Question Number 56772 by Hassen_Timol last updated on 23/Mar/19 | ||
$$\mathrm{Given}\:\mathrm{that}\:: \\ $$$$\:\:\:\:\frac{\mathrm{1}}{\:\Phi\:}\:\:=\:\:\frac{\Phi}{\:\mathrm{1}\:+\:\Phi\:} \\ $$$$ \\ $$$$\mathrm{Prove}\:\mathrm{that}\::\:{without}\:{using}\:{the}\:{exact} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{value}\:{of}\:\Phi... \\ $$$$\:\:\:\:\:\frac{\:\mathrm{1}\:}{\:\Phi\:}\:\:=\:\:\Phi\:−\:\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Thank}\:\mathrm{you} \\ $$ | ||
Answered by JDamian last updated on 23/Mar/19 | ||
$$ \\ $$$$\frac{\mathrm{1}}{\:\Phi\:}\:\:=\:\:\frac{\Phi}{\:\mathrm{1}\:+\:\Phi\:} \\ $$$$\mathrm{1}+\Phi=\Phi^{\mathrm{2}} \\ $$$$\mathrm{1}=\Phi^{\mathrm{2}} −\Phi \\ $$$$\mathrm{1}=\Phi\left(\Phi−\mathrm{1}\right) \\ $$$$\frac{\mathrm{1}}{\Phi}=\Phi−\mathrm{1} \\ $$ | ||
Commented by Hassen_Timol last updated on 23/Mar/19 | ||
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir} \\ $$ | ||