distribution of $\{na\}$ when $a$ is irrational number

Discussion in 'Mathematics' started by kvphxga, Oct 8, 2018.

1. kvphxgaGuest

(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.