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Please solve RD Sharma class12 Chapter Relation exercise 1 question 1 subquestion (i) maths textbook solution.

Answers (1)

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 ):\left ( a-b\right ) \: is\: divisible \: by\: 5\right \}\: is\: a\: relation\: de\! f\! ined\: on\: Z.
Explanation:
Let us check these properties on R.
Reflexivity:
Let a be an arbitrary element of the set Z
\Rightarrow a-a=0=0\times 5
\Rightarrow a-a\: is\: divisible\: by \: 5           
(a, a)\: \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\: 5
a-b=5p\: f\! or\: some\: p\: \epsilon \: Z
then\: b-a=5(-p)
Here, -p\: \epsilon \: Z \: \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; [\because p \: \epsilon \: Z]
b-a\: is\: divisible\: by\: 5
(b, a)\: \epsilon \: R \: f\! or\: all \: a, b \: \epsilon \: Z
So, R is symmetric on Z
Transitivity:
\text{Let (a, b) and (b, c)}\: \epsilon R
\Rightarrow a-b\: is\: divisible\: by\: 5
a-b=5p \: f\! or\: some\: p \: \epsilon \: Z                            ...(i)                                                                   
Also,b-c \: is\: divisible\: by\: 5...(ii)
b-c=5q\: f\! or\: some\: q \: \epsilon \: Z                                                                                                                                    
Adding eq. (i) and (ii)
a-b+b-c=5p+5q
a-c=5\left (p+q \right )
a-c\: is\: divisible\: by\: 5
Here, p+q \: \epsilon \: Z
(a, c) \: \epsilon \: R\: is\: transitive\: on\: Z
Therefore R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on Z.

 

 

 


 

 

 

 

 

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