1. Relation R in the set A = {1, 2, 3,……. , 13, 14} defined as R = {(x, y) : 3x – y = 0
2. Relation R in the set N of natural numbers defined as R = {(x, y) : y = x + 5 and x < 4}

Solution:

(i)R = {(x, y) : 3x – y = 0}

A = {1, 2, 3, 4, 5, 6, … 13, 14}

Subsequently, R = {(1, 3), (2, 6), (3, 9), (4, 12)} … (1)

According to reflexive property: (x, x) ∈ R, then, at that point R is reflexive) Since there is no such pair, so R isn’t reflexive.

According to symmetric property: (x, y) ∈ R and (y, x) ∈ R, then, at that point R is symmetric. Since there is no such pair, R isn’t symmetric

According to transitive property: If (x, y) ∈ R and (y, z) ∈ R, then, at that point (x, z) ∈ R. Subsequently R is transitive. From (1), (1, 3) ∈ R and (3, 9) ∈ R however (1, 9) ∉ R, R isn’t transitive.

Hence, R is neither reflexive, nor symmetric and nor transitive.

(ii) R = {(x, y) : y = x + 5 and x < 4} in set N of normal numbers.

Upsides of x are 1, 2, and 3

Along these lines, R = {(1, 6), (2, 7), (3, 8)}

According to reflexive property: (x, x) ∈ R, then, at that point R is reflexive Since there is no such pair, R isn’t reflexive.

According to symmetric property: (x, y) ∈ R and (y, x) ∈ R, then, at that point R is symmetric. Since there is no such pair, so R isn’t symmetric

According to transitive property: If (x, y) ∈ R and (y, z) ∈ R, then, at that point (x, z) ∈ R. Accordingly R is transitive. Since there is no such pair, so R isn’t transitive.

Accordingly, R is neither reflexive, nor symmetric and nor transitive.