Question Number 145273 by ArielVyny last updated on 03/Jul/21
$${show}\:{that}\:\forall{x}\in\mathbb{R}\:{x}−\mathrm{1}\leqslant{E}\left({x}\right)\leqslant{x} \\ $$
Answered by mathmax by abdo last updated on 04/Jul/21
![we have [x]≤x<[x]+1 ⇒[x]≤x and x−1<[x] ⇒ x−1<[x]≤x](Q145297.png)
$$\mathrm{we}\:\mathrm{have}\:\left[\mathrm{x}\right]\leqslant\mathrm{x}<\left[\mathrm{x}\right]+\mathrm{1}\:\Rightarrow\left[\mathrm{x}\right]\leqslant\mathrm{x}\:\mathrm{and}\:\mathrm{x}−\mathrm{1}<\left[\mathrm{x}\right]\:\Rightarrow \\ $$$$\mathrm{x}−\mathrm{1}<\left[\mathrm{x}\right]\leqslant\mathrm{x} \\ $$