# Mediant is Between

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## Contents

## 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 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 $\S 1.8 \ (2)$