Are there any ordered fields containing $\mathbb{Q}$ as a non-dense subfield? Does there exist a totally ordered field for which $\mathbb{Q}$ is not dense? The answer is yes, but I can't understand why. Can you provide an example?
 A: I assume that by a 'totally ordered field', you mean an ordered field i.e. a field $K$ equipped with a total order $\leq$ that is compatible with the field operations, by which I mean that $x\leq y\implies x+z\leq y+z$ and $x,y\geq 0\implies xy\geq 0$.
If we drop these compatibility assumptions, then it might be possible to put a (completely unnatural) total order on $\mathbb{C}$ making it a 'totally ordered field' containing $\mathbb{Q}$ as a proper dense subfield. But of course that would be a bit silly, because such an ordering would not respect the algebraic structure of $\mathbb{C}$ (we would need $i^2\geq 0$ for that to work).
Some important points to make about ordered fields: It is clear that $x^2\geq 0$ for all $x\in K$. If $K$ were to have positive characteristic, then $1+\cdots+1=1^2+\cdots+1^2=0$, yet $1^2>0$, so LHS must be positive. This shows that $K$ must have characteristic zero. $\mathbb{Q}$ then sits inside $K$ in a natural way: we identify each $q\in\mathbb{Q}$ with $q\cdot 1_K$. The image of this identification is the smallest subfield of $K$, called its prime subfield. It can be obtained by starting with $\{0_K,1_K\}$ and extending using the field operations.
For your question to make complete sense, there needs to be a topology on the ordered field $K$. Thankfully, there is a natural choice: the order topology. We can define open intervals same as with $\mathbb{R}$. The order topology is then defined to be the smallest topology containing the open intervals. With a natural topology on $K$, it then makes sense to talk about the closure of $\mathbb{Q}$ within $K$, and whether or not it is equal to the whole of $K$.
One example of an ordered field containing $\mathbb{Q}$ as a non-dense subfield would be the field of Hahn series with real coefficients. $\mathbb{Q}$ is then just the subfield of constant Hahn series with rational coefficients. The ordering is given by declaring a Hahn series to be strictly positive iff its leading coefficient is strictly positive. I suspect that $\mathbb{Q}$ is not dense in the field of formal Laurent series over $\mathbb{Q}$ and many other non-Archimedean extensions of $\mathbb{Q}$ and $\mathbb{R}$ also.
