JEE Main 2025 · 28 January · morning · Single correct · +4 / −1
The relation R={(x, y): x, y ∈ ℤ and x+y is even } is:
- (a) reflexive and transitive but not symmetric
- (b) reflexive and symmetric but not transitive
- (c) an equivalence relation
- (d) symmetric and transitive but not reflexive
Answer and solution
Answer (c) an equivalence relation
Solution
Maths sir · 4 steps
Here x+y is even exactly when x and y have the same parity, and sharing a parity passes all three tests.
- Reflexive. For any integer x, the sum x + x = 2x is even. So (x, x) ∈ R for every x ∈ ℤ, and R is reflexive.
- Symmetric. If x + y is even, then y + x is the same number, so it is also even. So (x, y) ∈ R gives (y, x) ∈ R, and R is symmetric.
- Transitive. Suppose x + y and y + z are both even. Add them and remove 2y, which is even:
x + z = (x + y) + (y + z) - 2y
- So. The right side is even + even − even, so x + z is even and R is transitive. R is reflexive, symmetric and transitive, so it is an equivalence relation, option (c).
The idea to keep. Rewrite a condition like "x+y is even" as "x and y have the same parity". Any relation of the form "same something" is an equivalence relation.
Answer from the NTA final answer key. Solution by PrepWiser.