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Question Number 156172 by MathSh last updated on 08/Oct/21

Commented by Rasheed.Sindhi last updated on 09/Oct/21

$Infinitely many sets .$
Commented by MathSh last updated on 09/Oct/21

$Thanks ser, but how$
Commented by Rasheed.Sindhi last updated on 09/Oct/21

$A r e y o u mathdanisur ?$
Commented by MathSh last updated on 09/Oct/21

$Yes dear Ser, I lost my other profile$
Commented by Rasheed.Sindhi last updated on 09/Oct/21

$Welcome with new identity ser!$
Commented by MathSh last updated on 09/Oct/21

$THANK YOU DEAR SER$
	Answered by Rasheed.Sindhi last updated on 09/Oct/21
	$x.y=xy−x−2y+3 x.y=x(y−1)−2(y−1)+1 =(x−2)(y−1)+1 Perticular cases: Let x∈A x.x=x^2 −x−2x+3=x^2 −3x+3∈A { ((x.(x^2 −3x+3)=x^3 −3x^2 +3x−x−2(x^2 −3x+3)+3_(=x^3 −5x^2 +8x+3∈A) )),(((x^2 −3x+3).x=x^3 −3x^2 +3x−(x^2 −3x+3)−2x+3_(=x^3 −4x^2 +4x∈A) )) :} In Same way..... For Example Let 0∈A Other members of A 0.0=0×0−0−2(0)+3=3 { ((0.3=0×3−0−2(3)+3=−3)),((3.0=3×0−3−2(0)+3=0)) :} { ((−3.0=−3×0−(−3)−2(0)+3=6)),((0.−3=0×−3−0−2(−3)+3=9)) :} { ((0.6=0−0−12+3=−9)),((6.0=0−6−0+3=−3)) :} { ((6.9=54−6−18+3=33)),((9.6=54−9−12+3=36)) :} Seems that A contains multiples of 3 Let 1,2∈A⇒1.2=2−1−4+3=0 Also 2.1=2−2−2+3=1∈A 1.0=0−1−0+3=2 0.1=0−0−2+3=1 ..... ... Fix some members of A:a,b,c,..(say) Calcuate all other possible members by the formula xy−x−2y+3 where x,y ∈ A.Keep to calculate members with new produced members. Recall that x.y≠y.x in general. In this way you can produce infinitely many sets.$
	$x . y = x y - x - 2 y + 3$ $x . y = x (y - 1) - 2 (y - 1) + 1$ $= (x - 2) (y - 1) + 1$ $P e r t i c u l a r c a s e s :$ $L e t x \in A$ $x . x = x^{2} - x - 2 x + 3 = x^{2} - 3 x + 3 \in A$ ${\begin{cases} x . (x^{2} - 3 x + 3) = \underset{= x^{3} - 5 x^{2} + 8 x + 3 \in A}{x^{3} - 3 x^{2} + 3 x - x - 2 (x^{2} - 3 x + 3) + 3} \\ (x^{2} - 3 x + 3) . x = \underset{= x^{3} - 4 x^{2} + 4 x \in A}{x^{3} - 3 x^{2} + 3 x - (x^{2} - 3 x + 3) - 2 x + 3} \end{cases}$ $I n S a m e w a y . . . . .$ $F o r E x a m p l e$ $L e t 0 \in A$ $O t h e r m e m b e r s o f A$ $0.0 = 0 \times 0 - 0 - 2 (0) + 3 = 3$ ${\begin{cases} 0.3 = 0 \times 3 - 0 - 2 (3) + 3 = - 3 \\ 3.0 = 3 \times 0 - 3 - 2 (0) + 3 = 0 \end{cases}$ ${\begin{cases} - 3.0 = - 3 \times 0 - (- 3) - 2 (0) + 3 = 6 \\ 0 . - 3 = 0 \times - 3 - 0 - 2 (- 3) + 3 = 9 \end{cases}$ ${\begin{cases} 0.6 = 0 - 0 - 12 + 3 = - 9 \\ 6.0 = 0 - 6 - 0 + 3 = - 3 \end{cases}$ ${\begin{cases} 6.9 = 54 - 6 - 18 + 3 = 33 \\ 9.6 = 54 - 9 - 12 + 3 = 36 \end{cases}$ $S e e m s t h a t A c o n t a i n s m u l t i p l e s$ $o f 3$ $L e t 1, 2 \in A \Rightarrow 1.2 = 2 - 1 - 4 + 3 = 0$ $A l s o 2.1 = 2 - 2 - 2 + 3 = 1 \in A$ $1.0 = 0 - 1 - 0 + 3 = 2$ $0.1 = 0 - 0 - 2 + 3 = 1$ $. . . . .$ $. . .$ $F i x s o m e m e m b e r s o f A : a, b, c, . . (s a y)$ $C a l c u a t e a l l o t h e r p o s s i b l e m e m b e r s$ $b y t h e f o r m u l a x y - x - 2 y + 3 w h e r e$ $x, y \in A . K e e p t o c a l c u l a t e m e m b e r s$ $w i t h n e w p r o d u c e d m e m b e r s .$ $R e c a l l t h a t x . y \neq y . x i n g e n e r a l .$ $I n t h i s w a y y o u c a n p r o d u c e i n f i n i t e l y$ $m a n y s e t s .$
	Commented by DavidSmath last updated on 09/Oct/21
	thanks for this sir.
	Commented by MathSh last updated on 09/Oct/21

	$Very nice Ser, cool thank you$
	Commented by Rasheed.Sindhi last updated on 09/Oct/21

	$T h a n X D a v i d S m a t h & M a t h S h!$
	Commented by Rasheed.Sindhi last updated on 09/Oct/21

	$Y o u s h o u l d c h a n g e y o u r q u e s t i o n$ $f o r^{'} a l l \underset{―}{f i n i t e} s e t s A \subset Z^{'}$
	Commented by Rasheed.Sindhi last updated on 09/Oct/21

	$T h e q u e s t i o n r e q u i r e s a l l s e t s A \subset Z$ $w h e t h e r t h e y a r e f i n i t e o r i n f i n i t e .$ $F o r f i n i t e s e t s t h e a n s w e r s e e m s$ $t r u e .$
	Commented by MathSh last updated on 09/Oct/21

	$S er, A = {1} or A = {3}$
	Commented by MathSh last updated on 09/Oct/21

	$Yes dear Ser, thankyou$
	Answered by Rasheed.Sindhi last updated on 10/Oct/21
	$x.y=xy−x−2y+3 Let A consist of a single member x According to the condition: ⇒x.x∈A⇒x.x=x x.x=x^2 −x−2x+3=x x^2 −4x+3=0 (x−3)(x−1)=0 x=3 ∨ x=1 A={1} or A={3} _(−) ^(−) ∣•∣ ^(−) _(−)$
	$x . y = xy - x - 2 y + 3$ $Let A consist of a single member x$ $According to the condition :$ $\Rightarrow x . x \in A \Rightarrow x . x = x$ $x . x = x^{2} - x - 2 x + 3 = x$ $x^{2} - 4 x + 3 = 0$ $(x - 3) (x - 1) = 0$ $x = 3 \lor x = 1$ $A = {1} or A = {3}$ $\underline{\overset{―}{}} ∣ ∙ ∣ \underline{\overset{―}{}}$
	Answered by Rasheed.Sindhi last updated on 10/Oct/21
	$x.y=xy−x−2y+3 Search for 2-membered A Let A consist of two members x & y According to the condition: x.x , y.y , x.y , y.x ∈ A For x,y: (x.y=x ∨ x.y=y).........(i) ∧ (y.x=x ∨ y.x=y).........(ii) (i): xy−x−2y+3=x_((ia)) ∨ xy−x−2y+3=y_((ib)) (ii):xy−y−2x+3=x_((iia)) ∨xy−y−2x+3=y_((iib)) (ia)&(iia) or (ib)&(iib) xy−x−2y+3=xy−y−2x+3 −x−2y=−y−2x x=y (ia)&(iib) xy−x−2y+3=x ∧ xy−y−2x+3=y xy=2x+2y−3 ∧ xy=2y+2x−3 (ib)&(iia) xy−x−2y+3=y ∧ xy−y−2x+3=x xy=x+3y−3 ∧ xy=3x+y−3 ⇒ x+3y−3=3x+y−3 x=y Hence conditions x.y∈A={x,y}∧ y.x∈A={x,y} only meets when x=y i-e A is a singleton consist of single member.$
	$x . y = xy - x - 2 y + 3$ $Search for 2 - membered A$ $Let A consist of two members x & y$ $According to the condition :$ $x . x, y . y, x . y, y . x \in A$ $For x, y :$ $(x . y = x \lor x . y = y) . . . . . . . . . (i)$ $\land$ $(y . x = x \lor y . x = y) . . . . . . . . . (ii)$ $(i) : \underset{(ia)}{\underset{⏟}{xy - x - 2 y + 3 = x}} \lor \underset{(ib)}{\underset{⏟}{xy - x - 2 y + 3 = y}}$ $(ii) : \underset{(iia)}{\underset{⏟}{xy - y - 2 x + 3 = x}} \lor \underset{(iib)}{\underset{⏟}{xy - y - 2 x + 3 = y}}$ $(ia) & (iia) or (ib) & (iib)$ $xy - x - 2 y + 3 = xy - y - 2 x + 3$ $- x - 2 y = - y - 2 x$ $x = y$ $(ia) & (iib)$ $xy - x - 2 y + 3 = x \land xy - y - 2 x + 3 = y$ $xy = 2 x + 2 y - 3 \land xy = 2 y + 2 x - 3$ $(ib) & (iia)$ $xy - x - 2 y + 3 = y \land xy - y - 2 x + 3 = x$ $xy = x + 3 y - 3 \land xy = 3 x + y - 3$ $\Rightarrow x + 3 y - 3 = 3 x + y - 3$ $x = y$ $Hence conditions$ $x . y \in A = {x, y} \land y . x \in A = {x, y}$ $only meets when x = y$ $i - e A is a singleton consist of single$ $member .$
	Commented by MathSh last updated on 11/Oct/21

	$Perfect my dear S er, thank you$
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