# One Question about Proof on Irrationality of e

• Apr 13th 2012, 07:19 PM
scherz0
One Question about Proof on Irrationality of e
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:

http://i40.tinypic.com/v8o2e0.jpg
• Apr 13th 2012, 07:30 PM
Prove It
Re: One Question about Proof on Irrationality of e
Quote:

Originally Posted by scherz0
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:

http://i40.tinypic.com/v8o2e0.jpg

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.