JEE Main 2023 · 1 February · morning · Single correct · +4 / −1
Let R be a relation on ℝ, given by R={(a, b): 3 a-3 b+√7 is an irrational number }. Then R is
- (a) an equivalence relation
- (b) reflexive and symmetric but not transitive
- (c) reflexive and transitive but not symmetric
- (d) reflexive but neither symmetric nor transitive
Answer and solution
Answer (d) reflexive but neither symmetric nor transitive
Solution
Maths sir · 4 steps
Reflexive is free here; for the other two, pick numbers that make the √7 cancel.
- Reflexive. For any real a, 3a - 3a + √7 = √7, which is irrational. So (a, a) ∈ R for every a, and R is reflexive.
- Not symmetric. Take a = √7/3, b = 0. Then 3a - 3b + √7 = 2√7, irrational, so (a, b) ∈ R. But 3b - 3a + √7 = -√7 + √7 = 0, rational, so (b, a) ∉ R.
- Not transitive. Take a = 0, b = √7, c = √7/3. Then 3a - 3b + √7 = -2√7 and 3b - 3c + √7 = 3√7, both irrational. But 3a - 3c + √7 = -√7 + √7 = 0, rational, so (a, c) ∉ R.
- So. R is reflexive but neither symmetric nor transitive, which is option (d).
The idea to keep. When a relation is defined by an expression being irrational, break symmetry or transitivity by choosing values whose irrational parts cancel to give 0. One counterexample is enough.
Answer from the NTA final answer key. Solution by PrepWiser.