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What is the simplest way to show that $\cos(r \pi)$ is irrational if $r$ is rational and $\displaystyle r \in \left(0,\frac{1}{2} \right)\setminus \left\{\frac{1}{3} \right\}$?

I proved it using the following sequence $x_1 = \cos(r \pi)$; $x_{k} = 2 x_{k-1}^2-1$ and periodicity of the cosine function. Is there any proof that is based on definition of rational numbers and trigonometric identities only? Thanks!

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    @Qiaochu Yuan: Thanks2012-06-24

1 Answers 1

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Let us try to rescue another poor question from Unanswered Questions' inferno, following the comment by Qiaochu:

For $\,\,\displaystyle{r=\frac{m}{n} \in\mathbb Q}\,\,\,$ we get that $\,\,\,2\cos r\pi:=e^{ri\pi}+e^{-ri\pi}\,\,$ , and since $\,\,1=e^{2mi\pi}=\left(e^{ri\pi}\right)^{2n}\,\,$ we

have that $\,2\cos r\pi\,$ is an algebraic integer (as it satisfies an integer monic polynomial), so if it were

also rational it'd have to be an integer (which is not if $\,\displaystyle{r\notin \mathbb Z\,,\,\frac{1}{2}\mathbb Z}\,$), so it must be irrational.

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    Thanks for the edit, @Generic Human.2012-06-23