geometric proof that the square root is irrational

[Kenneth A. Ribet]

Folks,

I drew the book's diagram (p. 91) on my white board in Evans. The
proof can be expressed in terms of Fermat's method of descent. If
the square root of 1 + m^2 is a rational number, there is a right
triangle with sides a and ma whose three sides are integers. Take a
as small as possible. Write the length of the hypotenuse as ma + b,
where b is a positive integer. The sum of the two sides ma + a is
bigger than ma + b, so b is a smaller integer than a. The claim is
that the right triangle whose two sides are b and mb has a hypotenuse
that's also of integral length. Because b is less than a, this
contradicts the supposed minimality of a.

To prove the claim, note that (am)^2 + a^ 2 = (ma+b)^2. This leads
to the relation a^2 - b^2 = 2amb. It follows that (mb)^2 + b^2 = (a-
mb)^2 by expanding out the square and doing a small amount of
algebraic manipulation. Thus the hypotenuse that we wanted to have
integral length has length a-mb or mb-a, whichever is positive. The
diagram tells us that a is bigger than mb, but we don't really need
to know this.

Ken R