Assume \(\sqrt{2}\) is rational. Then \(\sqrt{2} = \frac{p}{q}\) where \(\frac{p}{q}\) is an irreducible fraction.
Squaring both sides: \(2 = \frac{p^2}{q^2} \implies p^2 = 2q^2\).
This implies \(p^2\) is even, so \(p\) must be even. Let \(p = 2k\).
Substitute back: \((2k)^2 = 2q^2 \implies 4k^2 = 2q^2 \implies 2k^2 = q^2\).
This implies \(q^2\) is even, so \(q\) must be even.If both \(p\) and \(q\) are even, they have a common factor of 2, contradicting that the fraction is irreducible.
Thus, \(\sqrt{2}\) cannot be rational.