Approximation of an irrational number by rational numbers. What happens with the denominators?

Question

Given a sequence $(a_n)_n \subset \mathbb{Q}$ with

$$ \lim\limits_{n\rightarrow\infty} a_n = r \in \mathbb{R}\backslash\mathbb{Q} $$ I want to show that $$ \lim\limits_{n\rightarrow\infty} \frac{1}{q_n} = 0 $$

where $q_n$ is determined by $a_n = \frac{p_n}{q_n}$ with $p_n \in \mathbb{Z}, q_n \in \mathbb{N}$ mutually prime.

Answer 1

For any positive $m\in\mathbb{Z}$, define $$ d_m(r)=\min_{\substack{p,q\in\mathbb{Z}\\0\lt q\le m\\p\in\left[rq-\frac12,rq+\frac12\right]}}\left|\,r-\frac pq\,\right| $$ Since $r\not\in\mathbb{Q}$, and this is a minimum over a finite set, $d_m(r)\gt0$.

$d_m(r)$ is the closest to $r$ that any rational number with denominator not exceeding $m$ can get.

Since $\lim\limits_{n\to\infty}a_n=r$, there are rational numbers whose distance from $r$ is less than $d_m(r)$. Their denominators must be bigger than $m$.

Answer 2

Hint: For each $N$, there are only finitely many rationals in the interval $[a, b]$ that have a representation as $\frac{p}{q}$, where $1 \le q \le N$. Use this to prove that for each $N$, you can choose an Interval around $r$ in which all rational numbers have a denonimator greater than $N$.