Let A = { 1,2,3 }, then Match List-I with List-II
| List-I | List-II | ||
| (A) | \( R_1 \) = {(1, 1), (2, 2), (3, 3), (1, 2)} | (I) | Symmetric but neither reflexive nor transitive |
| (B) | \( R_2 \) = {(1, 1), (3, 3), (1, 3), (3, 1)} | (II) | Reflexive and transitive but not symmetric |
| (C) | \( R_3 \) = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2,3), (3,2)} | (III) | Symmetric and transitive but not reflexive |
| (D) | \( R_4 \) = {(2, 2), (3, 3), (1, 2), (2,1)} | (IV) | Reflexive and symmetric but not transitive |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (II), (B) - (III), (C) - (IV), (D) - (I)
On A = {1, 2, 3}, a relation is reflexive if it contains (1, 1), (2, 2) and (3, 3); symmetric if (b, a) is in it whenever (a, b) is; transitive if (a, c) is in it whenever (a, b) and (b, c) are.
(A) \( R_1 \) = {(1, 1), (2, 2), (3, 3), (1, 2)}
This matches (II) Reflexive and transitive but not symmetric.
Reflexive: all three pairs (a, a) are present.
Not symmetric: (1, 2) is present but (2, 1) is not.
Transitive: the only chains are with (a, a) pairs, such as (1, 1), (1, 2) giving (1, 2), which is present.
(B) \( R_2 \) = {(1, 1), (3, 3), (1, 3), (3, 1)}
This matches (III) Symmetric and transitive but not reflexive.
Not reflexive: (2, 2) is missing.
Symmetric: (1, 3) and (3, 1) are both present.
Transitive: (1, 3), (3, 1) give (1, 1), and (3, 1), (1, 3) give (3, 3); both are present.
(C) \( R_3 \) = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}
This matches (IV) Reflexive and symmetric but not transitive.
Reflexive: all three pairs (a, a) are present.
Symmetric: (1, 2), (2, 1) and (2, 3), (3, 2) occur in pairs.
Not transitive: (1, 2) and (2, 3) are present, but (1, 3) is not.
(D) \( R_4 \) = {(2, 2), (3, 3), (1, 2), (2, 1)}
This matches (I) 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.
So the matching is (A)-(II), (B)-(III), (C)-(IV), (D)-(I).