# Consistency of Peano axioms (Hilbert's second problem)?

(Putting aside for the moment that Wikipedia might not be the best source of knowledge.)

I just came across this Wikipedia paragraph on the Peano-Axioms:

The vast majority of contemporary mathematicians believe that Peano's axioms are consistent, relying either on intuition or the acceptance of a consistency proof such as Gentzen's proof.

And then went back to the article on mathematical induction:

The principle of mathematical induction is usually stated as an axiom of the natural numbers; see Peano axioms.

I'm not a math person, but I'm curious - how can something which sounds like it is believed to be consistent, be the base of such widely used proof technique?

(The Wikipedia article on Peano-Axioms also states: "When the Peano axioms were first proposed, Bertrand Russell and others agreed that these axioms implicitly defined what we mean by a "natural number".)

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In broad strokes: You can't prove a set of axioms consistent without other axioms. Thus we must always assume that some set of axioms is consistent in order to do anything. The problem is unavoidable. – Alex Becker Aug 24 '12 at 22:11
The vast majority of people in number theory are too busy exploring the properties of the integers and related structures to even give a thought about somebody's axioms. – André Nicolas Aug 24 '12 at 22:54
You might enjoy this paper by Peter Smith on "Induction and Predicativity". – MJD Aug 25 '12 at 2:33

The answer is relatively simple, but complicated.

We cannot prove that Peano axioms (PA) is a consistent theory from the axioms of PA. We can prove the consistency from stronger theories, e.g. the Zermelo-Fraenkel (ZF) set theory. Well, we could prove that PA is consistent from PA itself if it was inconsistent to begin with, but that's hardly helpful.

This leads us to a point discussed on this site before. There is a certain point in mathematical research that you stop asking yourself whether foundational theory is consistent, and you just assume that they are.

If you accept ZF as your foundation you can prove that PA is consistent, but you cannot prove that ZF itself is consistent (unless, again, it is inconsistent to begin with); if you want a stronger theory for foundation, (e.g., ZF+Inaccessible cardinal), then you can prove ZF is consistent, but you cannot prove that the stronger theory is consistent (unless... inconsistent bla bla bla).

However what guides us is an informal notion: we have a good idea what are the natural numbers (or what properties sets should have), and we mostly agree that a PA describes the natural numbers well -- and even if we cannot prove it is consistent, we choose to use it as a basis for other work.

Of course, you can ask yourself, why is it not inconsistent? Well, we don't know. We haven't found the inconsistency and the contradiction yet. Some people claim that they found it, from time to time anyway, but they are often wrong and misunderstand subtle point which they intend to exploit in their proof. This works in our favor, so to speak, because it shows that we cannot find the contradiction in a theory: it might actually be consistent after all.

Alas, much like many of the mysteries of life: this one will remain open for us to believe whether what we hear is true or false, whether the theory is consistent or not.

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Thanks for your very comprehensible answer! - I think I might imagine where this is headed. – miku Aug 24 '12 at 22:18
@Ben: Tarski could not have proved his first-order geometry consistent without using some foundation for his meta-level work. And since that foundation was, evidently, able to speak about consistency at all, it must have been stronger than the theory he proved to be consistent. The very fact that Tarski's geometry cannot even express arithmetic means that the metatheory he used must have been stronger, as far as my intuitive understanding of that word goes. – Henning Makholm Aug 25 '12 at 12:26
@HenningMakholm A metatheory that talks about the consistency of a theory $T$ surely needn't be outright stronger than $T$. For example, Gentzen's proof of PA's consistency is informative precisely because the relevant meta-theory's assumptions are weaker in some respects and stronger in others that those of PA. – Peter Smith Aug 25 '12 at 14:08
@PeterSmith: Not for arbitary $T$'s, but I do think a metatheory that talks about consistency has to be stronger (in some appropriately fuzzy sense) than Tarski's first-order axiomatization of geometry in particular. – Henning Makholm Aug 25 '12 at 14:10
@BenCrowell mathematical logic provides us with a number of different, but mathematically precise ways of calculating whether a theory $T$ is stronger than a theory $S$. They include consistency strength, (several flavours of) interpretability, reverse mathematical strength and the height of the respective proof theoretic ordinals. These different ways of cashing out the notion of a stronger theory often give different answers, so the question of which is the appropriate notion will depend on the circumstances. – Benedict Eastaugh Jan 6 at 12:18

Mathematicians believe in "the" induction principle for natural numbers not because of set theory or PA or their consistency, they believe in "the" induction principle because of their intuition about natural numbers. Axioms are means for expressing these intuitions.

Mathematicians believe that PA is consistent and in fact sound because it satisfies the standard model of natural numbers (0 and its successors). If you believe in natural numbers then the induction principle would follow automatically for all well defined "properties".

One should be careful here because "the induction principle" is used for various things, e.g. the informal intuition that

if a "property" $P$

• holds for 0, and
• if it holds for natural number $n$, then it holds for its successor $n+1$,

then it holds for all natural numbers.

Here "property" is an informal concept.

Induction is also used to refer to various formal axioms that try to capture this informal intuition e.g. the first-order induction axiom in the theory of arithmetic, the second-order induction axiom of Peano, etc.

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I would like to inform that commonly with Dr Teodor J. Stepien, I delivered a talk at the Conference "2009 European Summer Meeting of Association for Symbolic Logic, Logic Colloquium'09" (July 31 - August 5, 2009, Sofia, Bulgaria). In this talk, entitled "On consistency of Peano's Arithmetic System", we presented a sketch of the proof of the consistency of Peano's Arithmetic System (of course, the full proof was constructed by us before the mentioned Conference "Logic Colloquium 2009"). This proof is ABSOLUTELY ELEMENTARY, i.e. there are used ONLY the axioms of first-order logic and the axioms of Peano's Arithmetic System. Hence, from the construction of this proof, it follows that Gödel's Second Incompleteness Theorem is INVALID. The asbtract of this talk was published in "The Bulletin of Symbolic Logic": T. J. Stepien and L. T. Stepien, Bull. Symb. Logic 16, 132 (2010). It is accessible under the following link http://www.math.ucla.edu/~asl/bsl/1601-toc.htm and after clicking on: "2009 European Summer Meeting of the Association for Symbolic Logic, Logic Colloquium '09, Sofia, Bulgaria, July 31—August 5, 2009" (page 132).

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I have no plans to vote on this answer, but I think a comment is warranted to provide context. Assuming that the second incompleteness theorem is correct, as is commonly accepted, there must be an error in the proof that is claimed by Stepien and Stepien. As I understand, the Association for Symbolic Logic permits any member in good standing to contribute a talk to any meeting; there is no peer review of the contents of these talks. I am not aware of a published proof of the results of Stepien and Stepien in a peer reviewed journal. – Carl Mummert Sep 26 '12 at 16:05
Part I: At first, cit. "[...] the second incompleteness theorem is correct, as is commonly accepted [...]". Then, it suffices to see the papers: 1. Barzin, M., "Sur la Portee du Theoreme de M. Godel", Academie Royale de Belgique, Bulletin de la Classe des Sciences, Series 5, 26, pp. 230-239, (1940), – Lukasz Sep 27 '12 at 13:10
@Carl Mummert . From Lukasz: Part I. At first,cit."[...]the second incompleteness theorem is correct, as is commonly accepted[...]". Then, it suffices to see the papers: Barzin, M., Academie Royale de Belgique, Bulletin de la Classe des Sciences, Series 5, 26, 230,(1940); Kuczynski, J., Kwartalnik Filozoficzny 14, 74,(1938)-in Polish; Perelman, Ch., Academie Royale de Belgique, Bulletin de la Classe des Sciences, Series 5, 22, 730,(1936); Cattabriga, P.,"OBSERVATIONS CONCERNING GODEL'S 1931",arXiv:math/0306038,2009; Saa, D.,"GODEL'S THEOREM IS INVALID",arXiv:math/0510469,2005. – Lukasz Sep 27 '12 at 13:30
@Carl Mummert . From Lukasz: Part II. We have NOT investigated the proof of Gödel's Second Incompleteness Theorem, but we have proved the consistency of Peano's Arithmetic System, by using only the axioms of first-order logic and the axioms of Peano's Arithmetic System. Hence, the invalidity of Gödel's Second Incompleteness Theorem follows from the construction of our proof. – Lukasz Sep 27 '12 at 13:31
@Lukasz: It doesn*t seem tha ta puported prrof of consistency of PA (using only FOL and PA) show sthat Gödel's Second Incompleteness Theorem is invalid. Rather, both results together (would) show that PA is inconsistent. – Hagen von Eitzen Oct 17 '12 at 17:39
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Peano's "symmetric axiom of equality" states that if $a=b$, then $b=a$. From that, the "substitution axiom of equality", which states that we can always substitute $\left(\frac{a}{a}\right)$ for $\left(\frac{b}{b}\right)$, immediately follows. However, given $a\in\{1,2,3...\}$ and $b\in\{2,3,4...\}$, if $\left(\frac{a}{a}\right)\cdot a^{3}=\left(\frac{b}{b}\right)\cdot a^{3}$ where $a=b$, and the properties of logarithms allow the identity

$$\left(\frac{b}{b}\right)\cdot a^{3}=b\cdot \left(\frac{a}{b}\right)^{\frac{\frac{3\cdot \ln(a)}{\ln(b)}-1}{\frac{\ln(a)}{\ln(b)}-1}}$$ where $b\not=a$ and $\left(\frac{a}{a}\right)$ can't be substituted for $\left(\frac{b}{b}\right)$, then clearly, Peano's "symmetric and substitution axioms of equality" are in actuality neither consistent nor even true! Thus, Hilberts second problem, which asks for a proof that Peano's axioms are consistent, has been answered in the negative by a proof that Peano's axioms are inconsistent. Shockingly enough, these are by no means the only axioms that are inconsistent with these newly discovered properties of logarithms. There are many other such long held "notions" in mathematics that will also be found quite incompatible with the properties of logarithms, and since logarithms are simply a logical extention of exponentiation and cannot be dispensed with, constructs such as the one above must ultimately dictate the genuine foundations of mathematics. Lots to ponder! Sincerely, Don Blazys.

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What is your question exactly? – Hagen von Eitzen Jan 6 at 11:33
The above post is not a question. It's an answer. – Don Blazys Jan 6 at 12:54
Ah, sorry. I stumbled across this post while reviewing and didn't notice it was in fact intended as answer. So I should rather rephrase: What is your claim? The display equation is not valid in case $a=b$ because a zero denominator occurs. – Hagen von Eitzen Jan 6 at 13:48
It's obviously the right answer to Hilbert's second problem because clearly, we can't let a=b or substitute (a/a) for (b/b) like Peano's axioms say we can without also compromising the natural domains of the positive integer variables or in some other way sacrificing logic. – Don Blazys Jan 6 at 13:52
How does this form a contradiction? I only see one $\left(\frac bb\right)$ and if it is replaced by $\left(\frac aa\right)$, the result is fine. – robjohn Jan 6 at 20:39
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