Mediant is Between

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Theorem

Let $a, b, c, d \in \R$ be 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 1

Let $a, b, c, d \in \Z$ be integers such that:

\(\text {(1)}: \quad\) \(\ds b, d\) \(>\) \(\ds 0\)
\(\text {(2)}: \quad\) \(\ds \dfrac a b\) \(<\) \(\ds \dfrac c d\)


Then the mediant of $\dfrac a b$ and $\dfrac c d$ is between $\dfrac a b$ and $\dfrac c d$:

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


Corollary 2

Let $a, b, c, d \in \R$ be real numbers such that $b > 0, d > 0$.

Let $\dfrac a b = \dfrac c d$.

Then:

$\dfrac a b = \dfrac {a + c} {b + d} = \dfrac c d$


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

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