For the relations \( \mathrm{R_1},\mathrm{R_2},\mathrm{R_3},\mathrm{R_4} \) defined on the set A = { 1, 2, 3},
Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| A. | \( \mathrm{R_1} \) = { (1, 2), (2, 1), (3, 3) } | I. | Transitive only |
| B. | \( \mathrm{R_2} \) = { (1, 1), (2, 2), (3, 3) } | II. | Neither reflexive nor symmetric nor transitive |
| C. | \( \mathrm{R_3} \) = { (1, 2), (2, 3), (1, 3) } | III. | Symmetric but neither reflexive nor transitive |
| D. | \( \mathrm{R_4} \) = { (3, 3), (3, 1), (1, 2) } | IV. | Equivalence |
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
On A = {1, 2, 3} a relation is reflexive only if it contains (1, 1), (2, 2) and (3, 3).
(A) \( \mathrm{R_1} \) = { (1, 2), (2, 1), (3, 3) }
This matches (III) Symmetric but neither reflexive nor transitive.
Not reflexive: (1, 1) is missing. Symmetric: (1, 2) and (2, 1) are both present. Not transitive: (1, 2) and (2, 1) are present but (1, 1) is not.
(B) \( \mathrm{R_2} \) = { (1, 1), (2, 2), (3, 3) }
This matches (IV) Equivalence.
This is the identity relation: it is reflexive, and it is symmetric and transitive because every pair has the form (a, a).
(C) \( \mathrm{R_3} \) = { (1, 2), (2, 3), (1, 3) }
This matches (I) Transitive only.
Not reflexive: no (a, a) pair. Not symmetric: (1, 2) is present but (2, 1) is not. Transitive: the only chain is (1, 2), (2, 3), and (1, 3) is present.
(D) \( \mathrm{R_4} \) = { (3, 3), (3, 1), (1, 2) }
This matches (II) Neither reflexive nor symmetric nor transitive.
Not reflexive: (1, 1) is missing. Not symmetric: (3, 1) is present but (1, 3) is not. Not transitive: (3, 1) and (1, 2) are present but (3, 2) is not.
So the matching is A-III, B-IV, C-I, D-II.