In the question and solution below, I am wondering how to #$7$ it says "The monotonicity of $f$" implies that $2^u < 3^v$ if and only if $3^u < 6^v$, $u,v$ being positive integers." How does this even depend on the definition of $f$? And if it does how is it true?
Problem
Solution
On
Because $f(mn) = f(m)f(n)$, we have:
$$ f(m^k) = f(\underbrace{m m \dots m}_{\text{$k$ times}}) = \underbrace{f(m)f(m)\dots f(m)}_{\text{$k$ times}} = f(m)^k $$
Also, because $f$ is strictly increasing, we have:
$$ a > b \implies f(a) > f(b) $$
(indeed that's the definition of strictly increasing).
With these two together, the argument goes like this -- suppose we had $2^u < 3^v$. Then,
$$ \begin{aligned} 2^u &< 3^v\\ f(2^u) &< f(3^v)\\ f(2)^u &< f(3)^v\\ 3^u &< 6^v\\ \end{aligned} $$
(What a bizarre argument, may I say. Couldn't they have just calculated $\log_2(3)$ and $\log_3(6)$ once they got there, found that they were not equal, and called contradiction before doing any of the golden ratio business?)
It's just the property of a strictly increasing function $f$
$a<b$ iff $f(a)<f(b)$
$2^u < 3^v$ iff $f(2^u) < f(3^v)$
$f(2^u)=f(2)^u=3^u, f(3^v)=f(3)^v=6^v$ by definition.