Proving the irrationality of $e^n$

Question

Let $n$ be a positive integer. I know the traditional proof that $e$ is irrational. How do we show that $e^n$ is irrational in some sort of similar line? I am of course assuming it is but I would be astounded if not. It occurs to me that since $e$ is transcendental, of course $e^n$ is irrational, but I don't want to use that fact.

Googling gives me something for $e^2$, but I could not easily find anything for $e^3$.

Wikipedia gives a sketch of a proof that $e$ is transcendental — Henry, Aug 27 '13 at 01:04
Edited my question. I don't want to use the fact that $e$ is transcendental, just as we don't need to in order to show that $e$ is irrational. — abnry, Aug 27 '13 at 01:06
And as another note, I'm motivated by this question: http://math.stackexchange.com/questions/476899/linear-independence-of-the-numbers-1-e-e2-e3. — abnry, Aug 27 '13 at 01:12
There is a proof for $e^r$ with $r\in\mathbb{Q}$ in Proofs From the Book by Martin Aigner and Günter M. Ziegler. You may want to look it up. — EuYu, Aug 27 '13 at 01:16
There are some nice notes by Keith Conrad in which in particular it is shown that if $r$ is a non-zero rational then $e^r$ is irrational (that's equivalent to your question). The proof is "elementary" but challenging. — André Nicolas, Aug 27 '13 at 01:17
Indeed, if you want to show $e^n$ is irrational for any positive integer $n$, then this is equivalent to showing it for any non-zero rational $n$. — user2566092, Aug 27 '13 at 12:10
Yes, that's clear. If $e^n$ is irrational, then $e^(n/m)$ cannot be rational, because that would imply $e^n$ is rational. — abnry, Aug 30 '13 at 20:55
If you like continued fractions, there is a neat proof (not my own!) here — Old John, Nov 24 '13 at 18:39
Are you allowed to use the fact that $e=\sum_{n=0}^{\infty}\frac{1}{n!}$? — Gregory Grant, Dec 24 '15 at 12:34

felipeuni · Accepted Answer · 2013-08-29T07:04:58.343

15

Niven's polynomials

Let $f:[0,1]\longrightarrow \mathbb{R}$ , $\displaystyle f(x)=\frac{x^n(1-x)^n}{n!}$ then

$$f(x)=f(1-x)$$

$$\displaystyle 0\le f(x)<\frac{1}{n!}$$

$$f^{(j)}(0)\;,\;f^{(j)}(1) \in \mathbb{Z} \;,\; j\ge 0$$

Proposition. The number $e^3$ is irrational.

Proof: Suppose that $\displaystyle e^3=\frac{a}{b}$

$$\displaystyle F=3^{2n}f-3^{2n-1}f'+3^{2n-2}f''-\cdots + f^{(2n)}$$

$$\displaystyle F'+3F=3^{2n+1}f$$

$$\displaystyle \mathbb{Z^+} \ni aF(1)-bF(0)=b\Bigl[e^{3x}F(x)\Bigr]_0^1=b\int_0^1 3^{2n+1}e^{3x} f(x)dx \longrightarrow 0^+\;,\;n \longrightarrow\infty $$ Contradiction , analogously $e^h$ is irrational for $h \in \mathbb{Z}^+.$

edited Aug 29 '13 at 07:04

answered Aug 27 '13 at 02:57

felipeuni

5,080

Which book shall I read to learn more on these techniques? – High GPA May 29 '17 at 02:37
In the book by Ivan Niven, Irrational Numbers. – felipeuni Jul 02 '17 at 20:20
I'm confused on the step where you go from $b \left[ e^{3x}F(x) \right] ^1_0$ to $b \int _0 ^1 3^{2n+1} e^{3x} f(x) dx$. Could you perhaps explain your reasoning for that part? – Tim Hargreaves Aug 29 '17 at 21:05
Note that $\dfrac{d}{dx}[e^{3x}F(x)]=3^{2n+1}e^{3x}f(x)$ using $F'+3F=3^{2n+1}f$. – felipeuni Sep 05 '17 at 01:20

score 3 · Answer 2 · edited Dec 24 '15 at 12:46

3

I believe there's a proof of this in the book described here.

I have never personally looked into that, however.

How to show $e^2$ and $e^4$ are irrational is show here.

edited Dec 24 '15 at 12:46

wythagoras

25,026

answered Dec 24 '15 at 12:35

StephenG - Help Ukraine

725

score 3 · Answer 3 · answered Dec 24 '15 at 12:39

3

In fact, $e^n$ is transcendental for all $n \in \mathbb{Z}-\{0\}$. To see this it is sufficient to show that $e^n$ is transcendental for all $n \in \mathbb{Z}^+$. If $e^n$ were to be algebraic, this means that there exists a polynomial $p(x)$ with integer coefficients, such that $p(e^n) = 0$. This means that the polynomial $g(x) = p(x^n)$ has $e$ as its root, contradicting the transcendence of $e$.

answered Dec 24 '15 at 12:39

Adhvaitha

20,259

2

Sure, if you know the very general result that $\alpha$ is transcendental, then, specifically, $\alpha^n$ is not rational for any $n$. But that just shows that you are using a much more powerful result to prove this specific instance. It's like if someone here asked you to prove that $\pi$ was irrational and you said, "since we know $\pi$ is transcendental..." – Thomas Andrews Dec 24 '15 at 13:40

Wojowu · Answer 4 · 2015-12-24T13:47:56.440

You can find a pretty direct proof of this result here, in proposition 7.

Here is the essence of the argument: for $r(x)$ a polynomial of degree $k$ we can express $$\int_0^1r(x)e^{z-xz}dx=\frac{F(z,0)e^z-F(z,1)}{z^{k+1}}$$

where $F(z,x)$ is a polynomial in $z$ whose coefficients are successive derivatives of $r$ at $x$ (this is proposition 5). In particular, if $e^p=\frac{a}{b}$ is rational and we take $r(x)=x^n(1-x)^n$ for some large $n$ to be specified later, we can use this formula to estimate $aF(p,0)-bF(p,1)\leq ap^{2n+1}$. However, $aF(p,0)-bF(p,1)$ is clearly positive (by looking at the integral), and from the formula for $F(z,x)$ and the fact that many derivatives of $r(x)$ vanish at $0,1$ we can deduce $aF(p,0)-bF(p,1)$ is integer divisible by $n!$. Hence $n!\leq aF(p,0)-bF(p,1)\leq ap^{2n+1}$, so $n!\leq ap^{2n+1}$ which is contradictory for large $n$.

The link is broken. I think the paper is now at http://math.arizona.edu/~jeremybooher/expos/transcendence.pdf — Barry Cipra, May 28 '18 at 14:11

Proving the irrationality of $e^n$

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