# A question on approximating irrational numbers by rational numbers.

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. $$\lim \limits_{\epsilon \to0^+} m(\{x\in [0,1]:\exists\;\text{infinitely many}\;a,b \in \mathbb{Z},\text{ s.t.}~(a,b)=1\quad\text{and}\quad\left|\frac{a}{b}-x\right|<\frac{\epsilon}{b^2}\})$$

## 1 Answer

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.

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