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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 $e^3$.

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2  
Wikipedia gives a sketch of a proof that $e$ is transcendental – Henry Aug 27 '13 at 1:04
    
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. – nayrb Aug 27 '13 at 1:06
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And as another note, I'm motivated by this question: math.stackexchange.com/questions/476899/…. – nayrb Aug 27 '13 at 1:12
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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 1:16
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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 1:17
up vote 12 down vote accepted

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 $h \in \mathbb{Z}^+.$

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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.

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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 $e$.

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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

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 $n$.

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