Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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

2 Answers 2

up vote 14 down vote accepted

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:

$$ a_{n+1} = \frac{a_n}{2} + \frac{1}{a_n}$$

share|improve this answer

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, $\sqrt{a} \in \mathbb R$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.