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Name: Please Solve RD Sharma Class 12 Chapter Relation Exercise 1.2 Question 17 Maths Textbook Solution.
Uploaded: Mon, 2021-07-05 09:17
Description: Please Solve RD Sharma Class 12 Chapter Relation Exercise 1.2 Question 17 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Relation Exercise 1.2 Question 17 Maths Textbook Solution.

Answers (1)

Answer: Hence prove, R is an equivalence relation.
Hint: To prove an equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.
Explanation:
Given: $\text{C\: be\: the\: set\: o\! f\: all\: complex \: numbers.}$
$C_{0}\: \text{be\: the\: set\: o\! f\: all\: non-zero\: complex\: numbers.}$
$\text{Relation R on }C_{0} \text{ be defined as}$
$z_{1}\: R\: z_{2}\Leftrightarrow\frac{ z_{1}-z_{2}}{z_{1}+z_{2}} \text{ is real for all } z_{1}, z_{2}\: \epsilon \: C_{0}$
(i) Test for reflexivity
$Since, \frac{ z_{1}-z_{1}}{z_{1}+z_{1}}=0, which\: is\: a \: real \: number$
$So, (z_{1}, z_{1})\: \epsilon \: R$
Hence, R is a reflexive relation.
(ii) Test for symmetric
$\frac{z_{1}-z_{2}}{z_{1}+z_{2}}=x , where\: x\: is\: real$
$-\frac{ z_{1}-z_{2}}{z_{1}+z_{2}}= -x$
$\frac{z_{2}-z_{1}}{z_{2}+z_{1}}= -x \: is\: also\: a\: real\: number$
$So, (z_{2},z_{1}) \: \epsilon \: R$
Hence, R is a symmetric relation.
(iii) Test for transitivity
$Let (z_{1}, z_{2}) \: \epsilon \: R \: and\: (z_{2}, z_{3} ) \: \epsilon \: R$
$Then\: \frac{z_{1}-z_{2}}{z_{1}+z_{2}}= x$
$z_{1}-z_{2}=xz_{1}+xz_{2}$
$\frac{z_{1}}{z_{2}}=\frac{1+x}{1-x }$                                                                                                                                                 $...(i)$
$Also, \frac{ z_{2}-z_{3}}{z_{2}+z_{3}}=y$
$z_{2}-z_{3}=yz_{2}+yz_{3}$
$z_{2}\left ( 1-y \right )=z_{3}(1+y)$
$\frac{z_{2}}{z_{3}}=\frac{1+y}{1-y }$                                                                                                                                                     $...(ii)$
Dividing (i) and (ii), we get
$\frac{z_{1}}{z_{3}}=\frac{1+x}{1-x}\times \frac{1+y}{1-y}$
$=z\: where\: z\: is\: a\: real \: number.$
$\frac{z_{1}-z_{3}}{z_{1}+z_{3}}=\frac{z-1}{z+1}, which \: is\: real$
$(z_{1}, z_{3})\: \epsilon \: R$                                                                                                                                                               $...(iii)$
From (i), (ii) & (iii)
$\text{Hence, R\: is\: transitivity\: relation.}$
$\text{Hence\: proved, R \: is\: an\: equivalence \: relation.}$