Timeline for How is $\mathbb N$ actually defined?
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| when toggle format | what | by | license | comment | |
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| Sep 9, 2015 at 6:24 | audit | Reopen votes | |||
| Sep 9, 2015 at 14:03 | |||||
| Sep 8, 2015 at 22:44 | audit | Suggested edits | |||
| Sep 8, 2015 at 22:44 | |||||
| Sep 2, 2015 at 9:17 | history | edited | Asaf Karagila♦ |
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| Sep 1, 2015 at 19:35 | comment | added | Asaf Karagila♦ | @columbus8myhw: $\mathsf{ACF}_p$ is complete for every choice of characteristics (prime or $0$, of course). And the theory really just says that you have a field (of a certain characteristics), and that every polynomial has a root. This can be easily done with finitely many axioms and two schemata (one for $p>0$). So it's easily complete and recursively enumerable. But those two things together imply decidability! | |
| Sep 1, 2015 at 19:25 | comment | added | Akiva Weinberger | @AsafKaragila Really? It's complete and decidable? I need to go learn some model theory. | |
| Sep 1, 2015 at 19:23 | comment | added | Asaf Karagila♦ | @columbus8myhw: No, the theory of the complex numbers is complete, decidable and consistent. So it cannot possibly interpret enough of the arithmetic of the natural numbers. In any case, context is important of course. I'm just being a smartypants. :-) | |
| Sep 1, 2015 at 19:21 | comment | added | Akiva Weinberger | @AsafKaragila Meh, I wrote that I was working in $\Bbb Z$, but then I wanted to edit some stuff before pressing 'enter' and I accidentally got rid of that comment. Pfft. (In any case, $\Bbb N$ isn't first-order definable in $(\Bbb C,+,\times)$, is it?) | |
| Sep 1, 2015 at 19:16 | comment | added | Asaf Karagila♦ | @columbus8myhw: So every complex number is a natural number? :-) | |
| Sep 1, 2015 at 18:16 | comment | added | Akiva Weinberger | Perhaps I could say "a number is a natural number if it's the sum of four squares." (It would be odd to take it as a definition, but it could be taken as a definition if you want it to.) That particular definition is interesting in that it only uses the concepts of $+$ and $\times$, rather than something like $>$. (Of course, by this definition, $0$ is natural. You could easily change it if you don't want to count $0$ as a natural.) | |
| Sep 1, 2015 at 14:32 | answer | added | Asaf Karagila♦ | timeline score: 14 | |
| Sep 1, 2015 at 14:25 | answer | added | Mauro ALLEGRANZA | timeline score: 4 | |
| Sep 1, 2015 at 14:21 | answer | added | Lee Mosher | timeline score: 0 | |
| Sep 1, 2015 at 14:17 | answer | added | Adam Bartoš | timeline score: 2 | |
| Sep 1, 2015 at 13:53 | answer | added | Ennar | timeline score: 1 | |
| Sep 1, 2015 at 13:53 | answer | added | Thomas Andrews | timeline score: 4 | |
| Sep 1, 2015 at 13:41 | comment | added | Hagen von Eitzen | If $0$ occurs in your Peano axioms the $0\in\mathbb N$. If you replace $1$ for $0$ in them, then $0\notin \mathbb N$. | |
| Sep 1, 2015 at 13:41 | comment | added | walkar | I think you're going to run into a problem assuming there's a standard definition that anyone uses. Most mathematicians just run with either $\{1,2,...\}$ or $\{0,1,2,...\}$ based on what's most useful at the moment and don't think much more about it. | |
| Sep 1, 2015 at 13:40 | review | First posts | |||
| Sep 1, 2015 at 13:41 | |||||
| Sep 1, 2015 at 13:39 | comment | added | hardmath | The decision to include or exclude zero as a "natural number" is not a problem for the foundations of mathematics. Some authors prefer to include it, some exclude it, and some (like myself) are inconsistent in finding it convenient for some applications to have it, and other times not. As long as the author makes the decision clear for purposes of a discussion/publication, either approach is okay. | |
| Sep 1, 2015 at 13:33 | history | asked | user266061 | CC BY-SA 3.0 |