# How does ZFC describe addition?

Surprisingly, the Wikipedia article on addition doesn't contain the answer. I looked elsewhere online for it, but didn't find it.

Intuitively, the cardinal of the union of two sets seemed appealing. But that approach doesn't work, given that the union of 2 and 2 is 2. I wondered whether ordered sets might be essential to the answer.

So, how does ZFC describe addition?

• Addition of what kinds of objects? – Cameron Buie Feb 12 '14 at 15:35
• Natural numbers – Hal Feb 12 '14 at 15:37
• he probably means ordinal addition – Kaa1el Feb 12 '14 at 15:37
• I have answered this very question at least once on the site. This is not your first duplicate, either. Please try and search the site before posting questions (and not just Wikipedia!). – Asaf Karagila Feb 12 '14 at 15:39
• See at list the post on Set theory and addition – Mauro ALLEGRANZA Feb 12 '14 at 15:51

## 3 Answers

There are a few ways we can go about it, which ultimately amount to the same operation.

One is as the cardinal of the disjoint union of sets, which avoids the problem of union that you mentioned--and is simply cardinal addition, restricted to finite cardinals. Another is as the order type of the disjoint union, where the elements of the first set are supposed to be less than all the elements of the second, and the sets themselves keep the same orders--this is the non-recursive definition of ordinal addition. Yet another is to proceed recursively--for any ordinal $\alpha,$ we have $S(\alpha):=\alpha\cup\{\alpha\},$ and for any natural numbers $n,m,$ we define:

• $n+0:=n$
• $n+S(m):=S(n+m)$

Regardless, we end up with an addition operation having all the usual properties that we are used to.

Define $n++$ by $n++=n\cup\{n\}$, called the successor of $n$.

Define recursively,

$n+0=n$

$n+(m+1)=(n+m)++$

• While n++ works great for some programming language this is one of the ugliest, most horrifying, and un-mathematical ways to write the successor function. – Asaf Karagila Feb 12 '14 at 15:46

It is defined recursively by $$n + 0 = n\\ n + (a + 1) = (n + a) + 1$$ so to figure out $2+3$ (where $2 = 1 + 1$ and $3 = 2 + 1$), you do: $$2 + 3 = 2 + (2 + 1) = (2 + 2) + 1 = (2 + (1 + 1)) + 1 = \\((2 + 1) + 1) + 1 = (3 + 1) + 1 = 4 + 1 = 5$$

• But you never said how to define $n+1$ for a given $n$. – Clive Newstead Feb 12 '14 at 15:47
• @Clive Newstead: In ZFC, for any ordinal $\alpha$, $\alpha + 1$ is defined to be $\alpha \cup \{\alpha\}$. This is due to the von Neumann representation of ordinals. – Carl Mummert Feb 12 '14 at 15:49
• @CarlMummert: I know that, I was worried that the OP might not, and wouldn't be able to work it out from Arthur's answer. – Clive Newstead Feb 13 '14 at 0:42