# Dirichlet’s approximation theorem

Dirichlet’s approximation theorem is a result about rational approximation. First, let’s derive a trivial bound on the error of the best rational approximation so that we can compare it with Dirichlet’s theorem.

If we approximate a real number $α∈R$ by a rational $qp $ with $p∈Z$, $q∈N_{+}$, it is straightforward to see that there exists a $p∈Z$ such that

$qp ≤α≤qp+1 $If both $α−qp >2q1 $ and $qp+1 −α>2q1 $, we could add the inequalities to get

$(α−qp )+(qp+1 −α)>q1 $The left side is obviously equal to $q1 $, so this is a contradiction. So either $0≤qp −α≤2q1 $ or $0≤qp+1 −α≤2q1 $. So, we have proved the following lemma:

**Lemma 1**: *Let $α∈R$. For any $q∈N_{+}$ there exists a $p∈Z$ such that*

It follows from Dirichlet’s theorem that there exist infinitely many rationals $qp $ such that $∣∣∣∣ α−qp ∣∣∣∣ <q_{2}1 $. The quadratic order of the error turns out to be optimal; This is what Hurwitz’ theorem proves.

It seems that Dirichlet was fond of arithmetic progressions; Dirichlet’s most famous theorem is about arithmetic progressions, and they turn out to play an important role in the proof of this theorem as well.

Let $α∈R$. Now consider the sequence $0,{α},{2α},{3α},...,{Nα}$, where ${x}=x−⌊x⌋$ (so $0≤{x}<1$ for any $x∈R$). It turns out that if two fractional parts are close, we can find a good rational approximation to $α$.

Say we have $∣{rα}−{sα}∣<ϵ$ for $r,s∈N_{0}$. Expanding this with the definition of the fractional operator, we see

$∣(rα−⌊rα⌋)−(sα−⌊sα⌋)∣=∣qα−p∣<ϵ$Where $p=⌊rα⌋−⌊sα⌋,q=r−s$. Finally, dividing by $∣q∣$, we get

$∣∣∣∣∣ α−qp ∣∣∣∣∣ <qϵ $Noting that $q≤N$, we have proved the following lemma:

**Lemma 2**: *Let $α∈R$. If $∣{rα}−{sα}∣<ϵ$ for natural numbers $r,s∈N_{0}$, then there exists a rational $qp ∈Q$ such that*

*for some $q≤N$*

Now to prove Dirichlet’s theorem we just need to obtain a good bound $ϵ$ which we can then plug in this lemma. We can obtain such an $ϵ$ by splitting the interval $[0,1)$ up in $N$ equal parts. We then consider the $N+1$ terms of the sequence $0,{α},{2α},...,{Nα}$ and apply the pigeonhole principle: Since there are $N+1$ terms and only $N$ intervals, there is at least one interval with two terms in it, say ${rα}$ and ${sα}$, so that

$∣{rα}−{sα}∣<N1 $Now, by lemma 2 there exists a rational approximation $qp ∈Q$ of $α$ that satisfies $∣α−qp ∣<Nq1 $. So we have proved the following theorem:

**Theorem (Dirichlet)**: Let $α∈R$. For any positive integer $N∈N_{+}$, there exists a fraction $qp ∈Q$ such that

Now suppose that for a given $α∈R$ the set $S_{α}={qp :∣α−qp ∣<q_{2}1 }$, (with each fraction $qp $ irreducible) is finite. Then $N=max_{qp∈S_{α}}(q)$ exists. This implies that there is no $qp ∈Q$ such that $∣α−qp ∣<Nq1 $. This contradicts Dirichlet’s theorem, so we conclude that our assumption that $S_{α}$ is finite was wrong.

**Corollary**: *For any $α∈R$, there are infinitely many rational numbers $qp ∈Q$ such that*