This question arose in a discussion with Peter Doyle.

It is provable in ZF that one can divide by any positive finite cardinal $k$: if $X \times \{1,\ldots,k\} \simeq Y \times \{1,\ldots,k\}$ then $X \simeq Y,$ where I use $\simeq$ to denote the existence of a bijection between two sets.

Assuming the Axiom of Choice, the positive finite cardinals are precisely the cardinals that one can divide by since $1\cdot\mathfrak{m} = 2\cdot\mathfrak{m}$ for any infinite cardinal $\mathfrak{m}$. The ultimate question is whether this is actually provable in ZF:

If $\mathfrak{m}$ is an infinite cardinal, must there be two cardinals $\mathfrak{p} \neq \mathfrak{q}$ such that $\mathfrak{p}\cdot\mathfrak{m} = \mathfrak{q}\cdot\mathfrak{m}$?

As a candidate for a counterexample, I thought it might be possible to divide by the cardinal of a (strongly) amorphous set. Specifically, in the First Fraenkel model:

Is it true that if $X \times A \simeq Y \times A$ then $X \simeq Y$, where $A$ is the set of atoms?

  • $\begingroup$ There is Lovasz's unique factorization and unique roots for finite structures, and related results. It might be prudent (and useful if not tangential for this question) to see in which logical domains his arguments can be repeated. Gerhard "Forward Engineering For Reverse Mathematics" Paseman, 2016.04.09. $\endgroup$ Commented Apr 9, 2016 at 22:10
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    $\begingroup$ I like this question! $\endgroup$
    – Asaf Karagila
    Commented Apr 9, 2016 at 22:15
  • $\begingroup$ Can someone give me a hint on how to prove that in $\text{ZF}$ we can divide by finite cardinals? $\endgroup$
    – Lorenzo
    Commented Sep 11, 2022 at 16:30
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    $\begingroup$ @Lorenzo: A hint would imply that this is easy... it's not! Peter Doyle has written up division by three and division by four - math.dartmouth.edu/~doyle $\endgroup$ Commented Sep 11, 2022 at 20:28


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