Menu Close

Solve-the-following-equation-x-x-x-x-1-x-x-is-floor-of-x-




Question Number 211467 by mnjuly1970 last updated on 10/Sep/24
     Solve the following equation         ⌊ x ⌋ + (√(x −(√x) )) = ⌊ x+ (1/x) ⌋         ⌊ x ⌋ is floor of ,  x  ,
$$ \\ $$$$\:\:\:{Solve}\:{the}\:{following}\:{equation} \\ $$$$ \\ $$$$\:\:\:\:\:\lfloor\:{x}\:\rfloor\:+\:\sqrt{{x}\:−\sqrt{{x}}\:}\:=\:\lfloor\:{x}+\:\frac{\mathrm{1}}{{x}}\:\rfloor \\ $$$$ \\ $$$$\:\:\:\:\:\lfloor\:{x}\:\rfloor\:{is}\:{floor}\:{of}\:,\:\:{x}\:\:,\: \\ $$
Answered by Frix last updated on 10/Sep/24
(√(x−(√x)))∈R ⇒ x=0∨x≥1  ⌊x⌋∈N∧⌊x+(1/x)⌋∈N ⇒ (√(x−(√x)))∈N    ⌊x+(1/x)⌋−⌊x⌋=(√(x−(√x)))  1. ⌊x+(1/x)⌋=⌊x⌋       impossible because (√(x−(√x)))=0 ⇒       x=0∨x=1 ⇒ ⌊x+(1/x)⌋≠⌊x⌋  2. ⌊x+(1/x)⌋=⌊x⌋+1       ⇒ (√(x−(√x)))=1 ⇒ x=((3+(√5))/2)
$$\sqrt{{x}−\sqrt{{x}}}\in\mathbb{R}\:\Rightarrow\:{x}=\mathrm{0}\vee{x}\geqslant\mathrm{1} \\ $$$$\lfloor{x}\rfloor\in\mathbb{N}\wedge\lfloor{x}+\frac{\mathrm{1}}{{x}}\rfloor\in\mathbb{N}\:\Rightarrow\:\sqrt{{x}−\sqrt{{x}}}\in\mathbb{N} \\ $$$$ \\ $$$$\lfloor{x}+\frac{\mathrm{1}}{{x}}\rfloor−\lfloor{x}\rfloor=\sqrt{{x}−\sqrt{{x}}} \\ $$$$\mathrm{1}.\:\lfloor{x}+\frac{\mathrm{1}}{{x}}\rfloor=\lfloor{x}\rfloor \\ $$$$\:\:\:\:\:\mathrm{impossible}\:\mathrm{because}\:\sqrt{{x}−\sqrt{{x}}}=\mathrm{0}\:\Rightarrow \\ $$$$\:\:\:\:\:{x}=\mathrm{0}\vee{x}=\mathrm{1}\:\Rightarrow\:\lfloor{x}+\frac{\mathrm{1}}{{x}}\rfloor\neq\lfloor{x}\rfloor \\ $$$$\mathrm{2}.\:\lfloor{x}+\frac{\mathrm{1}}{{x}}\rfloor=\lfloor{x}\rfloor+\mathrm{1} \\ $$$$\:\:\:\:\:\Rightarrow\:\sqrt{{x}−\sqrt{{x}}}=\mathrm{1}\:\Rightarrow\:{x}=\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$
Commented by mnjuly1970 last updated on 10/Sep/24
$$\:\cancel{ } \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *