# Infinitely many rationals with $|a-\frac pq|<\frac1{q^2}$

If $a$ is irrational, there are infinitely many $\frac pq$ s.t $|a-\frac pq|<\frac1{q^2}\tag1$

I have the proof but don't understand it:

Take a finite set of rationals $S$ then for sufficiently large $Q$ the result of Dirichlet approximation theorem$^1$ does not hold for any $s\in S$ (this is clear), so the set of rational numbers satisfying $(1)$ is infinite

My Questions:

$\bullet$ Does that mean the following ?

Due the Dirichlet theorem for a given $Q$ you find $p/q$ satisfying the condition below, then you choose another $\tilde Q$ (much greater than $Q$) for which we cannot find a $q$ in the set $\{q:q\le Q\}$, but the Dirichlet theorem must hold, so there must be another $\tilde q$, so different from $q$ (actually $q<\tilde q)$ and proceeding in that manner there are infinitely many rationals.

$\bullet$ Why ''for sufficiently large $Q$ the result of Dirichlet approximation theorem does not hold for any $s\in S$''

($^1$Dirichlet approximation theorem states: for any $r\in\mathbb R$ and $Q\in \mathbb Z_{+}$ there are $p,q$ with $1\le q\le Q$ s.t. $|r-\frac pq|<\frac1{Qq}$)

Dirichlet approximation: for any $$\alpha$$, any $$N$$, there exists $$p,q$$ integers $$1\leq q\leq N$$ such that:
$$\mid q\alpha-p\mid <{1\over N}$$ this implies $$q\mid \alpha-{p\over q}\mid <{1\over N}$$, thus $$\mid \alpha-{p\over q}\mid <{1\over {Nq}}\leq {1\over{q^2}}$$， since $$q\leq N$$. In particular this applies if $$\alpha$$ is irrational, you can take a sequence $$N_n$$ going to infinity such that
$$\mid q_n\alpha-p_n\mid <{1\over {N_n}}$$. The sequence of $$q_n$$ has to go towards infinity also, so you have an infinite numbers of $$q_n$$ such that
$$\mid \alpha-{{p_n}\over {q_n}}\mid <{1\over {q_n^2}}$$.
• why should $q_n$ go to inifinity ? Mar 20, 2016 at 13:28
• Suppose that $q_n$ is bounded, if $p_n$ is bounded you have a finite values of $\mid q_n\alpha-p_n\mid$ and this can't go towards zero since $\alpha$ is irrational thus $q_n\alpha-p_n\neq 0$. Suppose that $p_n$ is not bounded, $\mid q_n\alpha-p_n\mid$ can't goes towards zero since you add an integer $-p_n$ to $q_n\alpha$ which can take its value in a finite set. So there exists a finite set of $p_n$ such that $\mid q_n\alpha-p_n\mid\leq 1$. Mar 20, 2016 at 13:35