There is no first order property of a totally ordered group $G$ which

(a) implies that $G$ is archimedean

(b) is satisfied by the real numbers (with the usual order and usual addition).

There is no first order property of a totally ordered group $G$ which

(a) implies that $G$ is archimedean

(b) is satisfied by the real numbers (with the usual order and usual addition).

EDIT: In view of Andreas Blass' interpretation and answer, this may be irrelevant now, but here are two proof sketches:

1. "logical proof": Take the first order theory of the reals, add constants $c,d$ to the language, and add the axioms $0\lt c\lt d$, $c+c\lt d$, $c+c+c\lt d$, etc. The resulting theory is consistent (by compactness) and hence has a model - the desired non-archimedean counterexample.

2. "Algebraic proof": Let $U$ be a non-principal ultrafilter on the natural numbers $\mathbb N$. Let $M$ be the ultrapower $\mathbb R^{\mathbb N}/U$. Compare the class of the identity function and any constant function (say: 1) to see that $M$ is not archimedean. By Łoś' theorem, $M$ satisfies the same first order theory as the real numbers.