Existence of Dyadic Rational between two Rationals

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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:

\(\ds m - 1\) \(\le\) \(\ds a 2^r\)
\(\ds \leadsto \ \ \) \(\ds m\) \(\le\) \(\ds a 2^r + 1\)
\(\ds \leadsto \ \ \) \(\ds \frac m {2^r}\) \(\le\) \(\ds a + \frac 1 {2^r}\)
\(\ds \) \(<\) \(\ds a + \paren {b - a}\)
\(\ds \) \(=\) \(\ds b\)


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

$\blacksquare$


Sources

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