Problem source

Let $a,b,c,d,p$ and $q$ be positive integers. Prove that:
\begin{questionparts}
\item if $b > a$ and $c > 1,$ then $bc\geqslant2c\geqslant2+c$; 
\item if $a < b$ and $d < c$, then $bc-ad\geqslant a+c$; 
\item if ${\displaystyle \frac{a}{b} < p < \frac{c}{d}}$, then $\left(bc-ad\right)p\geqslant a+c$; 
\item if ${\displaystyle \frac{a}{b} < \frac{p}{q}  < \frac{c}{d}},$ then ${\displaystyle p\geqslant\frac{a+c}{bc-ad}}$
and ${\displaystyle q\geqslant\frac{b+d}{bc-ad}}$. 
\end{questionparts}
Hence find all fractions with denominators less than 20 which lie between $8/9$ and $9/10$.

Solution source

\begin{questionparts}
\item If $b > a$ and $c > 1$ then $c \geq 2 \underbrace{\Rightarrow}_{\times c} bc \geq 2c = c+c \underbrace{\geq}_{c \geq 2} 2 + c$
\item If $a < b$ and $d < c$ then $b \geq a+1$ and $c \geq d+1$ so

\begin{align*}
bc - ad &\underbrace{\geq}_{b \geq a+1} (a+1)c - ad \\
&\underbrace{\geq}_{d \leq c-1} (a+1)c - a(c-1) \\
&= a+c
\end{align*}

\item If $\displaystyle \frac{a}{b} < p < \frac{c}{d}$ then $a < pb$ and $pd < c$ so by the previous part $(pb)c - a(pd) \geq a + c \Leftrightarrow (bc-ad)p \geq a+c$.

\item If $\displaystyle \frac{a}{b} < \frac{p}{q} < \frac{c}{d}$ then $\displaystyle \frac{qa}{b} < p < \frac{qc}{d}$ and so by the previous part we must have

$(bc-ad)qp \geq q(a+c) \Rightarrow p \geq \frac{a+c}{bc-ad}$. Similarly we have $\frac{d}{c} < \frac{q}{p} < \frac{b}{a}$ and so $q \geq \frac{b+d}{bc-ad}$

\end{questionparts}

Suppose $\frac{p}{q}$ is a fraction such that $q \leq 20$ and $\frac89 < \frac{p}q < \frac9{10}$ then:

\begin{align*}
q & \leq 20 \\
p & \geq \frac{17}{81-80} = 17 \\
q & \geq \frac{19}{81-80} = 19
\end{align*}

Therefore the only fraction
is $\frac{17}{19}$ since $\frac{18}{19} > \frac{18}{20} = \frac{9}{10}$