 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ib/paper-2/11g/b/i/solution

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

R

must also be a unit of

Q [X]

, hence a nonzero constant

q \in Q

. Membership of both

q

and

q^{- 1}

R

requires

q, q^{- 1} \in Z

, so

R^{\times} = {1, - 1} .

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