*You may wish to explore Farey Sequences and Mediant Madness before working on this problem.*
The Farey sequence $F_n$ is the list written in increasing order of all the rational numbers between $0$ and $1$ that have only the numbers $1, 2, 3, ... n$ as denominators. We have $$\eqalign{F_1&=\frac{0}{1}, \frac{1}{1}\cr F_2&=\frac{0}{1}, \frac{1}{2}, \frac{1}{1}.}$$

For the two rational numbers $\frac{a}{c}$ and $\frac{b}{d}$ the mediant is defined as $\frac{a+b}{c+d}$.

**Show that if $0< \frac{b}{d} < \frac{a}{c}< 1$ then $\frac{b}{d} < \frac{a+b}{c+d} < \frac{a}{c}$.**
*Each Farey sequence $F_{n+1}$ must contain all of the terms of $F_{n}$, along with some new terms. Each 'new' term in the Farey sequence $F_{n+1}$ is the mediant of two consecutive terms in $F_n$, but not all mediants of consecutive terms of $F_n$ are included: where the denominator of the mediant is greater than $n+1$ the mediant does not occur in $F_{n+1}$ and instead two consecutive terms
in $F_n$ are repeated in $F_{n+1}$.*
**Use the mediants to work out $F_3, F_4$ and $F_5$.**
*Two consecutive terms in a Farey sequence are known as Farey Neighbours.*
Let $\frac{b}{d}$ and $\frac{a}{c}$ be Farey Neighbours. Choose some pairs of Farey Neighbours from the Farey Sequences you have worked out, and calculate $ad-bc$.

**What do you notice?**
**Can you prove it using mathematical induction?**
*This is the key result that links Farey sequences to infinite sets of circles that are packed together, each circle touching its neighbours. See the problem Ford Circles.*