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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 $N$?

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2 Answers 2

up vote 14 down vote accepted

$\mathcal{Q}(x,\lambda)$ 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.

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I knew continued fractions had to appear somewhere... –  Miguel Apr 26 '10 at 7:00

Interesting. Intuitively, transcendental numbers are characterized by having very good rational approximations (that is how one usually prove that they ARE transcendental).

And since almost every number is transcendental, well, my reasoning ends here.

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Actually most transcendental numbers that "occur in nature" (to use a phrase of Serge Lang) are believed to have irrationality measure 2. It's known that e does. The examples of Liouville are explicitly constructed to have irrationality measure of infinity, but they are really exceptional. –  Victor Miller Apr 26 '10 at 4:22

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