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Show that the relation \(R\) on the set \(Z\) of integers, given by \(R=\left\{ \left( a,b \right) :2\ \text{divides}\ a-b \right\} \) is an equivalence relation.

Answer

The relation \(R\) on \(Z\) is given by \(R=\left\{ \left( a,b \right) :2\quad \text{divides}\quad a-b \right\} \)
We observe the following properties of relation \(R\)

Reflexivity: for any \(a\in Z\)
\(a-a=0=0\times 2\)
\(\Rightarrow\) \(2\) divides \(a-a\Rightarrow (a,a)\in R\)
So, \(R\) is relexive relation on \(Z\)

Symmetry: Let \(a,b\in Z\) be such that
\((a,b)\in R\)
\(\Rightarrow\) \(2\) divides \(a-b\)
\(\Rightarrow\) \(a-b=2\lambda\) for some \(\lambda \in Z\)
\(\Rightarrow\) \(b-a=2(-\lambda)\), where \(-\lambda \in Z\)
\(\Rightarrow\) \(2\) divides \(b-a\Rightarrow (b,a)\in R\)
Thus \((a,b)\in R\Rightarrow (b,a)\in R\). So, \(R\) is a symmetric relation on \(Z\).

Transitivity. Let \(a,b,c\in Z\) be such that \((a,b)\in R\) and \((b,c)\in R\). Then,
\((a,b)\in R\Rightarrow 2\) divides \(a-b\Rightarrow a-b=2\lambda\) for some \(\lambda \in Z\)
and \((b,c)\in R\Rightarrow 2\) divides \(b-c\Rightarrow b-c=2\mu\) for some \(\mu \in Z\)
\(\therefore\) \(a-b+b-c=2(\lambda +\mu)\)
\(\Rightarrow\) \(a-c=2(\lambda +\mu)\), where \(\lambda +\mu \in Z\)
\(\Rightarrow\) \(2\) divides \(a-c\)
\(\Rightarrow\) \((a,c)\in R\)
thus, \((a,b)\in R\) and \((b,c)\in R\Rightarrow (a,c)\in R\)
So, \(R\) is a transitive relation on \(Z\)
Hence, \(R\) is an equivalence relation on \(Z\)
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