# Is it possible to create division via Set Theory?

I've been reading a book on Set Theory (Charles C. Pinter), and it says,

...set theory is recognized to be the cornerstone of the "new" mathematics... [emph. added]

and that

...we can still form all the sets essential for mathematics... [emph. added]

# Background

The book also states that Set Theory can be used to form the framework for mathematics (I am regrettably unable to quote Pinter on that). For example, if $A$ and $B$ are two disjoint sets, and that $a=\# A$ and $b=\# B$ denote that $a$ and $b$ are the cardinal numbers representing $A$ and $B$ respectively, then:

$$a+b=\#(A\cup B),$$

and that

$$a\cdot b=\#(A\times B).$$

Exponentiation can be produced by:

... Let $a$ and $b$ be cardinal numbers, and let $I$ be a set such that $b=\#I$. If $a=a_i$,$\forall i\in I$, then:
...
ii) $\displaystyle\ a^b=\bigotimes_{i\in I}a_i$ [symbol changed].

Where

$\displaystyle\bigotimes_{i\in I}a_i=\#(\prod_{i\in I}A_i)$, when $A_i$ is a family of sets, $a_i=\# A_i, \forall i\in I$ [symbol changed].

Was originally an over-sized $\times$ symbol, but was changed due to the fact that this (to my knowledge) cannot be achieved

# Question

This, oddly enough, is where Pinter decided to stop. He did not say how to produce division, or even the subtraction of cardinal numbers. I guess the subtraction of a cardinal would be (assuming that $0=\varnothing$, $1=\{\varnothing\}$, $2=\{\varnothing,\{\varnothing\}\}$... or that $0=\varnothing$, $1=\{0\}$, $2=\{0,1\}$...), and if $a=\#A$ and $b=\#B$: $$a-b=A\setminus B$$ ...but this, to my knowledge, only works when $a\geq b$, for one cannot have a cardinal (let's say $x$) where $x<0$ (because that means the set is a proper subset of $\varnothing$, which is impossible).

How can I use Set Theory to represent functions like subtraction and division?

• You can formally construct the integers with an equivalence relationship atop the natural numbers, then construct the rationals using the integers you've just constructed. – Alex Nelson Dec 8 '14 at 2:32
• And, indeed, it is impossible to define subtraction or division on cardinal numbers in any meaningful way once you get to infinite sets (which is the point at which the true power of set theory manifests itself). – Peter Košinár Dec 8 '14 at 2:53
• I guess that those questions are related: - math.stackexchange.com/questions/146844/… - math.stackexchange.com/questions/140930/… - math.stackexchange.com/questions/920468/division-of-cardinals – MphLee Dec 8 '14 at 12:48
• Conor, have you looked at the links that @MphLee posted? If you're asking about divisions of cardinal numbers, and it seems that this is what you're asking about, then the topic was covered already. For infinite sets this notion becomes either too trivial or not well-defined; for finite sets, well we can formally define the non-negative rational numbers and now say that the cardinals are a subset of that; but what good comes out of this? This won't correspond to any natural notion of cardinality of a set anyway. – Asaf Karagila Dec 8 '14 at 21:22
• Those are just infinite cardinals. Assuming the axiom of choice these are all the infinite cardinals, too. But without the axiom of choice, everything about the structure of cardinals (including much of their arithmetic, except the basic definitions) can break down. – Asaf Karagila Dec 8 '14 at 21:33