$\frac{0}{4},\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{4}{4};\frac{0}{3},\frac{1}{3},\frac{2}{3},\frac{3}{3};\frac{0}{2},\frac{1}{2},\frac{2}{2};\frac{0}{1},\frac{1}{1}$

Obviously there's a bunch of duplicates here: all the fractions equal to zero or one are the same as each other, there's also $\frac{2}{4}=\frac{1}{2}$. We also need them to be in ascending order, so therefore the proper Farey Sequence for F₄ is the following:

$F_4=\frac{0}{1},\frac{1}{4},\frac{1}{3},\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{1}{1}$

F₄

$F_4=\frac{0}{1},\frac{1}{4},\frac{1}{3},\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{1}{1}$

Some properties of a Farey Sequence:

If a/b and c/d are successive terms of F_n then $b+d>n$. The mediant of the fractions is $\frac{a+c}{b+d}$ and we know it will always be in between the other two like so: $\frac{a}{c}<\frac{a+b}{c+d}<\frac{b}{d}$ this is because(from the Wikipedia page for mediant) $\frac{a+b}{c+d}-\frac{a}{c}=\frac{bc-ad}{c(c+d)}=d(c+d)(\frac{b}{d}-\frac{a}{c})$ and since b/d > a/c this will always be positive so we know the mediant is greater than a/c and $\frac{b}{d}-\frac{a+b}{c+d}=\frac{bc-ad}{d(c+d)}=\frac{c}{c+d}(\frac{b}{d}-\frac{a}{c})$ which will also be positive for the same reason so the mediant is less than b/d. It might be more intuitive if we look at an example, using 1/2 and 2/3. $\frac{1+2}{2+3}=\frac{3}{5}$ which is in between the two numbers. You can kind of think of it as smooshing the 2 numbers together and getting something in between. The significance of this property for us is that the denominator of the mediant is b+d and if it was less than or equal to n then it would be in F_n but since these are successive terms it can't be, so b+d must be greater than n.

If n > 1, then all successive terms of F_n have differing denominators. Obviously for n = 1, we would have 0/1 and 1/1 in sequence, but for any other n we can't have a/b and c/b as successive terms since a+1≤c<b and $\frac{a}{b}<\frac{a}{b-1}<\frac{a+1}{b}\le \frac{c}{b}$. So e.g. for 1/3 and 2/3 for F₄ they can't be successive since 1/2 comes in between.

If a/b and c/d are successive terms of F_n then we have $bc-ad=1$

e.g. for F₄ we have 1/2 followed by 2/3 so we get $2\times 2-3\times 1=1$

Additionally, if a/b, c/d and e/f are three successive terms of F_n then we have $\frac{c}{d}=\frac{a+e}{b+f}$

e.g. for F₄ we have 1/2, 2/3 and 3/4 successively so $\frac{2}{3}=\frac{1+3}{2+4}$

To prove these last two we first show that the two statements are equivalent to each other. If we assume $bc-ad=1$ then for three successive equations we can form $de-cf=1$ as well. Now we want to get equations for c and d without each other so we do the following: If we do a similar thing for d And if we put our first equation over the other equation we get $\frac{c}{d}=\frac{a+e}{b+f}$ as expected.

To prove they're equivalent we need to prove it goes the other way as well. i.e. a implies b and b implies a. Thus if we assume $\frac{c}{d}=\frac{a+e}{b+f}$ is true then we can show this means the other equation must be true by induction. To prove something by induction means to prove something is true for a specific case then prove that if it's true for a certain case it will be true for the next case. This is useful for us because if $bc-ad=1$ then we can do the following:

The base case is evident since the first two elements for any F_n will be 0/1 and 1/n and $1\times 1-0\times n=1$ Thus we have shown that the two equations are equivalent to each other but we have still yet to prove they're true.

To do this, we will also use induction; this time using F_n-1 to F_n as the inductive case. It's obviously true for n = 1 as the base case. Suppose that a/b and e/f are consecutive for F_n-1 but for F_n they're separated by c/d. If we set $ad-bc=r$ and $cf-de=s$ then we can do the following, remembering that $af-be=1$ from F_n-1

We can do a similar thing for d to get $d=bs+fr$ so that we can form $\frac{c}{d}=\frac{as+er}{bs+fr}$. What values of r and s can work for this? They have to be greater than zero since otherwise it would just be equal to a/b or e/f and wouldn't be separate for the Farey Sequence. To be a new value in F_n it must be the smallest possible value otherwise and so it must be r=s=1, and thus c/d is the mediant of the other two. Thus, $\frac{c}{d}=\frac{a+e}{b+f}$. As an example, for 2/3 and 1/1 in F_n, $\frac{c}{d}=\frac{2s+1r}{3s+1r}$ since this is a new value that was just added to F_n and is in between two values that were in F_n-1, if s or r were greater than 1 then the value for r=s=1 would've already been included in F_n-1 and 2/3 and 1/1 wouldn't be successive in F_n

This is a bit handwavy and I don't formalise it much but there are other proofs given in the book chapter I'm basing this off which I may return to.