Question Number 119048 by Hassen_Timol last updated on 21/Oct/20 | ||
$$\mathrm{Show}\:\mathrm{that}\:: \\ $$$$ \\ $$$$\:\:\:\:\lfloor{x}\rfloor\:=\:\lfloor\frac{\lfloor{nx}\rfloor}{{n}}\rfloor\:\:\:\:\mathrm{where}\:{n}\in\mathbb{N}^{\ast} \:\:\:{x}\in\mathbb{R} \\ $$$$ \\ $$ | ||
Answered by mr W last updated on 21/Oct/20 | ||
$${say}\:{x}={m}+{f}\:{with}\:\mathrm{0}\leqslant{f}<\mathrm{1} \\ $$$$\lfloor{x}\rfloor={m} \\ $$$$\lfloor{nx}\rfloor={nm}+\lfloor{nf}\rfloor \\ $$$$\mathrm{0}\leqslant{nf}<{n} \\ $$$$\mathrm{0}\leqslant\lfloor{nf}\rfloor<{n} \\ $$$$\mathrm{0}\leqslant\frac{\lfloor{nf}\rfloor}{{n}}<\mathrm{1} \\ $$$$\lfloor\frac{\lfloor{nf}\rfloor}{{n}}\rfloor=\mathrm{0} \\ $$$$\frac{\lfloor{nx}\rfloor}{{n}}={m}+\frac{\lfloor{nf}\rfloor}{{n}} \\ $$$$\lfloor\frac{\lfloor{nx}\rfloor}{{n}}\rfloor={m}+\lfloor\frac{\lfloor{nf}\rfloor}{{n}}\rfloor={m}=\lfloor{x}\rfloor \\ $$$$\Rightarrow\lfloor{x}\rfloor=\lfloor\frac{\lfloor{nx}\rfloor}{{n}}\rfloor\:{proved} \\ $$ | ||
Commented by Hassen_Timol last updated on 21/Oct/20 | ||
Thank you so much Sir, may god bless you | ||