Consider the following relations: R = \(x, y) | x, y R and x = wy for some rational number w\ ; S = \(m/n, p/q) | m, n, p, q Z, n, q ≠ 0 and qm = pn\ - Mathematics MCQ AIEEE 2026 | ExamTest.live

mediumMCQAIEEE2026Mathematics

1 mark

Consider the following relations: R = {(x, y) ∣ x, y \in R and x = w y for some rational number w}; S = {(\frac{m}{n}, \frac{p}{q}) ∣ m, n, p, q \in Z, n, q \neq = 0 and q m = p n} . Then:

Solution & Step-by-step Explanation

1. For R : x = w y . If x = 0, 0 = w \cdot y for any y . If y \neq = 0, then 0 = 0 \cdot y (rational). But if y = 0 and x = 1, 1 = w \cdot 0 is impossible. R is not symmetric (if x = 0, y = 1, 0 = 0 \cdot 1, but 1 \neq = w \cdot 0). Thus not an equivalence relation.2. For S : \frac{m}{n} S \frac{p}{q} ⟺ q m = p n ⟺ \frac{m}{n} = \frac{p}{q} . This is the standard equality relation on rational numbers, which is reflexive, symmetric, and transitive. Thus S is an equivalence relation.

Try it yourself before checking the explanation above.

Consider the following relations: R = {(x, y) ∣ x, y \in R and x = w y for some rational number w}; S = {(\frac{m}{n}, \frac{p}{q}) ∣ m, n, p, q \in Z, n, q \neq = 0 and q m = p n} . Then:

neither R nor S is an equivalence relation

S is an equivalence relation but R is not an equivalence relation

R and S both are equivalence relations

R is an equivalence relation but S is not an equivalence relation

Practice with timed mock tests and track your performance across Mathematics.