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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

$$sayx=m+fwith0⩽f<1$$$$\lfloor x\rfloor =m$$$$\lfloor nx\rfloor =nm+\lfloor nf\rfloor$$$$0⩽nf<n$$$$0⩽\lfloor nf\rfloor <n$$$$0⩽\frac{\lfloor nf\rfloor }{n}<1$$$$\lfloor \frac{\lfloor nf\rfloor }{n}\rfloor =0$$$$\frac{\lfloor nx\rfloor }{n}=m+\frac{\lfloor nf\rfloor }{n}$$$$\lfloor \frac{\lfloor nx\rfloor }{n}\rfloor =m+\lfloor \frac{\lfloor nf\rfloor }{n}\rfloor =m=\lfloor x\rfloor$$$$\Rightarrow \lfloor x\rfloor =\lfloor \frac{\lfloor nx\rfloor }{n}\rfloor proved$$

Commented by Hassen_Timol last updated on 21/Oct/20

Thank you so much Sir, may god bless you

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