Proof.
Suppose not. Then \(\sqrt 2\) is equal to a fraction \(\frac{a}{b}\text{.}\) Without loss of generality, assume \(\frac{a}{b}\) is in lowest terms (otherwise reduce the fraction). So,
\begin{equation*}
2 = \frac{a^2}{b^2}
\end{equation*}
\begin{equation*}
2b^2 = a^2\text{.}
\end{equation*}
Thus \(a^2\) is even, and as such \(a\) is even. So \(a = 2k\) for some integer \(k\text{,}\) and \(a^2 = 4k^2\text{.}\) We then have,
\begin{equation*}
2b^2 = 4k^2
\end{equation*}
\begin{equation*}
b^2 = 2k^2\text{.}
\end{equation*}
Thus \(b^2\) is even, and as such \(b\) is even. Since \(a\) is also even, we see that \(\frac{a}{b}\) is not in lowest terms, a contradiction. Thus \(\sqrt 2\) is irrational.