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Question-146017




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

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