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
±