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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:

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

Many thanks.

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The definition of powers is problematic ; how do you define $i^i$ for example ? –  Ewan Delanoy Mar 29 '11 at 14:43
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The definition of $\mathbb P$ 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
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Why stop at this point? Why not go on: Set $\mathbb A_0=\mathbb A$, $\mathbb A_1=\mathbb P$, define $\mathbb A_{n+1}$ as the set of all "polynomials" with coefficients in $\mathbb A_n$ and exponents in $\mathbb A_n\cap \mathbb R$? The union $\cup \mathbb A_n$ is then still enumerable and thus distinct from $\mathbb C$. –  Roland Bacher Mar 29 '11 at 14:59
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@Ewan, I took it as Andreas explains. Everything over the reals. @Andreas, thanks for the comment. –  Unknown Mar 29 '11 at 15:04
    
@Roland, that is a good observation. –  Unknown Mar 29 '11 at 15:48

1 Answer 1

up vote 18 down vote accepted

$\mathbb P$ 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 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 for these generalized polynomials along similar lines.

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@Emil, thank you. Sorry for confusing what $\mathbb R / \mathbb A_E$ meant. I have clarified it now. But it is already answered. Would it be hard to find examples? –  Unknown Mar 29 '11 at 15:21
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Proving explicit numbers transcendental is already quite tricky, and tends to be an unresolved problem even for simple numbers like $\pi+e$. I assume that here it gets much worse. However, now that I think about it, the Lindemann–Weierstrass theorem implies that at least $e$ is extra-transcendental. –  Emil Jeřábek Mar 29 '11 at 15:35
    
Thanks, again. –  Unknown Mar 29 '11 at 15:46

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