# Show that a rational number has no good rational approximations

This is homework question. The teacher proved that if $a$ is irrational, there are infinitely many rational numbers $\frac{x}{y}$ such that $|\frac{x}{y} - a|<\frac{1}{y^2}$.

What we need to prove is:

1. Let $a$ be a rational number. Show that $a$ has no good rational approximations in the following sense: there exists a real number $c > 0$ such that for any rational number $\frac{p}{q}$ other than $a$ itself, we have $|a-\frac{p}{q}|\geq \frac{c}{q}$.

2. Show that for a rational number $a$ the inequality $|a-\frac{p}{q}|<\frac{1}{q^2}$ has only finitely many solutions $(p, q)$.

Your help is appreciated because I'm clueless here... Thank you!

• I assume you mean $<\frac 1{q^2}$ in part 2. – Hagen von Eitzen Jun 15 '13 at 11:11

If $a=\frac rs$ (and both are in shortest terms and $s,q>0$) then $\left|\frac rs-\frac pq\right|=\frac{|rq-ps|}{sq}\ge\frac1{sq}$, so we can use $c=\frac 1s$.
From $\frac1{sq}\le \left|a-\frac pq\right|\le \frac 1{q^2}$ we get $q\le s$. There are only finitely many fractions $\frac pq$ with denominator $q\le s$ and(!) $|a-\frac pq|\le 1$.