Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $e^3$.

share|improve this question
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
1  
And as another note, I'm motivated by this question: math.stackexchange.com/questions/476899/…. –  nayrb Aug 27 '13 at 1:12
5  
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
2  
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

1 Answer 1

up vote 8 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 $h \in \mathbb{Z}^+.$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.