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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?

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Solution enter image description here

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Jacob Willis
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2 Answers2

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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.

ZTransformer
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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?)

Eli Rose
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  • Even simpler than taking logarithms, just find a counterexample. With $u=8,v=5$, we have $2^8 = 256 > 243 = 3^5$, but $3^8 = 6561 < 7776 = 6^5$. – Théophile Jan 02 '16 at 04:28
  • @Théophile: Even better -- and that way you don't have to use a black box to calculate your logs. – Eli Rose Jan 02 '16 at 04:30