#### Please Solve RD Sharma Class 12 Chapter Relation Exercise1.2 Question 2 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.

$Given: R=\left \{\left ( a, b \right ):2 \text{ divides } a-b \right \}is\: a\, relation \: def\! ined \: on \: Z.$
Explanation:
Let us check these properties on R.
Reflexivity:
Let a be an arbitrary element of the set Z
$Then, a-\! a=0=0\! \times \! 2$
$\Rightarrow 2\: divides\: a-a$
$\Rightarrow (a, a) \: \epsilon \: R\: f\! or \: all\: a \: \epsilon \: Z.$
So, R is reflexive on Z.
Symmetry:
$Let (a, b)\: \epsilon \: R$
$\Rightarrow 2 \: divides\: a-b$
$\Rightarrow \frac{a-b}{2}=\! p\: f\! or\: some\: p \: \epsilon \: Z$
$\Rightarrow \frac{b-a}{2}=-p$
$Here, -p \: \epsilon \: Z$
$\Rightarrow 2\: divides\: b-a$
$(b, a) \: \epsilon \: R \: f\! or \: all\: a,b \: \epsilon \: Z$
So, R is symmetric on Z
Transitivity:
$Let (a, b) and (b, c) \: \epsilon \: R$
$\Rightarrow 2\: divides\: a-b\: and\: 2 \: divides\: b-c$
$\Rightarrow \frac{a-b}{2}=p$                                                                                                                                                      $...(i)$
$\Rightarrow\frac{b-c}{2}=q$                                                                                                                                                      $...(ii)$
$f\! or\: some\: p, q\: \epsilon \: Z$
$\frac{a-b}{2}+\frac{b-c}{2}=p+q$
$\frac{a-c}{2}=p+q$
$Here, p+q \: \epsilon \: Z$
$\Rightarrow 2\: divided\: a-c$
$(a,c) \: \epsilon \: R\: f\! or\: all\: a, c \: \epsilon \: Z$
So, R is transitive on Z
Therefore, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on Z.