Is there an "algebraic" way to construct the reals?

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?

Solution 1:

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.

Solution 2:

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).