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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 $${\displaystyle \left|q\alpha -p\right|<{\frac {1}{N}}} .\tag1$$ 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 $q_n \to +\infty$?

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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.

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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 $k$.

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  • $\begingroup$ Thanks! I can understand you except for the former part. How you know that $p<p_1<p_2<\cdots<p_n$ and $q<q_1<q_2<\cdots<q_n$? $\endgroup$ – mengdie1982 Apr 19 at 11:45
  • $\begingroup$ 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 $q_{i+1}>q_i$. $\endgroup$ – Esteban Crespi Apr 19 at 18:48

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