniques of the solution to Exercise 5.40.

CHAPTER 6

6.13 The other formula IS also valid. These formulas are known asDeMorgan's laws.

6.14 If k is the larger of the two numbers 111 and n, then the answer may begiven in terms of an inequality involving k.

6.15 The answer is a formula involving k, m, and n.

6.16 Yes, the notationf(a) makes sense: If (a, b) Ef, thenf(a) = b.

6.19 The function f: R -4 R according to the rule f(x) = x 3 is sufficientlydifferent from Example 6.11.

6.21 If you hold the page on which Fig. 6.5 is printed up to the light, andlook at the other side of the page so that the x-axis is vertical and the positivey-axis is to the right, this has the effect of interchanging these two axes;thus, what you see is the graph off-I.


6.22 Here is one way to show thatf- l is one-to-one:Suppose that x and yare elements of B, and that f-l(X) = f-l(y).

Since f is a function, f(f-l(X)) = f(f-l(y)). Hence x = y. Therefore, ifx :F y, thenf-l(x) :F f-l(y). Therefore f- l is one-to-one.

6.27 The appropriate notation would be {f - 1(g - I)}.

6.29 Note that (x + 1)2 is not always equal to x 2 + 1.

6.30 Is the converse true?

6.32 Try f: N -+ E according to the rule

f(x) = 2x.

6.34 What about the function: 9 N -+ T by

g(x) = x + 9?

6.36 One possibility is

f(1) = 2,

f(2) = 1,

f(x) = x if x > 3.

Now find two more.

6.38 First devise the correspondence; then, if you have been sufficientlysystematic, you can devise a formula for the appropriate function.

6.39 Draw a triangle with base two units long. It has a parallel median whichmust be one unit long. How do you correspond the points of the base withthe points of the median?

6.40 Modify the answer to the first question to answer the second.

6.43 Since you want to put Band C into one-to-one correspondence, thetrick is to draw two circles and label them Band C. Inside of each draw asmaller circle, and determine what sets these two should represent in orderto be able to apply the Cantor-Schroeder-Bernstein Theorem.

6.45 One can let f: W -+ N according to the rule

f(x) = 2X

f(x) = 3- X

if

if

x > 1,

x ~ o.

6.52 You may wish to show this first in the special case in which A n B = 0,and then apply some of the previous exercises or theorems.

6.59 Remember that S must contain a denumerable subset.

6.72 How many three-element subsets has N? How many five-elementsubsets has N?

6.75 Compare the following two versions of the method of performingthe experiment:

First Method: At each stage, remove from the urn the three lowestnumbered balls not previously moved, then replace in the urn the two lowestnumbered balls outside the urn.

Second Method: At each stage, remove the three lowest-numberedballs in the urn, then replace the two highest-numbered balls then outsidethe urn.

Do you see a way to perform the experiment so that, at its conclusion,exactly thirty-seven balls are in the urn? It can be done!

CHAPTER 7

7.2 Yes; see the next exercise.

7.4 Check your answer with a few experiments.

7.5 Remember that integers may be negative as well as positive.

7.6 Remember that each prime is positive.

7.9 This is true under certain conditions, but not always true.

7.10 Yes; supply a proof.

7.12 Yes; supply an example.

7.17 The last n numbers in the sequence are composite.

7.18 The answer to each question is "no."

7.21 See Exercise 1.23 and the answer to it.

7.25 If a were the least positive real number, what about a12?

7.28 E is a subset of N.

7.29 What is the last positive rational number?

7.31 The number happens to be 65, but you should prove its existence usingthe Well-Ordering Axiom.

7.33 This is a moderately long problem. If you have studied Chapter 3,compare this exercise with Exercise 3.33.

7.34 If a < b, x < y, and all numbers involved are positive, then ax < by.Why?


7.35 Simplify the expression

(n + 1)3 - (n + I).

7.38 The second question is much easier than the first!

7.41 Use the Fundamental Theorem of Arithmetic.

7.42 No; but why not?

7.43 Handle 3m + nand m + 2n separately if you wish.

7.44 No; but why not?

7.45 Take n = 7, of course.

7.46 Use Wilson's Theorem to show that 20 has a proper divisor.

7.47 Note that 101 is prime.

7.48 Compare this with Exercise 7.9.

7.50 The formula is

«(Xl + 1)((X2 + 1)((X3 + 1)··· «(Xk + 1).

7.53 This one is not easy.

7.58 The number 4 is always a divisor of the left-hand side and never adivisor of the right-hand side.

7.59 First find integers m and n such that 12m + 13n = 1.

7.61 If (n, n + 3) = 2, then 2 would be a divisor of (n + 3) - n. Butwhy?

7.63 If n is odd, then n has the form n = 2k + 1 for some integer k.

7.64 Show that every common divisor of m and m + n is a divisor of n.It will then follow (why?) that (m, m + n) I n.

7.73 The author obtained the following solutions, but could persuade noone else to check his answer to this laborious problem: The triple (x, y, z)may be any of the following and no others.

(7, 1, 1) (3, 3, 1)

(5, 2, 1) (2, 2, 2)

(4, 1, 2) (1, 1, 3)

(1, 4, 1)

7.74 No way. It cannot be done.

7.75 If X is the number of new eggs, y the number of fresh eggs, and z thenumber of old eggs, then

X + Y + z = 100,and

lOx + 2y + z = 200.

If you solve the first equation for z and substitute the result in the secondequation, you obtain

9x + Y = 100,

which is no problem to solve. Of the ten positive solutions, only x = 10,y = 10, z = 80 satisfy the last condition of the problem.

7.76 Use the same sort of simplification as in the previous problem. Thereis only one solution.

7.77 There is no need to list the solutions in order to count them; however,there turn out to be 45 ways in which the check could have been written.

7.81 If there are b brass balls, c copper balls, and s silver balls, then oneobtains the equation

15b + 16c + I7s = 121.

This problem is quite long, and the author is reasonably sure that there isonly one solution. Hint: In that solution, no balls of one type were used.This is not ruled out as a possible solution.

7.82 You can get an easy proof if you know the following fact: If m > 2is a whole number, then there exists at least one prime p such that m <p < 2m. Observe that if

1/2 + 1/3 + ... + 1/11 = k,

where k is a whole number, then n must be much larger than k. However,there is a proof-hard to find-that does not use the above rather advancedresult of number theory.

7.83 If the prime p is of the form 411 + 3, and is the sum of two squares,then one of the squares must be odd and the other even.

7.84 A small number of cases should be considered.


7.90 One solution is


7.91 This is not so easy. The author believes that the smallest perimeterthat solves the problem is 480.

7.92 This is easy, and can be answered without much trouble for any wholenumber, not just 10.

7.93 There is only one solution.

7.94 This is quite difficult, unless you have found a short cut unknown to theauthor. The smallest solutions he obtained are

32 + 42 = 52,

(696)2 + (697)2 = (985)2,

(23,660)2 + (23,661)2 = (33,461)2.

Might this problem have infinitely many solutions?

7.95 This is just a matter of checking a few cases.

7.96 Unlike Exercise 7.94, this is quite easy.

7.100 Use Fermat's Theorem.

CHAPTER 8

8.1 The conditions of the problem indicate that k is positive. Hence N'is always positive, but as N increases, the value of N' gets closer to zero.

8.2 Since N' is constant, the graph of N(t) will be a straight line in eachcase.

8.3 The equation N' = B - D becomes

N' = bN - dN,so that

N' = (b - d)N.

Since b - d is constant, set k = b - d.

8.4 Note that

No = No' ekT

2

by definition of the half-life T.

8.6 The graph of N(t) will look very much like one of those shown in Fig.8.6; that is, taking the square root of the degree of realization term has littleeffect on the long-term behavior of N.

8.9 Since the water flows out at the rate V'et), and V'(t) is proportional tothe pressure at time t, and the pressure then is proportional to Vet) itselfbecause of the shape of the tank, we obtain the differential equation

V'et) = k Vet),

where k is a negative constant. Now see Exercise 8.4.

8.10 An equation giving a satisfactory interpretation of the conditions of theproblem would be

N ' = bN M - N - dM '

where band d are positive constants.

_~.11 You should obtain a stable critical point where the bluegill and redearlines cross; that is, the two species can coexist.

<8:14 Arrows should definitely be drawn on the coordinate axes; it can happenthat some curve of population trend does meet an axis, indicating the disappearance of one species.

8.16 If the three populations are A, B, and C, then one of the three equationswould be

A' = kA M - A - aB - f3CM '

where M is the maximum population of population A the pond will supportand a and f3 are positive constants. The behavior of the system could beexamined by means of a graph in the first octant in three-dimensional space;instead of lines where A I = 0, there would be planes, with A I positive beneaththe plane corresponding to it and negative above. There are a large numberof possibilities for the eventual behavior of the system; for example, onespecies might disappear, followed by the coexistence of the remaining two.

8.18 The system is critical for N = M and for N = 0. The former is stable,the latter unstable.

8.20 Compare this exercise with Exercise 8.10.

8.21 Wide variations in population may result in the elimination of onespecies from the pond.

8.23 One reasonable set of equations is

A' = (kA _ f3B) M ;; A ,

B' = (vB _ aA) C ~ B ,


where all constants are positive, M and C representing the maximum populations of A and B, respectively, that can be supported by the pond.

8.28 The qualitative behavior of the solution is unchanged.

8.29 The fact that ex and pare positive describes the condition that each of thetwo species contributes to the success of the other. A culture of yeast andslime mold on nutrient agar in a large dish will exhibit this sort of behavior.

8.33 You should not spray unless you can spray enough to completelyeliminate aphids.

8.35 Yes; what is the value of to in terms of the given constants?

8.37 One could obtain the solution of the differential equation, substituteenough values of the population at various times in order to evaluate allunknown constants, and then check the resulting solution with values of thepopulation at other times.

8.38 See the answer to the previous exercise.

CHAPTER 9

9.2 If and only if a = b.

9.3 Use the same definition.

9.5 Since p E [a, b], if p is different from a and different from b then thesame straight line contains {a, p} as contains {p, b}.

9.7 Use the previous two exercises if you wish.

9.8 See the next exercise.

9.9 This is Theorem 9.2.

9.10 No; give an example.

9.13 See the material on "proof by induction" in Chapter 7, or see Exercise1.45.

9.14 No; proof by induction can only show something true "for each naturalnumber n." However, although it does not follow from Theorem 9.2 andExercise 9.12 that the intersection of infinitely many convex sets is convex,this is nevertheless true.

9.15 To answer the last question of the exercise, you know that eitherA c B or B c A; there is no harm in supposing that the sets have been sonamed that A c B.

9.16 Yes, and the proof is the same.


9.17 Yes; try proving it.

9.18 Again, yes.

9.19 Of course; the proof is very simple.

9.20 Consider the collection of all circular regtons (including boundarypoints) centered at the origin in E 2

•

9.25 What if A consists of three noncollinear points in E 2 ?

9.27 The first three questions of the exercise have affirmative answers.

9.28 The number of applications of Amust be increased by one.

9.30 This exercise turns out to be very useful in some later problems.

9.31 No. Instead of triangular regions, what sort of sets should~ consist of?

9.32 The answer to the second question is too easy if 0 E CC, so try to givean example in which 0 does not belong to CC.

9.33 Some candidates for the property might be "being linear" and "containing no straight line segment of length exceeding one."

9.34 It is possible but not necessary; try finding a proof in which the axiomis used, just for practice.

9.35 It turns out that the only consistent interpretation of n CC is E 2• No

points p of E 2 have the property indicated in the exercise, since there are nosets-and thus no such sets-in CC. Try showing that if CC is the emptycollection of subsets of E 2

, then vCC = 0.

9.41 Yes. Hence our proof is really a disguised proof by induction. See theIndex for more on induction.

9.42 Helly's Theorem does apply here.

9.43 Although Helly's Theorem does not immediately apply, it is stillpossible to reach some conclusion by considering the circular regions boundedby the circles in CC.

9.44 Nothing, for the analogue of Helly's Theorem in E 3 requires that eachfour sets have a point in common. There is a connection with Exercise 9.28and the version of Exercise 9.30 for E 3

, a connection which explains why thenumber in the theorem must be increased by one.

9.45 This is not easy. See Exercise 9.43.

9.46 See Exercise 9.44 and the answer to it.

9.48 The easiest way to consider cases is to consider how many of the pointsare collinear.

9.52 In order to apply Helly's Theorem in E 3, the number of sets that in

tersect must be increased to four. Hence it would be necessary for each set offour pictures to be visible from some point in the gallery.

9.53 Yes; give an example.

9.55 Yes; one way is to divide E 2 into the disjoint convex sets E 2 and 0.There are other solutions; find them.

9.56 By Helly's Theorem, it is clear that k < 3.

9.57 Note that the equation of the straight line through (n, 0) with slope n is

y = nx - n2•

9.58 Examine a regular tetrahedron.

9.64 Use the axiom of Section 9.3.



9.72 This is possible even if the intersection is "connected"-see if you canfind such an example.

9.80 One of the implications is true, the other false.

CHAPTER 10

10.2 Use the fact that each rational number can be expressed in the formmin, where m and n are integers and n i= O.

10.4 No. Why not?

10.5 The decimal expansion given in the text can be so used; how?

10.6 How about 3.14159?

10.10 Every rational number has a finite continued fraction expanSIOn.This follows from the Euclidean Algorithm (Theorem 7.8).

10.12 Use the technique of Theorem 10.1.

10.15 8/9.

10.16 327/999. Examine the previous exercise; do you see a pattern?

10.17 Note that if 0.999 999 ... were less than 1, then there would be anumber r such that

0.999 999 ... < r < 1.


10.20 The techniques of Chapter 7 can be used to show that there areirrational numbers not solutions to any equation of the form

p(x) = 0,

where p(x) is a polynomial with rational coefficients. These are calledtranscendental numbers.

10.22 There is a connection with Exercise 3.48.

10.23 Suppose by way of contradiction that IX + r is a rational number.

10.25 You can conclude that b 2 - 4ac is the square of an integer.

10.27 The proof is similar to that given in the answer to Exercise 3.42.

10.29 Nothing can be done with either (d) or (e). Why not?

10.32 The sequence has no limit.

10.33 This is not easy.

10.34 Expand the product

(1 - r)(1 + r + r 2 + r 3 + . . . + rn).

10.39 See Section 10.5.

10.44 Let J 1 = /b J2 = /k+ l' and so on.

10.45 Treat the case when both real numbers are positive first. Define theproduct in the other cases by using absolute values.

10.59 If a set had two least upper bounds, one would have to exceed theother.

10.61 Clearly, 0 ~ m ::; 9. Or is it clear?

10.62 The construction differs in the choice of the first interval.

10.63 What about zero?

10.65 Both (0, 1) and (0, - 1) are solutions of the equation

x 2 + 1 = O.

10.68 For example, show that 1/4 E K.

10.69 The function has the same value at the two numbers.

Absolute value, 289Aleksandrov, A. D., 318Alexandroff, Paul, 317American Mathematical Monthly, 107Antoine, L., 51Approximations by continued frac-

tions, 63-65Archimedean property, 303Area, 21Art Gallery Theorem, 254Associative

law, 82operations in real number system,

301

Bach, J. S., 52, 76Banach, S., 21-23Batting average, 66Beck, Anatole, 318Beet virus molecule, 142Beiler, A. H., 218, 316Benade, Arthur H., 81Benson, R. V., 279Bers, Lipman, 317Birkhoff, Garrett, 106, 316Bleicher, Michael N., 318Boltyanskii, V. G., 23, 279

INDEX

Bolyai Farkas, 23-24Janos, 24

Bolyai-Gerwin Theorem, 2, 6Borromean Rings, 25

generalizations, 29formula, 41see also Brunnian links

Brunn, H., 29, 51Brunnian links, 29

(4, 2)-links, 45-46(n, k)-links, 43, 49

Cancellative operations, 95Cantor, Georg, 177, 182Cantor-Schroeder-Bernstein

Theorem, 164applications, 169-172, 176,181

Cantor Ternary Set, 312length, 313number of elements, 314

Cardinal numbers, 179existence, 180

Center of group, 104Cohen, Paul J., 182Cohn-Vossen, S., 143, 317Commutative

law, 34

345

346 Index

Commutative-continuedoperation in group, 87, 94see also Group, Abelian

Commutator, 45k-commutator,48

Complex number system, 310Composite numbers, 185

consecutive, 187prime factors, 186

Congruence motionsof disk, 90product, 84-85of square, 93of tetrahedron, 90of triangle, 83-87

Constructions with straightedge andcompass, 6, 19, 23

Continued fractions, 59-61and batting averages, 66-67construction, 75-76evaluation, 62-63and grade distributions, 67-69and irrational numbers, 286

Continuum, 179Convex hull, 261

kernel, 261polyhedron in Steinitz's Theorem,

116sets, intersection of, 258-260sets, tower of, 260

Convexity, 255generalizations, 276-279

Coset of subgroup, 96Coxeter, H. S. M., 143, 317Critical point, 240Cross-cancellative operation, 95

semigroup, 105Crowdis, David G., 318Crowe, Donald W., 318Crowell, R. H., 51Curve, knotted, 30

on torus, 30polygonal, 28simple closed, 27-30tame, 28wild, 28

Cycles in population, 249

Debrunner, H., 23, 51, 279, 317Decimal expansion, 308-309Dedekind, Richard, 183Dedekind Box Principle, 172, 175

applications, 177Dedekind infinite, 174Degree of realization, 230, 232

in competing populations, 234Dehn, M., 18Differential equations, 223

for competing populations, 234, 245in population growth, 224-226in radioactive decay, 227

Divisibility, 184Divisors, 184

product, 189number, 205see also Greatest common divisor

Dorrie, Heinrich, 317Dudley, Underwood, 218Duplication of cube, 18Dynkin, E. B., 143

Equidecomposable figures, 3, 22, 23Equivalence relation, 5, 162

between nested sequences, 293, 295Euclidean algorithm, 92, 196, 197,

219proof, 197-202

Euler, L., 118, 144Euler's formula, 108, 118

proof, 118-122Exponent of group element, 98

Factorization into primes, 196, 202,209-210

Fermat, Pierre de, 184Fermat's Theorem, 218Fifths, 73

improving, 77Five-Color Theorem, 144Fort, M. K., 51Fox, R. H., 51Fraleigh, John B., 318Frequencies, 70

of notes on piano, 73Functions, 154-162

Fundamental Theorem of Arithmetic, 202, 204

applications, 281proof, 202-203

Gause, G. F., 252Gauss, C. F., 218-219Gelfond, A., 218Generator of group, 102Geodesic dome, 142Geometric series, 283Goffman, Caspar, 315Golden Mean, 62Greatest common divisor, 205

computation, 205-206Greek alphabet, 319Griffin, Harriett, 218Group, 92

Abelian, 102associated with curves in space, 38center, 104cyclic, 102examples, 82-92of prime order, 102uniqueness of identity, 93

Grtinbaum, Branko, 143, 279

Hadwiger, H., 18, 23, 279, 317Half-plane, 273, 275Hall, M., 106Harmonics, 70Hausdorff, Felix, 182Helly's Theorem, 266, 268

applications, 270-271, 273, 276Herstein, I. N., 106Hilbert, D., 143, 317Hocking, John G., 317Homomorphism, 105Horn, Alfred, 279

Image of homomorphism, 105of function, 155-156

Induction Principle, 10, 190and well-ordering, 190-193applications, 10-11, 49, 193-195,

259Infinite series, 69-70, 295

Index 347

Integers, 184Intermediate fractions, 64Intervals, 288

length, 288nested sequences of, 293sum of, 297

Inverse of group element, 83uniqueness of, 93

Jacobson, Nathan, 106Jakobson, Roman, 317Jeans, Sir James, 81Join of point and set, 277

Kernel of homomorphism, 106see also Convex

Khinchine, A. Ya., 81Klee, Victor, 23, 279, 317Kolmogorov, A. N., 318Krasnoselskii, M. A., 271Krasnoselskii's Theorem, 254, 271Kripke, Bernard, 315Kurosh, A. G., 106

Lack, D. L., 253LaGrange, J. L., 107LaGrange's Theorem, 96Lavrent'ev, M. A., 318Law of Quadratic Reciprocity, 218Law of Small Whole Numbers, 72L-convexity, 277Least upper bound, 303Limit of sequence, 65, 69, 284, 289

291, 294-295Links, 30, 32

splittable, 30see also Brunnian links

Lobachevsky, N. I., 24Logarithms, 53-56Lomont, J. S., 317Lyusternik, L. A., 143, 279

MacLane, Saunders, 106, 316Mansfield, M. J., 314Map on Mobius strip, 117

on sphere, 116on torus, 116

348 Index

Mobius strip, 50, 117Moise, E. E., 23Mordell, L., 218

n-link,30completely splittable, 32splittable, 30sublink of, 32

Natural numbers, 184composite, 185prime, 185

Nested sequences of intervals, 293equivalence of, 295

Net of polyhedron, 118Niven, Ivan, 218, 316Nonmeasurable set, 21

Odum, Eugene, 253One-to-one correspondence, 156,

160-162Order of group, 96

infinite, 96, 99of group element, 99

Ordered pair, 159Ore, Oystein, 143

Parallelogram formed from triangle,11

reassembled into rectangle, 11Partial quotients, 62Passman, D. S., 106Polya, George, 192, 318Polygon, 2, 109

connected, 109edge, 2equidecomposable, 3vertex, 2

Polyhedron, 113edge, 115face, 1152-connected, 116vertex, 115

Primes, 102, 185Pythagorean right triangles, 217-218Pythagorean Theorem, 16

Rademacher, Hans, 218Radioactive decay, 227Rational numbers, 280Real numbers, 297

Archimedean property, 303as nondenumerable set, 178decimal expansions, 308-309order relation, 302

Rectangle formed from parallelogram, 11

reassembled into square, 12-14Regular solid, 123Rudin, Walter, 314

Schoenflies Theorem, 110Semigroup, 104Sets, 146

algebra, 153and Venn diagrams, 151Cartesian product, 160convex, 255denumerable, 176descriptive definition, 147difference, 154distributive laws, 152-153element, 146empty, 150equality, 148finite, 163inclusion, 149infinite, 163intersection, 150listing, 147maximal convex, 264maximal tower, 263-264maximal with respect to a property,

263nondenumerable, 177nonlinear, 263notation, 147-148number of elements, 154subset, 149tower, 260union, 149

Sierpinski, W., 23Sigmoid curve, 232Singer, I. M., 51

Slobodkin, Lawrence B., 253Snowflake curve, 19-21Spivak, Michael, 317Sprecher, David A., 315Square formed from rectangle, 12-14

formed from several squares, 14-16reassembled into given polygon,

16-17Squaring the circle, 18Stein, Sherman, 144, 318Steinitz, E., 127, 143Steinitz's Theorem, 127

proof, 128-133Stoll, Robert R., 182, 315, 317Straight line segment, 255Subgroup, 95

improper, 96normal, 104proper, 96

Tarski, A., 19, 22Ternary system, 311Texan rectangle, 7Thirds, 78

improving, 81Thomas, J. M., 144Thorpe, John A., 51

ABCDE79876S432

Index 349

Tietze, Heinrich, 143, 317Torus, 30, 116Transposition, 74Triangle reassembled into parallelo

gram, 11Trisection of angle, 18

Unbounded figure, 6Uspenskii, V. A., 143

Valentine, F. A., 279Venn diagrams, 151-152

Wall, H. S., 81, 317Well-Ordering Axiom, 190Well-tempering, 76-77Wheeler, Brandon W., 318Wilder, R. L., 315, 318Williams, J. D., 317Wilson's Theorem, 204, 205Wolf Interval, 74

Yaglom, Y. A., 279Young, G. S., 317

Zermelo Axiom, 263Zuckerman, H. S., 218, 316

Penney Perspectives in Mathematics 1972

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