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Question Number 119048 by Hassen_Timol last updated on 21/Oct/20

Show that :        ⌊x⌋ = ⌊((⌊nx⌋)/n)⌋    where n∈N^∗    x∈R

$$\mathrm{Show}\:\mathrm{that}\:: \\ $$$$ \\ $$$$\:\:\:\:\lfloor{x}\rfloor\:=\:\lfloor\frac{\lfloor{nx}\rfloor}{{n}}\rfloor\:\:\:\:\mathrm{where}\:{n}\in\mathbb{N}^{\ast} \:\:\:{x}\in\mathbb{R} \\ $$$$ \\ $$

Answered by mr W last updated on 21/Oct/20

say x=m+f with 0≤f<1  ⌊x⌋=m  ⌊nx⌋=nm+⌊nf⌋  0≤nf<n  0≤⌊nf⌋<n  0≤((⌊nf⌋)/n)<1  ⌊((⌊nf⌋)/n)⌋=0  ((⌊nx⌋)/n)=m+((⌊nf⌋)/n)  ⌊((⌊nx⌋)/n)⌋=m+⌊((⌊nf⌋)/n)⌋=m=⌊x⌋  ⇒⌊x⌋=⌊((⌊nx⌋)/n)⌋ proved

$${say}\:{x}={m}+{f}\:{with}\:\mathrm{0}\leqslant{f}<\mathrm{1} \\ $$$$\lfloor{x}\rfloor={m} \\ $$$$\lfloor{nx}\rfloor={nm}+\lfloor{nf}\rfloor \\ $$$$\mathrm{0}\leqslant{nf}<{n} \\ $$$$\mathrm{0}\leqslant\lfloor{nf}\rfloor<{n} \\ $$$$\mathrm{0}\leqslant\frac{\lfloor{nf}\rfloor}{{n}}<\mathrm{1} \\ $$$$\lfloor\frac{\lfloor{nf}\rfloor}{{n}}\rfloor=\mathrm{0} \\ $$$$\frac{\lfloor{nx}\rfloor}{{n}}={m}+\frac{\lfloor{nf}\rfloor}{{n}} \\ $$$$\lfloor\frac{\lfloor{nx}\rfloor}{{n}}\rfloor={m}+\lfloor\frac{\lfloor{nf}\rfloor}{{n}}\rfloor={m}=\lfloor{x}\rfloor \\ $$$$\Rightarrow\lfloor{x}\rfloor=\lfloor\frac{\lfloor{nx}\rfloor}{{n}}\rfloor\:{proved} \\ $$

Commented by Hassen_Timol last updated on 21/Oct/20

Thank you so much Sir, may god bless you

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