Question Number 161704 by HongKing last updated on 21/Dec/21
![let n∈N fixed , solve for real numbers [x]{x}=nx](Q161704.png)
$$\mathrm{let}\:\:\mathrm{n}\in\mathbb{N}\:\:\mathrm{fixed}\:,\:\mathrm{solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers} \\ $$$$\left[\mathrm{x}\right]\left\{\mathrm{x}\right\}=\mathrm{nx} \\ $$
Commented by mr W last updated on 21/Dec/21
$${two}\:{solutions}: \\ $$$${x}=−\frac{\mathrm{1}}{{n}+\mathrm{1}},\:\mathrm{0} \\ $$
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](Q161713.png)
$${x}=\left[{x}\right]+\left\{{x}\right\} \\ $$$${let}\:\left[{x}\right]={a},\left\{{x}\right\}={b} \\ $$$${ab}={n}\left({a}+{b}\right) \\ $$$${b}\left({a}−{n}\right)={na} \\ $$$${b}=\frac{{na}}{{a}−{n}}<\mathrm{1} \\ $$$${na}<{a}−{n} \\ $$$${a}\left({n}−\mathrm{1}\right)<{n} \\ $$$${a}<\frac{{n}}{{n}−\mathrm{1}}<\mathrm{2} \\ $$$${a}=\mathrm{0}\:\rightarrow\:{b}=\mathrm{0} \\ $$$${a}=\mathrm{1}\:\rightarrow\:{b}=\frac{{n}}{\mathrm{1}−{n}}<\mathrm{1}\:\nexists{n} \\ $$$$ \\ $$
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 ✓](Q161717.png)
$${say}\:{x}={k}+{f}\:{with}\:\mathrm{0}\leqslant{f}<\mathrm{1},\:{k}\:{is}\:{integer}. \\ $$$$\left[{x}\right]={k},\:\left\{{x}\right\}={f} \\ $$$${kf}={n}\left({k}+{f}\right) \\ $$$${f}=\frac{{nk}}{{k}−{n}} \\ $$$$\mathrm{0}\leqslant\frac{{nk}}{{k}−{n}}<\mathrm{1} \\ $$$${case}\:\mathrm{1}:\:\:{k}>{n}>\mathrm{0} \\ $$$${nk}<{k}−{n} \\ $$$${k}<−\frac{{n}}{{n}−\mathrm{1}}<\mathrm{0}\:\Rightarrow{contradiction} \\ $$$${case}\:\mathrm{2}:\:{k}<{n} \\ $$$${k}\leqslant\mathrm{0} \\ $$$$\frac{{nk}}{{n}−{k}}>−\mathrm{1} \\ $$$${nk}>−{n}+{k} \\ $$$${k}>−\frac{{n}}{{n}−\mathrm{1}}=−\mathrm{1}−\frac{\mathrm{1}}{{n}−\mathrm{1}} \\ $$$$−\mathrm{1}−\frac{\mathrm{1}}{{n}−\mathrm{1}}<{k}\leqslant\mathrm{0} \\ $$$$\Rightarrow{k}=−\mathrm{1},\:\mathrm{0} \\ $$$${with}\:{k}=−\mathrm{1}: \\ $$$${f}=−\frac{{n}}{−\mathrm{1}−{n}}=\frac{{n}}{{n}+\mathrm{1}} \\ $$$$\Rightarrow{x}=−\mathrm{1}+\frac{{n}}{{n}+\mathrm{1}}=−\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\checkmark \\ $$$${with}\:{k}=\mathrm{0}: \\ $$$${f}=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{0}\:\checkmark \\ $$
Commented by HongKing last updated on 21/Dec/21
$$\mathrm{cool}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$