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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.

$$\mathrm{Infinitely}\:\mathrm{many}\:\mathrm{sets}. \\ $$

Commented by MathSh last updated on 09/Oct/21

Thanks ser, but how

$$\mathrm{Thanks}\:\mathrm{ser},\:\mathrm{but}\:\mathrm{how} \\ $$

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

Are you mathdanisur?

$${Are}\:{you}\:\mathrm{mathdanisur}? \\ $$

Commented by MathSh last updated on 09/Oct/21

Yes dear Ser, I lost my other profile

$$\mathrm{Yes}\:\mathrm{dear}\:\mathrm{Ser},\:\mathrm{I}\:\mathrm{lost}\:\mathrm{my}\:\mathrm{other}\:\mathrm{profile} \\ $$

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

Welcome with new identity ser!

$$\mathrm{Welcome}\:\mathrm{with}\:\mathrm{new}\:\mathrm{identity}\:\mathrm{ser}! \\ $$

Commented by MathSh last updated on 09/Oct/21

THANK YOU DEAR SER

$$\mathrm{THANK}\:\mathrm{YOU}\:\mathrm{DEAR}\:\mathrm{SER} \\ $$

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.

$$ \\ $$$${x}.{y}={xy}−{x}−\mathrm{2}{y}+\mathrm{3} \\ $$$${x}.{y}={x}\left({y}−\mathrm{1}\right)−\mathrm{2}\left({y}−\mathrm{1}\right)+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:=\left({x}−\mathrm{2}\right)\left({y}−\mathrm{1}\right)+\mathrm{1} \\ $$$${Perticular}\:{cases}: \\ $$$${Let}\:{x}\in\mathrm{A} \\ $$$${x}.{x}={x}^{\mathrm{2}} −{x}−\mathrm{2}{x}+\mathrm{3}={x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{3}\in\mathrm{A} \\ $$$$ \\ $$$$\begin{cases}{{x}.\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{3}\right)=\underset{={x}^{\mathrm{3}} −\mathrm{5}{x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{3}\in\mathrm{A}} {{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}−{x}−\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{3}\right)+\mathrm{3}}}\\{\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{3}\right).{x}=\underset{={x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}\in\mathrm{A}} {{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}−\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{3}\right)−\mathrm{2}{x}+\mathrm{3}}}\end{cases} \\ $$$${In}\:\mathrm{S}{ame}\:{way}..... \\ $$$${For}\:{Example} \\ $$$${Let}\:\mathrm{0}\in\mathrm{A} \\ $$$${Other}\:{members}\:{of}\:\mathrm{A} \\ $$$$\:\:\:\:\mathrm{0}.\mathrm{0}=\mathrm{0}×\mathrm{0}−\mathrm{0}−\mathrm{2}\left(\mathrm{0}\right)+\mathrm{3}=\mathrm{3} \\ $$$$\:\begin{cases}{\mathrm{0}.\mathrm{3}=\mathrm{0}×\mathrm{3}−\mathrm{0}−\mathrm{2}\left(\mathrm{3}\right)+\mathrm{3}=−\mathrm{3}}\\{\mathrm{3}.\mathrm{0}=\mathrm{3}×\mathrm{0}−\mathrm{3}−\mathrm{2}\left(\mathrm{0}\right)+\mathrm{3}=\mathrm{0}}\end{cases} \\ $$$$\begin{cases}{−\mathrm{3}.\mathrm{0}=−\mathrm{3}×\mathrm{0}−\left(−\mathrm{3}\right)−\mathrm{2}\left(\mathrm{0}\right)+\mathrm{3}=\mathrm{6}}\\{\mathrm{0}.−\mathrm{3}=\mathrm{0}×−\mathrm{3}−\mathrm{0}−\mathrm{2}\left(−\mathrm{3}\right)+\mathrm{3}=\mathrm{9}}\end{cases} \\ $$$$\begin{cases}{\mathrm{0}.\mathrm{6}=\mathrm{0}−\mathrm{0}−\mathrm{12}+\mathrm{3}=−\mathrm{9}}\\{\mathrm{6}.\mathrm{0}=\mathrm{0}−\mathrm{6}−\mathrm{0}+\mathrm{3}=−\mathrm{3}}\end{cases} \\ $$$$\begin{cases}{\mathrm{6}.\mathrm{9}=\mathrm{54}−\mathrm{6}−\mathrm{18}+\mathrm{3}=\mathrm{33}}\\{\mathrm{9}.\mathrm{6}=\mathrm{54}−\mathrm{9}−\mathrm{12}+\mathrm{3}=\mathrm{36}}\end{cases} \\ $$$$\:{Seems}\:{that}\:\mathrm{A}\:{contains}\:{multiples} \\ $$$${of}\:\mathrm{3} \\ $$$${Let}\:\mathrm{1},\mathrm{2}\in\mathrm{A}\Rightarrow\mathrm{1}.\mathrm{2}=\mathrm{2}−\mathrm{1}−\mathrm{4}+\mathrm{3}=\mathrm{0} \\ $$$${Also}\:\mathrm{2}.\mathrm{1}=\mathrm{2}−\mathrm{2}−\mathrm{2}+\mathrm{3}=\mathrm{1}\in\mathrm{A} \\ $$$$\mathrm{1}.\mathrm{0}=\mathrm{0}−\mathrm{1}−\mathrm{0}+\mathrm{3}=\mathrm{2} \\ $$$$\mathrm{0}.\mathrm{1}=\mathrm{0}−\mathrm{0}−\mathrm{2}+\mathrm{3}=\mathrm{1} \\ $$$$..... \\ $$$$... \\ $$$${Fix}\:{some}\:{members}\:{of}\:\mathrm{A}:{a},{b},{c},..\left({say}\right) \\ $$$${Calcuate}\:{all}\:{other}\:{possible}\:{members} \\ $$$${by}\:{the}\:{formula}\:{xy}−{x}−\mathrm{2}{y}+\mathrm{3}\:{where} \\ $$$${x},{y}\:\in\:\mathrm{A}.{Keep}\:{to}\:{calculate}\:{members} \\ $$$${with}\:{new}\:{produced}\:{members}. \\ $$$${Recall}\:{that}\:{x}.{y}\neq{y}.{x}\:{in}\:{general}. \\ $$$${In}\:{this}\:{way}\:{you}\:{can}\:{produce}\:{infinitely} \\ $$$${many}\:{sets}. \\ $$$$ \\ $$

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

$$\mathrm{Very}\:\mathrm{nice}\:\mathrm{Ser},\:\mathrm{cool}\:\mathrm{thank}\:\mathrm{you} \\ $$

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

ThanX DavidSmath & MathSh!

$$\mathcal{T}{han}\mathcal{X}\:{DavidSmath}\:\&\:{MathSh}! \\ $$

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

You should change your question for ′ all finite sets A⊂Z ′

$${You}\:{should}\:{change}\:{your}\:{question} \\ $$$${for}\:'\:\boldsymbol{{all}}\:\underline{\boldsymbol{{finite}}}\:\boldsymbol{{sets}}\:\mathrm{A}\subset\mathbb{Z}\:' \\ $$

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

The question requires all sets A⊂Z whether they are finite or infinite. For finite sets the answer seems true.

$${The}\:{question}\:\:{requires}\:\boldsymbol{{all}}\:\boldsymbol{{sets}}\:\mathrm{A}\subset\mathbb{Z} \\ $$$${whether}\:{they}\:{are}\:{finite}\:{or}\:{infinite}. \\ $$$${For}\:{finite}\:{sets}\:{the}\:{answer}\:{seems} \\ $$$${true}. \\ $$

Commented by MathSh last updated on 09/Oct/21

Ser, A={1} or A={3}

$$\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{A}=\left\{\mathrm{1}\right\}\:\:\mathrm{or}\:\:\mathrm{A}=\left\{\mathrm{3}\right\} \\ $$

Commented by MathSh last updated on 09/Oct/21

Yes dear Ser, thankyou

$$\mathrm{Yes}\:\mathrm{dear}\:\mathrm{Ser},\:\mathrm{thankyou} \\ $$

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} _(−) ^(−) ∣•∣ ^(−) _(−)

$$\mathrm{x}.\mathrm{y}=\mathrm{xy}−\mathrm{x}−\mathrm{2y}+\mathrm{3} \\ $$$$\mathrm{Let}\:\mathrm{A}\:\mathrm{consist}\:\mathrm{of}\:\mathrm{a}\:\mathrm{single}\:\mathrm{member}\:\mathrm{x} \\ $$$$\mathrm{According}\:\mathrm{to}\:\mathrm{the}\:\mathrm{condition}: \\ $$$$\Rightarrow\mathrm{x}.\mathrm{x}\in\mathrm{A}\Rightarrow\mathrm{x}.\mathrm{x}=\mathrm{x} \\ $$$$\mathrm{x}.\mathrm{x}=\mathrm{x}^{\mathrm{2}} −\mathrm{x}−\mathrm{2x}+\mathrm{3}=\mathrm{x} \\ $$$$\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{3}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\left(\mathrm{x}−\mathrm{3}\right)\left(\mathrm{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{x}=\mathrm{3}\:\vee\:\mathrm{x}=\mathrm{1} \\ $$$$\mathrm{A}=\left\{\mathrm{1}\right\}\:\mathrm{or}\:\mathrm{A}=\left\{\mathrm{3}\right\} \\ $$$$\underset{−} {\overline {\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}}\mid\bullet\mid\underset{−} {\overline {\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}}\:\: \\ $$

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.

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}.\mathrm{y}=\mathrm{xy}−\mathrm{x}−\mathrm{2y}+\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{Search}\:\mathrm{for}\:\mathrm{2}-\mathrm{membered}\:\mathrm{A} \\ $$$$\mathrm{Let}\:\mathrm{A}\:\mathrm{consist}\:\mathrm{of}\:\mathrm{two}\:\mathrm{members}\:\mathrm{x}\:\&\:\mathrm{y} \\ $$$$\mathrm{According}\:\mathrm{to}\:\mathrm{the}\:\mathrm{condition}: \\ $$$$\mathrm{x}.\mathrm{x}\:,\:\mathrm{y}.\mathrm{y}\:,\:\mathrm{x}.\mathrm{y}\:,\:\mathrm{y}.\mathrm{x}\:\in\:\mathrm{A} \\ $$$$\mathrm{For}\:\mathrm{x},\mathrm{y}: \\ $$$$\left(\mathrm{x}.\mathrm{y}=\mathrm{x}\:\vee\:\mathrm{x}.\mathrm{y}=\mathrm{y}\right).........\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\wedge \\ $$$$\left(\mathrm{y}.\mathrm{x}=\mathrm{x}\:\vee\:\mathrm{y}.\mathrm{x}=\mathrm{y}\right).........\left(\mathrm{ii}\right) \\ $$$$\left(\mathrm{i}\right):\:\underset{\left(\mathrm{ia}\right)} {\underbrace{\mathrm{xy}−\mathrm{x}−\mathrm{2y}+\mathrm{3}=\mathrm{x}}}\:\:\vee\:\underset{\left(\mathrm{ib}\right)} {\underbrace{\mathrm{xy}−\mathrm{x}−\mathrm{2y}+\mathrm{3}=\mathrm{y}}} \\ $$$$\left(\mathrm{ii}\right):\underset{\left(\mathrm{iia}\right)} {\underbrace{\mathrm{xy}−\mathrm{y}−\mathrm{2x}+\mathrm{3}=\mathrm{x}}}\vee\underset{\left(\mathrm{iib}\right)} {\underbrace{\mathrm{xy}−\mathrm{y}−\mathrm{2x}+\mathrm{3}=\mathrm{y}}} \\ $$$$\left(\mathrm{ia}\right)\&\left(\mathrm{iia}\right)\:\:\mathrm{or}\:\left(\mathrm{ib}\right)\&\left(\mathrm{iib}\right) \\ $$$$\mathrm{xy}−\mathrm{x}−\mathrm{2y}+\mathrm{3}=\mathrm{xy}−\mathrm{y}−\mathrm{2x}+\mathrm{3}\:\: \\ $$$$\:\:−\mathrm{x}−\mathrm{2y}=−\mathrm{y}−\mathrm{2x} \\ $$$$\:\:\mathrm{x}=\mathrm{y} \\ $$$$\left(\mathrm{ia}\right)\&\left(\mathrm{iib}\right) \\ $$$$\mathrm{xy}−\mathrm{x}−\mathrm{2y}+\mathrm{3}=\mathrm{x}\:\wedge\:\mathrm{xy}−\mathrm{y}−\mathrm{2x}+\mathrm{3}=\mathrm{y} \\ $$$$\mathrm{xy}=\mathrm{2x}+\mathrm{2y}−\mathrm{3}\:\wedge\:\mathrm{xy}=\mathrm{2y}+\mathrm{2x}−\mathrm{3} \\ $$$$\left(\mathrm{ib}\right)\&\left(\mathrm{iia}\right) \\ $$$$\:\mathrm{xy}−\mathrm{x}−\mathrm{2y}+\mathrm{3}=\mathrm{y}\:\wedge\:\mathrm{xy}−\mathrm{y}−\mathrm{2x}+\mathrm{3}=\mathrm{x}\: \\ $$$$\mathrm{xy}=\mathrm{x}+\mathrm{3y}−\mathrm{3}\:\wedge\:\mathrm{xy}=\mathrm{3x}+\mathrm{y}−\mathrm{3} \\ $$$$\Rightarrow\:\mathrm{x}+\mathrm{3y}−\mathrm{3}=\mathrm{3x}+\mathrm{y}−\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\mathrm{x}=\mathrm{y}\:\:\:\: \\ $$$$\mathrm{Hence}\:\mathrm{conditions} \\ $$$$\:\mathrm{x}.\mathrm{y}\in\mathrm{A}=\left\{\mathrm{x},\mathrm{y}\right\}\wedge\:\mathrm{y}.\mathrm{x}\in\mathrm{A}=\left\{\mathrm{x},\mathrm{y}\right\} \\ $$$$\mathrm{only}\:\mathrm{meets}\:\mathrm{when}\:\mathrm{x}=\mathrm{y} \\ $$$$\mathrm{i}-\mathrm{e}\:\mathrm{A}\:\mathrm{is}\:\mathrm{a}\:\mathrm{singleton}\:\mathrm{consist}\:\mathrm{of}\:\mathrm{single} \\ $$$$\mathrm{member}. \\ $$

Commented by MathSh last updated on 11/Oct/21

Perfect my dear Ser, thank you

$$\mathrm{Perfect}\:\mathrm{my}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{thank}\:\mathrm{you} \\ $$

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