Question Number 85003 by M±th+et£s last updated on 18/Mar/20
![x≤[x]<x+1 is that right if (x) was negative](Q85003.png)
$${x}\leqslant\left[{x}\right]<{x}+\mathrm{1} \\ $$ $${is}\:{that}\:{right}\:{if}\:\left({x}\right)\:{was}\:{negative} \\ $$
Commented byM±th+et£s last updated on 18/Mar/20
$${please}\:{help}\:{me} \\ $$
Commented byM±th+et£s last updated on 18/Mar/20
![and this [x]= { ((−x x∈z)),((−[x]−1 x∉z)) :}](Q85010.png)
$${and}\:{this} \\ $$ $$\left[{x}\right]=\begin{cases}{−{x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\in{z}}\\{−\left[{x}\right]−\mathrm{1}\:\:\:\:\:\:\:\:{x}\notin{z}}\end{cases} \\ $$
Commented byMJS last updated on 18/Mar/20
![[π]=3 [−π]=−4 let x=i+f; i∈Z∧0≤f<1 [x]=[i+f]=i x=π ⇒ i=3∧f=.141592... x=−π ⇒ i=−4∧f=.858407... that′s what I learned in school back in ≈1980](Q85011.png)
$$\left[\pi\right]=\mathrm{3} \\ $$ $$\left[−\pi\right]=−\mathrm{4} \\ $$ $$\mathrm{let}\:{x}={i}+{f};\:{i}\in\mathbb{Z}\wedge\mathrm{0}\leqslant{f}<\mathrm{1} \\ $$ $$\left[{x}\right]=\left[{i}+{f}\right]={i} \\ $$ $${x}=\pi\:\Rightarrow\:{i}=\mathrm{3}\wedge{f}=.\mathrm{141592}... \\ $$ $${x}=−\pi\:\Rightarrow\:{i}=−\mathrm{4}\wedge{f}=.\mathrm{858407}... \\ $$ $$\mathrm{that}'\mathrm{s}\:\mathrm{what}\:\mathrm{I}\:\mathrm{learned}\:\mathrm{in}\:\mathrm{school}\:\mathrm{back}\:\mathrm{in}\:\approx\mathrm{1980} \\ $$
Commented byM±th+et£s last updated on 18/Mar/20
$${thank}\:{you}\:{sir}.\:{but}\:{i}\:{want}\:{to}\:{now}\:{if}\:{the}\:{tow}\:{rules} \\ $$ $${that}\:{i}\:{post}\:{is}\:{right}\:{or}\:{no} \\ $$
Commented byMJS last updated on 18/Mar/20
![[x]= { ((−x x∈z)),((−[x]−1 x∉z)) :} makes no sense. you cannot define a function using the function](Q85013.png)
$$\left[{x}\right]=\begin{cases}{−{x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\in{z}}\\{−\left[{x}\right]−\mathrm{1}\:\:\:\:\:\:\:\:{x}\notin{z}}\end{cases}\:\mathrm{makes}\:\mathrm{no}\:\mathrm{sense}.\:\mathrm{you} \\ $$ $$\mathrm{cannot}\:\mathrm{define}\:\mathrm{a}\:\mathrm{function}\:\mathrm{using}\:\mathrm{the}\:\mathrm{function} \\ $$
Commented byM±th+et£s last updated on 18/Mar/20
![sorry sir thereis a typo i mean [−x]](Q85016.png)
$${sorry}\:{sir}\:{thereis}\:{a}\:{typo}\:{i}\:{mean}\:\left[−{x}\right]\: \\ $$
Commented byMJS last updated on 18/Mar/20
![[−x]= { ((−x; x∈Z)),((−[x]−1; x∉Z)) :} ⇒ [−3]=−3; [−π]=−[π]−1=−3−1=−4 is true](Q85018.png)
$$\left[−{x}\right]=\begin{cases}{−{x};\:{x}\in\mathbb{Z}}\\{−\left[{x}\right]−\mathrm{1};\:{x}\notin\mathbb{Z}}\end{cases} \\ $$ $$\Rightarrow\:\left[−\mathrm{3}\right]=−\mathrm{3};\:\left[−\pi\right]=−\left[\pi\right]−\mathrm{1}=−\mathrm{3}−\mathrm{1}=−\mathrm{4} \\ $$ $$\mathrm{is}\:\mathrm{true} \\ $$
Answered by MJS last updated on 18/Mar/20
![x≤[x]<x+1 is only true for x∈Z 3≤[3]<3+1 ⇔ 3≤3<4 −3≤[−3]<−3+1 ⇔ −3≤−3<−2 π≤[π]<π+1 ⇔ 3.14...≤3<4.14... −π≤[−π]<−π+1 ⇔ −3.14...≤−4<−2.14...](Q85017.png)
$${x}\leqslant\left[{x}\right]<{x}+\mathrm{1}\:\mathrm{is}\:\mathrm{only}\:\mathrm{true}\:\mathrm{for}\:{x}\in\mathbb{Z} \\ $$ $$\mathrm{3}\leqslant\left[\mathrm{3}\right]<\mathrm{3}+\mathrm{1}\:\Leftrightarrow\:\mathrm{3}\leqslant\mathrm{3}<\mathrm{4} \\ $$ $$−\mathrm{3}\leqslant\left[−\mathrm{3}\right]<−\mathrm{3}+\mathrm{1}\:\Leftrightarrow\:−\mathrm{3}\leqslant−\mathrm{3}<−\mathrm{2} \\ $$ $$ \\ $$ $$\pi\leqslant\left[\pi\right]<\pi+\mathrm{1}\:\Leftrightarrow\:\mathrm{3}.\mathrm{14}...\leqslant\mathrm{3}<\mathrm{4}.\mathrm{14}... \\ $$ $$−\pi\leqslant\left[−\pi\right]<−\pi+\mathrm{1}\:\Leftrightarrow\:−\mathrm{3}.\mathrm{14}...\leqslant−\mathrm{4}<−\mathrm{2}.\mathrm{14}... \\ $$
Commented byM±th+et£s last updated on 18/Mar/20
$${thank}\:{you}\:{so}\:{much}\:{sir}\:{mjs}.\:{i}\:{learn}\:{a}\:{lot} \\ $$ $${from}\:{you} \\ $$ $$ \\ $$