Let $\mathbb{Z}$ be the set of all integers and let R be a relation on $\mathbb{Z}$, defined by
$R=\left\{ \left( a,b \right):a-b\text{ is even} \right\}$.
Show that R is an equivalence relation in $\mathbb{Z}$.

Hint: Recall the definition of an equivalence relation. A relation is said to be an equivalence relation if it is reflexive, symmetric and transitive. Prove that the given relation is reflexive, symmetric as well as transitive. Hence prove that the given relation is an equivalence relation.

Complete step-by-step answer:
Before solving the question, we need to understand what is an equivalence relation.
Reflexive relation: A relation R on a set “A” is said to be reflexive if $\forall a\in A$ we have $aRa$.
Symmetric relation: A relation R on a set “A” is said to be symmetric if $aRb\Rightarrow bRa$
Transitive relation: A relation R on a set “A” is said to be transitive if $aRb,bRc\Rightarrow aRc$.
Equivalence relation: A relation R on a set “A” is said to be an equivalence relation if the relation is reflexive, symmetric and transitive.

Reflexivity: We have $\forall a\in \mathbb{Z},a-a=0$ and since 0 is even, we have $aRa$. Hence the relation is reflexive.
Symmetricity: if 2 divides a-b, then 2 divides b-a. Hence if a-b is even, then b-a is even and hence if aRb, then bRa. Hence the relation is symmetric.
Transitivity: We have $aRb,bRc\Rightarrow a-b\text{ is even and }b-c\text{ is even}\text{.}$ Since the sum of two even numbers is even, we have $a-b+b-c\text{ is even}\Rightarrow a-c\text{ is even}\text{.}$ Hence aRc and hence the relation is transitive
Since the relation is reflexive, symmetric and transitive, the relation is an equivalence relation.

Note: [1] Students usually make a mistake while proving reflexivity of a relation. In reflexivity, we need all the elements of a to be related with themselves, and even if a single element is found such that it is not related with itself, then the relation is not reflexive.