Let $(G,+,0,<)$ be an ordered divisible group of uncountable dimension. Consider the subset $G^{<0}$ of $G$.

Question: Are $G$ and $G^{<0}$ isomorphic as ordered sets? Does there exists an order-preserving ismorphism of $G^{<0}$ onto $G$? Intuitively I would say, this is true.

My ideas:

1. In the case where $(K,+, \cdot, 0,1,<)$ is an divisible ordered field we can give an explicit order-preserving isomorphism of $K$ onto $K^{<0}$, e.g. $f(x):= \begin{cases} x-1 \quad \text{if} \; x \leq0, \\ -\frac{1}{x+1} \quad \text{else}. \end{cases} $

2. In the case where $G$ is in ordered set the statement is false.

How can we use the group structure and the divisibility of $G$ to construct an isomorphism or does anybody know a counterexample?

Let $(G,+,0,<)$ be an ordered divisible group of uncountable dimension. Consider the subset $G^{<0}$ of $G$.

Question: Are $G$ and $G^{<0}$ isomorphic as ordered sets? Does there exist an order-preserving isomorphism of $G^{<0}$ onto $G$? Intuitively I would say, this is true.

My ideas:

1. In the case where $(K,+, \cdot, 0,1,<)$ is an divisible ordered field we can give an explicit order-preserving isomorphism of $K$ onto $K^{<0}$, e.g. $f(x):= \begin{cases} x-1 \quad \text{if} \; x \leq0, \\ -\frac{1}{x+1} \quad \text{else}. \end{cases} $

2. In the case where $G$ is in ordered set the statement is false.

How can we use the group structure and the divisibility of $G$ to construct an isomorphism or does anybody know a counterexample?