TITLE: How to prove that construction of Farey sequence by mediant is coverage?
QUESTION [5 upvotes]: Farey sequence of order $n+1$ ($F_{n+1}$) can be construct by adding mediant value (${a+c \over b+d}$) into $F_{n}$, where ${a \over b}$ and ${c \over d}$ are consecutive term in $F_{n}$, and  $b+d = n+1$.
I've already prove that

${a \over b} < {a+c \over b+d} < {c \over d}$
$b+d$ always irreducible in ${a+c \over b+d}$.
the middle of any 3 consecutive term in any $F_{n}$ are mediant.
Number of new elements added is $\phi(n+1)$.

Now I wonder does the construction by mediant value coverage all elements in $F_{n+1}$. It's easy to show that ${1 \over n+1}$ and ${n \over n+1}$ does involve in it, but I have no idea how to show that other ${m \over n+a}$, where $gcd(m, n+1) = 1$, are going to be constructed.
Please give me a hint.

REPLY [3 votes]: The gap in Gerry Myerson's proof can be filled easily indeed. It is easy to show by induction that $ad-bc=1$ for any two adjacent fractions $\frac{a}{b} > \frac{c}{d}$in any $F_n$. The key property is then that if $\frac{a}{b} >\frac{p}{q} > \frac{c}{d}$ with $a,b,c,d,p,q>0$, $ad-bc=1$, then $q\geq b+d$.
In fact, this last property has already been explained (by me) in an older MSE question : see here.