Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

$(\Leftarrow)$: ???

share|improve this question
    
What is positive cone? –  Makoto Kato Jul 29 '12 at 21:16

1 Answer 1

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 $y<2^nx$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.