Density_of_$bb(Q)$in$bb(R)$

Theorem: For every two real numbers $a$ and $b$ with $a < b$, there exists a rational number $r$ satisfying $a < r < b$.

Proof: A rational number is a quotient of integers, so we must produce an $m \in \mathbb{Z}$ and $n \in \mathbb{N}$ such that

$$ a < \frac{m}{n} < b $$

The first step is to choose the denominator $n$ large enough so that the consecutive increments of size $1/n$ are too close together to “step over” the interval $\left. (a,b) \right.$.

Using the Archimedean_Property, we may pick $n \in \mathbb{N}$ large enough such that

$$ \frac{1}{n} < b - a $$

Inequality (1), which we are trying to prove, is equivalent to $na < m < ab$. With $n$ already chosen, the idea now is to choose $m$ to be the smallest integer greater than $na$. In other words, pick $m \in \mathbb{Z}$ such that

$$ \frac{m - 1}{n} \leq a \leq \frac{m}{n} $$

rather, that

$$ m - 1\overset{\left. (3) \right.}{\leq}na\overset{\left. (4) \right.}{<}m $$

Now, inequality (4) immediately yields $a < \frac{m}{n}$, which is half the battle. Keeping in mind the inequality (2) is equivalent to $a < b - \frac{1}{n}$, we can use (3) to write

$$ \begin{aligned} m - 1 & \leq na \\ m & \leq na + 1 \\ & < n\left. \left( b - \frac{1}{n} \right) \right. + 1 \\ & = nb \end{aligned} $$

Because $m < nb$ imples $\frac{m}{n} < b$, we have $a < \frac{m}{n} < b$ as desired.

So $\mathbb{Q}$ is dense in $\mathbb{R}$.