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Question Number 156463 by MathSh last updated on 11/Oct/21

Find: f : N^∗ → N^∗ ; 2(f(x) + f(y)) = x ∀x∈N^∗

$$\mathrm{Find}: \\ $$$$\mathrm{f}\::\:\mathbb{N}^{\ast} \:\rightarrow\:\mathbb{N}^{\ast} \:\:;\:\:\:\mathrm{2}\left(\mathrm{f}\left(\mathrm{x}\right)\:+\:\mathrm{f}\left(\mathrm{y}\right)\right)\:=\:\mathrm{x} \\ $$$$\forall\mathrm{x}\in\mathbb{N}^{\ast} \\ $$

Answered by FongXD last updated on 11/Oct/21

Given: 2(f(x)+f(y))=x (1) replace y with x and x with y we get: 2(f(y)+f(x))=y (2) (1)&(2), ⇒ x=y substituting y=x into (1) we get: 2(f(x)+f(x)) =x ⇒ f(x)=(1/4)x so. determinant (((f(x)=(1/4)x)))

$$\mathrm{Given}:\:\mathrm{2}\left(\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{y}\right)\right)=\mathrm{x}\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{replace}\:\mathrm{y}\:\mathrm{with}\:\mathrm{x}\:\mathrm{and}\:\mathrm{x}\:\mathrm{with}\:\mathrm{y} \\ $$$$\mathrm{we}\:\mathrm{get}:\:\mathrm{2}\left(\mathrm{f}\left(\mathrm{y}\right)+\mathrm{f}\left(\mathrm{x}\right)\right)=\mathrm{y}\:\:\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)\&\left(\mathrm{2}\right),\:\Rightarrow\:\mathrm{x}=\mathrm{y} \\ $$$$\mathrm{substituting}\:\mathrm{y}=\mathrm{x}\:\mathrm{into}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{we}\:\mathrm{get}:\:\mathrm{2}\left(\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}\right)\right)\:=\mathrm{x} \\ $$$$\Rightarrow\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{x} \\ $$$$\mathrm{so}.\:\begin{array}{|c|}{\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{x}}\\\hline\end{array} \\ $$

Commented by Rasheed.Sindhi last updated on 11/Oct/21

But f:N^∗ →N^∗ Here for x∈N^∗ , f(x)∈Q I think there′s a problem in the question.

$$\mathrm{But}\:\mathrm{f}:\mathbb{N}^{\ast} \rightarrow\mathbb{N}^{\ast} \\ $$$$\mathrm{Here}\:\mathrm{for}\:\mathrm{x}\in\mathbb{N}^{\ast} ,\:\mathrm{f}\left(\mathrm{x}\right)\in\mathbb{Q} \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{there}'\mathrm{s}\:\mathrm{a}\:\mathrm{problem}\:\mathrm{in}\:\mathrm{the}\:\mathrm{question}. \\ $$

Commented by MathSh last updated on 12/Oct/21

Yes Ser

$$\mathrm{Yes}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$

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