Fulfilling Pythagoras' Dream using Nonstandard Models of Arithmetic and/or Surreal Numbers - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T22:03:37Z http://mathoverflow.net/feeds/question/54526 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/54526/fulfilling-pythagoras-dream-using-nonstandard-models-of-arithmetic-and-or-surrea Fulfilling Pythagoras' Dream using Nonstandard Models of Arithmetic and/or Surreal Numbers Keshav Srinivasan 2011-02-06T15:24:51Z 2011-02-16T04:27:01Z <p>Pythagoras and his followers believed that the Universe was made of numbers. Specifically, they thought that if you compared any magnitudes of the same kind, say the lengths of two objects, you would always get a whole number ratio. Then someone came up with a proof that the side length of a square and its diagonal are not in a whole number ratio, thereby demonstrating that space could not be completely described by Number. This seemed catastrophic for the Pythagoreans, but then mathematicians eventually grew to accept "irrational" numbers, and the rest is history.</p> <p>But I'm curious how things would work out if Pythagoras and his peers had chosen to take a different route, one in which they didn't have to give up their dream of directly characterizing all magnitudes using whole numbers. The standard proof of sqrt(2) being irrational goes like this: suppose that sqrt(2)=a/b, where a and b contain no common factors. Then we show that a and b must have a common factor of 2. But what if the fraction a/b could NOT be written in simplest terms, but instead you could keep dividing a and b by 2 and never get them to be coprime? Then you'd be able to construct an infinite descending sequence of natural numbers, which contradicts the well-ordering principle/mathematical induction.</p> <p>So I'm trying to find a nonstandard model of arithmetic in which, among other things, sqrt(2) is rational. Countable nonstandard models of PA are not well-ordered, which is what I want, but first order induction is still sufficient to prove that sqrt(2) is irrational. So we would have to find nonstandard models of a weaker system, like Robinson's Q. Q also has the advantage of having computable nonstandard models. So I want to find a nonstandard model of Q, let's call it S, and then construct the real numbers using ordered pairs of elements of S. But the least upper bound axiom of the real numbers is actually sufficient to prove second-order induction for the natural numbers. So the system of real numbers I'm constructing should not satisfy LUB; it should only satisfy the first-order properties of the real numbers. </p> <p>So to sum up: I want a nonstandard model S of Robinson Arithmetic which is not well-ordered, and such that we can define operations on S^2 which would make S^2 a nonstandard model of the first-order theory of real closed fields. (EDIT: As François said, a simpler way to say this is that I want the field of fractions of S to be real closed.) Is that even possible, and if so would S even be a computable nonstandard model?</p> <p>I've been thinking about this problem a while, and I think Conway's Surreal Numbers might be promising. The analogue of the integers for Surreal numbers are called Omnific integers. Omnific integers are not well-ordered, and it turns out that you can find Omnific integers a and b such that a^2=2b^2, which are both hopeful signs. And even better, the set of ratios of Omnific integers actually satisfies the first-order theory of real closed fields! If Omnfic integers really do satisfy all the properties I want, then I have a few questions about them. Do they constitute a nonstandard model of Robinson arithmetic? Does there exist an axiomatization of them in first order logic? In other words, can we define them alone, without resorting to the full-blown definition (which involves set theory) of the Surreal numbers? On a related note, can they even be put in a set, or do they form a proper class? I think the latter may be true, in which case I may have to find some subclass of the Omnific integers which is an actual set (because I think it may not be legitimate to take the Cartesian product of proper classes). But in that case, will that subclass still satisfy all the properties I want?</p> <p>Any help would be greatly appreciated. Thank You in Advance.</p> http://mathoverflow.net/questions/54526/fulfilling-pythagoras-dream-using-nonstandard-models-of-arithmetic-and-or-surrea/54577#54577 Answer by François G. Dorais for Fulfilling Pythagoras' Dream using Nonstandard Models of Arithmetic and/or Surreal Numbers François G. Dorais 2011-02-06T23:45:29Z 2011-02-06T23:45:29Z <p>First note that the axioms of <a href="http://en.wikipedia.org/wiki/Robinson_arithmetic" rel="nofollow">Robinson Arithmetic</a> are usually straightforward to verify in any structure. Surely the nonnegative omnific integers satisfy these basic axioms. Note that a nonstandard model of Robinson Arithmetic cannot be wellordered because of the axiom which says that every nonzero element has an immediate predecessor.</p> <p>You can push a little farther than Robinson Arithmetic and still have recursive models. Open induction is the statement that induction holds for quantifier-free formulas (in the language of Robinson Arithmetic). Shepherdson (<em>A non-standard model for a free variable fragment of number theory</em>, <a href="http://www.ams.org/mathscinet-getitem?mr=161798" rel="nofollow">MR161798</a>) has shown that there are recursive models of open induction too. Furthermore, he showed that one can get such models where $\sqrt2$ is rational, for example. I described Shepherdson's construction in my previous answers <a href="http://mathoverflow.net/questions/21367/proof-that-pi-is-transcendental-that-doesnt-use-the-infinitude-of-primes/21389#21389" rel="nofollow">here</a> and <a href="http://mathoverflow.net/questions/19857/has-decidability-got-something-to-do-with-primes/20944#20944" rel="nofollow">here</a>. In this construction, the field of fractions of the model is not necessarily real closed, but if my memory serves correctly there is a variant of the construction does do that.</p>