Question Number 24944 by soufiane zarik last updated on 29/Nov/17
Commented by prakash jain last updated on 30/Nov/17
![x=n+f 0≤f<1, n∈Z,n≥0 [(n+f)^2 ]=[n+f]^2 ⇒[n^2 +2nf+f^2 ]=n^2 ⇒n^2 +[2nf+f^2 ]=n^2 [2nf+f^2 ]=0 0<2nf+f^2 <1 f^2 +2nf−1<0∧ 2nf+f^2 >0 the expression is less than 0 between the roots ((−2n±(√(4n^2 +4)))/2)=−n±(√(n^2 +1)) x=[k,(√(k^2 +1)))for k≥0 x=−n+f, f>0,n>0 [n^2 −2nf+f^2 ]=n^2 [f^2 −2nf]=0 0≤f^2 −2nf<1⇒f=0 x=k,k∈Z^− x=[k,(√(k^2 +1))],k≥0 x=k,k<0](Q24952.png)
$${x}={n}+{f}\:\:\mathrm{0}\leqslant{f}<\mathrm{1},\:{n}\in\mathbb{Z},{n}\geqslant\mathrm{0} \\ $$$$\left[\left({n}+{f}\right)^{\mathrm{2}} \right]=\left[{n}+{f}\right]^{\mathrm{2}} \\ $$$$\Rightarrow\left[{n}^{\mathrm{2}} +\mathrm{2}{nf}+{f}^{\mathrm{2}} \right]={n}^{\mathrm{2}} \\ $$$$\Rightarrow{n}^{\mathrm{2}} +\left[\mathrm{2}{nf}+{f}^{\mathrm{2}} \right]={n}^{\mathrm{2}} \\ $$$$\left[\mathrm{2}{nf}+{f}^{\mathrm{2}} \right]=\mathrm{0} \\ $$$$\mathrm{0}<\mathrm{2}{nf}+{f}^{\mathrm{2}} <\mathrm{1} \\ $$$${f}^{\mathrm{2}} +\mathrm{2}{nf}−\mathrm{1}<\mathrm{0}\wedge\:\mathrm{2}{nf}+{f}^{\mathrm{2}} >\mathrm{0} \\ $$$${the}\:{expression}\:{is}\:{less}\:{than}\:\mathrm{0} \\ $$$${between}\:{the}\:{roots} \\ $$$$\frac{−\mathrm{2}{n}\pm\sqrt{\mathrm{4}{n}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}=−{n}\pm\sqrt{{n}^{\mathrm{2}} +\mathrm{1}} \\ $$$${x}=\left[{k},\sqrt{{k}^{\mathrm{2}} +\mathrm{1}}\right){for}\:{k}\geqslant\mathrm{0} \\ $$$${x}=−{n}+{f},\:{f}>\mathrm{0},{n}>\mathrm{0} \\ $$$$\left[{n}^{\mathrm{2}} −\mathrm{2}{nf}+{f}^{\mathrm{2}} \right]={n}^{\mathrm{2}} \\ $$$$\left[{f}^{\mathrm{2}} −\mathrm{2}{nf}\right]=\mathrm{0} \\ $$$$\mathrm{0}\leqslant{f}^{\mathrm{2}} −\mathrm{2}{nf}<\mathrm{1}\Rightarrow{f}=\mathrm{0} \\ $$$${x}={k},{k}\in\mathbb{Z}^{−} \\ $$$${x}=\left[{k},\sqrt{{k}^{\mathrm{2}} +\mathrm{1}}\right],{k}\geqslant\mathrm{0} \\ $$$${x}={k},{k}<\mathrm{0} \\ $$
Commented by soufiane zarik last updated on 29/Nov/17
$$\mathrm{thanks}\:\mathrm{alot}\:\mathrm{sir}\:! \\ $$
Commented by prakash jain last updated on 30/Nov/17
$$\mathrm{I}\:\mathrm{have}\:\mathrm{updated}\:\mathrm{the}\:\mathrm{solution}.\:\mathrm{There} \\ $$$$\mathrm{were}\:\mathrm{few}\:\mathrm{mistakes}\:\mathrm{earlier}. \\ $$
Commented by soufiane zarik last updated on 30/Nov/17
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{sir}\:!! \\ $$