Let $\mathbb{A^+}$ be the set of non-negative [algebraic numbers](http://en.wikipedia.org/wiki/Algebraic_number). 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 [$2^\sqrt2$](http://en.wikipedia.org/wiki/Gelfond%25E2%2580%2593Schneider_constant) which are extra-algebraic but not algebraic and more by the [Gelfond–Schneider theorem](http://en.wikipedia.org/wiki/Gelfond%25E2%2580%2593Schneider_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 [$e^\pi$](http://en.wikipedia.org/wiki/Gelfond%2527s_constant) and $e^\frac{-\pi}{2}$. 


**Question**:


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


Many thanks.



Transcendental numbers: yet another classification
$\mathbb P$ is countable. Moreover, any $f\in\mathbb P$ is analytic, hence it has only countably many zeros. Thus $\mathbb A\_E$ is countable, and in particular, extra-transcendental reals exist, and $\mathbb R^+\smallsetminus\mathbb A\_E$ has the power of continuum.


For a concrete example, the [Lindemann–Weierstrass theorem](http://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem) implies that $e$ is extra-transcendental.


EDIT: To tie up a loose end, every nonzero $f\in\mathbb P$ has only *finitely* many positive real roots. Since $f(x)$ is eventually dominated by its nonzero term with the highest exponent, the roots are bounded. Similarly, $f(x)$ is dominated by the term with the smallest exponent when $x\to0+$, hence the roots are bounded away from $0$, i.e., they are contained in a compact subset of $(0,+\infty)$. However, choosing a branch of logarithm makes $f$ holomorphic in $U=\mathbb C\smallsetminus(-\infty,0]$, therefore it can have only finitely many roots in any compact subset of $U$.


In fact, if $f(x)=a\_0x^{r\_0}+a\_1x^{r\_1}+\cdots+a\_nx^{r\_n}$ (with $r\_i\in\mathbb R$ pairwise distinct, $a\_i\in\mathbb R\smallsetminus\{0\}$), then $f$ has at most $n$ positive real roots. 


We can prove this by induction on $n$. If $n=0$, then $f(x)=a\_0x^{r\_0}$ has no positive real root. Assume the statement holds for $n-1$. Put $g(x)=a\_0+a\_1x^{r\_1-r\_0}+a\_2x^{r\_2-r\_0}+\cdots+a\_nx^{r\_n-r\_0}=f(x)/x^{r\_0}$. Then every positive root of $f$ is also a root of $g$. Moreover, between each two consecutive roots of $g$, there is a root of its derivative $g'$. Since $g'$ has at most $n$ nonzero terms (the derivative of the constant $a\_0$ vanishes), it has at most $n-1$ positive real roots by the induction hypothesis, thus $f$ has at most $n$ such roots.


I guess that one could also prove a variant of [Decartes' rule of signs](http://en.wikipedia.org/wiki/Descartes%27_rule_of_signs) for these generalized polynomials along similar lines.


