Proof

According to the Pythagorean theorem, given a square with the side \(q\) for some integer \(q\in\mathbb N\), the length of the diagonal \(p\) is given by \[p=\sqrt 2 q,~\quad\text{or}\quad\sqrt 2=\frac pq.\]
Assume that \(\frac pq\) is a rational number \(\frac pq\in\mathbb Q\) and that the positive integers \(p\) (numerator) and \(q\) (denominator) have no common prime factors (otherwise cancel out these prime factors to get the reduced fraction).
Given these numbers, we can construct a smaller triangle (shaded) with the sides \(2q-p\), \(p-q\), \(p-q\), like shown in the below picture.

sqrt2_1

Since both triangles are similar, we conclude that \[\sqrt 2=\frac pq=\frac {2q-p}{p-q}.\]
Since the numerator and the denominator of \(\frac {2q-p}{p-q}\) are smaller then the numerator and the denominator of \(\frac pq\), the natural numbers \(2q-p\) and \(p\) as well as \(p-q\) and \(q\) have to have different factorizations.
But this is a conctradiction to how the fraction \(\frac pq\) was constructed (i.e. there were no more common prime factors we could cancel out from \(p\) and \(q\)).
On the other hand, only by such a cancellation, we would preserve the ratio of \(\sqrt 2\). Therefore, our assumption that \(p\) is a whole number must be wrong.
Consequently, \(\sqrt 2\) must be irrational.
∎

Thank you to the contributors under CC BY-SA 4.0!

Alsina, Claudi; Nelsen, Roger B.: "Bezaubernde Beweise", Springer Spektrum, 2013