$e^r$ is irrational for $r\in \mathbb{Q}\setminus \{0\}$

We here present a proof of the following theorem:

Theorem.

$e^r$ is irrational for all $r\mathrm{\in }\mathrm{Q}\mathrm{\setminus }\mathrm{\{}\mathrm{0}\mathrm{\}}$

To begin with, note that it is sufficient to show that $e^u$ is irrational for any positive integer (http://planetmath.org/NaturalNumber) $u$ (for if $e^r=e^{\frac{u}{v}}$ were rational, so would ${(e^{\frac{u}{v}})}^v=e^u$). Next, we look at some simple properties of polynomial $f_n(x):={\displaystyle \frac{x^n{(1-x)}^n}{n!}}$:

• $f_n(x)={\displaystyle \frac{1}{n!}}{\displaystyle \sum _{i=n}^{2n}}{c}_i{x}^i$, with $c_i\in \mathbb{Z}$ for all $i$.

• $f_n^{(k)}(0)$ and $f_n^{(k)}(1)$ are integers for all $k\in {\mathbb{N}}_0$: as $0$ is a root (http://planetmath.org/Root) of order $n$, $f_n^{(k)}(0)=0$ unless $n\le k\le 2n$, in which case $f_n^{(k)}(0)=\frac{k!}{n!}{c}_k$, an integer. Since $f_n^{(k)}(x)={(-1)}^k{f}_n^{(k)}(1-x)$, the same is true for $f_n^{(k)}(1)$.

• For all we have .

Now we can readily prove the theorem:

Proof.

Assume that $e^u=\frac{a}{b}$ for some $(a,b)\in {\mathbb{N}}^2$ and let

$$F_n(x):=\sum _{k=0}^{\mathrm{\infty }}{(-1)}^k{u}^{2n-k}{f}_n^{(k)}(x),$$

which is actually a finite sum since $f_n^{(k)}(x)=0$ for all $k>2n$. Differentiating $F_n(x)$ yields $F_n^{\prime }(x)=u^{2n+1}{f}_n(x)-u{F}_n(x)$ and thus:

$$\frac{d}{dx}\left[e^{ux}{F}_n(x)\right]=u{e}^{ux}{F}_n(x)+e^{ux}{F}_n^{\prime }(x)=u^{2n+1}{e}^{ux}{f}_n(x).$$

Now consider the sequence

$${(w_n)}_{n\in \mathbb{N}}:=b{\int }_0^1{u}^{2n+1}{e}^{ux}{f}_n(x)𝑑x=b{\left[e^{ux}{F}_n(x)\right]}_0^1=a{F}_n(1)-b{F}_n(0).$$

Given the remarks on $f_n(x)$, $w_n$ should be an integer for all $n\in \mathbb{N}$, yet it is clear that and so $\underset{n\to \mathrm{\infty }}{lim}{w}_n=0$, a contradiction.∎

The result could also easily have been obtained by starting with $w_n$ and integrating by parts $2n$ times. Note also that much stronger statements are known, such as “$e^a$ is transcendental for all $a\in 𝔸\setminus \{0\}$”. We conclude this entry with the following evident corollary:

Corollary.

For all $r\mathrm{\in }{\mathrm{Q}}^{\mathrm{+}}\mathrm{,}\log \mathit{}r$ is irrational.