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

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

Answered by FongXD last updated on 11/Oct/21

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

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

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

Commented by MathSh last updated on 12/Oct/21

$$\mathrm{Yes}\mathbf{S}\mathrm{er}$$

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