Non-archimedean group over the reals I have a totally ordered group $(\mathbb{R};\leq,\oplus,0,-)$ with the reals as base set satisfying monotonicity, i.e.
for all $x,y,z$ we have that if $y\leq z$ then $x\oplus y \leq x\oplus z$, and I am wondering if this already suffices to show that the group is archimedean. (i.e., that for all $a,b$ with $a \leq b$ there exists an $n$ such that $\underbrace{a\oplus\ldots\oplus a}_n \geq b$).
I strongly suppose it is, as I couldn't find any counterexample, but neither can I find a convincing proof. (My further goal is to show that the group is isomorphic to $(\mathbb{R};\leq,+,0,-)$ which should be immediate once I can proof archimedeanness.)
 A: I guess that the notation $(\mathbb R;\leq,\oplus,0)$ is intended to say that your ordered group is the set of reals with its standard ordering $\leq$ but with a possibly strange group operation $\oplus$ (whose identity element is nevertheless the standard 0).  In this case, archimedianness can be proved as follows.  Consider some positive $a$ and its multiples $na=\underbrace{a\oplus\ldots\oplus a}_n$; you want these to be cofinal in $\mathbb R$.  If they were not cofinal, they would have a least upper bound $b$ (because the ordering is standard).  Then $b-a$, being strictly smaller than $b$ (because $a$ is positive), is $<na$ for some $n$; but then $b<(n+1)a$, a contradiction.  
(If my guess is wrong and you intended to allow a non-standard interpretation of $\leq$, then archimedianness does not follow, simply because there are non-archimedian totally ordered groups of the cardinality of the continuum.  Furthermore, archimedian examples would not have to be isomorphic to the reals, since $\mathbb R$ has proper subgroups of the cardinality of the continuum.)
A: 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: 


*

*"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. 

*"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. 
