Is $\mathbb{R}$ the only complete ordered Abelian group? I know that the field of real numbers is the only complete, ordered field in the sense that any field satisfying these properties is isomorphic to $(\mathbb{R},+,\cdot,<)$.
Question 1. Is it true that any complete, ordered Abelian group is isomorphic to $(\mathbb{R},+,<)$? If not, is there an example of a complete, ordered Abelian group $(G,+,<)$ which is not isomorphic to $\mathbb{R}$?
I am using the following definition of complete: A partially ordered set $(P,\leq)$ is complete if each subset $E\subseteq P$ which is bounded above has a least upper bound in $P$. This is modeled after the completeness axiom for $\mathbb{R}$, but I can't find a good source for this more general definition.
Edit 1. It is false. A counterexample is $\mathbb{Z}$. It is complete since any subset has a maximum, and ordered as usual,  but not isomorphic to $\mathbb{R}$.
Question 2. What if we require $G$ to be "dense" in the following sense?
$$\forall a,b\in G, \quad \{g\in G\mid a<g<b\}\neq\emptyset$$
Does this additional condition imply $G$ is isomorphic to $\mathbb{R}$?
Edit 2. It is true. It was shown the only complete ordered Abelian groups are $\mathbb{Z}$ and $\mathbb{R}$, and the "dense" condition leaves only $\mathbb{R}$.
 A: There are in fact only two complete ordered nontrivial Abelian groups (up to isomorphism): $\mathbb{Z}$ and $\mathbb{R}$. The former is complete for a silly reason, namely that every nonempty bounded-above set has a maximal element (not just a unique supremum); the latter is the interesting one
(What about e.g. the group of integer multiples of ${1\over 2}$? That's just $\mathbb{Z}$ again, up to isomorphism.)
The proof goes roughly as follows (letting $G$ be our nontrivial complete ordered Abelian group):

*

*First we show that $G$ is Archimedean: fixing any positive element $a$, for every $g$ in the group there is some natural number $n$ such that $a+...+a$ ($n$ times) is greater than $g$.


*Now we ask, is there a minimal positive element in $G$? If so, we can show that $G\cong \mathbb{Z}$.


*We're now left with the case where there is no minimal positive element, and we want to show that $G\cong\mathbb{R}$. Fix some positive element $a\in G$, and let $A$ be the set of rational multiples of $a$: that is, $$A=\{g\in G:\exists k,l\in\mathbb{Z}(ka=lg)\},$$ where multiplication of a group element $h$ by an integer $m$ is defined as follows: if $m=0$ then $m\cdot h=e$ (the identity of $G$); if $m>0$ then $m\cdot h=h+...+h$ ($m$ times); and if $m<0$ then $m\cdot h$ is the inverse of $\vert m\vert \cdot h$.


*Having defined $A$ as above, we show that there is a natural injection $i$ of $A$ into $\mathbb{Q}$; in fact, $i$ is the unique embedding of $A$ into $\mathbb{Q}$ as ordered Abelian groups.


*Now we don't know a priori that the image of $i$ is all of $\mathbb{Q}$ - why should there be something in $G$ which is "a third of $a$"? However, we can show that the image of $i$ is dense in $\mathbb{R}$. Now using completeness, the Dedekind cut construction, and the Archimedean-ness of $G$, we can in fact extend $i$ to a (unique!) isomorphism between $G$ and $\mathbb{R}$.
