Exponential of Rational Number is Irrational

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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 $r \ne 0$.

Aiming for a contradiction, suppose $e^r$ 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:

\(\text {(1)}: \quad\) \(\ds \map f x\) \(:=\) \(\ds \frac {x^n \paren {1 - x}^n} {n!}\)
\(\ds \) \(=\) \(\ds \frac 1 {n!} \sum_{k \mathop = n}^{2 n} c_k x^k\) where the $c_k$'s are some integers


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:

\(\text {(3)}: \quad\) \(\ds \map F x\) \(:=\) \(\ds \sum_{k \mathop = 0}^{2 n} \paren {-1}^k p^{2 n - k} \map {f^{\paren k} } x\)
\(\ds \) \(=\) \(\ds p^{2 n} \map f x - p^{2 n - 1} \map {f'} x + p^{2 n - 2} \map {f''} x - \cdots - p \map {f^{\paren {2 n - 1} } } x + \map {f^{\paren {2 n} } } x\)


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


Next we have:

\(\text {(4)}: \quad\) \(\ds \map {\frac {\d} {\d x} } {e^{p x} \map F x}\) \(=\) \(\ds e^{p x} \paren {\map {F'} x + p \, \map F x}\)
\(\ds \) \(=\) \(\ds p^{2 n + 1} e^{p x} \map f x\) from $\map {F'} x + p \, \map F x$ based on $(3)$


$(4)$ leads to:

\(\ds b \int_0^1 p^{2 n + 1} e^{p x} \map f x \d x\) \(=\) \(\ds b \bigintlimits {e^{p x} \map F x} 0 1\)
\(\ds \) \(=\) \(\ds a \, \map F 1 - b \, \map F 0\)

which is an integer.


But from $(2)$:

\(\ds 0\) \(<\) \(\ds b \int_0^1 p^{2 n + 1} e^{p x} \map f x \rd x\)
\(\ds \) \(<\) \(\ds \frac {b p^{2 n + 1} e^p} {n!}\)
\(\ds \) \(=\) \(\ds b p e^p \frac {\paren {p^2}^n e^p} {n!}\)

The expression on the right hand side 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.

$\blacksquare$


Sources