## Incommensurability

### 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.