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

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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⩽[x]<x+1isthatrightif(x)wasnegative
Commented by M±th+et£s last updated on 18/Mar/20
please help me
pleasehelpme
Commented by M±th+et£s last updated on 18/Mar/20
and this [x]= { ((−x x∈z)),((−[x]−1 x∉z)) :}
andthis[x]={−xx∈z−[x]−1x∉z
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
[π]=3[−π]=−4letx=i+f;i∈Z∧0⩽f<1[x]=[i+f]=ix=π⇒i=3∧f=.141592…x=−π⇒i=−4∧f=.858407…that′swhatIlearnedinschoolbackin≈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
thankyousir.butiwanttonowifthetowrulesthatipostisrightorno
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
[x]={−xx∈z−[x]−1x∉zmakesnosense.youcannotdefineafunctionusingthefunction
Commented by M±th+et£s last updated on 18/Mar/20
sorry sir thereis a typo i mean [−x]
sorrysirthereisatypoimean[−x]
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
[−x]={−x;x∈Z−[x]−1;x∉Z⇒[−3]=−3;[−π]=−[π]−1=−3−1=−4istrue
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⩽[x]<x+1isonlytrueforx∈Z3⩽[3]<3+1⇔3⩽3<4−3⩽[−3]<−3+1⇔−3⩽−3<−2π⩽[π]<π+1⇔3.14…⩽3<4.14…−π⩽[−π]<−π+1⇔−3.14…⩽−4<−2.14…
Commented by M±th+et£s last updated on 18/Mar/20
thank you so much sir mjs. i learn a lot from you
thankyousomuchsirmjs.ilearnalotfromyou

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