\(\sqrt2\) is irrational
Theorem 6.1 (\(\sqrt 2\) is irrational). There are no integers \(a\) and \(b \neq 0\) with \((a/b)^2 = 2\).
Proof. Suppose there are, and argue by contradiction. Replacing \(a\) and \(b\) by \(-a\) and \(-b\) if necessary, we may take \(b > 0\). By well-ordering (Axiom 3.1), choose such a pair with \(b\) as small as possible. Then \(a^2 = 2b^2\) is even, so \(a\) is even by Theorem 4.4. Write \(a = 2c\). Then \(4c^2 = 2b^2\), so \(b^2 = 2c^2\) is even, and \(b = 2d\) is even too. Now \((c/d)^2 = 2\) with \(0 < d < b\), which contradicts the choice of \(b\).