Mediant is Between

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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
\(\displaystyle \) \(=\) \(\displaystyle \paren {a + c} b\)


From Reciprocal of Positive Real Number is 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.

$\blacksquare$


Sources