A geometric proof that $\sqrt{3}$, $\sqrt{5}$ and $\sqrt{7}$ are irrational

Abstract

We show geometrically that $\sqrt n$ is irrational for $n=3,5,7$ by adapting Tennenbaum's geometric proof that $\sqrt 2$ is irrational. We also show that this method cannot be used to prove the irrationality of $\sqrt n$ for a bigger $n$.