Divisibility in totally ordered, dense, groups without commutativity If $(G,+,<)$ is a commutative, totally ordered, dense group one can prove that for every $x>0$ there exists $y>0$ such that $y+y\le x$. In fact by density there exists $z$ such that $0<z<x$ then either $z+z \le x$ (and we are done) or $z+z>x$. In the latter case
we take $(x-z)$ and with commutativity we can prove that $(x-z)+(x-z)< x$.
What if $G$ is not commutative? Is it possible to prove the same thing?
With this property and adding completeness one can prove that the group is divisible (for each $x\in G$ and $n\in \mathbb N$ there exists $y$ such that $ny=x$). So if the group is also not trivial it has a subgroup isomorphic to $\mathbb Q$ and, by completeness it is actually isomorphic to $\mathbb R$.
So a second question is whether there exists a non-commutative, totally ordered, dense, complete group.

Since we are not assuming the group is commutative, I will write it multiplicatively rather than additively.  So, suppose $x>1$ and take $z$ such that $1<z<x$.  Let $y=xz^{-1}$, so $1<y<x$ as well.  If $y\leq z$, then we have $y^2\leq yz=x$ and we're done.  If $y>z$, then $z^2<yz=x$ and we're again done.
