Proof. Let $x$ be a real number, and let $S = \{ a \in \mathbb{N} : a \leq x \}$ . If $S$ is empty, let $n = 1$ ; note that $x < n$ (otherwise $1 \in S$ ).

Assume $S$ is nonempty. Since $S$ has an upper bound, $S$ must have a least upper bound; call it $b$ . Now consider $b - 1$ . Since $b$ is the least upper bound, $b - 1$ cannot be an upper bound of $S$ ; therefore, there exists some $y \in S$ such that $y > b - 1$ . Let $n = y + 1$ ; then $n > b$ . But $y$ is a natural, so $n$ must also be a natural. Since $n > b$ , we know $n \not\in S$ ; since $n \not\in S$ , we know $n > x$ . Thus we have a natural greater than $x$ .

Corollary 1   If $x$ and $y$ are real numbers with $x > 0$ , there exists a natural $n$ such that $nx > y$ .

Proof. Since $x$ and $y$ are reals, and $x \neq 0$ , $y/x$ is a real. By the Archimedean property, we can choose an $n \in \mathbb{N}$ such that $n > y/x$ . Then $nx > y$ .

Corollary 2   If $w$ is a real number greater than $0$ , there exists a natural $n$ such that $0 < 1/n < w$ .

Proof. Using Corollary 1, choose $n \in \mathbb{N}$ satisfying $nw > 1$ . Then $0 < 1/n < w$ .

Corollary 3   If $x$ and $y$ are real numbers with $x < y$ , there exists a rational number $a$ such that $x < a < y$ .

Proof. First examine the case where $0 \leq x$ . Using Corollary 2, find a natural $n$ satisfying $0 < 1/n < (y-x)$ . Let $S = \{ m \in \mathbb{N} : m/n \geq y \}$ . By Corollary 1 $S$ is non-empty, so let $m_0$ be the least element of $S$ and let $a = (m_0-1)/n$ . Then $a< y$ . Furthermore, since $y \leq m_0/n$ , we have $y - 1/n < a$ ; and $x < y - 1/n < a$ . Thus $a$ satisfies $x < a < y$ .

Now examine the case where $x < 0 < y$ . Take $a = 0$ .

Finally consider the case where $x < y \leq 0$ . Using the first case, let $b$ be a rational satisfying $-y < b < -x$ . Then let $a = -b$ .