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Question Number 114235 by Aina Samuel Temidayo last updated on 18/Sep/20

$$\mathrm{Let}\mathrm{A}=\{(\mathrm{n},2\mathrm{n}):\mathrm{n}\in \mathrm{N}\}\mathrm{and}$$$$\mathrm{B}=\{(2\mathrm{n},3\mathrm{n}):\mathrm{n}\in \mathrm{N}\}.\mathrm{Then}\mathrm{A}\cap \mathrm{B}\mathrm{is}$$$$\mathrm{equal}\mathrm{to}$$

Answered by 1549442205PVT last updated on 18/Sep/20

$$\mathrm{A}\cap \mathrm{B}=(0,0)\mathrm{since}$$$$(0,0)=(0;2.0)=(2.0;3.0)$$

Commented by Aina Samuel Temidayo last updated on 18/Sep/20

$$\mathrm{Not}\mathrm{correct}.$$

Commented by MJS_new last updated on 18/Sep/20

$$\mathrm{then}{\mathrm{what}}^{\prime }\mathrm{s}\mathrm{correct}?$$

Commented by 1549442205PVT last updated on 18/Sep/20

$$\mathrm{Since}\mathrm{N}=\{0,1,2,...\},(0,0)\mathrm{can}\mathrm{be}$$$$\mathrm{represented}\mathrm{under}\mathrm{two}\mathrm{forms}:(\mathrm{n},2\mathrm{n})$$$$\mathrm{and}(2\mathrm{n},3\mathrm{n})$$

Answered by malwaan last updated on 18/Sep/20

$$(n_1,2n_1)=(2n_2,3n_2)$$$$n_1=2n_2$$$$2n_1=3n_2\Rightarrow n_2=\frac{2n_1}{3}$$$$\therefore n_1=2\left(\frac{2n_1}{3}\right)=\frac{4n_1}{3}$$$$n_1(1-\frac{4}{3})=0$$$$(-\frac{1}{3})n_1=0\Rightarrow n_1=0\Rightarrow n_2=0$$$$\therefore A\cap B=(0,0)$$

Commented by Aina Samuel Temidayo last updated on 19/Sep/20

$$\mathrm{Exactly},\mathrm{it}\mathrm{was}\mathrm{stated}\mathrm{in}\mathrm{the}\mathrm{question}.$$

Commented by 1549442205PVT last updated on 18/Sep/20

$$\mathrm{Thank}\mathrm{for}\mathrm{your}\mathrm{detail}\mathrm{solution}.\mathrm{We}$$$$\mathrm{can}\mathrm{also}\mathrm{argue}\mathrm{in}\mathrm{following}\mathrm{way}:$$$$\mathrm{For}\mathrm{n}\ne 0\mathrm{then}(\mathrm{n},2\mathrm{n})\ne (2\mathrm{n},3\mathrm{n})$$$$\mathrm{For}\mathrm{n}=0\mathrm{then}(\mathrm{n},2\mathrm{n})=(2\mathrm{n},3\mathrm{n})$$

Commented by Rasheed.Sindhi last updated on 18/Sep/20

$$Perhapsthesourceofthe$$$$questionconsiders:$$$$\mathbb{N}=\{1,2,3,...\}$$$$Inthatcase\mathrm{A}\cap \mathrm{B}=\mathrm{\varnothing }$$

Commented by MJS_new last updated on 19/Sep/20

$$\mathbb{N}\mathrm{had}\mathrm{been}\{1,2,3,...\}\mathrm{and}{\mathbb{N}}_0=\{0,1,2,...\}$$$$\mathrm{a}\mathrm{few}\mathrm{years}\mathrm{ago}\mathrm{obviously}\mathrm{there}\mathrm{was}\mathrm{a}\mathrm{change}$$$$\mathrm{to}\mathbb{N}=\{0,1,2,...\}\mathrm{and}{\mathbb{N}}^{★}=\{1,2,3,...\}$$$$\Rightarrow \mathrm{confusion}$$

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