 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-4/7e/b/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Since

p / q

is not a root,

q^{d} f (p / q)

is a nonzero integer, and hence

∣ f (p / q) ∣ \geq q^{- d} .

Moreover

0 < θ - p / q < 1

implies

∣ p / q ∣ < ∣ θ ∣ + 1

. Applying part (i) to

s = θ

t = p / q

, and using

f (θ) = 0

gives

q^{- d} \leq A (∣ θ ∣ + 1)^{d - 1} θ - \frac{p}{q} .

Rearrangement proves

θ - \frac{p}{q} \geq \frac{1}{A} (\frac{1}{∣ θ ∣ + 1})^{d - 1} \frac{1}{q ^{d}} .

This is the core estimate behind the Liouville approximation theorem.

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