(by $\{x\}$ I mean the fraction part of the real number $x$) If $a$ is an irrational number and $n$ is a integral number, what is the distribution of $\{na\}$? I'm asking for some continuous function $f:[0,1]\to\mathbb R$ such $\int_{\alpha}^{\beta}f(x)\;dx$ gives the probability that $\{na\}$ falls between $\alpha$ and $\beta$. When I calculated it for a bunch of irrational number, from $n=1$ to $10000$, I found that it's very close to uniform distribution. It's well known that {na} with a proper choose of $a$ could be arbitrary close to any real number in the $[0,1]$ interval. But this claim is more than that and wants the distribution to be uniform. I think that a quite simple simple proof may exist: If $a$ was rational, say $p/q$, a uniform discrete distribution have been existed. I mean if $n$ goes to infinity the number would fall into $[i/q,(i+1)/q]$ interval with probability $1/q$. Now If we could approximate $a$ with a rational $p/q$, with "sufficiently small" error, the same would happen for a. That is, {na}s would also fall into the $[i/q,(i+1)/q]$ with probability $1/q$. if $q$ goes to infinity the distribution would become continuous. And at last ... I think $a = p/q + c/(q^2)$ where $c$ is smaller than or equal to one, is a sufficiently good approximation. Good in the sense that such an approximation causes a uniform distribution.

17$\begingroup$ en.wikipedia.org/wiki/Equidistribution_theorem $\endgroup$ – Qiaochu Yuan Dec 6 '10 at 23:06

3$\begingroup$ Kuipers and Niederreiter's Uniform Distribution of Sequences is another good textbook source for this and related material. $\endgroup$ – Ed Dean Dec 7 '10 at 0:43

$\begingroup$ many thanks ... are distributions like {ab^n} or similar distributions covered in the textbook? $\endgroup$ – kvphxga Dec 7 '10 at 9:16

7$\begingroup$ @kvphxga, there's one way to find out what's covered in that book.... $\endgroup$ – Gerry Myerson Dec 7 '10 at 11:49
The distribution is known to be uniform (a result due to Weyl, I believe). An excellent reference for this (and much else) is Dym and McKean's book on harmonic analysis.

1$\begingroup$ Graham, Knuth, and Patashnik's Concrete Mathematics (p. 87, second edition) says that the result was discovered independently by Bohl, Sierpinski, and Weyl at about the same time in 1909. $\endgroup$ – Mike Spivey Dec 6 '10 at 22:21
For rational $a$ the answer (explicit bound for the error term) is given by Ostrowski's theorem (Ostrowski A. Bemerkungen zur Theorie der Diophantischen Approximationen,Abh. Math. Sem Hamburg, 1922, 1, s. 7798). It depends on the sum of partial quotients in continued fraction expansion of number $a$. For real number it is sufficient to take good rational approsimation (one of convegents). See also Khintchine A. Ya. Ein Satz uЁber KettenbruЁche, mit arithmetischen Anwendungen. — Mathematische Zeitschrift, 18: 1 (1923), 289–306.
For what it's worth, in the language of measures one can reformulate your statement as $$\mu_n:=\frac{1}{n}\sum_{k=1}^n\delta_{k\alpha}\rightharpoonup\mathcal{L}^1$$ on the circle $\mathbb R/\mathbb Z$ (in the weak* topology, meaning $\int f\,d\mu_n\to\int f\,d\mathcal{L}^1$ for any continuous $f$). Any limit measure $\mu_\infty$ has total mass $1$ and is invariant under the translation $\tau_\alpha$ (as $\(\tau_\alpha)_*\mu_n\mu_n\\to 0$), so $$\int e^{2\pi i nx}\,d\mu_n=:\widehat{\mu_\infty}(n)=\widehat{(\tau_\alpha)_*\mu_\infty}(n)=e^{2\pi i n\alpha}\widehat{\mu_\infty}(n)$$ and thus $\widehat{\mu_\infty}(n)=0$ for $n\neq 0$, $\widehat{\mu_\infty}(0)=1$. By uniqueness, $\mu_\infty=\mathcal{L}^1$. By compactness of measures we are done.