The Irrationality of the Square Root of 2

#Setup

Suppose the opposite. Then $\sqrt{2} = \tfrac{a}{b}$ for integers $a$ and $b$ with $b \neq 0$, and I may take this fraction in lowest terms, so $\gcd(a, b) = 1$. Squaring and clearing the denominator,

#A parity lemma

An odd integer squares to an odd integer, since $(2k+1)^2 = 2(2k^2 + 2k) + 1$. Contrapositively, an even square forces an even root.

#The contradiction

The equation $a^2 = 2\,b^2$ makes $a^2$ even, so $a$ is even. Write $a = 2c$. Substituting gives $4c^2 = 2\,b^2$, that is $b^2 = 2\,c^2$, so $b^2$ is even and $b$ is even as well. Now 2 divides both $a$ and $b$, which contradicts $\gcd(a, b) = 1$.

The assumption was the only thing that could fail, so no such $a$ and $b$ exist and $\sqrt{2}$ is irrational.

#A wider view

The contradiction is infinite descent in disguise. The step that took $(a, b)$ to $(b, c)$ produces a strictly smaller solution of the same equation, and no positive integers descend forever. Nothing used the number 2 beyond its being prime, because a prime dividing a square divides its root, so $\sqrt{p}$ is irrational for every prime $p$. Splitting any non-square into primes carries this further, and $\sqrt{m}$ is irrational whenever $m$ is not a perfect square.