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Here, $\langle a \rangle$ denotes the smallest subgroup of $\mathbb{Q}_{>0}$ containing $a$.

I've already proved this by showing that $\langle 2 \rangle$ and $\langle 3 \rangle$ are both isomorphic to $\mathbb{Z}$, but I'd like to prove it by showing that a well defined function $f: \langle 2 \rangle \to \langle 3 \rangle$ is bijective and operation-preserving. Since $\langle 2 \rangle = \{2^n:n\in\mathbb{Z}\}$ and $\langle 3 \rangle = \{3^n:n\in\mathbb{Z}\}$, I defined $f$ such that $f(2^n) = 3^n$. How do I now show that $f$ is well defined by writing $f(x) = \dots$? I know it probably includes a log and I can perhaps guess what the explicit function might be, but I was wondering if there was a process like 'solving for x' but in the context of writing an explicit function.

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    $\begingroup$ The function is defined is perfectly fine. $\endgroup$ – Fimpellizieri Nov 8 '16 at 1:15
  • $\begingroup$ I know that it can be explicitly shown that $f(2^n) = 3^n$ is well-defined, but my question is: how can one write $f(x) = \dots$ given $f(2^n) = 3^n$? $\endgroup$ – playitright Nov 8 '16 at 1:18
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You don't need a complicated expression but if you want one, here it is: $$ f(x) = 3^{\log_2(x)} $$

However, this expression only makes sense in $\langle 2 \rangle$, because for the other rationals, the value of $f$ will be irrational.

Bottom line: stick to the simple $f(2^n)=3^n$.

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  • $\begingroup$ Thank you! How did you achieve this expression, if not by simply thinking about it? Is there a step-by-step process similar to how one 'solves for x'? $\endgroup$ – playitright Nov 8 '16 at 1:23
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    $\begingroup$ @playitright, the expression implements this: take the exponent of the power of $2$ and raise $3$ to that exponent. $\log_2$ takes the exponent. $\endgroup$ – lhf Nov 8 '16 at 1:26
  • $\begingroup$ I've taken your advice to stick with $f(2^n) = 3^n$, but I'm having trouble proving surjectivity of this function. Would you do it algebraically? $\endgroup$ – playitright Nov 8 '16 at 1:45
  • $\begingroup$ @playitright, subjectivity is clear from the definition but if you want, consider $g(3^n)=2^n$ and prove that $f$ and $g$ are inverses. $\endgroup$ – lhf Nov 8 '16 at 1:48
  • $\begingroup$ Apologies for another follow-up question--I've never done set theory with a function that wasn't of the form $f(x) = \dots$, so this is relatively unintuitive to me. I'm currently trying to show that $f \circ g = id_{\langle 3 \rangle}$, but I don't really know where to start. I have the following: $(f \circ g)(3^n) = f(g(3^n)) = f(2^n) = 3^n$. I don't think this shows anything, though. $\endgroup$ – playitright Nov 8 '16 at 2:00

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