Mediant is Between

Theorem
Let $r, s \in \Q$, i.e. let $r, s$ be rational numbers.

Then the mediant of $r$ and $s$ is between $r$ and $s$.

Real Numbers
Let $a, b, c, d$ be any real numbers such that $b > 0, d > 0$.

Let $r = \dfrac a b < \dfrac c d = s$.

Then:
 * $r < \dfrac {a + c} {b + d} < s$

Proof
The same proof can apply to both.

Let $r, s \in \R$ be such that $r < s$ and $r = \dfrac a b, s = \dfrac c d$, where $a, b, c, d$ are real numbers such that $b > 0, d > 0$.

Because $b, d > 0$, it follows that $b d > 0$ from Real Number Ordering is Compatible with Multiplication. Thus we have:

Then:

From Inverse of Positive Real is Positive, $\left({a + c}\right)^{-1} > 0$ and $b^{-1} > 0$.

It follows from Ordering is Compatible with Multiplication that $\dfrac a b < \dfrac {a + c} {b + d}$.

The other half is proved similarly.