I'm trying to prove that an Archimedian field is a subfield of the real numbers, my plan is to use the fact that the rationals are dense within the field and their Dedekind completion is the real numbers.
However, I seem to run into some trouble writing a neat proof for the fact that the two properties of Archimedian fields are equivalent:
Let $F$ be an ordered field, the following are equivalent:
1. For every $x\in F$ there exists a natural number $n$ such that $x < n$
2. The rationals are dense in $F$
(In fact, I only need $(1)\implies (2)$ to be proved)
I have the essence of the proof, assume by contradiction some $a<b$ are two numbers that have no rational numbers in between, wlog we can assume $a,b\in (0,1)$, take some $n$ which is large enough and show some $\frac{k}{n}$ has to be between $a$ and $b$ otherwise there is some $x$ which is an upper bound of $\mathbb{N}$ in the field. However it seems that I can't write it elegantly.