Proof that the group of rotations of the circle has elements of infinite order.

Firstly, I am aware that this question has been asked twice before. First Time . Second Time.

My question is really about the validitiy of these answers. Note that in the first answer the proof relies on the irrationality of $\pi$, while the second proof only relies on the irrationality of $\sqrt{2}$.

It feels to me that there should be some flaw in this second proof, but I can't say exactly what it is.We see that if she/he were to replace "360" with "$2\pi$" the proof would not work. Any help towards a resolution would be awesome.

A followup question (assuming proof two is wrong): Is there any proof of this that does not rely on the irrationality of $\pi$?

• True $\pi$ is transcendental, while $\sqrt 2$ is "merely" irrational. Nonetheless, $n r \ne q$ for irrational $r$ and rational $n, q$ – Doug M Dec 14 '16 at 23:50
• The first proof is not relying on the irrationality of $\pi$. It is relying on the fact that there are irrational numbers. Rotation by any irrational multiple of $\pi$ will work. I think part of your confusion is due to the fact that the first proof is in radians and the second is in degrees. – D Wiggles Dec 14 '16 at 23:50
• You're right, the proof does not rely on the irrationality of $\pi$. I think this fixes the original misunderstanding but now I have a new one haha: If angle $\alpha$ not a rational multiple of $\pi$ why is it true that $k \alpha \neq 2m\pi$ for $m, k \in \mathbb{Z}$. – Sean Haight Dec 14 '16 at 23:55
• Contrapositive: if $k\alpha = 2m \pi$, then $\alpha = \frac{2m}{k} \pi$ where $\frac{2m}{k} \in \Bbb Q$ since $m, k \in \Bbb Z$. Thus $\alpha$ is a rational multiple of $\pi$. – pjs36 Dec 14 '16 at 23:59