Show that the relation R defined by R = {(a, b): a – b is divi

1.	Show that the relation R defined by R = {(a, b): a – b is divisible by 3; a, b ∈ Z} is an equivalence relation.
Answer» Given as R = {(a, b): a – b is divisible by 3; a, b ∈ Z} is a relation To prove equivalence relation, the given relation should be reflexive, symmetric and transitive. We have to check these properties on R. Reflexivity: Let a be an arbitrary element of R. Then, a – a = 0 = 0 × 3 ⇒ a − a is divisible by 3 ⇒ (a, a) ∈ R for all a ∈ Z Therefore, R is reflexive on Z. Symmetry: Let (a, b) ∈ R ⇒ a − b is divisible by 3 ⇒ a − b = 3p for some p ∈ Z ⇒ b − a = 3 (−p) Here, −p ∈ Z ⇒ b − a is divisible by 3 ⇒ (b, a) ∈ R for all a, b ∈ Z Clearly, R is symmetric on Z. Transitivity: Let (a, b) and (b, c) ∈ R ⇒ a − b and b − c are divisible by 3 ⇒ a – b = 3p for some p ∈ Z And b − c = 3q for some q ∈ Z On adding above two equations, we get a − b + b – c = 3p + 3q ⇒ a − c = 3(p + q) Here, p + q ∈ Z ⇒ a − c is divisible by 3 ⇒ (a, c) ∈ R for all a, c ∈ Z Thus, R is transitive on Z. ∴ R is reflexive, symmetric and transitive. Clearly, R is an equivalence relation on Z.

Show that the relation R defined by R = {(a, b): a – b is divisible by 3; a, b ∈ Z} is an equivalence relation.

Answer»

Given as R = {(a, b): a – b is divisible by 3; a, b ∈ Z} is a relation

To prove equivalence relation, the given relation should be reflexive, symmetric and transitive.

We have to check these properties on R.

Reflexivity:

Let a be an arbitrary element of R.

Then, a – a = 0 = 0 × 3

⇒ a − a is divisible by 3

⇒ (a, a) ∈ R for all a ∈ Z

Therefore, R is reflexive on Z.

Symmetry:

Let (a, b) ∈ R

⇒ a − b is divisible by 3

⇒ a − b = 3p for some p ∈ Z

⇒ b − a = 3 (−p)

Here, −p ∈ Z

⇒ b − a is divisible by 3

⇒ (b, a) ∈ R for all a, b ∈ Z

Clearly, R is symmetric on Z.

Transitivity:

Let (a, b) and (b, c) ∈ R

⇒ a − b and b − c are divisible by 3

⇒ a – b = 3p for some p ∈ Z

And b − c = 3q for some q ∈ Z

On adding above two equations, we get

a − b + b – c = 3p + 3q

⇒ a − c = 3(p + q)

Here, p + q ∈ Z

⇒ a − c is divisible by 3

⇒ (a, c) ∈ R for all a, c ∈ Z

Thus, R is transitive on Z.

∴ R is reflexive, symmetric and transitive.

Clearly, R is an equivalence relation on Z.