ATRACTOR

Let's see first how we can prove the irrationality of \(\sqrt{2}\). Given an arbitrary nonzero point on the line \(x=\sqrt{2}y\), we saw that the sequence of its iterates is a sequence of distinct points on that line converging to the origin. Supposing that the coordinates of that initial point are integer numbers, the sequence of the abscissas (or the ordinates) of its iterates would be a bounded  sequence of distinct integer numbers, which is absurd (indeed, in any bounded subset of the real line there exists only a finite number of points with integer abscissas). But if we had \(\sqrt{2}=\frac{a}{b}\), with \(a\) and \(b\) positive integers, then we would get \(a=\sqrt{2}b\) and the point \((a,b)\) would be a point in the first quadrant with integer coordinates and belonging to the line \(x=\sqrt{2}y\), which cannot happen. Hence, \(\sqrt{2}\) is irrational (and, therefore, so is \(-\sqrt{2}\)).

Assuming now that the diagonal \(d\) and the side \(l\) of the square are commensurable, we have \(\frac{d}{l}\in\mathbb{Q}\), that is, \(d=r.l\) for some \(r \in \mathbb{Q}\). Since \(\pm \sqrt{2} \notin \mathbb{Q}\), we have \(r\neq \pm \sqrt{2}\) and the point \((d,l)\) belongs to neither of the previous two lines, so its successive iterates get closer to the line \(x=-\sqrt{2}y\) and away from the line \(x=\sqrt{2}y\). Therefore, beyond a certain order, at least one of the coordinates of the iterates will be negative and so cannot be the length of a line segment, which contradicts the fact that, geometrically, we can go on indefinitely constructing new squares. Therefore, \(d\) and \(l\) are incommensurable.

Note that the points of the first quadrant on the line \(x=\sqrt{2}y\) are precisely the points corresponding to pairs \((d,l)\) where \(d\) is the diagonal of a square and \(l\) is its side. Indeed, that happens if and only if \(d=\sqrt{2}l\), with \(d,l>0\). So, starting with an arbitrary square, and applying the construction procedure of new squares used in the proof of incommensurability, we obtain a sequence of points in the first quadrant on the line \(x=\sqrt{2}y\) corresponding to ordered pairs of lengths of diagonals and sides of those squares. Will these be the only points for which the successive iterates always stay in the first quadrant?

The app below seems to suggest so. For each value of \(n\), the grey area represents the set of points for which at least the first \(n\) iterates stay in the first quadrant. Increasing the value of \(n\), we can see that the grey area gets smaller and smaller, which suggests that only the points in the first quadrant belonging to the line \(x=\sqrt{2}y\) lie in the grey area for all values of \(n\) (note that the points whose successive iterates always stay in the first quadrant are precisely those that belong to the grey area for all values of \(n\)).

Instructions:
Click on the red dot to change the initial point; you can also choose if you want to see the lines joining the iterates shown in the graphic, restrict the initial point to the points inside the grey area, change the value of \(n\) corresponding to the grey area or change the regular polygon on whose geometric proof of incommensurability this dynamic process is based.

Why does this happen?

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