## Incommensurability

### 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.