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This arises from the standard question at the beginning of any basic number theory course: determining whether $ab$ is always irrational, given that $a$ and $b$ both are.

As far as I can tell, the counterexamples to the above statement fall into the following two categories (given $p \in \mathbb{Q}, q \in \mathbb{Q}$):

  • $a = b \cdot q \text{ and } b = \sqrt{p}$
  • $a = b^{-1} \cdot q$

The answers to this question appear to me to be a case of the latter.

Are there any examples beyond those listed above?

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    $\begingroup$ By definition, the second bullet-point has to cover all ways for $ab$ to be rational with $a,\,b$ irrational. $\endgroup$
    – J.G.
    Sep 21, 2019 at 20:04
  • $\begingroup$ $\sqrt8\cdot\sqrt2=2$ $\endgroup$
    – Andrew Chin
    Sep 21, 2019 at 20:06
  • $\begingroup$ @AndrewChin: $\sqrt(8)=2 \cdot \sqrt(2)$ $\endgroup$
    – Marcel
    Sep 21, 2019 at 20:08
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    $\begingroup$ @AndrewChin $\sqrt8=4/\sqrt2$ $\endgroup$
    – almagest
    Sep 21, 2019 at 20:08
  • $\begingroup$ I will point out that there are more types of irrational numbers than just ones of the form $\sqrt{p}$. As such, every example falls under the second, and only a few restrictive examples fall under the first (but would also fall under the second) $\endgroup$
    – JMoravitz
    Sep 21, 2019 at 20:16

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No, there aren't. If $ab=q$, with $q\in\mathbb Q$, then $a=b^{-1}q$,

Note that if $a=bq$ and $b=\sqrt p$ for some $p\in\mathbb Q$, then$$a=b^{-1}b^2q=b^{-1}pq.$$Therefore, the first case is a particular type of the second one.

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