Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
8  
The title doesn't seem to ask the same question as the body. –  Qiaochu Yuan Aug 3 '11 at 20:31
    
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 $\cos(\pi x) \in \mathbb{Q} \text{ where } x \in \mathbb{Q}$. –  Henry Aug 3 '11 at 20:49
1  
@rajah9: I still don't think you are asking the question you meant to ask. Don't you want $y = \cos (\pi x)$? –  Qiaochu Yuan Aug 3 '11 at 21:10
    
@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. –  rajah9 Aug 4 '11 at 14:09
    
@Americo Tavares, than you for editing the question. –  rajah9 Aug 4 '11 at 14:09

3 Answers 3

up vote 6 down vote accepted

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.

share|improve this answer
1  
I think the question is asking for $x$, not $x/\pi$, to be rational. –  Rahul Aug 3 '11 at 20:50
2  
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. –  Robert Israel Aug 3 '11 at 21:03
    
Thanks, the reference to Lindemann's theorem and your proof were what I was looking for. –  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:

http://www.uni-math.gwdg.de/jahnel/Preprints/cos.pdf

http://www.mathpages.com/home/kmath460/kmath460.htm

http://groups.google.com/group/sci.math/msg/9a4a0e0fe9e2f8e6

http://www.oberlin.edu/faculty/jcalcut/tanpap.pdf

share|improve this answer

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.

share|improve this answer
    
I think the question changed a bit; you may want to revisit this. –  Willie Wong Aug 3 '11 at 21:12
    
@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. –  Ross Millikan Aug 3 '11 at 21:18

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.