(i) N = (1, 2, 3, …} 

The relation from N to N is given by: {(1, 3), (2, 5), (3, 7), (4, 9), …} 

when, x = 1 then y = 3 

x = 2 then y = 5 

x = 3 then y = 7 

x = 4 then y = 9 

3, 5, 7, 9, … is an A.P. 

Hence, its nth term = a + (n – 1 ).d, 

where a is first term and d, is a common difference. 

T_n = 3 + (n – 1) × 2 = 3 + 2n – 2 = 2n + 1 

Here, we get the required rule by putting n = x and T_n = y

 {(x, y) | x, y ∈ N and y = 2x + 1}. 

(ii) Relation in N is expressed as : 

{(2, 3), (4, 2), (6, 1)} = {(6, 1), (4, 2), (2, 3)} 

Here, 6, 4, 2 are in an A.P. 

Its general term T_n = 6 + (n – 1) × (-2) 

T_n = 6 – 2n + 2 

T_n = 8 – 2n 

Here, we get the required rule by putting x = y and T_n = x 

{(x, y) | x, y ∈ N, x = 8 – 2y or x + 2y = 8} and y < 4

(iii) Relation in N is expressed as: 

{(2, 1), (3, 2), (4, 3), (5, 4), …} 

Here, 2, 3, 4, 5, … are in an A.P. 

So, nth term T_n = 2 + (n – 1) × 1 = 2 + n – 1 = n + 1

Here, by putting n = x and T_n = y 

Required rule = {(x, y) | x, y ∈ N, x = y + 1 or y = x – 1}