If $a$ and $b$ are positive integers, show that the inequality $e < \frac{a}{b} < \frac{87}{32}$ implies that $b \geq 39$.
This is exercise 15.3.7 b) from the book Elementary Number Theory by David M. Burton.
To our help we have the theorem (among many), $|x - a/b| < \frac{1}{b^2}$, if $a/b$ is the n:th convergent of the irrational number $x$.
A trivial computation shows that the 6th and 7th convergents of $e$ are $C_6=\frac{{87}}{{32}}$ and $C_7=\frac{{106}}{{39}}$. The result then follows due to the convergents being the best rational approximations to an irrational number (in this case $e$) in the sense that every other rational number with the same or smaller denominator differs from the irrational number by a greater amount. In other words, $C_6$ shows that $b$ is greater than 32 and if $b$ were strictly between 32 and 39, then there would have been an intermediate convergent between $C_6$ and $C_7$ [ERROR: this was my mistake; please see Daniel Fischer's solution].
This demonstration suffices to answer the posed question. Note, however, that we can actually pass to the 8th convergent ($C_8 = \frac{{193}}{{71}}$) to obtain the stronger restriction $b\ge71$ by exploiting the fact that $e$'s even numbered convergents $C_{2n}$ are all greater than $e$ (the problem stipulates that we want a rational approximation greater than $e$) while $e$'s odd numbered convergents $C_{2n+1}$ are all less than $e$. Note that I'm starting the convergent indexing at $C_1$ rather than $C_0$.