Theorem (Archimedean property).
Let $\varepsilon$ and $x$ be two positive real numbers. Then $n\varepsilon > x$ for some $n\in \mathbb N$.
Although the Archimedean property seems obvious by taking $n=\left\lceil\frac{x}{\varepsilon} \right\rceil$, a formal proof is given in the next section by the completeness property of $\mathbb R$. The following are some equivalent statements of the Archimedean property:
Corollary.
The following are equivalent.
(a) For positive real numbers $\varepsilon$ (however small) and $x$ (however large), we have $n\varepsilon > x$ for some $n\in \mathbb N$.
(b) For a positive real numbers $\varepsilon$ (however small), we have $\frac{1}{n}<\varepsilon$ for some $n\in \mathbb N$.
(c) For a positive real number $x$ (however large), we have $n > x$ for some $n\in \mathbb N$. (i.e., $\mathbb N$ is unbounded).
$(a)\implies (b)$ Set $x=1$ in (a).
$(b)\implies (c)$ Set $\varepsilon=1/x$ in (b).
$(c)\implies (a)$ Set $x=x/\varepsilon$ in (c).
Note that statement (c) means that there is no largest real number.
We can use the Archimedean property to prove that $\mathbb Q$ is dense in $\mathbb R$, i.e., there is a rational number between any two real numbers.
Theorem (The density of $\mathbb Q$ in $\mathbb R$). Let $a$ and $b$ be two real numbers such that $a < b$. Then there exists a rational number $r$ such that $a < r < b$.
When $a < 0 < b$, $r=0\in \mathbb Q$. Assume $a > 0$. By the Archimedean Property on $\varepsilon=b-a >0$ and $x=1$, we find a natural number $n$ such that $n(b-a) > 1$ which implies $na+1 < nb$. Similarly we find a natural number $m$ such that $m > na$. By the Well-ordering Property, there exists a smallest integer $m'$ such that $m' > na$. Then $m' > na\geq m'-1$ which implies
\[na < m'\leq na+1 < nb.\]
Dividing by $n$ we get $a < \frac{m'}{n} < b$. Here $r=\frac{m'}{n}\in \mathbb Q$.
When $a < b < 0$, we can similarly find $r=-\frac{m'}{n}\in \mathbb Q$ between $a$ and $b$ where
\[-b < \frac{m'}{n} < -a.\]
