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Is $\tan^{-1}2$ an irrational number or a rational number? How to show that?

Or generally how to show $\tan^{-1}3, \tan^{-1}4, \tan^{-1}5...$ is irrational or rational?

Akiva Weinberger
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esege
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  • Looking at a calculation, that number seems irrational. I only know the approach that is used for the irrationality of $\sqrt{2}$, assuming $x$ to be rational and looking for an argument why it will not work with integer nominator and denominator. In this case there is the additional problem of what representation to choose - a power series, some trigonometric relation, a continued fraction? – mvw Feb 09 '16 at 12:02
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    It's known that the only rational values of $x$, $0\le x<1/2$, such that $\tan \pi x$ is rational are $x=0$ and $x=1/4$. – Gerry Myerson Feb 09 '16 at 12:11

1 Answers1

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Trasform this into $\cos x = \frac{1}{\sqrt{5}}$. Now, you have an equation $$e^{ix}+e^{-ix}=\frac{2}{\sqrt{5}}$$ or, with $z=e^{ix}$, $$z^2-\frac{2}{\sqrt{5}}z+1=0$$ Now, above is an algebraic equation, so $z$, the solution of this equation, must be algebraic. By Lindemann-Weirstrass theorem, if $e^{ix}$ is algebraic, then $ix$ must be transcendental, except if $x=0$.

Of course, $\tan^{-1}1 = \frac{\pi}{4}$ is also transcendental (and therefore irrational). If you want to prove that it's an irrational multiple of $\pi$, you have to proceed a bit differently.

Consider an equation $z+z^{-1}=2a$, $|a|<1$, which has solutions $$z=a\pm i\sqrt{1-a^2}$$ Now we require $z=e^{i\pi p/q}$. Raise this to the power of $q$: $$e^{i\pi p}=\pm 1=(a\pm i\sqrt{1-a^2})^q$$ There must be such $q$, so that the right hand side is an integer ($\pm 1$). If you are given $a$, then you can just check that particular case. In general, you are basically looking for roots of unity in terms of their cartesian components. For example, you can set $a=\sqrt{n}/2$ and check for which $n$ this has a solution for $q$.

orion
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