Exponential of Rational Number is Irrational

Theorem
Let $r$ be a rational number such that $r \ne 0$.

Then:
 * $e^r$ is irrational

where $e$ is Euler's number.

Proof
Let $r = \dfrac p q$ be rational such that $r \ne 0$.

$e^r$ is rational.

Then $\paren {e^r}^q = e^p$ is also rational.

Then if $e^{-p}$ is rational, it follows that $e^p$ is rational.

It is therefore sufficient to derive a contradiction from the supposition that $e^p$ is rational for every $p \in \Z_{>0}$.

Let $e^p = \dfrac a b$ for $a, b \in \Z_{>0}$.

Let $n \in \Z_{>0}$ be a strictly positive integer.

Let $\map f x$ be the function defined as:

Note that:
 * $(2): \quad 0 < x < 1 \implies 0 < \map f x < \dfrac 1 {n!}$

We also have:
 * $\map f 0 = 0$


 * $\map {f^{\paren m} } 0 = 0$ if $m < n$ or $m > 2 n$

and:
 * $n \le m \le 2 n \implies \map {f^{\paren m} } 0 = \dfrac {m!} {n!} c_m$

and this number is an integer.

Thus at $x = 0$, $\map f x$ and all its derivatives are integers.

Since $\map f {1 - x} = \map f x$, the same is true for $x = 1$.

Let $\map F x$ be the function defined as:

Because of the properties of $\map f x$ and its derivatives above, $\map F 0$ and $\map F 1$ are integers.

Next we have:

$(4)$ leads to:

which is an integer.

But from $(2)$:

The expression on the tends to $0$ as $n$ tends to $\infty$.

Hence:
 * $0 < a \, \map F 1 - b \, \map F 0 < 1$

for sufficiently large $n$.

But there exists no integer strictly between $0$ and $1$.

From this contradiction it follows that our original assumption, that is, that $e^r$ is rational, must have been false.