(related to Proposition: Discovery of Irrational Numbers)
The following proof is based on arguments from the elementary number theory. * Assume, $\sqrt {2}=\frac pq$ is a rational number. * Without loss of generality, we can choose $p$ and $q$ coprime. * Then $$2=\frac{p^2}{q^2}\Longleftrightarrow p^2=2q^2.$$ * It follows from the fundamental theorem of arithmetic that $p^2$ is even. * Therefore, $p$ is also even, since it has $2$ as a prime factor, or $p=2n$ for a natural number $n\in\mathbb N.$ * Since $p^2=4n^2=2q^2$ it follows that $q^2$ is even, since $q^2=2n^2$ as $2$ as a prime factor. * But then, also $q$ is even, following a similar argument as above for $p.$ * This is a contradiction since $p$ and $q$ were chosen coprime.
