InterviewSolution
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InterviewSolution
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A relation `R` on the set of complex numbers is defined by `z_1 R z_2` if and only if `(z_1-z_2)/(z_1+z_2)` is real. Show that R is an equivalence relation. |
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Answer» (i) For Z in C `_(z)R_(z)hArr (Z-Z)/(Z+Z)` is a real number. `rArr` o is a real number which is true. `:.` R is symmetric. (ii) Let `Z_(1), Z_(2)in C and _(Z_(1))R_(Z_(2))` Now, ` _(Z_(1))R_(Z_(2))rArr (Z_(1)-Z_(2))/(Z_(1)+Z_(2))`is a real number. `rArr-((Z_(2)-Z_(1))/(Z_(2)+Z_(1)))` is a real number. `rArr(Z_(2)-Z_(1))/(Z_(2)+Z_(1))` is a real number. ` rArr_(Z_(2))R_(Z_(1))` `:.` R is symmetric. (iii) `Let Z_(1), Z_(2), Z_(3) in C and _(Z_(1))R_(Z_(2)) and _(Z_(2))R_(Z_(3))` . Now, `_(z_(1))R_(z_(2))rArr (Z_(1)-Z_(2))/(Z_(1)+Z_(2))` is a real number. `rArr ((x_(1)+iy_(1))-(x_(2)+iy_(2)))/((x_(1)+iy_(1))+(x_(2)+iy_(2)))` is a real number. Where `Z_(1) = x_(1) + iy_(1) and Z_(2)=x_(2)+iy_(2)` `rArr ((x_(1)-x_(2))+i(y_(1)-y_(2)))/((x_(1)+x_(2))+i(y_(1)+y_(2)))` is a real number. `rArr([(x_(1)-x_(2))+i(y_(1)-y_(2))][(x_(1)+x_(2))-i(y_(1)+y_(2))])/([(x_(1)+x_(2))+i(y_(1)+y_(2))][(x_(1)+x_(2))-i(y_(1)+y_(2))])` is a real number `((x_(1)^(2)-x_(2)^(2))+(y_(1)^(2)-y_(2)^(2))+i{(x_(1)+x_(2))(y_(1)-y_(2))-(x_(1)-x_(2))-i(y_(1)+y_(2))})/((x_(1)+x_(2))^(2)+(y_(1)+y_(2))^(2))` is a real number. `rArr` Coefficient of i = 0 `rArr (x_(1)+x_(2))(y_(1)-y_(2))-(x_(1)-x_(2))(y_(1)+y_(2))= 0` `rArr x_(2)y_(1)-x_(1)y_(2)=0` `rArr (x_(1))/(y_(1))=(x_(2))/(y_(2))` `:. ""_(z_(1))R_(z_(2)) rArr (x_(1))/(y_(1))=(x_(2))/(y_(2))` Similarly, ` ""_(z_(1))R_(z_(2)) rArr (x_(2))/(y_(2))=(x_(3))/(y_(3))` `:.""_(z_(1))R_(z_(2)) rArr (x_(2))/(y_(2))=(x_(3))/(y_(3))` and `""_(z_(2))R_(z_(2)) ` `rArr (x_(1))/(y_(1))-(x_(2))/(y_(2)) and (x_(2))/(y_(2))=(x_(3))/(y_(3))` `rArr (x_(1))/(y_(1))= (x_(3))/(y_(3))rArr "" _(z_(1))R_(z_(3))` `:.` R is transitive. Thus, the given relation R is reflexive, symmetric. and transitive. i.e., R is an equivalence relation. Hence Proved. |
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