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Under which conditions will successive rotations of $(0,1)$ by an angle $\theta$ guarantee that given $\delta > 0$ and some point $p$ on the unit circle, there exists some $n$ such that rotating $(0,1)$ by $n\theta$ yields a point within $\delta$ radians of $p$? That is, are there certain angles that will guarantee that the set generated by rotating $(0,1)$ by integral multiples of $\theta$ is dense in the unit circle?

More specifically, I am working with the angle $\arccos(\frac{1}{3})$ and wish to show that this property is true, but I'm not sure how to go about proving this. It seems likely I'm missing something simple here... Does it hold for any angle that is an irrational multiple of $\pi$? If so, why?

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To complete Arturo's answer, $\theta=\arcos(1/3)$ is irational with respect to $\pi$, if you need help proving it check the little Lemma I posted here: math.stackexchange.com/questions/4845/… –  N. S. May 16 '11 at 3:45

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up vote 11 down vote accepted

Yes, a necessary and sufficient condition is that the angle not be a rational multiple of $\pi$.

The problem is equivalent to considering the subgroup generated by $\theta$ in the additive group $\mathbb{R}/\{2k\pi\}$, viewed as the interval $[0,2\pi)$. Under the map $x\mapsto \frac{x}{2\pi}$, we can consider $\mathbb{R}/\mathbb{Z}$, and the subgroup generated by $r+\mathbb{Z}$. The assertion is that the subgroup is dense if and only if $r$ is irrational.

If $r$ is rational, then $\langle r+\mathbb{Z}\rangle$ is finite, so it is not dense.

That the subgroup is dense when $r$ is irrational is a consequence of the Equidistribution theorem. Added. Or more simply, a consequence of Dirichlet's approximation theorem (as noted by Gerry Myerson).


Here is Dirichlet's Theorem:

Theorem. (Dirichlet) For any real number $\alpha$ and positive integer $N$, there exist integers $p$ and $q$, $1\leq q\leq N$, such that $$\left|q\alpha - p\right|\lt\frac{1}{N+1}.$$

To use it to derive the desired result, let $r$ be an irrational number in $[0,1]$ (corresponding to an angle which is not a rational multiple of $\pi$), and let $\beta\in[0,1]$ be any real number We want to show that for every $\epsilon\gt0$ there exists an integer $n$ such that $nr - \lfloor nr\rfloor$ is within $\epsilon$ of $\beta$ (that is, there is a multiple of the rotation which will land $(0,1)$ within $\epsilon$, along the circle, of $\beta$). Find an integer $M$ such that $\frac{1}{M+1}\lt\epsilon$. By Dirichlet's Theorem, you can find integers $p$ and $q$ such that $|q r - p|\lt \frac{1}{M+1} \lt \epsilon$. That means that if you take $qr$ modulo $1$, you lie within $\frac{1}{M+1}$ of either $0$ or $1$. Either way, by taking multiples of $qr$ you divide the circle into segments of length at most $\frac{1}{M+1}$, and by the Archimedean property some multiple of $qr\bmod 1$ will be strictly larger than $\beta$ (Note that $qr-p$ cannot be equal to $0$, because we are assuming that $r$ is irrational). The distance from $\beta$ to the smallest multiple which is strictly larger is less than $\frac{1}{M+1}$, which shows that there is some multiple of $r\bmod 1$ which is within $\epsilon$ if $\beta$, as desired.

(That is: use Dirichlet's Theorem to get a multiple of the rotation that is smaller in absolute value than $\frac{2\pi}{M+1}$, and note that since $\theta$ is not a rational multiple of $\pi$, this small angle cannot be $0$; then taking multiples of this rotation you get within $\frac{2\pi}{M+1}$ of any point in the circle; pick $M$ sufficiently large so that $\frac{2\pi}{M+1}\lt \epsilon$, and you are done).


So the answer to your question is "yes": the images of $(0,1)$ under successive rotations by an angle $\theta$ are dense in the unit circle if and only if $\theta$ is not a rational multiple of $\pi$.

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It's also a consequence of much weaker (and much easier to prove) theorems than equidistribution. –  quanta May 15 '11 at 23:33
    
In particular, it follows from Dirichlet's Theorem on diophantine approximation. –  Gerry Myerson May 15 '11 at 23:37
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@Gerry: Ah, indeed it is; cannonballs swatting flies strikes again. –  Arturo Magidin May 15 '11 at 23:41
    
@quanta; quite so. thanks. –  Arturo Magidin May 15 '11 at 23:41
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@Rachel: I added an argument. Use Dirichlet's theorem to get that there is a multiple of the rotation by $\theta$ which maps $(1,0)$ to within $\frac{1}{N+1}$ radians of $(0,1)$; if $\theta$ is irrational, it cannot be exactly $0$; now taking multiples of that rotation you can get within $\frac{1}{N+1}$ of any point on the circle, by the Archimedean property. –  Arturo Magidin May 16 '11 at 0:38

It's dense if and only if $\frac{\theta}{\pi}$ is irrational

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