(related to Proposition: Not all Cauchy Sequences Converge in the set of Rational Numbers)
Our prove will be in two steps: \((1)\) to prove that the sequence \((x_n)_{n\in\mathbb N}\) of (rational numbers) defined recursively by \[\begin{array}{ll} x_0 > 0 \in\mathbb Q\quad\text{(arbitrarily chosen)}\\ x_{n+1}:=\frac 12\left(x_n + \frac 2{x_n}\right) \end{array}\] is a (rational) Cauchy sequence and \((2)\) that it is not convergent in the metric space \((\mathbb Q,|~|)\).
We have to show that for any rational number \(\epsilon > 0\), (no matter how small it is), there exists an \(N\in\mathbb N\) with \[|x_n-x_m| < \epsilon\quad\quad\text{ for all }n,m\ge N.\]
(i) We have to make sure that the sequence is well-defined, in particular that there are no divisions by 0, when calculating the sequence members. We prove by induction that \(x_{n} > 0\) for all \(n \ge 0\). Since \(x_0 > 0\) by hypothesis, we can assume that \(x_n > 0\) is proven for all \(n\), which are lower or equal some \(n_0\ge 0\). Therefore (induction step) \(x_{n+1}\) is a sum of positive values. Thus, \(x_{n+1} > 0\).
(ii) Next, we observe that \(x_n^2 \ge 2\) for all \(n\ge 1\). This is because \[x^2_n-2=\frac 14\left(x_n +\frac 2{x_n}\right)^2-2=\frac 14\left(x^2_n +4+\frac 4{x^2_n}\right)-\frac 84=\frac 14\left(x^2_n -4+\frac 4{x^2_n}\right)=\frac 14\left(x_n -\frac 2{x_n}\right)^2 \ge 0.\]
(iii) From (ii), it follows that \(1/{x_n^2} \le 1/2\) for all \(n\ge 1\). After multiplying the inequality by \(2^2\), we observe that \(\left(\frac 2{x_{n}}\right)^2\le 2\) for all \(n \ge 1\).
After showing (i) - (iii), we can start to study to compare the distances between the sequence members.
(iv) We will show that \(x_{n+1}\le x_n\) for all \(n\ge 1\). This is because \[x_n-x_{n+1}=x_n-\frac 12\left(x_n + \frac 2{x_n}\right)=\frac{2x^2_n}{2x_n}-\frac{x^2_n}{2x_n}-\frac{2}{2x_n}=\frac{x^2_n - 2}{2x_n} \ge 0\]
(v) From (iii) and (iv) it follows for \(n\ge 1\) that \[\frac 2{x_{n}}\le \frac 2{x_{n+1}}.\]
(vi) We have for all \(n\ge 1\) that \[\frac 2{x_{n}}\le x_n\] Otherwise, we would have \((2/x_n)^2 > x_n^2\), which by (ii) leads to \((2/x_n)^2 > 2\), in contradiction to (iii).
We observe that all consecutive sequence members are rational numbers (since they start from a rational number \(x_0\) and are created by the arithmetic mean of the rational numbers \(x_n\) and \(2/x_n\), which is also a rational number). Moreover, from (vi), (v) and (vi) it follows that \(x_n\) and \(2/x_n\) are getting closer and closer to each other in each step of calculation so we get a chain of inequalities \[\frac 2{x_1}\le\ldots\le\frac 2{x_{n-1}}\le\frac 2{x_n}\le\ldots \le x_n\le x_{n-1}\le\ldots\le x_1\quad\quad( * ).\] More precisely, given an arbitrarily small rational number \(\epsilon > 0,\epsilon \in\mathbb Q\), we are always able to find an index \(N\in\mathbb N\), for which the distance between the two rational numbers \(x_n\) and \(2/x_n\) will get at least as small as the rational number \(\epsilon\), if the index \(n\) exceeds \(N\): \[\left|x_n - \frac 2{x_n}\right| < \epsilon\quad\text{ for all }n\ge N\quad\quad( * * ).\] To realize this, let's see what happens when we build the consecutive sequence members. Because \(x_{n+1}\) is always the arithmetic mean of \(x_n\) and \(2/x_n\), its distance to these numbers is always the \(\frac 12\) the distance from \(x_n\) and \(2/x_n\). Thus, we have \[\begin{array}{ccc} \left|x_2-\frac 2{x_2}\right|&\le&\frac 12\cdot \left|x_1-\frac 2{x_1}\right|\\ \left|x_3-\frac 2{x_3}\right|&\le&\frac 1{2^2}\cdot \left|x_1-\frac 2{x_1}\right|\\ \left|x_4-\frac 2{x_4}\right|&\le&\frac 1{2^3}\cdot \left|x_1-\frac 2{x_1}\right|\\ \vdots&\le&\vdots\\ \left|x_{n}-\frac 2{x_{n}}\right|&\le&\frac 1{2^{n-1}}\cdot \left|x_1-\frac 2{x_1}\right|\\ \end{array} \] Since \(Q:=|x_1 - 2/x_1|\) is a constant positive rational number, we will finally achieve \(( * * )\), if the index \(n\) exceeds an \(N\), which we can calculate only depending from the originally chosen \(\epsilon\) - just chose an \(N\) as big, as the inequation \[2^{N-1} > \frac Q\epsilon\] is fulfilled.
Recall the inequations \( ( * ) \) above and observe that also all sequence members \(x_m\) lie between \(x_n\) and \(2/x_n\), if \(m \ge n\), which we can assume without any loss of generality. Thus, we finally get the required result \[|x_n-x_m|\le\left|x_{n}-\frac 2{x_{n}}\right| \le \frac 1{2^{n-1}}\cdot \left|x_1-\frac 2{x_1}\right| \le \frac 1{2^{N-1}}\cdot \left|x_1-\frac 2{x_1}\right| < \epsilon\quad\quad\text{ for all }n,m\ge N,\] which means that \((x_n)_{n\in\mathbb N}\) is a rational Cauchy sequence.
We have seen in \((1\)) that the rational numbers \(x_n\) and \(\frac 2{x_n}\) get arbitrarily close to some limit \(x\) such that \(x^2=2\), or \(x=\sqrt 2\). However, \(\sqrt 2\) proves to be irrational. In other words, the sequence \((x_n)_{n\in\mathbb N}\) is a rational Cauchy sequence, but it does not converge to a rational limit.
