# $e^{r}$ 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 $r\in\mathbb{Q}\setminus\{0\}$

To begin with, note that it is sufficient to show that $e^{u}$ is irrational for any positive integer (http://planetmath.org/NaturalNumber)11In this entry, $\mathbb{N}:=\{1,2,3,\ldots\}$ and $\mathbb{N}_{0}:=\mathbb{N}\cup\{0\}$. $u$ (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 $\displaystyle f_{n}(x):=\frac{x^{n}(1-x)^{n}}{n!}$:

• $\displaystyle f_{n}(x)=\frac{1}{n!}\sum_{i=n}^{2n}c_{i}x^{i}$, 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 $n\leq k\leq 2n$, 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 $0 we have $0.

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

 $F_{n}(x):=\sum_{k=0}^{\infty}(-1)^{k}u^{2n-k}f_{n}^{(k)}(x),$

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

 $\frac{d}{dx}\left[e^{ux}F_{n}(x)\right]=ue^{ux}F_{n}(x)+e^{ux}F^{\prime}_{n}(x% )=u^{2n+1}e^{ux}f_{n}(x).$
 $(w_{n})_{n\in\mathbb{N}}:=b\int_{0}^{1}u^{2n+1}e^{ux}f_{n}(x)\,dx=b\left[e^{ux% }F_{n}(x)\right]^{1}_{0}=aF_{n}(1)-bF_{n}(0).$

Given the remarks on $f_{n}(x)$, $w_{n}$ should be an integer for all $n\in\mathbb{N}$, yet it is clear that $w_{n} and so $\lim\limits_{n\to\infty}w_{n}=0$, 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\mathbb{A}\setminus\{0\}$22$\mathbb{A}$ denotes the set of algebraic numbers  .. We conclude this entry with the following evident corollary:

###### Corollary.

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

## References

• 1 M. Aigner & G. M. Ziegler: Proofs from THE BOOK, 3${}^{\mathrm{rd}}$ edition (2004), Springer-Verlag, 30–31.
• 2 G. H. Hardy & E. M. Wright: An Introduction to the Theory of Numbers, 5${}^{\mathrm{th}}$ edition (1979), Oxford University Press, 46–47.
Title $e^{r}$ is irrational for $r\in\mathbb{Q}\setminus\{0\}$ ErIsIrrationalForRinmathbbQsetminus0 2013-03-22 15:07:46 2013-03-22 15:07:46 Cosmin (8605) Cosmin (8605) 12 Cosmin (8605) Theorem msc 11J72 $e^{r}$ is irrational for non-zero rational r irrationality of the exponential function   on $\mathbb{Q}$ Irrational EIsIrrationalProof EIsIrrational EIsTranscendental