## Incommensurability

### Regular Octagon

Consider a regular octagon for which the points $$A$$, $$B$$, $$C$$ and $$D$$ are consecutive vertices, $$d$$ is the length of its second shortest diagonal and $$l$$ the length of its side. If we assume 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 second shortest diagonal and the side of the octagon. Note that each side is a chord of the circle circumscribing the polygon, corresponding to an arc measuring $$\frac{1}{8}.360\,^{\circ}=45\,^{\circ}$$. Choose the point $$E$$ on the diagonal $$\left[ AD\right]$$ such that $$\overline{AE}=\overline{AB}=l$$. Then, as $$\left[ AEB\right]$$ is isosceles and $E\hat{A}B=D\hat{A}B=\frac{1}{2}.2.45\,^{\circ}=45\,^{\circ},$ we get $A\hat{E}B=A\hat{B}E=\frac{1}{2}.(180\,^{\circ}-45\,^{\circ})=67,5\,^{\circ}.$ On the other hand, $C\hat{B}E=C\hat{B}A-A\hat{B}E=\frac{1}{2}.6.45\,^{\circ}-67,5\,^{\circ}=135\,^{\circ}-67,5\,^{\circ}=67,5\,^{\circ}=A\hat{E}B,$ so the diagonal $$\left[ AD\right]$$ is parallel to the side $$\left[ BC\right]$$. Now choose the point $$F$$ on the diagonal $$\left[ AD\right]$$ such that $$\overline{EF}=\overline{BC}=l$$. Then, since $$\left[ EF\right]$$ is parallel to $$\left[ BC\right]$$, $$[EBCF]$$ is a parallelogram, so $D\hat{F}C=D\hat{E}B=180\,^{\circ}-A\hat{E}B=112,5\,^{\circ}.$ Since $A\hat{C}D=\frac{1}{2}.5.45\,^{\circ}=112,5\,^{\circ}=D\hat{F}C$ and $$A\hat{D}C=C\hat{D}F$$, the triangles $$\left[ ADC\right]$$ and $$\left[ CDF\right]$$ are similar, so $$\left[ DC\right]$$ is the second shortest diagonal of a regular octagon with side $$\left[ DF\right]$$. This being so, we can construct a regular octagon passing through the points $$D$$, $$F$$ and $$C$$, and whose side and second shortest diagonal have lengths given by: $l_{1}=\overline{DF}=\overline{AD}-\overline{AE}-\overline{EF}=d-2l$ $d_{1}=\overline{DC}=l$

As $$d=mx$$ and $$l=nx$$, we have: $l_{1}=mx-2nx=(m-2n)x=n_{1}x$ $d_{1}=nx=m_{1}x,$ where $$n_{1}=m-2n$$ and $$m_{1}=n$$ are two positive integer numbers, with $$m_{1}<m$$. Indeed, $\begin{array}{ccl} m_{1}\geq m & \Rightarrow & n\geq m\\ & \Rightarrow & l\geq d, \end{array}$ which is absurd, since any diagonal in a regular polygon is always larger than its side.

Proceeding in the same way, we obtain a new regular octagon whose second shortest diagonal is $$d_{2}=m_{2}x$$ and whose side is $$l_{2}=n_{2}x$$, where $$n_{2}=m_{1}-2n_{1}$$ 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 octagons, we shall 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, since 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 second shortest diagonal and the side of the regular octagon are incommensurable magnitudes.