# Product of two non-related transcendental numbers again transcendental?

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.
• @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! – TonyK Jun 1 at 10:27
• @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? – Algebear Jun 1 at 10:51
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.
• 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 $\beta\neq\frac{1}{\alpha\gamma}$? – Algebear Jun 1 at 10:07