## Incommensurability

### Regular Hexagon

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.