# Mathematics based only on real numbers [closed]

I'm aware that >90% will outright reject this, so feel free to ignore it. I'd much appreciate those trying to figure out in which way this question (or rather its eventual answer) would make sense.

Is there a way to construct the foundations of mathematics from just the real numbers? If not, why not?

Background: The real world (space, time) is arguably best described as a continuum. So it would make sense to take that as a foundation. Integers would only occur later e.g. as winding numbers, not as fundamental. All discrete things would be secondary. A continuous version of sets would probably be part of the foundation.

Rant: This construction from set theory to integers to rational numbers only to find that real numbers are then kind of problematic in the framework. I just can't see any sense in that.

• If you want an axiomatization of just the real numbers then Book 5 of Euclid (or similar) might do. If you want $\mathbf R$-based “foundation for all mathematics” your problem would be to translate from the language (sets) most of them are already written in. May 22, 2020 at 21:44
– user155396
May 22, 2020 at 21:57
• So how do you define, say, a function... Do you want to eliminate sets? May 22, 2020 at 21:59
• "real numbers are then kind of problematic in the framework" What do you mean? May 22, 2020 at 22:07
• @DoctorNuu Up to isomorphism, there is a unique Dedekind-complete ordered field, and this is what we mean when we talk about the real numbers. Dedekind cuts and Cauchy sequences show that one can construct (or “implement”) such a field starting from the rationals. May 23, 2020 at 18:29

$$\mathbb{Z}$$ is not first-order definable in $$\mathbb{R}$$ (as an ordered field) or $$\mathbb{R}^n$$ (as a vector space). That is, they cannot even "talk about" the integers.
• 1. Obviously we don't stop at $\mathbb{R}$ but also consider $\mathbb{R}^n$ Also no reason for integers to be "in" $\mathbb{R}$ 2. This is a soft question, so maybe explain what decidability is good for.
• @DoctorNuu $\mathbb{R}^n$, as a plain vector space, has an even weaker structure. “Also no reason for integers to be "in" ℝ.” It’s not about them being in $\mathbb{R}$, but about being able to define them (talk about them) from $\mathbb{R}$. And decidability is a sign that the theory is too weak, since even Robinson arithmetic is undecidable. May 22, 2020 at 23:20
• @DoctorNuu Put another way: the fact that the field of real numbers is decidable means that there is no way of "interpreting" (actually a technical term, but for now take it somewhat vaguely) the ring of integers in it. Going to $\mathbb{R}^n$ (even treated as an $\mathbb{R}$-algebra, which gives much more structure than as a mere vector space) doesn't help with this either; indeed, that's still interpretable in $\mathbb{R}$. May 23, 2020 at 0:07
• Although the OP doesn't seem to like sets, continuous functions $R\to R^2$ seem to be OK since their winding numbers are mentioned. So it's unclear how much of second-order logic might be allowed. It doesn't take much to produce the integers that way. Of course, continuous functions can be coded as reals, but the coding doesn't seem to be the sort of thing the OP would approve. May 23, 2020 at 1:21