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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
$$\mathrm{let}\:\:\mathrm{n}\in\mathbb{N}\:\:\mathrm{fixed}\:,\:\mathrm{solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers} \\ $$$$\left[\mathrm{x}\right]\left\{\mathrm{x}\right\}=\mathrm{nx} \\ $$
Commented by mr W last updated on 21/Dec/21
two solutions:  x=−(1/(n+1)), 0
$${two}\:{solutions}: \\ $$$${x}=−\frac{\mathrm{1}}{{n}+\mathrm{1}},\:\mathrm{0} \\ $$
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
$${x}=\left[{x}\right]+\left\{{x}\right\} \\ $$$${let}\:\left[{x}\right]={a},\left\{{x}\right\}={b} \\ $$$${ab}={n}\left({a}+{b}\right) \\ $$$${b}\left({a}−{n}\right)={na} \\ $$$${b}=\frac{{na}}{{a}−{n}}<\mathrm{1} \\ $$$${na}<{a}−{n} \\ $$$${a}\left({n}−\mathrm{1}\right)<{n} \\ $$$${a}<\frac{{n}}{{n}−\mathrm{1}}<\mathrm{2} \\ $$$${a}=\mathrm{0}\:\rightarrow\:{b}=\mathrm{0} \\ $$$${a}=\mathrm{1}\:\rightarrow\:{b}=\frac{{n}}{\mathrm{1}−{n}}<\mathrm{1}\:\nexists{n} \\ $$$$ \\ $$
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 ✓
$${say}\:{x}={k}+{f}\:{with}\:\mathrm{0}\leqslant{f}<\mathrm{1},\:{k}\:{is}\:{integer}. \\ $$$$\left[{x}\right]={k},\:\left\{{x}\right\}={f} \\ $$$${kf}={n}\left({k}+{f}\right) \\ $$$${f}=\frac{{nk}}{{k}−{n}} \\ $$$$\mathrm{0}\leqslant\frac{{nk}}{{k}−{n}}<\mathrm{1} \\ $$$${case}\:\mathrm{1}:\:\:{k}>{n}>\mathrm{0} \\ $$$${nk}<{k}−{n} \\ $$$${k}<−\frac{{n}}{{n}−\mathrm{1}}<\mathrm{0}\:\Rightarrow{contradiction} \\ $$$${case}\:\mathrm{2}:\:{k}<{n} \\ $$$${k}\leqslant\mathrm{0} \\ $$$$\frac{{nk}}{{n}−{k}}>−\mathrm{1} \\ $$$${nk}>−{n}+{k} \\ $$$${k}>−\frac{{n}}{{n}−\mathrm{1}}=−\mathrm{1}−\frac{\mathrm{1}}{{n}−\mathrm{1}} \\ $$$$−\mathrm{1}−\frac{\mathrm{1}}{{n}−\mathrm{1}}<{k}\leqslant\mathrm{0} \\ $$$$\Rightarrow{k}=−\mathrm{1},\:\mathrm{0} \\ $$$${with}\:{k}=−\mathrm{1}: \\ $$$${f}=−\frac{{n}}{−\mathrm{1}−{n}}=\frac{{n}}{{n}+\mathrm{1}} \\ $$$$\Rightarrow{x}=−\mathrm{1}+\frac{{n}}{{n}+\mathrm{1}}=−\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\checkmark \\ $$$${with}\:{k}=\mathrm{0}: \\ $$$${f}=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{0}\:\checkmark \\ $$
Commented by HongKing last updated on 21/Dec/21
cool my dear Sir thank you so much
$$\mathrm{cool}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$

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