Approximation of reals by rationals

Question

Is this fact true: For all real $\alpha$, and natural numbers $n$, there exists some $a, q$ with $n^{2/3} \le q \le n$ and $(a, q) = 1$ such that $\left\lvert\alpha - \frac{a}{q}\right\rvert \le \frac{1}{q^2}$. It should be true in most cases, but I am unsure what those cases are.

Answer 1

It is wrong for rational $\alpha$. In fact for $\alpha=0$ you demand that $|a|\le \frac1q$, and that is only possible if $a=0$ and $q=1$ (as $\gcd(a,q)=1$), so there is no solution for $\alpha=0$ and $n> 1$. More generally, if $\alpha=\frac uv\in\mathbb Q$ with $\gcd(u,v)=1$, then $|\alpha-\frac aq|=\frac{|uq-av|}{vq}\ge \frac1{vq}$ unless $\frac aq=\frac uv$ (and henec $q=v$). This is $>\frac1{q^2}$ as soon as $q>v$. Hence for large $n$, no solution exists.

If $\alpha$ is irrational, consider the set $$A=[\alpha,\alpha+1]\cap\bigcup_{k=1}^n \frac1k{\mathbb Z}$$ of fractions "slightly above" $\alpha$ and having denominator $\le n$. The set $A$ is finite (contains at most $k$ elements of $\frac1k\mathbb Z$) and is nonempty because $\lceil\alpha\rceil\in A$. Let $\frac ab$ be the smallest element of $A$. Since $\gcd(a,b)=1$ there exist integers $d,c$ with $ad-bc=1$. With $(c,d)$ also $(c\pm a,d\pm b)$ is such a pair. Therefore let $(c,d)$ be such a pair with $d\le n$, but as large as possible, i.e. $d+b>n$. By minimality of $\frac ab$ we conclude $\frac cd<\alpha$. From $$\frac ab-\frac cd=\frac{ad-bc}{bd}=\frac1{bd}<\max\left\{\frac1{b^2},\frac1{d^2}\right\}$$ we see that one of $\frac ab$, $\frac cd$ is a good approximation to $\alpha$ with denominator $\le n$.

But is the denominator $\ge n^{2/3}$?

In general, no. Especially by considering numbers with rapidly growing continued fractions, we can construct $\alpha$ that has infinitely many good approximations, but their denominators grow too fast for the $\n^{2/3}\le q\le n$ constraint.