Let A = set of straight lines in a plane.
(i) R : {(a, b): a is perpendicular to b]
Let a and b be two perpendicular lines.
(a) If line a is perpendicular to b, then b is perpendicular to a ⇒ R is symmetric.
(b) But a is not a perpendicular to itself.
∴ R is not reflexive.
(c) If a is perpendicular to b and & is perpendicular to c, then a is not perpendicular to c.
∴R is not transitive.
Thus, R is symmetric but neither reflexive nor transitive.
(ii) Let A = set of real numbers and R = {(a, b): a > b}
(a) An element is not greater than itself.
∴ R is not reflexive.
(b) If a > b, then b is not greater than a.
⇒ R is not symmetric.
(c) If a > b and b> c, then a > c.
Thus, R is transitive.
Hence, R is transitive but neither reflexive nor symmetric.
(iii) The relation R in the set {1, 2, 3}, is given by R = [(a, b) : a + b ≤ 4}
R = {(1, 1), (1, 2), (2, 1), (1, 3), (3, 1), (2, 2)}
Here (1,1), (2,2) ∈ R ⇒ R is reflexive.
(1, 2), (2, 1), (1, 3), (3, 1) ∈ R ⇒ R is symmetric.
But it is not transitive, since (2, 1) ∈ R and (1, 3) ∈ R does not imply (2, 3) ∈ R.
(iv) The relation R in the set {1, 2, 3}, given by
R = {(a, b): a < b} = (1, 1), (1, 2), (2, 2), (3, 3), (2, 3), (1, 3)
(a) (1, 1), (2, 2), (3, 3) ∈ R ⇒ R is reflexive.
(b) (1, 2) ∈ R, but (2, 1) ∉ R ⇒ R is not symmetric.
(c) (1,2) ∈ R, (2,3) ∈ R. Also, (1, 3) ∈ R
⇒ R is transitive.
(v) The relation R in the set {1,2, 3}, given by R = [(a, b) : 0 < | a – b | ≤ 2}
= {(1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)} .
(a) R is not reflexive,
since, (1, 1), (2, 2), (3, 3) do not belong to R.
(b) R is symmetric.
∵ (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2) ∈ R
(c) R is transitive because (1, 2) ∈ R, (2, 3) ∈ R and also (1, 3) ∈ R.
