 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ia/paper-4/5f/i/c/solution

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

a e + b / e

is transcendental, hence irrational. If it were algebraic,

e

would solve

a x^{2} - (a e + b / e) x + b = 0

over the algebraic numbers, contradicting transcendence of

e

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