Lesson 4

A real number \(x\) is called rational if there exist integers \(p,q\) with \(q\ne0\) such that \(x=\frac{p}{q}\). It is known that the number \(\sqrt 2\) is irrational: there do not exist integers \(p,q\) with \(q\ne0\) such that \(\sqrt{2}=\frac{p}{q}\).

Problem. Is this a correct proof?

Assume \(\sqrt{2}\) is rational. Then, \(\sqrt{2}= \frac{a}{b}\). Without loss of generality, one may assume that this fraction is irreducible, that is, \(a\) and \(b\) do not share non-trivial divisors (that is, \(\gcd(a, b) = 1\)). Squaring both sides, one gets \(2 = \frac{a^2}{b^2}\), hence \(a^{2} = 2b^{2}\). This implies that \(a^{2}\) is even and hence \(a\) is even, that is, \(a = 2k\) for some \(k\in \mathbb{Z}\). Then, \(a^{2} = 4k^{2} = 2b^{2}\), hence \(b^{2} = 2k^{2}\) and \(b^{2}\) is even, and thus, \(b\) is also even. But then, \(\gcd(a, b) \ge 2\), a contradiction.