# 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:

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

### Corollary

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

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

## 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:

 $\displaystyle \frac a b$ $<$ $\displaystyle \frac c d$ $\displaystyle \leadsto \ \$ $\displaystyle \frac a b \paren {b d}$ $<$ $\displaystyle \frac c d \paren {b d}$ Real Number Ordering is Compatible with Multiplication $\displaystyle \leadsto \ \$ $\displaystyle a d$ $<$ $\displaystyle b c$

Then:

 $\displaystyle a \paren {b + d}$ $=$ $\displaystyle a b + a d$ $\displaystyle$ $<$ $\displaystyle a b + b c$ Real Number Ordering is Compatible with Addition, as $a d < b c$ from above $\displaystyle$ $=$ $\displaystyle \paren {a + c} b$
$\paren {a + c}^{-1} > 0$

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.

$\blacksquare$