Rational Numbers are Densely Ordered

Theorem

Let $a, b \in \Q$ such that $a < b$.

Then $\exists c \in \Q: a < c < b$.

That is, the set of rational numbers is densely ordered.

Proof

From the definition of rational numbers, we can express $a$ and $b$ as $a = \dfrac {p_1} {q_1}, b = \dfrac {p_2} {q_2}$.

Thus from Mediant is Between:

$\dfrac {p_1} {q_1} < \dfrac {p_1 + p_2} {q_1 + q_2} < \dfrac {p_2} {q_2}$

From Rational Numbers form Field:

$\dfrac {p_1 + p_2} {q_1 + q_2} \in \Q$

Hence $c = \dfrac {p_1 + p_2} {q_1 + q_2}$ is an element of $\Q$ strictly between $a$ and $b$.

$\blacksquare$

Sources

• 1996: Winfried Just and Martin Weese: Discovering Modern Set Theory. I: The Basics ... (previous) ... (next): Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations