 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-1/16h/d/solution

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

α = ω^{δ_{1}} n_{1} + ω^{δ_{2}} n_{2} + \dots + ω^{δ_{r}} n_{r},

where

δ_{1} > \dots > δ_{r}

and

0 < n_{i} < ω

. Taking

δ = δ_{1}

n = n_{1}

, and

η = ω^{δ_{2}} n_{2} + \dots + ω^{δ_{r}} n_{r}

gives

α = ω^{δ} n + η

with

η < ω^{δ}

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