Question -
Answer -
A relation R in A is said to be reflexive if aRa for all a∈A, R is symmetric if aRb ⇒ bRa, for all a, b ∈ A and it is said to be transitive if aRb and bRc ⇒ aRc for all a, b, c ∈ A.
• x > y, x, y ϵ N
(x, y) ϵ {(2, 1), (3, 1).......(3, 2), (4, 2)....}
This is not reflexive as (1, 1), (2, 2)....are absent.
This is not symmetric as (2,1) is present but (1,2) is absent.
This is transitive as (3, 2) ϵ R and (2,1) ϵ R also (3,1) ϵ R ,similarly this property satisfies all cases.
• x + y = 10, x, y ϵ N
(x, y)ϵ {(1, 9), (9, 1), (2, 8), (8, 2), (3, 7), (7, 3), (4, 6), (6, 4), (5, 5)}
This is not reflexive as (1, 1),(2, 2)..... are absent.
This only follows the condition of symmetric set as (1, 9)ϵR also (9, 1)ϵR similarly other cases are also satisfy the condition.
This is not transitive because {(1, 9),(9, 1)}ϵR but (1, 1) is absent.
• xy is square of an integer, x, y ϵ N
(x, y) ϵ {(1, 1), (2, 2), (4, 1), (1, 4), (3, 3), (9, 1), (1, 9), (4, 4), (2, 8), (8, 2), (16, 1), (1, 16)...........}
This is reflexive as (1,1),(2,2)..... are present.
This is also symmetric because if aRb ⇒ bRa, for all a,bϵN.
This is transitive also because if aRb and bRc ⇒ aRc for all a, b, c ϵ N.
• x + 4y = 10, x, y ϵ N
(x, y) ϵ {(6, 1), (2, 2)}
This is not reflexive as (1, 1), (2, 2).....are absent.
This is not symmetric because (6,1) ϵ R but (1,6) is absent.
This is not transitive as there are only two elements in the set having no element common.