Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 1G / Solution

Apply the continued-fraction algorithm. Since $3<\sqrt{11}<4$,The successive complete quotients areandTaking one more reciprocal returns to $x_1$, so the continued fraction of the square root of eleven is

WriteThe first convergents are $p_0/q_0=3$ and $p_1/q_1=10/3$, whenceThe continued fraction convergent recurrence iswhere $a_{n+1}=3$ for even $n$ and $a_{n+1}=6$ for odd $n$.

SetIf $n$ is even and $z_{n+1}=\beta z_n$, thenbecause ${\beta }^{-1}=\sqrt{11}-3$. If $n$ is odd and $z_{n+1}=\alpha z_n$, thenbecause ${\alpha }^{-1}=(\sqrt{11}-3)/2$. The base case and induction prove the alternating multiplier recurrence for convergents of the square root of eleven:

Solved by gpt-5.6-sol high.