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This link gives the cases of $n \in \mathbb{N}$*

Proving the irrationality of $e^n$

What about cases of rational exponents $q \in \mathbb{Q}$?

Sil
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High GPA
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4 Answers4

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Write $q=a/b$.

If $e^q$ is rational, then $e^a=(a^q)^b$ is rational.

Similarly, if $e^q$ is algebraic, then $e^a=(e^q)^b$ is algebraic.

So if you know that $e$ is transcendental, you get it for all $e^q$.

Thomas Andrews
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  • Are you sure that the irrationality of $e^q$ follows immediately from the transcendence of $e$? – High GPA May 28 '17 at 06:34
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    Yes, if $\alpha$ is transcendental, then so is $\alpha^q$ for any rational $q\neq 0$. And transcendentals are always irrational. – Thomas Andrews May 28 '17 at 06:36
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    If $e^{r/s}=n/m $ then $e $ is a solution to $m^sx^r-n^s=0$. So $e $ is algebraic. So yes, the irrationalty of $e^q $ follows directly from the transcendence of $e $. – fleablood May 28 '17 at 06:39
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Irrationality of $e^q$ for nonzero $q\in\mathbb{Q}$ follows immediately from irrationality of $e^n$ for positive integers $n$. Indeed, suppose $q=\frac{a}{b}$ where $a$ and $b$ are integers. We then have $(e^q)^b=e^a$, so if $e^q$ were rational, $e^a$ would also be rational. Taking the reciprocal if $a<0$, we get that $e^{|a|}$ is rational. But if $q\neq 0$, then $|a|$ is a positive integer, so $e^{|a|}$ is irrational. Thus $e^q$ must be irrational as well.

The exact same argument works for transcendence, replacing "rational" with "algebraic".

Eric Wofsey
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Let suppose $e$ is a rational number, and it can be written by $2$ integer which are pride to other ;

$$e=p/q \quad p,q\in\mathbb Z\quad (p\nmid q)$$

$$e=\lim\limits_{n\to\infty}\displaystyle\sum_{i=0}^n\dfrac{1}{i!}$$

To gain a national number, let's multiply each side by $q!$

$$eq!=\lim\limits_{n\to\infty}\displaystyle\sum_{i=0}^n\dfrac{q!}{i!}=\underbrace{\sum_{i=0}^q\dfrac{q!}{i!}}_R+\underbrace{\lim\limits_{n\to\infty}\displaystyle\sum_{i=q+1}^n\dfrac{q!}{i!}}_Q$$ Since $eq!$ is a national number, right side of the equation is a national number, as well. Moreover, $R$ is a national number too, so $Q$ must be a national number, but..

$$Q=\lim\limits_{n\to\infty}\displaystyle\sum_{i=1}^n\dfrac{q!}{(i+q)!}=\lim\limits_{n\to\infty}\displaystyle\sum_{i=1}^n\dfrac{1}{(1+q)(2+q)...(i+q)}$$

Due to this inequality,

$$i,q \in \mathbb N\\ (1+q)(2+q)...(i+q)>(1+q)^i$$

and we now see that this assumption is impossible, because;

$$0<Q<\lim\limits_{n\to\infty}\sum_{i=i}^n\dfrac1{(1+q)^i}=\dfrac{1}{1-\frac{1}{1+q}}-1<1$$

There isn't any element of national numbers that can be $$(n\in\mathbb N>0)\quad\wedge\quad(n\in\mathbb N<1)$$ $\Box$

Micheal Brain Hurts
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I assume you have seen it proven that $e $ is transcendental.

So $m^r*e^s-n^r \ne 0$ for any integers $r,s,n,m;s\ne 0$.

So $e^{s/r}\ne n/m $ for any rational $\frac sr,\frac nm $.

fleablood
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