# subfield of Archimedean ordered field that is isomorphic to $\mathbb{Q}$ is dense in the field

If F is Archimedean ordered field, the subfield of F isomorphic to $\mathbb{Q}$ is dense in F ?

How can i approach to this question i have no idea and the field is new to me.

• What does density mean here? Elements arbitrarily close? – Jacob Wakem Mar 31 '17 at 7:12
• Assume Q is not dense around f. That means there is some distance between f and the nearest interval of rational numbers. – Jacob Wakem Mar 31 '17 at 7:32
• @Alephnull. Perhap referring to order-denseness,i.e. a member of Q lying between any 2 members of F. – DanielWainfleet Mar 31 '17 at 7:39
• @user254665 this is probably the same as topological density. – Jacob Wakem Mar 31 '17 at 7:51
• @Alephnull. Yes. If a subset A is order-dense it is dense in the order topology unless A is empty and the space has just 1 point. – DanielWainfleet Mar 31 '17 at 20:04

Let $x \in F$ and $\epsilon >0$ in $F$. Because $F$ is archimedian, we may find $b \in \mathbb{N}^*$ such that $b. 1_F>\frac{1}{\epsilon}$ (so that $\frac{1}{b.1_F} < \epsilon$). Then let $a \in \mathbb{Z}$ the smallest integer such that $a. 1_F > b.x$ (once again, this integer exists because $F$ is archimedian). Then from the minimality of $a$, the distance between $a.1_F$ and $b.x$ is smaller than $1_F$, so dividing by $b. 1_F >0$, we get that the distance between $\frac{a.1_F}{b.1_F}$ and $x$ is smaller than $\frac{1}{b.1_F} < \epsilon$.
Wouldn't $\mathbb{R}$ work? Or $\mathbb{Q}[\sqrt{2}]$ or any subfield of the real numbers.