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How to construct a bijection between $\{0,1,\dots,2017\}×\Bbb N$ and $\Bbb N$?

I've tried to come up with different combinations, but no luck. Any ideas how to solve this problem?

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Perhaps the simplest such map $f:\{0,1,\dots,2017\}×\Bbb N\to\Bbb N$ is defined by $$f((a,b))=2018b+a$$ (Assuming $0\in\Bbb N$. If not, then use $2018b+a-2017$ instead.)

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  • $\begingroup$ This is correct or not depending on whether you consider that $0\in\mathbb N$. (I think most people don’t.) Please edit as appropriate. $\endgroup$ – user491874 Dec 6 '17 at 6:08
  • $\begingroup$ @user8734617 Most do take $0\in\Bbb N$ because it simplifies the Peano axioms. In addition, 0 is already listed in the finite set in the product... $\endgroup$ – Parcly Taxel Dec 6 '17 at 6:10
  • $\begingroup$ Fair enough, I concede about that claim. Where I studied maths 30 years ago they used $\mathbb N$ for $\{1,2,3,\ldots\}$ and $\mathbb N_0$ if they wanted to add 0; however this may be different elsewhere, and I’ve just found out that there is even an ISO standard (ISO 80000-2:2009) which prescribes that $\mathbb N=\{0,1,2,\ldots\}$... $\endgroup$ – user491874 Dec 6 '17 at 6:25
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You can simply map (0,1) to 1, (1,1) to 2.. (2017,1) to 2017 and then (0,2) to 2018 and so on..

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Always try to think on a smaller scale, like $f:\{0, 1, 2, 3\} \times \Bbb N \to \Bbb N$

Something like $f(x,y)=x+4y-3$ would work. Now you can think on a larger scale.

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