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.