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Question Number 126845 by mnjuly1970 last updated on 24/Dec/20

$$...calculus\left(\mathrm{I}\right)...$$$$prove::$$$$i::\lfloor 2x\rfloor \stackrel{?}{=}\lfloor x\rfloor +\lfloor x+\frac{1}{2}\rfloor$$$$ii::\lfloor 3x\rfloor =\lfloor x\rfloor +\lfloor x+\frac{1}{3}\rfloor +\lfloor x+\frac{2}{3}\rfloor$$$$$$

Answered by floor(10²Eta[1]) last updated on 24/Dec/20

$$i::$$$$\lfloor \mathrm{x}\rfloor =\mathrm{n}\in \mathbb{Z}$$$$\mathrm{x}=\mathrm{n}+\alpha ,0⩽\alpha <1$$$$\lfloor 2\mathrm{n}+2\alpha \rfloor =\mathrm{n}+\lfloor \mathrm{n}+\alpha +\frac{1}{2}\rfloor$$$$\left(\mathrm{I}\right)\lfloor 2\mathrm{n}+2\alpha \rfloor =2\mathrm{n}\mathrm{if}\alpha <\frac{1}{2}$$$$\left(\mathrm{II}\right)\lfloor 2\mathrm{n}+2\alpha \rfloor =2\mathrm{n}+1\mathrm{if}\alpha ⩾\frac{1}{2}$$$$\mathrm{cheking}\mathrm{the}\mathrm{cases}\mathrm{in}\mathrm{the}\mathrm{original}\mathrm{equation}$$$$\mathrm{both}\mathrm{of}\mathrm{them}\mathrm{will}\mathrm{work}\mathrm{which}\mathrm{prove}\mathrm{the}\mathrm{question}$$$$$$$$\mathrm{same}\mathrm{logic}\mathrm{for}\mathrm{question}2$$

Answered by mathmax by abdo last updated on 24/Dec/20

$$\left[\mathrm{x}\right]=\mathrm{n}\Rightarrow \mathrm{n}⩽\mathrm{x}<\mathrm{n}+1\Rightarrow 2\mathrm{n}⩽2\mathrm{x}<2\mathrm{n}+2\mathrm{we}\mathrm{have}$$$$[2\mathrm{n},2\mathrm{n}+2[=[2\mathrm{n},2\mathrm{n}+1[\cup [2\mathrm{n}+1,2\mathrm{n}+2[$$$$\mathrm{if}2\mathrm{x}\in [2\mathrm{n},2\mathrm{n}+1[\Rightarrow \left[2\mathrm{x}\right]=2\mathrm{n}\mathrm{and}\mathrm{x}\in [\mathrm{n},\mathrm{n}+\frac{1}{2}[\Rightarrow \left[\mathrm{x}\right]=\mathrm{n}$$$$\mathrm{x}+\frac{1}{2}\in [\mathrm{n}+\frac{1}{2},\mathrm{n}+1[\Rightarrow [\mathrm{x}+\frac{1}{2}]=\mathrm{n}\Rightarrow \left[\mathrm{x}\right]+[\mathrm{x}+\frac{1}{2}]=\mathrm{n}+\mathrm{n}=2\mathrm{n}=\left[2\mathrm{x}\right]$$$$\mathrm{if}2\mathrm{x}\in [2\mathrm{n}+1,2\mathrm{n}+2[\Rightarrow \left[2\mathrm{x}\right]=2\mathrm{n}+1\mathrm{and}\mathrm{x}\in [\mathrm{n}+\frac{1}{2},\mathrm{n}+1[\Rightarrow \left[\mathrm{x}\right]=\mathrm{n}$$$$\mathrm{x}+\frac{1}{2}\in [\mathrm{n}+1,\mathrm{n}+\frac{3}{2}[\Rightarrow [\mathrm{x}+\frac{1}{2}]=\mathrm{n}+1\Rightarrow \left[\mathrm{x}\right]+[\mathrm{x}+\frac{1}{2}]=\mathrm{n}+\mathrm{n}+1=2\mathrm{n}+1$$$$=\left[2\mathrm{x}\right]\mathrm{in}\mathrm{all}\mathrm{cases}\mathrm{the}\mathrm{dquality}\mathrm{is}\mathrm{proved}..$$

Answered by mindispower last updated on 24/Dec/20

$$\left[x\right]=n$$$$\Rightarrow n⩽x<n+1$$$$x\in [n,n+\frac{1}{3}[\Rightarrow$$$$\left[x\right]=[x+\frac{1}{3}]=[x+\frac{2}{3}]=n$$$$3x\in [3n,3n+1[\Rightarrow \left[3x\right]=3n=\left[x\right]+[x+\frac{1}{3}]+[x+\frac{2}{3}]$$$$x\in [n+\frac{1}{3};n+\frac{2}{3}[$$$$\Rightarrow \left[x\right]=[x+\frac{1}{3}]=n$$$$[x+\frac{2}{3}]=n+1$$$$3x\in [3n+1,3n+2[\Rightarrow \left[3x\right]=[x[+[x+\frac{1}{3}]+[x+\frac{2}{3}]=3n+1$$$$andsamforx\in [n+\frac{2}{3},n+1[$$

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