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Please Solve RD Sharma Class 12 Chapter Relation Exercise 1.2 Question 1 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 )\! :\! a\! -\! b\: is\: divisible\: by\: 3;a,\! b\: \epsilon \: Z \right \}
Explanation:
Let us check the properties on R.
Reflexivity:
Let a be an arbitrary element of R.
Then      a-a= 0= 0\times 3
\Rightarrow a-a \: is\: divisible\: by\: 3
\Rightarrow \left ( a,a\right )\: \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\: 3
a\! -\! b\! =\! 3p\: f\! or\: some\: p\: \epsilon \: Z
Here,
\Rightarrow b\! -\! a=3(\! -p) is\: divisible\: by \: 3\: f\! or\: some\:\! \! -p\: \epsilon \: Z
(b\! ,a)\: \epsilon \: R \: f\! or\: all \: a,\! b\: \epsilon \: Z
So, R is symmetric on Z.
Transitivity:
Let \left ( a,b \right )\epsilon\: R
\Rightarrow a-b\: is \: divisible\: by \: 3 \! \! \: \: f\! or\: some \: p \: \epsilon \: Z
Here,   
\Rightarrow b-a=3(-p) \: is\: divisible\: by\: 3 , f\! or\: some \: -p\: \epsilon \: Z
(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
a\! -\! b\: and \: b\! -\! c \:\: are\: divisible\, by\: 3\: f\! or\: some\: p\: \epsilon \: Z                          ....(i)
b\! -\! c=3q\: f\! or\: some\: q\: \epsilon \: Z                                                                              ....(ii)
Adding eq. (i) and (ii)
a-b+ b-c= 3p+3q
a-c= 3\left ( p+q \right )
Here,p+q\: \epsilon \: Z
\Rightarrow a-c \: is \: divisible\: by\: 3
\Rightarrow (a,c)\epsilon \: R\: for \: all\: a,c\: \epsilon \: Z
So, R is transitive on Z
Therefore, R is reflective, symmetric and transitive.
Hence, R is an equivalence relation on Z.

 

 

 

 

 

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