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This question arose while trying to prove a theorem in real analysis.

Let $B=\{2^k\cdot 3^n:k,n\in\mathbb{N}\}$.

Define $f:B\to \mathbb{N}$ by $f(2^k\cdot 3^n)=nk$ with $n,k\in\mathbb{N}$.

How do we know that this is indeed a function and that we don't have two different sets of values $n,k$ that generate the same $2^k\cdot 3^n$?

What I came up with was

$$2^{k_1}3^{n_1}=2^{k_2}3^{n_2}$$

$$2^{k_1-k_2}3^{n_1-n_2}=1$$

$$k_1-k_2+\log{3^{n_1-n_2}}=0$$

$$\frac{n_1-n_2}{\log_3{2}}=k_2-k_1$$

$$n_1-n_2=(k_2-k_1)\log_3{2}$$

The lefthand side is an integer, and the righthand side does not seem to be an integer, though I don't know how to prove this adequately.

It seems that there is no solution and so $f$ is indeed a function. But how to make the argument rigorous?

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    $\begingroup$ Use the Fundamental Theorem of Arithmetic. $\endgroup$ Nov 30, 2023 at 1:19

1 Answer 1

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Uniqueness of prime factorization...

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  • $\begingroup$ Quickest response to a qn ever seen. $\endgroup$ Nov 30, 2023 at 1:10

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