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My question is related to the following one: Is $\{\tan(x) : x\in \mathbb{Q}\}$ a group under addition?

It was shown that the above set is not closed under addition using the Lindemann-Weierstrass theorem. That tells us that $\tan(x) + \tan(y)$ need not be of the form $\tan(z)$ where $x,y,z \in \mathbb{Q}$.

An initial attempt in trying to show that the above set is not closed involved trying to find $x, y \in \mathbb{Q}$ such that $\tan(x) + \tan(y) \in \mathbb{Q}$, because $\tan(x)$ is irrational when $x$ is a non-zero rational. However, I have not managed to find any rational numbers that do satisfy this property. I'm looking for non-trivial examples, so I'm discounting the case $x=-y$ for which $\tan(x) + \tan(y) = 0$.

The primary difficulty that I faced was that the decimal expansion gave me no clues whether I was looking at a rational number or an irrational number. I also tried playing with the formula for $\tan(x+y)$ but I was not able to derive anything from that either. I checked a few of my guesses on Wolfram|Alpha and it says that they are all transcendental.

I'm beginning to suspect that $\tan(x) + \tan(y)$ is never rational when $x$ and $y$ are rational, but I don't know how to prove that either. Does anyone have any ideas on the best way to proceed? Thank you for your help.

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  • $x=y=0$ then $\tan x+\tan y$ is rational –  Nov 04 '15 at 18:05
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    "Discounting $x = -y$" –  Nov 04 '15 at 18:06
  • In other words, you're asking if ${\tan x+\tan y\mid x,y\in\Bbb Q}\cap\Bbb Q$ equals the empty set. – Akiva Weinberger Nov 04 '15 at 18:10
  • @AkivaWeinberger Yes, that's a neater way to put it. I want to know if it's not the set ${ 0 }$. –  Nov 04 '15 at 18:34
  • The answer would be no if it is known that tan x must be transcendental for $x \neq 0$ rational. Is that known? – Aravind Nov 04 '15 at 18:39
  • @Aravind It is known that $\tan(x)$ is transcendental when $x$ is a non-zero algebraic number. This result is stated in Niven's monograph Irrational Numbers as an application of the Generalized Lindemann Theorem. –  Nov 04 '15 at 18:49
  • @Aravind Could you clarify how the answer would then be known? –  Nov 04 '15 at 18:55
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    Let $x=\dfrac{p}{q}$ and $y=\dfrac{r}{s}$ and let $z= tan \dfrac{1}{qs}$. Then it appears that tan x+tan y $\in \mathbb{Q}$ yields a polynomial equation for $z$. – Aravind Nov 04 '15 at 18:56
  • @Aravind That looks like a legitimate solution to me! You should write that up as an answer. (Short version: by induction, $\tan(nz)$ is a rational function of $\tan(z)$ for all $n$; therefore, $\tan(p/q)$ and $\tan(r/s)$ are rational functions of $\tan(1/qs)$, etc etc.) – Steven Stadnicki Nov 04 '15 at 19:57
  • I don't understand how that works. If I have a rational function of an irrational, would that be rational? –  Nov 05 '15 at 02:09
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    @BrahadeeshS. if Aravind hasn't written up his argument by tomorrow morning I'll give it a quick writeup to make things clear. (It's not just 'irrational' but 'transcendental' that's essential here). – Steven Stadnicki Nov 05 '15 at 02:21

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Let $x=\dfrac{p}{q}$ and $y=\dfrac{r}{s}$ and $z=\dfrac{1}{qs}$.

We observe that tan x and tan y are polynomials in tan z; further tan x + tan y is a non-constant polynomial if $x+y \neq 0$. Here some calculation should show that tan $A\theta$ + tan $B\theta$ is a non-constant polynomial for $A + B \neq 0$.

Hence we have:if tan x + tan y is rational, then tan z is the root of a polynomial with rational coefficients, contradicting the Generalized Linedmann Theorem, as pointed out by the OP.

Aravind
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    Wouldn't $\tan(x)$ and $\tan(y)$ would be ratios of polynomials in $tan(z)$? –  Nov 07 '15 at 04:42