1
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

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

$\endgroup$
  • $\begingroup$ @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 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 at 10:51
  • 1
    $\begingroup$ @Algebear: No, of course we don't know that! Please try and figure this out for yourself. It's trivial! $\endgroup$ – TonyK Jun 1 at 17:50
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 at 10:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.