Question about the proof of $e$ irrationality I found an article with the following elementary proof of $e$ irrationality
https://people.ohio.edu/diao/papers/Irrationality%20of%20e.pdf.
But I am having trouble understanding one step.
It states that, from:
\begin{equation}
 \frac{p_{n+1}}{q_{n+1}} = \frac{n p_{n} - q_{n}}{q_n}
\end{equation}
follows:
\begin{equation}
 q_n \geq q_{n+1}
\end{equation}
I can not understand why this is so. Can anyone explain in more detail?
 A: Both fractions$$\frac{p_{n+1}}{q_{n+1}}\quad\text{and}\quad\frac{np_n-q_n}{q_n}$$are equal to the same rational number, but the first one is an irreducible fraction. Therefore, and since both numerators and both denominators are natural numbers, $np_n-q_n\geqslant p_{n+1}$ and $q_n\geqslant q_{n+1}$. Actually, not only we have $q_n\geqslant q_{n+1}$ as, in fact, $q_n=Nq_{n+1}$ for some $N\in\Bbb N$.
A: If (as in the paper you cite) you have:
$$
 \frac{p_{n+1}}{q_{n+1}} = \frac{n p_{n} - q_{n}}{q_n}
$$
where $p_{n+1}$ and $q_{n+1}$ are coprime, you must have
$$
q_{n+1} = \frac{q_n}{\gcd(np_n - q_n, q_n)} \le q_n
$$
A: The proof states that $p_n$ and $q_n$ are relatively prime.
So, if you rewrite $$\frac{p_{n+1}}{q_{n+1}} = \frac{n p_{n} - q_{n}}{q_n}$$
as $$p_{n+1}q_{n} = (np_{n}-q_{n})q_{n+1}$$ you can see that $p_{n+1}$ has to divide $np_{n}-q_{n}$ in words $np_{n}-q_{n} \ge p_{n+1}$ 
Now just plug in this estimation to the first equation to the what you need.
By the way, nice proof, thanks for bringing it up.
