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 $y\otin S$ , a contradiction. Thus it must be that $L^{2}=x$ .∎