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It is known that the Dedekind-finite cardinals are closed under addition and multiplication, so one may do arithmetic in them, as opposed to only natural numbers.

How much can those two arithmetics be different? For example, can there be a Diophantine equation which is not solvable in the natural numbers but solvable in the Dedekind-finite cardinals? Can there be two nonempty Dedekind finite sets $A,B$ such that $|A|^2=2|B|^2$?

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    $\begingroup$ Well, Dedekind-finite cardinals need not be linearly ordered, so they can be very different. Sageev proved that assuming an inaccessible cardinal exists, there is an extension of the universe where all Dedekind-finite cardinals are comparable (in a nontrivial way, of course) and the axiom of choice for families of finite sets hold. Then, by a theorem of Ellentuck, it follows that the Dedekind-finite cardinals are a model of true arithmetic, if I remember correctly. $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 7, 2015 at 9:57
  • $\begingroup$ @AsafKaragila Interesting. Since exponentiation is definable in arithmetic, I guess that means that exponentiation of Dedekind-finite cardinals is defined in Sageev's model. But it can't be cardinal exponentiation, becase D-finite cardinals aren't closed under cardinal exponentiation. So what is it? $\endgroup$
    – bof
    Commented Apr 7, 2015 at 10:15
  • $\begingroup$ @bof: Yes, the exponentiation is not cardinal exponentiation, but of course it coincides with it when the power is finite. What is it exactly? Some binary operation I suppose. I haven't investigated Sageev's model yet. My guess is that you somehow quantify this as some sort of limit exponentiation (e.g. one might notice that the set of all finite injective sequences of a Dedekind-finite set, is also Dedekind-finite, so there is some weak exponentiation to be done here. But it seems to be closer to factorial than to exponentiation. So in short, I have no idea.) $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 7, 2015 at 10:20
  • $\begingroup$ @AsafKaragila, why not post your comments as an answer? $\endgroup$
    – Joel David Hamkins
    Commented Apr 7, 2015 at 12:36
  • $\begingroup$ @Joel: Because it's not quite an answer. In particular, I have no idea if it is consistent that there is such Diophantine equation. $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 7, 2015 at 12:42

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A joint paper by David Blair, Paul Howard, and me, "Divisibility of Dedekind-Finite Sets" (available at http://www.math.lsa.umich.edu/~ablass/ddiv.pdf and published in J. Math. Logic 5 (2005) 49-85) contains information of the sort you want about some rather specific linear diophantine equations.

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