Let \(A = \{1, 2, 3\}\). Consider the following relations on A.
Match List-I with List-II
| List-I | List-II | ||
| (A) | \(\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}\) | (I) | Reflexive and transitive but not symmetric |
| (B) | \(\{(1, 1), (2, 2), (1, 2), (2, 1)\}\) | (II) | Neither reflexive, nor symmetric, nor transitive |
| (C) | \(\{(1, 1), (2, 2), (3, 3), (1, 2)\}\) | (III) | An equivalence relation |
| (D) | \(\{(1, 2), (2, 3)\}\) | (IV) | Symmetric and transitive but not reflexive |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
(A) \(\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}\)
This matches (III) An equivalence relation.
All of (1, 1), (2, 2), (3, 3) are present; (1, 2) comes with (2, 1); the chains (1, 2), (2, 1) and (2, 1), (1, 2) give (1, 1) and (2, 2), which are present. Its classes are {1, 2} and {3}.
(B) \(\{(1, 1), (2, 2), (1, 2), (2, 1)\}\)
This matches (IV) Symmetric and transitive but not reflexive.
(3, 3) is missing, so not reflexive. It contains every pair made from 1 and 2, so every pair asked for by symmetry or by a chain is present.
(C) \(\{(1, 1), (2, 2), (3, 3), (1, 2)\}\)
This matches (I) Reflexive and transitive but not symmetric.
(1, 2) is present but (2, 1) is not, so not symmetric. The only chains use a pair (a, a) and give back a pair already present, so transitive.
(D) \(\{(1, 2), (2, 3)\}\)
This matches (II) Neither reflexive, nor symmetric, nor transitive.
(1, 1) is missing; (2, 1) is missing; (1, 2) and (2, 3) are present but (1, 3) is not.