If $a$ and $b$ are positve integers, show that the inequality $e <\frac{a}{b}<\frac{87}{32} $ imples that $b \geq 39$. If $a$ and $b$ are positve integers, show that the inequality $e <\frac{a}{b}<\frac{87}{32} $ imples that $b \geq 39$.
Use the continued fraction $e=[2:1,2,1,1,4,1,1,6,...]$

Note that $e>\frac{19}7$ and that $\frac{19}7<\frac ab<\frac{87}{32}$ implies that the numerators of the differences $\frac{a}{b}-\frac{19}{7}=\frac{7a-19b}{7b}$ and $\frac{87}{32}-\frac ab=\frac{87b-32a}{32b}$ are integers $\ge 1$.
Therefore
$$b=(7\cdot 87-32\cdot19)b=32(7a-19b)+7(87b-32a)\ge 32+7=39 .$$

Why is $e>\frac{19}7$?
$$\begin{align}[2:1,2,1,1,4,1,1,6,\ldots]>[2:1,2,2] &\iff [1:2,1,1,4,1,1,6,\ldots]<[1:2,2]\\
&\iff[2:1,1,4,1,1,6,\ldots]>[2:2]\\
&\iff [1:1,4,1,1,6,\ldots]<2 \end{align}$$
