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:

\(\ds \frac a b\)

\(<\)

\(\ds \frac c d\)

\(\ds \leadsto \ \ \)

\(\ds \frac a b \paren {b d}\)

\(<\)

\(\ds \frac c d \paren {b d}\)

Real Number Ordering is Compatible with Multiplication

\(\ds \leadsto \ \ \)

\(\ds a d\)

\(<\)

\(\ds b c\)

Then:

\(\ds a \paren {b + d}\)

\(=\)

\(\ds a b + a d\)

\(\ds \)

\(<\)

\(\ds a b + b c\)

Real Number Ordering is Compatible with Addition, as $a d < b c$ from above

\(\ds \)

\(=\)

\(\ds \paren {a + c} b\)

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.

$\blacksquare$

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 1$: Real Numbers: Exercise $\S 1.8 \ (2)$