Does a transcendental number raised to the power of an irrational algebraic number, always result in an algebraic number

exponentiationirrational-numberstranscendental-numbers

If $b$ is transcendental number and $a$ is an irrational algebraic number, is it safe to conclude that $b^a$ must be an algebraic number?

My reasoning is:

  1. Write the result of the Gelfond–Schneider theorem as $$b = d^c $$ where $d$ is algebraic, $c$ is an irrational algebraic number and $b$ is transcendental.
  2. Raise both sides to the power of $(1/c)$ to obtain: $$b^{(1/c)} = d.$$
  3. Since $c$ is an irrational algebraic number, $1/c$ must also be an irrational algebraic number. Let us write $ a = 1/c$ where a is also an irrational algebraic number, then the equation becomes: $$b^{(a)} = d.$$

This would appear to imply that if $b^{x}$ is algebraic, then x must be an irrational algebraic number and if $ \ X^{a}$ is algebraic, then $X$ must be transcendental.

Is this reasonable or have I messed up somewhere?

Best Answer

You've misapplied the quantifiers for the Gelfond-Schneider theorem. In particular, when you make the statement $b = d^c$ for $b$ transcendental, $d$ algebraic, and $c$ algebraic irrational, it doesn't automatically hold for all such $b$, $c$, and $d$

The theorem states that if $d \neq 0,1$ is algebraic and $c$ is algebraic irrational, then $b$ must be transcendental.

But for your reasoning to work, you would need that if $b$ is transcendental and $c$ algebraic irrational, then $d$ is algebraic. And that's not true at all, so your reasoning is based on a false assumption.

For an illustrative analogy, it is true that a rational number added to an irrational number is always irrational. Thus you could write the equation

$a = b + c$

For $b$ rational and $a$ and $c$ irrational. You could then subtract $c$ from both sides to get

$a - c = b$

and claim that the difference of irrational numbers is always rational - clearly false, for example $\sqrt{2} - \sqrt{3}$ isn't rational. It's the same mistake - the equation holds for all rational $b$ and irrational $c$, but that doesn't mean it holds for all irrational $a$ and $c$.

Edit: I realized I didn't actually answer your question as asked, but the answer is no. As has been pointed out in the comments, one can easily see that the set of "transcendental numbers raised to irrational algebraic numbers" is uncountable, while the set of algebraic numbers is countable.

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