ATRACTOR

In the case of the regular hexagon, the ratio between the diagonal and the side is \(\sqrt{3}\). Assuming there exist positive integers \(m\) and \(n\) such that \(\frac{m}{n}=\sqrt{3}\), we have: \[\frac{m}{n}=\sqrt{3}\] \[\left(\frac{m}{n}\right)^{2}=3\] \[\frac{m^{2}}{n^{2}}=3\] \[m^{2}=3n^{2}\]

Then, as \(m^{2}\) is a multiple of \(3\), so is \(m\), so there exists a positive integer \(m_{1}\) such that \(m=3m_{1}\). Substituting above, we get: \[(3m_{1})^{2}=3n^{2}\] \[9m_{1}^{2}=3n^{2}\] \[3m_{1}^{2}=n^{2}\]

As \(n^{2}\) is a multiple of \(3\), so is \(n\), so there exists a positive integer \(n_{1}\) such that \(n=3n_{1}\). Substituting, we get: \[3m_{1}^{2}=(3n_{1})^{2}\] \[3m_{1}^{2}=9n_{1}^{2}\] \[m_{1}^{2}=3n_{1}^{2}\]

Analogously, we deduce that \(m_{1}\) and \(n_{1}\) are also multiples of \(3\), so there exist positive integers \(m_{2}\) and \(n_{2}\) such that \(m_{1}=3m_{2}\) and \(n_{1}=3n_{2}\). Substituting above, we get: \[m_{2}^{2}=3n_{2}^{2}\]

In the same way, we see that \(m_{2}\) and \(n_{2}\) are also  multiples of \(3\), so there exist positive integers \(m_{3}\) and \(n_{3}\) such that \(m_{2}=3m_{3}\) and \(n_{2}=3n_{3}\), hence \(m_{3}^{2}=3n_{3}^{2}\), 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 third of the previous one. But this is absurd, since there cannot be an infinite number of positive integers less than \(m\). Therefore, \(\sqrt{3}\) is irrational, from which we conclude the incommensurability between the diagonal and the side of the regular hexagon.