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I am following Understanding Analysis by Stephen Abott. I read a well bit into the book but I decided to go through and do the exercises through the book because I felt as if I wasn’t being rigorous.

I reached “Prove there does not exist rational number r such that $$ 2^{r} = 3$$ I started with the goal of reaching contradiction. $$let\ r=\frac pq \ such\ that\ p,q\in \mathbb Z$$ $$2^{\frac pq} = 3$$ Raise to q power $$2^p = 3^q$$ I understand where this is the typical endpoint where the conclusion is that there exists no integer multiple such that these two coprime numbers will be equal. However I was stumped there because I am still trying to get used to thinking how I should for these proofs.
So I took the natural log which felt wrong because I feel it is something that takes tip toeing around because I’d have to prove a statement I’ll denote in a following line.

$$pLn(2)=qLn(3) $$ $$\implies \frac pq = \frac{ln(3)}{ln(2)} $$

I know that ln(2) and ln(3) are both transcendental, and will result in some number that is not rational and thus cannot be equal to $\frac pq$ which is as they do not belong to the same set. Plus that neither the numerator nor denominator are integers. $$\therefore \nexists r \ S.T \ 2^r = 3, r \in \mathbb{Q}$$

I feel as if this is a weaker proof and resultant of not being horribly familiar with numbers that are relatively prime. Any advice on the validity of this proof as well as advice for moving forward into proof based math is deeply appreciated. Thanks!

Solution: I just need to get better at analysis and the way of thinking required for it.

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  • $\begingroup$ How do you know that $\ln(3)/\ln(2)$ is transcendental? $\endgroup$ Commented Jun 20, 2024 at 0:18
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    $\begingroup$ It's not a proof at all. It's possible for $\ln 2$ and $\ln 3$ to both be transcendental but for their quotient to be rational. For example, both $\pi$ and $2 \pi$ are transcendental yet their quotient, $2$, is an integer. As another example, $\ln 2$ and $\ln 4$ are both transcendental, but their quotient is $2$. $$$$ To solve your problem, I think you need to use the uniqueness of prime factorizations. $\endgroup$ Commented Jun 20, 2024 at 0:18
  • $\begingroup$ The argument actually goes in the opposite direction: uniqueness of prime factorization implies that $\frac{\ln 3}{\ln 2}$ is irrational. It is also possible to prove that it's transcendental but this is 1) harder and 2) not at all necessary here. $\endgroup$ Commented Jun 20, 2024 at 1:03
  • $\begingroup$ Thank you three, my question was regarding the validity and and has been answered, it’s not valid. I acknowledge that the proper way is by showing that the coprimes shouldn’t share any multiples for all of Z not including 0 and thus making the contradiction. Apologies for it not being typed well, it was all done on a phone at work. But that should be all, I’ll try to mark as closed, have a good day, thanks! $\endgroup$ Commented Jun 20, 2024 at 3:39
  • $\begingroup$ Plus to add, the original assumption was p and q to be integers and so I feel it could be shown that no integer multiple of ln(2) = some multiple of ln(3) disregarding 0 of course. However it was all unnecessary. I simply just need to get better at analysis, something that can only be done through experience and failure. $\endgroup$ Commented Jun 20, 2024 at 3:43

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