most recent 30 from http://mathoverflow.net 2013-05-19T21:38:02Z http://mathoverflow.net/feeds/question/78300 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/78300/extension-of-the-peano-axioms Lawrence Wickert 2011-10-17T00:54:59Z 2011-10-17T13:09:55Z

<p>The Peano Axioms (partially) formalize our intuitive notion of arithmetic. Partially because they also describe the behaviour of nonstandard models and there are some theorems that they can not prove that might seem, at first sight, to be within their domain e.g. Goodstein's Theorem and some of Harvey Friedman's combinatorial theorems <a href="http://arxiv.org/abs/math/9811187" rel="nofollow">http://arxiv.org/abs/math/9811187</a>. So there is room for some clever soul to find additional axioms which would be independent of PA and also convey a natural intuitive property of the integers. These additional axioms would prove new theorems and also restrict the class of nonstandard models.</p> <p>Could such a process ever be complete in the sense that all nonstandard models would be excluded?</p> <p>PS I read the FAQ and am not sure if this is a suitable question. My apologies if it is not.</p> <p>Larry Wickert, Truth or Consequences, New Mexico</p>

http://mathoverflow.net/questions/78300/extension-of-the-peano-axioms/78301#78301 François G. Dorais 2011-10-17T01:17:02Z 2011-10-17T01:17:02Z

<p>There is no hope for a <em>first-order</em> theory to eliminate non-standard models. If a first-order theory over a countable language has an infinite model then it has models of all infinite cardinalities (<a href="http://en.wikipedia.org/wiki/L%C3%B6wenheim%E2%80%93Skolem_theorem" rel="nofollow">Löwenheim-Skolem Theorem</a>). In the case of a theory $T$ of arithmetic. If the standard model satisfies $T$, then $T$ must have an uncountable model, which has to be non-standard (though there will be countable non-standard models of $T$ as well).</p> <p>If you formulate PA in <a href="http://en.wikipedia.org/wiki/Second-order_logic" rel="nofollow">second-order logic</a>, then it is already complete with respect to the standard semantics.</p>

http://mathoverflow.net/questions/78300/extension-of-the-peano-axioms/78303#78303 Jeff Burdges 2011-10-17T02:03:33Z 2011-10-17T02:03:33Z

<p>Peano's axioms are categorical if you take the second order formulation, i.e. no non-standard models. If you replace the second order axiom by a first order axiom scheme, they cease being categorical by the compactness theorem, as Francois stated. If you embed the second order version into a first-order model of set they, they remain categorical relative to the set theory, but the set theory itself has non-standard models.</p> <p>Non-standard models are useful because they help us formalize some features of infinitary processes, i.e. non-standard analysis, geometric group theory, etc. In practice, model theorists would usually have way more structure than simply Peano's axioms, such as a real closed field or hyperbolic group. </p>

http://mathoverflow.net/questions/78300/extension-of-the-peano-axioms/78333#78333 Richard Rast 2011-10-17T13:09:55Z 2011-10-17T13:09:55Z

<p>There is a stronger result than the ones stated. As noted above, by compactness (and the L\"owenheim-Skolem theorems) we have a lot of nonstandard models.</p> <p>However, by G\"odel's First Incompleteness Theorem, we have that no recursive (consistent) extension of PA can be complete. So, no finite set of axioms (or even axiom schemes, like induction) can ever even nail down the complete theory of arithmetic. Not only will there be nonstandard models, but there will be models which satisfy statements which are not "true."</p>