List of fractions $\dfrac{a}{b}$ with $ab < 1988$ 
All irreducible positive rational numbers such that the product of the numerator and the denominator is less than $1988$ are written in increasing order. Prove that any two adjacent fractions $\dfrac{a}{b}$ and $\dfrac{c}{d}$, $\dfrac{a}{b} < \dfrac{c}{d}$ satisfy the equation $bc-ad = 1$.

I am wondering how to relate the above condition, $ab < 1988$, to the definition about Farey sequences below which says $0 \leq a \leq b \leq n$. Also, $1988$ should be able to be replaced by any positive integer.
Definition. For any positive integer $n$, the Farey sequence $F_n$ is the sequence of rational numbers $\dfrac{a}{b}$ with $0 \leq a \leq b \leq n$ and $\gcd(a,b) = 1$ arranged in increasing order.
Theorem. If $\dfrac{p_1}{q_1}$ and $\dfrac{p_2}{q_2}$ are two successive terms in a Farey sequence, then $p_2q_1-p_1q_2 = 1$.
 A: It's actually quite related to Farey sequences. Imagine all the fractions/pairs
$$M_n=\left \{ \frac{a}{b} \mid (a,b) \in \mathbb{N}^2, 1 \leq a\cdot b \leq n, \gcd(a,b)=1\right \}$$
Which also means $1 \leq a \leq n$ and $1 \leq b \leq n$. We can split $M_n$ into 2 subsets:
$$M_n=\left \{ \frac{a}{b} \in M_n \mid a \leq b\right \} \bigcup \left \{ \frac{a}{b} \in M_n \mid a > b\right \} $$
The first one is literally Farey sequence (well, except $0$ but it should be easy to deal with this case), the other contains all the inverses of the Farey sequence except $0,1$, i.e.
$$M_n=F_n \bigcup \left \{ \frac{1}{\alpha} \mid \alpha \in F_n \setminus \left \{ 0,1 \right \}\right \} $$
The property of $bc - ad=1$ holds for $F_n$ individually. 
And so it does for $\left \{ \frac{1}{\alpha} \mid \alpha \in F_n \setminus \left \{ 0,1 \right \}\right \}$ because $\frac{a}{b} < \frac{c}{d}$ implies $\frac{d}{c} < \frac{b}{a}$ which still yields $bc - ad=1$.
The only remaining case to address is the "transition" from the last element of $F_n$, which is always $1$, to the first element of $\left \{ \frac{1}{\alpha} \mid \alpha \in F_n \setminus \left \{ 0, 1 \right \}\right \}$, which is $\frac{n}{n-1}$.
