Playing with some algebra using only the assumption that the algebraic numbers are closed under algebraic functions, I discovered it's easy to show that for any pair of transcendental numbers $t_1,t_2\in \mathbb{T}$ and the algebraic numbers $\overline{\mathbb{Q}}$, if we define the set $X$ by:

$X=\{t_1\times(t_2+\overline{q}): \overline{q}\in \overline{\mathbb{Q}}\}$

then $\lvert X \cap \overline{\mathbb{Q}}\rvert\leq1$

This implies for example if we take $\{e\times (\pi+q): q\in\overline{\mathbb{Q}}\}$, at most one element of this infinite set is algebraic - giving some measure of just how unlikely it is that $e\times\pi$ is algebraic.

I thought this was kind of interesting and I was able to extend it in various ways such as showing $e+\pi$ is also almost certainly transcendental (since $\lvert\{t_1+\overline{q}t_2: q\in\overline{\mathbb{Q}}\}\cap\overline{\mathbb{Q}}\rvert\leq1$).

I'm not an advanced mathematician; I just like to work things out for myself so I often discover things like this which are already known, which I presume is the case here. So I presume this is a special case of some theorem already in use. What theorem is it, and what field of maths is this? I'm guessing it's something like transcendental extensions of algebraic fields.

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    $\begingroup$ The field of math is elementary number theory. The result is so close to the definitions that it probably hasn't attracted any attention. If $t_1(t_2+q_1)$ and $t_1(t_2+q_2)$ are both algebraic, then so is their difference, which is $t_1q_3$, where $q_3=q_1-q_2$ is algebraic, contradiction. $\endgroup$ – Gerry Myerson Aug 9 '16 at 13:08
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    $\begingroup$ @RobertFrost: I know that you didn't say that such probabilistic reasoning is valid, but anyway just for the record, the problem is that the answer is already fixed and absolute as long as you grant the existence of natural numbers and real numbers, so we can't apply probability to conclude anything. For all we know there might be some subtle deep connection between $π$ and $e$ that makes their sum algebraic! See en.wikipedia.org/wiki/Schanuel%27s_conjecture for a possible avenue to resolving the question about $π+e$. $\endgroup$ – user21820 Aug 11 '16 at 12:49
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    $\begingroup$ @RobertFrost: Yes for a randomly selected element, but no for $π,e$, which are in no way random. By the way, your use of "$\frac1{\infty}$" is at best misleading and at worst meaningless, but I'm ignoring that. Also, absence of information has absolutely nothing to do with probability. $\endgroup$ – user21820 Aug 11 '16 at 13:46
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    $\begingroup$ Bayesian interpretation is irrelevant to your question. Probability is a very specific concept defined in mathematical terms and has zero relation to whether we know something or not. We might choose to assume that things in the real world (such as a shuffled pack of cards) are drawn randomly from some distribution, and calculate the likelihood that something (verifiable) is true, but the fact remains that the outcome is fixed and also unchanged by whether we know it. $\sum_{k=1}^\infty \frac1{k^2} = \frac{π^2}{6}$ with probability $1$ whether or not anyone knows it. $\endgroup$ – user21820 Aug 11 '16 at 14:02
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    $\begingroup$ Sorry but you're getting it all wrong. You should not confuse between the mathematical structure, which we can define via measure theory, and the attempt to create a meaningful interpretation of theorems to assist us in handling/making real world decisions. Let me answer your questions in your second-to-last comment. (1) You didn't show me that the dice roll obeys the mathematical theorems. What if the universe is deterministic? (2) It is 1 if you know it is 6. It is also 1 if it is a 6 even if you don't know it. $\endgroup$ – user21820 Aug 11 '16 at 14:05

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