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A totally ordered abelian group is an abelian group (G,+) with a total order $\leq$ such that for all $a,b,c \in G$ if $a \leq b$ then $a+c \leq b+c$.

We will say that an ordered abelian group is dense in itself if for all $a<b$ exist $c$ such that $a<c<b$.

I want to prove that, if for all $g>0$ exist $c$ such that $0<c<g$, then for all $g>0$ and for all $n \in \mathbb{N}$ exist $h$ such that $0<nh<g$.

I have tried this for induction on $n$, but I can't prove the $n+1$-case.

Any suggestion will be welcomed.

HeMan
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    Let's say that $0<nh<g$. We want to find $h'$ s.t. $0<(n+1)h'<g$. If $(n+1)h<g$, then pick $h'=h$. If $(n+1)h\ge g$, then pick $h'$ s.t. $0<h'<g-nh$. – Batominovski Feb 08 '20 at 00:25
  • how do you conclude from this that $0<(n+1)h'<g$? – HeMan Feb 08 '20 at 00:41
  • I truly don't follow, can you please elaborate a bit more please @Berci – HeMan Feb 08 '20 at 00:50
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    Forget it. So, we have $0<nh<g$ and assume $(n+1)h\ge g$, then $nh+h\ge g$, thus $h\ge g-nh$. By hypothesis, for $g-nh>0$, there is an $h'$ such that $0<h'<g-nh$. Then $h'<g-nh\le h$, so $(n+1)h'=nh'+h'<nh+h'<g$. Draw it. – Berci Feb 08 '20 at 01:01

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