A History of Mathematics From Mesopotamia to Modernity

54 A History ofMathematics

However, there is a simpler proof which is generally thought to be a probable reconstruction, as

follows. If you have a ratio of honest whole numbers, say 42 to 15, then you can reduce it in steps,

as follows:

42

15

= 2 +

12

15

15

12

= 1 +

3

12

12

3

= 4

ending with an exact whole number. At each stage, you have a remainder less than 1; you invert

this at the next stage, and get a whole number and a new remainder. This is Euclid’s way of finding

the ‘common measure’ or greatest common divisor for two numbers, in the above case 3, which

divides both 42 and 15. The process stops because the denominators of the fractions get smaller,

and so must finally reach 1. You cannot do this with the golden ratio, because of the way in which

it is defined. The equation

1

x

= 1 +x

which is equivalent to the one above shows that the ‘fraction’ procedure (invert the remainder, take

the whole number partaway, keep theremainder, and restart), never stops.

And indeed this can be stated geometrically, as even ‘the Pythagoreans’ could have done.

Exercise 8.Explain why the fraction procedure cannot terminate.

Solutions to exercises

1. I do not think it is quite obvious. Clearly the four sides are equal, since they are diagonals
   of equal squares. The other point to note is that all the triangles are isosceles (the diagonals
   bisect each other by symmetry, for example). So all their small angles are just half a right
   angle—as we would say 45◦—and the angles at the corners of the figure, each being
   composed of two such small angles, are 90◦. This makes the figure a square. Wehaveused
   the result on angle sum of a triangle, which does not work in non-Euclidean geometry (see
   Chapter 8); but then the idea of ‘square’ becomes dubious anyway.


2. (Trying to use Greek language as far as possible.) ‘Diagonal is commensurable with side’
   means that there is a line L such that the side S is a multiple, sayq.L, and the diagonal D
   is also a multiple, sayp.L. Now clearly (using arguments like theMeno), the area of the
   square on S isq^2 times the square on L, and that on D isp^2 times the square on L; and, by
   Pythagoras’s theorem, the square on D is twice the square on S. Sop^2 is equal to twiceq^2 ,
   or, the square of the ratio ofptoqequals the ratio of 2 to 1 (i.e. 2). Check that all these
   implications go both ways.


3. The usual proof (see, for example, Fowler for speculation on which proof the Greeks, in
   particular Aristotle, used), is the following. Suppose(p/q)^2 =2; we can assume thatp
   andqhave no common factor. Thenp^2 = 2 q^2. It follows thatp^2 is even, and sopis even.