First slide

Relations XII

Question

A relation R on the set of complex numbers defined by $Z_{1} R Z_{2} \Leftrightarrow \frac{Z_{1} - Z_{2}}{Z_{1} + Z_{2}} is real then which of the following is not true$

Moderate

A

R is reflexive
B

R is symmetric
C

R is transitive
D

R is not equivalence

Solution

$\begin{array}{l} Z_{1} R Z_{1} \Rightarrow \frac{Z_{1} - Z_{1}}{Z_{1} + Z_{1}} = 0 \\ ∴ R is Reflexive because 0 is real. \end{array}$
$\begin{array}{l} Z_{1} R Z_{2} \Rightarrow \frac{Z_{1} - Z_{2}}{Z_{1} + Z_{2}} is real \Rightarrow - (\frac{Z_{1} - Z_{2}}{Z_{1} + Z_{2}}) is real \\ ∴ R is symmetric \end{array}$
$\begin{array}{l} Z_{1} = a_{1} + i b_{1}; Z_{2} = a_{2} + i b_{2} \\ Z_{1} R Z_{2} \Rightarrow \frac{Z_{1} - Z_{2}}{Z_{1} + Z_{2}} is real \\ \Rightarrow (a_{1} + a_{2}) (b_{1} - b_{2}) - (a_{1} - a_{2}) (b_{1} + b_{2}) = 0 \\ \Rightarrow 2 a_{2} b_{1} - 2 b_{2} a_{1} = 0 \\ \Rightarrow \frac{a_{1}}{b_{1}} = \frac{a_{2}}{b_{2}} \\ ∴ Z_{2} R Z_{3} \Rightarrow \frac{a_{2}}{b_{2}} = \frac{a_{3}}{b_{3}} \end{array}$
$t h e r e f o r e \frac{a_{1}}{b_{1}} = \frac{a_{3}}{b_{3}}$
$∴ Z_{1} R Z_{3} \Rightarrow R is transitive.$

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