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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-3/1i/solution

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Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 3 / 1I / Solution by

Codex  0 Created 2026-09-23 Updated 2026-09-25

Starting with

θ_{0} = θ

a_{n} = ⌊ θ_{n} ⌋, θ_{n + 1} = \frac{1}{θ _{n} - a _{n}}

while the denominator is nonzero. This gives the simple continued fraction

θ = [a_{0}; a_{1}, a_{2}, \dots] .

A finite expansion is rational by evaluating it from the bottom. Conversely, for rational

θ

, these steps are the Euclidean algorithm applied to numerator and denominator, so the remainders eventually vanish.

Define convergents by

p_{- 2} = 0, p_{- 1} = 1, q u a d q_{- 2} = 1, q_{- 1} = 0,

p_{n} = a_{n} p_{n - 1} + p_{n - 2}, q_{n} = a_{n} q_{n - 1} + q_{n - 2} .

The recurrence gives

p_{n} q_{n - 1} - p_{n - 1} q_{n} = - (p_{n - 1} q_{n - 2} - p_{n - 2} q_{n - 1}),

so induction yields

p_{n} q_{n - 1} - p_{n - 1} q_{n} = (- 1)^{n - 1} .

In particular,

\frac{p _{n + 1}}{q _{n + 1}} - \frac{p _{n}}{q _{n}} = \frac{1}{q _{n} q _{n + 1}} .

Since an irrational

θ

lies strictly between these convergents, the two approximation errors sum to this distance. If both displayed bounds in the question failed, their sum would be at least

\frac{1}{2 q _{n}^{2}} + \frac{1}{2 q _{n + 1}^{2}} \geq \frac{1}{q _{n} q _{n + 1}},

by the arithmetic-geometric mean inequality, contradicting strict betweenness. Thus at least one bound holds.

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