Let \(a,b,c,d,p\) and \(q\) be positive integers. Prove that:
if \(b > a\) and \(c > 1,\) then \(bc\geqslant2c\geqslant2+c\);
if \(a < b\) and \(d < c\), then \(bc-ad\geqslant a+c\);
if \({\displaystyle \frac{a}{b} < p < \frac{c}{d}}\), then \(\left(bc-ad\right)p\geqslant a+c\);
if \({\displaystyle \frac{a}{b} < \frac{p}{q} < \frac{c}{d}},\) then \({\displaystyle p\geqslant\frac{a+c}{bc-ad}}\)
and \({\displaystyle q\geqslant\frac{b+d}{bc-ad}}\).
Hence find all fractions with denominators less than 20 which lie between \(8/9\) and \(9/10\).
Solution:
If \(b > a\) and \(c > 1\) then \(c \geq 2 \underbrace{\Rightarrow}_{\times c} bc \geq 2c = c+c \underbrace{\geq}_{c \geq 2} 2 + c\)
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*}
If \(\displaystyle \frac{a}{b} < p < \frac{c}{d}\) then \(a < pb\) and \(pd < c\) so by the previous part \((pb)c - a(pd) \geq a + c \Leftrightarrow (bc-ad)p \geq a+c\).
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 \(q \geq \frac{b+d}{bc-ad}\)
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 \(\frac{18}{19} > \frac{18}{20} = \frac{9}{10}\)