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I am trying to prove that $\cos n$, $n\in \mathbb{N}$, is dense in $[-1,1]$ and would like to know if my answer is rigorous and correct enough.

Given any $x,y \in [-1,1]$ with $y>x$, there are $a,b \in [0,\pi)$ with $a>b$ such that $\cos a = x$ and $\cos b = y$, since the cosine function is decreasing on $[0,\pi)$.
Using the fact that $\{m+n\pi\mid m,n\in\mathbb{Z}\}$ is dense on $\mathbb{R}$, there exists $m+2n\pi$ $\in(b,a)$. Hence there is a $\cos(m+2n\pi)=\cos (m)\in (y,x) $ with $m\in \mathbb{N}$.

Is it correct?

Ѕᴀᴀᴅ
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jnyan
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2 Answers2

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Arguments don't just exist in vacuums waiting to be determined "correct" or otherwise. In this case, comment by HCP16 is appropriate: If you are willing to assume that $\{m + n\pi\}_{m,n \in \mathbb{Z}}$ is dense in $\mathbb{R}$, then a proof like this ought to work easily.

Taking what you've written at face value, one question that comes to my mind is: How do you go from $m + n \pi$ being dense to $m + 2n\pi \in (b,a)$?

SBK
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  • I meant to say, $m+2n\pi$ is dense in reals – jnyan May 20 '18 at 16:41
  • OK, sure. Still, the main point here is that assuming this result would seem to beg the question. – SBK May 20 '18 at 20:18
  • i understand. but for this specific problem, even after the assumed density, other answers used continuity of cosine, mod $2\pi$ function and other things. So I wanted to get the later part verified. – jnyan May 21 '18 at 03:57
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As others have said, how good your argument is depends greatly on what you are assuming has already been shown, but there are a couple of genuine minor gaps. If you are willing to assume that the cosine function is continuous and is decreasing on $[0,\pi]$, you can appeal to the intermediate value theorem to justify the existence of your $a$ and $b$, but you do need to use the closed interval $[0,\pi]$, not $[0,\pi)$: otherwise you can’t handle the possibility that $x=-1$.

If you are willing to assume that $\{m+n\alpha:m,n\in\Bbb Z\}$ is dense in $\Bbb R$ when $\alpha$ is irrational, then certainly $\{m+n\pi:m,n\in\Bbb Z\}$ is dense in $\Bbb R$, but that does not obviously justify your conclusion that there are $m,n\in\Bbb Z$ such that $m+2n\pi\in(b,a)$. However, $2\pi$ is also irrational, so $\{m+2n\pi:m,n\in\Bbb Z\}$ is also dense in $\Bbb R$, and we can indeed conclude that there are $m,n\in\Bbb R$ such that $m+2n\pi\in(b,a)$. (And I now see that you corrected that oversight in the comments, though not yet in the question itself.)

However, the fact that $\{m+n\alpha:m,n\in\Bbb Z\}$ is dense in $\Bbb R$ when $\alpha$ is irrational is really the heart of the argument and its most difficult part1: the rest is comparatively trivial, assuming that the intermediate value theorem and basic facts about the cosine function are already available (and I do think that a reasonable assumption). I would therefore expect a proof of the desired result to include (or, in a textbook, directly refer to) a proof of this fact.

That can actually be quite short: in this answer I used the pigeonhole principle to give a short, easy proof that $\{n\alpha\bmod 1:n\in\Bbb Z\}$ is dense in $[0,1)$ if $\alpha$ is irrational, where $x\bmod 1=x-\lfloor x\rfloor$ is the fractional part of $x$, and from this it follows immediately that $\{m+n\alpha:m,n\in\Bbb Z\}$ is dense in $\Bbb R$. In this answer I point out that the same argument works if we replace $\Bbb Z$ by $\Bbb Z^+$ and show how to extend the argument to show that in fact $\{m+n\alpha:m,n\in\Bbb Z^+\}$ is dense in $\Bbb R$.

1 The proof that $\pi$ is irrational is actually harder, but in most contexts that fact qualifies as known even if no proof has been exhibited.

Brian M. Scott
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  • Actually, the hardest part of the argument is the proof of $\pi$ being irrational, I think. – Aphelli Apr 25 '20 at 15:49
  • Density of ${m+\alpha n | n,m \in\mathbb{Z}}$ has an official name, Kroneckers Approximation Theorem. Also, OP insists on $n\in\mathbb{N}$. The link I posted in comments is for $\sin$ and can be copy/pasted 1:1 for $\cos$. – rtybase Apr 25 '20 at 16:15
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    @Mindlack: True, if one is starting from scratch, though I think that in most contexts it qualifies as known (or assumed) even if a proof hasn’t been exhibited. – Brian M. Scott Apr 25 '20 at 16:24
  • @rtybase: You mean a commonly used name: there is no such thing as an official name, since there is no authority deciding such things. It’s not clear whether the OP wants $n\in\Bbb N$, $m\in\Bbb N$, or both, but I did indeed overlook the places where he appears to make such restrictions. I’ve now added a bit to accommodate both. – Brian M. Scott Apr 25 '20 at 16:28
  • It was introduced by Kronecker, e.g. here. I learned it as such back in 1993. There are good books on this subject, my favourite one, having density of ${m+\alpha n | n\in\mathbb{N}, m\in\mathbb{Z}}$ given as an exercise. I think Kronecker deserves his name being mentioned, it also helps younger people to google for the right subject. – rtybase Apr 25 '20 at 17:32
  • Such a fine answer! Clear and beautiful exposition. Thanks Prof. Scott!! – James Apr 26 '20 at 19:53
  • Prof. can you help us with this question: https://math.stackexchange.com/questions/3645378/estimating-the-difference-between-an-irrational-and-a-rational-number – James Apr 26 '20 at 20:05