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Comment by zzless | Hacker News Reader

In fact, he did not miss anything. Using the language he started with (variables range over 'numbers', and the relations are <, >, +, and *), the reals and the rationals indeed have the same properties (elementary theory as logicians would put it). The reason things like \sqrt2 present no problems is that it is simply impossible to define such 'sequences of numbers' in this theory (you are only allowed to 'refer' to numbers by your variables, not ordinary sets and the usual language for sets is missing).

If I remember right, the fact he was referring to was proved by Tarsky.

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> he reason things like \sqrt2 present no problems is that it is simply impossible to define such 'sequences of numbers' in this theory (you are only allowed to 'refer' to numbers by your variables, not ordinary sets and the usual language for sets is missing).

Doesn't that mean that you can't even define the reals using the language he started with? If your language doesn't even let you express the difference between the reals and the rationals, it seems to me that the thing to do is to extend your language until it can.

> it seems to me that the thing to do is to extend your language until it can.

This depends on what you mean by define. If you mean a unique model than this is impossible. The reason is compactness theorem (every theory in which every finite set of formulas has a model has a model). The basic idea is to add constants and introduce axioms stating they are different. This will allow models of, say reals where there are plenty of reals that are not real reals (sorry for the pun, I could not resist). Nonstandard analysis takes it a bit further and makes it a bit more precise and useable.

If you mean you want to deal with (naively) definable reals only then intuitively there are only countably many of those that you can define (by formulas, etc) and you are missing a huge chunk of the real line again.

> The reason is compactness theorem

As I understand it, there is no compactness theorem in second-order logic, only in first-order logic. So your objection would not apply to extending one's language by using second-order logic.

True, the first order logic is unique in that it satisfies compactness and L-S. Extending the language to second order language (although this is not quite standard terminology) is a whole new ball of wax since the concept of a model changes as well. You can introduce a quantifier over 'unique' reals but this is a rather hollow extension since the nature of that quantifier remains just as vague. I also fail to see why the uniqueness of reals is so important. Using second order language you would have to forcefully / 'declare' every such object.

You're right that it is not possible to define the reals using the language he started with, but it's worse than that. It's also not possible to define the natural numbers using any first order theory. There is no way to extend a first order theory so that it defines the natural numbers and only the natural numbers and furthermore there is no way to define a first order theory that defines the reals and only the reals.

Having said that, you were originally right that no theory of the reals can be satisfied by the rationals, but that's for a fairly unrelated reason.

> It's also not possible to define the natural numbers using any first order theory.

Yes, agreed.

> you were originally right that no theory of the reals can be satisfied by the rationals, but that's for a fairly unrelated reason.

Can you elaborate?

At a minimum any theory of real numbers will imply a theorem of the form "There exists an x such that x * x = 2". The set of rational numbers doesn't contain any such x and hence the rational numbers will not satisfy any theory of real numbers.

You are correct, I should have omitted the * from the language.

The reals and the rationals do not have the same elementary theory over the language (>,+,*).

You are correct, one would have to exclude the multiplication for that.

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