The History Of Mathematics

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The History of

MATHEMATICS AN INTRODUCTION

SECOND EDITION

David M. Burton

Some Important Historical Names, Dates* and Events Mathematical

General

Early Beginnings (Before the 6th Century B.C.

30000 8000 2500

Notched wolf bone

B.C.

Ishango bone Table tablets from Nippur

1900

Plimpton 322

1850

Moscow Papyrus

1650

Rhind Papyrus

3300 2600 2100 500 200 700

B.C.)

Menes

Egypt

unites

Great Pyramid at Gizeh

Code

of

Hammurabi

1

Phoenician alphabet

1

Trojan

War

Homer: The Odyssey

Classical Period (6th Century B.C. to 5th Century) B.C.

622-547

Thales of Miletus

585-501

Pythagoras of Samos Theodorus of Cyrene Hippocrates of Chios

470 460-380 ca. 420 408-355 323-285 287-212 262 - 90 ca. 240 ca. 230 ca.

1

A.D.

ca. 75 ca.

100

85

160

ca. ca.

250 260

300 365 395 d. 415 410 4X> ca.

B.C.

558 -486 485 -430

Battle of

469 -399

Socrates

431

388 -323 356

Euclid

Herodotus

480

Hippias of Elis

Eudoxus of Cnidos

Darius the Great

Thermopylae

War Academy

Peloponnesian Plato founds

Alexander the Great

370

Eudemus

Apollonius of Perga

331

Foundation of Alexandria

Nicomedes

213

Books burned

Eratosthenes of Cyrene

212

Fall of

Heron of Alexandria Nicomachus of Gerasa Claudius Ptolemy Diophantus Liu Hui Pappus of Alexandria Theon of Alexandria

195

Rosetta Stone engraved

Archimedes

Hypatia Proclus

*Most dates before 600

B.

123456789

may

instead of being indicated by a group of five vertical strokes.

20

40

30

^ The

60

50

—

A

70

-t

80

>i

j>>

90

9

10

1000

100

^

? $

numbers is as shown in the table. Note that and 1000 are essentially abbreviations for the pictographs used

hieratic system used to represent

the signs for

1,

10, 100,

earlier. In hieroglyphics, the

number 37 had appeared

as

ooo UN III but

in hieratic script

it is

replaced by the less

cumbersome

Ax The

number of symbols called for in this notation imposed an annoying tax on the memory, but the Egyptian scribes no doubt regarded this as justified by its speed and conciseness. The idea of ciphering is one of the decisive steps in the development larger

of numeration, comparable in significance to the Babylonian adoption of the positional principle.

Around

the

fifth

century

B.C..

the Greeks of Ionia also developed a ciphered numeral

They ciphered their numbers by means of the 24 letters of the ordinary Greek alphabet, augmented by 3 obsolete Phoenician letters (the digamma^ for 6, the koppa? for 90, and the sampi'X. for 900). The resulting 27 letters were used as follows. The initial nine letters were associated with the numbers from to 9; the next nine letters represented the first nine system, but with a more extensive set of symbols to be memorized.

1

integral multiples of 10; the final nine letters tiples of 100.

The accompanying

table

were used

shows how

for the first nine integral

mul-

the letters of the alphabet (including

the special forms) were arranged for use as numerals.

Number Recording of the Egyptians and Greeks

15

la

10

2

20

k

7

30 X

4 5 6 >

40 n 50 v 60 £

7 f

70

o

80

7T

3

5e

8

77

still

999 could be represented by

300 r 400 v 500 600 x 700 ^ 800 a;

900X

90?

9

Since the Ionic system was

100 p 200 a

at

a system of additive type, all numbers between most three symbols. The principle is shown by

iM =

700

+

80

+

4

=

1

and

784.

For larger numbers, the following scheme was used. An accent mark placed to the left and below the appropriate unit letter multiplied the corresponding number by 000; 1

thus ,0 represents not 2 but 2000. Tens of thousands were indicated by using a new letter M, from the word myriad (meaning "ten thousand"). The letter placed either next to or below the symbols for a number from 1 to 9999 caused the number to be

M

multiplied by 10,000, as with i

5M,

or

M =

40,000

pv

pvM,or

M =

1,500,000.

With these conventions, the Greeks wrote TMeM,0pM

-

2

The dimensions of encompasses a

4

3

the rectangle are

1

5

+2

+

3+-+/iby«+l,so

+

n)(n

dots. This gives

+

2

+

3

one side of the desired

+

•

•

•

identity.

[1

For the other side, we add the dots

in

+ n2 + + 22 + 32 + + (1 + 2) + (1 + 2 + 3) + •

•

•

+

1)

sum

consecutive squares and rows, to arrive at the 2

it

total of (1

(I

that

)

•

•

+

(1

+

2

+

3

+

•

•

•

+

«)].

Pythagorean Mathematics

101

In algebraic form, our identity (l 2

we

+

22

+

[1

If

+

+

(1

+

32

+

2)

+

•

•

+

(1

n2 )

+

+

3)

2

+

•

2

•

l

+

[1

+

(1

The foregoing first

•

2

/j

+

(1

+

---

+

n)(n

2)

-(1+2 + 3 +

•

+ +

2

+

+

(1

2

+

3

+

+

2

+

3

•

•

•

+

•

•

»)]

becomes

,

3)

+

•

•

= (l+2 + 3 + --- + n)(n + 1). let S = + 22 + 32 + + this S +

the

•

is

•

+

•

(1

+

•

+

*)]

1).

expression can be simplified by appealing to the fact that the

k integers

|

k{k

is

+

making the appropriate

l)/2; after

il? + 111 + 111 +

.

+

.

.

n(< n

substitutions,

sum of we get

n{n+ \Y

+ ^\

which can be written

+

5 + y[l(l

1)

+

+

2(2

1)

+

+

3(3

+

I)

•

•

•

+

n(n

+

= *!±J>_

1)]

.

This yields

S +

4"[( 12

+

22

+

32

+

•

+

•

n2 )

+

+

(1

2

3

+

n(n

+

+

+

•

n)]

=

"(

"

^

^

whence

S + It

now becomes

T

+

n(n

s +

1)

l)

2

a matter of solving for S: _

3_

n(n

""

2

+

l)

2

n(n

+

1)

4

2

n(n

+

l)(2n

+

1)

4

which leads

at

once to

S = All in in

all,

we have shown

that the

n(n

+

sum

of the

1)(2>J

+

first

1)

n squares has a simple expression

terms of n; namely 2 l

+

22

+

3

2

+

•

"

•

+

,

n2

=

n(n

+

l)(2/i

r

+

1) "

•

Chapter

102

The Beginnings of Greek Mathematics

3

A

strikingly original proof of the last result, propounded by the thirteenth-century mathematician Fibonacci, comes from the identity

+

k(k

By

putting

k

= -

1

1, 2, 3,

2

-

.

.

=

3

1)(2*

6

•

.

l

n

,

-

+

ik

-

-

\)k{2k

in turn into this

+

1)

6k 2

.

formula, one gets the set of equations

2

2-3-5 = l-2-3 + 6-2 2 3-4-7 = 2-3-5- + 6-3 2

- 0*(2/! n(n + l)(2/2 + (/i

What

is

important

1) 1)

it is

easily

n(n

(n

When

sides of successive equations.

cancelled,

~ -

(«

2)(/i

-

\)n(2n

common term

that a

is

= =

0(2/i

-

1)

- 3) + + 6/i 2

6(/i

-

l) 2

.

appears on the left-hand and right-hand

these n equations are added and

common

terms

shown that

+

+

0(2/i

=

2

6(1

+

22

+

3

+

2

•

•

•

4-

n 2 ),

leading to the desired conclusion.

3.2

1.

Problems ad

Plutarch (about triangular

number

is

100) stated that

if

multiplied by

and

added, then the result that this 2.

is

fact

and

is

a square

illustrate

it

8,

a

6. 1

Verify that 1225 and 41,616 are simultaneously

square and triangular numbers. [Hint: Finding

is

number. Prove

an integer n such that

geometrically.

Prove that the square of any odd multiple of 3

,.

is

*P

-

the difference of two triangular numbers, specifically that

is

[3(2n+

OP"

W. +

4-

+

Prove that 9/„

4.

+

1

is

if

tm

is

+

-

n

=

2450

0.]

1-

7. 3.

.225

equivalent to solving the quadratic equation

n1 '3.

-

a triangular number, then

also triangular.

An

oblong number counts the number of dots

has columns; the

first

Write each of the following numbers as the sum

5.

(b) 69;

56;

For n

>

1,

(c)

185;

(d) 287.

establish the formula

(2n+

1)>

= (4/.+ 1)'- (40». Oi

=

2

o2

=

6

it

few of these numbers are

of three or fewer triangular numbers: (a)

in

a rectangular array having one more row than

03

=

12

o4

=

20

Pythagorean Mathematics and

oblong number

in general, the nth

=

o„

+

n(n

103

(c)

Prove algebraically and

1).

---

l-2 + 2-3 + 3-4 +

+

n(n

+

l)(/i

+ n(/i+l)

2)

geometrically that:

=

(a) o„

(b)

Any

(c)

o,

+

2

+

4

+

6

oblong number

•

•

+

•

the

is

2«.

Use

[Hint:

sum

of two equal

=

+

k2

+

the identity k(k

1)

k and collect the squares.]

triangular numbers.

+ n 2 = ln o„ - n 2 = n. n 2 + 2o H + (n + l) 2 - (2/i + I) 2 + 2 + 3 + + n + (n + o„ = + (*- 1) + («-2) + •• + 3 +

(d) (e)

1

.

1

1

-(-••

1

-i

5-7

3-5

•3

1

n

.

(f)

8.

t

•

1

1)

is

sum

the

number

of a square

-

(2/i

+1)

l)(2n

'

+

2n

\

2.

Lebesgue proved that (1) every positive

In 1872,

integer

-

* 1

•

Hint: Use the identity

(possibly 1

2

and two triangular numbers, and

)

positive integer

numbers and

sum

the

is

a triangular

(2) every

(2*

of two square

and (e)

Display the consecutive integers

through n

1

in

3

+

3

\

2k

+

(2n

-

1

+ 53 + = n 2 (2n 2 - 1). l

1) 1

-

V2*

2

100.

two rows as

+

1)(2*

1/1

number. Confirm these

results in the cases of the integers 9, 44, 81,

9.

"

+

3

1/ l)

3

follows:

[Hint: Separate the left-hand side of the 2

1

n

3

-

n

n

n

1

-

-

2

identity

n

1

2

1

l

3

+

sum obtained by adding the n columns is set equal to the sum obtained by

If the

+

23

+

33

(2/i)(2/j

+

•

•

•

"

+

(2/z)

3

1)

vertically

adding the two rows horizontally, what well-

known formula

results?

into

odd and even terms and solve

sum

of odd cubes.]

for the

10. Derive the identity 12.

+

[n(n

-

1)

+

-

1)]

=

n\

-

1)

+

1]

+

[n(n

-

1)

+

{In

[n(n

+

3]

(a)

Prove that the

sum

of a finite arithmetic

series equals the product of the

•

number of

terms and half the sum of the two extreme terms; in symbols, this reads

where n 11. For (a)

any positive

is

any integer n 1

+

+

2

(n

(b)^ 1-2

^ 23

+

(a

>

+ 3 + - 1) +

integer.

1,

prove that:

•••

+

+

•

(/!3

+

1)

+

+

2

1

+

n

=

n

2 .

(b)

+

(a

+

-

nd)

+

=

n

2d) [ (a

+ (a + + d) +

3d) (a

1

3-4

+

1

+ n(n

+

1)

1

+ 7 + n(3n - 1)

+

4

•

+

(3«

•

+

(2/i

-

2)

-

1)

and

splitting identity

l/(k

+

1)

=

\/k(k

to rewrite the left-hand side.]

+

1)

1

+

3

+

+ nd)

result of part (a) to confirm the

2

Use the

l/k

(a

identities

+1)

[Hint:

+

d)

Use the

n (li

+

+

5

+

•

'

Chapter

104 13.

The

identity (1

+

=

3

16.

2 4- 3 4-

4

l

2 3 4-

•

•

+

•

3 3 4-

•

was known as early as the a derivation of

+

•

The

tetrahedral numbers count the

n>

3

/i

,

1

century. Provide

first

If the base

is

pyramid

is

formed by placing similarly situated

triangles

upon

the triangle of side

it,

h + h + h +

sides than that

r,

=

Hindu

+

+

n(n

=

tm

+

l)(n

pairs, replacing

_

tk

+

i

even.]

4

where

+

22

3

sum

for the

+

2

•

•

of squares.

=

n2

4-

•

Fill in

4

n{n

\)(2n

+

1)

any missing 17.

+ +

k

2

-

(n

2k(n

+

k)

2

-

(n

*) 2

l ,

l

let

n

k take on the successive values

—

\.

Add

the resulting n

1, 2, 3,

l .

2

2

.

2

—

together with the identity 2n 2

l

equations,

1

=

2n 2 to arrive at ,

2 l

(*)(n

=

l)n 2

-I-

+ 3 + + n2 + 2[l(n - 1) + 2(n - 2) + (n + 3(* - 3) +

2(1

2

+

22

2

•

•

sum

Next, A:

2

let

=

k

Ac

+

2(1

n

3

1, 2,

+

2

+

•

•

•

•

3

+

in

2 I

4-

22

+

32

+

)

1)1].

+

•

+

••

(k

-

1)]

to get

n2

= (l+2 + 3+-+n) 4 2[l(n -1)4 2(n - 2) 4 3(n - 3) 4 + («- 1)1]. •

•

The desired

n(n

+

!)(/!

+

+

n

2)

t.

+

1

-(2/.

facts to derive the

of the squares of the

first

+

n).

formula for

n integers:

result follows on

1

22

•

2

combining (•) and

f

,

•

2

•

t

the formula

and add the n equations so obtained (••)

+

= fi, = 5 = 342-1= 2 4 27/,, 2 2 4 3 2 = 14 = 6 4 2 4 = 3 4 2T2 10 2 2 4 3 2 4 4 = 30 = 10 4- 2 = U 4- 27V 4 2 4 3 2 4 4 2 4 5 2 = 55 = 15 4 2 20 = 5 4 2T4

= 4 + 4

,

and Tk are the kth triangular and tetrahedral numbers, respectively.

where

=

t3

2

k))

-

4

the kth triangular number. Prove that

Use the following the

[k

t2

[Hint: See Problem 14.]

the formula

= =

t„ is

r.=

details in the following sketch of his proof. In

n2

= h +

derived the

B.C.) also

formula

2

is

given by the formula

;

is

number T„

In general, the nth tetrahedral

by k 2 consider the

tk

odd and n

Archimedes (287-212

l

less in

it.

i

T„ 15.

which precedes

A

2)

the terms on the left-hand side in

is

each of which has one

its

mathematician Aryabhata (circa 500):

two cases n

then the

formula for the sum of

triangular numbers, given by the

Group

n,

it.

14. Prove the following

[Hint:

number of

dots in pyramids built up of triangular numbers.

n) 2

•

The Beginnings of Greek Mathematics

3

tk

The Pythagorean Problem 3.3

105

The Pythagorean Problem While

tradition

unanimous

is

ascribing the so-called Pythagorean theorem to the

in

we have seen

great teacher himself,

specific triangles at least a

that the Babylonians

millennium

We

earlier.

knew the

recall the

the square built upon the hypotenuse of a right triangle

is

result for certain

theorem as "the area of

equal to the

sum

of the areas

of the squares upon the remaining sides." Since none of the various Greek writers attributed the theorem to Pythagoras lived within five centuries of him, there

who

is little

convincing evidence to corroborate the general belief that the Master, or even one of his

immediate

disciples,

gave the

first

rigorous proof of this characteristic property of

when Pythagoras had discovered Muses in gratitude for the inspiration

right triangles. Moreover, the persistent legend that

the theorem, he sacrificed a hundred oxen to the

appears an unlikely

story, since the

What

blood was shed.

is

certain

is

Pythagorean

forbade any sacrifice

ritual

that the school Pythagoras founded did

in

which

much

to

increase the interest in problems directly connected with the celebrated result that

bears his name. Still

more are we

in

doubt about what

line of

demonstration the Greeks originally

offered for the Pythagorean theorem. If the methods of

were used, then

it

was probably a

large square of side a

+

b

is

Book

II

of Euclid's Elements

dissection type of proof similar to the following.

A

divided into two smaller squares of sides a and b respec-

and two equal rectangles with sides a and b; each of these two rectangles can two equal right triangles by drawing the diagonal c. The four triangles can be arranged within another square of side a + b as shown in the second figure. tively,

be

split into

c

Now

the area of the

same square can be represented

in

two ways: as the sum of the

areas of two squares and two rectangles, (a

+

b) 2

-

a2

+

b1

and as the sum of the areas of a square and four (a

When

+

b) 2

=

c2

+

+

lab;

triangles,

4[

^

the four triangles are deducted from the larger square in each figure, the re-

2 sulting areas are equal; or equivalently, c

equal to the

sum

of the squares on a and

b.

=

a2

+

b 2 Therefore, the square on c .

is

Chapter

106

3

The Beginnings of Greek Mathematics

Such proofs by addition of areas are so simple that they may have been made earlier and independently by other cultures (no record of the Pythagorean theorem appears, however, in any of the surviving documents from ancient Egypt). In fact, the contemporary Chinese civilization, which had grown up in effective isolation from both the had a neater and possibly much earlier proof than the one just cited. This is found in the oldest extant Chinese text containing formal mathematical theories, the Arithmetic Classic of the Gnomon and the Circular Paths of Heaven. Assigning the date of this work is difficult. Astronomical evidence suggests that the oldest parts go back to 600 B.C., but there is reason to believe that it has undergone considerable change since first written. The first firm dates that we can connect with it are over a century later than the dates for Nine Chapters on the Mathematical Art. A diagram in the Arithmetic Classic represents the oldest known proof of the Pythagorean theorem.

Greek and Babylonian

The proof later tician

found

civilizations,

inspired by this figure

its

way

was much admired

for its simple elegance,

into the Vijaganita (Root Calculations) of the

Bhaskara, born

in

and

it

Hindu mathema-

1114. Bhaskara draws the right triangle four times in the

square of the hypotenuse, so that

in

the middle there remains a square whose side

equals the difference between the two sides of the right triangle. This last square and the four triangles are then rearranged to

make up

the areas of two squares, the lengths

of whose sides correspond to the legs of the right triangle. "Behold," said Bhaskara,

without adding a further word of explanation. b

a - b

The Pythagorean Problem

107

The geometrical

we

discovery that the sides of a right triangle were connected by a law

numbers

expressible in

led naturally to a corresponding arithmetical problem,

which

Pythagorean problem. This problem, one of the earliest problems in all right triangles whose sides are of integral

shall call the

the theory of numbers, calls for finding length, that

finding

is,

all

solutions in the positive integers of the Pythagorean equation

x2

A triple (x, y,

+

y

=

2

z2

.

z) of positive integers satisfying this equation

is

said to be a Pythagorean

triple.

Ancient tradition attributes to Pythagoras himself a partial solution of the problem, expressed by the numbers

=

x where n

>:

=

y

1,

+

In 2

an arbitrary integer. As

is

1

+

In

z

In,

In 2

+

In

+

1,

perhaps more often the rule than the ex-

is

ception in such instances, the attribution of the

Pythagoras presumably arrived at

=

name may

readily be questioned.

by a relation that produces a square number from the next smaller square number, namely

-

(2k

1)

The

strategy

was

we

1)

+

— 2k —

to suppose that 2A:

often; for instance, if for k,

his solution

k

=

5,

then

(k is

1

-

=

k2

.

3

2

.)

Letting 2k

—

1

= m2

and solving

get

m2 +

m k- = — 2 2

1

and

k

these values are substituted in (1),

m2 +

1

it

m2 - V i

(

2

1

1

2

When

2

a perfect square. (This happens infinitely

=

1

l)

t

follows that

NT)'

*

whence

X

2)

satisfy the

m = 2

3)

2k

=

m2 m,

y

Pythagorean equation

—

1

x

is

odd).

-

2n

1,

y

-

m + 2

1

z

t

any odd integer 2n + 1, where

for

When m =

+

-

=

o

m >

n>

2n 2

+

2n,

z

1

=

which is Pythagoras's result. Some of the Pythagorean from (3) are given in the accompanying table.

1,

1

(m must be

odd, since

the numbers in (2)

2n 2

+

2n

+

triples that

become

1,

can be obtained

Chapter

108

As one

sees,

The Beginnings of Greek Mathematics

3

n

X

y

z

1

3

4

5

2

5

12

13

3

7

24

25

4

9

40

41

5

11

60

61

Pythagoras's solution has the special feature of producing right triangles

having the characteristic that the hypotenuse exceeds the larger leg by one. to

Another special solution in which the hypotenuse and a the Greek philosopher Plato, to wit,

x

4)

=

y

2n,

=

—

n2

=

z

1,

leg differ

+

n2

by 2

is

ascribed

1.

This formula can be obtained, like the other, with the help of the relation (1); but now, we apply it twice: (k

+

l)

2

=

+ (2k + = [(* - l) 2 + k2

Substituting n 2 for k in order to

1)

-

(2k

make 4k

+

(2n) 2

(n 2

-

+

1)]

2k

+

1

=

(k

-

l)

2

+

4k.

a square, one arrives at the Platonic formula l)

-

2

(n 2

+

l)

2 .

Observe that from (4) it is possible to produce the Pythagorean triple (8, 15, 17), which cannot be gotten from Pythagoras's formula (3). Neither of the aforementioned rules accounts for all Pythagorean triples, and it was not until Euclid wrote his Elements that a complete solution to the Pythagorean problem appeared. In Book X of the Elements, there is geometric wording to the effect that

x

5)

=

2mn,

y

= m2 —

n2

z

,

= m2 +

n2

,

and n are positive integers, with m > n. In his Arithmetica, Diophantus (third century) also stated that he could get right triangles "with the aid of" two numbers m and n according to the formulas in (5). Diophantus seems to have arrived at these formulas by the following reasoning. Given the equation x 2 + y 2 = z 2 put y = kx — z, where k is any rational number. Then

where

m

,

z2

which leads

-

x2

=

y

2

-

=

(kx

=

k 2x2

z) 2

-

2kxz

to

— x2 =

k 2x 2

—

2kxz

k 2x

—

2kz.

or

x

=

+

z2

The Pythagorean Problem

When

this

109 equation

is

we

solved for x,

get

Ik

The

implication

that

is

y

=

But k

m/n, with

m

one

sets z

kx

—

z

=

and n integers (there

X If

=

= m2 +

n2

in

'

1

no harm

is

2mn

= "

+

k2

in

m — ~ m2 + 2

_

Z

m + 2

n2

y

'

order to obtain a solution

in

taking

m>

n), so that

n2 n2

*'

the integers,

it

is

found im-

mediately that

—

x

Our argument

2mn,

y

= m — 2

n2

= m2 +

z

,

n2

.

indicates that x, y, z as defined by the preceding formulas satisfy the

Pythagorean equation. The converse problem of showing that any Pythagorean triple necessarily of this form is much more difficult. The details first appeared in Fibo-

is

nacci's treatise Liber

Quadratorum (1225).

The most important achievement of the Pythagorean school in its influence on the number concept was the discovery of the "irrational." The Pythago-

evolution of the

reans

felt intuitively

starting with

two

that any two line segments had a

line

common measure;

that

is

to say,

segments, one should be able to find some third segment, perhaps

very small, that could be marked off a whole number of times on each of the given

segments. From this

it

would follow that the

ratio of the lengths of the original line

segments could be expressed as the ratio of integers or as a rational number. (Recall that a rational number is defined as the quotient of two integers a/b, where b ^ 0.)

One can imagine

the shattering effect of the discovery that there exist

whether

it

Who

some

ratios that

was that

first

established this, or

was done by arithmetical or geometric methods,

will

probably remain a

cannot be represented

in

terms of integers.

it

mystery forever.

The

oldest

known proof dealing with incommensurable line segments corresponds the modern proof that \fl is irrational. This is the proof of the

in its essentials to

incommensurability of the diagonal of a square with the tenth book of Euclid's Elements.

A

its

side,

and

it

is

to be found in

reference in one of Aristotle's works, however,

was known long before Euclid's time. As in most classical demonstrations, the method of argument was indirect. Thus, the negation of the desired conclusion is assumed, and a contradiction is derived from the assumption. The reasoning goes as follows. If the diagonal AC and side AB of the square ABCD have a common measure, say b, then there exist positive integers m and n satisfying

makes

it

clear that the proof

AC = The

ratio of these

segments

AB =

mb,

is

AC AB

_ ~

m n

nb.

Chapter

110

To make matters

The Beginnings of Greek Mathematics

3

simpler, let us suppose that

any common factors of m and n have been

Now

cancelled.

(Acy (AB¥

m*

Applying the Pythagorean theorem to the triangle ABC, one gets (AC) 2 that the displayed equation becomes

2 or In 1

= m2

Now

2n

m itself?

2 ,

.

The

since

task it is

is

to

show that

a multiple of 2,

m

=

2(AB) 2 so ,

A

-T,

this

is

=

cannot happen.

an even integer; hence

m2

even.

is

What

about

m were odd, then m would be odd, because the square of any odd integer must be odd. Consequently, m even; say m = 2k. Substituting this value in the 2 = equation m 2n and simplifying, we get 2

If

is

2

2k 2

=

n2

.

By an argument similar to the one above, it can be concluded that n is an even number. The net result is that m and n are both even (that is, each has a factor of 2), which contradicts our initial assumption that they have no

common

factor whatsoever.

The Pythagoreans cannot be given full credit for discovering the irrationality of \/2. An old cuneiform tablet, now in the Yale Babylonian collection, contains the diagram of a square with its diagonals, as shown herewith.

In sexagesimal notation, the

number 1,24,51,10

1+

24

60

is

equal to

+ !L2 + iO 3 60

'

60

which gives 1.414213 when translated into the decimal system. You should a very close approximation to \[2

familiar, for

it is

of the other

numbers

30.

The

result

is

in

the diagram

=

1.414213562.

.

V2-

its

.

find this

The meaning

becomes clear when we multiply 1;24,51,10 by Thus the

42;25,35, the length of the diagonal of a square of side 30.

Babylonians not only seemed to know that the diagonal of a square length of

.

side,

is

\/2 times the

but also had the arithmetic techniques to accurately approximate

The Pythagorean Problem

\ \ \

Theon of Smyrna (circa 130) devised a procedure for reaching closer and closer approximations of \/2 by rational numbers. The computations involve two sequences of numbers, the "side numbers" and the "diagonal numbers." We begin with two numbers, one called the first side and denoted by X\ and the other the first diagonal and t

indicated by

y t The second

and diagonal (x 2 and y 2 ) are formed from the first, the third side and diagonal (x 3 and y 3 ) from the second, and so on, according to the .

side

scheme x2

=

x,

x3

=

x2

lt

y2

=

2x,

+

>>,

,

y*

=

2x 2

+

y2

+y + y2

In general, x„ and y„ are obtained from the previous pair of side and diagonal numbers by the formulas

x„ If

we take

The names

Xi

= y% —

side

associated pairs

square to

x„ 1

_,+>>„_

y„

,,

=

2xm _

i

+ ym _

i.

as the initial values, then

x2

=1 + 1=2

x3

=

2

+

3

=

5

^2=2-1 + 1=3 = 2-2 + 3 = 7 3

x4

=

5

+

7

=

12

y4

>>

=

2

5

+

7

=

17

numbers and diagonal numbers hint that the quotients y„/x„ of the of these numbers come to approximate the ratio of the diagonal of a

its side:

zi=i

?i

xj

x2

=

—

y±

2

x3

=

2-

y±

5

x4

=

11 '

1

2

This follows from the relation

yS -

(*)

2x„ 2

±

1;

for the relation, if true, implies that

\x„

x„l Since the value of (l/x„) 2

is

made

as small as desired by taking n large enough,

appears that the ratio

x.

V

x. 1

it

Chapter

112

number

tends to stay near some fixed \/2.

is

You can

how

see

The Beginnings of Greek Mathematics

3

for large n. It can be shown that the fixed "limit" works by considering the case n = 4. Here,

this

yiY = (11

xj

\12 - 2?? " 144

144

\12

144

whence

v^u

y*

x4

The

y 4 /x 4

ratio

from the true value of \Jl by less than \ of 1 percent. — 2x„ 2 = ± 1, can be justified by

differs

Now

condition (*), which can be written y„ 2 using the algebraic identity

(**) If

x

we

+

(2x

=

x

,

y

=

assert that

-

2

+

2(x

=

+

x

y

-

2

yo, y —

2x

2

2x

+

yo

=

(2x

=

-W- 2x

+

-

2

y

)

— +

means that if y„ 2 must also hold for n

this

then

it

see that

ym

2

—

2x

2

= ±

2x„ 2 1,

y

2 .

y

—

2

in

1,

then

by virtue of (**),

2 )

= +

1)

it is

1.

possible to find infinitely

which side and diagonal numbers are happens

1

™

= ±

2x 2

to hold for a certain value of

™ y\ —

but with opposite sign. Setting Xi

when n

holds

1

known,

is

and hence

1,

every value of n thereafter. In consequence, (*) It is

-

>-

= -(±

)

= ±

+

2(x

2

manner

In the present situation, by the

n.

2x 2

also a solution. For

is

Thus, when one solution of y 2 — 2x 2 = ± 1 many more solutions by using identity (**). formed,

=

2

y)

are any two numbers satisfying the equation

>'o

x

y)

is

this

equation

a correct identity for

is

all

n

1,

wc

valid for

>

1.

natural to raise the question whether the notion of side numbers and diagonal

numbers can be used

approximations to an arbitrary square

to obtain rational

root.

Theon's original rule of formation was \„

For 2

in

=

x„ -

i

+

v, _

the second equation,

let

y„

i,

=

2x n

-

\

+

ym

.

n

u

us substitute a positive integer a (which

square) to develop the following scheme:

x2

=

X|

+

i

*j

"

xi

+

v2

x4

=

x3

+

yy

x„

=

xm

-

>

,

y2

=

ax,

+

yu

.

.V3

=

ax 2

+

y%,

,

y4

=

ax 3

+

y3

v.

-

axm - 1

,.

+ ym -

1,

,

+ >,-

1,

2.

is

not a perfect

The Pythagorean Problem

1

13

Notice that

y ax and

so,

2

2

=

(ax„

=

a(x„ _

_,+>•„_

+

,

2

,)

2

_

ym

,)

=

a 2x

=

ax\

-

a)x\ _

2

_

_

+

,

+

,

2ax„ -

2ax„

_

-

tf>.

x

y„ _

+ >£ _ „

1

+

,

a^J _ „

on subtraction,

y* - ax? =

(a 2

=

+

,

-

(1

*)>£ _

,

-«)W-i -a^-i)-

(1

The import of this relation is that we have represented y„ 2 — ax„ 2 by an expression of the same form, but with n replaced by n — 1. Repeating this transformation for the next expression, we evidently arrive at the chain of equalities

and as a

ax

2

=

(1

-a)(yj_, ~ ax H2

=

(1

~

.

2

=

(1

-aV(yl_

3

=

(1

-aY-*W-ax?)

-

a)" -

xn this,

it

)

(1

=

For an

=

x„

tt

-

)

x

ax H2

_

2)

-ax 2n

_

3)

axx 2 )

: z

n

,

closely

>

2.

approach the irrational number

illustration, consider the case of \/3; that

is,

a

=

3. If

we take

Xi

\fa.

=

1,

2 as the initial side and diagonal numbers, then the foregoing formula reduces

^-i

successive rational approximations of \/3 are

X\

A

1

'

x2

3

one solution, say x

2

(y

=

x

'

theme

variation of the above

(***) If

-

2

_

can be concluded that as n increases, the right-hand term tends to zero,

ter— The

W

a H

whence the values y H /x H more and more y\

o) 2 iy

result,

y„Y — From

-

2

y

+ ,

y

3x 2 ) 2

=

x3 is

4

x4

'

1 1

x5

15'

afforded by starting with the algebraic identity

-

3(2xy) 2

yo, of the

y

'

2

-

=

(y

equation

3x 2

=

1

2

-

3x 2 ) 2

.

Chapter

114

is

x

The Beginnings of Greek Mathematics

3

known, then (***) indicates that a second solution can be found simply by 2x yo, y = yo 2 + 3x 2 Indeed, on substitution,

=

letting

.

y

-

2

= (V + ~ =

3x 2

W

Thus we have a machine

By

lution.

3x

2 2

3x

2 2

)

)

-

3(2x yo) 2

=

l

for generating solutions of

—

2

y

2

1.

=

3x 2

1

from a single

so-

the rule of formation,

x„ a fresh solution x„,

=

2x„ _ iy n _

=

y„

i,

y n can be derived from

y\ _

+

,

3xJ _ „

a previous one x„ _

i,

y„ _

Since x„,

|.

j>„

satisfy >>„

or

what amounts

same

to the

-

2

=

3x„ 2

1,

thing,

*/

xn 2

the successive values (y„/x„) 2 will approach 3 increasingly closely; that

y„/x„ provides a "very good"

(in

some

sense) approximation of

\/2>

is,

the sequence

by rational num-

bers. It is

clear that the equation

namely

integers,

is

X\

on.

irrational

We

=

*i

Let us

y

=

}

—

2

x2

=

>'2

=>-i

2xi>i

+

2

\/3

=

2x2^2

y3

=

y2

2

+

2

'

now view

the positive

see then that

=

2

•

2

=

4,

22

+

3

•

1

=

3x, 2

=

in

=

2

•

l

7

2

4

•

3x 2 2

=

for the

•

7

72

=

The next one

is

56,

+

3

•

42

=

97

sequence of rational approximations of the

just 7

yi

1

has at least one solution

1

solutions are generated out of given ones.

x3

is

=

3x 2

We

2.

have almost finished;

number y±

y

\,

Thus new

also a solution.

and so

=

4

x,

'

V3

_

x3

""

97 '

56

18,817

y4 x4

10,864'

a strictly geometric proof of the incommensurability of the diagonal

and side of a square. This argument, apparently older than the first, is in the spirit of the arguments found in Euclid's Elements. The basic idea is to show that we can build onto an arbitrary square a sequence of smaller and smaller squares. In the

diagonal

which i

run

it

Ics

square

ABCD. draw

AC

d\.

intersects

BAF

and

Now draw

arc equal.

the

x

BC. By one of the congruence theorems,

FAE

are congruent; consequently,

gruent sides. Furthermore

FE

BE in order to lay off the side AB = s on the line EF perpendicular to d\, with F the point at

the arc

CEF

is

it is

FB =

easy to prove that the

FE, since they are conits legs CE and

an isosceles right triangle, whence

The Pythagorean Problem

115

D I.

A>

/

\

-

~

//

E/ V. 5 2

d\

TV

\\ \ \

S

\\

S F

X

\

s/

""

*"

^

| |

triple

having n as one of

y

2

-

=

z 2 and

3 a given integer, find a its

(n,

— -

y(« 2 n 2 /4

l,/i

7.

+

/4

5.

(a)

+

+

x

1,

and show that x

2)

=

Establish that there are infinitely

+

(3x

+ (b)

x

five

+

=

3x a

y„

=

4x„

+

y.

=

3.

_

.

+

2y„ _

,,

i

+

3y„ _

,

for n

>

2.

Write out the

(b)

Show

first five

numbers

in

each of

the above sequences. that

3.]

many

y.

which x

2

-

2x, 2

=

y\ _

=

y,

-

2x 2

-

ii

whence y.

2

- 2x

2

2

-

2x, 2 =

1.

is 1

,

3x

+

2z

+

2,

4x

+

3z

(c)

From

part (b), conclude that successive

values of y m /x„ are nearer and nearer

2).]

Find (x,

2z

XH

,

then so

triple,

2,

(a)

Pythagorean and y are consecutive integers. [Hint: If z) happens to be a Pythagorean (x, x + triples (x, y, z) in

1

=

Pythagorean

[Hint: Consider the Pythagorean triple

x

x,

triple

triple involving consecutive positive integers.

(x,

in

Pythagorean

1).]

4. Verify that (3, 4, 5) is the only

and below,

\/2.

Let two sequences of numbers be formed accordance with the rule given below.

n even, consider the triple

1)); for 2

+y = 8.]

x

4)

members. [Hint:

For n an odd integer, consider the (n,

-

4)(y

hence y„/x„ approximates

Pythagorean l,z).

triples of the

form

approximations of \/2.

Chapter

118

8.

The Beginnings of Greek Mathematics

3

Consider the sequence of numbers defined by the

and then

rule given below.

1351

...

xi

=

x„

= —I

+

J

for n

X„ -

>

Write out the sequence

(b)

/

1

\

in

first

\/I

>TjV26-

1

l

(b) (a)

1

2,

four terms of this

Obtain the right-hand bound

manner by replacing

^

decimal form. 12. Since \/3

Assuming that the terms x„ approach a number L as n increases, show that L = \/2. [Hint: The number L satisfies L = \(L + 2/L).)

y/3 (a)

=

is

(y

approximately

+

1/x), where

x

in a similar

with jp

-§-,

is

one can put

unknown.

Square both sides of this expression, neglect 2 1 /x and solve the resulting linear equation ,

9.

Prove that \/3 and \fl are irrational by

assuming that each a contradiction 10.

Replace 2 by

is

and arguing

rational

(b)

until

13.

Theon's definition of side

3 in

to get a second approximation of \/3.

Repeat

this

numbers and diagonal numbers, so that the

(a)

Given a positive integer n that

rule

y„

=

x» _

+ ym _

i

3x„ _

,

(a) Starting with *i first six

numbers

below

to n (above or

so that n

—

+

I,

yt

>

n

,

»

a

=

a2

± 2a

2.

2, write out the

(b)

Use

each of the resulting

in

not a

±

b.

n,

may

as the case

be),

Prove that

j,

y„ _

=

is

perfect square, let a 2 be the nearest square

of formation becomes

x„

procedure once more to find a

third approximation.

reached.

is

x

for

±