Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

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.

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.

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?

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?

Any help would be greatly appreciated. Thank You in Advance.

share|improve this question
4  
The Shepherdson models are what you're looking for! See my old answers mathoverflow.net/questions/21367/… and mathoverflow.net/questions/19857/… –  François G. Dorais Feb 6 '11 at 15:36
    
I have to admit, I can't understand a lot of what you wrote (e.g. I have no idea what T is).I also looked up Puiseux series, but the Wikipedia article went above my head. So let me just ask you if this model meets my conditions: 1.Is S well-ordered? 2.Can you construct a real closed field out of S^2? 3.Is there an isomorphism between S and points of the form (a,1) in this real closed field? 4.Would this real closed field obey the Least Upper Bound axiom of the real numbers? If the answers are No, Yes, Yes, No, I'm good to go! I also want to know if it's a computable, but it's fine if it's not. –  Keshav Srinivasan Feb 6 '11 at 19:48
    
François, why not post your comment as an answer? –  Joel David Hamkins Feb 6 '11 at 22:18
    
@Joel: Done. (I was in a hurry earlier today.) –  François G. Dorais Feb 6 '11 at 23:46

1 Answer 1

up vote 10 down vote accepted

First note that the axioms of Robinson Arithmetic 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.

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 (A non-standard model for a free variable fragment of number theory, MR161798) 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 here and here. 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.

share|improve this answer
    
You said "if my memory serves correctly there is a variant of the construction does do that." If such a variant of Shepherdson's model does indeed exist, would it be there in the paper you cited? I don't have access to MathSciNet, so is there any other way I could read it? Also, what does this model actually look like? I assume it contains the standard model of arithmetic as its initial segment (or is that just a result for models of PA?). What is the order type of the model? And this may be really trivial, but what's the difference between open induction and primitive recursive arithmetic? –  Keshav Srinivasan Feb 7 '11 at 21:18
    
I don't think Shepherdson does the more general construction in his paper. You can probably find the paper in a library. Yes, the model does contain the standard model as an initial segment. The order type of the model is $\omega+\eta\zeta$. Primitive recursion is much stronger than open induction. –  François G. Dorais Feb 8 '11 at 10:49
    
Where would I find the more general construction? Because having a real closed field is critical to what I'm trying to do. I know that eta is the order type of the rational number, but what is zeta? Finally, why is primitive recursion stronger than open induction? They both only allow induction on quantified-free formulas. –  Keshav Srinivasan Feb 8 '11 at 14:57
    
I couldn't find a reference for the construction, but I'll keep looking. I use $\zeta$ for the order type of $\mathbb{Z}$. PRA is much stronger because has a lot more functions than just addition and multiplication. –  François G. Dorais Feb 9 '11 at 12:32
    
François, were you able to find the construction? –  Keshav Srinivasan Feb 16 '11 at 4:15

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.