# 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}. $S$ is nonempty, for if $0, then $y^{2}, 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}. 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}} (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. 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, which gives $x<\big{(}L-1/n\big{)}^{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 ExistenceOfSquareRootsOfNonnegativeRealNumbers 2013-03-22 16:32:42 2013-03-22 16:32:42 PrimeFan (13766) PrimeFan (13766) 8 PrimeFan (13766) Theorem msc 11A25 AxiomOfAnalysis ArchimedeanProperty Supremum ExistenceOfNthRoot