Suppose $x\in \mathbb{R}$ is irrational, with irrationality measure $\mu=\mu(x)$; this means that the inequality $|x-\frac{p}{q}|< q^{-\lambda}$ has infinitely many solutions in integers $p,q$ if and only if $\lambda < \mu$. A beautiful theorem of Roth asserts that algebraic numbers have irrationality measure $2$. For $\lambda<\mu$, let $\mathcal{Q}(x,\lambda) \subset \mathbb{N}$ be the (infinite) set of all $q$ occuring in solutions to the aforementioned inequality.

Question: For which pairs $(x,\lambda)$ does $\mathcal{Q}(x,\lambda)$ have positive relative density in the positive integers? For which pairs $(x,\lambda)$ does the cardinality of $\mathcal{Q}(x,\lambda) \cap [1,N]$ grow like a positive power of $N$?


2 Answers 2


$\mathcal{Q}(x,\lambda)$ has positive relative density if and only if $\lambda\le 1$. This follows from Weyl's Theorem on Uniform Distribution. (There is a nice concise proof in Cassels' "Diophantine Approximation".)

Weyl's Theorem: Let $I\subset \mathbb{R}$ be an interval of length $\epsilon \le 1$. Let $S_N(I)$ be the set of all integers $q$ in the interval $[1,N]$ such that for some integer $p$, it holds that $xq-p\in I$. Then

$$\frac{Card(S_N(I))}{N} \to \epsilon \text{ as } N\to\infty.$$

Here's a proof-sketch, using Weyl's Theorem, that if $\lambda > 1$ then $\mathcal{Q}(x,\lambda)$ has relative density zero:

Fix $\epsilon > 0$, and take $I$ (in Weyl's Theorem) to be the interval $(-\epsilon,\epsilon)$. Suppose $\lambda>1$. Let $q\in \mathcal{Q}(x,\lambda)$; so for some $p\in \mathbb{Z}$, $$|xq-p| < q^{1-\lambda}.$$

There is an integer $M$, depending only on $\epsilon$, such that $|xq-p| < \epsilon$ whenever $p$ and $q$ satisfy the above inequality and $q\ge M$. Therefore $$\mathcal{Q}(x,\lambda)\cap [M,N]\subset S_N(I).$$ It follows from Weyl's Theorem that the relative density of $\mathcal{Q}(x,\lambda)$ does not exceed $2\epsilon$. Since $\epsilon$ is arbitrary, the relative density of $\mathcal{Q}(x,\lambda)$ must be zero.

This can be proved in a more elementary but laborious way using the "Ostrowski Number System", which is explained in the Rockett and Szusz book on continued fractions.

  • $\begingroup$ I knew continued fractions had to appear somewhere... $\endgroup$
    – Miguel
    Apr 26, 2010 at 7:00

An answer to your second question may be found in an old paper of Erdos: Some results on diophantine approximation. Acta Arith. 5 1959 359–369 (1959). (MathSciNet)

He proves that for $1 < \lambda < 2$, for almost every $x$, the set $\mathcal{Q}(x,\lambda)$ grows like $n^\frac{1}{2-\lambda}$. More precisely, Erdos proves that a one-sided version of your set $$\mathcal{Q}'(x,\lambda) = \big\{ q \geq 1 \ \big| \ \text{there exists } p, \ (q,p) = 1, \text{ such that } 0 < x-\frac pq < q^{-\lambda} \big\}$$ satisfies $$\lim_{n \to \infty} \frac{\big |\mathcal{Q}'(x,\lambda) \cap [1,n] \big| }{\sum_{q=1}^n q^{1-\lambda}} = \frac{12}{\pi^2}.$$ That the set $\mathcal{Q}(x,\lambda)$ grows like $n^\frac{1}{2-\lambda}$ is consistent with the fact that $\mathcal{Q}(x,1)$ is all of $\mathbb{N}$ while the numbers in $\mathcal{Q}(x,2)$ coming from continued fraction convergent denominators grow exponentially.

  • $\begingroup$ What is meant by a polynomial of fractional degree? $\endgroup$ Aug 4, 2015 at 2:15
  • $\begingroup$ Thanks for the comment; I've updated the post to make it more precise. $\endgroup$ Aug 5, 2015 at 13:32

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