NT-B1-M06-P004 Irrationality of \(\sqrt{2}\)

Assume \(\sqrt{2}=\frac{a}{b}\) in lowest terms.

Let \(\sqrt{2}=\frac{a}{b}\), where \(\gcd(a,b)=1\). Then \(a^2=2b^2\). Hence \(a\) is even, \(a=2c\). Then \(4c^2=2b^2\), so \(b^2=2c^2\), and \(b\) is even. Thus \(a\) and \(b\) have common divisor \(2\), a contradiction.

The classical proof should be kept in complete form.