Talk:Peano axioms/Archive 1

message: I deleted this:
:This was a proof using logic alone, but of course infinite. It gives an algorithm for simplifying a :possible proof of contradiction by a series of simple transformations, until it becomes very short. :To see that it becomes very short we need transfinite induction. Gentzen's proof has been :humorously called "assuming the dubious to prove the obvious".
since (a) it was not a proof using 'infinite logic' but a straightforward mathematical proof; (b) the procedure makes the proofs longer, not shorter; (c) whoever called the proof that doesn't understand it.
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Someone was getting confused about Peano's axioms vs. first order Peano Arithmetic.  I(different I than the above: this I's a PhD student who's
area is models of PA) changed
Although natural numbers satisfy these axioms, there are other, nonstandard models of arbitrary large cardinality - by Compactness theorem the existence of infinite natural numbers cannot be excluded in any axiomatization.
to
Dedekind proved, in his 1888 book Was sind und was sollen die Zahlen, that any model of the second order Peano axioms is isomorphic to the natural numbers.
On the other hand, the last axiom listed above, the mathematical induction
axiom, is not itself expressible in the first order language of arithmetic.
If one replaces the last axiom with the schema:
#If P(0) is true and for all n P(x) implies P(x+1)
for each first order property P(x) (an infinite number of axioms) then although natural numbers satisfy these axioms, there are other, nonstandard models of arbitrary large cardinality - by the Compactness theorem the existence of infinite natural numbers cannot be excluded in any axiomatization; by the Lowenheim-Skolem theorem there exist models of all cardinalities.
:Thanks for the clear-up and help! Revolver 07:04, 30 Sep 2004 (UTC)
replies: text: "For all x, P(x) implies P(x + 1)."
Michael Hardy 00:07, 25 Oct 2003 (UTC)
thread_title: Older posts

message: The article claims that  "Peano axioms (or Peano postulates) are a set of first-order axioms"
can someone please tell how is it possible to postulate:
"if a property is possessed by 0 and also by the successor of every natural number which possesses it, then it is possessed by all natural numbers."
in a first order logic? i think you will have two use second order for that claim.(it seems to quantify over properties(definition of second order logic)) 
:(post moved to bottom of page)
:As an axiom, induction is second-order.  However it can be formulated in a first-order manner by adding not a single axiom, but an axiom schema with infinitely many instances.  Since the schema defines a(primitive) recursive set, from the point of view of proof theory, it is acceptable. ---- Charles Stewart 07:46, 26 Jan 2005 (UTC)
replies: text: I see, for example we can get around induction by using no-cycles claim, (no size 1 cycle, no size 2 cycle..etc up to infinity).Still, In this case the text of the artcile should be changed to this particular(infinitary) axiomatization, because as it stands it is confusing as to why is it first-order.--Hq3473 08:02, 26 Jan 2005 (UTC)

text: If I'm not mistakened, forbidding cycles is both too weak and can't be captured by a first-order axiomatisation.  I agree the text needs (much) improvement, but I guess the issue won't be clear until we give an actual axiomatisation in a Hilbert-style proof theory - then we can talk about this point properly.  I'll put it on my overburdened to do list ---- Charles Stewart 10:22, 26 Jan 2005 (UTC)
replies: text: i see, cycles will not rule out higher orders. Any way, can you maybe give me link or a refernce to riogorous first order peano axiomatization? i want to see how is the deduction overcome.--Hq3473 21:44, 26 Jan 2005 (UTC)

text: I agree with the original objection that the induction axiom is second order. The fact that it can be replaced by an infinite number of first order formulae does not change this statement. Having arbitrarily large sets of formulae is not usually meant when speaking of a "first order theory" and it is commonly said that the natural numbers are not first order axiomatizable (meaning "by a single formula (or a finite set of formulae)"). Allowing for infinite sets of first-order axioms gives you quite a lot of expressive power, e.g. you can describe the natural numbers and any subset thereof. I think it is misleading to speak of first-order in the intro, especially since the section on Peano arithmetic also speaks of "the restriction of Peano axioms to a first-order theory". --Markus Krötzsch 20:19, 15 November 2005 (UTC)
replies: text: Peano arithmetic is definitely a first-order theory; this is completely standard usage. The fact that it's not finitely axiomatizable makes no difference. (Actually, "true arithmetic", the set of all true first-order statements about the natural numbers, is also a first-order theory; the difference is that it doesn't have a computable or even computably enumerable set of axioms.). The full Peano axioms are second order because you quantify over properties, not just over individual natural numbers. --Trovatore 20:28, 15 November 2005 (UTC)
thread_title:  first order? 

message: I asked a vague question at Talk:Preintuitionism ... are there any examples of systems in which the first four axioms are kept, the fifth (induction) is intentionally broken or discarded, and yet are not finite, and somehow manage to achieve some sort of "infinity"? (By "intentionally broken", I mean, are there different set of axioms, in which induction is a theorem that may be proven false?) (By requiring "infinity", I want to exclude "obvious" systems like finite groups and fields (and finite state machines?) which seem like they might not need induction as an axiom).  linas 05:50, 7 September 2005 (UTC)
replies: text: Yes, this is an active area of study. If you limit the induction schema to formulas all of whose quantifiers are bounded, you get a theory called IΔ0, which is insufficient to prove that the exponential function (the function that sends n to 2n) is total. (No warranties on the exact definition of IΔ0). There's a whole hierarchy of theories weaker than PA. --Trovatore 20:49, 16 October 2005 (UTC)
replies: text: Looks like I didn't read your question carefully enough; this isn't what you were after. But it still might be of interest to you.... --Trovatore 23:24, 16 October 2005 (UTC)
thread_title:  Discardng induction? 

message: The following part of the article badly needs to be reworded and cleared up:
:But in 1931, Kurt Gödel in his celebrated second incompleteness theorem showed such a proof cannot exist. It is even impossible to prove consistency of Peano arithmetic while assuming the axioms themselves. Furthermore, we can never prove that any axiom system is consistent within the system itself, if it is at least as strong as Peano's axioms. In 1936, Gerhard Gentzen proved the consistency of Peano's axioms, using transfinite induction.
:Most mathematicians assume that Peano arithmetic is consistent, although this relies on intuition only.
For a layman without a mathematical background, this basically reads "It's impossible to prove the consistency of PA. Gentzen proved the consistency of PA in 1936. It's impossible to prove the consistency of PA, but we assume it anyway", which of course is highly problematic. -- Schnee (cheeks clone) 21:55, 26 October 2005 (UTC)
thread_title:  Consistency 

message: To the anonymous user who tried to put this in, it is not needed.  You can write to me about
why this is true, but don't try to edit it this way again.  The form of Peano's axioms is a matter
of history and mathematical practice.
1 is defined as S(0) (S is a primitive notion; it is not defined as "adding 1".)  Your counterexample is not a counterexample:  the structure 0, pi, 2*pi, etc (non-negative integer multiples of pi in the usual real number system)
is a perfectly good Peano-Dedekind structure:  the "1" of this structure is the usual pi, and "adding
1" in this structure is adding pi in the subset of the real numbers which is its domain.
Randall Holmes 23:01, 25 December 2005 (UTC)
thread_title:  0 and 1 do not need to be posited separately 

message: Did Peano actually have 0 as the first natural number?  The axioms are also sometimes presented
with 1 as first, and I suspect this may be historically the original form.  Of course, both are
fine mathematically -- but one defines addition and multiplication differently in each case.  I'm not going to edit the article on this point without historical sources.  Randall Holmes 05:58, 27 December 2005 (UTC)
replies: text: I suggest the article on Zermelo set theory as a model: someone has taken the trouble there to go back to the original text, and what he finds is rather interesting.  I'm pretty sure that here the original text will bear out Lethe's claim that the original axiom set was second-order, but who knows what else we might find? Randall Holmes 06:00, 27 December 2005 (UTC)
thread_title:  historical question about 0 vs. 1, quite distinct from the previous issue 

message: It's all in the title, really. Randall Holmes 17:45, 17 January 2006 (UTC)
thread_title:  Reverted all changes by Evildictaitor; will continue to do this as necessary 

message: This link in the bottom group on the article page:
would not work for me at this time - perhaps a bad omen :-) Carrionluggage 21:38, 10 February 2006 (UTC)
thread_title: Dead Link to Gentzen's  Proof

message: The article uses  \forall x[\varphi(0) \wedge (\varphi(x)\rightarrow\varphi(Sx)) \rightarrow \varphi(x)], but this doesn't seem a correct formulation.  For example, let \varphi(x) denote the statement x = 0.  Then consider the case of x = 3.  \varphi(0) \wedge (\varphi(3) \rightarrow \varphi(4)) \rightarrow \varphi(3) = T \wedge (F \rightarrow F) \rightarrow F = T \wedge T \rightarrow F = T \rightarrow F = F, so the statement is false.
I would formalize induction instead as (\varphi(0) \wedge \forall x[\varphi(x) \rightarrow \varphi(Sx)]) \rightarrow \forall x[\varphi(x)]. 67.166.242.232 06:05, 16 April 2006 (UTC)
thread_title:  Formalization of induction 

message: The subsection Formalized atoms appeared recently, with no prose and an unclear comment in the page source code.  I don't think it belongs here at all, but if it does it ought to be expanded somehow (the induction axiom is particularly tricky to formalize...).  The entire point of the "Peano axioms" is that they are informally stated; the formalized version is Peano arithmetic. 
In fact, the entire section The axioms seems to repeat itself three times. And it is certainly false that Dedekind defined a "Dedekind-Peano structure" in 1888. Any objections to a rewrite of that section?  CMummert 03:26, 3 January 2007 (UTC)
replies: text: No objection here. I TeX'd it up, but anything you can do to improve it would be welcome. CRGreathouse (t | c) 05:01, 4 January 2007 (UTC)
thread_title:  Formalized atoms? 

message: Why are the references indented as they are. As a non-mathematician, not familiar with this topic, it appears to me that something is mixed up with this indentation. I believe that Wikipedia works best when it is clear to the non-specialist. Thanks. Nwbeeson 15:40, 3 January 2007 (UTC)
replies: text: If you read the wording of the references, you'll see the reason for the indentation. However, the vertical spacing was inconsistent, which made the lay-out somewhat unclear. I've fixed it now. --Zundark 16:01, 3 January 2007 (UTC)
thread_title:  References Question 

message: The following statement (very last in the article), is not "intuitively obvious" (and perhaps incorrect [?]):
:"The small number of mathematicians who advocate finitism reject Peano's axioms because the axioms require an infinite set of natural numbers."
I do not believe this is correct --  the axioms do not provide for a completed infinity, or even "an infinity" whatever that means, but simply show how numbers get created. We need an axiom of infinity to "go there." I've not read that intuitionists, e.g. Brouwer's bunch, reject the Peano axioms. Maybe some do, but then they won't have induction available to them. They don't allow for Cantor's "completed infinity", but they do allow for a never-ending progression of "successor events," and in particular they are great fans of induction:
:"A generality statement all natural numbers n have the property P, or briefly for all n, P(n), is understood by the intuitionist as an hypotheitical assertion to the effect that, if any particular natural number were given to us, we couldb e sure that that number n has the property P. This is a meaning which does not require us to take into view the classicaal completed infinity of the natural numbres.
:"Mathematical induction is an example of an intuitionistic method for proving generality propostions abou the natural numbers. A proof by induction of the proposition for all n, P(n) shows that any given n would have to have the property P, by reasoning which uses only the number from 0 up to n ... Of course, for a particular proof by induction to be intuitionistic, also the reasonings used within its basis and induction step must be intuitionistic." (Kleene 1952, 1971 p. 49)
I suggest the line in question should go away or get an in-line reference, if someone knows of one. wvbaileyWvbailey 17:34, 25 January 2007 (UTC)
replies: text: You are correct that ordinary intuitionism has no problem with the infinite set of natural numbers or with induction. But see for example this paper by Vladmir Sazonov, where he studies what he calls "feasible numbers".  He does not accept the fact that 2^x is a total function on the natural numbers, and gives the specific value of 2^1024 as one that does not exist because it is too large to be physically computed. So not only does he reject a completed infinity of natural numbers, he rejects even a potential infinity of natural numbers, because he rejects (at least) the existence of any number bigger than 2^1024. This viewpoint is sometimes called "ultrafinitism".  It is not the same meaning of the word finitism that is used in the context of Hilbert's program. The number of mathematicians who subscribe to this viewpoint is exceedingly small. CMummert · talk 17:55, 25 January 2007 (UTC)
thread_title:  Last sentence 

message: Substituted at 06:49, 7 May 2016 (UTC)
thread_title: Assessment comment