Question Number 190625 by mnjuly1970 last updated on 07/Apr/23
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{solve}\:\mathrm{in}\:\:\:\mathbb{R}\:\:\:: \\ $$$$\:\:\:\:\:\:\:\:\lfloor\:\frac{\mathrm{1}}{{x}}\:\rfloor\:\:+\:\lfloor\:{x}\:\rfloor\:=\:\mathrm{2}\:\:\:\:\: \\ $$$$ \\ $$
Answered by mr W last updated on 07/Apr/23
$${if}\:{x}>\mathrm{1}: \\ $$$$\lfloor\frac{\mathrm{1}}{{x}}\rfloor=\mathrm{0} \\ $$$$\Rightarrow\lfloor{x}\rfloor=\mathrm{2} \\ $$$$\Rightarrow\mathrm{2}\leqslant{x}<\mathrm{3}\:\checkmark \\ $$$${if}\:{x}=\mathrm{1}: \\ $$$$\mathrm{1}+\mathrm{1}=\mathrm{2}\:\checkmark \\ $$$${if}\:{x}<\mathrm{1}: \\ $$$$\lfloor{x}\rfloor=\mathrm{0} \\ $$$$\Rightarrow\lfloor\frac{\mathrm{1}}{{x}}\rfloor=\mathrm{2} \\ $$$$\Rightarrow\mathrm{2}\leqslant\frac{\mathrm{1}}{{x}}<\mathrm{3} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{3}}<{x}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\:\checkmark \\ $$$${summary}: \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}<{x}\leqslant\frac{\mathrm{1}}{\mathrm{2}},\:{x}=\mathrm{1},\:\mathrm{2}\leqslant{x}<\mathrm{3} \\ $$
Commented by mnjuly1970 last updated on 07/Apr/23
$${thank}\:{you}\:{so}\:{much}\:{sir}\:{W}… \\ $$