Show that the relation R on the set Z of integers, given by
R = {(a, b) : 2 divides a – b}, is an equivalence relation.
We observe the following properties of relation R.
Reflexivity:
Let $a$ be an arbitrary element of the set $Z$. Then,
$a \in R$
$\Rightarrow a-a=0=0 \times 2$
$\Rightarrow 2$ divides $a-a$
$\Rightarrow(a, a) \in R$ for all $a \in Z$
So, $R$ is reflexive on $Z$.
Symmetry:
Let $(a, b) \in R$
$\Rightarrow 2$ divides $a-b$
$\Rightarrow \frac{a-b}{2}=p$ for some $p \in Z$
$\Rightarrow \frac{b-a}{2}=-p$
Here, $-p \in Z$
$\Rightarrow 2$ divides $b-a$
$\Rightarrow(b, a) \in R$ for all $a, b \in Z$
So, $R$ is symmetric on $Z$.
Transitivity:
Let $(a, b)$ and $(b, c) \in R$
$\Rightarrow 2$ divides $a-b$ and 2 divides $b-c$
$\Rightarrow \frac{a-b}{2}=p$ and $\frac{b-c}{2}=q$ for some $p, q \in Z$
Adding the above two, we get
$\frac{a-b}{2}+\frac{b-c}{2}=p+q$
$\Rightarrow \frac{a-c}{2}=p+q$
Here, $p+q \in Z$
$\Rightarrow 2$ divides $a-c$
$\Rightarrow(a, c) \in R$ for all $a, c \in Z$
So, $R$ is transitive on $Z$.
Hence, R is an equivalence relation on Z.