Are there multiple (small class) extensions of the naturals?
Suppose we have the natural numbers $\mathbb N = \left\{0,\,1,\,2,\,\cdots\right\}$, and the usual addition operator $+$. Suppose also that we want to extend the natural numbers, so we have more numbers to play with. We can define the set $\mathbb Z$ as equivalence classes of pairs of naturals $(a, b)$ and $(c, d)$ under the relation $a+d=b+c$: these are (isomorphic to) the integers. It makes sense to identify the integer 3 with the natural number 3, and so on, treating $\mathbb N \subset \mathbb Z$, in which case $\mathbb Z$ can be thought of as the closure of $\mathbb N$ under additive inverses.
However, if we want to extend this further, we have a choice to make. We can either take the field of fractions of the integers, giving us $\mathbb Q$, or we can choose a prime $p$ and construct the $p$-adic integers. From there, we can construct either the reals ($\mathbb R$), the algebraic numbers, or the $p$-adic numbers. Both the reals and the algebraic numbers can be seen as subsets of the complex numbers ($\mathbb C$), but there are $p$-adic numbers with no complex equivalent, and to my knowledge it is not possible to unify these two branches. There is, therefore, no unambiguous extension of the integers, and it does not necessarily make sense to unify $-2$ with the rational $\frac{-2}{1}$ or with the $3$-adic $\cdots22221$.
This makes me wonder: can we necessarily identify the integer 3 with the natural number 3? Might there be a natural (no pun intended) extension of the natural numbers which does not admit analogues of the integers? Can this notion be firmed up enough to prove the non-existence of such an integer-less extension / generalisation of the natural numbers, under the usual axioms of mathematics?
I know we can consider the natural numbers as equivalent to the finite ordinals, or the finite cardinals (which can be identified with a subset of the ordinals, iff we assume the Axiom of Choice), but the ordinals and cardinals are both proper classes, so there's a sense in which they don't really count. (And there are negative surreals, anyway, so it's only cardinals-without-choice that are suspect.)
1 answer
No, at least in some sense. Any structure that plays the "role" of the integers will be either the integers or a homomorphic image thereof (i.e. $\mathbb{Z}/n\mathbb{Z}$ for some $n$).
This is because the integers are the free group containing the naturals. What this means is that there is a natural correspondence between monoid homomorphisms $\mathbb{N} → G$ and group homomorphisms $\mathbb{Z} → G$. (Much as there is a natural correspondence between set homomorphisms $X → G$ and group homomorphisms $F(X) → G$ where $F(X)$ is the free group on $X$; this is a prototypical example of a free/forgetful adjunction.)
More concretely: any group containing the naturals will contain the integers. There's no way to have incompatible extensions of the naturals to truly different structures playing the role of integers: there is only one structure of integers given the structure on the natural numbers. In fact, this uniqueness property is even stronger than the corresponding property with the rational numbers, where every injection of the naturals into a field has to contain the rational numbers: every morphism, whether injective or not, has to contain the integers!2
1We often think of free algebraic structures as being those "generated" by objects (e.g. of the free group on 3 elements as being the group generated by the letters $a$, $b$, $c$, and their inverses, with no other relations between them), but another way to think about a free structure is that it has a universal property of mapping uniquely to a class of other structures. For example, the free group on 3 elements can be defined as the unique (up to isomorphism) group $F_{\mathbf{3}}$ such that, for any mapping from $\mathbf{3} = \{0,1,2\}$ to a group $G$, there is a unique homomorphism from $F_{\mathbf{3}}$ to $G$ that extends the mapping from $\mathbf{3}$ to $G$.)
2 That this is not true of the rational numbers can be seen from the example of finite fields: there is a homomorphism from the natural numbers into $\mathbb{Z}/p\mathbb{Z}$ the finite field of characteristic $p$, but clearly not all the rational numbers are present in $\mathbb{Z}/p\mathbb{Z}$---only those whose denominator is not divisible by $p$. But there is no monomorphism/injective homomorphism from $\mathbb{N}$ to $\mathbb{Z}/p\mathbb{Z}$, and hence the rationals are the unique smallest extension of $\mathbb{N}$ to a field.
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