Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Do we know if the axioms of the real numbers are consistent, complete or neither of both?

And if so, is it a consequence of Godel's theorem or of something else?

share|improve this question
    
I'm quite sure the axioms (complete field with Archimedean order) are complete as any two such fields are canonically isomorphic. –  Hagen von Eitzen Apr 15 '13 at 22:21
    
In its essence the real numbers are consistently built from the natural numbers. The axioms you mention are just a shortcut to the proper way of doing things. –  Git Gud Apr 15 '13 at 22:21
2  
The theory of real-closed fields is even decidable, there is a computer program that will answer any question. –  André Nicolas Apr 15 '13 at 22:29

1 Answer 1

up vote 18 down vote accepted

The real numbers can be axiomatized in many different ways. I will focus on the main interpretation of a first-order theory of an ordered field. I will remark on other alternatives at the end of my answer.

Yes, the theory of the real numbers also known as the theory of real closed fields (RCF) is complete. In fact we can write a recursively enumerable theory which is already enough to prove everything which is true in that structure.

This theory is also consistent, because the real numbers exhibit a model of that theory, and a theory in first-order logic is consistent if and only if it has a [set] model. The real numbers form a set, and therefore witness that the theory is consistent. This is in fact a consequence of Godel's completeness theorem.

However if we combine both these we can get a very nice result about the real numbers, that they are not a structure in which we can develop the basic laws of arithmetics of the integers. Indeed the integers themselves cannot be defined as a subset of the real numbers in the first-order theory of real-closed fields.

Of course, one can discuss alternative methods of characterizing the real numbers. Perhaps in second-order logic. Perhaps as a particular structure of a particular language (e.g. exponential field; an ordered field with extra predicates for the integers, and so on). In different languages, or different logics, we will be able to write different axioms, and the theory of the real numbers will end up being very different.

In any case, however, if we limit ourselves to first-order logic (where the completeness theorem holds), if we consider "the theory of the real numbers" simply as the set of all sentences which are true in the structure of the real numbers, then the answers are both yes: it is a complete theory and it is consistent. This is because every "theory of a structure" is a complete and consistent theory.

The actual, and interesting, question is whether or not there is a "nice" theory which is sufficient to prove everything which is true -- as in the case of the ordered field, where RCF is enough to prove everything else. And of course the answer here will vary depending on the language we chose and so on.


Couple of words on the incompleteness theorem, while mainly known to the laymathematician as "a theory cannot be both consistent and complete", or even worse "a consistent mathematical theory cannot prove everything", this is not what the theorem actually says.

The theorem states that if we have a theory in first-order logic whose axioms can be recognized by a computer program (namely, a program will halt when it is given an axiom), and that theory is sufficient to express basic arithmetical statements, then the theory cannot be both consistent and complete.

However there are plenty of mathematical theories which are consistent and complete. These theories fail to satisfy the two assumptions needed for the incompleteness theorem to kick in. For example consider the theory made of all the sentences which are true in the natural numbers. This theory is complete, because every sentence is either true or false in a fixed structure; and this theory is consistent because it has a model (the natural numbers). But this theory is not recursively enumerable, there is no computer program which will be able to recognize the axioms.

On the other end, the theory of real-closed fields, and a closely related theory of algebraically closed fields in a fixed characteristics, both are consistent and complete, and both are recursively enumerable.

It is a very nice consequence of the incompleteness theorem in the case of these two theories, that the integers cannot be defined within $\Bbb R$ or within $\Bbb C$. If one follows into deeper model theoretical territories then one can show that in fact both these models have very little subsets that they can actually define.

But it is very important to remember that the incompleteness theorem is nothing more than what it is. And it is certainly not the statement that no theory is both consistent and complete.


Also related:

share|improve this answer
2  
@Myke: The incompleteness theorem has four conditions: (1) consistency; (2) completeness; (3) recursively enumerable; and (4) can express basic arithmetical statements. The incompleteness theorem tells us that a theory cannot have all four properties. In the case of RCF the fourth property fails. –  Asaf Karagila Apr 15 '13 at 22:56
4  
@Myke: To clarify, Asaf means basic integer arithmetical statements. While RCF can, for example, tell you about solutions to the polynomial equation $x^3 + y^3 = z^3$ -- and there are lots, such as $(1, 1, \sqrt[3]{2})$ -- it cannot even formulate the question of whether any of those solutions are integer solutions. –  Hurkyl Apr 15 '13 at 23:11
1  
@Myke: You are using a lot more than just ordered field property in calculus. Often you don't even use a first-order theory at all. –  Asaf Karagila Apr 16 '13 at 8:54
1  
@Myke: That would depend on the axioms themselves. Generally if a theory has a model then it is consistent. The real numbers are a model of the axioms, so the theory is consistent. If you are considering the second-order axiomatization of a complete ordered field, then it is in fact the only model (up to isomorphism) of these axioms; however second-order logic does not have a completeness theorem, so you cannot conclude that the theory is complete (although these are two different kinds of "complete"). –  Asaf Karagila Apr 16 '13 at 11:28
1  
@Myke: Yes. The fact that we use second-order logic means that we can no longer conclude that having a unique model implies completeness. Ordinary calculus is consistent (relative to one foundation of mathematics or another) because we construct it mainly by hands. We then prove general theorems, but they apply to our constructions in particular. –  Asaf Karagila Apr 16 '13 at 11:56

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.