Why does the monotonicity imply $2^u < 3^v$ if and only if $3^u < 6^v$? 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



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?)
