# Irrationality of $\frac{1}{\pi} \arccos{\frac{1}{\sqrt{n}}}$

This paper arxiv.org/pdf/0911.1933 discusses, regarding the irrationality of certain trigonometric functions. Recently, i encountered this problem which says states the given function, $$f(n)=\frac{1}{\pi} \arccos{\frac{1}{\sqrt{n}}}$$ is irrational for every odd $n \geq 3$. But i couldn't find the proof anywhere. Can anyone provide me with the proof.

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You could find the proof in this book: Proofs from the Book by Martin Aigner, Günter M. Ziegler, Karl H. Hofmann, pages 40-41

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Don't you think it's more appropriate to write that as a comment? – Asaf Karagila Sep 17 '10 at 11:17
@Asaf: It really doesn't matter! I got what i wanted, and Baudrillard deserves credit for that! – anonymous Sep 17 '10 at 11:26
@Asaf Karagila In one of the previous posts by Chandru1 I've written the source in the comments, but since the question gathered a lot of comments, it got lost in "see n other comments" line. I think that the source of the problem with a complete solution is important enough to be left as an answer. I've made it community wiki, as you can see. – Baudrillard Sep 17 '10 at 11:29
This is a perfectly valid answer. No idea why people insist on this being a comment. So what if Baudrillard didn't come up with it him/herself? – Aryabhata Sep 17 '10 at 16:18

If you want to prove that a certain angle is irrational with respect to $\pi$, most of the times it follows from the following simple result:

Lemma: If $n, k \in {\mathbb Z}$ and $n>0$ then $2 \cos( \frac{k \pi}{n})$ is an algebraic integer.

Proof: We show that there exists a $P_n \in {\mathbb Z}[x]$ monic of degree $n$ so that

$$2 \cos(nx) = P_n(2 \cos(x)) (*) \,.$$

$P_0(X)=1, P_1(X)=X$ and using

$$\cos((n+1)x)+ \cos((n-1)x)=2 \cos(x) \cos(nx) \,,$$

we get the recurence:

$$P_{n+1}(x) = XP_n(x)-P_{n-1}(X) \,.$$

Now $(*)$ follows by induction, and then Lemma follows immediately.

Now, back to the problem. Suppose by contradiction $\frac{1}{\pi} \arccos{\frac{1}{\sqrt{n}}}$ is rational. Then we can find $n, k \in {\mathbb Z}$ and $n>0$ so that

$$\frac{1}{\pi} \arccos{\frac{1}{\sqrt{n}}}=\frac{k}{n} \,.$$

Hence

$$\frac{1}{\sqrt{n}}= \cos(\frac{2k}{n}) \,.$$

Then, by Lemma, $2\frac{1}{\sqrt{n}}$ is an algebraic integer, and hence is its square. But then $\frac{4}{n}$ is an algebraic integer and rational, thus integer.

This shows that $4$ divides $n$.

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Very nice! $\text{ }$ – Matt E May 16 '11 at 3:49
@user9176: if you want boldface, just put it between double asterisks: **this is bold** creates this is bold – Ross Millikan May 16 '11 at 4:02