TITLE: Set of Elementary Real Numbers Without Elementary Combination
QUESTION [19 upvotes]: Find a set of $n$ real numbers such that none can be created by combining the others with elementary operations ($+, -, \times, /$).


This question came up in my attempt to prove an $n$ dimensional version of the fundamental theorem of algebra; however, I am interested in the answer to this regardless of the relevance it has to the associated question.

I have thought about this a bit, and so far my best guess is the square roots of the first $n$ primes. Clearly, this fulfills the above property on just multiplication, but I am not sure how to prove this for addition, and especially for a combination of the two. It would probably involve some very strong statements about sums of square roots, but I'm not sure how to proceed. Any help would be appreciated. (I'm also not sure how to tag this, as I don't know what branch of math questions like this belong to. If anyone could help there that'd be appreciated as well.)

REPLY [9 votes]: AC is trivially not needed to get a finite set; use induction and the fact that there are only countably many arithmetic expressions involving $n$ reals, to find some real that is not equal to any them.
It turns out that no choice is needed to get a countable set either. Let $x_n$ be the (unique) real in $[0,1)$ whose $k$-th binary digit is $1$ iff $k$ is in the $n$-th Turing jump. By the unsolvability of the halting problem relative to any jump, it is clear that $x_n$ cannot even be computed from $x_{0..n-1}$, much less be equal to some arithmetic expression involving them. Furthermore, if any $x_k$ is equal to some arithmetic expression involving $x_{0..k-1}$ and $x_{k+1..n}$, for some minimal $n$, then $x_n$ is a root of some non-zero polynomial with coefficients computable from $x_{0..n-1}$, and thus is computable from them, which is impossible. Therefore none of the $x_n$ can be equal to some arithmetic expression involving the others.