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Please Solve RD Sharma Class 12 Chapter Relation Exercise1.2 Question 2 Maths Textbook Solution.

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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.
Let us check these properties on R.
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.
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
 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
Adding eq. (i) and (ii)
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.

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