A PROOF OF THE CONSISTENCY OF GEOMETRY

by

MICHAEL JAY JACOBS

A THESIS

submitted to

OREGON STATE UNIVIRSITY

in partial fulfillment of the requirements for the

degree of

MASTER OP SCIENCE

June 1961

Redacted for Privacy

Professor ,f Nathematics

In Charf.e of ilajor

Redacted for Privacy______________ iimáii öf Dópartment of Natheraati es

Redcted for Privacy

Chairman o 1oo1 G±uate Committee

Redacted for Privacy ean of Graduate Ecboo1

Date thesis is presented

TypeJ by Lilah N. Potter

2n

TABLE OF CONTENTE

i-age

INTRODUCTION. . . . . . . . . . . . . . . . i

I. AXIOItz3 OF CONNECTION. . . . . . . . . . . . 2

II. AXIONS OF ORDER . , . . , 15

III. . . . . . . . . . . . . . . . . . . . . . . 23

IV. PIAYFAIi'S AXIOM. . . . . . . . . . . . . . 44

V. ARCHINEDES' AXIOM . . . . . . . . . . . . . 46

VI. THE CONELETENESS AXIOM. . . . . . . . . . . 48

BIBLIOGRAPHY. . . . . . . . . . . . . . . . 49

A PROOF OF THE CONSISTENCY OF GEOMETRY

INTRODUCTION

It is our purpose to prove the following Theorem:

If the axioms of arithmetic are consistent (i.e.,

don't lead to a contradiction) then so are the axioms of

geometry.

1or the axioms of arithmetic we take the well

known axioms of reano. For the axioms of geometry, we

take the system of axioms given by Hubert.

The proof of the theorem consists of exhibiting

an arithmetic model for geometry. Once such a model is

constructed, then any contradiction in geometry neces-

sarily will involve a contradiction in arithmetic. Thus

the theorem will follow.

The following notations will be used throughout

the text:

Iletadefinition -- A definition which assigns a

geometric name to an arithmetic entity.

Definition -- A definition wbich assigns a name

to a complex of geometric entities, or a definition

which assigns a name to a complex of arithmetic entities.

Theorem -- A deductive result of arithmetic which

has the same wording as an axiom of geometry.

Lemma -- All other deductive results.

2

I. AXIONS OF CONNECTION

Metadef 1. A point is a triple (x,y,z) of real numbers.

Netadef 2. A line is an equivalence class of

pairs of equations equivalent to a pair of equations, A1x+B1y+C1z =0 A2x + B2y + C2z + D2 = O

where A1, 2' B1, B2, C1, C2, D1 and D2 are real con-

stants and x, y and z are real variables and

A B C

rank 1 1 = 2.

2 B2 C2

I'íetadef 3. A plane is an equivalence class of

equations equivalent to an equation,

Ax + By + Cz + D = O

where À, B, C and D are real constants and x, y and z

are real variables and

A2 + B2 + c2 0.

Remark: We will sometimes say that 'a plane is an

equation' or that 'a line is a pair of equations' when

the equation or pair of equations in question are rep- resentatives of the plane or the line. The equations are used as naine for the plane or line.

Netadef 4. A line,

A1x + B1y + C1z + O

+ B2y + C2z +1)2 = O

is incident upon a pair of distinct points, (x1, y1,

z1), (x2, y2, z2) 1ff,

+ B1y1 + C1z1 +

= O

A1x2 + B1y2 + C1z2 + D1= O

2l + B2y1 + C2z1 + D2 = O

À2x2 + B2y2 + C2z2 + D2 = O

Lemnia 1. Two points, P1 = (x1y1z1) and P2 =

(x2y2z2), where y1 , y2, have a unique line incident

upon them.

4e will first show the uniqueness part of the

lemma. Suppose therefore that a line represented by

(ï) . A1x + B1y + C1z + D1 = O

and (2) . . . A2x + B2y + C2z + D2 = O

is incident upon P1 and P2.

Then we have,

(3) . . . + B1y1 + C1z1 + D1 = O

(4) . . . A1x2 + B1y2 + C1z2 + 1)1

(5) . . . A2x1 + B2y1 + 02z1 + D2 = O

(6).. . A2x2+B2y2+C2z2+D2=O From (3) and. (4) and (5) and (6) respectively,

(7) 1(x1-x2) + B1(y1-y2) + C1(z1-z2) = O

(8) . . . A2(x1-x2) + B2(y1-y2) + C2(z1-z2) = O

4

From (7) and (8),

(9) . s . (A1C2-A2C1)(x1-x2) = (B1C2-C1B2)(y2-y1)

and

(lo). . . (A1C2-A2C1)(z1-z2) = (A2B1-A1B2)(y2-y1)

From (3) and (5) and (4) and (6) respectively,

(11). . . (A1C2-C1A2)x1 + (C2B1-C1B2)y1 + (C2D1-C1D2)

=0

(12). . . (A2B1-A1B)y2 + (C1A2-C2A1)z2 + (À2D1-A1D2)

=0

Since y1 y2, from equations (9) and (10),

(l). . . (B1C2-C1B2) = (A1C2-A2C1)(x3-x2)/(y1-y2)

(14). . . (A9B1-A1B2) = (A1C2-A201)(z1-z2)/(y1-y2)

Now we notice using (13) and (14) that if

A1C0-A2C1 = C, the rank of a B1 is equal to zero. A2 B2 02

ience

(15)..... A1C2 - A2C1 o

Hence from (13), (14), (15),

(16). .

B1C2-C1B2 x1-x2

A1C2-A2C1

(17). A2B1-A1B2 z1-z2

A1CA2C1 = y1-y2

From (11) and (16) and (12) and (17)

5

(18). . . x1 + x1-x2 y1 + C2D1-C1D2 = O

A1C2-C1A2

(19). . . z1-z2 y2 + z2 + A2D1-A1D2 = O

C1A202A1

ie now write a pair of equations which are

equivalent to (1) and (2),

(20). . . (Á102-A2C1)x + (B1C2-B201)y + (D1C2-D2C1) = O

(21). . . (A2B1-A1B2)y + (A2C1-A1C2)z + (D1A2-D2A1) = o

From (16), (17), (18), (19), (20) and (21) we get

another pair of equivalent equations,

(22). .

X y y1 - X1 = O

y1-y2 y1-y2

(23). . . z1-z z -z 2 2

yl2 y2i 2 - = o

This shows uniqueness. To show that such a line

exists we write a pair of equations which are satisfied

by the given triples.

(2'.). . . x(y1-y2) + (x1-x2)y + (x2-x1)y1 + x1(y2-y1) = O

(25). . . y(z1-z2) + (y1-y2)z + (z2-z1)y2 + z2(y2-y1) = O

Since y1 y2 the rank of the equations = 2 and.

thus (24) and (25) are a line incident upon P1 and P2.

Theorem 1. Two distinct points have a unique line

incident upon them.

Proof. If y1 y2 then either z1 z2 or x2. Hence

proceed with a parallel argument to that of the previous

le nun a.

Definition 1. A point F, is a point a line

iff there exists a point , such that the line is mci-

dent upon P and .. e sometimes say the point is on or

the line.

Theorem 2. Two distinct points P and Q of a line

i have the line 1 incident upon them.

Froof'. If P = (x',y',z'), ., = (xu,yu,zhl)

1 = Cz+D=O

o then we have that

Ax' + By' + Cz' + D' = O Ax" + By" + Cz" + D = O A'x' + 13'y' + C'z' + D' = O A'x" + B'y" + C'z" + D' = O

Theorem a. There exist two distinct points on

any line.

Proof. Suppose we are given a line,

A1x + B1y + C1z D1 O

A2x+B2y+C2z D20 (A B c\

In order for the rank of A B to be 2 one of

\ 2 2 2)

the following must be different from zero:

A1B2 - A2B1 A1C1 - A2C1 B1C2 - B2C1

Suppose that A1B2 - A2B1 O. Then we define,

X0 = B1D - B2D1/À1B2 - A2B1

y0 = D1A2 - D2A1/A1B2 - A2B1

i = B1C2 - B2C1

in = C1A - C2A1

n A1B2 A2B1

t = any non-zero real number.

Then the points (x0, y0, o) and (x0-s-lt, y0+int, z0+nt)

are distinct and are on the line.

A parallel proof is required for the assumption

that either i or m is different from zero.

¡'letadef 5. A plane represented by x + By + Cz +

D = O is incident upon three distinct points (x1y1z1)

(x2y2z2) (x3y3z3) iff

Ax1 + By1 + Cz1 + D = O

Ax2 + By2 + Cz2 + 1) = O

Ax3 + By3 + Cz3 + D = O

Definition 2. A point is a point of a plane iff

there exist 2 other distinct points such that the plane

is incident upon the 3 points. e sometimes say that

the point is on or in the plane. e say that a line is

a line of or in a plane iff every point of the line is

in the plane.

roi

Theorem 3b. There exist three distinct points

on every plane such that the three are not in any one

line.

Froof. Given plane, Ax + By + Cz + D = O, where

A2 + B2 + C2 O, then one of À, B and C must be differ-

ent from zero. Assume A O. Then

Ax + By + Cz + D = O y=O

and

Ax + By + Cz + D = O y+l=O

are lines. Cince no triple with y = O is a solution to

the second pair of equations, the lines are distinct.

Call the lines li

and 12s From theorem 3a we have that

there exist two points P1 and P1' on the line 1 and

P ' = (a', o, b') = (a, o, b), 1.

where a' ji( a, or b' b.

There exists a point l2 (c, -1, d) on 12

pl', P1, 2

are points of the plane since they must

satisfy the equation of the plane.

Theorem 3c. There exists a plane.

Froof. x = O is a plane.

Theorem k. Three points which are not in any

one line have a unique plane incident upon them.

rroof. Let the three given points be: (x1, y1, z1),

(x2, y2, z,..), (x, y7, z7)

Case 1.

Assume: X1 y1 z1

X2 y2 z2 = O

X3 y3 z3

therefore

Ax1 + By1 + Cz1 = O

Ax2 + By2 + Cz2 = O

Ax3 + By3 + Cz3 = O

C)

for some values of A, B, C such that A2 + B2 + C2 O.

(for if A2 4- B + C2 =0 then A= B= C = O)

Hence the points are on the plane Ax + By + Cz = O.

In order to show that this is the unique plane we

must only show that the rank of (x1 y1 z1\ is 2.

x2 y2 z2

\x3 Y3 Z3

The rank is less than 3 since the determinant is

zero. The rank is not zero since then

(x1, y1, z1) = (x2, y2, z.b) = (x3, y3, z3)

= (O, O, o)

If the rank is i then we may write without loss

of generality,

(x2, y2 z2) = k (x1, y1, z1),

(x3, y3, z3) = k'(x1, y1, z1),

in the usual vector notation.

Now suppose (x1, y1, z1) and. (x2, y2, z2) satisfy

an equation

then

and

Ax + By + Cz + D = O

Ax1 + By1 + Cz1 + D = O

+Bky1 +Ckz1 +D=O

hence -D Ax1 + By1 + Cz1 = k(Ax1 + By1 + Cz1).

If (x2, y2, z2) = (O, O, O) then D = O. If (x2, y2, z2)

(o, O, O) then k O and then + By1 + Cz1 = O and

hence D = O.

Therefore in any case, D G.

1-lence a line,

Ax + By + Cz = O A'x + B'y + C'z = O

is incident upon (x1 y1 z1) and (x2 y2 z2).

But since (x, y3, z3) = k(x1, y1, z1),

A kx, + B ky1 + C kz1 = O

A'kx1 + B'ky1 + C'kz1 = O

hence (x3, y3, z3) is on the line of (x1, y1, z1) and

(x2, y2, z2) which is a contradiction.

11

Case 2

Then

Xl yl z1

X2 2 Z2 C

X3 Y3 Z3

+By +Cz1 +0=0 + By2 + Cz2 + I) = 0

Ax3 + By3 +0z3 +D=0

has a solution for some D 0.

{ience it must be that

But since

-4_x1 + + ---z1 + i = o

-4--x2 + --2 + + i =

A B C ----x3 + + Z3 + i = O

xl yl zl

X2 y2 z2 AO

X3 y3 z3

this solution is unique for A/D, B/D, C/i).

Theorem 5. Any 3 points of a plane a which are

not in the same straight line have x incident upon them.

Proof. The 3 points (x1, y1, z1) (x2, y2, z2) (x3, y3, z3)

satisfy

1K

+ By1 + Cz1 + i) = O

Ax2 + By2 + Cz2 + D = O

Ax3 + By3 + Cz3 + D = O

where Ax +By +Cz +D=O is the plane in question.

Theorem 6. If a plane contains two points of a

line then it contains every point of the line.

Proof. Suppose a plane Ax + By + Cz + D = O contains

two points, (x1, y1, z1) and (x2, y2, z2), then

Ax1 + By1 + Cz1 + D = O

Ax2 + By2 + Cz2 + D 0.

Now there exists a line,

A'x+B'y+C'z+D' =0

A"x + B"y + C"z + " = O

incident upon (x1 y1 z1) and (x2 y2 z2) and

(A' B' c'\ Rank =2.

\ÀtI B" C")

Therefore

¡A B

Rank A' B' C' 2

A" B" C"

It is impossible that the rank be 3 since the

equations

13

Ax + By + Cz + 1) = O

A'x+B'y+C'z+D' =0 A"x + B"y + Cttz + D" = O

have more than one solution. Jience the rank is 2.

Therefore there exists a k 0 such that either (kA, kB, ko) = (n', B', C')

or

(kA, kB, kC) = (ATI, ti ,

3uppose, without loss of generality, that it is the first of these condition which is true. Then

-D' = kAx1 + kBy1 + kCz1 = -kD

hence

D' = kA)

A'x + B'y + C'z + i)' = O

and

.x + By + C7 + D = O

are equivalent. hence every solution to the line is a

solution to the plane.

Theorem 7. If two distinct planes have a point in common then there exists at least one other point in couimon with the planes. i>roof. Suppose planes

Ax + By + Cz + D = O

anJ

A'x + B'y + C'z + U' = O

have a point (x0, y0, z0) in common.

L$ince the planes have a point in common

(A B Rank

( t =2

A' B' C')

And therefore there exists a line

Ax + By + Cz + i) = O 'x+B'y+C'z+D' =0

such that each point of the line is a point of both

planes. Then by theorem 3a, the present theorem is

proved.

lL.

Theorem 8. There exist 4 points which are not

in any one plane.

froof. Consider the points, Pi = (0, 0, 0), P2 =

(1, 0, 0), P3 = (0, 1, o), P4 = (o, o, 1).

If there were a plane

Ax + By + Cz + D = O

which contained E1, then D = O. If it contained P2,

then A = -D = 0. If it contained P3, then B = -D = O.

If it contained .L4 then C = -D = O. But then

A2 + B2 + C2 o.

15

II. AXIOMS OF ORDER

Lemma 2. The set of all points on a line

x+By+Cz+D =0

A'x+B'y+C'z+D' =0

is the saine as the set of all points (x, y, z) such that

X = X0 + lt

y = y0 + mt

z z + nt, o

where (x0, y0, z0) is any point of the line, 1, in, n are

real constants, i + in + n = i and t takes on all real

values.

Proof. It was shown in the proof of theorem a that

there exists a point (r', y', z') on the line such that

every point (x, y, z) where

X = X' + i't'

y = y' + iiì't'

z = z' + n't'

1', ni', n' are real constants and t' assumes every real

value, is a point of the line and that

k = (lt)2 + (m')2 + (n')2 O

Now let t" = t'/f

i= in =

n = n'/J

16

then x = x' + it"

y = y' + mt"

z = z' + nt"

where t" assumes all real values, is an equivalent set

of equations.

Also i2 + + 1.

Now since (x0, y0, z0) is on the line, there exists a

t such that,

1ow if

then

X = X' + its, o o

yo = Y' + mt

z =z' +nt" o o

xl = X' + lt:'Ji

= y' + mt

z1 = y' + nt

x = x + l(t -t) 1 o

y1 y0 + m(t -t) z1 = z + n(t -t)

o

Hence if we let T = t "-T " i i o

xl = Xo + ltl

y1 = y + mt1

z1 = z0 + nt1

Now assume that (x9 y2 z2) is on the line. Then

Ax2+By2+Cz2+D =0 + B'y2 + C'z2 + D' O

and since

we have

Also

Hence

Ax +By +Cz +D =0 o o o

A'x0 + B'y0 + C'z0 + D' = O

ì(x -x ) + B(y0-y2) + C(z0-z2) = O 02 A'(x0-x2) +B'(y0-y2) +C'(z0-z2) = O

+ B(y1-y0) + C(z1-z) = o

-x ) + B'(y1-y0) + C'(z1-z0) = O lo

Al + Bin + Cn = O

i.'l + B'm + C'ri = O

fA B c\ Now since

( has rank 2, there exists a

IA' B' C'

t2 O such that

Hence

xo-x2 = t2l

yo-y2 = -t2m

z0-z2 = -t2n

X =X +tl 2 o 2

y2 y0 + t2in

z2 = z0 + t2n

17

Since the steps can be reversed, the lemma is proved in

both directions.

Remark: These equations, which are equivalent

tOILX+ By +Cz D= O, A'x+B'y+C'z +D' =Oare referred to as the parametric equations of the line with

respect to the point (x0, y0, z0).

Jietadef 6. Let

X = X0 + lt

y = y0 + mt

z = z0 + nt

be the parametric equations of a line with respect to

(x0 y0 z0). e say that three distinct points on the

line, A, B and C, are such that B is between A and C

(written ABC) 1ff one of the following is true,

t\> tB'> t or tK tB< ta,

where tA tB t are the values of t corresponding to

A, B and C in the above equations.

Note that if (x1, y1, z1) (x0, y0, z0) is on

the line, we will have

X = X1 + it'

y = y1 + mt'

z = z1 + nt'

is another set of parametric equations for the line.

1f t], t1, t are the values corresponding to A, B and

19

C then it is obvious that if tB is between tA and t0 in

the real ordering then t is between tk and t in the

real ordering.

Theorem 9. If A, B and C are points, then BC

implies OBA.

Proof. By symmetry.

Theorem 10. If A, B and C are points, then at

most one of ABC, BAC and ACB hold.

iroof. At most one of three real numbers is between

the other two in the ordering of the real numbers.

Theorem 11. If A and B are distinct points of

a line, then there exists a point C of the line such

that B is between A and C.

Proof. By lemma 2, the parametric equations in the

metadefinition of 'between' exist. uppose tA corre-

sponds to point A, tB corresponds to point t3 (with

respect to some point (x0, y0, z0)), then

tA < tß implies that there exists t such

that tA < tB t0

tA > tB implies that there exists t such

that tA > tB > to.

20

Definition 3. A line segment AB is the set of

all points between A and B, where A and B are distinct

points.

Definition 4. A closed line segment is the

union of AB and A y B , where A and B are distinct

points.

Definition 5. i. triangle L. ABC is defined for

3 points A, B, C such that they are not all in any one

line and is the set y y

Theorem 12. If a line 1, in the plane a, of the

points P, R (P, Q and R are not in the same straight

line) has a point of PQ y QR y RF on it, then it has

another distinct point of the triangle fR on it. Proof. Assume that the equation of the plane a is

A'x + B'y + C'z + D' = 0. Since line i has all its

points in the plane a, it may be represented by

A'x+B'y+C'z+ L)' =0

Ax + By + Cz + D = O

(A' B' '

where rank Ç

= 2, as was shown in the proof A B C

of theorem 6.

Now let

21

P = (x , y, z ) P p

= (Xq Yq z ) q

R = (Xr r' Zr

Then by lemma 2 there exists a representation of the

line of f and as follows:

X = X + lt

y = y + mt

Z = Z + nt p

3ince i has a point in common with the line of

P<,, the function

f(t) = i.(x0 + it) + B(y0 + mt) + C(z0 + nt)

+D

must have the value zero at some point of PQ. We have

two cases, since f(t) is a linear function of t.

(1) f(t) is identically zero.

Then the theorem is proved since there must exist

another point on PQ in common with 1.

(2) f(t1) = o for exactly one value of t. This

value of t corresponds to a point of P. and therefore

o=t (t <'dt p l\ q

where tq is the value of t for point .

Therefore, since f is linear on t, f(t) and

f(tq) have opposite sign.

22

Now in exactly symmetrical fashion we define

g(t') and h(t") to be the corresponding functions with

respect to the lines of and RF respectively.

Then we have,

h(ttI) = f(t) ß(t'q) = f(tq)

and

h(t") =

Hence h(tI') and g(t'q) have opposite signs. Now

since h(t") = g(t') it must be that there is a zero of

h(t") over the segment lIP or a zero of g(t') over the

segment ¿R or that h(t"r) = (t') = O. In any case the

theorem follows.

Remark:

X,

(1)... y' =

z'

III.

A transformation

¡a11 a12 a13\

(

&)] a22 a23 y

\a31 a32 a33 z

is denoted

(aij), (ai)

where

(a11 a12 a13

(aij) = f

a a22 a23

\a31 a32 a33

a1

(ai) = a2

a3

a1

a

a7

23

The statement of equation (1) is written

() =

(aij), (ai)1

() The product of two transformations (aij), (ai)' and

(bij), (bi) is written

(bij), (bi) (aij), (ai)

= (bij)(aij), (bij)(ai) + (bi)

24

Lemma 5.

(bij), (bi) ' (aij), (ai) ()

= (bij), (bi) (ai), (ai) (y)]

Proof. Follows from the above definition.

Remark: (aij), (ai)1 is a riRid motion if f

(1) all the a's in (aij) and (ai) are real.

(2) (aij)(aijì I

where (aij)T = the transpose of (aij)

and I = the 3 x 3 identity matrix.

() Iaii\ = i

Lemma 4. The set of all rigid motions with the

prouct mapping forms a group, G.

:roof. (1) The operation is associative.

(2) To establish closure we must show

(B)(BA)T i

where A = (aij) B = (bij)

and (aij),(ai)' , (bij), (bi) are rigid motions.

If we denote the right inverse of a matrix X by X1, we

have

(BA) = A1B1 ÀTBT = (BÂ)T

25

iso we note that

det (BA) = det Bdet A = 11 1.

(3) The identity is

" (

(4) (aij), (ai) -1 = (aij)(aij)T(ai)

where -1 stands for the right inverse of a rigid motion

with respect to the ;roup operation and identity.

Lemma 5. A transformation (aij), (ai)

is a member of the group G only if aij\ = i and

(1)

= i for i = 1, 2, 3

(2)

= O for all i Ai'

,j=l

(3)

a12 = i for i = 1, 2, 3

(4)

= O for all j A j'

j j;

all hold. ither conditions i and 2 or conditions 3 and

4 are sufficieLt for (aij), (ai) to be in G if

f

aij =

26

Proof. (aij)(ai)1 conditions i and 2

(aid), (ai) G (aij)T - (alj)T(ai) G

T ..TT (aij) (aij) = (aij) (aij) =

conditions 3 and 4.

The sufficiency statement is obtained by rever-

sing the steps in the above arguments.

Netadef 7. An ordered pair of distinct points

(F], 12) is congruent to an ordered pair of distinct

points (1Di, £2 ) , (written i] 2) E (.E ' ,

P2' ) where

Ii = (x1 y1 z1) P2 = (x2 y2 z2)

'ji' = (x1' y1' z1 ) P2' = (x2' y2' z2')

iff that there exists a rigid motion (aij), (ai)5

such that

and

I xi'

I ±

y1'1=

{(aij), (ai) (

y\

z1,) z])

1X2\ f X\

1y2

(aij), (ai) ( yj

z2'

Remark. We will later change the above nietadef.

Lemnia 6.

1. (i:i, I2) is congruent to (iIi' , 12 (' )

27

is congruent to (, k2). 2. (p1, P2) E (P1', Pj)

and

implies (P1, P2) (P1', 12)

l' Fi' P2)

Proof. Follows from the fact that G is a group.

Lemma 7. (Fi, (P1',p2t)irr

2 2 + + (z1-z2) = (x1'-x2')2 + (y1'-y2')2

+ (z1'-z2')2

2 2 Proof. Let u = x1-x2 , r = u + u X X y

uy =

uz = zl_z2 = r2 + 2

z

then we have that at least one of u, u, u is not

zero since the points are distinct. Assume either

uy O or u O and hence r O and b O. 'e define

a transformation, A, as follows:

( r/ O u/\ o\

( ui'r u/r O

A = O i O , O . u/r u/r O

u/ O -r/L o )

o o i

(X2\

, I2 Z2/)

It is easy to verify that is a rigid motion and

o X2'\

O A y2

o z2

= fi

o z1

If both x1 = X2 and y1 = y2 then the same result

is easily obtained by interchanging the second and third

rows of each matrix in A, and writing r = u2 + u2

and interchanging u and u in each matrix.

Hence by repeated application of the previous

lemma,

°l' '2 ((o, O, O), (c, O, O)

Similarly we can show that

(P1', F21)a ((o, o, O), (c', O, O)

where b' = (x1'-x2) + (y1'-y2')2 + (z1'-z2')2

But we have by hypothesis

(J' = b.

Hence by the previous lemma,

(F1, k2) (E1' , F2' )

The steps are reversible so the implication holds in the

other direction.

29

Leinìa 8. (Pi, k2) = 2' r1) where P1 and k2

are arbitrary distinct points.

Proof. Follows from the previous lemma since

+ (y1-y2)2 + (z1-z2)2

+ (y2-y1)2 + (z2-z1)2

where P (x, y, z) P2 = (x2, y2, z2)

ietadef 7'. The metadef 7 for congruence is now

changed by striking the word 'ordered' wherever it

appears.

Theorem 1. Congruence of pairs of points is

transitive and symmetric.

Proof. Follows from the preceeding lemmas.

Definition 6. A ray from a point P is a set con-

taming a given point B and all points Y such that YPB

is not true and Y is on the line of IB.

Lemma 9. t set of points X is a ray from a point

(x0, y0, z0) iff there exists a line 1 represented by

X = X1 + lt (where x1, y1, z1 is

y = y1 4 mt any point of the line)

z = z1 + nt

and

30

X = X + lt o i o

yo = Yi + mt

z o =z

i +nt

and (x, y,, z), (XB, B' ZB) are members of X imply

that eì.ther

t > to and tB > to

tA Kt0 and tBK t o

where XA = X1 + itA

YA=Yl+mtA

ZA = Z1 + fltA

and XB=Xl+ltB

= yl + mt3

ZB = Z1 + fltB

i'roof. Follows from lemma 2, metadef 6 and definition 6.

Theorem 14. In a ray from a point I there exists

a unique point such that F E RS where R and are

arbitrary distinct points.

iroof. Let F = (x y, z, = (Xq Yq Zq) R =

(Xr r' zr) S = (x5, '

za). !e have for any point

(x1, y1, z1) on a ray from P that

X = X + lt 1 p 1

y1 = y + mt1

z1 = z + nt1

Then if (x2, y2, z2) is another point on this ray

31

X = X + lt 2 p 2

y2 = y + mt2

z = z + nt 2 p 2

and it follows from lemma 9 that t1 and t2 have the

same non-zero sign. hence

(x1-x)2 + (y1-y)2 + (z1z0) =

+ z2-ç2 = \t2\

since i2 + m2 + = 1. Hence if

it follows that there exists a uni'ue point on the ray

such that

x = x ± 1(t\

y = y +

z = z ± n\t

where either the + or the - sign applies according to

whether the ray is characterized by positive or negative

values of t.

Theorem 15. If ABC on a line i and if A'B'C' on

a line 1' and AB A'B', BC B'C' then AC A'C'.

Proof. Let

= a ) '' (a', a' a' it =(a,

B =(b, y) B' = (b', b' b's)

=(c, c, c) ' = (c'y, c'i, c's)

Then the line i is represented by

X = b + lt X

=b+mt z=b +nt

z

and line 1' by

X' =b' +l't' X

y' =b' +m't' y

z' =b' +n't' z

Let ta tb t and ta tb t' represent the

values of t and t' corresponding to ., B, C and .', B',

C' on lines 1 and 1' respecLively. Then

t =t' =0 b

and we may suppose that ta O and hence t O. Now,

in this case, if t'a O we let ni" = -ni', n" = -n' and

1" = -1'. iLvidently this new set of equations if equi-

valent to the old set. So in any case we have

ta > O , t'a> O

tc<o , t'c<o

ïlso, it is evident that

Then

AB A'B' implies t'a = ta

BC B'C' implies t' = t

(a-c) + (a_c)2 + (a-c)2 = ta -

= t' - t' - J(a' -c' )2

+ (a'-c')2+(a'-a')2 a c x x

Jefinition 7. If we have two distinct rays r1 and

r2 from a point A we define an angle 4 (ri, A, r2) as

the ordered set (r1, A, r2).

hemark. e will later change the definition of

angle.

Definition 8. A straight angle is an angle 3 (r1, A, r2) such that the points of r1 and r2 are on the

same line.

rietadef 8. 4 (r1, ¡, r2) 2. (r1',A', r2')

(read congruent) iff there exists a rigid motion which

takes points of r1 to points of r1' and points of r, to

points of r2' and to '.

Lemma 10. A rigid motion takes points of lines

into points of lines. Therefore if a rigid motion takes

two distinct points of a line into two image points, the

rigid motion takes the line of the points into the line

3L.

of the image points.

Proof. A rigid motion (aij), (ai) takes points of

the line

A1 B1 Cl'\ x D1

O J

A2 B2 02/ y

z

into points of the line

(o A1 B1 C1 Ç(aij), (ai) y +

A B2 02 )

The 1econd part of the lemma follows from theorem 2.

Lemma 11. (r1 P1 r2) 4 (r2 r1)

Proof. Let P1 = (x1 y1 z1) and P2 be a point of r1 and

let k2 = (x2, y2, z2).

Now the rigid motion 'A' in the proof of lemma 7

takes P3 into (O, O, O) and P2 into (ci, O, C). Hence

by the preceding lemma, it takes r1 into a ray r1' from

(O, O, o) and containing all points of the form (x, O O)

where x as is obviously verifiable.

Now on r2 there is a point Z such that

(P1, z) ( (O, O, o), (O, o, l)

hence the transformation A takes Z into a point Z' =

(1, m, n) such that i2 + in2 + n2 = i. Let the image of

r2 under be r21. Then (1, m, n)E r2'.

35

If we can find a rigid motion B, sucí that B

takes r1' into r2' and. vice versa, the present lemma

would follow since ABÀ will take r into r1 and vice

versa.

íe want to find a rigid motion B such that

I/o o

(c B O

\; O

i

m = B O

n O

i

= B O m

O n

Such a rigid motion is,

n

B -1-n2

= In

n\ O

rim u

-l-m2 O 1+1

Definition 7'. henceforth definition 7 should

omit the word 'ordered'.

)

Theorem 16. An angle is congruent to itself.

Proof. Follows from the previous lemma and the fact

r

t}iat , O transforms rays and points into them-

O) selves.

Definition 9. A side of a line in a plane is a

set of points containing a given point A not on the line,

and all points X in the plane of A and the line, such

that there does not exist a point of the line such that

XQA.

Remark. Je will use the expression "A rigid mo-

tion takes planes into planest' to mean that all the points

of a plane are taken into all the points of an image plane

by the rigid motion. Similarly we will sa,y that 'lines

are taken into lines' to mean that points of lines are

taken into points of lines.

Lemma li. A rigid motion takes a plane to an

image plane.

Proof. If A + B + C + D = O is a plane, and

(aii)(ai) is a rigid motion and

Ax0+By+Cz+f)0 and

() = {(ai)(ai)

37

then we have

where

A'x1 + B'y1 + C'z1 + D' = O

(aij)

=

and D' = -A'a1 - B'a2 - C'a3 + D.

Lemma 12. A set of points X is a side of a line

A'x+B'y+C'z+D' =0 Ax + By + Cz + i) = O

in a plane

Â'x+B'y+C'z +D' =0

if f

(x1, y1, z1) X and (x2, y2, z2) ( X

imp li es

Ax1 + By1 + Cz1 + 1)

has the same non-zero sign as does

+ By2 + Cz2 + D.

froof. If the points (x1, y1, z1) and (x2, y2, z2) have

no point of 1 between them and the line of (x1, y1, z1)

and (x2, y2, z2) is represented by

X = X0 + lt

y = y0 + mt

z = z0 + nt

we define

f(t) = .(x0+lt) + í3(y0-i-mt) + C(z0+nt) +

then in the range t1 t t2 or t2

ever applies), f(t) has no zero values.

a linear function of t, f(t1) and .f(t2)

same sign. Reversing the steps we see

follows.

j)

t t1 (which

.5ince f(t) is

must have the

that the converse

Lemma 13. A rigid motion takes a side of a

plane with respect to a line into a side of the image

plane with respect to the image of the line.

Iroof. The proof follows from the fact that lines are

transformed into lines and hence the line determined by

two points on a given side of the plane goes into an

image line. This image line does not intersect the image

of the line defining the side of the plane because

Uistances' are preserved and hence a non-zero 'distance'

cannot go into a zero 'distance' as would be the case if

the image lines had a common point.

Theorem 17. If 1 is a line in a

point of 1, r1 is a ray from P on 1, an

1 in a, there exists a ray r2 in X such

4 (r1, A, r2) is congruent to a given

B, r2') in plane .

plane cx, P is a

X is a side of

that the angle

angle (r1',

39

Froof. Let the equation of cx. be

Ax + By + Cz + Ii = O

where

A2 + B2 + C2 O.

ì'ow consider the matrix,

O O A/k

A = i O B/k

o i C/k

where k = A2 + B2 + C2 O.

Obviously the columns of A are linearly independent and

(A/k)2 + (B/k)2 + (C/k)2 = i.

Hence by the Gram-Schmidt process we can find a matrix

(a11

B

i

a12 /k

a22 B/k

a32 C/k

where the columns are mutually orthogonal and are normal.

Je notice that

B O B

i C

Further the determinant of B is either ±1 and (if

it is -1) can be made +1 by multiplying one of the first

two columns by -1. Hence by lemma 11, since

o

i

the rigid motion

into the plane

z = K

J.

= B

B' (\

40

takes tx + By + Cz + D

where K is some constant. Further, the rigid motion

( I, (-K\ will take z = K into the plane z = 0.

\-KI)

vie also note that sides of the plane are taken into sides

of the image plane, z = 0, by lemma l. Let 1' be the

image of the line 1 in the plane z = O and let the image

of the point i' be (x', y', 0). Then the ri!id motion

I, (:;

takes points of z = O into points of z = O and the point

(x', y', 0) into (0, 0, o). Now the image of 1' under

this last transformation is a line 1". There exists a

point (x", y", O) on i" such that + y"2 = i.

Now the rigid motion

¡x' Y' O (o

(-i' X' o

\ O i O

14.1

takes points of z = O to points of z = O, (O, O, O)

into itself and (x", y", O) into (1, 0, 0).

Hence in the proof of the theorem we will assume

that these transformations have ail been applied to the

orií"inal plane and that the image of line i is y = O

z=O

(by lemnia 10) and that the image of F is (O, O, O).

Then if we can prove the theorem for this particular line

and this particular point on the line and in the plane

(z = 0), we can then apply the inverse transformations

one at a time and prove the theorem in general.

Je first notice that there exist exactly two

sides of the plane with respect to the line y = O

z =0

The one side is given by the points (x, y, o) where

y > O and the other by the points (x, y, O) where

y 'Z.° That these sets of points each constitutes a

side of the plane is obvious. To show that there cannot

exist another side of the plane with respect to the given

line we note that two distinct sides of a plane with

respect to a given line cannot have a non-empty inter-

section. This is obvious when we consider the definitions

involved.

hence let us suppose that the side of the plane

in question is given by the set (x, y, 0) y > O

\e may also suppose that the ray in question is the ray

(x, O, O): x > O since the aruinents will be sym-

metric for all four cases.

Now it is obvious that the plane 1 may be trans-

formed by rigid motions into the plane z = O and the

ray r1' into the ray (x, O, O) x> O . Now if the

image of r2' is in the side (x, y, O): y> O we

can pause for a moment. If the image ray of r2' is in

the other side of the plane, the rigid motion

ç(i o o o

(o -i o o

\ \O o -i O

will take it into (x, y, O) : y O and leave the

other ray and the plane of the two rays unchanged.

Hence the existence part of the theorem is proved since

an angle is congruent to itself.

The uniqueness part is next proved. 3uppose

there is another distinct ray r3 from (O, O, O) in the

given side of the plane such that the two angles are

congruent. Then there exists a point (x, y, O) where

x2 + y2 i on this ray. It is obvious that any rigid

motion which leaves the ray (x, O, O) x > O fixed

must leave ______________

+ »2

fixed where x' and yt are the imace points of x and y

4.3

and also we have x + y = 1. It is ap:arent that

these two conditions completely determine the point

(x', y', O) and further that (x', y', O) = (x, y, O).

This proves the theorem since the line of r3 must be

the line of r2.

Derinition. In a triangle ABC we define A3C

as the angle 4 rBr where r1 contains A and r2 con-

tains B.

that

Theorem 18. If in A's ABC and A'B'C' we have

ABC A'B'C'

AB E A'B'

and BC B'C'

then BAC E ß'A'C'.

Iroof. B,y the same series of transformations used in

theorem 17, we can transform 4 's ABC and A'B'C' into

the angle 3 (r1, P, r2) where i = (O, O, O), r1 =

(- (x, O, O)t X > O and r2 is a ray from i in the

plane z = O. Under this set of transformations it is

obvious that the image points of A and A', of B and 13'

will coincide. Hence BaC and - B'A'C' coincide.

IV. PLAYFAIR'S AXIOM

Theorem 19. If

and a point I- not on 1

containing P such that

of' 1. (Flayfair's xi

.iroof. Let Ax + By +

x + By +

'x + B'y

represent 1. ¡ny line

by

44

in a plane cx we have a line 1

then there exists a unique line

it is in cx and contains no point

)Ifl)

Cz + J = O represent cx and 1

Cz + J = O

+ C'z + D' = O

1' which is in a can be represented

Ax + By + Cz + t) = O

A"x+B"y+C"z+D" =0.

Now is 1' does not have a point in common witi 1, then

(A B C

Rank A' B' C' 2.

A B" CT

Hence since Rank (A B = 2, the rank

\A' B' c'i

of the first matrix must be exactly 2. Therefore (since

Rank (A B C = 2) we have that (A', B", C") and

" c"l

(A', B', C') must be linearly dependent. iience

( , I, ' ' I '\ i ( t -i: T I

b , = , i. ,

where k 0. Low if 1' is to contain (x0, y0, z0) then

Ll.5

DU = - - 13??y - =

o o o

-k(Á'x0 + B'y0 + C'z).

Therefore a unique line 1' exists which satisfies the

stated conditions.

L46

V. ARCHIMED.ES' AXIOM

Definition. A set of points B1, B2

on a line is liiearly ordered iff i < i Ç k is equi-

valent to B.. Bk.

Theorem 20. For any seent XI there exists a

sequence B1, B2, B3 .... of points on a line 1, such

that the sequence is linearly ordered and BB+i XY

for all i, and such that for any point Z on i there

exists a B such that B1 'Z1 B.

roof. For any line i with parametric equations

X = X0 + lt

y = y0 + mt

z = z0 + nt,

let B1 = (x0, y0, z0). Then there exists a t1 O

such that the points (x0 ± it1, y0 mt1, z0 ± nt1)

and (x0, y0, z0) are congruent to X and Y. It follows

that the sequences

= X0 ± lit1, y0 ± lit1, z0 ± lit1

are each linearly ordered. Now if the point ¿ is

(xi, y, z) then there exists at Z such that

4.7

X z

= X o

+ lt z

yz = Y0 + lflt

zz = Yo + fltz

If te.) O then the sequence contains

a point Bk = (x0 + lkt1, y0 + inkt1, z0 + nkt1) where

kt1 ) t) O. If t < O a symmetric proof is required.

VI. THE CO1PLETENESS AXION

The completeness axiom of geometry reads as

follows:

If we have a system of points, lines and planes

which satisfy the previous axioms, then it is impossible

to add additional points to the set of points on a line

and still have a system which satisfies the axioms.

It then follows as a theorem of geometry that

the set of points on a line is the set of real numbers.

since the set of points on a line in the present model

is the set of real numbers, it follows that we cannot

add points to this set without contradicting the above

theorem of geometry. Hence the completeness axiom is

satisfied.

4.9

BIBLIOGRAIHY

1. Hubert, David. Grundlagender Geometrie 8. Auf 1, mit Revisionen and rganzungen von iaul Bernays, L;tuttgart, Teubner, 1956. 271 p. (Followed in lecture notes of Dr. Harry Gohoen, OSU Nathematics Dept., 1961.)

2. Landau, Edmund Georg Hermann. Foundations of analysis. Trans. by Steinhar3t. New York, Chelsea, 1951. 134 p.