It is a well known fact that for $\alpha$ irrational that $\langle n\alpha\rangle$ is dense on the unit circle. I want to know what the result is for computing $a_n=\min_{1\le N \le n} |\langle N \alpha\rangle|$.
Update Answer Due to Guest:
The precise decay of $a_N$ depends heavily on the Diophantine properties of $\alpha$, and in particular on its continued fraction expansion. In general $a_N<\frac{1}{N}$, as Michael Hardy notes; we can write this as $\limsup(Na_N)<1$. This is a consequence of the pigeonhole principle and known as Dirichlet's theorem in Diophantine approximation.
For algebraic irrational $\alpha$, Roth's theorem tells us that for every $\epsilon>0$ and any $N>N(\epsilon,\alpha)$, $$a_N > C_\epsilon N^{-1-\epsilon}.$$
Numbers for which $\liminf(Na_N)>0$ are called badly approximable. A lot is known about these numbers, but many questions about them remain open as well. They are uncountable in number. Numbers which are not badly approximable, i.e. which decay faster than $1/N$ at infinitely many scales, are also uncountable and form a full measure set in $\mathbb{R}$.
Then there are numbers for which $a_N$ not only decays much faster than $1/N$, but the decay can even be exponential for infinitely many $N$. By Roth's theorem, all these numbers are transcendental. It is instructive to try and construct such a number; the answer can be found on Mathoverflow with a little googling.
For more information about these things consult any book on Diophantine Approximation; good sources are W.M. Schmidt's book and Kuipers and Niederreiter (Uniform distribution of sequences).
I will take $\langle \beta\rangle$ to mean the fractional part of $\beta$, and I will construe absolute value on $\mathbb R\bmod 1$ in such a way that, for example, $|0.9|=0.1$ since $0.9$ differs from $1\equiv 0\bmod 1$ by $0.1$.
Trigger warning: I will use the same notation for three different things:
According to the equidistribution theorem, for every interval $J\subseteq \mathbb R\bmod 1$, $$ \lim_{n\to\infty} \frac {|\{\langle k\alpha\rangle : k\in\{1,\ldots,n\}\} \cap J| } n = |J|. $$ Hence for every $\varepsilon>0$, $$ \lim_{n\to\infty} \frac {|\{\langle k\alpha\rangle : k\in\{1,\ldots,n\}\} \cap (-\varepsilon,\varepsilon)| } n = 2\varepsilon. $$
. . . and now I will make some hand-waving comments suggesting a direction in which to go to seek the solution of this problem: Try to show that as $n$ grows, the distance between neighboring points of the sequence in the long run approximates $1/n$; therefore the distance from $0$ to the nearest point of the sequence behaves like $1/n$.