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Question
Show that the relation R defined by R = {(a, b) : a – b is divisible by 3; a, b ∈ Z} is an equivalence relation.
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Solution
We observe the following relations of relation 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
So, 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
So, 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
Adding the above two, 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
So, R is transitive on Z.
Hence, R is an equivalence relation on Z.
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