 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-4/11i/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

M (a_{0}) \dots M (a_{n}) = (p_{n} q_{n} p_{n - 1} q_{n - 1}) .

Each

M (a_{j})

is a symmetric matrix, so reversing the order gives the transpose:

M (a_{n}) \dots M (a_{0}) = (p_{n} p_{n - 1} q_{n} q_{n - 1}) .

[a_{n}, \dots, a_{0}] = \frac{p _{n}}{p _{n - 1}}, [a_{n}, \dots, a_{1}] = \frac{q _{n}}{q _{n - 1}} .

These are respectively the

n

th and

(n - 1)

th convergents of the reversed continued fraction.

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