Is Dedekind completion of ${}^{\ast}\Bbb R$ a Archimedean field? Here's Theorem 1.2 on page 6, Martin Andreas Väth's Nonstandard Analysis(See here on googlebooks) 

The Dedekind completion $\overline{X}$ of a totally ordered field $X$
  is a complete Archimedean field with $\Bbb{Q}_{\overline{X}}$ as the canonical copy of $\Bbb{Q}_{X}$.

$X$ has the Archimedean property. For each $x \in X$ there is some $n \in \Bbb{N}_{X}$
such that $n > x$.
Each totally ordered field X contains a “canonical copy” of the set $\Bbb{N}_{X}$, namely $\{1_X, 1_X +1_X, 1_X +1_X +1_X, \ldots\}$.
${}^{\ast}\Bbb R$ is a totally ordered field without Archimedean property. Isn't it the case that its Dedekind completion doesn't have Archimedean property?

The Dedekind-completion is an order completion, and if the field is non-Arcihmedean then its Dedekind-completion is not a field at all.
To see this simply note that in the completion, there is a point $t$ which is the realization of the cut $R=\{x\mid\exists n\in\Bbb N. x<n\}$. And $t-1$ cannot exist.
