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Question
Show that the relation R on the set Z of integers, given by R = {(a,b):2divides (a - b)} is an equivalence relation.
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Solution
The relation R on Z is given by R = {(a,b) :2divides a - b}.
We observe the following properties of relation R.
Refelxivity : For any a ∈ Z
a - a = 0 = 0 × 2
⇒ 2 divides a - a
⇒ (a, a) ∈ R
So, R is a reflexive relation on Z.
Symmetry: Let a,b ∈ Z be such that
(a,b) ∈ R
⇒ 2 divides a - b
⇒ a - b = 2λ for some λ ∈ Z
⇒ b - a = 2(- λ ),where - λ ∈ Z
⇒ 2 divides b - a
⇒ (b, a) ∈ R
Thus, (a,b) ∈ R ⇒ (b, a) ∈ R. So, R is a symmetric relation on Z.
Transitivity: Let a,b, c ∈ Z be such that (a,b) ∈ R and (b, c) ∈ R. Then,
(a,b) ∈ R ⇒ 2 divides a - b ⇒ a - b = 2λ for some λ ∈ Z
and (b, c) ∈ R ⇒ 2 divides b - c ⇒ b - c = 2 μ for some μ ∈ Z
a - b + b - c = 2( λ + μ )
2 divides a - c
⇒ (a, c) ∈ R
Thus, (a,b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R.
So, R is a transitive relation on Z.
Since R is symmetric and transitive
reflexive therefore an equivalence relation
Hence, R is a transitive relation on Z.
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