Existence of Dyadic Rational between two Rationals

Theorem
Let $a$ and $b$ be rational numbers such that $a < b$.

Then there exist integers $m$ and $r$ such that:
 * $a < \dfrac m {2^r} < b$

That is, there exists a dyadic rational between any pair of rational numbers.

Proof
As $a < b$ it follows that $a \ne b$ and so $b - a \ne 0$.

Thus:
 * $\dfrac 1 {b - a} \in \R$

By the Axiom of Archimedes:
 * $\exists r \in \N: r > \dfrac 1 {b - a}$

Notice that $2^r > r$.

Thus we also have:
 * $2^r > \dfrac 1 {b - a}$

Let $M := \set {x \in \Z: x > a 2^r}$.

By Set of Integers Bounded Below has Smallest Element, there exists $m \in \Z$ such that $m$ is the smallest element of $M$.

That is:
 * $m > a 2^r$

and, by definition of smallest element:
 * $m - 1 \le a 2^r$

As $2^r > \dfrac 1 {b - a}$, it follows from Ordering of Reciprocals that:
 * $\dfrac 1 {2^r} < b - a$

Thus:

Thus we have shown that $a < \dfrac m {2^r} < b$.