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






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?

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

1 Answer 1


Uniqueness of prime factorization...

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

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