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If anyone could me any advice/direction of how to approach this problem, I'd greatly appreciate it. I couldn't find anything on the internet that looked or was similar to proving this type of proof.

The fact that that there are irrational numbers a,b such that ab is rational was proved in Problem 1.5. Unfortunately, that proof was nonconstructive: it didn’t reveal a specific pair, a,b, with this property. But in fact, it’s easy to do this:

We know √2 is irrational, and obviously a=3. Finish the proof that this a,b pair works, by showing that 2 log base 2 of 3 is irrational.

Woody
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  • What does it mean for an $a,b$ pair to "work"? Is $b=\sqrt2$? I can only guess that what it means for the pair to work is for $ab$ to be rational despite $a,b$ being irrational, but this doesn't make sense since $a=3$ is rational. – YiFan Tey Jan 30 '20 at 23:04
  • What does $2\log_23$ have to do with this? What is $b$ in "this $a,b$ pair"? – saulspatz Jan 30 '20 at 23:06
  • I don't understand. Your example is incomplete (and you appear to choose $a=3$, contrary to the requirement that $a$ be irrational.) And if all you need is an example: Why not $a=\sqrt 2$, $b=\frac 1{\sqrt 2}$? – lulu Jan 30 '20 at 23:07
  • I assume that $2\log_23$ arises because the claim is that the equation $\sqrt{2}^{2\log_23}=3$ exemplifies an irrational to an irrational power being rational. This requires the post (or original question) to accidentally use $a$ in two different ways, and to be missing a superscript. – Mark S. Jan 31 '20 at 01:34

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