Question

Prove that the relation $R$ on $Z$ defined by $(a,b)\in R\iff$ $5$ divides $a-b$, is an equivalence relation on $Z$.

Answer 1

Let the relation be $R$ on $Z$ is given by $R=\{(a,b):5\text{divides}a-b\}$.
We observe the following properties of relation $R$
Reflexivity: for any $a\in Z$
$a-a=0=0\times 5$
$\Rightarrow$ $5$ divides $a-a\Rightarrow (a,a)\in R$
So, $R$ is relexive relation on $Z$
Symmetry: Let $a,b\in Z$ be such that
$(a,b)\in R$
$\Rightarrow$ $5$ divides $a-b$
$\Rightarrow$ $a-b=5\lambda$ for some $\lambda \in Z$
$\Rightarrow$ $b-a=5(-\lambda )$, where $-\lambda \in Z$
$\Rightarrow$ $5$ 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,
$(a,b)\in R\Rightarrow 5$ divides $a-b\Rightarrow a-b=5\lambda$ for some $\lambda \in Z$
and $(b,c)\in R\Rightarrow 5$ divides $b-c\Rightarrow b-c=5\mu$ for some $\mu \in Z$
$\therefore$ $a-b+b-c=5(\lambda +\mu )$
$\Rightarrow$ $a-c=5(\lambda +\mu )$, where $\lambda +\mu \in Z$
$\Rightarrow$ $5$ 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 $Z$.