Timeline for What is the canonical way to extend Peano's axioms to the set of all integers?
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25 events
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Sep 19, 2022 at 9:17 | answer | added | memorial | timeline score: 2 | |
Sep 17, 2022 at 22:46 | comment | added | kaya3 | Is it not true that every $\mathbb{Z}/n\mathbb{Z}$ satisfies all of these axioms? | |
Sep 17, 2022 at 13:08 | vote | accept | CommunityBot | ||
Sep 17, 2022 at 13:08 | comment | added | user155516 | @JoelDavidHamkins I am referring to the first order theory that includes S (see Formulario Mathematico) | |
Sep 17, 2022 at 11:06 | comment | added | Joel David Hamkins | Some people refer to Dedekind arithmetic as "Peano arithmetic" or as "second-order Peano arithmetic", in light of Peano's brilliant presentation of the theory, but Peano himself credits Dedekind. | |
Sep 17, 2022 at 11:04 | comment | added | Joel David Hamkins | I think there is some confusion in the comments and the question between Peano arithmetic and Dedekind arithmetic. Dedekind defined his (second-order) arithmetic in the signature with only 0 and successor S, and then proves that + and * are second-order definable by a recursive definition. What is called Peano arithmetic PA, in contrast, is a first-order theory in the signature with +,*,0,1,< (and not S), but one needs the defining axioms for + and *, and the induction axiom stated as a scheme over first-order formulas. | |
Sep 17, 2022 at 11:00 | answer | added | Joel David Hamkins | timeline score: 15 | |
Sep 17, 2022 at 10:41 | answer | added | Andrej Bauer | timeline score: 2 | |
Sep 17, 2022 at 6:20 | answer | added | Emil Jeřábek | timeline score: 13 | |
Sep 17, 2022 at 6:08 | comment | added | Emil Jeřábek | Axiom 3 follows from axiom 4. The extra axiom $Px+y=P(x+y)$ follows from $Sx+y=S(x+y)$ and Axiom 4. You need a primitive ordering predicate to state the induction axiom. | |
Sep 17, 2022 at 5:57 | history | became hot network question | |||
Sep 17, 2022 at 2:22 | answer | added | user44143 | timeline score: 22 | |
Sep 17, 2022 at 0:56 | comment | added | Gerald Edgar | I do not know of a "standard" axiom system of this kind. I would guess that you need to consider two sets, $\mathbf Z$ and a subset $\mathbf N$. This is closer to the Peano axioms than including operations like $+$ among the axioms. | |
Sep 16, 2022 at 23:11 | review | Close votes | |||
Sep 22, 2022 at 4:01 | |||||
Sep 16, 2022 at 22:54 | comment | added | user155516 | @LSpice Well it's precisely what I've just done | |
Sep 16, 2022 at 22:53 | history | edited | user155516 | CC BY-SA 4.0 |
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Sep 16, 2022 at 22:53 | comment | added | LSpice | I think you need to state the entire axiom system you have in mind, rather than modifying the question each time I comment. Peano's axioms as usually stated do not, and do not need to, include addition, since it can be defined in terms of $S$ using induction. | |
Sep 16, 2022 at 22:49 | history | edited | user155516 | CC BY-SA 4.0 |
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Sep 16, 2022 at 22:47 | comment | added | user155516 | @LSpice Actually $0$ would be singled out by the additive axiom $\forall x,x+0=0$ | |
Sep 16, 2022 at 22:45 | comment | added | LSpice | Unless you single out $0$ as a distinguished constant, any structure of the form you define is at best a torsor under $\mathbb Z$. (The Peano axioms do not need to single out $0$, since it is the unique natural number that is not the successor of any natural number.) Anyway, I think currently the question is probably not sufficiently well formed for MO, although I can imagine that it is possible to make it so. | |
Sep 16, 2022 at 22:42 | history | edited | user155516 | CC BY-SA 4.0 |
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Sep 16, 2022 at 22:36 | comment | added | user155516 | I edited the question to require $SPx=x$. I have no other goal than extending Peano to $\mathbf Z$. I am not sure about the best way to formalize the induction axiom, maybe I have to include extra axioms | |
Sep 16, 2022 at 22:35 | history | edited | user155516 | CC BY-SA 4.0 |
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Sep 16, 2022 at 22:03 | comment | added | LSpice | What would your exact proposed induction axiom be? I imagine that, among other things, you would run into trouble by not requiring that $S P x = x$. Regardless, what is your goal? | |
Sep 16, 2022 at 21:54 | history | asked | user155516 | CC BY-SA 4.0 |