existence of square roots of non-negative real numbers

Theorem.

Every non-negative real number has a square root.

Proof.

Let $x\geq 0\in\mathbb{R}$ . If $x=0$ then the result is trivial, so suppose $x>0$ and define $S=\{y\in\mathbb{R}:y>0\text{ and }y^{2}<x\}$ . $S$ is nonempty, for if $0<y<\min\{x,1\}$ , then $y^{2}<y<x$ , and $y\in S$ . $S$ is also bounded above, for if $y>\max\{x,1\}$ , then $y^{2}>y>x$ , so such a $y$ is an upper bound of $S$ . Thus $S$ is nonempty and bounded, and hence has a supremum which we denote $L$ . We will show that $L^{2}=x$ . First suppose $L^{2}<x$ . By the Archimedean Principle there exists some $n\in\mathbb{N}$ such that $n>(2L+1)/(x-L^{2})$ . Then we have

$\bigg{(}L+\dfrac{1}{n}\bigg{)}^{2}=L^{2}+\dfrac{2L}{n}+\dfrac{1}{n^{2}}<L^{2}+% \dfrac{2L}{n}+\dfrac{1}{n}<x\text{.}$

(1)

So $L+1/n$ is a member of $S$ strictly greater than $L$ , contrary to assumption. Now suppose that $L^{2}>x$ . Again by the Archimedean Principle there exists some $n\in\mathbb{N}$ such that $1/n<(L^{2}-x)/2L$ and $1/n<L$ . Then we have

$\bigg{(}L-\dfrac{1}{n}\bigg{)}^{2}=L^{2}-\dfrac{2L}{n}+\dfrac{1}{n^{2}}>L^{2}-% \dfrac{2L}{n}>x\text{.}$

(2)

But there must exist some $y\in S$ such that $L-1/n<y<L$ , which gives $x<\big{(}L-1/n\big{)}^{2}<y^{2}$ , so that $y\notin S$ , a contradiction. Thus it must be that $L^{2}=x$ . ∎

Title	existence of square roots of non-negative real numbers
Canonical name	ExistenceOfSquareRootsOfNonnegativeRealNumbers
Date of creation	2013-03-22 16:32:42
Last modified on	2013-03-22 16:32:42
Owner	PrimeFan (13766)
Last modified by	PrimeFan (13766)
Numerical id	8
Author	PrimeFan (13766)
Entry type	Theorem
Classification	msc 11A25
Related topic	AxiomOfAnalysis
Related topic	ArchimedeanProperty
Related topic	Supremum
Related topic	ExistenceOfNthRoot