Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 2 / 2F (Topics in Analysis)

The Liouville approximation theorem states that if $\alpha$ is a real irrational algebraic number of degree $d$, there is a constant $C(\alpha )>0$ such that, for all integers $p,q$ with $q\ne 0$,Reduction of the fraction is not necessary: an unreduced denominator only weakens this bound. More generally the same type of bound holds for an algebraic number of degree $d$ whenever $p/q\ne \alpha$. The exception is essential for a rational $\alpha$, since an exact rational representation gives zero error. A rational number of degree one obeys the bound directly: if $\alpha =a/b$ in lowest terms and $p/q\ne \alpha$, then $|\alpha -p/q|=|aq-bp|/(|bq|)\ge 1/(|bq|)$.

Set $r=p/q$. If $r\notin [1,2]$, then $|\sqrt{2}-r|\ge \sqrt{2}-1>1/4\ge 1/(4q^2)$. If $1\le r\le 2$, irrationality of the square root of two implies that the integer $2q^2-p^2$ is nonzero. Factoring a difference of squares therefore givesThus the requested bound holds for every integer $p$ and every nonzero integer $q$, without using the Liouville approximation theorem. For completeness, irrationality follows because $p^2=2q^2$ in lowest terms would make $p$ even and then $q$ even, contradicting coprimality.