# Proof

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 $$(\mathbb Q,|~|)$$.

### $$(1)$$ $$(x_n)_{n\in\mathbb N}$$ is a rational Cauchy sequence.

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.

### $$(2)$$ The rational Cauchy sequence $$(x_n)_{n\in\mathbb N}$$ does not converge in $$(Q,|~|)$$.

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.

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### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983
2. Forster Otto: "Analysis 2, Differentialrechnung im $$\mathbb R^n$$, Gewöhnliche Differentialgleichungen", Vieweg Studium, 1984