(i) A = {1,2,3,.....,13,14}
R = {x.y) : 3 x – y ≠} = {(x, y) : y = 3 x}
= {(1,3), (2, 6), (3, 9), (4, 12)}
(a)    R is not reflexive as (x, x) ∉ R    [ ∵ 3 x – x ≠ 0]
(b)    R is not symmetric as (x,y) ∈ R does not imply (y, x) ∈ R
[ ∴ (1, 3) ∈ R does not imply (3. 1) ∈ R]
(c)    R is not transitive as (1.3) ∈ R , (3, 9) ∈ R but (1.9) ∉ R.

Let A = {1, 2, 3, 4, 5, 6}
R = {(a, b) : b = a + 1} = {(a, a + 1)}
= {(1, 2), (2, 3), (3, 4), (4,5), (5,6)}
(i) R is not reflexive as (a, a) ∉ R ∀ a ∈ A
(ii) (a,b) ∈ R ⇏ (b,a) ∈ R [∵ (a, b) ∈ R ⇒ b = a + 1 ⇒ a = b –1]
∴ R is not symmetric.
(iii) (a, b) ∈ R, (b, c) ∈ R    ⇏ (a, c) ∈ R
[∵ (a, b), (b, c) ∈ R ⇒ b = a + 1, c = b + 1 ⇒ c = a + 2]
∴ R is not transitive.

R = {(a, b) : a ≤ b}
(i) Since (a, a) ∈ R ∀ a ∈ R    [∵ a ≤ a ∀ a ∈ R]
∴ R is reflexive.
(ii) (a, b) ∈ R ⇏ (b, a) ∈ R    [∵ if a ≤ b. then b ≤ a is not true]
∴ R is not symmetric.
(iii) Let (a, b), (b, c) ∈ R ∴ a ≤ b, b ≤ c ∴ a ≤ c ⇒ (a, c) ∈ R ∴ (a, b), (b. c) ∈ R ⇒ (a, c) ∈ R ∴ R is transitive.

R = {(a, b) : a ≤ b²}
(i) Since (a, a) ∉ R



∴ R is not reflexive.
(ii) Also (a, b) ∈ R ⇏ (b, a) ∈ R
[Take a = 2 ,b = 6, then 2 ≤ 6² but (6)² < 2 is not true]
∴ R is not symmetric.
(iii) Now (a, b), (b, c) ∈ R ∉ (a, c) ∈ R
[Take a = 1, b = – 2, c = – 3 ∴ a ≤ b² . b ≤ c² but a ≤ c² is not true) ∴ R is not transitive.

R = {(a, b) : a ≤ b³}
(i) Since (a, a) ∉ R as a ≤ a³ is not always true
[Take a = 1/3. then a ≤ a³ is not true]
∴ R is not reflexive.
(ii) Also (a, b) ∈ R ⇏ (b, a) ∈ R
[Take a = 1, b = 4, ∴ 1 ≤ 4³ but 4 ≰ (l)³ ]
∴ R is not symmetric.
(iii) Now (a, b) ∈ R, (b, c) ∈ R ⇏ (a, c) ∴ R
[Take a = 100, b = 5, c = 3, ∴ 100 ≤ 5³, 5 ≤ 3³ but 100 ≥ 3³] R is not symmetric.