Question Number 146017 by Khalmohmmad last updated on 10/Jul/21
Answered by Olaf_Thorendsen last updated on 10/Jul/21
$${F}\left({x}\right)+{F}\left({y}\right)\:=\:{F}\left({x}\right){F}\left({y}\right)\:\:\:\left(\mathrm{1}\right) \\ $$$${x}\:=\:{y}\:\Rightarrow\:\mathrm{2}{F}\left({x}\right)\:=\:{F}^{\mathrm{2}} \left({x}\right)\:\Leftrightarrow\:{F}\left({x}\right)\left({F}\left({x}\right)−\mathrm{2}\right)\:=\:\mathrm{0} \\ $$$$\forall{x},\:{F}\left({x}\right)\:=\:\mathrm{0}\:\mathrm{or}\:\mathrm{2}\::\:\mathrm{impossible} \\ $$$$\mathrm{because}\:\mathrm{F}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{incompatible}\:\mathrm{conditions}\:! \\ $$$$\mathrm{We}\:\mathrm{verifly}\:\mathrm{that}\:\mathrm{if}\:\mathrm{F}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\frac{\mathrm{3}}{\mathrm{2}}\:: \\ $$$$\left(\mathrm{1}\right)\:\mathrm{F}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{F}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:\left(=\mathrm{3}\right)\:\neq\:\mathrm{F}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{F}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:\left(=\frac{\mathrm{9}}{\mathrm{4}}\right) \\ $$