Show that R is an equivalence relation on the set Z of integers given by R = {(a, b) : 2 divides a – b}. Write the equivalence class [0].
Given R = {(a, b) : 2 divides a – b}
For equivance relation we have to check three parameters:
(i) Reflexive:
If (a-b) is divisible by 2 then,
⇒ (a-a)=0 is also divisible by 2
⇒ (a,a) ∈ R
Hence R is Reflexive ∀ (a,b) ∈ Z
(ii)Symmetric:
If (a-b) is divisible by 2 then,
⇒ (b-a)=-(a-b) is also divisible by 2
⇒ (a,b) ∈ R and (b,a) ∈ R
Hence R is Symmetric ∀ (a,b) ∈ Z
(iii)Transitive:
If (a-b) and (b-c) are divisible by 2 then,
⇒ a-c=(a-b)+(b-c) is also divisible by 2
⇒ (a,b) ∈ R, (b,c) ∈ R and (a,c) ∈ R
Hence R is Transitive ∀ (a,b) ∈ Z
⇒ As Relation R is satisfying all the three parameters, hence R is an equivalence relation.
Now equivalence class [0] is the set of all those elements in A which are related to 0 under relation.
Now,
(a,0) ∈ R
⇒ a – 0 is divisible by 2 and a ∈ A.
⇒ a ∈ A such that 2 divides a.
⇒ a = 0, 2,4
Thus [0] = {0,2,4}