Farey Approximation
Stage: 5 Short Challenge Level: 
In 1816 the British geologist John Farey defined the Farey
sequence $F_n$ as 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 can do the same
thing for rational numbers between any two positive numbers.
For example we can consider sequences betweeen $1$ and $2$
where we have
\[ \eqalign{F_1&=\frac{1}{1}, \frac{2}{1}\cr
F_2&=\frac{1}{1}, \frac{3}{2}, \frac{2}{1}.} \]
What would $F_3$ and $F_4$ be in this case?
For the two positive rational numbers $\frac{b}{d}$ and
$\frac{a}{c}$ the mediant is defined as $\frac{a+b}{c+d}$. The
mediant has the nice property that it is always in between the
two fractions giving rise to it: if $0< \frac{b}{d} <
\frac{a}{c}$ then $\frac{b}{d} < \frac{a+b}{c+d} <
\frac{a}{c}$. You might want to verify this for yourself.
Clearly each Farey sequence $F_{n+1}$ must contain all of the
terms of $F_{n}$, along with some new terms. Mediants also have
the nice property that each 'new' term in the Farey sequence
$F_{n+1}$ is the mediant of two consecutive terms in $F_n$. Why
not test this out on $F_2, \dots, F_5$?
Farey sequences have all sorts of interesting properties. One
is that they give an interesting way of approximating rational
numbers.
In order to find a rational approximation to an irrational
number using Farey fractions you need to pick the interval
between Farey fractions that contains the target number and
narrow the interval at each step. If the target number is
between $\frac{b}{d}$ and $\frac{a}{c}$ then at the next step
you have to decide which of the two intervals
\[ \Big[\frac{b}{d}, \frac{a+b}{c+d}\Big], \quad
\Big[\frac{a+b}{c+d}, \frac{a}{c}\Big]. \]
contains the target number.
Carry out this numerical approximation process for $\sqrt 2$,
and for $\pi$, to get rational approximations that are correct
to 4 decimal places. You may like to use a spreadsheet. Compare
your results to the approximations obtained by using a
straightforward interval-halving method. What do you
notice?
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