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 $r \ne 0$.
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
- 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.17$: More About Irrational Numbers. $\pi$ is Irrational