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Let-A-n-2n-n-N-and-B-2n-3n-n-N-Then-A-B-is-equal-to-

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Question Number 114235 by Aina Samuel Temidayo last updated on 18/Sep/20
Let A={(n,2n):n∈N} and B={(2n,3n):n∈N}. Then A∩B is equal to
$$\mathrm{Let}\:\mathrm{A}=\left\{\left(\mathrm{n},\mathrm{2n}\right):\mathrm{n}\in\mathrm{N}\right\}\:\mathrm{and} \\ $$$$\mathrm{B}=\left\{\left(\mathrm{2n},\mathrm{3n}\right):\mathrm{n}\in\mathrm{N}\right\}.\:\mathrm{Then}\:\mathrm{A}\cap\mathrm{B}\:\mathrm{is} \\ $$$$\mathrm{equal}\:\mathrm{to} \\ $$
Answered by 1549442205PVT last updated on 18/Sep/20
A∩B=(0,0)since (0,0)=(0;2.0)=(2.0;3.0)
$$\mathrm{A}\cap\mathrm{B}=\left(\mathrm{0},\mathrm{0}\right)\mathrm{since} \\ $$$$\left(\mathrm{0},\mathrm{0}\right)=\left(\mathrm{0};\mathrm{2}.\mathrm{0}\right)=\left(\mathrm{2}.\mathrm{0};\mathrm{3}.\mathrm{0}\right) \\ $$
Commented by Aina Samuel Temidayo last updated on 18/Sep/20
Not correct.
$$\mathrm{Not}\:\mathrm{correct}. \\ $$
Commented by MJS_new last updated on 18/Sep/20
then what′s correct?
$$\mathrm{then}\:\mathrm{what}'\mathrm{s}\:\mathrm{correct}? \\ $$
Commented by 1549442205PVT last updated on 18/Sep/20
Since N={0,1,2,...},(0,0)can be represented under two forms:(n,2n) and (2n,3n)
$$\mathrm{Since}\:\mathrm{N}=\left\{\mathrm{0},\mathrm{1},\mathrm{2},…\right\},\left(\mathrm{0},\mathrm{0}\right)\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{represented}\:\mathrm{under}\:\mathrm{two}\:\mathrm{forms}:\left(\mathrm{n},\mathrm{2n}\right) \\ $$$$\mathrm{and}\:\left(\mathrm{2n},\mathrm{3n}\right) \\ $$
Answered by malwaan last updated on 18/Sep/20
(n_1 ,2n_1 )=(2n_2 ,3n_2 ) n_1 =2n_2 2n_1 =3n_2 ⇒n_2 =((2n_1 )/3) ∴ n_1 =2(((2n_1 )/3))=((4n_1 )/3) n_1 (1−(4/3))=0 (−(1/3))n_1 =0⇒n_1 =0⇒n_2 =0 ∴ A∩B = (0 , 0)
$$\left({n}_{\mathrm{1}} ,\mathrm{2}{n}_{\mathrm{1}} \right)=\left(\mathrm{2}{n}_{\mathrm{2}} ,\mathrm{3}{n}_{\mathrm{2}} \right) \\ $$$${n}_{\mathrm{1}} =\mathrm{2}{n}_{\mathrm{2}} \\ $$$$\mathrm{2}{n}_{\mathrm{1}} =\mathrm{3}{n}_{\mathrm{2}} \Rightarrow{n}_{\mathrm{2}} =\frac{\mathrm{2}{n}_{\mathrm{1}} }{\mathrm{3}} \\ $$$$\therefore\:{n}_{\mathrm{1}} =\mathrm{2}\left(\frac{\mathrm{2}{n}_{\mathrm{1}} }{\mathrm{3}}\right)=\frac{\mathrm{4}{n}_{\mathrm{1}} }{\mathrm{3}} \\ $$$${n}_{\mathrm{1}} \left(\mathrm{1}−\frac{\mathrm{4}}{\mathrm{3}}\right)=\mathrm{0} \\ $$$$\left(−\frac{\mathrm{1}}{\mathrm{3}}\right){n}_{\mathrm{1}} =\mathrm{0}\Rightarrow{n}_{\mathrm{1}} =\mathrm{0}\Rightarrow{n}_{\mathrm{2}} =\mathrm{0} \\ $$$$\therefore\:{A}\cap{B}\:=\:\left(\mathrm{0}\:,\:\mathrm{0}\right) \\ $$
Commented by Aina Samuel Temidayo last updated on 19/Sep/20
Exactly, it was stated in the question.
$$\mathrm{Exactly},\:\mathrm{it}\:\mathrm{was}\:\mathrm{stated}\:\mathrm{in}\:\mathrm{the}\:\mathrm{question}. \\ $$
Commented by 1549442205PVT last updated on 18/Sep/20
Thank for your detail solution.We can also argue in following way : For n≠0 then (n,2n)≠(2n,3n) For n=0 then (n,2n)=(2n,3n)
$$\mathrm{Thank}\:\mathrm{for}\:\mathrm{your}\:\:\mathrm{detail}\:\mathrm{solution}.\mathrm{We} \\ $$$$\mathrm{can}\:\mathrm{also}\:\mathrm{argue}\:\mathrm{in}\:\mathrm{following}\:\mathrm{way}\:: \\ $$$$\mathrm{For}\:\mathrm{n}\neq\mathrm{0}\:\mathrm{then}\:\left(\mathrm{n},\mathrm{2n}\right)\neq\left(\mathrm{2n},\mathrm{3n}\right) \\ $$$$\mathrm{For}\:\mathrm{n}=\mathrm{0}\:\mathrm{then}\:\left(\mathrm{n},\mathrm{2n}\right)=\left(\mathrm{2n},\mathrm{3n}\right) \\ $$
Commented by Rasheed.Sindhi last updated on 18/Sep/20
Perhaps the source of the question considers: N={1,2,3,...} In that case A∩B=∅
$${Perhaps}\:{the}\:{source}\:{of}\:{the} \\ $$$${question}\:{considers}: \\ $$$$\:\:\:\:\mathbb{N}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},…\right\} \\ $$$${In}\:{that}\:{case}\:\mathrm{A}\cap\mathrm{B}=\emptyset \\ $$
Commented by MJS_new last updated on 19/Sep/20
N had been {1, 2, 3, ...} and N_0 ={0, 1, 2, ...} a few years ago obviously there was a change to N={0, 1, 2, ...} and N^★ ={1, 2, 3, ...} ⇒ confusion
$$\mathbb{N}\:\mathrm{had}\:\mathrm{been}\:\left\{\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:…\right\}\:\mathrm{and}\:\mathbb{N}_{\mathrm{0}} =\left\{\mathrm{0},\:\mathrm{1},\:\mathrm{2},\:…\right\} \\ $$$$\mathrm{a}\:\mathrm{few}\:\mathrm{years}\:\mathrm{ago}\:\mathrm{obviously}\:\mathrm{there}\:\mathrm{was}\:\mathrm{a}\:\mathrm{change} \\ $$$$\mathrm{to}\:\mathbb{N}=\left\{\mathrm{0},\:\mathrm{1},\:\mathrm{2},\:…\right\}\:\mathrm{and}\:\mathbb{N}^{\bigstar} =\left\{\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:…\right\} \\ $$$$\Rightarrow\:\mathrm{confusion} \\ $$

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