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Question Number 191926 by Rupesh123 last updated on 04/May/23

	Answered by deleteduser1 last updated on 04/May/23

	$[x] + {x} = x$ $[x] . [y] = [x] + [y] + {x} + {y} < [x] + [y] + 2$ $\Rightarrow C a s e I : [x] . [y] = [x] + [y] o r I I : [x] . [y] = [x] + [y] + 1$ $I \Rightarrow [x] ([y] - 1) = [y] \Rightarrow [x] = \frac{[y]}{[y] - 1} = 1 + \frac{1}{[y] - 1}$ $\Rightarrow ([y] - 1) ∣ 1 \Rightarrow [y] - 1 = 1 o r [y] - 1 = - 1$ $I f [y] - 1 = 1, t h e n [y] = 2 a n d [x] = 2$ $\Rightarrow 4 = x + y = [x] + [y] + {x} + {y} \Rightarrow {x} + {y} = 0$ $\Rightarrow y = x = [y] + {y} = [x] + {x} = 2$ $\Rightarrow y = x = 2 i s a s o l u t i o n$ $I f [y] - 1 = - 1 \Rightarrow [y] = 0; [x] = 0$ $\Rightarrow 0 = x + y = [x] + [y] + {x} + {y} \Rightarrow {x} + {y} = 0$ $\Rightarrow y = x = 0 i s a s o l u t i o n$ $I I \Rightarrow [x] . [y] = [x] + [y] + 1 \Rightarrow [x] = \frac{[y] + 1}{[y] - 1}$ $\Rightarrow ([y] - 1) ∣ [y] + 1 \Rightarrow ([y] - 1) ∣ [y] - 1 + 2 \Rightarrow ([y] - 1) ∣ 2$ $\Rightarrow [y] - 1 = 1, - 1, 2, - 2$ $[y] - 1 = 1 \Rightarrow [y] = 2 \Rightarrow [x] = 3$ $\Rightarrow 6 = x + y = [x] + [y] + {x} + {y}$ $\Rightarrow {x} + {y} = 1$ $U p t o s y m m e t r y \Rightarrow x = 3 + {x}, y = 2 + {y} s u c h$ $t h a t {x} + {y} = 1 a r e s o l u t i o n s .$ $\Rightarrow T h e r e a r e i n f i n i t e l y m a n y s o l u t i o n s t o t h e$ $e q u a t i o n .$
	Commented by Rupesh123 last updated on 04/May/23
	Perfect ��
	Answered by mehdee42 last updated on 04/May/23
	$suppose f(x)=[x][y]−x−y if x,y∈Z ⇒xy=x+y⇒(x,y)=(0,0)&(2,2) if x=[x]+α ; 0<α<1 & y=[y]+β ; 0<β<1 ⇒[x][y]=[x]+[y]+α+β ; 0<α+β<2 ⇒^(α+β=1) [y]=(([x]+1)/([x]−1)) ∈Z ⇒[x]−1≠0⇒1≰x≮2 & [x]=3⇒3<x<4 & [x]=2⇒2≤x<3 & [x]=0⇒0≤x<1 & [x]=−1⇒−1<x<0 ⇒D_f =(−1,1)∪[2,3)∪(3,4) therefor,the solution set of the given equation is in the domain of function f the function rule is as follows if −1<x <1 → f(x)=−x if 2≤x <3 or 3<x<4 → f(x)=−x+6 the graph of the function is as follows sccorfing to the set diagram,the answer the equation is as follows {(x,y) ∣ −1<x<1 & −1<y<1 ; y=−x }∪{(x,y) ∣ 2<x<3 {(x,y) ∣ −1<x,y<1 ; y=−x}∪{(x,y)∣ 2≤x<3 & 3<y<4 ∨ 3<x<4 & 2≤y<3 ; y=−x+6}$
	$s u p p o s e f (x) = [x] [y] - x - y$ $i f x, y \in Z \Rightarrow x y = x + y \Rightarrow (x, y) = (0, 0) & (2, 2)$ $i f x = [x] + α; 0 < α < 1 & y = [y] + β; 0 < β < 1$ $\Rightarrow [x] [y] = [x] + [y] + α + β; 0 < α + β < 2 \overset{α + β = 1}{\Rightarrow} [y] = \frac{[x] + 1}{[x] - 1} \in Z$ $\Rightarrow [x] - 1 \neq 0 \Rightarrow 1 ⪇ x ≮ 2$ $& [x] = 3 \Rightarrow 3 < x < 4$ $& [x] = 2 \Rightarrow 2 ⩽ x < 3$ $& [x] = 0 \Rightarrow 0 ⩽ x < 1$ $& [x] = - 1 \Rightarrow - 1 < x < 0$ $\Rightarrow D_{f} = (- 1, 1) \cup [2, 3) \cup (3, 4)$ $t h e r e f o r, t h e s o l u t i o n s e t o f t h e g i v e n e q u a t i o n i s i n t h e$ $d o m a i n o f f u n c t i o n f$ $t h e f u n c t i o n r u l e i s a s f o l l o w s$ $i f - 1 < x < 1 \to f (x) = - x$ $i f 2 ⩽ x < 3 o r 3 < x < 4 \to f (x) = - x + 6$ $t h e g r a p h o f t h e f u n c t i o n i s a s f o l l o w s$ $s c c o r f i n g t o t h e s e t d i a g r a m, t h e$ $a n s w e r t h e e q u a t i o n i s a s f o l l o w s$ ${(x, y) ∣ - 1 < x < 1 & - 1 < y < 1; y = - x} \cup {(x, y) ∣ 2 < x < 3$ ${(x, y) ∣ - 1 < x, y < 1; y = - x} \cup {(x, y) ∣ 2 ⩽ x < 3 & 3 < y < 4 \lor 3 < x < 4 & 2 ⩽ y < 3; y = - x + 6}$
	Commented by mehdee42 last updated on 04/May/23

	Commented by Rupesh123 last updated on 04/May/23
	Perfect ��
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