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.



Integer multiples of a irrational dense in R/Z ?
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.


