Set of Elementary Real Numbers Without Elementary Combination 
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.)
 A: Well you can make a diagonalization argument to get $\omega_1$ many such numbers strictly through counting.
Choose an arbitrary $a_0$ and close off under $(*,+,-,/)$ this gives you a set $A_1$ that has countable size. Now assume that you have already chosen $a_i$ independent for $i< k$. Let $A_k$ be the closure of $\{a_i\}_{i<k}$ under $(*,+,-,/)$ then $A_k$ is still countable (countable iterations over countable base set with finite signature). So choose $a_{k}\in \mathbb{R}\setminus A_k$.
The above shows you can get a countable set. But it's not hard to continue. The successor stage works just as above. At a limit $\beta$ you just take $A_\beta=\bigcup_{\alpha<\beta}A_\alpha$ and then choose $a_\beta\in\mathbb{R}\setminus A_\beta$.
You can actually go all the way to $\mathfrak{c}$ in this way.
The existence can also be proved in a non-constructive manner by noticing that if we have some set or real numbers $A$ such that $|A|<\mathfrak{c}$ then it's closure still has size less than $\mathfrak{c}$.
Note: I'm assuming choice (AC) throughout here.
A: Your idea works, which is because of the non-trivial fact $\sqrt{p_n}\notin\Bbb Q[\sqrt 2,\sqrt 3,\ldots,\sqrt{p_{n-1}}]$. To show this, I'd resort to Galois theory, but maybe this special case allows something more elementary?
Let $p$ be a prime and $S$ be the set of real numbers of the form $a_1\sqrt{m_1}+\ldots+a_n\sqrt{m_N}$, where $a_i\in\Bbb Q$ and $1,m_1,\ldots, m_n$ are distinct square-free integers not divisible by $p$. The set $S$


*

*contains $\sqrt q$ for all primes $q\ne p$, obviously

*is closed under $+$ and $-$, obviously

*is a bit less obviously closed under multiplication, but noting that $a\sqrt m\cdot a'\sqrt{m'}=aa'\gcd(m,m')\sqrt{\frac m{\gcd(m,m')}\frac {m'}{\gcd(m,m')}}$ helps

*is even less obviously closed under division: Let $q_1,\ldots, q_r$ be the primes occuring in the radicands $m_i$. We can write $a_1\sqrt{m_1}+\ldots +a_n\sqrt{m_n}$ as $\alpha+\beta\sqrt {q_r}$, where only $q_1,\ldots, q_{r-1}$   occur in the radicands used for $\alpha$ or $\beta$. Then $(\alpha+\beta\sqrt {q_r})(\alpha-\beta\sqrt {q_r})=\alpha^2-q_r\beta^2$ and by induction on $r$, we may assume that $\alpha^2-q_r\beta^2$ has an inverse in $S$; multiplying it with $\alpha-q_r\sqrt \beta$ gives us an inverse of the original number


We conclude that $\sqrt p$ cannot be obtained from all other $\sqrt q$ by means of $+,-,\cdot,/$.
EDIT: After the fact, I noticed that one also needs to prove (once more, by induction) that $a_1\sqrt{m_1}+\ldots +a_n\sqrt{m_n}=0$ only if all $a_i=0$. Without this, it may happen that $\alpha-\beta\sqrt{q_r}=0$. 
A: 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. 
