# On the set of good approximators in the sense of Dirichlet's theorem

This question came up when thinking about an older question that hasn't been answered as of now.

Let $$\mathbb{N}$$ be the set of positive integers. If $$\alpha\in\mathbb{R}$$, we say $$q\in\mathbb{N}$$ is good for approximating $$\alpha$$ if there is $$p\in\mathbb{Z}$$ such that $$\left|\alpha - \frac{p}{q}\right|< \frac{1}{q^2},$$

and denote the set of those positive integers by $$G_\alpha$$. The approximation theorem of Dirichlet states that $$G_\alpha$$ is infinite for any $$\alpha\in\mathbb{R}$$.

Clearly, if $$\alpha\in\mathbb{Z}$$, the set $$G_\alpha$$ is as large as it can get, that is $$G_\alpha = \mathbb{N}$$.

Question. Does $$G_\alpha = \mathbb{N}$$ imply $$\alpha\in \mathbb{Z}$$?

• Great question! Along the lines of a 'local to global' principle – Sandeep Silwal Dec 19 '19 at 18:19
• Thanks @SandeepSilwal! – Dominic van der Zypen Dec 19 '19 at 19:21

The answer is yes. For an irrational number $$\alpha$$, the inequality says that the fractional part $$\{q\alpha\}$$ lies in $$(0,1/q)\cup(1-1/q,1)$$. In particular, $$\{q\alpha\}\not\in[1/3,2/3]$$ when $$q\geq 3$$. However, it is easy to show by Dirichlet's approximation theorem that the fractional parts $$\{q\alpha\}$$ are dense in $$(0,1)$$, and even uniformly distributed by Weyl's theorem. Hence $$G_\alpha=\mathbb{N}$$ implies that $$\alpha$$ is a rational number, say $$\alpha=r/s$$ with $$s\in\mathbb{N}$$ coprime to $$r\in\mathbb{Z}$$. Then, for every $$q\in\mathbb{N}$$, there is $$p\in\mathbb{Z}$$ such that $$\left|\frac{r}{s}-\frac{p}{q}\right|<\frac{1}{q^2},\qquad\text{that is}\qquad |rq-ps|<\frac{s}{q}.$$ For $$q>s$$ this forces $$rq-ps=0$$, hence also $$s\mid q$$. In particular, taking $$q=s+1$$, we conclude that $$s\mid s+1$$, hence $$s=1$$. So $$\alpha=r$$ is an integer.