### 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.