Let R₁ and R₂ be two relations defined on ℝ by
a R₁ b ⇔ a b ≥ 0 and a R₂ b ⇔ a ≥ b
Then,

Solution

Maths sir · 3 steps

Test reflexive, symmetric and transitive one by one, and watch what zero does to R₁.

1. Checking R1. a · a = a² ≥ 0, so R₁ is reflexive. ab = ba, so it is symmetric. For transitivity take a = 2, b = 0, c = -2: 2 × 0 = 0 ≥ 0 and 0 × (-2) = 0 ≥ 0, but 2 × (-2) = -4 < 0.
2. Checking R2. a ≥ a always holds, so R₂ is reflexive. If a ≥ b and b ≥ c, then a ≥ c, so it is transitive. But 3 ≥ 1 while 1 ≥ 3 is false, so it is not symmetric.

   Relation	Reflexive	Symmetric	Transitive
   R₁: ab ≥ 0	yes	yes	no: (2, 0), (0, -2) but not (2, -2)
   R₂: a ≥ b	yes	no: (3, 1) but not (1, 3)	yes

3. So. An equivalence relation needs all three properties. R₁ fails transitivity and R₂ fails symmetry, so neither is an equivalence relation. The answer is (d).

The idea to keep. To kill a property you need just one counterexample. For relations built on products or signs, always try 0 as the middle element: it relates to everything.

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