# Nonexistence of a strongly multiplicative increasing function with $f(2)=3$

Show that there does not exist a strictly increasing function $f:\mathbb{N}\rightarrow\mathbb{N}$ satisfying

$$f(2)=3$$ $$f(mn)=f(m)f(n)\forall m,n\in\mathbb{N}$$

Progress: Assume the function exists. Let $f(3)=k$ Since $2^3 < 3^2$, $$3^2=f(2)^3=f(2^3)<f(3^2)=f(3)^2=k^2$$ so $k>5$ and since $3^3 < 5^2$, then $$k^3=f(3)^3=f(3^3)<f(2^5)=f(2)^5=3^5=243<343=7^3$$ so $k<7$ therefore $k=6$.

I've messed around with knowing $f(3)=6$ and $f(2)=3$ but I am stuck.

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Ah, sorry. Typo: I have edited the question to include 'strictly increasing' – rckrd Jun 6 '12 at 18:38
Have you made any attempt to solve the problem yourself? – Ben Millwood Jun 6 '12 at 18:47
I have updated my progress so far – rckrd Jun 6 '12 at 18:52

Hint: suppose not. You know what $f(2^k)$ must be. You'll show that $f(3)$ can't be any natural number. You have bounds since $f(2) < f(3) < f(4)$. Start considering $f(3^j) = f(3)^j$ for some small values of $j$ and compare to $f(2^k)$ for some small values of $k$ to eliminate all possibilities for $f(3)$.
I've found that $f(3)=6$ and this implies that $2^x < 3^y \iff 3^x < 6^y$ – rckrd Jun 6 '12 at 18:54
Got it! Using logarithms: The final step and contradiction was $65536 = 2^{5 \times 3.2} < 2^{5(1+\sqrt{5})}=9^5=59049$. Thanks for giving me a chance to work it out. – rckrd Jun 6 '12 at 19:05