Let R₁ = (a, b) ∈ N × N : |a - b| ≤ 13 and
R₂ = (a, b) ∈ N × N : |a - b| ≠ 13. Then on N :

Solution

Maths sir · 4 steps

Both relations pass reflexive and symmetric easily, so hunt for one triple that breaks transitivity in each.

1. Reflexive and symmetric. For any a, |a - a| = 0, which is both ≤ 13 and ≠ 13. So both relations are reflexive. Also |a - b| = |b - a|, so both are symmetric.
2. Transitivity of R1. Take (1, 14) and (14, 27): each difference is 13, so both pairs are in R₁. But |1 - 27| = 26 > 13, so (1, 27) ∉ R₁. R₁ is not transitive.
3. Transitivity of R2. Take (1, 2) and (2, 14): the differences are 1 and 12, neither equal to 13, so both pairs are in R₂. But |1 - 14| = 13, so (1, 14) ∉ R₂. R₂ is not transitive.
4. So. Each relation fails transitivity, so neither is an equivalence relation. The answer is (b).

The idea to keep. Reflexive and symmetric are usually quick to check. For transitivity, do not try to prove it: look for two small steps that chain into a forbidden pair.

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