Scroll down \(\sqrt 2\) is irrational. We proceed by contradiction. Suppose \(\sqrt 2\) is rational. There exist some integers \(a\) and \(b\) such that: \(\sqrt 2 = \frac a b\) irreducible By squaring each side, we obtain: \(2 = \left(\frac {a} b\right)^2\) \(2 = \frac {a^2} {b^2}\) \(2b^2 = a^2\) \(a^2\) is even. \(a^2\) even implies \(a\) even. By contrapositive. Suppose \(a\) is odd. \(a = 2k+1\) for some integer \(k\) By squaring \(a^2 = 4k^2+4k+1\) \(a^2\) is odd \(a\) is even. \(a = 2k\) for some \(k\) \(2b^2 = 4k^2\) \(b^2 = 2k^2\) \(b^2\) is even \(b\) is even Contradiction Scroll up