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Question : Show that $\arctan(n)$ is irrational for all $n \in \mathbb{N}$.

Hint:

My solution doesn't use continued fraction.

I am interested in other possible proofs for this question.

1 Answers1

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Suppose that $\arctan n=r\in\Bbb Q$, where $n$ is a non-zero integer. Then $r$ is not zero, so $2r$ is not zero, and $$\cos2r=\frac{\cos^2r-\sin^2r}{\cos^2r+\sin^2r} =\frac{1-\tan^2r}{1+\tan^2r}=\frac{1-n^2}{1+n^2}$$ which is rational. But this contradicts the result that the cosine of a non-zero rational number is irrational.


As for the proof of this result, it is usually done by taking an integral such as $$\int_0^r f(x)\sin x\,dx\ ,$$ where $f(x)=x^n(a-bx)^{2n}(2a-bx)^n$ and $r=a/b$, and showing that if $n$ is large we get contradictory estimates for the integral. See, for example, my lecture notes, starting at page 20.
David
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  • My solution is equal to what you did: I suppose $arctg(n)$ is rational, so $2arctg(n) = c = \frac{a}{b} \in \Bbb Q$ . So, $tg(\frac{c}{2})=n$. Define $f(x) = x^n \frac{(a-bx)^n}{n!}$. Take $\int_0^{\frac{a}{b}} f(x)\sin x dx = F'(a/b)sin(a/b) +F(0)(1- cos(a/b)) > 0$, where $F(0)$ and $F'(a/b)$ are integers. But $\frac{(1-cos(x))}{sen(x)}=tg(\frac{x}{2})$. So, $F'(a/b) +F(0)tg(\frac{a}{2b}) = F'(a/b) +F(0)tg(\frac{c}{2}) > 0$ ... – João Conceição Jul 09 '14 at 01:53
  • By an identical proof: $\arctan \frac 1n$ is irrational for all $n \in \mathbb{N}$. – Descartes Before the Horse Jan 15 '20 at 03:04