Question Number 185981 by mnjuly1970 last updated on 30/Jan/23
$$ \\ $$$$\:\:\:\:\:\:\:{solve}\:\:{in}\:\:\:\mathbb{R} \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\:\lfloor\:\:\mathrm{2}{x}\:\rfloor\:\:+\:\lfloor\:\:\mathrm{3}{x}\:\rfloor\:=\:\mathrm{1} \\ $$$$\: \\ $$$$\:\:\:\:\lfloor\:{x}\:\rfloor\::=\:{greatest}\:{integer}\:{number} \\ $$$$\:\:\:\:{more}\:{than}\:\:{or}\:{equal}\:\:{to}\:\:''\:{x}\:'' \\ $$
Answered by mr W last updated on 30/Jan/23
$${x}={n}+{f} \\ $$$$\mathrm{5}{n}+\lfloor\mathrm{2}{f}\rfloor+\lfloor\mathrm{3}{f}\rfloor=\mathrm{1} \\ $$$$\mathrm{1}−\mathrm{5}{n}=\lfloor\mathrm{2}{f}\rfloor+\lfloor\mathrm{3}{f}\rfloor\geqslant\mathrm{0}\:\Rightarrow{n}\leqslant\mathrm{0} \\ $$$$\mathrm{1}−\mathrm{5}{n}=\lfloor\mathrm{2}{f}\rfloor+\lfloor\mathrm{3}{f}\rfloor<\mathrm{5}\:\Rightarrow{n}\geqslant\mathrm{0} \\ $$$$\Rightarrow{n}=\mathrm{0}\:\Rightarrow\mathrm{0}\leqslant{x}<\mathrm{1} \\ $$$$\mathrm{2}{x}<\mathrm{1}\:\Rightarrow{x}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{1}\leqslant\mathrm{3}{x}<\mathrm{2}\:\Rightarrow\frac{\mathrm{1}}{\mathrm{3}}\leqslant{x}<\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{3}}\leqslant{x}<\frac{\mathrm{1}}{\mathrm{2}}\:\checkmark \\ $$
Commented by mnjuly1970 last updated on 31/Jan/23
$$\:\:\:{mercey}\:{sir}\:{W}.. \\ $$