Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 3 / 12F (Topics in Analysis)

The Liouville approximation theorem says that for a real algebraic number $\xi$ of degree $d\ge 2$, there is a constant $C(\xi )>0$ such thatfor all integers $p$ and $q\ge 1$ with $p/q\ne \xi$. In particular a quadratic irrational cannot have arbitrarily accurate rational approximations of order $o(q^{-2})$.

The matrix recurrence gives $p_{n+1}{q}_n-p_n{q}_{n+1}=(-1{)}^n$ and $q_{n+1}=a_{n+1}{q}_n+q_{n-1}$. Successive continued fraction convergents bracket $x$, henceThe infinite continued fraction is irrational: if $x=p/q$ were rational, a distinct convergent would be at distance at least $1/(q{q}_n)$, contradicting the first bound once $q_{n+1}>q$. If $x^2$ were rational, this irrational $x$ would have degree two. The Liouville approximation theorem would then give $|x-p_n/q_n|\ge C/q_n^2$. Comparing with the displayed bound would force $a_{n+1}\le 1/C$ for every $n$, contrary to the unbounded coefficients. Thus $x^2$ is irrational, by the unbounded continued-fraction coefficients exclude quadratic irrationality argument.