#### Please Solve RD Sharma Class 12 Chapter Relation Exercise 1.2 Question 1 Maths Textbook Solution.

Answer: R is an equivalence relation on Z.
Hint:To prove equivalence relation it is necessary that the given relation should be reflexive,symmetric and transitive.

$Given\! :\! R\! =\! \left \{ \left ( a,b \right )\! :\! a\! -\! b\: is\: divisible\: by\: 3;a,\! b\: \epsilon \: Z \right \}$
Explanation:
Let us check the properties on R.
Reflexivity:
Let a be an arbitrary element of R.
Then      $a-a= 0= 0\times 3$
$\Rightarrow a-a \: is\: divisible\: by\: 3$
$\Rightarrow \left ( a,a\right )\: \epsilon \: R \: f\! or all, a\: \epsilon\: Z$
So, R  is reflexive on Z.
Symmetry:
$Let\: a,\! b\: \epsilon \: R$
$\Rightarrow a\! -\! b\: is\: divisible\: by\: 3$
$a\! -\! b\! =\! 3p\: f\! or\: some\: p\: \epsilon \: Z$
Here,
$\Rightarrow b\! -\! a=3(\! -p) is\: divisible\: by \: 3\: f\! or\: some\:\! \! -p\: \epsilon \: Z$
$(b\! ,a)\: \epsilon \: R \: f\! or\: all \: a,\! b\: \epsilon \: Z$
So, R is symmetric on Z.
Transitivity:
$Let \left ( a,b \right )\epsilon\: R$
$\Rightarrow a-b\: is \: divisible\: by \: 3 \! \! \: \: f\! or\: some \: p \: \epsilon \: Z$
Here,
$\Rightarrow b-a=3(-p) \: is\: divisible\: by\: 3 , f\! or\: some \: -p\: \epsilon \: Z$
$(b, a) \: \epsilon \: R\: f\! or\: all\: a, b \: \epsilon \: Z$
So,          R is symmetric on Z.
$Let (a,b) and (b,c)\: \epsilon \: R$
$a\! -\! b\: and \: b\! -\! c \:\: are\: divisible\, by\: 3\: f\! or\: some\: p\: \epsilon \: Z$                          $....(i)$
$b\! -\! c=3q\: f\! or\: some\: q\: \epsilon \: Z$                                                                              $....(ii)$
$a-b+ b-c= 3p+3q$
$a-c= 3\left ( p+q \right )$
Here,$p+q\: \epsilon \: Z$
$\Rightarrow a-c \: is \: divisible\: by\: 3$
$\Rightarrow (a,c)\epsilon \: R\: for \: all\: a,c\: \epsilon \: Z$
So, R is transitive on Z
Therefore, R is reflective, symmetric and transitive.
Hence, R is an equivalence relation on Z.