consecutive convergents Problem: Let $\phi=\frac{1+\sqrt{5}}{2}$ be the golden ratio and let $a$, $b$, $c$, $d$ be positive integers so that $\frac{a}{b}>\phi>\frac{c}{d}$. It is also known that $ad-bc=1$. Prove that $a/b$ and $c/d$ are consecutive convergents of $\phi$. 
Numerical experimentations point towards the validity of this statement.
The converse is well known (and easy to show) but I cannot seem to prove the direct statement. 
This is not a homework question; I came across it while investigating the geometric discrepancy of a certain lattice point set. Any help would be appreciated.
 A: Note that $\phi=1+\frac{1}{1+\frac{1}{1+\ldots}}$ has convergents $\frac{f_{n+1}}{f_n}$, i.e. ratios of consecutive Fibonacci numbers. Note that I have used lower case for the Fibonacci numbers so as to avoid confusion with the Farey sequence $F_n$.

The main idea is to appeal to the properties of Farey sequences.


*

*Suppose $a, b, c, d$ are positive integers with $a \leq b, c \leq d, \frac{c}{d}<\frac{a}{b}$. Then $\frac{a}{b}$ and $\frac{c}{d}$ are consecutive members of the Farey sequence $F_n$ for some $n$ if and only if $ad-bc=1$. 

*If $\frac{a}{b}<\frac{c}{d}$ are consecutive members of the Farey sequence $F_n$ for some $n$, then either they are consecutive members in $F_{n+1}$, or $\frac{a}{b}, \frac{a+c}{b+d}, \frac{c}{d}$ are consecutive members in $F_{n+1}$, in which case we have $b+d=n+1$. In other words, as we increase the order of the Farey sequence, $\frac{a+c}{b+d}$ is the first term to appear between $\frac{a}{b}$ and $\frac{c}{d}$.

Consider $0<\frac{1}{\phi}<1$. 
For each $m$, suppose that the Farey sequence $F_m$ is given by $\frac{0}{1}=a_{m, 0}<a_{m, 1}< \ldots <a_{m, |F_m|-1}=\frac{1}{1}$. We partition the interval $[0, 1)$ into $|F_m|-1$ intervals $[a_{m, i}, a_{m, i+1})$. Note that $\frac{1}{\phi}$ must belong to exactly one such interval. Also note that $\frac{1}{\phi}$ is irrational and so cannot be equal to $a_{m, i}$. 
Thus for each $m$, there is a unique pair of rational numbers $r_m, s_m$ s.t. $r_m, s_m$ are consecutive members of the Farey sequence $F_m$ and $r_m<\frac{1}{\phi}<s_m$.

Observe that $\frac{1}{\phi}$ has convergents $\frac{f_{n-1}}{f_n}$. We observe that for $f_n \leq m<f_{n+1}$, we have that $\frac{f_{n-2}}{f_{n-1}}$ and $\frac{f_{n-1}}{f_n}$ are consecutive members of $F_m$. 
Explanation: This is because we have the identity $f_{n-2}f_n-f_{n-1}^2=(-1)^{n-1}$, so by property $1$ $\frac{f_{n-2}}{f_{n-1}}$ and $\frac{f_{n-1}}{f_n}$ are consecutive members of some Farey sequence $F_k$. Note that we necessarily have $k \geq f_n$, since $\frac{f_{n-1}}{f_n}$ is a member of $F_k$. Therefore $\frac{f_{n-2}}{f_{n-1}}$ and $\frac{f_{n-1}}{f_n}$ will be consecutive members in $F_{f_n}$. (Removing elements doesn't affect the fact that they are consecutive) Now, as we increase the order of the Farey sequence, the first term that appears between them is $\frac{f_{n-2}+f_{n-1}}{f_{n-1}+f_n}=\frac{f_n}{f_{n+1}}$, which cannot appear for $m<f_{n+1}$. Therefore $\frac{f_{n-2}}{f_{n-1}}$ and $\frac{f_{n-1}}{f_n}$ remain as consecutive members in the Farey sequence $F_m$, for $f_n \leq m<f_{n+1}$.
Also, as the convergents of $\frac{1}{\phi}$ are alternately greater and smaller than $\frac{1}{\phi}$, we see that $\frac{1}{\phi}$ is strictly between $\frac{f_{n-2}}{f_{n-1}}$ and $\frac{f_{n-1}}{f_n}$. 
Therefore $\frac{f_{n-2}}{f_{n-1}}$ and $\frac{f_{n-1}}{f_n}$ must be $r_m$ and $s_m$ in some order, i.e. $\{\frac{f_{n-2}}{f_{n-1}},\frac{f_{n-1}}{f_n}\}=\{r_m, s_m\}$.

Finally, from the question, $\frac{a}{b}>\phi>\frac{c}{d}$ so $\frac{d}{c}>\frac{1}{\phi}>\frac{b}{a}$. Since $ad-bc=1$, $\frac{b}{a}$ and $\frac{d}{c}$ are consecutive members of some Farey sequence $F_m$. Thus $\{\frac{b}{a}, \frac{d}{c}\}=\{r_m, s_m\}$
Clearly we have $f_n \leq m<f_{n+1}$ for some $n$ so by above $\{r_m, s_m\}=\{\frac{f_{n-2}}{f_{n-1}},\frac{f_{n-1}}{f_n}\}$, so $\{\frac{b}{a}, \frac{d}{c}\}=\{\frac{f_{n-2}}{f_{n-1}},\frac{f_{n-1}}{f_n}\}$. 
Therefore $\{\frac{a}{b}, \frac{c}{d}\}=\{\frac{f_{n-1}}{f_{n-2}},\frac{f_n}{f_{n-1}}\}$ are consecutive convergents of $\phi$. 
