Question Number 191926 by Rupesh123 last updated on 04/May/23
Answered by deleteduser1 last updated on 04/May/23
![[x]+{x}=x [x].[y]=[x]+[y]+{x}+{y}<[x]+[y]+2 ⇒Case I:[x].[y]=[x]+[y] or II:[x].[y]=[x]+[y]+1 I⇒[x]([y]−1)=[y]⇒[x]=(([y])/([y]−1))=1+(1/([y]−1)) ⇒([y]−1)∣1⇒[y]−1=1 or [y]−1=−1 If [y]−1=1,then[y]=2 and [x]=2 ⇒4=x+y=[x]+[y]+{x}+{y}⇒{x}+{y}=0 ⇒y=x=[y]+{y}=[x]+{x}=2 ⇒y=x=2 is a solution If [y]−1=−1⇒[y]=0;[x]=0 ⇒0=x+y=[x]+[y]+{x}+{y}⇒{x}+{y}=0 ⇒y=x=0 is a solution II⇒[x].[y]=[x]+[y]+1⇒[x]=(([y]+1)/([y]−1)) ⇒([y]−1)∣[y]+1⇒([y]−1)∣[y]−1+2⇒([y]−1)∣2 ⇒[y]−1=1,−1,2,−2 [y]−1=1⇒[y]=2⇒[x]=3 ⇒6=x+y=[x]+[y]+{x}+{y} ⇒{x}+{y}=1 Up to symmetry⇒x=3+{x},y=2+{y} such that {x}+{y}=1 are solutions. ⇒There are infinitely many solutions to the equation.](Q191929.png)
$$\left[{x}\right]+\left\{{x}\right\}={x} \\ $$$$\left[{x}\right].\left[{y}\right]=\left[{x}\right]+\left[{y}\right]+\left\{{x}\right\}+\left\{{y}\right\}<\left[{x}\right]+\left[{y}\right]+\mathrm{2} \\ $$$$\Rightarrow{Case}\:{I}:\left[{x}\right].\left[{y}\right]=\left[{x}\right]+\left[{y}\right]\:{or}\:{II}:\left[{x}\right].\left[{y}\right]=\left[{x}\right]+\left[{y}\right]+\mathrm{1} \\ $$$${I}\Rightarrow\left[{x}\right]\left(\left[{y}\right]−\mathrm{1}\right)=\left[{y}\right]\Rightarrow\left[{x}\right]=\frac{\left[{y}\right]}{\left[{y}\right]−\mathrm{1}}=\mathrm{1}+\frac{\mathrm{1}}{\left[{y}\right]−\mathrm{1}} \\ $$$$\Rightarrow\left(\left[{y}\right]−\mathrm{1}\right)\mid\mathrm{1}\Rightarrow\left[{y}\right]−\mathrm{1}=\mathrm{1}\:{or}\:\left[{y}\right]−\mathrm{1}=−\mathrm{1} \\ $$$$ \\ $$$${If}\:\left[{y}\right]−\mathrm{1}=\mathrm{1},{then}\left[{y}\right]=\mathrm{2}\:{and}\:\left[{x}\right]=\mathrm{2} \\ $$$$\Rightarrow\mathrm{4}={x}+{y}=\left[{x}\right]+\left[{y}\right]+\left\{{x}\right\}+\left\{{y}\right\}\Rightarrow\left\{{x}\right\}+\left\{{y}\right\}=\mathrm{0} \\ $$$$\Rightarrow{y}={x}=\left[{y}\right]+\left\{{y}\right\}=\left[{x}\right]+\left\{{x}\right\}=\mathrm{2} \\ $$$$\Rightarrow{y}={x}=\mathrm{2}\:{is}\:{a}\:{solution} \\ $$$${If}\:\left[{y}\right]−\mathrm{1}=−\mathrm{1}\Rightarrow\left[{y}\right]=\mathrm{0};\left[{x}\right]=\mathrm{0} \\ $$$$\Rightarrow\mathrm{0}={x}+{y}=\left[{x}\right]+\left[{y}\right]+\left\{{x}\right\}+\left\{{y}\right\}\Rightarrow\left\{{x}\right\}+\left\{{y}\right\}=\mathrm{0} \\ $$$$\Rightarrow{y}={x}=\mathrm{0}\:{is}\:{a}\:{solution} \\ $$$$ \\ $$$${II}\Rightarrow\left[{x}\right].\left[{y}\right]=\left[{x}\right]+\left[{y}\right]+\mathrm{1}\Rightarrow\left[{x}\right]=\frac{\left[{y}\right]+\mathrm{1}}{\left[{y}\right]−\mathrm{1}} \\ $$$$\Rightarrow\left(\left[{y}\right]−\mathrm{1}\right)\mid\left[{y}\right]+\mathrm{1}\Rightarrow\left(\left[{y}\right]−\mathrm{1}\right)\mid\left[{y}\right]−\mathrm{1}+\mathrm{2}\Rightarrow\left(\left[{y}\right]−\mathrm{1}\right)\mid\mathrm{2} \\ $$$$\Rightarrow\left[{y}\right]−\mathrm{1}=\mathrm{1},−\mathrm{1},\mathrm{2},−\mathrm{2} \\ $$$$\left[{y}\right]−\mathrm{1}=\mathrm{1}\Rightarrow\left[{y}\right]=\mathrm{2}\Rightarrow\left[{x}\right]=\mathrm{3} \\ $$$$\Rightarrow\mathrm{6}={x}+{y}=\left[{x}\right]+\left[{y}\right]+\left\{{x}\right\}+\left\{{y}\right\} \\ $$$$\Rightarrow\left\{{x}\right\}+\left\{{y}\right\}=\mathrm{1} \\ $$$${Up}\:{to}\:{symmetry}\Rightarrow{x}=\mathrm{3}+\left\{{x}\right\},{y}=\mathrm{2}+\left\{{y}\right\}\:{such} \\ $$$$\:{that}\:\left\{{x}\right\}+\left\{{y}\right\}=\mathrm{1}\:{are}\:{solutions}. \\ $$$$\Rightarrow{There}\:{are}\:{infinitely}\:{many}\:{solutions}\:{to}\:{the} \\ $$$${equation}. \\ $$
Commented by Rupesh123 last updated on 04/May/23
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Answered by mehdee42 last updated on 04/May/23
![suppose f(x)=[x][y]−x−y if x,y∈Z ⇒xy=x+y⇒(x,y)=(0,0)&(2,2) if x=[x]+α ; 0<α<1 & y=[y]+β ; 0<β<1 ⇒[x][y]=[x]+[y]+α+β ; 0<α+β<2 ⇒^(α+β=1) [y]=(([x]+1)/([x]−1)) ∈Z ⇒[x]−1≠0⇒1≰x≮2 & [x]=3⇒3<x<4 & [x]=2⇒2≤x<3 & [x]=0⇒0≤x<1 & [x]=−1⇒−1<x<0 ⇒D_f =(−1,1)∪[2,3)∪(3,4) therefor,the solution set of the given equation is in the domain of function f the function rule is as follows if −1<x <1 → f(x)=−x if 2≤x <3 or 3<x<4 → f(x)=−x+6 the graph of the function is as follows sccorfing to the set diagram,the answer the equation is as follows {(x,y) ∣ −1<x<1 & −1<y<1 ; y=−x }∪{(x,y) ∣ 2<x<3 {(x,y) ∣ −1<x,y<1 ; y=−x}∪{(x,y)∣ 2≤x<3 & 3<y<4 ∨ 3<x<4 & 2≤y<3 ; y=−x+6}](Q191952.png)
$${suppose}\:\:{f}\left({x}\right)=\left[{x}\right]\left[{y}\right]−{x}−{y} \\ $$$${if}\:\:{x},{y}\in\mathbb{Z}\:\Rightarrow{xy}={x}+{y}\Rightarrow\left({x},{y}\right)=\left(\mathrm{0},\mathrm{0}\right)\&\left(\mathrm{2},\mathrm{2}\right) \\ $$$$\:{if}\:\:{x}=\left[{x}\right]+\alpha\:\:\:;\:\mathrm{0}<\alpha<\mathrm{1}\:\:\:\&\:\:\:{y}=\left[{y}\right]+\beta\:\:;\:\:\mathrm{0}<\beta<\mathrm{1}\: \\ $$$$\Rightarrow\left[{x}\right]\left[{y}\right]=\left[{x}\right]+\left[{y}\right]+\alpha+\beta\:\:\:;\:\:\mathrm{0}<\alpha+\beta<\mathrm{2}\:\:\overset{\alpha+\beta=\mathrm{1}} {\Rightarrow}\:\:\left[{y}\right]=\frac{\left[{x}\right]+\mathrm{1}}{\left[{x}\right]−\mathrm{1}}\:\in\mathbb{Z} \\ $$$$\Rightarrow\left[{x}\right]−\mathrm{1}\neq\mathrm{0}\Rightarrow\mathrm{1}\nleqslant{x}\nless\mathrm{2}\:\:\: \\ $$$$\&\:\:\left[{x}\right]=\mathrm{3}\Rightarrow\mathrm{3}<{x}<\mathrm{4} \\ $$$$\&\:\:\left[{x}\right]=\mathrm{2}\Rightarrow\mathrm{2}\leqslant{x}<\mathrm{3} \\ $$$$\&\:\:\left[{x}\right]=\mathrm{0}\Rightarrow\mathrm{0}\leqslant{x}<\mathrm{1} \\ $$$$\&\:\:\left[{x}\right]=−\mathrm{1}\Rightarrow−\mathrm{1}<{x}<\mathrm{0} \\ $$$$\Rightarrow{D}_{{f}} =\left(−\mathrm{1},\mathrm{1}\right)\cup\left[\mathrm{2},\mathrm{3}\right)\cup\left(\mathrm{3},\mathrm{4}\right)\: \\ $$$${therefor},{the}\:{solution}\:{set}\:{of}\:{the}\:{given}\:{equation}\:{is}\:{in}\:{the}\: \\ $$$${domain}\:{of}\:{function}\:{f} \\ $$$${the}\:{function}\:{rule}\:\:{is}\:{as}\:{follows} \\ $$$${if}\:\:\:−\mathrm{1}<{x}\:<\mathrm{1}\:\rightarrow\:{f}\left({x}\right)=−{x} \\ $$$${if}\:\:\:\:\:\mathrm{2}\leqslant{x}\:<\mathrm{3}\:{or}\:\mathrm{3}<{x}<\mathrm{4}\:\rightarrow\:{f}\left({x}\right)=−{x}+\mathrm{6} \\ $$$${the}\:{graph}\:{of}\:\:{the}\:{function}\:\:{is}\:{as}\:{follows} \\ $$$${sccorfing}\:{to}\:{the}\:{set}\:{diagram},{the}\: \\ $$$${answer}\:{the}\:{equation}\:{is}\:{as}\:{follows} \\ $$$$\left\{\left({x},{y}\right)\:\mid\:−\mathrm{1}<{x}<\mathrm{1}\:\&\:−\mathrm{1}<{y}<\mathrm{1}\:;\:{y}=−{x}\:\right\}\cup\left\{\left({x},{y}\right)\:\mid\:\mathrm{2}<{x}<\mathrm{3}\right. \\ $$$$\left\{\left({x},{y}\right)\:\mid\:−\mathrm{1}<{x},{y}<\mathrm{1}\:;\:{y}=−{x}\right\}\cup\left\{\left({x},{y}\right)\mid\:\mathrm{2}\leqslant{x}<\mathrm{3}\:\&\:\mathrm{3}<{y}<\mathrm{4}\:\vee\:\mathrm{3}<{x}<\mathrm{4}\:\&\:\mathrm{2}\leqslant{y}<\mathrm{3}\:;\:{y}=−{x}+\mathrm{6}\right\} \\ $$
Commented by mehdee42 last updated on 04/May/23
Commented by Rupesh123 last updated on 04/May/23
Perfect ��