## Incommensurability

### Regular Pentagon

Let $$[ABCDE]$$ be a regular pentagon, $$d$$ the length of its diagonal and $$l$$ the length of its side. Assuming that $$d$$ and $$l$$ are commensurable, we have $$d=mx$$ and $$l=nx$$, where $$m$$ and $$n$$ are positive integers and $$x$$ is a common measure to the diagonal and the side of the pentagon.

Constructing the circle circumscribing the pentagon, we note that the points $$A$$, $$B$$, $$C$$, $$D$$ and $$E$$ divide the circle into arcs of equal length, given by $$\frac{1}{5}.360\,^{\circ}=72\,^{\circ}$$. Let $$F$$ be the point of intersection of the diagonal $$\left[ AC\right]$$ with the diagonal $$\left[ BD\right]$$. Since $F\hat{B}C=D\hat{B}C=\frac{1}{2}.72\,^{\circ}=36\,^{\circ}$ and $F\hat{C}B=A\hat{C}B=\frac{1}{2}.72\,^{\circ}=36\,^{\circ}$ (angles inscribed in a circle have measure equal to one half of the respective arc), we have that $$\left[ FBC\right]$$ is isosceles, with $$\overline{BF}=\overline{CF}$$. Since $F\hat{C}D=A\hat{C}D=\frac{1}{2}.144\,^{\circ}=72\,^{\circ}$ and $C\hat{F}D=F\hat{B}C+F\hat{C}B=36\,^{\circ}+36\,^{\circ}=72\,^{\circ},$ we get that $$\left[ FCD\right]$$ is also isosceles, with $$\overline{CD}=\overline{FD}$$. On the other hand, $B\hat{F}C=180\,^{\circ}-C\hat{F}D=180\,^{\circ}-72\,^{\circ}=108\,^{\circ},$ where this is the measure of the interior angle of a regular pentagon. So we can construct a new regular pentagon $$[BFCGH]$$, passing through the points $$B$$, $$C$$ and $$F$$. Denoting by $$d_{1}$$ the length of its diagonal and $$l_{1}$$ the length of its side, we have: $l_{1}=\overline{BF}=\overline{BD}-\overline{FD}=\overline{BD}-\overline{CD}=d-l$ $d_{1}=\overline{BC}=l$

Since $$d=mx$$ and $$l=nx$$, we get: $l_{1}=mx-nx=(m-n)x=n_{1}x$ $d_{1}=nx=m_{1}x,$ where $$n_{1}=m-n$$ and $$m_{1}=n$$ are two positive integer numbers, with $$m_{1}<m$$. Indeed, the largest angle in the triangle $$\left[ BCD\right]$$ is the angle $$\measuredangle BCD$$, so the largest side is opposed to this angle and, therefore, $$d=\overline{BD}>\overline{BC}=l$$, that is, $$m>n$$.

Proceeding in the same way with the pentagon $$[BFCGH]$$, we obtain a new regular pentagon whose diagonal is $$d_{2}=m_{2}x$$ and whose side is $$l_{2}=n_{2}x$$, where $$n_{2}=m_{1}-n_{1}$$ and $$m_{2}=n_{1}$$ are two positive integer numbers, with $$m_{2}<m_{1}<m$$.

Since we can go on indefinitely constructing new regular pentagons, we obtain a strictly decreasing sequence of positive integer numbers: $m_{1}>m_{2}>m_{3}>m_{4}>...$ such that $$m_{i}<m,\:\forall i\in\mathbb{N},$$ which is absurd, as there cannot be an infinite number of distinct positive integers less than $$m$$ (there exist exactly $$m-1$$ such numbers: $$1$$, $$2$$, $$3$$,...,$$m-2$$ and $$m-1$$). Therefore, the diagonal and the side of $$[ABCDE]$$ are incommensurable magnitudes.