Mediant is Dependent upon Representation

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

Let $r$ and $s$ be expressed as:
 * $r = \dfrac a b$
 * $s = \dfrac c d$

where $a, b, c, d$ are integers such that $b > 0, d > 0$.

Then the mediant of $r$ and $s$ is dependent upon the specific integers chosen for $a, b, c, d$.

Proof

 * Proof by Counterexample

Let $r = \dfrac 1 2$ and $s = 1$.

We have:


 * $r = \dfrac 1 2 = \dfrac 2 4 = \dfrac 3 6$

Then the mediant of $r = \dfrac 2 4$ and $s = \dfrac 1 1$ gives:


 * $\dfrac {2 + 1} {4 + 1} = \dfrac 3 5$

but the mediant of $r = \dfrac 1 2$ and $s = \dfrac 1 1$ gives:


 * $\dfrac {1 + 1} {2 + 1} = \dfrac 2 3$

which is not the same as $\dfrac {2 + 1} {4 + 1}$.

Also see

 * Applesellers' Problem