# Proof

(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:

### $$(1)$$ We show that $$\lim_{n\rightarrow\infty} x_n=\sqrt a$$.

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 $$\sqrt a$$.

### $$(2)$$ With $$\sqrt a$$ also $$-\sqrt a$$ is a solution of $$( * )$$.

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

### $$(3)$$ There are no other solutions of $$( * )$$.

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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### References

#### Bibliography

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