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Question Number 33733 by rahul 19 last updated on 22/Apr/18
![Solve : (x−2) × [x] = {x} −1 . • [.]= greatest integer function • {.}= fractional part function.](Q33733.png)
$${Solve}\:: \\ $$$$\left({x}−\mathrm{2}\right)\:×\:\left[{x}\right]\:=\:\left\{{x}\right\}\:−\mathrm{1}\:. \\ $$$$\bullet\:\left[.\right]=\:{greatest}\:{integer}\:{function} \\ $$$$\bullet\:\left\{.\right\}=\:{fractional}\:{part}\:\:{function}. \\ $$
Commented by MJS last updated on 23/Apr/18
![x=(n/(n−1)) ∧ n=[x] ⇒ x=2 so something went wrong... ...you forgot the “−1” on the right side (x −2)[x]= x −[x]−1 let put [x]=n (x−2)n−x=n−1 (n−1)x=n−1 x=1∨n=1 n=[x]=1 ⇒ x∈[1;2[](Q33750.png)
$${x}=\frac{{n}}{{n}−\mathrm{1}}\:\wedge\:{n}=\left[{x}\right]\:\Rightarrow\:{x}=\mathrm{2} \\ $$$$\mathrm{so}\:\mathrm{something}\:\mathrm{went}\:\mathrm{wrong}... \\ $$$$ \\ $$$$ \\ $$$$...\mathrm{you}\:\mathrm{forgot}\:\mathrm{the}\:``−\mathrm{1}''\:\mathrm{on}\:\mathrm{the}\:\mathrm{right}\:\mathrm{side} \\ $$$$\left({x}\:−\mathrm{2}\right)\left[{x}\right]=\:{x}\:−\left[{x}\right]−\mathrm{1}\:\mathrm{let}\:\mathrm{put}\:\left[{x}\right]={n} \\ $$$$\left({x}−\mathrm{2}\right){n}−{x}={n}−\mathrm{1} \\ $$$$\left({n}−\mathrm{1}\right){x}={n}−\mathrm{1} \\ $$$${x}=\mathrm{1}\vee{n}=\mathrm{1} \\ $$$${n}=\left[{x}\right]=\mathrm{1}\:\Rightarrow\:{x}\in\left[\mathrm{1};\mathrm{2}\left[\right.\right. \\ $$
Commented by caravan msup abdo. last updated on 24/Apr/18
$${yes}\:{yes}\:{i}\:{have}\:{forgotten}\:−\mathrm{1}\:{i}\:{will}\:{delete} \\ $$$${this}\:{solution}\:{and}\:{take}\:{another}\:{try}... \\ $$
Answered by MJS last updated on 23/Apr/18
![0≤{x}<1 ⇒ −1≤{x}−1<0 ⇒ ⇒ −1≤(x−2)[x]<0 x∈[z;z+1[ ⇒ ⇒ (x−2)[x]=(x−2)z∈[z^2 −2z;z^2 −1[ z^2 −2z=−1∧z^2 −1=0 ⇒ z=1 ⇒ we could also try z^2 −2z=0∧z^2 −1=−1 but in this case −1<(x−2)z≤0 but −1≤{x}−1<0 ⇒ 1≤x<2](Q33740.png)
$$\mathrm{0}\leqslant\left\{{x}\right\}<\mathrm{1}\:\Rightarrow\:−\mathrm{1}\leqslant\left\{{x}\right\}−\mathrm{1}<\mathrm{0}\:\Rightarrow \\ $$$$\Rightarrow\:−\mathrm{1}\leqslant\left({x}−\mathrm{2}\right)\left[{x}\right]<\mathrm{0} \\ $$$${x}\in\left[{z};{z}+\mathrm{1}\left[\:\Rightarrow\right.\right. \\ $$$$\Rightarrow\:\left({x}−\mathrm{2}\right)\left[{x}\right]=\left({x}−\mathrm{2}\right){z}\in\left[{z}^{\mathrm{2}} −\mathrm{2}{z};{z}^{\mathrm{2}} −\mathrm{1}\left[\right.\right. \\ $$$${z}^{\mathrm{2}} −\mathrm{2}{z}=−\mathrm{1}\wedge{z}^{\mathrm{2}} −\mathrm{1}=\mathrm{0}\:\Rightarrow\:{z}=\mathrm{1}\:\Rightarrow \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{we}\:\mathrm{could}\:\mathrm{also}\:\mathrm{try}\:{z}^{\mathrm{2}} −\mathrm{2}{z}=\mathrm{0}\wedge{z}^{\mathrm{2}} −\mathrm{1}=−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{but}\:\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:−\mathrm{1}<\left({x}−\mathrm{2}\right){z}\leqslant\mathrm{0}\:\mathrm{but} \\ $$$$\:\:\:\:\:\:\:\:\:\:−\mathrm{1}\leqslant\left\{{x}\right\}−\mathrm{1}<\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{1}\leqslant{x}<\mathrm{2} \\ $$
Commented by rahul 19 last updated on 23/Apr/18
$${pls}\:{explain}\:{the}\:{last}\:{step}. \\ $$
Commented by MJS last updated on 23/Apr/18
$$\mathrm{changed}\:\mathrm{my}\:\mathrm{post} \\ $$