## 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.

Let $$F$$, $$G$$, $$H$$, $$I$$ and $$J$$ be the points of intersection of the diagonals.  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 $$\frac{1}{5}.360\,^{\circ}=72\,^{\circ}$$. Considering, for example, the triangle $$\left[ FBC\right]$$, we have that $F\hat{B}C=F\hat{C}B=\frac{1}{2}.72\,^{\circ}=36\,^{\circ}$ (angles inscribed in a circle have measure equal to one half of the respective arc), so the triangle is isosceles. Analogously, the triangles $$\left[ GCD\right]$$, $$\left[ HDE\right]$$, $$\left[ IEA\right]$$ and $$\left[ JAB\right]$$ are also isosceles. Additionally, these triangles are all congruent, as they have one side and the adjacent angles equal. Therefore, $\overline{AJ}=\overline{BJ}=\overline{BF}=\overline{CF}=\overline{CG}=\overline{DG}=\overline{DH}=\overline{EH}=\overline{EI}=\overline{AI}$ and, since $I\hat{A}J=J\hat{B}F=F\hat{C}G=G\hat{D}H=H\hat{E}I=\frac{1}{2}.72\,^{\circ}=36\,^{\circ},$ the triangles $$\left[ IAJ\right]$$, $$\left[ JBF\right]$$, $$\left[ FCG\right]$$, $$\left[ GDH\right]$$ and $$\left[ HEI\right]$$ are also isosceles and congruent. We then have that $\overline{FG}=\overline{GH}=\overline{HI}=\overline{IJ}=\overline{JF}$ and $F\hat{G}H=G\hat{H}I=H\hat{I}J=I\hat{J}F=J\hat{F}G=180\,^{\circ}-36\,^{\circ}-36\,^{\circ}=108\,^{\circ},$ so $$[FGHIJ]$$ is a regular pentagon.

Since $F\hat{C}D=\frac{1}{2}.144\,^{\circ}=72\,^{\circ}$ 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 isosceles, with $$\overline{FD}=\overline{CD}$$. In the same way, $$\left[ BCG\right]$$ is also isosceles, with $$\overline{BG}=\overline{BC}$$. On the other hand, as $$\overline{IJ}=\overline{JF}$$, we have that $$\left[ IJF\right]$$ is isosceles and $J\hat{I}F=J\hat{F}I=\frac{1}{2}(180\,^{\circ}-I\hat{J}F)=36\,^{\circ}=J\hat{B}F,$ so $$\left[ BFI\right]$$ is also isosceles, with $$\overline{FI}=\overline{BF}$$. Denoting by $$d_{1}$$ the length of the diagonal of $$[FGHIJ]$$ and by $$l_{1}$$ the length of its side, we have: $d_{1}=\overline{FI}=\overline{BF}=\overline{BD}-\overline{FD}=\overline{BD}-\overline{CD}=d-l$ $l_{1}=\overline{FG}=\overline{BG}-\overline{BF}=\overline{BC}-\overline{BF}=l-(d-l)=2l-d$

Since $$d=mx$$ and $$l=nx$$, we get: $l_{1}=2nx-mx=(2n-m)x=n_{1}x$ $d_{1}=mx-nx=(m-n)x=m_{1}x,$ where $$n_{1}=2n-m$$ and $$m_{1}=m-n$$ are two positive integer numbers, with $$m_{1}<m$$. Proceeding in the same way with the pentagon $$[FGHIJ]$$, we obtain a new regular pentagon whose diagonal is $$d_{2}=m_{2}x$$ and whose side is $$l_{2}=n_{2}x$$, where $$n_{2}=2n_{1}-m_{1}$$ and $$m_{2}=m_{1}-n_{1}$$ are two positive integer numbers, with $$m_{2}<m_{1}<m$$.

Since we can go on indefinitely building 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.