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 by a rational with , , it is straightforward to see that there exists a such that

If both and , we could add the inequalities to get

The left side is obviously equal to , so this is a contradiction. So either or . So, we have proved the following lemma:

Lemma 1: Let . For any there exists a such that

It follows from Dirichlet’s theorem that there exist infinitely many rationals such that . 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 . Now consider the sequence , where (so for any ). It turns out that if two fractional parts are close, we can find a good rational approximation to .

Say we have for . Expanding this with the definition of the fractional operator, we see

Where . Finally, dividing by , we get

Noting that , we have proved the following lemma:

Lemma 2: Let . If for natural numbers , then there exists a rational such that

for some

Now to prove Dirichlet’s theorem we just need to obtain a good bound which we can then plug in this lemma. We can obtain such an by splitting the interval up in equal parts. We then consider the terms of the sequence and apply the pigeonhole principle: Since there are terms and only intervals, there is at least one interval with two terms in it, say and , so that

Now, by lemma 2 there exists a rational approximation of that satisfies . So we have proved the following theorem:

Theorem (Dirichlet): Let . For any positive integer , there exists a fraction such that

Now suppose that for a given the set , (with each fraction irreducible) is finite. Then exists. This implies that there is no such that . This contradicts Dirichlet’s theorem, so we conclude that our assumption that is finite was wrong.

Corollary: For any , there are infinitely many rational numbers such that