(related to Proposition: Square Roots)

Let \(a > 0\) be a real number. We want to find all solutions of the equation \[x^2=a.\quad\quad( * )\]

The key to find a solution is the observation that if \(x\) is a solution of the equation \( ( * ) \), then \(x=\frac ax\), otherwise \(x\neq\frac ax\). Therefore, if \(x_0\) is any approximation of the solution of \( ( * ) \), the mean value

\[x_1:=\frac 12\left(x_0+\frac a{x_0}\right)\]

will be a better approximation to that solution.

Therefore, we define a sequence of real numbers. \[\begin{array}{ll} x_0 > 0 \in\mathbb R\quad\text{(arbitrarily chosen)}\\ x_{n+1}:=\frac 12\left(x_n + \frac a{x_n}\right) \end{array}\]

The proof consists of three steps:

First, we gather some extensions about the behavior of the sequence members of \((x_n)_{n\in\mathbb N}\).

(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 \(x_n^2 \ge a\) for all \(n\ge 1\). This is because \[x^2_n-a=\frac 14\left(x_n +\frac a{x_n}\right)^2-a=\frac 14\left(x^2_n + 2a +\frac {a^2}{x^2_n}\right)-\frac {4a}4=\frac 14\left(x^2_n -2a+\frac 4{x^2_n}\right)=\frac 14\left(x_n -\frac a{x_n}\right)^2 \ge 0.\]

(iii) From (ii), it follows that \(1/{x_n^2} \le 1/a\) for all \(n\ge 1\). After multiplying the inequality by \(a^2\), we observe that \(\left(\frac a{x_{n}}\right)^2\le a\) 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 \[x_n-x_{n+1}=x_n-\frac 12\left(x_n + \frac a{x_n}\right)=\frac{2x^2_n}{2x_n}-\frac{x^2_n}{2x_n}-\frac{a}{2x_n}=\frac{x^2_n - a}{2x_n} \ge 0\]

(v) From (iii) and (iv) it follows for \(n\ge 1\) that \[\frac a{x_{n}}\le \frac a{x_{n+1}}.\]

(vi) We have for all \(n\ge 1\) that \[\frac a{x_{n}}\le x_n\] Otherwise, we would have \((a/x_n)^2 > x_n^2\), which by (ii) leads to \((a/x_n)^2 > a\), in contradiction to (iii).

From (iii), (iv) and (vi), it follows that \((x_n)_{n\in\mathbb N}\) is a bounded sequence with

\[\frac{a}{x_1} \le x_n \le x_1.\]

and that it is monotonically decreasing. According to the monotone convergence theorem it has a limit. Thus, it must be a number \(x\) satisfying the equation \(x=\frac ax\). We denote \(x\) by \(\sqrt a\).

This follows from the rule that \((-x)(-y)=xy\) holds for all \(x,y\in\mathbb R\).

By construction, the sequence \((x_n)_{n\in\mathbb N}\) converges to the number \(x\) satisfying \(x=\frac ax\). The proposition follows from the fact that the limit of a convergent sequence is unique.

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**Forster Otto**: "Analysis 1, Differential- und Integralrechnung einer VerĂ¤nderlichen", Vieweg Studium, 1983