Pretty simple question. Does there exist a ordered field smaller than (i.e. is a strict subset of) $\mathbb{Q}$? It seems like we can't go any smaller than $\mathbb{Q}$. Is this true? Why?
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4$\begingroup$ Well, you have to argue that no finite field can be ordered. More broadly, try to argue that no field of finite characteristic $p$ can be ordered. Beyond that, if $\mathbb F$ is a subfield of $\mathbb Q$ it must contain $1$ hence it contains $\mathbb Z$, hence it contains $\mathbb Q$. $\endgroup$ – lulu Nov 9 at 1:02
