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. $|r-\frac pq|<\frac1{Qq}$)


1 Answer 1


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

$\mid \alpha-{{p_n}\over {q_n}}\mid <{1\over {q_n^2}}$.

  • $\begingroup$ why should $q_n$ go to inifinity ? $\endgroup$
    – ketum
    Mar 20, 2016 at 13:28
  • 2
    $\begingroup$ 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 $\mid q_n\alpha-p_n\mid\leq 1$. $\endgroup$ Mar 20, 2016 at 13:35

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