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Is there a set theoretic construction of the natural numbers or integers such that the product of two numbers is their Cartesian product? What I mean is, e.g., if $25 = A$ and $2 = B$ then $50 = A\times B = \{(a,b)\mid a\in A \land b\in B\}$.

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    $\begingroup$ Almost certainly not, since from $X\times Y$ we can determine $X$ and $Y$, so you'd have trouble with $3\times 4=2\times 6$, or even $1\times 6=2\times 3$ $\endgroup$ – Thomas Andrews Jan 10 '13 at 16:34
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    $\begingroup$ @Thomas: In fact, we can determine $X$ and $Y$ from $X \times Y$ whenever $X$ and $Y$ are nonempty. But of course we can only use the empty set once. $\endgroup$ – Chris Eagle Jan 10 '13 at 16:36
  • $\begingroup$ Can someone clarify the question please? It looks like A and B are being used as numbers $and$ sets. A number is not a set. $\endgroup$ – Adam Rubinson Jan 10 '13 at 19:07
  • $\begingroup$ @AdamRubinson in this case, numbers are sets. That's what is meant by "set theoretic construction". $\endgroup$ – Massey Cashore Jan 10 '13 at 19:23
  • $\begingroup$ Alright fair enough. I have no idea about it then $\endgroup$ – Adam Rubinson Jan 10 '13 at 19:59
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If this property holds, then at most one number is represented by the empty set. Let $n,k$ be two numbers which are not empty sets.

We have that $n\times k=k\times n$. Therefore $(a,b)$ appears in both and so $a\in n$ implies that $a\in k$ and similarly $b\in k$ implies that $b\in n$. So $n=k$.

This means that all the numbers which are non-empty sets are equal, which is impossible.

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