### 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.