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A student asked me the following today :

Is $S:= \{\tan(x) : x\in \mathbb{Q}\}$ a group under addition?

I am quite perplexed by it. Clearly, the only non-trivial part is to check

For any $x, y\in \mathbb{Q}$, does there exist $z \in \mathbb{Q}$ such that $$ \tan(z) = \tan(x) + \tan(y) $$

A couple of things I have learnt from various sources that appear to be relevant are :

  1. (Source) If $x \in \mathbb{Q}\setminus\{0\}$, then $\tan(x) \in \mathbb{Q}^c$.
  2. (Source) $\arctan(\alpha)$ is a rational multiple of $\pi$ iff $\exists n \in \mathbb{Z}$ such that $(1+i\alpha)^n \in \mathbb{R}$

My guess is that it is not a group, and my initial goal was to find two $x,y \in \mathbb{Q}$ such that $\tan(x) + \tan(y) \in \mathbb{Q}$, but that seems to be going nowhere.

Perhaps I am missing something obvious, but it doesn't seem to strike me. Does anyone have a suggestion on how to tackle this? Thanks for your help.

Prahlad Vaidyanathan
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1 Answers1

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It follows rather easily from this fact: for every $0\neq n\in\mathbb N$ the number $e^{i/n}$ is transcendental. If we suppose it: if $\tan y=2\tan x$ then $(e^{2iy}-1)(e^{2ix}+1)=2(e^{2ix}-1)(e^{2iy}+1)$. If $x\neq0$ and $y$ are rational and $n$ is the least common denominator of $x$ and $y$, this becomes a polynomial equation for $e^{i/n}$, hence we get a contradiction.

[the used result is a special case of the Lindemann-Weierstrass theorem which says, in particular, that $e^\alpha$ is transcendental for every algebraic $\alpha\neq0$. There is probably a simpler proof in this case.]

user8268
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