# Questions on Dirichlet's approximation theorem

In Wikipedia, the entry Dirichlet's approximation theorem states as follows：

In number theory, Dirichlet's theorem on Diophantine approximation, also called Dirichlet's approximation theorem, states that for any real number $$\alpha$$ and any positive integer $$N$$, there exists integers $$p$$ and $$q$$ such that $$1 \leq q \leq N$$ and $$\left|q\alpha -p\right|<{\frac {1}{N}}} .\tag1$$ This is a fundamental result in Diophantine approximation, showing that any real number has a sequence of good rational approximations: in fact an immediate consequence is that for a given irrational $$\alpha$$, the inequality $$\left|\alpha -{\frac {p}{q}}\right|<{\frac {1}{q^{2}}}\tag2$$ is satisfied by infinitely many integers $$p$$ and $$q$$.

Let $$(1)$$ be divided by $$q$$. We have $$\left|\alpha -{\frac {p}{q}}\right|<{\frac {1}{qN}}\leq \frac{1}{q^2},$$ which is just $$(2)$$. But my question is that:

(1) how to know that there exist infinitely many $$p$$ and $$q$$ for a given irrational $$\alpha$$?

(2) if $$\alpha$$ is rational, is the number of such $$p$$ and $$q$$ finite?

(3) if there exist infinitely many $$p$$ and $$q$$ for a given irrational $$\alpha$$, can we choose a sequence $$q_1,q_2,q_3,\cdots,q_n,\cdots$$ such that $$q_n \to +\infty$$?

## 2 Answers

Such approximations can be found by continued fractions In particular, Theorem 5 on the Wikipedia page shows that the convergents of the simple continued fraction of $$\alpha$$ will give infinitely many approximations of the desired form.

To answer your first and third question, just use the theorem pick $$N$$, and find $$p$$ and $$q$$ satisfying (1). Then pick $$N_1$$ such that $$\vert q\alpha-p\vert > \frac{1}{N_1}$$ and use the theorem to find $$q_1,p_1$$ satisfying $$\vert q_1\alpha-p_1\vert < \frac{1}{N_1}$$, and again find $$N_2$$ with $$\vert q_1\alpha-p_1\vert > \frac{1}{N_2}$$ and so on. You get an infinite series $$p,q,p_1,q_1,p_2,q_2,\dots$$ that, as you have stated, verify $$\left \vert \alpha - \frac{p_n}{q_n} \right \vert < \frac{1}{q_n^2}$$

You can convince yourself easily that if you chose your $$q$$'s positive (and $$N>2$$) then the sequence of $$q$$ is strictly increasing and so tends to infinity.

To answer the second question there are only finitely many coprime integers $$p,q$$ when $$\alpha=\frac{P}{Q}$$ is rational with $$P$$ and $$Q$$ coprime. Suppose $$p,q$$ is one pair satisfying (2) with $$q>Q$$, then as you can show that $$qP-pQ \ne 0$$ and so $$\left\vert \frac{P}{Q}-\frac{p}{q}\right\vert = \frac{\vert qP-pQ\vert}{qQ} \ge \frac{1}{qQ} > \frac{1}{q^2}$$ so there are only finitely many with $$p$$ and $$q$$ coprime, if you remove this condition then there are infinitely many for instance $$p=kP, q=kQ$$ for any integer $$k$$.

• Thanks! I can understand you except for the former part. How you know that $p<p_1<p_2<\cdots<p_n$ and $q<q_1<q_2<\cdots<q_n$? – mengdie1982 Apr 19 '19 at 11:45
• You have the liberty to chose the sequence of $N_i$, so you can just take the $N_{i+1}$ big enough to ensure that $\vert q'\alpha -p'\vert >1/N_{i+1}$ for every $q'\le q_i$ and every integer $p$, this is possible since there is only a finite number of such $q'$. In this way you are sure that $q_{i+1}>q_i$. – Esteban Crespi Apr 19 '19 at 18:48