### Square

Let \([ABCD]\) be a square, \(d\) the length of its diagonal and \(l\) the length of its side. Suppose the two magnitudes are commensurable, that is, \(d=mx\) and \(l=nx\) for some common measure \(x\) and for some positive integer numbers \(m\) and \(n\).

Choose the point \(E\) on the diagonal \(\left[ AC\right] \) such that \(\overline{BC}=\overline{EC}\), and the point \(F\) on the side \(\left[ AB\right] \) such that \(EF\perp AC\). Since \(\overline{AB}=\overline{BC}\), the triangle \(\left[ ABC\right] \) is isosceles and, as \(A\hat{B}C=90\,^{\circ} \), we have \(C\hat{A}B=A\hat{C}B=45\,^{\circ} \). Since \(E\hat{A}F=C\hat{A}B=45\,^{\circ} \) and \(A\hat{E}F=90\,^{\circ} \), we get \(A\hat{F}E=45\,^{\circ} \), so \(\left[ AFE\right] \) is isosceles, with \(\overline{AE}=\overline{EF}\). As \(\left[ FBC\right] \) and \(\left[ FEC\right] \) are right triangles with the same hypotenuse and a common small side (\(\overline{BC}=\overline{EC}\)), by the Pythagorean Theorem the other side is also equal in both, \(\overline{BF}=\overline{EF}=\overline{AE}\).

Marking \(G\), the point of intersection of the line parallel to \(AE\) and passing through \(F\) with the line parallel to \(EF\) and passing through \(A\), we obtain a new square \([AEFG]\). Denoting by \(d_{1}\) the length of its diagonal and by \(l_{1}\) the length of its side, we have: \[l_{1}=\overline{AE}=\overline{AC}-\overline{EC}=\overline{AC}-\overline{BC}=d-l\] \[d_{1}=\overline{AF}=\overline{AB}-\overline{BF}=\overline{AB}-\overline{AE}=l-(d-l)=2l-d\]

Since \(d=mx\) and \(l=nx\), we get: \[l_{1}=mx-nx=(m-n)x=n_{1}x\] \[d_{1}=2nx-mx=(2n-m)x=m_{1}x,\] where \(n_{1}=m-n\) and \(m_{1}=2n-m\) are two positive integer numbers, with \(m_{1}<m\). Indeed, \[\begin{array}{ccl} m_{1}\geq m & \Rightarrow & 2n-m\geq m\\ & \Rightarrow & n\geq m\\ & \Rightarrow & l\geq d, \end{array}\] which is absurd, as the diagonal of any square is always larger than its side.

Proceeding in the same way with the square \([AEFG]\), we obtain a new square 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 \(m_{2}=2n_{1}-m_{1}\) are two positive integer numbers, with \(m_{2}<m_{1}<m\).

Since we can go on indefinitely constructing new squares, 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 \([ABCD]\) are incommensurable magnitudes.