Question Number 49640 by maxmathsup by imad last updated on 08/Dec/18
![if x ∈[p,(√(p^2 +2))] calculate [x]](https://www.tinkutara.com/question/Q49640.png)
$${if}\:{x}\:\in\left[{p},\sqrt{{p}^{\mathrm{2}} \:+\mathrm{2}}\right]\:\:{calculate}\:\left[{x}\right] \\ $$
Commented by maxmathsup by imad last updated on 08/Dec/18
$${p}\:{from}\:{N} \\ $$
Commented by Abdo msup. last updated on 23/Dec/18
![we have p≤x≤(√(p^2 +2))≤p+1 ⇒[x]=p](https://www.tinkutara.com/question/Q50982.png)
$${we}\:{have}\:{p}\leqslant{x}\leqslant\sqrt{{p}^{\mathrm{2}} \:+\mathrm{2}}\leqslant{p}+\mathrm{1}\:\Rightarrow\left[{x}\right]={p} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 09/Dec/18
![consider p=+ve integer p=I (√(p^2 +2)) =I+f when x lies between [p,(√(p^2 +2)) ] =[I,I+f] [x]=I that implies [x]=p](https://www.tinkutara.com/question/Q49677.png)
$${consider}\:{p}=+{ve}\:{integer} \\ $$$${p}={I} \\ $$$$\sqrt{{p}^{\mathrm{2}} +\mathrm{2}}\:={I}+{f} \\ $$$${when}\:{x}\:{lies}\:{between}\:\left[{p},\sqrt{{p}^{\mathrm{2}} +\mathrm{2}}\:\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left[{I},{I}+{f}\right] \\ $$$$\left[{x}\right]={I}\:\:{that}\:{implies}\:\left[{x}\right]={p} \\ $$