Let $\mathbb{A^+}$ be the set of non-negative algebraic numbers. Consider the set of "polynomials" : $$\mathbb{P} = \lbrace a_0 + a_1x^{r_1} + a_2x^{r_2} + a_3x^{r_3} +\cdots + a_nx^{r_n}| a_0, a_i, r_i \in \mathbb{A}, r_i > 0, i= 1,2,\cdots,n\rbrace$$ We call $\alpha \in \mathbb{R}, \alpha \geq 0$ *extra-algebraic* if there exists a polynomial in $\mathbb{P}$ satisfying $f(\alpha)=0$. Denote the set of all extra-algebraic numbers by $\mathbb{A}_E$. So, $\mathbb{A} \subset \mathbb{A}_E$.(The strict inclusion is because of numbers like $2^\sqrt2$ which are extra-algebraic but not algebraic and more by the Gelfond–Schneider theorem). We call $\beta \in \mathbb{R}, \beta > 0$ *extra-transcendental* if it is not extra-algebraic. Candidates for examples of extra-transcendental numbers are $e^\pi$ and $e^\frac{-\pi}{2}$.

**Question**:

- Do extra-transcendental numbers exist?
- Is $\mathbb{R^+} - \mathbb{A}_E$ uncountable?

Many thanks.

`$\mathbb P$`

included the requirement that`$r_i>0$`

, so the question involves only real exponents. Since he also asked only about real roots of such "polynomials", the OP probably intended everything in the problem to be real. (In particular, $x$ should probably range only over positive real numbers, to avoid difficulties with non-integer exponents on negative real bases.) I'd expect that each such "polynomial" has only countably many roots, in which case affirmative answers to both parts of the question would follow immediately. – Andreas Blass Mar 29 '11 at 14:54