Let R = {(a, b) : a, b ∈ Z and (a - b) is divisible by 5}.
Show that R is an equivalence relation on Z.
In order to show R is an equivalence relation, we need to show R is Reflexive, Symmetric and Transitive.
Given that, ∀ a, b ∈Z, aRb if and only if a – b is divisible by 5.
Now,
R is Reflexive if (a,a) ∈ R ∀ a ∈ Z
aRa ⇒ (a-a) is divisible by 5.
a-a = 0 = 0 × 5 [since 0 is multiple of 5 it is divisible by 5]
⇒ a-a is divisible by 5
⇒ (a,a) ∈ R
Thus, R is reflexive on Z.
R is Symmetric if (a,b) ∈ R ⇒ (b,a) ∈ R ∀ a,b ∈ Z
(a,b) ∈ R ⇒ (a-b) is divisible by 5
⇒ (a-b) = 5z for some z ∈ Z
⇒ -(b-a) = 5z
⇒ b-a = 5(-z) [∵ z ∈ Z ⇒ -z ∈ Z ]
⇒ (b-a) is divisible by 5
⇒ (b,a) ∈ R
Thus, R is symmetric on Z.
R is Transitive if (a,b) ∈ R and (b,c) ∈ R ⇒ (a,c) ∈ R ∀ a,b,c ∈ Z
(a,b) ∈ R ⇒ (a-b) is divisible by 5
⇒ a-b = 5z1 for some z1∈ Z
(b,c) ∈ R ⇒ (b-c) is divisible by 5
⇒ b-c = 5z2 for some z2∈ Z
Now,
a-b = 5z1 and b-c = 5z2
⇒ (a-b) + (b-c) = 5z1 + 5z2
⇒ a-c = 5(z1 + z2 ) = 5z3 where z1 + z2 = z3
⇒ a-c = 5z3 [∵ z1,z2 ∈ Z ⇒ z3∈ Z]
⇒ (a-c) is divisible by 5.
⇒ (a, c) ∈ R
Thus, R is transitive on Z.
Since R is reflexive, symmetric and transitive it is an equivalence relation on Z.