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.
Irrationality of $ \frac{1}{\pi} \arccos{\frac{1}{\sqrt{n}}}$
2 Answers
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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6This 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? – 2010-09-17
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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1@user9176: if you want boldface, just put it between double asterisks: \*\*this is bold\*\* creates **this is bold** – 2011-05-16