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.)



Non-archimedean group over the reals
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.)


