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. | |