# Show that the relation S in the set $A= \left \{ x\: \epsilon\: Z:0\leq x\leq 12 \right \}$  given by $S= \left \{ \left ( a,b \right ):a,b\, \epsilon \, Z,\left | a-b \right | \right$  is divisible by 3} is an equivalence relation.

Reflexive
$S= \left \{ \left ( a,b \right ):a,b\, \epsilon A,\left | a-b \right |is \; divisible\; by\; 3 \right \}$
is defined on the set $A= \left \{ x\, \epsilon \, z:0\leq x\leq 12 \right \}$
clearly, S is reflexive as 3 divides $\left | a-a \right |= 0$
$\psi a\, \epsilon \, A$
symmetric
further if $\left ( a,b \right )\epsilon \, S$  then 3 divides $\left | a-b \right |.$  therefore 3 divides $\left | -\left ( b-a \right ) \right |.= \left | b-a \right |$   as well
Hence $\left ( b,a \right )\, \epsilon S$  which follows that S is symmetric
Transitive
similarly, if $\left ( a,b \right )\, \epsilon S$  and $\left ( b,c \right )\, \epsilon S$  then $\left | a-b \right |$  and $\left | b-c \right |$   are both divisible by 3.
that is $\left | a-b \right |= 3m$     i.e $a-b= \pm 3m$
and    $\left | b-c \right |= 3n$    i.e  $b-c= \pm 3n$
so $\left ( a-b \right )+\left ( b-c \right )= a-c= \pm 3\left ( m+n \right )$
$\Rightarrow \left | a-c \right |= 3\left ( m+n \right )$
$\therefore \left | a-c \right |$  is divisible by 3
This shows that S is transitive as $\left ( a,c \right )\, \epsilon \, S$
Then S is an equivalence relation in the set A

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