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Sep 9, 2015 at 6:24 audit Reopen votes
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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