Monday, July 4, 2016

calculus - approximate irrational numbers by rational numbers




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?


Answer



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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