Can rationals be completed to reals? It's a well-known theorem that any metric space can be completed. Its proof essentially uses the completeness of the reals.
The rationals $\mathbb{Q}$ can be treated as an uncomplete metric space with the metrics $d:(x,y)\rightarrow |x-y|$.
The question arises: can the metric space $(\mathbb{Q},d)$  be completed without using the reals?
If I correctly remember it, I saw that construction in a course of mathematical analysis lectured at Moscow University
in 70s of the 20-th century. A proof and/or reference are welcome.
 A: First of all, completions don't need the real numbers.
Look in Kelley's book for the chapter on uniform spaces, where it is hinted that they can be described by a set of axioms not so different from the axiom of a topology. You can find the system in Bourbaki, but it's not difficult to derive it. The axiom system is basically a description of entourages, special subsets of $X\times X$ containing the diagonal: an entourage for a metric space $(X,d)$ would be a subset of $X\times X$ containing a set of the form $\{(x,y):d(x,y)<r\}$, for some $r\in\mathbb{R}$. However, the axiom system never mentions the real numbers.
It is possible to define the concept of convergence and of Cauchy sequence only using entourages and to show that every uniform space has a completion using the set of Cauchy sequences modulo an equivalence relation, which set can be endowed with a uniform space structure and every Cauchy sequence in it converges.
Metric spaces are uniform spaces having a countable basis of entourages (basis in the sense of filter basis).

About getting the reals from the rational numbers, here's a construction similar to the one described above, which is a formalization of Cantor’s idea.
A C-sequence in the rational numbers is a sequence $(x_n)$ such that, for every $\varepsilon\in\mathbb{Q}$, $\varepsilon>0$, there exists $k$ with the property that, for every $m,n>k$, $|x_m-x_n|<\varepsilon$.
Real numbers are not used, as you see.
Now it's not difficult to show that the set $\mathscr{C}$ of C-sequence can be made into a ring, with pointwise addition and multiplication: the sum and product of C-sequences are C-sequences.
A Z-sequence in the rational numbers is a sequence $(x_n)$ such that, for every $\varepsilon\in\mathbb{Q}$, $\varepsilon>0$, there exists $k$ with the property that, for every $n>k$, $|x_n|<\varepsilon$.
It's easy to see that every Z-sequence is a C-sequence and that the set $\mathscr{Z}$ of Z-sequences is an ideal of $\mathscr{C}$.
Now the quotient ring $\mathscr{R}=\mathscr{C}/\mathscr{Z}$ can be shown to be complete in the sense that every C-sequence in it converges. And, guess what? It is a field that's isomorphic to the real numbers.
Basic steps.

*

*$\mathscr{Z}$ is a maximal ideal in $\mathscr{C}$, so $\mathscr{R}$ is a field.


*Define an order relation on $\mathscr{C}/\mathscr{Z}$ by declaring an element of the form $[(x_n)]=(x_n)+\mathscr{Z}$, where $(x_n)$ is a C-sequence, to be positive if there exist $\varepsilon\in\mathbb{Q}$ and $k$ such that $\varepsilon>0$ and $x_n>\varepsilon$, for every $n>k$. Prove that $\mathscr{R}$ becomes an ordered field.


*Embed $\mathbb{Q}$ into $\mathscr{R}$ in the unique possible way.


*Define a C-sequence in $\mathscr{R}$ as at the beginning (only rational numbers).


*Define a sequence $(x_n)$ to be convergent in $\mathscr{R}$ if there exists $x\in\mathscr{R}$ such that, for every $\varepsilon\in\mathbb{Q}$, $\varepsilon>0$, there exists $k$ with the property that, for all $n>k$, $|x_n-x|<\varepsilon$.


*Prove that all C-sequences converge in $\mathscr{R}$.
What's the idea in the last step? Take a sequence $(x_n)$ in $\mathscr{R}$. Each $x_n$ is the residue class of a sequence $(x_n^{(1)},x_n^{(2)},\dotsc)$ in $\mathbb{Q}$. Show that the sequence $y$ defined by $y_n=x_n^{(n)}$ is a C-sequence in $\mathbb{Q}$ and that the given C-sequence converges to $[y]$.
