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Question Number 24944 by soufiane zarik last updated on 29/Nov/17

Commented by prakash jain last updated on 30/Nov/17

$$x=n+f0⩽f<1,n\in \mathbb{Z},n⩾0$$$$\left[{(n+f)}^2\right]={[n+f]}^2$$$$\Rightarrow [n^2+2nf+f^2]=n^2$$$$\Rightarrow n^2+[2nf+f^2]=n^2$$$$[2nf+f^2]=0$$$$0<2nf+f^2<1$$$$f^2+2nf-1<0\wedge 2nf+f^2>0$$$$theexpressionislessthan0$$$$betweentheroots$$$$\frac{-2n\pm \sqrt{4n^2+4}}{2}=-n\pm \sqrt{n^2+1}$$$$x=[k,\sqrt{k^2+1})fork⩾0$$$$x=-n+f,f>0,n>0$$$$[n^2-2nf+f^2]=n^2$$$$[f^2-2nf]=0$$$$0⩽f^2-2nf<1\Rightarrow f=0$$$$x=k,k\in {\mathbb{Z}}^{-}$$$$x=[k,\sqrt{k^2+1}],k⩾0$$$$x=k,k<0$$

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

$$\mathrm{thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{sir}!!$$

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