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Question
Prove that the relation R on Z defined by
(a, b) ∈ R ⇔ a − b is divisible by 5
is an equivalence relation on Z.
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Solution
We observe the following properties of relation 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
So, 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
So, 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
Adding the above two, 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
So, R is transitive on Z.
Hence, R is an equivalence relation on Z.
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