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  1. Is the cardinality of an infinite set divided by the cardinality of another infinite set indeterminate?
  2. And what if it is divided by itself?
  3. Have these results been proven or are they unprovable?

Thank you!

Andrés E. Caicedo
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user5103512
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  • I don't know of anyone who has even defined cardinal division. – André Nicolas Jun 13 '16 at 13:25
  • @AndréNicolas See W. Sierpiński, Cardinal and Ordinal Numbers Second Edition Revised, Chapter IX "Difference of cardinal numbers", especially section 6, "Quotient of cardinal numbers". – bof Jun 14 '16 at 02:36

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Division is not well defined for cardinal numbers at all.

For finite numbers $a/b$ means the $x$ such that $x\cdot b=a$ -- but for infinite cardinal numbers $\lambda$ and $\kappa$ we have $\lambda\cdot\kappa = \max(\lambda,\kappa)$ (assuming the axiom of choice holds).

So if $\lambda<\kappa$ there is no cardinal that would qualify to be $\lambda/\kappa$. If $\lambda>\kappa$ we could have $\lambda/\kappa=\lambda$, but that is not very interesting, so there's no real point in defining it at all.

hmakholm left over Monica
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  • If I have understood correctly, does that mean that if $k$ denotes the cardinality of an infinite set, $\frac{k}{k} = k$? 2) And what if we have $\frac{a^n}{b^n}$ where $a$, and $b$ are natural numbers and $n$ is the cardinality of an infinite set?
    • – user5103512 Jun 13 '16 at 13:29
  • @user347464: You have not understood correctly unless what you have understood is that division of cardinal numbers is not defined at all. The rest of the answer is an attempt to explain why it is not defined. – hmakholm left over Monica Jun 13 '16 at 14:00
  • Division of cardinal numbers is well defined in some cases; it is even of some interest if the axiom of choice is not assumed. Sierpiński's Cardinal and Ordinal Numbers (Second Edition Revised) has a whole chapter on "Difference of cardinal numbers" which contains a 2-page section on "Quotient of cardinal numbers". For example he quotes this ZF-result of Lindenbaum and Tarski: "If a cardinal number is divisible by $2$ and by $3,$ then it is also divisible by $6.$" – bof Jun 14 '16 at 02:42