#### Please Solve RD Sharma Class 12 Chapter Relation Exercise 1.2 Question 6 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, symmetry and transitive.
$Given: \! R=\left \{\left (m, n \right ) \! :\! m-n\: is\: divisible\: by\: 13 \right \} be\: a \: relation\: on\: Z$
Explanation:
Let us check these properties on R.
Reflexivity:
Let m be an arbitrary element of Z.
$Then, m-m=0=0\times 13$
$\Rightarrow m-m \: is \: divisible\: by\: 13.$
$\Rightarrow (m, m) \: \epsilon \: R$
Hence, R is reflexive on Z.
Symmetry:
$Let (m, n) \: \epsilon \: R$
$Then, m-n\: is\: divisible\: by \: 13$
$m-n=13p$
$Here, p \: \epsilon \: Z$
$n-m=13(-p)$
$Here, -p \: \epsilon \: Z$
$n-m\: is\: divisible \: by\: 13$
$(n, m) \: \epsilon \: R \: f\! or\: all\: m, n \: \epsilon \: Z$
So, R is symmetric on Z.
Transitivity:
$Let (m, n) and (n, o) \: \epsilon \: R$
$m-n\: and\: n-o \: are\: divisible\: by \: 13$
$m-n=13p\: f\! or\: some\: p \: \epsilon \: Z \: \: \: \: \: \: \: \: \: ...(i)$
$n-o=13q \: f\! or\: some\: q \: \epsilon \: Z \: \: \: \: \: \: \: ...(ii)$
$m-n+n-o=13p+13q$
$m-o=13(p+q)$
$Here, p+q \: \epsilon \: Z$
$m-o\: is \: divisible\: by \: 13$
$(m, o) \: \epsilon \: R\: f\! or\: all\: m, o \: \epsilon \: Z$
So, R is transitive on Z
Therefore, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on Z.