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Basically, the question is: Is $e^a$ rational? Since $a$ is an irrational number (if it depends on $a$ being algebraic or transcendental, let me know also). I have seen something like this before, however I cannot remember the answer at all and I'm not finding it either.

Thanks

Mason
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Mr. N
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    don't you know $e^{2 \pi i} = 1$? – WhatsUp Nov 28 '19 at 00:57
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    and $e^{\ln 2}=2$ – J. W. Tanner Nov 28 '19 at 00:58
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    Seriously, I guess what you have seen is https://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem – WhatsUp Nov 28 '19 at 00:58
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    Just to be clear, it can't always be rational, just by counting arguments. But, as others have remarked, there are examples for which it is rational. – lulu Nov 28 '19 at 01:00
  • Thank you all. Yes, I have been looking for this theorem. Yes, I know this examples but as I said I have been looking for a general rule if that had existed. – Mr. N Nov 28 '19 at 01:18
  • Also, do you know if there is a easier proof for this theorem. Unfortunately, I am not used to these proofs with so much rigour.... – Mr. N Nov 28 '19 at 01:20
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    Here's a sketch of some of the important arguments which I think you might enjoy. Mathologer does a good job of showing the important elements without letting the details (or a commitment to rigor) obsfucate the big picture. – Mason Nov 28 '19 at 01:36
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    There are only countable many rational number and uncountably many irrational. So there are at most countably many $a$ so that $e^a$ is rational but uncountably many so the $e^a$ is irrational. $e^a$ is rational is possible via $e^{\ln q} = q$. – fleablood Nov 28 '19 at 02:10

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In this answer I will give two major theorems and one conjecture which speak to the question of whether $a^b$ is in the algebraic numbers which we let $\bar{\mathbb{Q}}$ denote.

There is a powerful theorem called the Lindemann–Weierstrass theorem which allows to make some sweeping statements about $e^x$. But the statements are about the transcendentality of $e^x$ and not that rationality. Of course if a number is not algebraic it is not rational.
For example, by LW Theorem we have

For non-zero algebraic numbers $A,B$ and any algebraic numbers $x,y$ we have
$Ae^x-Be^y \ne 0$.

By taking $A=1, y=0$ we have that $e^x \neq B$ meaning that $e^x$ is not algebraic for any algebraic $x$. And through contrapositive we have have that when $e^x$ is algebraic $x$ must be some non-algebraic number. As appears in the comments $e^{\ln 2}, e^{2\pi i} $ are both in the rationals. This implies that $2\pi i$ (and therefore $\pi)$ and $\ln 2$ are transcendental numbers.

So for example: $\color{red}{e^{\sqrt 2},\ln(e^3+1),\ln(2) \notin \bar{\mathbb{Q}}}$

The other thing that you might be (mis)recalling is the Gelfond Schneider Theorem:

$a^b$ is transcendental whenever $a\neq 0$ and $b$ are algebraic but $b$ is irrational. We also interpret $b$ as irrational when $Im(b)\neq 0$

This isn't about $e^x$ per se but it does have an irrational exponent which is maybe what you were getting at with your question. We can fiddle with it a tiny bit and make it a statement about $e^x= e^{\ln(a^b)}=e ^{b\ln(a)}$ is transcendental and therefore irrational whenever $a$ and $b$ are algebraic but $b$ irrational.

So for example: $\color{red}{{\sqrt{5}}^{\sqrt 7} \notin \bar{\mathbb{Q}}}$

Also this theorem together with Euler's identity implies $e^{\pi z}$ is transcendental whenever $z$ is an algebraic number with non zero real part. Note $$e^{\pi z}= e^{i\pi \times-i z}=(e^{\pi i})^{-iz}=(-1)^{-iz}$$ In this case $a=(-1)$ is a non zero algebraic number and $b=-iz$ is algebraic and is to be considered irrational. So we have met the criteria of the theorem.

So for example: $\color{red}{{e}^{\pi\sqrt 2} \notin \bar{\mathbb{Q}}}$

The last thing to discuss is Schanuel's conjecture. Assuming Schanuel's conjecture is true we also can say $\color{red}{{e}^{e} \notin \bar{\mathbb{Q}}}$. My understanding is that we don't have currently a proof that $e^e \notin \mathbb{Q}$.

Mason
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  • First of all, thanks for your attetion and answers. Well, I'm gonna watch it for sure very soon, however, just let me reply this answer with another question: Do we you know which algebraic xs make e^x algebraic? Besides zero, of course. – Mr. N Nov 30 '19 at 22:38
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    Done! Rapid fire response! The Gelfond Schneider Thm allows us to claim when they are not algebraic (so that may be a tool that's helpful depending on what you're up to). The simple response to your question is that $e^{\ln(Algebraic)}$ is Algebraic. I am not sure that's really what you are getting at. – Mason Nov 30 '19 at 22:42
  • Well, thanks again. These theorems are quite helpful and handy to me, but since you have "asked" what I'm up to, I'm gonna tell you. I was wondering what numbers a (any kind, except 1) and b (irrational and algebraic) make a^b RATIONAL. – Mr. N Nov 30 '19 at 22:54
  • Ok. I organized it a little more to speak to this broader question. – Mason Dec 01 '19 at 01:07
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    In response t the question you asked in comment: That implies $a$ is transcendental. That is, $b\notin{\mathbb{Q}},a^b \in \mathbb{Q}\implies b\notin{\mathbb{Q}}, a^b \in \bar{\mathbb{Q}} \implies a\notin \bar{\mathbb{Q}}$. This follows from GS Theorem which I wrote about above. I am not sure much more is known about that but if you find something be sure to let us know. – Mason Dec 01 '19 at 01:19
  • Sorry, I did not have proper time to respond you. First of, thanks for this enlightment that you brought to me since now I really understand what I wasn't. Even if I am still gonna read it, so far, I am without questions and thankfull for the answers. Thanks again. – Mr. N Dec 09 '19 at 22:48
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    Indeed, according to this post: $e^{\alpha \pi + \beta}$ is transcendental for algebraic nonzero $\alpha$ and $\beta$. – Mason Oct 05 '21 at 04:17
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    Here's a nice list of tools to demonstrate a number transcendental. – Mason Jul 14 '22 at 19:07
  • I'm not sure if $(e+1)^\sqrt{2}$ is demonstrably transcendental or irrational. I suspect $P(e)^\alpha$ for a (nontrivial) polynomial $P$ and an algebraic number $\alpha$ is beyond our tools currently. – Mason Jul 31 '22 at 05:04