[Math] How to prove $2+2=4$ using Zermelo–Fraenkel Set Theory

first-order-logiclogicset-theory

I'm familiar with the famous excerpt from Principia Mathematica by Bertrand Russel and Alfred Whitehead. However, as Zermelo–Fraenkel Set Theory is today's most used foundation of mathematics (or ZF if you believe the axiom of Choice is implied), I was wondering if there is a postulation expressed in the symbolism of first-order logic for $2 + 2 = 4$ following ZFC. I'm not a logician, not even an undergraduate yet, and I was unable to find such a proof online, maybe I'm just overlooking some elementary understanding of what ZFC is.

Best Answer

Zermelo–Fraenkel set theory is formulated in a language that has a binary relation $\in$, and that's it. The symbols $1,2,3,4$ and $+$ make "no sense" from a syntactic point of view.

So first one needs to explain what is $2$ in the context of set theory. This means that you need to define some sets that will behave like you would expect of the natural numbers. Normally, these are the finite von Neumann ordinals, defined recursively as $0=\varnothing$ and $n+1=n\cup\{n\}$. There are other ways to define the natural numbers, or you could be thinking about the real numbers, and then you need to define those as well first, but for now let's stick with the von Neumann ordinals and see where it takes us.

Note that even in the case of Peano arithmetic or ring theory, the symbol for $2$ and the symbol for $4$ are not part of the language. They are used as shorthand for either repeated sums of $1$ or repeated application of the successor function to $0$, or so on.

So if $2$ is defined as $1+1$ and $4$ is defined as $((1+1)+1)+1$, and you have an axiom saying that $(x+y)+z=x+(y+z)$, then you're practically done: $(1+1)+(1+1)=((1+1)+1)+1$, so $2+2=4$. But okay, let's go back to set theory.

We have the natural numbers, so $2=\{\varnothing,\{\varnothing\}\}$ and $4$, well, let's just write $4$. After this you need to ask what is $+$. Do you think about $+$ as a recursive definition of applying the successor function? Even the Peano axioms which usually form the basis for arithmetic have $+$ as a standalone object, and there is a connection between $+$ and the successor function. There are two standard ways to define what is $+$ on the natural numbers in the context of set theory:

  1. Successive application of the successor function, so $x+0=x$ and $x+S(y)=S(x+y)$. In that case, $2+2=S(1)+S(1)=S(S(0))+S(S(0))=\dots=S(S(S(S(0))))$. And now we can also translate back into sets to get "a fully set theoretic statement". But it's awful, and I don't want to do it.

  2. Using cardinality of sets. Note that $n$ is a set with exactly $n$ elements. $0$ has no elements and $1$ has exactly one element (which is $0$, as it turns out). So we can define $n+m=k$ if and only if the number of elements in a disjoint union of a copy of $n$ and a copy of $m$, is $k$. Or, in modern terms, there is a bijection between the two sets.

    In that case, one needs to write down a function from $\{\langle 0,0\rangle,\langle 0,1\rangle\}\cup\{\langle 1,0\rangle,\langle 1,1\rangle\}$, or $\{0\}\times2\cup\{1\}\times2$, and $4$ which is $\{0,1,2,3\}$. Of course, this starts with defining what is an ordered pair from a set theoretic view, what is a function, etc., and then write this as a set theoretic expression again. Which, as before, is an awful exercise in futility.

The easiest way, at the end, is to prove this "by blocks". First prove there is a way to formalize the natural numbers, then formalize addition, then show that no matter what formalization you chose, as long as it satisfies certain properties (which you would expect from the natural numbers), it has to be the case that the object corresponding to $2+2$ is the object corresponding to $4$.


Or just be annoying and declare that you interpret $2,+$ and $4$ in a non-standard way so $2+2=5\neq 4$. Whatever floats your boat.

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