Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 4 / 12H / Solution

Multiplying the matrices inductively gives columns $(p_n,q_n{)}^T$ and $(p_{n-1},q_{n-1}{)}^T$. Their determinant is $(-1{)}^{n-1}$, and writing $x=(p_n{x}_{n+1}+p_{n-1})/(q_n{x}_{n+1}+q_{n-1})$ givesIf $A{y}^2+By+C=0$, then for rationals $p/q$ near $y$, factorization against the conjugate root gives $|p/q-y|\ge M/q^2$; rationals away from $y$ are handled by reducing $M$. Finally $q_{n+1}\ge a_{n+1}{q}_n$, so the upper bound is at most $1/(a_{n+1}{q}_n^2)$. Unbounded partial quotients contradict the fixed lower bound for a quadratic irrational.