Given a range of the rational numbers, $x$, between $0$ and $2\pi$\, what is the set of rational numbers $ y = \cos(x) $?

I was inspired by the stackoverflow question Can $\cos(a)$ ever equal $0$ in floating point? (The irrational number $\frac{\pi}{2}$ does not translate well into a computer representation.)

I looked for rational cosines, and came up with the likes of $$ 0, \frac{\pi}{3},\frac{\pi}{2}, \pi, \frac{3\pi}{2}$$ Following this rabbit hole, I wondered if there were any rational (Floating Point) numbers (besides $0$) that yielded rational cosines.

One respondent opened a different question, on english.stackexchange.com, What is the upper bound on “several”? which involves the size of the set in question.

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    $\begingroup$ The title doesn't seem to ask the same question as the body. $\endgroup$ – Qiaochu Yuan Aug 3 '11 at 20:31
  • $\begingroup$ All floating point numbers are rational, so the floating point cosine of a floating point number might answer your question, if that is what you are asking. But I think you are actually asking if there are rational non-zero exact solutions for $\cos(x) \in \mathbb{Q} \text{ where } x \in \mathbb{Q}$ and if so what they are. I don't see why it has to be restricted to $(0,2\pi]$ unless you are looking for solutions where $\cos(\pi x) \in \mathbb{Q} \text{ where } x \in \mathbb{Q}$. $\endgroup$ – Henry Aug 3 '11 at 20:49
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    $\begingroup$ @rajah9: I still don't think you are asking the question you meant to ask. Don't you want $y = \cos (\pi x)$? $\endgroup$ – Qiaochu Yuan Aug 3 '11 at 21:10
  • $\begingroup$ @Henry, I agree, it does not have to be restricted to $(0, 2\pi] $ but I was looking for solutions on the unit circle. Yes, I am looking for rational, non-zero exact solutions. $\endgroup$ – rajah9 Aug 4 '11 at 14:09
  • $\begingroup$ @Americo Tavares, than you for editing the question. $\endgroup$ – rajah9 Aug 4 '11 at 14:09

The only cases where $x/\pi$ and $\cos(x)$ are both rational are the obvious ones, where $2\cos(x)$ is an integer. The slick way to show this uses the following facts:

1) when $r$ is rational, $e^{\pm i r \pi}$ are algebraic integers

2) the sum of algebraic integers is an algebraic integer

3) the only algebraic integers that are rational numbers are (ordinary) integers.

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    $\begingroup$ I think the question is asking for $x$, not $x/\pi$, to be rational. $\endgroup$ – Rahul Aug 3 '11 at 20:50
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    $\begingroup$ It's not clear, since rajah9 mentioned $0, \pi/3, \pi/2, \pi,3 \pi/2$. If $x$ is algebraic and nonzero, $\cos(x)$ is transcendental by Lindemann's theorem. $\endgroup$ – Robert Israel Aug 3 '11 at 21:03
  • $\begingroup$ Thanks, the reference to Lindemann's theorem and your proof were what I was looking for. $\endgroup$ – rajah9 Aug 4 '11 at 14:03

Robert Israel has already answered your specific question. For some deeper issues related to your question, see the following web pages:






Given the rationals are dense and $\cos$ is continuous, you can say that $\{y=\cos{x}\mid x\in \mathbb{Q} \cap [0,2\pi]\}$ is dense in $[-1,1]$ and countable.

  • $\begingroup$ I think the question changed a bit; you may want to revisit this. $\endgroup$ – Willie Wong Aug 3 '11 at 21:12
  • $\begingroup$ @Willie Wong: thanks. I think Robert Israel's answer is squarely on point for the new question. I think I would leave this up, but could be convinced to delete it. $\endgroup$ – Ross Millikan Aug 3 '11 at 21:18

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