Theorem 1   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 $S=\{y\in\mathbb{R}:y> 0{ and }y^2<x\}$ . $S$ 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 \begin{equation} \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{.} \end{equation}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 \begin{equation} \bigg(L-\dfrac{1}{n}\bigg)^2=L^2-\dfrac{2L}{n}+\dfrac{1}{n^2} >L^2-\dfrac{2L}{n}>x\text{.} \end{equation}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 $L^2=x$ .