### Square

In the case of the square, the ratio between the diagonal and the side is \(\sqrt{2}\). If we assume there exist positive integers \(m\) and \(n\) such that \(\frac{m}{n}=\sqrt{2}\), we have: \[\frac{m}{n}=\sqrt{2}\] \[\left(\frac{m}{n}\right)^{2}=2\] \[\frac{m^{2}}{n^{2}}=2\] \[m^{2}=2n^{2}\]

Then, as \(m^{2}\) is even, so is \(m\), so there exists a positive integer \(m_{1}\) such that \(m=2m_{1}\). Substituting above, we get: \[(2m_{1})^{2}=2n^{2}\] \[4m_{1}^{2}=2n^{2}\] \[2m_{1}^{2}=n^{2}\]

Since \(n^{2}\) is even, so is \(n\), so there exists a positive integer \(n_{1}\) such that \(n=2n_{1}\). Substituting, we get:\[2m_{1}^{2}=(2n_{1})^{2}\] \[2m_{1}^{2}=4n_{1}^{2}\] \[m_{1}^{2}=2n_{1}^{2}\]

Analogously, we deduce that \(m_{1}\) e \(n_{1}\) are also even, so there exist positive integers \(m_{2}\) and \(n_{2}\) such that \(m_{1}=2m_{2}\) and \(n_{1}=2n_{2}\). Substituting, we get: \[m_{2}^{2}=2n_{2}^{2}\]

In the same way, we get that \(m_{2}\) and \(n_{2}\) are also
even, so there exist positive integers \(m_{3}\) and \(n_{3}\)
such that \(m_{2}=2m_{3}\) and \(n_{2}=2n_{3}\), where
\(m_{3}^{2}=2n_{3}^{2}\), and so on.

We therefore obtain a sequence of positive integer numbers \[m_{1}>m_{2}>m_{3}>m_{4}>...\] which is strictly decreasing, since each term is one half of the previous one. But this is absurd, since there cannot be an infinite number of positive integers less than \(m\). Therefore, \(\sqrt{2}\) is irrational, from which we conclude the incommensurability between the diagonal and the side of the square.