Is this a correct alternative proof to "Prove there is no rational $r$ satisfying $2^r=3$"

Question

I wanted to check if an alternative proof made any sense to this question:

prove there is no rational r satisfying $2^r=3$

I am just starting out with Stephen Abbott's for self study, and came to this solution for this question trying to use proof by contradiction.

Let $r$ be a rational number written as $\frac{p}{q}$ where $p$ and $q\in\mathbb{Z}.$

Then we can write $2^r = 3$ as $2^\frac{p}{q} = 3$, then by taking the log of both sides, bringing the power down, and multiplying by $q$ we get $$p\ln(2) = q\ln(3)\\ p=q\frac{\ln(3)}{\ln(2)}$$ The contradiction I thought I found was that here $p$ must be an integer, but the RHS is clearly not, which is a contradiction. The solution in the other post makes much more sense, but I am curious if this also works.

Answer 1

As pointed out in the comments, your proof has a slight hole in it, in that you end by saying that $q\frac{\ln 3}{\ln 2}$ is not an integer, but you haven't shown why $q\frac{\ln 3}{\ln 2}$ is not an integer.

If we try to fix this gap, we'll end up going through the proofs given in the answers to your linked question. Here's an example of how this might go.

We want to show $$q\frac{\ln 3}{\ln 2}$$ is not an integer. Observe that by log rules, we can rewrite this number as $$\log_2 3^q .$$

In order for this not to be an integer, we must have that $3^q\ne 2^a$ for any $a\in\Bbb{Z}$ (note that $q> 0$ wlog).

This gets us to the same place as many of the proofs on the other question.

At this point, we might say, since $q> 0$, $3^q$ is an odd positive integer larger than 1, so it can't be a power of 2, since those are all even. Then we're done.

Answer 2

hint

$$2^{\frac pq}=3 \implies 2^p=3^q$$

with $\frac pq=\frac{\ln(3)}{\ln(2)}>1$.

we assume that $p>q>0$.

then

$$2.2...2=3.3...3$$ which is not possible since and odd integer cannot be even.

Answer 3

The problem is, "The RHS clearly is not an integer" isn't very clear at all. One thing I was taught was to never use the word "clearly" in a proof, especially when starting out -- if it is clear, then explain it! If it isn't clear, then well hmm.

How to finish the question is from one of your intermediary steps -- $$2^p=3^q$$ I know you didn't write this explicitly, but that's what it said. Unless $p=q=0$, this directly contradicts the fundamental theorem of arithmetic (an easier approach would be to note that the LHS is odd and the RHS is even) and thus $2^r=3$ isn't solvable in the rationals.