## Incommensurability ### Regular Pentagon

In the case of the regular pentagon, the ratio between the diagonal and the side is $$\frac{1+\sqrt{5}}{2}$$. If we assume there exist positive integers $$m$$ and $$n$$ such that $$\frac{m}{n}=\frac{1+\sqrt{5}}{2}$$, we have: $\frac{m}{n}=\frac{1+\sqrt{5}}{2}$ $\left(\frac{2m}{n}-1\right)^{2}=5$ $\frac{4m^{2}}{n^{2}}-\frac{4m}{n}-4=0$ $m^{2}-mn-n^{2}=0$ $\left(m-n\right)\left(m+n\right)=mn$

If $$m$$ and $$n$$ were both odd, then $$mn$$ would also be odd, but $$m-n$$ and $$m+n$$ would both be even, so $$\left(m-n\right)\left(m+n\right)$$ would be even. If $$m$$ were even and $$n$$ were odd (or vice-versa), then $$mn$$ would be even, but $$m-n$$ and $$m+n$$ would both be odd, so $$\left(m-n\right)\left(m+n\right)$$ would be odd. Therefore, $$m$$ and $$n$$ are both even, so there exist positive integers $$m_{1}$$ and $$n_{1}$$ such that $$m=2m_{1}$$ and $$n=2n_{1}$$. Substituting, we get: $\left(2m_{1}-2n_{1}\right)\left(2m_{1}+2n_{1}\right)=2m_{1}.2n_{1}$ $4\left(m_{1}-n_{1}\right)\left(m_{1}+n_{1}\right)=4m_{1}n_{1}$ $\left(m_{1}-n_{1}\right)\left(m_{1}+n_{1}\right)=m_{1}n_{1}$

Analogously, we deduce that $$m_{1}$$ and $$n_{1}$$ 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: $\left(m_{2}-n_{2}\right)\left(m_{2}+n_{2}\right)=m_{2}n_{2}$

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 $$\left(m_{3}-n_{3}\right)\left(m_{3}+n_{3}\right)=m_{3}n_{3}$$, 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, $$\frac{1+\sqrt{5}}{2}$$ is irrational, from which we conclude the incommensurability between the diagonal and the side of the regular pentagon.