# Is quantifying over natural numbers non first order?

Note that ‘x is an infinitesimal’ is not first order, because it requires you to quantify over the naturals.

Whats's non first order about quantifying over natural numbers?

• When we write statements in the first order language of rings, the intended interpretation is that the quantifiers range over the elements of a given ring. For example, the first-order formula $\forall x\forall y(x\cdot y=y\cdot x)$ is true when interpreted in $\mathbb Z$ (since multiplication of integers is commutative), but not true when interpreted in the ring of $2\times 2$ real matrices.
– Joe
Commented Jul 17 at 17:45
• I would read the statement that you have linked to as asserting that there is no way to express the assertion that $x$ is infinitesimal in the first order language of rings.
– Joe
Commented Jul 17 at 17:50
• @Joe: well, the first-order language of rings doesn't even have a way to express inequalities. You really want the first-order language of ordered fields at a minimum. Commented Jul 17 at 18:05

The issue is that in first-order logic you only quantify over the model, which here is a real closed field, say. So there's no obvious way to write down the condition that $$x$$ is infinitesimal as a first-order statement in the language of real closed fields, since the obvious way to write it down is something like "$$\forall n \in \mathbb{N} : 0 < x < \frac{1}{n}$$" but we cannot actually write this quantification down since we have no obvious way of discussing "$$\mathbb{N}$$" in the language of real closed fields.
This means it's possible for a real closed field like $$\mathbb{R}$$, which has no infinitesimals, to have an ultrapower which does have infinitesimals, without violating Łoś's theorem. In fact, in the previous paragraph we just said there's no obvious way to define infinitesimals, but Łoś's theorem implies that there's no way to do it at all, and so also implies that there's no way to isolate the copy of $$\mathbb{N}$$ sitting inside a real closed field $$R$$ in the language of real closed fields.
You might try to get around this is by introducing a second sort for the natural numbers, and axiomatizing pairs $$(R, N)$$ of a real closed field and a model of Peano arithmetic, for example. Then the issue is that after taking ultrapowers your model of Peano arithmetic also becomes nonstandard! So you can try to postulate the non-existence of infinitesimals with a statement like $$\forall r > 0 \in R : \exists n \in N : r \ge \frac{1}{n}$$, but the meaning of this statement changes if $$N$$ is nonstandard; it will still be true for an ultrapower of $$(\mathbb{R}, \mathbb{N})$$ but it no longer rules out infinitesimals in the ordinary sense, since the numbers $$\frac{1}{n}$$ can themselves be infinitesimal if $$N$$ is nonstandard.