Question Number 49639 by maxmathsup by imad last updated on 08/Dec/18
![find a relation betwen [x]^2 and [−x]^2](https://www.tinkutara.com/question/Q49639.png)
$${find}\:{a}\:{relation}\:{betwen}\:\left[{x}\right]^{\mathrm{2}} \:{and}\:\left[−{x}\right]^{\mathrm{2}} \\ $$
Commented by Abdo msup. last updated on 23/Dec/18
![if x∈Z [x]^2 =x^2 and [−x]^2 =(−x)^2 =x^2 ⇒ [x]^2 =[−x]^2 if x∈R−Z let [x]=p ⇒p<x<p+1 ⇒ −p−1<−x<−p ⇒[−x]=−p−1=−1−[x] ⇒ [−x]^2 =(1+[x])^2 =1+2[x] +[x]^2 .](https://www.tinkutara.com/question/Q50983.png)
$${if}\:{x}\in{Z}\:\:\:\left[{x}\right]^{\mathrm{2}} ={x}^{\mathrm{2}} \:\:{and}\:\left[−{x}\right]^{\mathrm{2}} =\left(−{x}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} \:\Rightarrow \\ $$$$\left[{x}\right]^{\mathrm{2}} =\left[−{x}\right]^{\mathrm{2}} \:\:{if}\:{x}\in{R}−{Z}\:\:{let}\:\left[{x}\right]={p}\:\Rightarrow{p}<{x}<{p}+\mathrm{1}\:\Rightarrow \\ $$$$−{p}−\mathrm{1}<−{x}<−{p}\:\Rightarrow\left[−{x}\right]=−{p}−\mathrm{1}=−\mathrm{1}−\left[{x}\right]\:\Rightarrow \\ $$$$\left[−{x}\right]^{\mathrm{2}} \:=\left(\mathrm{1}+\left[{x}\right]\right)^{\mathrm{2}} \:=\mathrm{1}+\mathrm{2}\left[{x}\right]\:+\left[{x}\right]^{\mathrm{2}} \:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 09/Dec/18
