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x-x-lt-x-1-is-that-right-if-x-was-negative-




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

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