Born within the range of 325 B.C in Greece and died around 265

B.C. Euclid most likely came from affluent family because it was

widely known that he actually enrolled and finished from the school

of Plato in the old Greece. Following his education, he got teaching

appointment in Alexandria, Egypt. He was in there when he wrote his

popular book “The elements”. To buttress this fact Euclid of

Alexandria is usually mistaken or confused as Euclid of Megara,

another Socratic philosopher. His book “The elements” is what really

distinguished Euclid from the rest. Euclidean Geometry is the content

of the book and is very useful in the elementary and middle class.

GEOMETRY AROUND US

Our daily life is filled with geometry—the pure mathematics of points, lines, curves and surfaces. We can observe various shapes and angles in the objects that surround us. Observe, for example, this table and its rectangular surface; the boomerang and its angular shape; the bangle and its circular shape.

Euclid, an ancient Greek mathematician, observed the various types of objects around him and tried to define the most basic components of those objects. He proposed twenty-three definitions based on his studies of space and the objects visible in daily life. Let us go through this lesson to learn each of Euclid’s definitions.

3

Eu

clid

`s G

eo

me

try

PREFACE

Euclidean geometry is a mathematical system attributed to the

Alexandrian Greek mathematician Euclid , which he described in his

textbook on geometry : the Elements Euclid's method consists in

assuming a small set of intuitively appealing axioms , and deducing

many other propositions (theorems ) from these. Although many of

Euclid's results had been stated by earlier mathematicians, Euclid

was the first to show how these propositions could fit into a

comprehensive deductive and logical system . The Elements begins

with plane geometry, still taught in secondary school as the first

axiomatic system and the first examples of formal proof . It goes on

to the solid geometry of three dimensions . Much of the Elements

states results of what are now called algebra and number theory

,explained in geometrical language .This entire project aims

at the explanation of the very complicated Euclid geometry , so that the

students are able to see and understand it in a better way.

4

Eu

clid

`s G

eo

me

try

INTRODUCTION TO

EUCLID`S GEOMETRY

Eu

clid

`s G

eo

me

try

5

Mat

hem

atics

pre

sent

atio

n

CONTENTS

Definitions of Euclid

Introduction to axioms

Axioms-

Axiom i and ii

Axiom iii

Axiom iv and v

Axiom vi and vii

Introduction to

Postulates

Postulates-

Postulate 1

Postulate 2

Postulate 3

Postulate 4

Postulate 5

6

Eu

clid

`s G

eo

me

try

DEFINITIONS OF EUCLID

Euclid gave the definitions of a few very basic attributes of objects that are normally around us. These

definitions are listed below.

1. A point is that which has no part.

2. A line is a breadth-less length.

3. The extremities of a line are called points.

4. A straight line is one that lies evenly with the points on itself.

5. A surface is that which has length and breadth only.

6. The edges of a surface are lines.

7. A plane surface is one that lies evenly with the straight lines on itself.

7

Eu

clid

`s G

eo

me

try

8. A plane angle is the inclination to each other of two lines in a plane, which meet each other and do not lie in a straight line.

9. When the lines containing the angle are straight, the angle is called rectilinear.

10. When a straight line set up on another straight line makes the adjacent angles equal to each other, each of the equal angles is right and the straight line standing on the other is called a perpendicular to that on which it stands.

11. An obtuse angle is an angle greater than the right angle.

12. An acute angle is an angle less than the right angle.

13. A boundary points out the limit or extent of something.

14. A figure is that which is contained by any boundary or boundaries.

8

Eu

clid

`s G

eo

me

try

INTRODUCTION TO AXIOMS

Euclid’s Axioms

Euclid assumed certain properties to be universal truths that did not need to be proved. He classified these properties as axioms and postulates. The properties that were not specific to geometry were referred to as common notions or axioms.

He compiled all the known mathematical works of his time into the Elements. Each book of the Elements contains a series of propositions or theorems, varying in number from about ten to hundred. These propositions or theorems are preceded by definitions. In Book I, twenty-three definitions are followed by five postulates. Five common notions or axioms are listed after the postulates.

In this lesson, we will study some of Euclid’s axioms.9

Eu

clid

`s G

eo

me

try

AXIOMS:

Things that are equal to the same thing are also

equal to one another (Transitive property of

equality).

If equals are added to equals, then the wholes are

equal.

If equals are subtracted from equals, then the

remainders are equal.

Things that coincide with one another are equal to

one another (Reflexive Property).

The whole is greater than the part.

10

Eu

clid

`s G

eo

me

try

AXIOMS I AND II

Let us start with the first axiom which states that things that are equal to the same thing are also equal to one another.

Let us suppose the area of a rectangle is equal to the area of a triangle and the area of that triangle is equal to the area of a square. Then, according to the first axiom, the area of the rectangle is equal to the area of the square. Similarly, if a = b and b = c, then we can say that a = c.

Now, the second axiom states that if equals are added to equals, then the wholes are equal.

Let us take a line segment AD in which AB = CD.

Let us add BC to both sides of the above relation (‘equals are added’). Then, according to the second axiom, we can say that AB + BC = CD + BC, i.e., AC = BD. 11

Eu

clid

`s G

eo

me

try

AXIOM III

The third axiom states that if equals are subtracted from

equals, then the remainders are equal.

Let us consider the following rectangles ABCD and PQRS.

Suppose the areas of the rectangles are equal. Now, let us

remove a triangle XYZ (as shown in the figure) from each

rectangle. Then, according to the third axiom, we can say that

the area of the remaining portion of rectangle ABCD is equal

to the area of the remaining portion of rectangle PQRS.

12

Eu

clid

`s G

eo

me

try

AXIOM IV AND V

The fourth axiom states that things that coincide with one another are

equal to one another.

This axiom is sometimes used in geometrical proofs.

Let us consider a point Q lying between points P and R of a line

segment PR, as is shown in the figure.

We can see that (PQ + QR) coincides with the line segment PR. So,

as per the fourth axiom, we can say that PQ + QR = PR.

Now, the fifth axiom states that the whole is greater than the part.

Let us again consider the line segment PR shown above. We can

see that PQ is a part of PR. So, as per the fifth axiom, we can say

that PR (i.e., the whole) is greater than PQ (i.e., the part).

Mathematically, we write it as PR > PQ.

13

Eu

clid

`s G

eo

me

try

AXIOM VI AND VII

The sixth and seventh axioms are interrelated. The former states that things that are

double of the same things are equal to one another, while the latter states that things that

are halves of the same things are equal to one another.

Let us consider two identical circles with radii r1 and r2. Also, suppose their diameters

are d1 and d2 respectively.

As the circles are identical, their radii are equal.

∴ r1 = r2

Now, as per the sixth axiom, we can say that 2r1 = 2r2

∴ d1 = d2

Hence, we can say that if two circles have equal radii, then their diameters are also

equal.

Now, instead of taking the radii as equal, let us say that the diameters of the two circles

are equal. Then, as per the seventh axiom, we can say that the radii of the two circles are

also equal.

14

Eu

clid

`s G

eo

me

try

INTRODUCTION TO POSTULATES

Certain things are considered universal truths that

need not be proved. Consider, for example, the

following: the sun rises from the east; Sunday

comes after Saturday; March has 31 days. These

things are universally true; hence, they do not need

to be proven.

Similarly, certain geometrical properties are

regarded as universal truths. Euclid identified and

presented such properties in the Elements. The

properties specific to geometry were classified by

him as postulates. In Book I, twenty-three

definitions are followed by five postulates. Let us

learn these postulates in this lesson.15

Eu

clid

`s G

eo

me

try

POSTULATES

"To draw a straight line from any point to any point."

"To produce [extend] a finite straight line continuously in a straight line."

"To describe a circle with any centre and distance [radius]."

"That all right angles are equal to one another."

The parallel postulates : "That, if a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles."

16

Eu

clid

`s G

eo

me

try

1ST POSTULATE

Postulate 1: A straight line may be drawn from any point to any other point.

Proof: Finally proved only yesterday, we must refer to the third and second postulate in order to fully prove this one. In order

to prevent accusations of lack of rigor, I will use the still incomplete third postulate only in those cases where it may be

applied.

Take any two collinear points A and B, where collinear means it is possible to draw a straight line between them. It is possible

therefore to draw a straight line between them.

Now any points on the line AB must also be collinear, for otherwise a straight line could not have been drawn. Hence, it is

also possible to draw a line from A to any point upon the line.

Now, let us rotate the line, such that the collinear point A is the centre of the circle so produced.

Now, it is possible to draw a straight line from A to any point in the circle. This is because the radius of the circle is a straight

line, and upon rotation, it covers all the points in the circle, implying that a straight line can be drawn from all the points

covered by the radius to the center of the circle, which is A.

Therefore all points in the circle are collinear to A i.e. they produce a straight line to A.

It is easy to show that all points in the plane are collinear: merely extend the radius infinitely, so the resultant circle

encompasses the entire region.

Repeating the above for any point in the circle, we see that it is possible to draw a straight line from that point to any other

point in its circle, and so on.

From the information above, we can deduce that all points are collinear to each other, or

It is possible to draw a straight line from any one point to any other point.

Hence proved.

17

Eu

clid

`s G

eo

me

try

2ND POSTULATE

Postulate 2: A finite straight line may be extended indefinitely.

Proof: There are an infinite number of points in a region.

This implies that there are an infinite number of collinear

points, as any operation with infinity that does not involve

another infinity results in infinity. By collinear, I mean points

between which a straight line may be drawn. ( I clarify this in

order to prevent accusations of using a circular argument with

the first postulate)

This implies that a line may be extended infinitely.

18

Eu

clid

`s G

eo

me

try

3RD POSTULATE

Postulate 3: A circle may be drawn with any center and any radius.

Note: By the term "collinear", I mean that it is possible to draw straight line from it another specific point.

Proof: This is a little trickier to prove, so I divided the problem down into two parts. I will first prove that a circle may have any radius.

Taking point A as centre, we may look at the radius as a line. By Postulate 2, we know that line may be extended indefinitely.

Therefore, the radius may be extended indefinitely.

This implies that a circle may have any radius.

The second part is to prove that a circle may have any center.

Taking any collinear point, we see that it is possible to draw a straight line between this and any other straight line.

By rotating the line by 360 degrees, we obtain a circle.

This implies that any collinear point may be the center of a circle, as the straight line that can be drawn may be considered a radius, and rotating the radius produces a circle.

19

Eu

clid

`s G

eo

me

try

4TH POSTULATE

Postulate 4: All right angles are equal to each other.

Proof: Let us assume that this is not true, and all right angles are not equal to each other.

This instantly leads to a contradiction, as it implies that a triangle may have more than one right angle.

Therefore, by reduction ad absurdum, we see that all right angles must be equal to each other.

Another proof is by looking at the definition of a right angle. A right angle is any angle equal to 90 degrees.

We know that 90 = 90 = 90 ...

We see therefore that all right angles are equal to 90 degrees and as 90 degrees is equal to 90 degrees,

this implies that all right angles are equal to each other.

Hence proved. 20

Eu

clid

`s G

eo

me

try

5TH POSTULATE

Postulate 5: If a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which the angles are less than two right angles.

Proof: This means to say that the two lines meet on the side whose sum is less than 180 degrees.

From the diagram, we see that if we extend the lines indefinitely, we eventually get a triangle.It has been proved that the sum of the angles of a triangle sum to 180 degrees. This implies that the two angles formed by the third line which goes through the other two lines) cannot be equal to 180 degrees, as it would then violate the angle sum property of that triangle.The triangle is only possible if the two angles are not equal to 180 degrees or more. This implies that the lines may only meet on those sides where the angles together sum up to less than 180 degrees.

Therefore, this implies that no line which produces an obtuse or acute angle may be parallel to l. Therefore, the only other angle possible is a right angle, which taken in conjunction with the other interior angle forms 180 degrees, implying that any line which produces 90 degrees with the perpendicular is always parallel to line l.

21

Eu

clid

`s G

eo

me

try

Euclid’s division algorithm is a technique to compute the Highest Common Factor(HCF) of two given positive integers. Recall that the HCF of two positive integers aand b is the largest positive integer d that divides both a and b.

Let us see how the algorithm works, through an example first. Suppose we needto find the HCF of the integers 455 and 42. We start with the larger integer, that is,455. Then we use Euclid’s lemma to get455 = 42 × 10 + 35Now consider the divisor 42 and the remainder 35, and apply the division lemmato get42 = 35 × 1 + 7Now consider the divisor 35 and the remainder 7, and apply the division lemmato get35 = 7 × 5 + 0Notice that the remainder has become zero, and we cannot proceed any further.We claim that the HCF of 455 and 42 is the divisor at this stage, i.e., 7. You can easilyverify this by listing all the factors of 455 and 42. Why does this method work? It works because of the following result.

To obtain the HCF of two positive integers, say c and d, with c > d, follow

the steps below:

Step 1 : Apply Euclid’s division lemma, to c and d. So, we find whole numbers, q

and

r such that c = dq + r, 0 ≤ r < d.

Step 2 : If r = 0, d is the HCF of c and d. If r ≠ 0, apply the division lemma to d and

r.

Step 3 : Continue the process till the remainder is zero. The divisor at this stage will

be the required HCF

This algorithm works because HCF (c, d) = HCF (d, r) where the symbol

HCF (c, d) denotes the HCF of c and d, etc.

Example 1 : Use Euclid’s algorithm to find the HCF of 4052 and 12576.Solution :Step 1 : Since 12576 > 4052, we apply the division lemma to 12576 and 4052, to get12576 = 4052 × 3 + 420Step 2 : Since the remainder 420 ≠ 0, we apply the division lemma to 4052 and 420, toget4052 = 420 × 9 + 272Step 3 : We consider the new divisor 420 and the new remainder 272, and apply thedivision lemma to get420 = 272 × 1 + 148We consider the new divisor 272 and the new remainder 148, and apply the divisionlemma to get272 = 148 × 1 + 124We consider the new divisor 148 and the new remainder 124, and apply the divisionlemma to get148 = 124 × 1 + 24We consider the new divisor 124 and the new remainder 24, and apply the divisionlemma to get124 = 24 × 5 + 4

We consider the new divisor 24 and the new remainder 4, and apply the division lemma to get

24 = 4 × 6 + 0

The remainder has now become zero, so our procedure stops. Since the divisor at this

stage is 4, the HCF of 12576 and 4052 is 4.

Notice that 4 = HCF (24, 4) = HCF (124, 24) = HCF (148, 124) =

HCF (272, 148) = HCF (420, 272) = HCF (4052, 420) = HCF (12576, 4052).

Euclid’s division algorithm is not only useful for calculating the HCF of very

large numbers, but also because it is one of the earliest examples of an algorithm that a computer

had been programmed to carry out.

1. Euclid’s division lemma and algorithm are so closely interlinked that people oftencall former as the division algorithm also.2. Although Euclid’s Division Algorithm is stated for only positive integers, it can beextended for all integers except zero, i.e., b ≠ 0. However, we shall not discuss thisaspect here.