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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) so $k>5$ and since $3^3 < 5^2$, then $k^3=f(3)^3=f(3^3) 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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    I have updated my progress so far2012-06-06

1 Answers 1

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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 $f(3)$.

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    No problem - we learn most from the struggle.2012-06-06