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Please Solve RD Sharma Class 12 Chapter Relation Exercise1.2 Question 5 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, symmetry and transitive.
Explanation:
Given:\! Z\: be\: set\: o\! f\: integers\: and\: R=\left \{\left ( a,b \right ):a,b \: \epsilon \: Z\: and\: a+b\: is\: even\right \}
Let us check these properties on R.
Reflexivity:
Let a be an arbitrary element of Z.
Then, a+a=2a\: is\: even\: f\! or\: all\: a \: \epsilon \: Z.
(a, a) \: \epsilon \: R\: f\! or\: all\: a \: \epsilon \: Z.
So, R is reflexive on Z.
Symmetry:
Let (a, \! b) \: \epsilon \: R
then, a+b\: is\: even
\Rightarrow b+a\: is\: even
(b, a) \: \epsilon \: R\: f\! or\: all\: a, b \: \epsilon \: Z
So, R is symmetric on Z.
Transitivity:
Let (a, b) and (b, c) \: \epsilon \: R
then, a+b\: and\: b+c\: are\: even
Now, let\: a+b=2x\: f\! or\: some\: x \: \epsilon \: Z \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(i)
b+c=2y\: f\! or\: some\: y \: \epsilon \: Z \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ...(ii)
Adding (i) and (ii)
\Rightarrow a+b+b+c=2x+2y
\Rightarrow a+2b+c=2x+2y
\Rightarrow a+c=2x+2y-2b
\Rightarrow a+c=2(x+y-b), which\: is\: even\: f\! or\: all\: x, y, b \: \epsilon \: Z
Thus, (a, c) \: \epsilon \: R
So, R is transitive on Z
Therefore, R is reflexive, symmetry and transitive.
Hence, R is an equivalence relation on Z.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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