# Prove that the relation R on the set A = {1,2,3,4,5} given by R = { (a,b) : | a - b | is even } is an equivalence relation.

Toolbox:
• A relation R in a set A is an equivalence relation if R is reflexive, symmetric and transitive.
• A relation R in a set A is called reflexive. if $(a,a) \in R\;for\; all\; a\in A$
• A relation R in a set A is called symmetric. if $(a_1,a_2) \in R\;\Rightarrow \; (a_2,a_1)\in R \;$ for $\;a_1,a_2 \in A$
• A relation R in a set A is called transitive. if $(a_1,a_2) \in\; R$ and $(a_2,a_3)\in R \Rightarrow \;(a_1,a_3)\in R\;$for all $\; a_1,a_2,a_3 \in A$
Given the set $A=\{1,2,,3,4,5\}$ and the relation $R=\{(a,b):|a-b| \;is\; even\}$:
Let $a=b$, $(a,a) \in R \rightarrow |a-a|=0$ which is even. Therefore $R$ is reflexive.
For $R$ to be symmetric, if $(a,b) \in R \rightarrow (b,a) \in R$.
$(a,b) \in R \rightarrow |a-b|=even$
$(b,a) \in R \rightarrow |b-a|=even$
$(a-b)=-(b-a);$ therefore $|(a-b)|=|(b-a)|$
Therefore $(b,a)\in R$. Hence, $R$ is symmetric
Let $(a,b) \in R \; and \;(b,c)\in R$
$\Rightarrow |a-b|\; is\; even$ and $|b-c|\;is \;even$
If $(a,c) \in R \rightarrow |a-c| = even$
Now, $a-c=a-b+b-c$, which is an even number, as the sum of even numbers is even.
Hence $R$ is transitive.
Since $R$ is reflexive, symmetric and transitive. $R$ is an eqivalence relation.