Question

Express the following relations in the form of sets or orders pairs: 

(i) R₁ is relation from set A = {1, 2, 3, 4, 5, 6} to set B = (1, 2, 3} such that “x = 2y”. 

(ii) R₂ is a relation set A = {8, 9, 10, 11} to set B = {5, 6, 7, 8} such that “y = x – 2”. 

(iii) R₃ is a relation in set A = {0, 1, 2,…, 10} defined by 2x + 3y = 12. 

(iv) R₄ is a relation from set A = {5, 6, 7, 8} to set B = {10, 12, 15, 16, 18} is defined such that “x is divisor of y”

Answer 1

(i) Given relation R₁ is defined as “x = 2y” 

and A = {1, 2, 3, 4, 5, 6} 

and in B = {1, 2, 3} 

x = 2y 

y = x/2

when x = 2 then y = 1. 

So, (2, 1) ∈ R₁ 

when x = 4 then y = 2. 

So, (4, 2) ∈ R₁ and x = 6 then y = 3. 

So, (6, 3) ∈ R₁ 

R₁ = {(2, 1), (4, 2), (6, 3)} 

(ii) Given relation R₂ is defined as “y = x – 2” 

and A = {8, 9, 10, 11} 

and B= {5,6, 7,8} 

So, from y = x – 2

when x = 8 then y = 8 – 2 = 6 

So, (8, 6) ∈ R₂ 

when x = 9 then y = 9 – 2 = 7 

So, (9, 7) ∈ R₂ 

when x = 10 then y = 10 – 2 = 8 

So, (10, 8) ∈ R₂ and 

when x = 11 then y = 11 – 2 = 9 

So, (11, 9) ∈ R₂ 

R₂ = {(8, 6), (9, 7), (10, 8)}

(iii) Given relation R₃ is defined as

(iv) Given relation R₄ is defined as “x, y” 

A = {5, 6, 7, 8} 

and B = {10, 12, 15, 16, 18} 

So, xR₄y ⇔ x is divisor of y 

x ∈ A, y ∈ B 

when x = 5 then 5 is a divisor of 10 and 15 

So, (5, 10), (5, 15) ∈ R₄ 

when x = 6 then 6 is a divisor of 12 and 18 

So, (6, 12), (6, 18) ∈ R₄ 

when x = 8 then 8 is a divisor of 16 

So, (8, 16) ∈ R₄ 

R₄ = {(5, 10), (5, 15), (6, 12), (6, 18), (8, 16)}