Consider the relations R₁ and R₂ defined as a R₁ b ⇔ a²+b²=1 for all a, b ∈ R and (a, b) R₂(c, d) ⇔ a+d=b+c for all (a, b),(c, d) ∈ N × N. Then :

Solution

Maths sir · 3 steps

Test each relation for reflexive, symmetric and transitive; one failed property knocks it out.

1. Test R1 reflexive. Reflexive needs a R₁ a for every real a, that is 2a² = 1. Take a = 1: 1² + 1² = 2 ≠ 1, so R₁ is not reflexive. It is symmetric, but that does not save it.
2. Rewrite R2. Move terms across: a + d = b + c is the same as a - b = c - d. So two pairs are related exactly when they have the same difference.

   (a, b) R₂ (c, d) ⇔ a - b = c - d

   Property	R₁: a² + b² = 1	R₂: a - b = c - d
   Reflexive	Fails: 1² + 1² = 2 ≠ 1	Holds: a - b = a - b
   Symmetric	Holds: a² + b² = b² + a²	Holds: a - b = c - d gives c - d = a - b
   Transitive	Fails: 1 R₁ 0 and 0 R₁ 1, but 1² + 1² ≠ 1	Holds: a - b = c - d and c - d = e - f give a - b = e - f
   Equivalence?	No	Yes

3. So. R₂ passes all three tests and R₁ fails reflexivity. Only R₂ is an equivalence relation, which is option (c).

The idea to keep. Check reflexivity first; one counterexample like a = 1 ends the matter. And when a condition mixes the pairs, regroup it so each side depends on one pair only.

Answer from the NTA final answer key. Solution by PrepWiser.