Question

Are the computable numbers one example?

Answer 1

$\mathbb{R}(x)$ with the "behavior at infinity" order is an example. $\frac{1}{x}$ is infinitesimal w.r.t. $1$.

Answer 2

Here is a more detailed explanation of Qiaochu's example. Suppose that $F$ is any ordered field. Make the polynomial ring $F[X]$ into an ordered integral domain by declaring that a polynomial is positive if its leading coefficient is positive. Then make the quotient ring $F(X)$ of rational functions an ordered ring by declaring a quotient $P(X)/Q(X)$ to be positive if $P(X)$ and $Q(X)$ "have the same sign", i.e., both are positive or both negative. (Note that this is exactly the way we go from the ordering on the integers to the ordering on the rationals.) It is now clear that the rational function $X$ is greater than every constant polynomial, i.e., the elements of the original field $F$ are bounded above in $F(X)$, and the Archimedean property for $F(X)$ is equivalent to the statement that the integers are unbounded in $F(X)$ (or, as Qiaochu says, $1 \over X$ is an infinitesimal in $F(X)$).

Answer 3

An ordered field is Archimedean iff it can be embedded, as an ordered field, into the real numbers (with their unique ordering). Therefore, a cheap way to get non-Archimedean ordered fields is to find an ordered field of larger cardinality than the real numbers. Indeed there are ordered fields of all (edit: infinite!) cardinalities, as one can see e.g. by applying the L\"owenheim-Skolem theorem.

In fact, that's overkill here. A basic (and elementary) theorem of Artin-Schreier says that a field admits an ordering iff it is formally real: i.e., iff $-1$ is not the sum of any finite number of squares in the field. It is easy to see that if $k$ is a formally real field (e.g. $\mathbb{Q}$), then so is any purely transcendental extension of $k$: i.e., we adjoin an arbitrary set of independent indeterminates.

For some more information on ordered fields and formally real fields, you could consult Chapter 11 of

http://math.uga.edu/~pete/FieldTheory.pdf

However, in this case I must in good conscience recommend something else as well, because these notes are quite incomplete (in particular Section 11.3 on Archimedean orderings exists only as a placeholder; all the material is missing). I find the treatment in Jacobson's Basic Algebra II to be more than satisfactory.

Answer 4

Non-archimedian fields can be obtained as follows:

Let $c$ be a constant symbol.
Let $T$ be the theory containing all axioms of ordered fields.
Let $\phi_n$ be the sentence that says that $c$ is greater than the sum of $n$ $1$'s.
Every finite subset of $T\cup\{\phi_n:n\in\mathbb N\}$ has a model: Take the reals and interpret $c$ by a sufficiently large real number.

Now the compactness theorem of first order logic says that
$T\cup\{\phi_n:n\in\mathbb N\}$ has a model $M$. In this model, $c$ is interpreted by an element that is larger than any finite sum of $1$'s. Hence $M$ is an ordered field that is not archimedian.

Another standard way to construct non-archimedian ordered fields is taking so-called ultraproducts of archimedian fields.

As pointed out in a comment above, the computable numbers are a subfield of the reals and hence archimedian.