# Prove that $\sqrt{2}$ is a real number. [closed]

I remember I saw this question somewhere in Lang's undergraduate real analysis.

Given any real number $\ge0$, show that it has a square root.

## closed as off-topic by user21820, Did, Jendrik Stelzner, user91500, HoloSep 9 '18 at 11:22

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• You can give an iteration sequence for it and show that it is Cauchy. Then use the completeness of $\mathbb R$. – AlexR Sep 24 '13 at 8:21

It depends on how you define a real number. If you use Dedekind cuts, then you should show that the set $\{ x \in \mathbb{Q}^+: x^2<2\}$ is a Dedekind cut. If you use Cauchy sequences to define a real number, you can prove that the the sequence that is obtained from Newton's method is Cauchy:
$$a_{n+1} = \frac{a_n}{2} + \frac{1}{a_n}$$
This means to show that for $a\geq 0$, the Polynomial $x^2-a$ has at least one real root. Chose $x_0 := 1$ and use Newton's method with $$x_{n+1} = x_n - \frac{x_n^2 - a}{2x_n}$$ Then since $(x_n)$ is cauchy and $a \geq 0$ is a fixpoint of the iteration $$\lim_{n\to\infty} x_n^2 = a$$ $x_n \to \sqrt{a}$ and since $\mathbb R$ is complete, $\sqrt{a} \in \mathbb R$.