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I have seen in many textbooks on analysis that the Archimedean property of reals is a consequence of the completeness axiom. However I am not convinced that we need to use such a powerful axiom (as the completeness axiom) to prove a very basic property like Archimedean Property. To me it looks simple enough as follows:

1) If a and b are natural numbers (like 1, 2, 3, ..., but not 0) and there is another natural number c such that b = a + c, then we write b > a. Using this we can see that given any natural number we can find another greater than it.

2) If a & b are positive rationals then a/b is rational and we can write a/b = p/q (p, q positive integers) and by division algorithm we have a non-negative integer p = qx + r with 0 <= r < q. Then clearly we have a positive integer (x + 1) > p/q = a/b. So that field of rationals possesses the Archimedean property.

3) If a, b are positive reals then a/b is also real. Any definition of real numbers (Dedekind's or Cauchy's for example) will lead to the fact that given a real number there is a rational greater than it and a rational less than it. So we have rational c > a/b. And clearly by Archimedean Property of rationals (point 2 above) we have a positive integer 'n' greater than 'c'. Thus we finally have a/b < n or a < nb.

I am really not sure why textbooks try to prove Archimedean property via completeness axiom. If it was really the case Archimedean property is a consequence of completeness property, then how come rational possess Archimedean property but not completeness axiom.

Please clarify if there is any mistake in above reasoning or the textbooks are treating this in high handed manner.

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closed as off-topic by Ricardo Andrade, Andrés E. Caicedo, Ramiro de la Vega, Andrey Rekalo, Steven Landsburg Oct 4 '13 at 12:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Ricardo Andrade, Andrés E. Caicedo, Ramiro de la Vega, Andrey Rekalo, Steven Landsburg
If this question can be reworded to fit the rules in the help center, please edit the question.

If you give a concrete definition of the reals (e.g. using Dedekind cuts) then you can prove that they have all the properties they have just from the definition. If you don't want to give a concrete definition, you may instead start with a collection of axioms that completely characterize the reals and then prove the other properties from this axioms. – Ramiro de la Vega Aug 26 '12 at 11:37
C(ompleteness) implies A(rchimedean property) but A does not imply C. It happens... What is your question? – Wadim Zudilin Aug 26 '12 at 11:38
@Qiaochu: Any Cauchy sequence of rationals (i.e. any real number) is bounded by a rational and therefore has a constant rational sequence (i.e. a rational number) above it. – Ramiro de la Vega Aug 26 '12 at 11:44
@Ramiro: ah, right. My apologies. It's late where I am. – Qiaochu Yuan Aug 26 '12 at 11:54
up vote 13 down vote accepted

One cannot prove the Archimedean property for the reals by appealing only to first-order algebraic truths of the ordered real field and the subring of the integers sitting inside it. The reason is that those statements are all first-order expressible in the structure $\langle\mathbb{R},{+},{\cdot},0,1,\lt,\mathbb{Z}\rangle$, and there are nonstandard models $\langle\mathbb{R}^\ast,{+},{\cdot},0,1,\lt,\mathbb{Z}^\ast\rangle$, which are not Archimedean and yet which satisfy exactly the same first-order truths as the standard model. In this sense, there cannot be a truly elementary proof of the Archimedean property.

What this observation shows is that the Archimedean property is fundamentally a second-order property of the reals, and to establish it one must appeal to the fact that one is using the actual standard integers instead of merely some first-order property of the integers and how they relate to the reals.

Your argument, which is fine, does this by appealing to the explicit constructions of the reals, say, as Dedekind cuts, rather than to its characterization by the completeness property.

So while I agree that there are other proofs of the Archimedean property that don't appeal to completeness---and as you point out there certainly are incomplete Archimedean ordered fields---nevertheless I don't mind the proof from completeness, since this is a defining property characterizing the reals, and the proof from completeness is both easy and clear.

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I'm confused. If $\mathbb{Z}$ is a symbol in your language, then the Archimedean property is first order: $\forall x \in \mathbb{R} \exists n \in \mathbb{Z} : n > x$. – David Speyer Aug 26 '12 at 12:55
In the structure I mentioned, $\mathbb{Z}$ is a predicate on the reals. And although the standard model has the standard interpretation for this predicate, a nonstandard model does not. In particular, a non-standard model thinks the statement you make is true, but still it is not Archimedean, because it has a non-standard interpretation of the predicate. And since that nonstandard model still has all the same first-order assertions being true about this interpretation, one cannot find a proof of the Archimedean property appealing only to those assertions. – Joel David Hamkins Aug 26 '12 at 13:11

As Ramiro says, if you know that reals are well approximable by rationals, or even if you know that every real is within $1$ of an integer, then you can prove the Archimedean property immediately. From most definitions of reals, this is easy to check. The issue comes up when you try to axiomatize the reals.

If all you know is that $\mathbb{R}$ is an ordered field which has a subring isoromorphic to the integers, then you can't deduce that $\mathbb{R}$ is Archimedean. Let $K$ be the field $\mathbb{R}(x)$, equipped with the ordering that $p(x) \prec q(x)$ if $p(t) < q(t)$ for $t$ a sufficiently large real. Then there is no integer larger than $x$.

So, if you want to prove the Archimedean property, you need some other axiom to do it, and completeness is a reasonable choice. Another reasonable choice is to impose the Archimedean property as an axiom and impose, in addition, that every Cauchy sequence converges. But (Dedekind) completeness is a reasonable choice, because then one axiom accomplishes both goals.

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+1. Or you can use $K=\mathbb{R}^\ast$, which not only has a subring isomorphic to $\mathbb{Z}$, but satisfies exactly the same first-order theory as $\mathbb{R}$ as an ordered field, with any extra algebraic structure you like, such as exponential function, etc. – Joel David Hamkins Aug 26 '12 at 13:16

This is an amplification, from a somewhat different point of view, of something that David Speyer briefly mentioned in his answer. When people put the study of some mathematical structure on an axiomatic basis, several considerations influence the choice of axioms. First, and most important, the axioms should be true for the intended structure and should suffice to imply the other known facts about the structure. (For example, Euclid needed the parallel postulate in order to deduce numerous known facts of geometry.) Second, the axioms should be reasonably simple to state and their truth should be reasonably evident. (This is why we usually think of the parallel postulate as asserting the existence and uniqueness of parallels, rather than Euclid's version about the sum of angles formed by a transversal of two lines.) Third, there should not be (easily seen) redundancies in the axioms. (After all, one purpose of axiomatization is to show how the whole subject can be deduced from a small number of simple assumptions, so we don't want extraneous assumptions.) In the case of the ordered field of real numbers, the first two considerations might very well lead (and perhaps did historically lead --- I don't know about the history) to an axiomatization consisting of the axioms for ordered fields plus the Archimedian property, plus some sort of completeness. But when one sees that the Archimedian property is easily deducible from (suitable versions of) completeness, the third consideration applies and leads one to remove the Archimedian property from the list of axioms and make it a theorem instead.

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Completeness is a sufficient condition for the Archimedean property, but as everyone, including the poser has observed, it is hardly necessary. I suspect that the real reason people want that complete implies Archimedean is to be able to prove the theorem that, up to isomorphism, the reals are the only complete ordered field. I think that is what Andreas was saying.

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When one is teaching, one is not only teaching results, one is also developing theory, i.e. a language in which to embed those results. When doing this, there is a high value put on parsimony. Therefore, if one can get a result using a basic property like completeness instead of approaching it some other way, then one does that. A teacher has to at some point trust the student to ferret out the rest of the truths of the universe for himself.

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