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Jan 14, 2013 at 2:41 comment added Todd Trimble Certainly there are many partial orders on $\mathbb{Q}$ compatible with the group structure; all you need is a submonoid $P \subseteq \mathbb{Q}$ with the property that $x \in P$ and $-x \in P$ implies $x = 0$. For example, $P = \mathbb{N}$ would work: define $x \leq y$ to mean there is a nonnegative integer $n$ such that $y = n + x$. "Torsion free" refers only to group structure, not order structure; $\mathbb{Q}$ is of course torsion free.
Jan 14, 2013 at 2:30 comment added Rajnish Is there ordered structure on the group of rational numbers $\mathbb Q$ which is not totally ordered? If the answer is yes, is that torsion free or not?
Jan 14, 2013 at 0:29 history edited Todd Trimble CC BY-SA 3.0
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Jan 13, 2013 at 23:04 comment added Rajnish @Trimble, thank you for the nice idea.
Jan 13, 2013 at 22:54 vote accept Rajnish
Jan 13, 2013 at 16:26 history answered Todd Trimble CC BY-SA 3.0