Set theory definition of addition, negative numbers, and subtraction? [closed]

Using the definition of natural numbers $0 = \emptyset$ and $S(n) = n \cup \lbrace n \rbrace$ where S is the successor function, what is the definition of addition on natural numbers?

Concerning the definition of negative integers, the wikipedia entry is a bit ambigous: http://en.wikipedia.org/wiki/Integer#Construction

It seems to claim that the integers are defined such that, for example, the natural number 2 is not equal to the integer 2, which is defined as $\lbrace n \in \mathbb{N}^2 | \exists m \in \mathbb{N}:n = (m + 2,m) \rbrace$ or simply $\lbrace (2,0), (3,1), (4,2), ... \rbrace$, whereas the natural number 2 would be $\lbrace 0,1 \rbrace = \lbrace \emptyset , \lbrace \emptyset \rbrace \rbrace$. It was to my understanding that the above definition for integers was reserved for the negative integers, for example $-1 = \lbrace (0,1), (1,2), ... \rbrace$, and the integers would be defined as the union of the sets of negative integers and natural numbers. Is wikipedia wrong and natural 2 = integer 2, or is it the other way around? And in the latter case, are rational 2, real 2 and complex 2 also distinct? (And the ordinal number 2, and the cardinal number 2...)

Assuming the former case, is there an "official" way of defining things next? It seems to me like the easiest way of doing things would be defining first subtraction for $(n,m)$ where $n \ge m$; that is, $n-m$ is the natural number $d$ such that $n = m + d$. Then defining additive inversion, i.e. the unary operation $-$ s.t. $0 \mapsto 0$, $n \mapsto -n$ for nonzero natural $n$ (here $-n$ is the already defined negative integer and not the $-$ operation on $n$), and for negative integers $n$, $n \mapsto m$ where $m$ is the natural number s.t. $(0,m) \in n$. From there we can define addition and subtraction more generally for all combinations of natural numbers and negative integers, using additive inversion whenver we get a negative result, for example, $3 - (-4) = 3 + 4$, $3 + (-4) = 3 - 4 = - (4-3)$, $(-3) - 4 = (-3) + (-4) = - (3+4)$.

Tl;dr: what is the definition of addition on naturals, is natural 2 = integer 2 or are they distinct elements, and how do we define addition and subtraction on integers?

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closed as off topic by Andres Caicedo, Douglas Zare, Andreas Blass, Goldstern, Andy PutmanNov 7 '12 at 4:08

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Perhaps you should read Benacerraf's classic 'What numbers could not be' (scribd.com/doc/56939539/Benacerraf-What-Numbers-Could-Not-Be). The point of numbers is not that they are coded in a certain way, but that they are models of the axioms of arithmetic, ring theory etc. If you are worried about whether $2\in\mathbb{N}$ is equal to $2\in\mathbb{Z}$, then what's going to really bake your noodle is the statement $3\in 5$ (and many others, like $\pi\in \sin x$, which is a legitimate assertion in ZFC). $3\in 5$ is true in some codings of the natural numbers and false in others. –  David Roberts Nov 6 '12 at 6:19
@David Thanks for this comment. That was what I wanted to point out in my answer, but you also give very good (counter?)examples of why not asking these questions. –  jmc Nov 6 '12 at 6:24

There are many different ways of defining the natural numbers, integers, fractions, reals and complex numbers. I for myself do not think there is a canonical way. Thus, Wikipedia is not wrong, and there is not a way to do it *more right".

You certainly do not want to think of all these numbers as their underlying sets. One could start thinking about the intersection of $\frac{2}{3}$ (as fraction) with $\pi$ (as real), but it would make absolutely no sense.

What really matters is the algebraic structure. No matter which definition you give of $\mathbb{N}, \mathbb{Z}, \mathbb{Q}, \mathbb{R}, \mathbb{C}$, you can identify them in a natural way. If $\mathbb{Z}_{1}$ is your first set-theoretic definition of the integers, and $\mathbb{Z}_{2}$ is another one, then there is a canonical function $f \colon \mathbb{Z}_{1} \to \mathbb{Z}_{2}$ such that:

• $f(m +_{1} n) = f(m) +_{2} f(n)$
• $f(m \times_{1} n) = f(m) \times_{2} f(n)$
• $f(0_{1}) = 0_{2}$
• $f(1_{1}) = 1_{2}$.

So, what really matters is this algebraic structure of these sets.

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Oh, and those natural identifications that I wrote about also preserve things like natural orderings (if available) and such. You might want to take a look at www.en.wikipedia.org/wiki/Category_theory . I hope that it might help you to look at your questions from a different (in my eyes, more satisfying) view. –  jmc Nov 6 '12 at 4:04

The set of natural numbers $\mathbb N$, together with its natural addition, is a commutative semigroup. There is a standard way (*) to extend a commutative semigroup $(S,+)$ to a commutative group $(G,+)$: G is the quotient space of $S\times S$ by the relation $(a,b)\sim (a',b')$ if and only if $a'+b=a+b'$, with addition $[a,b]+[a',b']=[a+a',b+b']$. All the formulas with sign are easily shown.

The semi group $S$ is not a subset of $G$ but embeds naturally as a semigroup by the injective map $a\mapsto [a,0]$.

The construction of ($\mathbb Z$,$+)$ from ($\mathbb N$,$+)$ is exactly the construction above.

(*) but I don't think we can call it "official"

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Well, it is pretty natural in the sense that that it is adjoint to the inclusion of commutative groups into commutative semigroups (/monoids). I would call it official. But does it answer his question? The poster seems pretty concerned about underlying sets for these objects. –  jmc Nov 6 '12 at 3:59
@Johan: I don't think Taladris meant that the adjoint isn't official but rather that this particular construction of it, as a quotient of the product, isn't official. –  Andreas Blass Nov 6 '12 at 12:29
@Taladris: In order to conclude that $S$ embeds naturally in $G$, one needs to assume more about $S$ than just that it's a commutative semigroup. One needs cancellation. So it's OK for $\mathbb N$, but in the general case one only has a natural homomorphism $S\to G$ which might not be one-to-one. –  Andreas Blass Nov 6 '12 at 12:31
An addendum to my preceding comment: In the absence of cancellation, the alleged equivalence relation defined in this answer can fail to be transitive. The general construction would identify $(a,b)$ with $(a',b')$ if there is some $c\in S$ such that $a'+b+c=a+b'+c$. (As in my previous comment, this is unnecessary in the construction of $\mathbb Z$ from $\mathbb N$ because there we have cancellation. –  Andreas Blass Nov 6 '12 at 12:35
@Andreas: you're perfectly right. Actually, the definition of semigroup seems to differ from one author to another. "Semigroup" in my answer is what is called in Wikipedia a "monoid with cancellation property", i.e. a set with an associative binary operation, an identity element for this operation and which has the cancellation property. –  Taladris Nov 6 '12 at 13:31