Please Solve RD Sharma Class 12 Chapter Relation Exercise1.2 Question 4 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.$Given\! : \! \! R=\left \{\left ( a, b \right ) :a-b \: is\: divisible\: by\: n \right \}\: is\: a \: relation \: de\! f\! ined\: on\: Z$
Explanation:
Let us check these properties on R
Reflexivity:
$Let \: a \: \epsilon \: N$
$Here, a-a=0\times n$
$a-a\: is\: divisible\: by\: n$
$(a, a) \: \epsilon \: R\: f\! or\: all\: a \: \epsilon \: Z$
So, R is reflexive on Z
Symmetry:
$Let (a, b) \: \epsilon \: R$
$Here, a-b\: is \: divisible\: by\: n$
$a-b=np\: f\! or \: some \: p \: \epsilon \: Z$
$b-a=n(-p)$
$b-a\: is\: divisible\: by\: n \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; [i\! f p \: \epsilon \: Z\:, then -p \: \epsilon \: Z]$
$(b, a) \: \epsilon \: R$
So, R is symmetric on Z.
Transitivity:
$Let (a, b) and (b, c) \: \epsilon \: R$
$Here, a-b\: is\: divisible\: by\: n \: and\: b-c\: is\: divisible\: by \: n$
$a-b=np \: f\! or\: some\: p \: \epsilon \: Z \: \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ...(i)$
$b-c=nq\: f\! or\: some\: q \: \epsilon \: Z \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ....(ii)$
$a-b+b-c=np+nq$
$a-c=np+q \; \; \; \; \; \; \; \; \; \; \; \; \; [Here, p+q \: \epsilon \: Z]$
$(a, c) \: \epsilon \: R \: f\! or\: all \: a, c \: \epsilon \: Z$
So, R is transitive on Z.
Therefore, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on Z.

Crack CUET with india's "Best Teachers"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support