 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ia/paper-2/2d/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

This relation is reflexive because choosing

n = 1

gives

∣ x / x ∣ = 1

. It need not be symmetric: take positive real numbers

y = e

and

x = e^{2}

. Then

∣ x / y^{2} ∣ = 1

is rational, so

x \sim y

. But

\frac{y}{x ^{m}} = e^{1 - 2 m}

is transcendental, hence irrational, for every integer

m \geq 1

. Thus

y \neq \sim x

, and the relation is not an equivalence relation.

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