# Yet another product of irrational numbers

Let $~\alpha~$ and $~\beta~$ be irrational numbers such that

$$~\alpha \notin \{\beta, -\beta\}$$ and $$~\alpha \notin \left\{\frac{1}{\beta}, -\frac{1}{\beta}\right\}$$

I suppose that in this case their product $~\alpha\beta~$ should also be irrational number. But is it true? And if so, how can we prove it?

EDIT The initial hypothesis is too weak as Michael Hardy has shown. So the following question arises: is there any criterion for the product of irrational numbers to be irrational?

Thanks in advance for any ideas.

• Say, would a good qualifier here be: α and β are both bigger than one. in this way you can't use the "trick" of one being basically the inverse of the other – Fattie Apr 26 '16 at 14:31

If $\alpha=\sqrt{2}$ and $\beta = 10/\sqrt{2}$, then $\alpha\beta$ is rational. Maybe you should think about strengthening the hypothesis.
• The natural strengthening of the hypothesis is that $\alpha\beta$ is irrational! – André Nicolas Sep 4 '14 at 21:58
• What about the following case: $~\alpha \notin \{k\cdot\beta~|~k \in \mathbb{Z}\}\cup \{k\cdot\frac{1}{\beta}~|~k \in \mathbb{Z}\}~$ ? – Igor Sep 4 '14 at 22:03
• @Igor: Take $\alpha=\frac{2}{3}\pi$ and $\beta=\frac{1}{\pi}$. – André Nicolas Sep 4 '14 at 22:54