TITLE: Is there an example of an ordered ring that is not isomorphic to any subring of the real numbers?
QUESTION [14 upvotes]: Somebody told me it's possible to show the existence of rings non-isomorphic to any subring of the real numbers using model theory, but they couldn't get an example.

REPLY [6 votes]: Another example (of a non-Archimedian ordered field) worth checking out is David Hilbert's example in Section 12 of Foundations of geometry. 
It is, to paraphrase, what you get by taking $\mathbb Q$, a single variable $t$, and then generating a field that is closed under the additional operation $x\mapsto \sqrt{1+x^2}$.
Viewed as real-valued functions, it can be proven that each such expression is eventually positive or eventually negative, and then the ordering is that $f\leq g$ if there exists an $b\in \mathbb R$ such that $f(x)\leq g(x)$ for all $x\in [b,\infty)$.
Hilbert's original description is a little archaic to a modern reader, but it's not too bad, and I'm sure there are more modern versions of it lying around somewhere.