Mediant is Between

Theorem
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
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 from Real Number Ordering is Compatible with Multiplication that:
 * $b d > 0$

Thus:

Then:

From Reciprocal of Strictly Positive Real Number is Strictly Positive:
 * $\paren {a + c}^{-1} > 0$

and:
 * $b^{-1} > 0$

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

The other half is proved similarly.