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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]={xxz[x]1xz

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;iZ0f<1[x]=[i+f]=ix=πi=3f=.141592...x=πi=4f=.858407...thatswhatIlearnedinschoolbackin1980

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]={xxz[x]1xzmakesnosense.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;xZ[x]1;xZ[3]=3;[π]=[π]1=31=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+1isonlytrueforxZ3[3]<3+133<43[3]<3+133<2π[π]<π+13.14...3<4.14...π[π]<π+13.14...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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