# Talk:Peano axioms

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## Powers in Peano arithmetic.

Seeing the recursive definitions of the addition and the multiplication, it is quite natural to ask why there isn't a similar recursive scheme for the power operation a^b (and for the factorial a!).

The answer is that the theory with + and x has enough expressive power to allow a derivation of those operations (whereas Presburger arithmetic wih + only would be unable to "create" x). More precisely, one can write a first order formula P(a,b,c) (written in the language of PA) equivalent to a^b=c -which means that for any integers a,b there exists an unique integer c such that P(a,b,c) and that the unique c' such that P(a,b+1,c') is bc (and, of course, P(a,1,a) is true). It would probably add too much weight to the article to give such a formula P but a reference would be welcome.

Alain Gen — Preceding unsigned comment added by 176.128.111.170 (talk) 12:54, 7 October 2015 (UTC)

There are two possible theorems you could be looking for. First, any recursive function is representible in PA (or, in fact in Robinson arithmetic). We don't have even:
${\displaystyle \vdash _{RA}\forall a\forall b\exists !cP(a,b,c);}$
but we do have
${\displaystyle \forall a,b,c\in \mathbb {N} \vdash _{RA}(P(S^{a}0,S^{b}0,S^{c}0)\leftrightarrow (a^{b}=c))}$
The other theorem relates to implementation of recursion in PA; essentially, in this context, given a "function" f(a,b,c) [= a c, in this case] and a function g(a) [= a], one can construct a formula P, such that
${\displaystyle \vdash _{PA}\forall a\forall b\exists !cP(a,b,c)}$
${\displaystyle \vdash _{PA}\forall a\forall cg(a)=c\rightarrow P(a,1,c)\leftrightarrow }$
${\displaystyle \vdash _{PA}\forall a\forall b\forall cP(a,b,c)\rightarrow P(a,b+1,f(a,b,c))}$
That is, a function h such that
${\displaystyle h(a,1)=g(a)}$
${\displaystyle h(a,b+1)=f(a,b,h(a,b))}$
This is of the form
${\displaystyle h(1)=g(a)}$
${\displaystyle h(n+1)=f(a,n,h(n))}$
In PA, one can construct (and prove) the existence of a pairing function, ${\displaystyle F:\mathbb {N} \times \mathbb {N} {\overset {\underset {\text{1-1}}{}}{\mapsto }}\mathbb {N} }$, and then a sequencing function, ${\displaystyle H:\mathbb {N} {\overset {\underset {\text{onto}}{}}{\mapsto }}\mathbb {N} ^{<\mathbb {N} }}$, with the length and all coordinate functions definable.
It follows then that you can do simple recursion using the induction axiom; given a "function" H (n, "F"), which depends only on the values of F on arguments less than n, you can construct (and prove, in PA) a function F consistent with that, so that F(n) = H(n, F). I don't know how to express that simply, but that might be what you're looking for.  :We have to be careful, because the induction step in Goodstein's theorem is definable in PA. — Arthur Rubin (talk) 16:55, 8 October 2015 (UTC)

Thank you for your answer. I had in mind the second statement, concerning the implementation of recursion. The fact that primitive recursive functions can be implemented in PA (with a reference to the article detailing Gödel's trick ) would be welcome, especially for the curious reader who would want to compare PA with primitive recursive arthmetic.

Alain Gen, 10 October 2015 — Preceding unsigned comment added by 176.128.111.170 (talk) 13:41, 11 October 2015 (UTC)

See the recursive definition of exponentiation in my posting, "The Modern Version of Peano's Axioms." It can't be original to me, though I can't find a source.
Exponentiation is just repeated multiplication. We have:
${\displaystyle x^{2}=x\cdot x}$
${\displaystyle x^{3}=x\cdot x\cdot x}$
${\displaystyle x^{4}=x\cdot x\cdot x\cdot x}$
${\displaystyle \vdots }$
${\displaystyle x^{S(n)}=x^{n}\cdot x}$ for ${\displaystyle n\geq 2}$
${\displaystyle \vdots }$
There are infinitely many binary functions on all of ${\displaystyle \mathbb {N} }$ that satisfy the above conditions and the definitions of addition and multiplicataion, but they differ only on the value of ${\displaystyle 0^{0}}$.
--Danchristensen (talk) 01:40, 22 March 2016 (UTC)
Dan, exponentiation is not part of the language of Peano arithmetic. I see no particular need to mention it in this article. It might be worth mentioning that it's definable, I suppose. But we don't need to go into details. --Trovatore (talk) 04:29, 22 March 2016 (UTC)

## The modern version of Peano's Axioms

With the formalization of logic and set theory since Peano first published his axioms for the natural numbers in 1889, his axioms or the natural numbers have been reduced today to just five. Today, the axioms of equality are found in the rules of logic (e.g. FOL). Addition, multiplication and exponentiation can also be constructed using the Cartesian product, subset and function rules/axioms from set theory, so their definitions, too, are now separate from Peano's Axioms.

We have ${\displaystyle (\mathbb {N} ,S,0)}$ such that:

1. ${\displaystyle 0\in \mathbb {N} }$

2. ${\displaystyle S:\mathbb {N} \to \mathbb {N} }$

3. ${\displaystyle S}$ is injective

4. ${\displaystyle \forall x\in \mathbb {N} :S(x)\neq 0}$

5. ${\displaystyle \forall P\subseteq \mathbb {N} :[0\in P\land \forall x\in P:S(x)\in P\rightarrow P=\mathbb {N} ]}$

Sources: Britannica, Wolfram

1. ${\displaystyle \forall x\in \mathbb {N} :[x+0=x]}$

2. ${\displaystyle \forall x,y\in \mathbb {N} :[x+S(y)=S(x+y)]}$

#### Multiplication

1. ${\displaystyle \forall x\in \mathbb {N} :[x\cdot 0=0]}$

2. ${\displaystyle \forall x,y\in \mathbb {N} :[x\cdot S(y)=x\cdot y+x]}$

#### Exponentiation

1. ${\displaystyle \forall x\in \mathbb {N} :[x\neq 0\rightarrow x^{0}=1]}$

2. ${\displaystyle 0^{1}=0}$

3. ${\displaystyle \forall x,y\in \mathbb {N} :[x^{S(y)}=x^{y}\cdot x]}$

--Danchristensen (talk) 22:14, 21 March 2016 (UTC)

Dan, what is your purpose for posting this here? Are you suggesting incorporating this into the article? --Trovatore (talk) 23:31, 21 March 2016 (UTC)
Yes. Incredibly, the article makes no mention of what seems to be the most widely used version of Peano's Axioms today. It is a MAJOR oversight.
--Danchristensen (talk) 00:55, 22 March 2016 (UTC)
I'm not too excited about what particular formulation we use. I think exponentiation should be left out; it's not essential, just one of many functions you can define. -Trovatore (talk) 04:33, 22 March 2016 (UTC)
Those axioms are already in the article (apart from exponentiation, which is not usually included in axiomatizations of arithmetic). For the first five, your #1 is Peano's #1; your #2 is Peano's #6; your #3 is Peano's #7; your #4 is Peano's #8; your #5 is Peano's #9. So these are not a "modern version"; they are just Peano's original axioms. For the later ones. we have sections on "Addition" and "Multiplication" which discuss those. — Carl (CBM · talk) 11:28, 21 July 2016 (UTC)

## Peanos Original Formulation

It seems that peano first only considered the positive integers, i.e. N+
and formulated his axioms with one instead of zero. See Page 55 here:

Hubert Kennedy, Life and Works of Giuseppe Peano, 2002
http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.201.8126

This is much better expressed in the German wiki about Peano axioms,
there we find a better introduction to the axioms:
https://de.wikipedia.org/wiki/Peano-Axiome#Urspr.C3.BCngliche_Formalisierung
Peano betrachtete ursprünglich 1 als kleinste natürliche Zahl. Jan Burse (talk) 18:07, 7 October 2016 (UTC)

Yes, Peano did start with 1 instead of 0. The German wikipedia also seems to start with 0 instead of 1, as does this article, with an explanation. Peano's axioms included axioms not stated in the German wikipedia or the article mentioned above. There is a much more direct version of Peano's original article in From Frege to Gödel, or see page XVII of [1] for the full list of 9 axioms. It is common in modern treatments to omit several of Peano's axioms which are now viewed as rules of logic, but this is anachronistic if we are trying to describe Peano's original axioms. — Carl (CBM · talk) 00:02, 8 October 2016 (UTC)

Peano did not consider 0 as a natural number. There is a very good reason for that. I understand, that original Peano axiom can be presented today in slightly different form, but considering 0 as a natural number is not that kind of reasonable "modernization". Wikipedia is becoming a criteria of truth, people are using it as a final reference. Providing here information that is, at least, suspicious is not acceptable. That are not Peano axioms and this interpretation contradicts many other sources as well as common sense: a natural number can be found in nature as a "measure" of finite set, while 0 is a measure of an empty set that dies not exist. I can add many links supporting my opinion, but do not see such an option here. And do not see much of reason to do that. AlexTalalai (talk) 17:40, 12 January 2017 (UTC)

## Kaye's Axiomatization

I don't see why we should stick to Kaye's axiomatization, since that axiom can be proven from the other axioms without induction. We first get x*0+x*0 = x*(0+0) = x*0 = x*0+0. Then by trichotomy x*0 < 0 or x*0 = 0 or x*0 > 0. If x*0 < 0 then x*0+x*0 < x*0+0 and hence x*0+0 < x*0+0, contradicting irreflexivity. Similarly if x*0 > 0. Therefore x*0 = 0. Since each step is completely verifiable, I think we should delete the redundant axiom, but make a comment-note containing this proof for reference — Preceding unsigned comment added by Lim Wei Quan (talkcontribs) 2017-04-03T10:00:56 (UTC)

The reason to stick with some printed axiomatization is that, once we start changing things, we can end up in a position where we have a list of axioms which we see are equivalent, but which we can't reference as being equivalent, which leads to problems when others start asking for a source for the axioms. This can happen when a sequence of small changes are made, so that the resulting axioms drift farther and farther from the original axioms. A separate issue can happen when an editor claims that they are changing the axioms to something equivalent, but others don't agree about the equivalence. The easiest solution to all this is to pick an "authoritative" source, like Kaye's book, and take the axioms from it verbatim. — Carl (CBM · talk) 17:17, 4 April 2017 (UTC)

I understand that. But this is one of the major reasons the current attitude on Wikipedia is bad for mathematics. I agree that there must be a clear justification, but it is in my opinion hypocritical to allow people to paraphrase viewpoints in the arts and humanities (which necessitates a significant amount of editor interpretation) but forbid people from giving rigorous and indisputable mathematical justification to simplify mathematical objects. In fact, many good Wikipedia articles in mathematics do not care about sticking to every tiny dot in the cited references. Since you disagree with changing the current axiomatization, I shall leave it as it is, but add in a paragraph stating that it is superfluous. For the sake of mathematics, please do not remove it just due to lack of citation. I am not going to waste my time trying to find a reference (is there even one?) for a trivial proof. Thanks! Lim Wei Quan (talk) 07:14, 17 April 2017 (UTC)
The results are not a problem, but I am not sure why the non-independence of one of Kaye's axioms is important enough to bother spending a paragraph on. It seems to me that it is just a distraction from the main point of that section, and that there is really not likely to be much interest in whether the axioms are independent. — Carl (CBM · talk) 12:04, 20 April 2017 (UTC)
Well there are a few reasons I want to make the fact explicit and without doubt. Firstly, this is not the first time that the second half of that axiom has been noticed to be independent, and it would save everyone the time and trouble to have the reasoning spelled out there. Secondly, the reason given for not changing the original set of axioms was precisely so that everything could be verified without expert knowledge, so this is precisely why I included sufficient detail. Thirdly, it is mathematically elegant to have the axioms follow the ring axioms. Fourthly, minimality is a reasonable mathematical reason, but following Kaye is not, unless there is a reason Kaye had that has not been stated so far. Lim Wei Quan (talk) 07:08, 30 April 2017 (UTC)
I agree with CBM. A footnote at axiom 6 may be sufficient; I'm not sure whether the proof should be given there (maybe in some abbreviated form) or whether it should be completely omitted. Apart from that, I'd prefer an axiomatization that only uses independent axioms - isn't there one in the textbooks? - Jochen Burghardt (talk) 14:15, 26 April 2017 (UTC)
I do not think a footnote is sufficient for the reasons I state above. If you find a proper reference with the same axioms minus the redundant one, I strongly prefer removing the redundant one, adding citation to your reference, and moving that paragraph I had written into a HTML comment just for record, in case any editor comes along and wonders whether the discrepancy between the references suggests an error. Without such a reference, I think the Wikipedia article is most helpful if it explicitly contains all the information that is of interest to its readers, and certainly the people who read the article on PA are going to be interested in this kind of detail. It is most frustrating to read Wikipedia articles on Mathematics that make claims that are vague or cannot be verified easily, such as if the citations are difficult to access. Lim Wei Quan (talk) 07:08, 30 April 2017 (UTC)
Why would you prefer that? My view is that it doesn't matter much. It's important that the axioms be intuitive; it's not so important to make them formally minimalistic. (We see this problem in a worse way with ZFC, where for example some prefer the less intuitive axiom of collection to the more natural axiom of replacement, just because separation formally follows from replacement.) --Trovatore (talk) 17:28, 27 April 2017 (UTC)
I agree that it is important that the axioms be intuitive. For your argument, the group axioms are a better example than ZFC; they can be reduced with some work to asserting closure, associativity, right-identity and right-inverses, but it is just silly to use that axiomatization since that is totally missing the point of a group. However, it is mathematically of interest to know what is a minimal axiomatization. And anyone with just a bit of familiarity with logic would be surprised at seeing Kaye's axiomatization with that extra half axiom. I, for instance, wondered whether it was necessary somehow, and even after writing down my proof I checked multiple times to see whether perhaps I made some careless mistake. All that unnecessary effort could have been avoided by making everything explicit. Lim Wei Quan (talk) 07:08, 30 April 2017 (UTC)

We should keep in mind that the article is mainly about Peano axioms, not about First-order theory of arithmetic, let alone about Kaye's first-order axioms. So properites of the latter are a rather unimportant detail here. (However, they would certainly be important in an article on Kaye's first-order axioms.) For this reason, I'm still inclined to move the independence issue to a footnote which also gives a proof sketch. This should meet Lim Wei Quan's arguments from above. I suggest to add the footnote at the end of axiom 6, with the following text:

The second half, ${\displaystyle \forall x\in \mathbf {N} (x\cdot 0=0)}$, can be proven from the other axioms: ${\displaystyle x\cdot 0+x\cdot 0=x\cdot 0+0}$ by 5 and additive identity. Using 11, 2, and 9, ${\displaystyle x\cdot 0<0}$ leads to a contradition, and ${\displaystyle 0 does, too. Hence ${\displaystyle x\cdot 0=0}$ by 10.

(Note that some Leibniz properties, like ${\displaystyle a, are needed, too, to infer e.g. ${\displaystyle x\cdot 0+x\cdot 0 from ${\displaystyle x\cdot 0+x\cdot 0<0+x\cdot 0}$ and ${\displaystyle 0+x\cdot 0=x\cdot 0+0}$; probably, they go without saying for Kaye?) - Jochen Burghardt (talk) 12:11, 30 April 2017 (UTC)

That's fine. I (mistakenly) thought by "footnote" you meant just a claim of redundancy without any justification. I've shortened it and transferred it to a footnote. Hope everyone is satisfied now! =) Lim Wei Quan (talk) 12:43, 30 April 2017 (UTC)