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Any number $x$ is called algebraic if there exist integer coefficients $a_0, a_1, ...,a_{n-1},a_n$ and integer exponents $b_0, b_1,...,b_{n-1},b_n$ where $b_0 = 0, b_1 = 1$ etc. such that $a_nx^{b_n}+a_{n-1}x^{b_{n-1}}+...+a_1x^{b_1}+a_0x^{b_0} = 0$. If such coefficients and exponents do not exist, we say that $x$ is transcendental.

If we allow those coefficients $a_0, a_1, ...,a_{n-1},a_n$ to be rational and not necessarily integers, we also get an algebraic number after clearing the denominators.

Even if we allow the coefficients $a_0, a_1, ...,a_{n-1},a_n$ to be algebraic and not necessarily rational, we should also get an algebraic number since the field of complex algebraic numbers is algebraically closed.

But what if we allow both the coefficients $a_0, a_1, ...,a_{n-1},a_n$ and the exponents $b_0, b_1,...,b_{n-1},b_n$ to be algebraic?

Let's call roots of such "polynomials" "superalgebraic". For example, the number ${\sqrt 2}^{\sqrt 2}$ is transcendental via Gelfond-Schneider theorem but it is "superalgebraic" since it satisfies the equation $x^{\sqrt 2}-2x^0 = 0$.

My question is: Does there exist any "supertranscendental" number which does not satisfy even any such equation? Intuitively there should be "supertranscendental" numbers, but how to formally prove this? Can we identify any specific "supertranscendental" number? Is it possible to determine whether $e$ and $π$ are "superalgebraic" or "supertranscendental"?

Thanks.

Jan
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    The set of superalgebraic numbers is countable. So nearly all real numbers are supertranscendental. – TonyK Feb 05 '19 at 11:23
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    @TonyK How do we know that every "superpolynomial" can only have countable many roots ? – Peter Feb 05 '19 at 11:50
  • Since we do not even know whether $e$ and $\pi$ are linear independent over $\mathbb Q$, I doubt that we can determine whether they are "supertranscendental", my guess would be "yes" in both cases. – Peter Feb 05 '19 at 11:53
  • @Peter: Hmmm. Good question. – TonyK Feb 05 '19 at 11:59
  • Well, your superpolynomials are holomorphic functions away from 0, hence they can't have too many zeros. – Ivan Neretin Feb 05 '19 at 13:22

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