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Question Number 211467 by mnjuly1970 last updated on 10/Sep/24

$$$$$$Solvethefollowingequation$$$$$$$$\lfloor x\rfloor +\sqrt{x-\sqrt{x}}=\lfloor x+\frac{1}{x}\rfloor$$$$$$$$\lfloor x\rfloor isfloorof,x,$$

Answered by Frix last updated on 10/Sep/24

$$\sqrt{x-\sqrt{x}}\in \mathbb{R}\Rightarrow x=0\vee x⩾1$$$$\lfloor x\rfloor \in \mathbb{N}\wedge \lfloor x+\frac{1}{x}\rfloor \in \mathbb{N}\Rightarrow \sqrt{x-\sqrt{x}}\in \mathbb{N}$$$$$$$$\lfloor x+\frac{1}{x}\rfloor -\lfloor x\rfloor =\sqrt{x-\sqrt{x}}$$$$1.\lfloor x+\frac{1}{x}\rfloor =\lfloor x\rfloor$$$$\mathrm{impossible}\mathrm{because}\sqrt{x-\sqrt{x}}=0\Rightarrow$$$$x=0\vee x=1\Rightarrow \lfloor x+\frac{1}{x}\rfloor \ne \lfloor x\rfloor$$$$2.\lfloor x+\frac{1}{x}\rfloor =\lfloor x\rfloor +1$$$$\Rightarrow \sqrt{x-\sqrt{x}}=1\Rightarrow x=\frac{3+\sqrt{5}}{2}$$

Commented by mnjuly1970 last updated on 10/Sep/24

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