Let R be a relation on Z × Z defined by (a, b) R(c, d) if and only if a d-b c is divisible by 5. Then R is

Solution

Maths sir · 4 steps

Test each property in turn, and look for a sneaky middle pair like (0, 0) that relates to everything.

1. Reflexive. For any (a, b), the test gives ab - ba = 0, and 0 is divisible by 5. So every pair is related to itself.
2. Symmetric. If (a, b) R (c, d), then 5 | ad - bc. For (c, d) R (a, b) we need 5 | cb - da, and cb - da = -(ad - bc). The negative of a multiple of 5 is still a multiple of 5, so R is symmetric.
3. Transitive fails. Take (0, 0) as the middle pair. (1, 2) R (0, 0) because 1 × 0 - 2 × 0 = 0, and (0, 0) R (1, 3) because 0 × 3 - 0 × 1 = 0. But for (1, 2) and (1, 3) we get 1 × 3 - 2 × 1 = 1, which is not divisible by 5.
4. So. R is reflexive and symmetric but not transitive, which is option (b).

The idea to keep. To break transitivity, hunt for a middle element that relates to everything, like (0, 0) or 0. It joins two pairs that have nothing to do with each other.

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