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Premise

If you have the set of all rationals, and the ability to add, subtract, multiply, and divide, then you will only be able to make more rationals. However, if you are also able to use exponentiation, you can immediately make several constants that are imaginary and or irrational:

$(-1)^{1/2}=i$

$2^{1/2}=1.414…$

$(-1)^{-(-1)^{1/2}}=e^\pi$

Question

My question is as follows: what other constants are needed to be included with the set of rationals in order for them to be closed under addition, subtraction, multiplication, division, and exponentiation?

Fine Print

Notes: principle exponent is used (the square root of 1 is 1, not -1). Limits and series’s are not allowed. Indeterminate forms ($\frac{0}{0}, 0^0$) are not allowed. I will be referring to the answer as $Q^{pemdas}$ for the remainder of this post, as the answer is the set of rationals (Q) extended to be closed under the operations of pemdas.

Previous Attempts

One person I asked suggested that $Q^{pemdas}$ would be defined using the following pattern:

$a*b^c*d^{e^f}*g^{h^{i^j}}*k^{l^{m^{n^o}}}$

Where each letter is a rational number and the number of stacked exponents would be analogous to the “degree” of the constant. However, we were unable to prove or disprove that this set is closed under the five operations.

Another suggestion was that $Q^{pemdas}$ would be the set of algebraic numbers. A commenter on this question, @mr_e_man, disproved this by showing that neither all of the algebraic numbers would be included (no quintic formula), nor would the algebraic numbers contain all the elements in $Q^{pemdas}$ (transcendental numbers can be reached).

Another suggestion was the entire complex plane. However, @mr_e_man stated that $Q^{pemdas}$ was countably infinite, while the complex plane was uncountably infinite. @joriki proved that $Q^{pemdas}$ was countably infinite by stating that Q was countably infinite and the operations performed on Q were finite.

@DaveL.Renfro pointed out that this would be a subset of $\mathbb{E}$ (Q extended by pemdas and logarithms), but it is currently unclear as to whether $Q^{pemdas}$ is a proper subset or the same set. In other words, we don’t know whether $Q^{pemdas}$ is also closed under logarithms. If it is, then $Q^{pemdas}$ is $\mathbb{E}$, which answers the question.

What We Know

The rationals extended by exponentiation is a proper subset of $Q^{pemdas}$, which is a subset of $\mathbb{E}$.

Clarification Regarding Similar but Distinct Questions

Note: this question is not a duplicate, as the other similar questions (on this site and others) do not allow for addition, subtraction, multiplication, and addition to be used in conjunction with exponentiation.

Matthew Smith
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    How is this question not the same as https://math.stackexchange.com/questions/330962/extending-the-rationals-using-exponentiation ? – Gerry Myerson Jan 02 '23 at 03:57
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    Your field is countably infinite, while $\mathbb C$ is uncountably infinite. And it probably doesn't contain all algebraic numbers, since some quintic polynomials can't be solved with radicals. On the other hand, it does contain some non-algebraic numbers, such as $2^{\sqrt2}$: https://math.stackexchange.com/a/177986/472818 – mr_e_man Jan 02 '23 at 04:12
  • a) It's not the algebraic numbers, because Gelfond's constant $\mathrm e^\pi$ is transcendental. b) It's a countable set (since the elements are obtained by finite operations on a countable set). – joriki Jan 02 '23 at 05:35
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    I might be overlooking something, but I think what you're describing is a certain proper subfield (since you are not including the logarithm operations) of the (real) EL numbers, introduced and defined at the bottom of p. 441 of What Is a closed-form number? by Timothy Y. Chow (1999). Note, for example, that $x^{y^z} = \exp(\exp(u)),$ where $u=z\ln y + \ln \ln x.$ Related is Closed forms: what they are and why we care by Borwein/Crandall (2013); (continued) – Dave L. Renfro Jan 02 '23 at 06:29
  • also the MSE question What does closed form solution usually mean? and these questions linked to it. – Dave L. Renfro Jan 02 '23 at 06:30
  • @GerryMyerson that question wouldn’t allow √2+1 since that can’t be made using only exponentiation, you’d also need addition. – Matthew Smith Jan 03 '23 at 02:45
  • The exact meaning of that question is open to speculation, as OP really doesn't pin it down. But as it talks about using addition to get the integers, and then multiplication to get the rationals, and then exponentiation to extend to ... whatever, it could be argued that OP still wants to have a field with all four arithmetical operations, and exponentiation. – Gerry Myerson Jan 03 '23 at 03:02
  • @GerryMyerson the question clearly states that they are “extending Q by raising rationals to rational powers [paraphrased]” as a completely separate action from “extending integers using division [paraphrased]”, which is also a completely separate action from “extending natural numbers using subtraction [paraphrased]”. Noticing that addition/subtraction is not carried into extending the integers to rationals, we can assume that the four operations are not carried into extending the rationals. The language isn’t precise, but it doesn’t need to be because English is semi-contextual. – Matthew Smith Jan 03 '23 at 09:17
  • @DaveL.Renfro $Q^{pemdas}$ would be the same as $\mathbb{E}$ iff $Q^{pemdas}$ is closed under logarithms. $Q^{pemdas}$ is definitely a subset of $\mathbb{E}$, so your answer is the closest answer I’ve seen so far. – Matthew Smith Jan 03 '23 at 10:29
  • I can't imagine anyone would be interested in the rationals if you couldn't add and subtract them. – Gerry Myerson Jan 03 '23 at 14:37
  • @GerryMyerson whether the asker was interested or even intended to ask doesn’t matter, it’s what they asked. It’s what the text of the question is asking. I’ve shown that with non-cherry-picked evidence. I know because I used all relevant text that could possibly indicate whether the user was asking about just exponentials, or all the other operations as well. – Matthew Smith Jan 04 '23 at 00:34
  • What the heck is "Defense against the dark arts" supposed to mean? It sounds very combative & thus out-of-place on this website. – Gerry Myerson Jan 04 '23 at 04:11
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    @GerryMyerson it was supposed to be a touch of humor but if it’s not welcome I’ll rename it. I mostly added it because it helps clarify exactly what I’m talking about, but the fact that it helps prevent the question from being mistakenly marked as a duplicate was a bonus. – Matthew Smith Jan 04 '23 at 07:01

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