Question
Let X = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Let R1 be a relation in X given by R1 = {(x, y) : x – y is divisible by 3} and R2 be another relation on X given by R2 = {(x, y) : {x, y} ⊂ {1,4, 7 }} or ⊂ {2, 5, 8} or {x, y,}⊂ {3, 6, 9}}. Show that R1 = R2.
Solution
The characteristic of sets {1, 4, 7 }, {2, 5, 8} and {3, 6, 9} is that difference between any two elements of these sets is a multiple of 3.
∴ (x,y) ∈ R1 ⇒ x – y is a multiple of 3
⇒ {x,y} ⊂ {1,4,7} or {x, y} ⊂ {2, 5, 8} or {x,y} ⊂ {3, 6, 9}
⇒ (x,y) ∈ R2.
Hence R1 ⊂ R2.
Similarly, { x, y} ∈ R2
⇒ {x, y} ⊂ {1, 4, 7} or {x,y} ⊂ {2, 5, 8} or {x, y} ⊂ {3, 6, 9} ⇒ x – y is divisible by 3 ⇒ {x ,y} ∈ R1.
∴ R2 ⊂ R1.
Hence, R1 = R2.
