This question is a continuation of my previous question.

I was wondering about products of transcendental numbers where I exclude the cases where we take products of transcendental numbers $\alpha$ and $\frac{1}{\alpha\gamma}$, with $\gamma$ algebraic, that is algebraic. Of course, the product $\alpha\cdot\frac{1}{\alpha\gamma}=\frac{1}{\gamma}$ must again be algebraic because the algebraic numbers form a field (so inverses of algebraic numbers are again algebraic).

Suppose $\alpha$ and $\beta$ are both transcendental and let $\beta\neq\frac{1}{\gamma\alpha}$ where $\gamma$ is algebraic. Is it then the case that $\alpha\beta$ is always transcendental?

I'm a bit familiar with the theory of transcendental numbers and showing they are. I know how to prove transcendence of $e$ and $\pi$ and the sum and product of it, but not of anything else, but maybe we can try the same technique used for that product and sum.

Attempted proof: Suppose both $\alpha$ and $\beta\neq\frac{1}{\gamma\alpha}$, where $\gamma$ is algebraic, are transcendental but both $\alpha+\beta$ and $\alpha\beta$ are algebraic. Then it must be the cases that $$ (\alpha+\beta)^2-4\alpha\beta=(\alpha-\beta)^2 $$ is algebraic (field of algebraic numbers) and since "the sum and product of two complex numbers are algebraic iff those two complex numbers are algebraic", we must also have that $\alpha-\beta$ is algebraic. But then it holds that $$ \frac{(\alpha+\beta)-(\alpha-\beta)}{2}=\beta $$ is algebraic, which is a contradiction, so both $\alpha\beta$ and $\alpha+\beta$ must be transcendental.

Is this correct?

EDIT: Since there seemed to be some misconceptions, I must clarify a bit more what I mean. I take $\alpha$ transcendental and $$ \beta\in\mathbb{T}-\left\{\beta\in\mathbb{C}\;\bigg|\;\beta=\frac{1}{\gamma\alpha}\right\}. $$

EDIT2: I now get that my claim makes it instantly true and the "proof" was superfluous. Does this now imply that e.g. $\frac{e}{\pi\sqrt{2}}$ and $\frac{\pi\sqrt[5]{\pi}}{ e\sqrt{57}}$ are transcendental numbers?


If you claim (what you do) that $\dfrac1\gamma=\alpha\beta $ is not algebraic number then it is certainly (by definition) transcendental one.

  • $\begingroup$ @Aglebear: you said that $\alpha\beta$ is not equal to $1/\gamma$, where $\gamma$ is algebraic. So $\alpha\beta$ must be transcendental. That is very simple! $\endgroup$ – TonyK Jun 1 '19 at 10:27
  • $\begingroup$ @TonyK That is indeed way simpler. I saw some threads that were discussing whether the product of two transcendental numbers is again transcendental and that was false by a simple counter-example. So I thought that by excluding those counter-examples, the product must be transcendental. Is that proven this easily? It doesn't even need the aglebra;-) I.e. do we now know that e.g. $\frac{e}{\sqrt{2}\pi}$ is transcendental? $\endgroup$ – Algebear Jun 1 '19 at 10:51
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    $\begingroup$ @Algebear: No, of course we don't know that! Please try and figure this out for yourself. It's trivial! $\endgroup$ – TonyK Jun 1 '19 at 17:50

If it's true for all $\gamma$, then yes your proof is fine. However, it seems like you're claiming that there exists $\gamma$ such that $\alpha\neq\frac{1}{\beta\gamma}$ which makes it false.

  • $\begingroup$ If $\alpha=\pi$ and $\beta=e\neq\frac{1}{\sqrt{2}\pi}$, then $e\pi$ is transcendental. Why can't I claim that there are two transcendental numbers $\alpha$ and $\beta\neq\frac{1}{\alpha\gamma}$? $\endgroup$ – Algebear Jun 1 '19 at 10:07

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