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I want to prove this below:

(1) For any irrational number $\alpha$, there exist infinitely many rational numbers $\frac{m}{n}$ such that $\left| {\alpha - \frac{m}{n}} \right| < \frac{1}{{{n^2}}}$.

I got a hint from somewhere to prove this below:

(2) For any irrational number $\alpha$ and any positive integer $n$, there exist positive integers $k,m$ such that $\left| {\alpha - \frac{m}{k}} \right| < \frac{1}{{{kn}}}$, where $k \leq n$.

I tried to prove (2), but still can't find out how to deal with (1).

Can you help me?

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  • $\begingroup$ $\frac{1}{kn} \leq \frac{1}{k^2}$, have you proved $(2)$? $\endgroup$ – clark Jun 17 '16 at 13:33
  • $\begingroup$ I did think of this. But it seems that we cannot assure "infinite". An extreme case is that $k=1$ always holds, then it's obvious true. $\endgroup$ – gžd15 Jun 17 '16 at 13:45
  • $\begingroup$ There should be finitely many $\frac{m}{k}$ by taking $n\rightarrow \infty$ $\endgroup$ – clark Jun 17 '16 at 13:51
  • $\begingroup$ Fix $q$ a rational number. Assume there were infinitely many $n_m$ that they give $q$ as in $|\alpha - q| \leq \frac{1}{n_mk_m}$ then since $\frac{1}{n_m k_m}\rightarrow 0$ we get a contradiction since $\alpha $ is irrational. $\endgroup$ – clark Jun 17 '16 at 14:00
  • $\begingroup$ Now since for each $n$ we get $q_n$ with the desired approximation, and since $q_n$ is obtain for a finite number of different $n$, we see there must be infinitely many different $q_n$. $\endgroup$ – clark Jun 17 '16 at 14:02
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In this answer, the following Lemma is proven using a pigeonhole argument:

Lemma: Let $x$ be any real number and $N$ be a positive integer. Then there are integers $p$ and $q$ with $0\lt q\le N$ so that $\left|p−qx\right|\lt\frac1N$.

Since $q\le N$, this gives that $$ \left|\,\frac pq-x\,\right|\le\frac1{Nq}\le\frac1{q^2}\tag{1} $$ Since $(1)$ says $$ N\le\frac1{\left|p−qx\right|}\tag{2} $$ we can get a larger $p'$ and $q'$ by applying the Lemma to any $N'$ greater than $$ \max\left\{\frac1{\left|a-bx\right|}:a,b\in\mathbb{Z},0\lt b\le q\right\}\tag{3} $$ Thus, we can find an infinite sequence of $p$ and $q$ that satisfy $(1)$.

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  • $\begingroup$ Your pigeonhole argument is enlightening. Could you please explain why we can get an infinite sequence of such approximations more explicitly? It still confuses me. $\endgroup$ – gžd15 Jun 17 '16 at 15:01
  • $\begingroup$ @gžd15: I have tried to improve the explanation of the infinitude of such approximations. $\endgroup$ – robjohn♦ Jun 17 '16 at 15:44
  • $\begingroup$ ♦: Why $\max\left\{\frac1{\left|a-bx\right|}:a,b\in\mathbb{Z},0\lt b\le q\right\}$ exists? Also why do those $\frac{p}{q}$ we get differ from each other? $\endgroup$ – gžd15 Jun 17 '16 at 16:31
  • $\begingroup$ @gžd15: The max exists since it is the maximum of a finite set of numbers. Note that if $x$ is irrational, for each value of $b$, there are two values of $a$ that make $\frac1{\left|a-bx\right|}\gt1$; thus, this is the max of $2q$ numbers. Applying the Lemma to an $N'$ greater than this max ensures that $q'\gt q$, since the Lemma guarantees $p'$ and $q'$ so that $\left|p'-q'x\right|\le\frac1{N'}$ and for $0\lt b\le q$, $\left|a-bx\right|\gt\frac1{N'}$. $\endgroup$ – robjohn♦ Jun 18 '16 at 7:47

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