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let-n-N-fixed-solve-for-real-numbers-x-x-nx-




Question Number 161704 by HongKing last updated on 21/Dec/21
let  n∈N  fixed , solve for real numbers  [x]{x}=nx
letnNfixed,solveforrealnumbers[x]{x}=nx
Commented by mr W last updated on 21/Dec/21
two solutions:  x=−(1/(n+1)), 0
twosolutions:x=1n+1,0
Answered by TheSupreme last updated on 21/Dec/21
x=[x]+{x}  let [x]=a,{x}=b  ab=n(a+b)  b(a−n)=na  b=((na)/(a−n))<1  na<a−n  a(n−1)<n  a<(n/(n−1))<2  a=0 → b=0  a=1 → b=(n/(1−n))<1 ∄n
x=[x]+{x}let[x]=a,{x}=bab=n(a+b)b(an)=nab=naan<1na<ana(n1)<na<nn1<2a=0b=0a=1b=n1n<1n
Answered by mr W last updated on 21/Dec/21
say x=k+f with 0≤f<1, k is integer.  [x]=k, {x}=f  kf=n(k+f)  f=((nk)/(k−n))  0≤((nk)/(k−n))<1  case 1:  k>n>0  nk<k−n  k<−(n/(n−1))<0 ⇒contradiction  case 2: k<n  k≤0  ((nk)/(n−k))>−1  nk>−n+k  k>−(n/(n−1))=−1−(1/(n−1))  −1−(1/(n−1))<k≤0  ⇒k=−1, 0  with k=−1:  f=−(n/(−1−n))=(n/(n+1))  ⇒x=−1+(n/(n+1))=−(1/(n+1)) ✓  with k=0:  f=0  ⇒x=0 ✓
sayx=k+fwith0f<1,kisinteger.[x]=k,{x}=fkf=n(k+f)f=nkkn0nkkn<1case1:k>n>0nk<knk<nn1<0contradictioncase2:k<nk0nknk>1nk>n+kk>nn1=11n111n1<k0k=1,0withk=1:f=n1n=nn+1x=1+nn+1=1n+1withk=0:f=0x=0
Commented by HongKing last updated on 21/Dec/21
cool my dear Sir thank you so much
coolmydearSirthankyousomuch

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