Dear all,

I'm looking at this particular proof that $\displaystyle e $ is irrational. Please find it beneath my writing.

This proof by contradiction assumes that $\displaystyle p,q \in \mathbb{Z} $. I understand this.

However, while proving that $\displaystyle q! \cdot b \in \mathbb{Z}$ is the contradiction, the proof uses the second boxed inequality

where

$\displaystyle \frac{1}{q + 1} > \frac{1}{2} \Longleftrightarrow q + 1 > 2 \Longleftrightarrow q > 1 $

and $\displaystyle q > 1 \Longleftrightarrow q + 2 > 3 > 2 \Longleftrightarrow q + 2 > 2 \Longleftrightarrow \frac{1}{q + 2} < \frac{1}{2}$

and so on...

However, although $\displaystyle e > 0$, it is entirely possible for $\displaystyle p \in \mathbb{R^-} $ AND $\displaystyle q \in \mathbb{R^-}$ since $\displaystyle p,q \in \mathbb{Z} $.

In this case then the $\displaystyle q \ngtr 1 $, so the above steps and the proof itself fail?

Thank you very much for your assistance.

__Proof:__