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Question Number 7326 by sou1618 last updated on 23/Aug/16

Is it correct?    floor(x)=⌊x⌋=[x]      floor(x)={n ∣ n∈Z , x−1<n≤x}  e.g....  floor(3)=3  floor(3.4)=3  floor(−2.5)=−3  floor(−π)=−4

$${Is}\:{it}\:{correct}? \\ $$ $$\:\:{floor}\left({x}\right)=\lfloor{x}\rfloor=\left[{x}\right] \\ $$ $$ \\ $$ $$ \\ $$ $${floor}\left({x}\right)=\left\{{n}\:\mid\:{n}\in\mathbb{Z}\:,\:{x}−\mathrm{1}<{n}\leqslant{x}\right\} \\ $$ $${e}.{g}.... \\ $$ $${floor}\left(\mathrm{3}\right)=\mathrm{3} \\ $$ $${floor}\left(\mathrm{3}.\mathrm{4}\right)=\mathrm{3} \\ $$ $${floor}\left(−\mathrm{2}.\mathrm{5}\right)=−\mathrm{3} \\ $$ $${floor}\left(−\pi\right)=−\mathrm{4} \\ $$

Commented byYozzia last updated on 24/Aug/16

For 2≤x<3, ⌊x⌋=2   For −2≤x<−1, ⌊x⌋=−2  ⌊x⌋=n∈Z for n≤x<n+1.

$${For}\:\mathrm{2}\leqslant{x}<\mathrm{3},\:\lfloor{x}\rfloor=\mathrm{2}\: \\ $$ $${For}\:−\mathrm{2}\leqslant{x}<−\mathrm{1},\:\lfloor{x}\rfloor=−\mathrm{2} \\ $$ $$\lfloor{x}\rfloor={n}\in\mathbb{Z}\:{for}\:{n}\leqslant{x}<{n}+\mathrm{1}. \\ $$ $$ \\ $$

Commented bysou1618 last updated on 24/Aug/16

thanks

$${thanks} \\ $$

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