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 $$r = \frac a b < \frac c d = s$$.

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

Proof
The same proof can apply to both.

Let $$r, s \in \R$$ be such that $$r < s$$ and $$r = \frac a b, s = \frac c d$$, 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 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 $$\frac a b < \frac {a + c} {b + d}$$.

The other half is proved similarly.