Statement: Prove that $\mathbb{Q}$ is dense in $\mathbb{R}$

Proof: To prove $\mathbb{Q}$ is dense in $\mathbb{R}$, it will suffice to show that for any $r_1,r_2 \in \mathbb{R}$, $r_2 \gt r_1 => \exists$ $q \in \mathbb{Q}$ such that $r_1 \lt q \lt r_2$. i.e We can always find a rational between any two reals.

If $r_1 \lt 0 \lt r_2$, we have our proof because $0 \in Q$. On the other hand if $r_1 \lt r_2 \lt 0$, this is same as the case $-r_1 \gt -r_2$, therefore without loss of generality, we can only need to solve the condition where $0 \lt r_1 \lt r2$.

Since $\mathbb{R}$ is an ordered field and $r_2 - r_1 \gt 0$, the Archimedean Property (proved below) tells us $\exists$ $n \in \mathbb{N}$ where $n(r_2 - r_1) \gt 1$. Now consider the set $S = \{s \in \mathbb{N_o} : s \leq nr_1\}$, clearly $S \subset \mathbb{N_o}$, therefore $S \subset \mathbb{R}$ and therefore by definition $S$ has a supremum $\sup(S) \in S$, and is bounded by $nr_1$ so that $\sup(S) \lt nr_1$. Therefore $\sup(S) + 1 \gt nr_1$.

We also have (by definition) $\sup(S) \lt nr_1$, therefore $\sup(S) \lt nr_1 + 1$, therefore we have $nr_1 \lt \sup(S) \lt nr_2$; set $\sup(S) = m$, which leads us to the statement we seek: $r_1 \lt \frac{m}{n} \lt r_2$; since $m,n \in \mathbb{N}$, $\frac{m}{n} \in \mathbb{Q}$

(Generic) Statement: An ordered field $F$ has the Archimedian Property if for $x, y \in \mathbb{F}$, $\exists$ $n \in \mathbb{N}$ where $nx \gt y$

Proof (for $\mathbb{R}$): $nx \gt y$ is same as $n \gt \frac{y}{x}$. Assume this statement is false and no such $n$ exists. This is same as saying $\frac{y}{x}$ is the $\sup(\mathbb{N})$. Let $M$ be the maximal element of $\mathbb{N}$ bounded by $\frac{y}{x}$, clearly $M \lt \frac{y}{x}$ and by definition $M + 1 \gt \frac{y}{x}$. We found an integer $M+1$ that is bigger than our supposed $\sup(\mathbb{N})$ that satisfies the property $(M+1)x \gt y$.