Prove that the relation R on Z defined by (a, b) ∈ R ⇔ a �

1.	Prove that the relation R on Z defined by (a, b) ∈ R ⇔ a − b is divisible by 5 is an equivalence relation on Z.
Answer» Given as R on Z defined by (a, b) ∈ R ⇔ a − b is divisible by 5 To prove equivalence relation, the 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 × 5 ⇒ a − a is divisible by 5 ⇒ (a, a) ∈ R for all a ∈ Z Therefore, R is reflexive on Z. Symmetry: Let (a, b) ∈ R ⇒ a − b is divisible by 5 ⇒ a − b = 5p for some p ∈ Z ⇒ b − a = 5(−p) Here, −p ∈ Z [Since p ∈ Z] ⇒ b − a is divisible by 5 ⇒ (b, a) ∈ R for all a, b ∈ Z Thus, R is symmetric on Z. Transitivity: Let (a, b) and (b, c) ∈ R ⇒ a − b is divisible by 5 ⇒ a − b = 5p for some Z Also, b − c is divisible by 5 ⇒ b − c = 5q for some Z On adding two equations above, we get a − b + b − c = 5p + 5q ⇒ a − c = 5(p + q) ⇒ a − c is divisible by 5 Here, p + q ∈ Z ⇒ (a, c) ∈ R for all a, c ∈ Z Clearly, R is transitive on Z. ∴ R is reflexive, symmetric and transitive. Hence, R is an equivalence relation on Z.

Prove that the relation R on Z defined by (a, b) ∈ R ⇔ a − b is divisible by 5 is an equivalence relation on Z.

Answer»

Given as R on Z defined by (a, b) ∈ R ⇔ a − b is divisible by 5

To prove equivalence relation, the 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 × 5

⇒ a − a is divisible by 5

⇒ (a, a) ∈ R for all a ∈ Z

Therefore, R is reflexive on Z.

Symmetry:

Let (a, b) ∈ R

⇒ a − b is divisible by 5

⇒ a − b = 5p for some p ∈ Z

⇒ b − a = 5(−p)

Here, −p ∈ Z [Since p ∈ Z]

⇒ b − a is divisible by 5

⇒ (b, a) ∈ R for all a, b ∈ Z

Thus, R is symmetric on Z.

Transitivity:

Let (a, b) and (b, c) ∈ R

⇒ a − b is divisible by 5

⇒ a − b = 5p for some Z

Also, b − c is divisible by 5

⇒ b − c = 5q for some Z

On adding two equations above, we get

a − b + b − c = 5p + 5q

⇒ a − c = 5(p + q)

⇒ a − c is divisible by 5

Here, p + q ∈ Z

⇒ (a, c) ∈ R for all a, c ∈ Z

Clearly, R is transitive on Z.

∴ R is reflexive, symmetric and transitive.

Hence, R is an equivalence relation on Z.