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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.

$(\Leftarrow)$: ???

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What is positive cone? – Makoto Kato Jul 29 '12 at 21:16
up vote 1 down vote accepted

$\Rightarrow$: for every $x>1$ and $n\in\mathbb{N}\setminus 0$ we have $x^n>1$ and therefore $x^n-x^{n-1}=(x-1)x^{n-1}>x-1$. Inductively this yields $x^n>n(x-1)$.

Now suppose $x,y>1$ are given and we want to show that $y<x^n$ for suitable $n\in\mathbb{N}$. By assumption there exists $n\in\mathbb{N}$ such that $y<n(x-1)$, hence $y<x^n$.

$\Leftarrow$: Suppose $x,y>1$ are given and we want to show that $y<nx$ for suitable $n\in\mathbb{N}$. We know $2>1>0$ and either $y<x$ (in which case there is nothing to prove) or $1<\frac{y}{x}$. By assumption there exists $n\in\mathbb{N}\setminus 0$ such that $\frac{y}{x}<2^n$ thus $y<2^nx$.

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Aha, for $(\Leftarrow)$, you've used the weak Archimedean property (equivalent to Archimedean property, for lattice ordered groups): Let $a,b\!\in\!K_{\geq0}$, $\mathbb{N} a\!\leq\!b$, and suppose that $a\!\neq\!0$. Then $1\!\leq\!2,\frac{b}{a}$ and $2^\mathbb{N}\!\leq\!\frac{b}{a}$, so by assumption, $2\!=\!1$, $\rightarrow\leftarrow$. Thank you very much! – Leon Jul 30 '12 at 21:03

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