1. ## One Question about Proof on Irrationality of e

Dear all,

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

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

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

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

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

and so on...

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

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

Thank you very much for your assistance.

Proof:

2. ## Re: One Question about Proof on Irrationality of e

Originally Posted by scherz0
Dear all,

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

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

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

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

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

and so on...

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

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

Thank you very much for your assistance.

Proof:

It should be pretty obvious that q is a positive integer, because of how q is defined in a and b.

And since the proof involves showing that e = p/q, and it can easily be seen that e > 0, since q > 0, then so must be p.