The incommensurability of side and diagonal in a regular pentagon
In this regular pentagon, call the side length s and the length of the diagonal d. Our job is to show that s and d are incommensurable.
First notice that in a regular pentagon, a diagonal is parallel to the opposite side. This follows from the symmetry of the pentagon.
Using this fact for two adjacent sides shows that this green quadrilateral is a parallelogram, with opposite sides equal: all four sides of this parallelogram have length s.
Now suppose that the side and the diagonal are commensurable: there exists a length h which fits exactly a whole number of times into the diagonal and into the side. The length h is symbolized by the distance between two adjacent dots in this picture.
The length of the highlighted segment must also be a whole number of h's, since it is the difference of two such numbers, d and s. Call this length d*. d* = d - s.
The length of this highlighted segment is d - 2d*. So this length must also be equal to a whole number of h's. Call this length s*. s* = d - 2d* = 2s - d.
Now notice that the five diagonals determine a smaller regular pentagon inside our original one. The side length of this pentagon is the s* we just considered.
In the smaller pentagon, the diagonals are again parallel to the opposite sides. It follows that the highlighted figure is a parallelogram, and that the diagonal in the smaller pentagon has length d*.
So in the smaller pentagon, side and diagonal are both multiples of our original h.
We can repeat the construction with the smaller pentagon.
Since s* is strictly shorter than s, and they are both whole multiples of h, they must differ by at least h. We can repeat the process again and again, and obtain an infinite decreasing sequence of side lengths s > s* > s** ... each of which is at least h smaller than the one before. But this can't work, because there was only a finite number of h's in s to start! This contradiction shows that the hypothesis, that s and d are commensurable, must be false.