Show that the relation $R$ defined by $R=\{(a, b): a-b$ is divisible by $3 ; a, b \in Z\}$ is an equivalence relation.
We observe the following relations of relation R.
Reflexivity:
Let $a$ be an arbitrary element of $R$. Then,
$a-a=0=0 \times 3$
$\Rightarrow a-a$ is divisible by 3
$\Rightarrow(a, a) \in R$ for all $a \in Z$
So, $R$ is reflexive on $Z$.
Symmetry:
Let $(a, b) \in R$
$\Rightarrow a-b$ is divisible by 3
$\Rightarrow a-b 3 p$ for some $p \in Z$
$\Rightarrow b-a=3(-p)$
Here, $-p \in Z$
$\Rightarrow b-a$ is divisible by 3
$\Rightarrow(b, a) \in R$ for all $a, b \in Z$
So, $R$ is symmetric on $Z$.
Transitivity:
Let $(a, b)$ and $(b, c) \in R$
$\Rightarrow a-b$ and $b-c$ are divisible by 3
$\Rightarrow a-b=3 p$ for some $p \in Z$
and $b-c=3 q$ for some $q \in Z$
Adding the above two, we get
$a-b+b-c=3 p+3 q$
$\Rightarrow a-c=3(p+q)$
Here, $p+q \in Z$
$\Rightarrow a-c$ is divisible by 3
$\Rightarrow(a, c) \in R$ for all $a, c \in Z$
So, $R$ is transitive on $Z$.
Hence, R is an equivalence relation on Z.
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