Let $f: N \to N$, $f(2) = 3$, and $f(ab) = f(a)f(b)$, that is, f is a multiplicative function. f is also strictly increasing. Show that no such function exists.
Progress: Apparently, this is proven by contradiction. So I used $f(2) = f(2 * 1) = f(2) * f(1)$. This gives me $3 = 3 * f(1)$, which means $f(1) = 1$.
I see no contradiction though. f is strictly increasing, not decreasing, so it makes sense that $f(1) = 1$ but $f(2) = 3$. Help please?
