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It's possible to construct $\mathbb{Q}$ from $\mathbb{Z}$ by constructing $\mathbb{Z}$'s field of fractions, and it's possible to construct $\mathbb{C}$ from $\mathbb{R}$ by adjoining $\sqrt{-1}$ to $\mathbb{R}$.

In both cases, the construction is done purely algebraically. I.e. we only rely on the operations of our given structure to build the new structure. But at no point do we have to rely on the order properties of $\mathbb{Z}$ or $\mathbb{R}$ to get to $\mathbb{Q}$ or $\mathbb{C}$.

Every construction of $\mathbb{R}$ that I'm familiar with ultimately comes down to endowing $\mathbb{Q}$ with its usual order, and then imposing the completeness axiom on it to recover the rest of the real numbers.

Is it possible to get to $\mathbb{R}$ from $\mathbb{Q}$ without relying on the ordering properties of $\mathbb{Q}$?

Alternatively (relatedly?): There is the notion of a greatest common divisor for an arbitrary ring. This notion doesn't rely on any ordering properties; just algebraic ones. Is it possible to recover an order relation on $\mathbb{Q}$ using the GCD relation on $\mathbb{Z}$, then to impose completeness on $\mathbb{Q}$ and obtain $\mathbb{R}$, and then subsequently re-cast completeness in some algebraic manner? Thus defining $\mathbb{R}$ in purely algebraic terms?

Bears
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    Sounds unlikely to me. There are real-closed fields other than $\Bbb R$ and I cannot see how one would distinguish them from $\Bbb R$ "algebraically". – Angina Seng Jul 18 '20 at 23:40
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    equivalence classes of rational Cauchy sequences? – janmarqz Jul 19 '20 at 00:38
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    Cauchy sequences are not "algebraic". But one can use the fact that complex field is the algebraic clousure of the transcendental extension of the rationals of degree continuum. The tanscendental extension is simply the field of rational functions in continuum variables. And the algebraic clousure is certainly an algebraic operation. Now it only remains to algebraically extract reals from the field of complex numbers. Leaving as an excercise. – markvs Jul 19 '20 at 00:44

2 Answers2

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The "order vs. algebra" issue is really a red herring here: in each of the structures $\mathbb{Z},\mathbb{Q},\mathbb{R}$, the order can in fact be recovered from the algebra alone!

  • In $\mathbb{R}$ we have $a\ge b$ iff there is some $c$ such that $c^2+b=a$.

  • In $\mathbb{Z}$ we have that $a\ge b$ iff there are $w,x,y,z$ such that $w^2+x^2+y^2+z^2+b=a$ (via Legendre).

  • In $\mathbb{Q}$ we first use the definability of $\mathbb{Z}$ inside $\mathbb{Q}$ (which is quite nontrivial). From that define the nonnegative integers as those which can be written as the sum of the squares of four integers, and then observe that $a\ge b$ iff for some positive integer $c$ the product $c(a-b)$ is a nonnegative integer. (Actually I'm pretty sure there's an easier way to algebraically define the ordering on $\mathbb{Q}$, but meh.)

Each of the definitions above is a definition in the sense of first-order logic; I'm ignoring the technicalities here, but the term is worth mentioning. Interestingly, $\mathbb{R}$ - despite its mathematical complexity in many senses - is actually quite simple from the logical perspective, and for instance neither $\mathbb{Z}$ nor $\mathbb{Q}$ are definable in $\mathbb{R}$. Bigger $\not=$ more structurally complicated!

The real issue is a "sets vs. objects" issue: in each of the constructions $\mathbb{Z}\leadsto\mathbb{Q}$ and $\mathbb{R}\leadsto\mathbb{C}$ we basically have that members of the new structure correspond to "simple combinations" of members of the old structure (e.g. appropriate ordered pairs perhaps modulo an appropriate equivalence relation), whereas in the construction $\mathbb{Q}\leadsto\mathbb{R}$ something weirder happens - objects in the new structure are "one type higher" than objects in the old structure. This is unavoidable on pure cardinality grounds: there are more reals than there are finite tuples of rationals. And this cardinality obstacle turns into a serious logical distinction via the downwards Lowenheim-Skolem theorem, which implies that there is no way to build $\mathbb{R}$ from $\mathbb{Q}$ via the machinery of first-order logic alone.

So there is indeed something essentially new about the construction $\mathbb{Q}\leadsto\mathbb{R}$, but it's not really about the ordering per se - it's more subtle than that. Rather, it's about the more general fact that (topological) completeness of any kind is fundamentally about sets/sequences rather than individual (or finite tuples of) elements of the structure.

Noah Schweber
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    @JCAA "To define an order on rationals in a way you propose you need first to define integers inside rationals. How are you going to do that?" The integers are first-order definable inside the rationals, as the link in that bulletpoint demonstrates! So I've explicitly addressed this. – Noah Schweber Jul 19 '20 at 02:00
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    @JCAA "elementary embeddings of rationals into reals are not possible. " Of course not - so what? I have no idea what elementary embeddings have to do with any of this. No two of the structures under consideration are elementarily equivalent to each other - so what? At no point have I mentioned elementary embeddings. I think you may not have carefully read my answer. – Noah Schweber Jul 19 '20 at 02:01
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    @JCAA " what does it have to do with rhe question?" The question implicitly assumes that what makes the $\mathbb{Q}\leadsto\mathbb{R}$-construction fundamentally different from the $\mathbb{Z}\leadsto\mathbb{Q}$- and $\mathbb{R}\leadsto\mathbb{C}$-constructions is the use of ordering. The point of my answer is that this is not accurate, since in each relevant case the ordering is in fact purely algebraically definable. The real specialness is the role of sets vs. (finite tuples of) elements. I thought my answer was pretty clear about this. – Noah Schweber Jul 19 '20 at 02:06
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    @JCAA "As I understand, the question is whether there is an algebraic way to define the field of real numbers." And as I've shown in my answer, there's a framing issue there: invoking the ordering does not make something non-algebraic, since the ordering itself is algebraically definable. (Did you look at the link above re: defining $\mathbb{Z}$ in $\mathbb{Q}$?) "Dedekind cuts are less "algebraic" than Cauchy sequences." Great - so? Each approach still falls prey to the "sets vs. elements" issue. That's the real non-algebraicity. – Noah Schweber Jul 19 '20 at 02:10
  • @JCAA Great - so? That doesn't contradict anything I've said. Using transcendence degree still requires us to talk about higher cardinalities, so we still get the "type issue." (I'm really not sure which part of my answer you're disagreeing with.) – Noah Schweber Jul 19 '20 at 02:16
  • @JCAA "What type issues?" In technical language, both the $\mathbb{Z}\leadsto\mathbb{Q}$ and $\mathbb{R}\leadsto\mathbb{C}$ constructions are examples of interpretations in the sense of first-order logic. However, for cardinality reasons alone $\mathbb{Q}$ can't interpret $\mathbb{R}$; and Lowenheim-Skolem rules out even more complicated ways of building $\mathbb{R}$ from $\mathbb{Q}$ in a "first-order way" (which I'm not going to go into here for length reasons). If you want to build $\mathbb{R}$ from $\mathbb{Q}$, you need to bring in something morally equivalent to sets of integers. – Noah Schweber Jul 19 '20 at 02:20
  • @JCAA Actually, to set my mind at rest let's go back to the beginning of your initial comment for a second. Do you accept my claim that in each of $\mathbb{Z},\mathbb{Q}$, and $\mathbb{R}$ the standard ordering is definable from the ring structure alone? – Noah Schweber Jul 19 '20 at 02:23
  • This is exactly the kind of answer I was looking for. Thanks a lot.

    So, one of the things I asked about was defining the order on $\mathbb{Q}$ in terms of algebraic relations on $\mathbb{Z}$, then imposing completeness and attempting to "translate" the order relation on $\mathbb{R}$ back in terms of some requisite algebraic relations on $\mathbb{Q}$. If I understand your post correctly, it's possible to do the first part (get the order of $\mathbb{Q}$ algebraically from $\mathbb{Z}$), but it's not possible to "translate" $\mathbb{R}$ back into algebraic relations on $\mathbb{Q}$?

     – Bears Jul 19 '20 at 17:39
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    @Bears Yes - although of course "algebraic" here is a bit subjective, and you'll notice a disagreement between me/Andreas Blass and JCAA on what exactly counts as algebraic (see the comments below Andreas' answer). – Noah Schweber Jul 21 '20 at 02:56
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Noah Schweber has already explained that $\mathbb R$ cannot be obtained from $\mathbb Q$ or $\mathbb Z$ by any purely algebraic process or even by any first-order definable process. Any construction of $\mathbb R$ must involve some second-order notions like arbitrary subsets or arbitrary infinite sequences (from $\mathbb Q$ or $\mathbb Z$).

There is, however, a construction of the reals that, though still using arbitrary sets (as it must for the reasons in Noah's answer) may look more algebraic and may be more to your liking. It's due to Steve Schanuel, who called it the Eudoxus reals, and you can read about it and find further references at https://ncatlab.org/nlab/show/Eudoxus+real+number# .

One nice property of this construction is that it starts with $\mathbb Z$ and does not first construct $\mathbb Q$ on the way to $\mathbb R$. Another is that it defines multiplication in a way that isn't obviously commutative (though commutativity isn't very hard to prove).

Andreas Blass
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    +1. For those interested in the Eudoxus construction, let me also mention this brief paper of Arthan on a generalization of the construction (and see also here). Note meanwhile that the Eudoxus construction does rely on the notion of boundedness of sets of integers - but, since the ordering on $\mathbb{Z}$ is definable from the algebra alone, this isn't a "true" non-algebraicity. – Noah Schweber Jul 19 '20 at 02:37
  • @JCAA Indeed, there are some notions that depend on the non-first-order concept of finiteness yet are generally deemed algebraic. Polynomials are perhaps the simplest example. I think this issue was looked at in an old paper of Barwise and Eklof, "Lefschetz's princple" [J. Alg. 13 (1969) 554-570]. And I think the relevant constructions, like first-order ones, won't get uncountable structures from countable ones. – Andreas Blass Jul 19 '20 at 03:00
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    I would not consider "continuum many variables" to be an algebraic notion. @JCAA – Andreas Blass Jul 19 '20 at 03:11
  • It is up to you of course. Emma Noether would disagree but she is dead. – markvs Jul 19 '20 at 03:15
  • @AndreasBlass This is actually an interesting read in its own right, but does it circumvent the 'ordering' issue in my initial post? It seems like the construction here still relies on the order of $\mathbb{Z}$, since the 'almost linear' functions used in this construction are required to be bounded. – Bears Jul 20 '20 at 16:25
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    @Bears A function into $\mathbb Z$ is bounded if and only if its range is a finite set. So ordering isn't really involved. – Andreas Blass Jul 20 '20 at 16:27
  • @Bears And as my answer says, the ordering is entirely algebraic: $a\le b$ iff there are $c_1,c_2,c_3,c_4$ such that $a+c_1^2+c_2^2+c_3^2+c_4^2=b$.. – Noah Schweber Jul 20 '20 at 19:20
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    @JCAA How do you propose to define "continuum many"? (I don't see how "the set of all infinite binary sequences" is algebraic in any sense.) – Noah Schweber Jul 20 '20 at 19:20
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    @JCAA There's no problem about ${0,1}^{\mathbb Z}$ being nonempty; it contains the constant 0 function, the characteristic function of the set of primes, and lots of other functions. The problem is in the claim that this is a purely algebraic construction. – Andreas Blass Jul 20 '20 at 20:36