I will use $\mathbf{\eta}$ and $\mathbf{\lambda}$ to respectively denote the order types of the rationals and the reals. In the book Linear Orderings, by Joseph Rosenstein (1982), he defines:
Definition 1: An order type $\mathbf{\alpha}$ is dense in an order type $\mathbf{\beta}$ if any linear ordering $B$ of type $\mathbf{\beta}$ has a subset $A$ of type $\mathbf{\alpha}$ such that between any two elements of $B$ there is an element of $A$.
I assume by 'between' he means 'strictly between' as that is how he defined dense linear orders. However, letting $C(\alpha)$ denote the order type of a completion of a linear ordering of type $\alpha$, in one of the exercises he asks the reader to show that if $\alpha$ is dense in $\beta$ then the respective completions $C(\alpha)$ and $C(\beta)$ are in fact the same (in the preceding exercise, where '$\preceq$' denotes the embeddability relation, we've already shown that $C(\alpha)\preceq C(\beta)$ when $\alpha\preceq \beta$).
The reason I consider this problematic is that this exercise implies that $C(\eta)=C(1+\eta)$, ($1+\eta$ being the same as $\eta$ but with a least element). Am I mistaken and just missing something or is there something legitimately wrong here? If the latter how can this be resolved?
I think you're right, this is an error in the book. But new endpoints are the only obstruction. You can prove that if:
Then $C(\alpha) = C(\beta)$.