Theorem: The number \(\sqrt{2}\) is irrational.

Proof:

Suppose, for the sake of contradiction, that \(\sqrt{2}\) is rational. Then there exist integers \(p\) and \(q\) (\(q \neq 0\)) such that:

\[ \sqrt{2} = \frac{p}{q} \]

We may assume the fraction is in simplest form, such that \(\gcd(p, q) = 1\). Squaring both sides yields:

\[ 2 = \frac{p^2}{q^2} \Longrightarrow p^2 = 2q^2 \]

This implies that \(p^2\) is even, and thus \(p\) is even (\(p = 2k\)). Substituting this in:

\[ (2k)^2 = 2q^2 \Longrightarrow 4k^2 = 2q^2 \Longrightarrow 2k^2 = q^2 \]

This implies that \(q^2\) is even, so \(q\) is even.

If both \(p\) and \(q\) are even, they share a factor of \(2\), contradicting our initial assumption that \(\gcd(p, q) = 1\). Therefore, the assumption that \(\sqrt{2}\) is rational must be false.

Hence, \(\sqrt{2}\) is irrational. ■