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# 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 $L^{2}=x$. ∎

## Mathematics Subject Classification

11A25*no label found*

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