Let $G$ be an ordered abelian group, $b\geq 0$. Suppose $k\geq 1$ and $-b\leq k\cdot a\leq b$. Does $-s\cdot b\leq a\leq s\cdot b$ hold for some $s$?

Question

Let $G$ be an ordered abelian group, $b\geq 0$. Suppose $k\geq 1$ and $-b\leq k\cdot a\leq b$. Does $-s\cdot b\leq a\leq s\cdot b$ hold for some $s$?

An ordered group is a group $G$ with a semigroup of positive elements $G'$ sucn that $G'\cap -G'=0$ and $G'-G'=G$.

I don't think the proposition is true but I can't find a counter example.

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I found one. Let $G=\mathbb Z$ and $G'=2\mathbb Z\cup 3\mathbb Z$. Then $-6\leq 6\leq 6$ while $-6\cdot t\not\leq 1\not\leq 6\cdot t$ for any $t$.