229.

The following three relations are defined on the set of natural numbers :
R = {(x, y) : x < y, x ∈ N, y ∈ N}
S = { (x,y) : x + y = 10, x ∈ N, y ∈ N}
T = { (x, y) : x = y or  x – y = 1, x ∈ N, y ∈ N } Explain clearly which of the above relations are (i) Reflexive (ii) Symmetric (iii) Transitive.

We are given that
R = { (x, y) : x < y, x ∈ N, y ∈ N }
S = { (x, y) : x + y = 10, x ∈ N, y ∈ N }
T = { (x. y) : x = y or x – y = 1, x ∈ N, y ∈ N } (i) R is not reflexive as (x ,x) ∉ R ∀ x ∈ N
R is not symmetric as 1 < 2 ⇒ (1, 2) ∈ R but 2 < 1 ⇒ (2, 1) ∉ R
R is transitive as x and so ( x, y ) ∈ R and ( y, z ) ∈ R ⇒ ( x, z) ∈ R.
(ii) S is not reflexive as (x, x) ∉ R ∀ x ∈ IN

S is symmetric as x + y = 10 ⇒ y + x =10 and so (x, y) ∈ R ⇒ (y, x ) ∈ R ∀ x, y ∈ N S is not transitive as 3 + 7 = 10 ⇒ (3, 7) ∈ R and 7 + 3 = 10 ⇒ (7, 3) ∈ R but 3 + 3 ≠ 10 ⇒ (3,3) ∉ R (iii) T is not reflexive as x = x ∀ x ∈ N ⇒ (x, x) ∈ R

Now 2 – 1 = 1 as (2, 1) ∈ R but 1 ≠ 2  or 1 – 2 ≠ 1 ⇒ (1,2) ∉ R T is not symmetric.

Again 4 – 3=1 ⇒ (4,3) ∈ R and 3 – 2 = 1 ⇒(3,2) ∈ R but 4 ≠ 2 or 4 – 2 ≠ 1 ⇒ (4, 2) ∉ R.

T is not transitive.