Can every field be "completed" I don't know if this is the right wording of this question, but just like $\mathbb{Q}$ has a superset $\mathbb{R}$ with the least upper bound property, can any other field, say the set of all rational functions (with the usual operations for addition and multiplication of functions), for example, be "extended" or "completed" into a field that has this property?
Thanks and sorry if this has already been asked.
 A: Short answer: The "least upper bound property" is a property defining the field $\mathbb{R}$ up to isomorphism. In general we cannot expect an ordered field to have an embedding into $\mathbb{R}$, so we cannot expend ordered fields to have a 'completion' in the sense you want it.
Now for the long answer.
Before it makes sense to speak of a field with "the least upper bound property" one needs to define what it means to have an ordering on a field. The most commonly used notion for an ordering (as on Wikipedia): a total order relation $\leq$ on the field such that for all $x, y, z$ in the field one has
$$
x \leq y \Rightarrow x + y \leq x + z \qquad \& \qquad x \leq y, 0 \leq z \Rightarrow xz \leq xy.
$$
I would like to contest Arturo Magidin's claim that there are not many ordered fields. A famous theorem due to Artin and Schreier states that a field has orderings if and only if $-1$ is a not a sum of squares in the field.
As an example of an ordering you might not know: consider the field $K = \mathbb{Q}(T)$ of rational functions over $\mathbb{Q}$. This field has many orderings. One of them is given as follows: for $f, g \in K$, we define
$$
f \leq g \quad \Leftrightarrow \quad (\exists t_0 \in \mathbb{Q}) (\forall t \geq t_0)(f(t) \leq g(t)).
$$
You can check for yourself that this defines an ordering on the field $K$. However, whatever ordered field extension $(L, \leq)$ of $(K, \leq)$ you take, $(L, \leq)$ will not be complete. Indeed, take such an ordered extension $(L, \leq)$ of $(K, \leq)$, and define the set
$$
S = \lbrace x \in L \mid \exists n \in \mathbb{N} : x \leq n \rbrace \subseteq L.
$$
This subset has an upper bound. For any $n \in \mathbb{N}$ we have that $n < T$ in $K$, hence the element $T$ of $K \subseteq L$ is an upper bound for $S$. But $S$ cannot have a least upper bound. Indeed, if $b \in L$ is an upper bound for $S$, then so is $b -1 < b$.
In fact, as said before, one can show that the least upper bound property is a property uniquely defining the field $\mathbb{R}$ up to isomorphism. Hence, the only ordered fields which can have a completion in the sense you describe, are the subfields of $\mathbb{R}$.
Finally, there is another, weaker sense of completeness, sometimes also called Cauchy completeness, which is the property that every Cauchy sequence converges. I am not 100% sure at the moment and can't immediately find a reference, but I believe that every ordered field $(K, \leq)$ does have a field extension which is Cauchy complete.
