Author: Ed Bueler

Date: 1 September 2013

Exercise 6.66. The square root of 2 is irrational.

Proof. Suppose to the contrary that \sqrt{2} is rational. Then there are integers a and b with no common factors such that

\sqrt{2} = \frac{a}{b}.

Squaring this equation we find that

(1)  2b^{2} = a^{2}.

Hence 2 divides a^{2} and therefore a^{2} is even. If a were odd, then a^{2} would also be odd, which it is not, so we conclude that a is even. So a = 2k for some integer k. It follows from equation (1) that

2b^{2} = (2k)^{2}
= 4k^{2}.

Hence

b^{2} = 2k^{2}.

Arguing as before we see that b^{2}, and thus also b itself, must be even. So 2 is a common factor of a and b, which is a contradiction.