JEE Main 2023 · 29 January · evening · Single correct · +4 / −1
Let R be a relation defined on ℕ as aRb if 2a+3b is a multiple of 5,a,b∈ ℕ. Then R is
- (a) an equivalence relation
- (b) non reflexive
- (c) symmetric but not transitive
- (d) transitive but not symmetric
Answer and solution
Answer (a) an equivalence relation
Solution
Maths sir · 3 steps
Rewrite 2a+3b as 5b + 2(a-b), and the relation turns into plain congruence modulo 5.
- Rewrite the condition. Split 3b as 5b - 2b. Then 2a + 3b = 5b + 2(a-b), and 5b is already a multiple of 5.
5 | (2a+3b) ⇔ 5 | 2(a-b) ⇔ 5 | (a-b)
- Why 2 drops out. 5 is prime and does not divide 2, so if 5 divides 2(a-b) it must divide a-b. So aRb exactly when a and b leave the same remainder on division by 5.
Property Check Holds? Reflexive 2a + 3a = 5a, a multiple of 5 Yes Symmetric 2b + 3a = 5(a+b) - (2a+3b), a multiple of 5 Yes Transitive 5 | (a-b) and 5 | (b-c) give 5 | (a-c) Yes - So. R is reflexive, symmetric and transitive, so it is an equivalence relation. The answer is (a).
The idea to keep. When a relation says some combination of a and b is a multiple of n, try to reduce it to n | (a-b). If it reduces, you have congruence mod n, which is always an equivalence relation.
Answer from the NTA final answer key. Solution by PrepWiser.