If $a,b \in P$ which is the positive cone of the ordered field $F$, show that there exists an $n \in \mathbb{N}$ such that $\underbrace{a+a+ \ldots + a}_{n \text{ times}} > b$.
For full context, this is the question
This problem is trivial in the real numbers, but how to show it in an arbitary ordered field? This is what I have tried.
If $a>b$, then we are done, so suppose $a<b$. Then let's suppose that for all $n$, $na<b$. Since $b^{-1} \in P$, we can mutiply through to get $nab^{-1}<u \implies ab^{-1}<u/n$, but this doesn't work, since $u \in F$.
Intuitively I know that I should choose an $n$ such that $n>b/a$, but since $b/a \in F$ and $n \in \mathbb{N}$, there's no way to compare them.
If we want to use the least upper bound property of a field, doesn't that require the field to be Archimedean (edit: complete)? But we can't make that assumption here, so proofs like this wouldn't work.
Hints appreciated.