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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 $\alpha\in \mathbb{Z}$?

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

P.S. After writing the above answer, I realize that the key step is already present in Fedor Petrov's response to the OP's earlier question.

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    $\begingroup$ Thanks for this clear and well-written answer! $\endgroup$ – Dominic van der Zypen Dec 19 '19 at 19:22

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