Let R₁ and R₂ be two relation defined as follows :
R₁ = (a, b) ∈ R² : a² + b² ∈ Q and
R₂ = (a, b) ∈ R² : a² + b² ∉ Q,
where Q is the set of all rational numbers. Then :

Solution

Maths sir · 4 steps

Both relations are symmetric, so go from a to b and back to a, then check whether (a, a) survives.

1. The shortcut. a² + b² = b² + a², so both relations are symmetric. If (a, b) is in the relation, so is (b, a). Transitivity would then force (a, a) in, which needs 2a² to pass the test.
2. Break R1. Take a = 2^(1/4) and b = √(2 - √2), so a² = √2 and b² = 2 - √2. Then a² + b² = 2 ∈ ℚ, so (a, b) and (b, a) are in R₁. But a² + a² = 2√2 ∉ ℚ, so (a, a) ∉ R₁.
3. Break R2. Take a = 1 and b = π. Then 1 + π² ∉ ℚ, so (1, π) and (π, 1) are in R₂. But 1² + 1² = 2 ∈ ℚ, so (1, 1) ∉ R₂.
4. So. Each relation has a chain (a, b), (b, a) without (a, a), so neither R₁ nor R₂ is transitive, which is option (a).

The idea to keep. To test transitivity of a symmetric relation, take (a, b) and (b, a): transitivity then demands (a, a). One pair of numbers where (a, a) fails settles it.

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