Show that the rationals are an incomplete metric space without reference to reals I know that you can create rational sequences that converge to irrationals, but is there a simple way to do this without explicit assumption of the existence of the reals?
I'm thinking of something along the lines of

1) Show that there exists a particular Cauchy sequence of rationals
2) Assume that the cauchy sequence converges to a rational.
3) Find a contradiction.
Hence rationals are incomplete.

I had an idea about using the property discussed in this question
Choice of $q$ in Baby Rudin's Example 1.1 but couldn
t make anything of it.
I would like to know this because I feel that this is an inherent property of the rationals and that a proof of it should not need to reference anything else.
 A: Your approach is spot on, and you don't need to assume the reals exist to do it.  Define a Cauchy sequence of rationals such that each one squared is closer to $2$ than the one before.  Note that every number in this argument is rational.  Then if the rationals are complete, the sequence converges to a rational.  Now prove that no rational squared equals $2$.  You have a Cauchy sequence that does not converge to an element of the space.
A: You don’t even have to deal directly with rationals (except as radii of metric balls): you can prove the superficially more general result that a countable metric space without isolated points is not complete. (The generality is only superficial, because such a space is homeomorphic to $\Bbb Q$.)
Let $\langle X,d\rangle$ be a countable metric space without isolated points, and index $X=\{x_k:k\in\Bbb N\}$. Let $m_0=0$, $r_0=1$, $B_0=\{x\in X:d(x_{m_0},x)<r_0\}$, $C_0=\operatorname{cl}B_0$, and $M_0=\{k\in\Bbb N:x_k\in B_0\}$; $M_0$ is infinite, since $X$ has no isolated points, and $m_0=\min M_0$.
Suppose that $n\in\Bbb N$, and we have an $m_n\in\Bbb N$, an open set $B_n$ containing $x_{m_n}$, $C_n=\operatorname{cl}B_n$, and an infinite $M_n=\{k\in\Bbb N:x_k\in B_n\}$ such that $m_n=\min M_n$. Let
$$m_{n+1}=\min(M_n\setminus\{m_n\})\;;$$
then $x_{m_{n+1}}\in B_n\setminus\{x_{m_n}\}$. We can therefore choose a positive rational $$r_{n+1}\le\min\left\{d(x_{m_n},x_{m_{n+1}}),r_n-d(x_{m_n},x_{m_{n+1}}),2^{-(n+1)}\right\}$$ and set
$$\begin{align*}
&B_{n+1}=\{x\in X:d(x_{m_{n+1}},x)<r_{n+1}\}\;,\\
&C_{n+1}=\operatorname{cl}B_{n+1}\;,\text{ and}\\
&M_{n+1}=\{k\in\Bbb N:x_k\in B_{n+1}\}\;.
\end{align*}$$
It’s not hard to verify that $C_{n+1}\subseteq B_n$, $M_{n+1}$ is infinite, and $m_{n+1}=\min M_{n+1}$, so the recursive construction can continue. Note that we always have $m_{n+1}>m_n$.
In the end we have a sequence $\langle C_n:n\in\Bbb N\rangle$ of closed sets such that $C_n\supset C_{n+1}$ for each $n\in\Bbb N$. Moreover, $\operatorname{diam}C_n\le 2r_n\le2\cdot2^{-(n+1)}=2^{-n}$, so if $X$ were complete, the Baire category theorem would ensure that $\bigcap_{n\in\Bbb N}C_n=\varnothing$. Suppose that some $x_k\in\bigcap_{n\in\Bbb N}C_n$. Then $k\in\bigcap_{n\in\Bbb N}M_n$. But $\langle m_n:n\in\Bbb N\rangle$ is strictly increasing, so there is an $n\in\Bbb N$ such that $k<m_n=\min M_n$, and hence $x_k\notin B_n$. Finally, $C_{n+1}\subseteq B_n$, so $x_k\notin C_{n+1}$, and $X$ therefore cannot be complete.
A: Construct a sequence
of rationals that would converge
to, say, $\sqrt{2}$.
For example,
$x_1 = 1$,
$x_{n+1}
=\frac{x_n + 2/x_n}{2}
=\frac{x_n^2+2}{2 x_n}
$.
Then, if this converges to a rational,
let the limit be
$r = a/b$,
and let
$x_n = a_n/b_n$.
Then
$a_{n+1}/b_{n+1}
=\frac{a_n/b_n + 2b_n/a_n}{2}
=\frac{a_n^2 + 2b_n^2}{2a_nb_n}
$
so
$a_{n+1} = a_n^2 + 2b_n^2$
and
$b_{n+1} = 2a_nb_n$.
$\begin{array}\\
a_{n+1}^2-2b_{n+1}^2
&=(a_n^2 + 2b_n^2)^2-2(2a_nb_n)^2\\
&=a_n^4 +4a_n^2b_n^2+ 4b_n^4-8a_n^2b_n^2\\
&=a_n^4 -4a_n^2b_n^2+ 4b_n^4\\
&=(a_n^2 - 2b_n^2)^2\\
\end{array}
$
If we start with $x_1 = 1$,
then
$a_1 = b_1 = 1$,
so
$(a_1^2 - 2b_1^2)^2
= 1
$.
Therefore
$a_n^2  -2b_n^2
= 1
$
for all $n > 1$.
Since $r$ is rational,
$r = a/b$
and
$2 = r^2
= a^2/b^2
$.
Therefore
$\begin{array}\\
1
&= a_n^2  -2b_n^2\\
&= a_n^2  -(a^2/b^2)b_n^2\\
&= \frac{a_n^2b^2  -a^2b_n^2}{b^2}\\
&= \frac{(a_nb  -ab_n)(a_nb  +ab_n)}{b^2}\\
&\ge \frac{(a_nb  +ab_n)}{b^2}
\quad\text{since } (a_nb  -ab_n) \ge 1\\
&> \frac{a_nb}{b^2}\\
&= \frac{a_n}{b}\\
\end{array}
$
Therefore,
$a_n \le b$.
But, since
$a_{n+1} = a_n^2 + 2b_n^2$,
$a_n$ gets arbitrarily large.
This is a contradiction,
so the assumption that
the sequence converges to a rational
is false.
Therefore,
the rationals are not complete.
Note:
This is a rewrite of my answer
to this question of mine:
What is the most unusual proof you know that $\sqrt{2}$ is irrational?
A: The line of reason you suggested is exactly "create a rational sequence that converge to an irrational" .
If you want an explicitly sequence, you can create a Cauchy sequence of rationals that converges to $\sqrt2$, the decimal approaches to it works.
