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Question-198939




Question Number 198939 by Mingma last updated on 26/Oct/23
Answered by witcher3 last updated on 26/Oct/23
ϕ:2N→Z  ϕ(2n)= { ((k if n=2(2k+1))),((−k if n=2.(2k))) :}  ϕ(m)=ϕ(n) ⇔m=n  ϕ injective  if  n∈Z if n≥0   n=ϕ(2(2n+1))  −n=ϕ(2.2n)  ϕ surjective   injective surjective?⇔bijection  ⇒2N≈Z
$$\varphi:\mathrm{2}\mathbb{N}\rightarrow\mathbb{Z} \\ $$$$\varphi\left(\mathrm{2n}\right)=\begin{cases}{\mathrm{k}\:\mathrm{if}\:\mathrm{n}=\mathrm{2}\left(\mathrm{2k}+\mathrm{1}\right)}\\{−\mathrm{k}\:\mathrm{if}\:\mathrm{n}=\mathrm{2}.\left(\mathrm{2k}\right)}\end{cases} \\ $$$$\varphi\left(\mathrm{m}\right)=\varphi\left(\mathrm{n}\right)\:\Leftrightarrow\mathrm{m}=\mathrm{n} \\ $$$$\varphi\:\mathrm{injective} \\ $$$$\mathrm{if}\:\:\mathrm{n}\in\mathbb{Z}\:\mathrm{if}\:\mathrm{n}\geqslant\mathrm{0}\: \\ $$$$\mathrm{n}=\varphi\left(\mathrm{2}\left(\mathrm{2n}+\mathrm{1}\right)\right) \\ $$$$−\mathrm{n}=\varphi\left(\mathrm{2}.\mathrm{2n}\right) \\ $$$$\varphi\:\mathrm{surjective}\: \\ $$$$\mathrm{injective}\:\mathrm{surjective}?\Leftrightarrow\mathrm{bijection} \\ $$$$\Rightarrow\mathrm{2}\mathbb{N}\approx\mathbb{Z} \\ $$

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