2
$\begingroup$

It is known from the theory of continued fractions that if $\epsilon<1/2$ then the only $a,b\in\mathbb{Z}$ such that $$(a,b)=1\quad\text{and}\quad\left|\frac{a}{b}-x\right|<\frac{\epsilon}{b^2}$$ are the convergents of the simple continued fraction of $x$. Hence not all $x$ are in the set measured. I am interested how many irrational number can be approximated by rational numbers arbitrarily, in a certain sense. That's why I am interested in the following expression, in which $m$ denotes the Lebesgue's measure. $$\lim \limits_{\epsilon \to0^+} m(\{x\in [0,1]:\exists\;\text{infinitely many}\;a,b \in \mathbb{Z},\text{ s.t.}~(a,b)=1\quad\text{and}\quad\left|\frac{a}{b}-x\right|<\frac{\epsilon}{b^2}\})$$

$\endgroup$
3
$\begingroup$

The expression inside the limit is equal to $1$ for every $\epsilon > 0$. In fact a much more precise statement is true.

Theorem (Khinchin): Let $\phi(q) : \mathbb{N} \to \mathbb{R}$ be a monotonically decreasing function. For almost all real numbers $\alpha$, the number of pairs of positive integers $(q, p)$ satisfying

$$\left| p - q\alpha \right| < \phi(q)$$

is infinite if $\sum \phi(q)$ diverges, and finite if $\sum \phi(q)$ converges.

$\endgroup$
  • $\begingroup$ Didn't you learn that on mathoverflow? I vaguely remember this theorem. $\endgroup$ – user22190 Jan 3 '12 at 20:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.