(This is from a proof by contradiction, so that's why the equality does not actually hold. Edited for brevity; I don't think I've omitted anything pertinent to my questions.)
[...] The monotonicity of $f$ implies that $2^u<3^v$ if and only if $3^u<6^v$, $u$, $v$ being positive integers. Taking logarithms this means that $\frac{v}{u}>\log_2 3$ if and only if $\frac{v}{u}>\log_3 6$. Since rationals are dense, it follows that $\log_2 3 = \log_3 6$. This can be written as $\log_2 3 = \frac{1}{\log_2 3}+1$, and so $\log_2 3$ is the positive solution of the quadratic equation $x^2−x −1=0$ [...]
- Question 1
When he concludes that:
Since rationals are dense, it follows that $\log_2 3 = \log_3 6$.
What is the reason for this? Wouldn't this only be true if $\frac{v}{u}$ goes to $0$?
- Question 2
[$\log_2 3 = \log_3 6$] can be written as $\log_2 3 = \frac{1}{\log_2 3}+1$.
This one is the most confusing. How did he jump from $x=y$ to $x = \frac{1}{x}+1$?