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The precise question is this: given a irrational $r\in \mathbb{R}$, is it true that $r\mathbb{Z}$ is dense (topologically) in $\mathbb{R}/\mathbb{Z}$?

The reason this came up was actually a teaching moment for Calc II; I wanted to use $sin(n)$ as an example of a bounded divergent sequence, but I was actually not sure (and certainly not sure how to explain to Calc II students) that it doesn't converge to some limit. I suspect the question above is true, which would imply that the example works (among other, more interesting, things).

Edit: The question was originally phrased in terms of transcendentals, but given the answer below, this is overkill and deceptive.

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You don't have to prove anything like this to show that $\sin n$ diverges. $\sin (n+1) - \sin (n-1) = 2 \sin 1 \cos n$. If this is not close to $0$, then $\sin n$ can't be close to both. If it is close to $0$, then $\cos n$ must be close to $0$, which implies that $\sin n$ is close to $1$ or $-1$, so $\sin (n+1) = \sin n \cos 1 + \sin 1 \cos n$ is close to $\pm \cos 1$, hence far from $\sin n$. You can also find the minimum of $(\sin (x-1)-\sin x)^2 + (\sin x - \sin(x+1))^2$ (at $x=\pi/2 + k \pi$) and check that it is above $0$. – Douglas Zare Apr 16 '13 at 19:04
Another trig-identity-based approach starts with $\sin 2n = 2\sin n\cos n$ and $\cos 2n = 2\cos^2n - 1$ to argue that the limit $S$, if it existed, would have to be 0: If $S\ne0$, then $S\approx2S\cos n$ implies $\cos n$ tends to the limit $C=1/2$, which does not satisfy $C=2C^2-1$. But now $\sin(n+1) = \sin n\cos1+\cos n\sin1$ implies $\cos n$ tends to $0$ (since $\sin1\ne0$), and this contradicts the identity $\sin^2n+\cos^2n=1$. – Barry Cipra Apr 16 '13 at 21:04
I just noticed, the question about the divergence of $\sin n$ was asked last year at… – Barry Cipra Apr 16 '13 at 21:18
up vote 6 down vote accepted

Yes. For elementary reasons.

Suppose it weren't dense. Then there would be some little interval not hit, of some positive length say $1/N$.

But this cannot happen. Divide the circle into N little equal intervals and consider $0,r,2r,...,Nr$. By pigeonhole some two of them lie in the same interval, say $ar,br$ (and these elements are distinct since $r$ is not rational). Then modulo $\mathbb{Z}$, $-1/N < (a-b)r < 1/N$, and so some multiple of this will lie in the interval we supposed contained no element.

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Clever! I was thinking about special properties of transcendental numbers (and blanking) when irrationality was enough. I've edited the question accordingly. – Richard Rast Apr 16 '13 at 18:21

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