Question Number 161704 by HongKing last updated on 21/Dec/21
![let n∈N fixed , solve for real numbers [x]{x}=nx](https://www.tinkutara.com/question/Q161704.png)
Commented by mr W last updated on 21/Dec/21

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](https://www.tinkutara.com/question/Q161713.png)
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 ✓](https://www.tinkutara.com/question/Q161717.png)
Commented by HongKing last updated on 21/Dec/21
