NT-B1-M06-P016 Irrationality of \(\sqrt{6}\) by Descent

From \(a^2=6b^2\), derive that both \(a\) and \(b\) are even.

Let \(\sqrt{6}=\frac{a}{b}\) in lowest terms. Then \(a^2=6b^2\), so \(a\) is even: \(a=2c\). We get \(4c^2=6b^2\), so \(2c^2=3b^2\). The right-hand side is even, hence \(b^2\) is even, and \(b\) is even. This contradicts lowest terms.

It can also be proved using prime \(3\); here the parity route is shown deliberately.