Show that the relation S in the set

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.

Answers (1)

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