Definitions: Let $(K,\leq)$ be a totally ordered field and $(G,\leq)$ a totally ordered abelian group (written additively). If we denote $\mathbb{Z}a\!=\!\{na;\, n\!\in\!\mathbb{Z}\}$, then $G$ is Archimedean when $\mathbb{Z}a\!\leq\!b\text{ implies }a\!=\!0,$ for all $a,b\!\in\!G$. Furthermore, $K$ is Archimedean when its additive group is Archimedean.
Question: Let $(K,\leq)$ be a totally ordered field. How can I prove that $K$ is an Archimedean field iff $K_{>0}\!:=\!\{a\!\in\!K; a\!>\!0\}$ is an Archimedean group (w.r.t. multiplication)?
In other words, t.f.a.e.: (i) $\mathbb{Z}a\!\leq\!b$ implies $a\!=\!0$, for $a,b\!\in\!K$; (ii) $a^\mathbb{Z}\!\leq\!b$ implies $a\!=\!1$, for $a,b\!\in\!K_{>0}$.
$(\Rightarrow)$: If $K$ is Archimedean, then by a theorem, $K$ is isomorphic as an ordered field, to some field $\mathbb{Q}\!\subseteq\!K'\!\subseteq\!\mathbb{R}$. Since in $\mathbb{R}$, the inequality $\mathbb{Z}a\!\leq\!b$ implies $a\!=\!0$, we are done.
$(\Leftarrow)$: ???