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I stumbled upon the following problem, but I can't seem to find a way to prove it.

Show that there does not exist a strictly increasing function that maps the set of natural numbers to natural numbers, satisfying $f(2)=3$ and $f(a\cdot b)=f(a)\cdot f(b)$ for any $a,b$ in the naturals.

The only information I was able to obtain from the probably are trivialities, such as $f(2)=3,f(1)=1,f(4)=9, 9<f(5)<27, f(6)=3f(3)\,$ if it is assumed that the function is actually increasing. Altough, I can't find a pattern nor disprove by contraddiction. How would I go about this?

DMH16
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  • Probably a noteworthy clarification: do the naturals in this context include $0$ or no? Depending on the context or professor/author, people seem to differ on whether $0 \in \mathbb{N}$. If it is allowed, then it makes finding such a counterexample fairly easy, I think, but best to ask first. – PrincessEev Nov 20 '18 at 00:18

1 Answers1

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Hint: Let $f(3)=k$. Then

$$f(2^m)=3^m$$

and

$$f(3^n)=k^n.$$

Can you find a pair $(m,n)$ so $2^m<3^n$ but $3^m>k^n$ or vice versa?

Carl Schildkraut
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