What are natural numbers? What are the natural numbers?
Is it a valid question at all?
My understanding is that a set satisfying Peano axioms is called "the natural numbers" and from that one builds integers, rational numbers, real numbers, etc. But without any uniqueness theorem how can we call it "the" natural numbers? If any set satisfying the axioms is called natural numbers then maybe there is no "natural numbers" as a definite object as we know it in real world, but just a class of objects satisfying certain axioms, like Noetherian rings?
 A: In a "realist" perspective the natural numbers are the counting numbers, something that we can perceive through our rational faculties.  In that view the natural numbers have an existence independent of any axiomatization of their properties.
A "formalist" perspective cannot promise much, if anything, about a definite system of natural numbers.
Indeed the Peano axiomatization specifically does not characterize the natural numbers, as it is an essentially incomplete first-order theory (if consistent, Godel-Rosser).
Nor can it be proven that two set-based models which satisfy the Peano axioms are isomorphic, since a first-order theory with a model of one infinite cardinality has a model of any infinite cardinality (Lowenheim-Skolem).
A: Actually the Natural numbers are 0,1,2,3, ... up to $\infty$.
These numbers are also whole numbers, not fractions or decimals, and can be used for counting or ordering.
A: We often say that Godel's incompleteness theorem finds a theorem that is "true but unprovable in Peano arithmetic."  That's because there are models of Peano where it isn't true, but our intuition says it is true.  So, in that sense, the "natural numbers" are an idea beyond the Peano axioms.
For example, lets say you have an integer polynomial $p(x_1,...,x_n)$, and you want to know whether $p=0$ has solutions with $x_i \in \mathbb{Z}$.  So, suppose we could show this question was undecidable in Peano arithmetic.  Our intuition is that in that case, we could obviously never find a specific numeric solution, so we'd need a "non-standard" model for this equation to have a solution.  In other words, we'd have reason to say that the undecidability of this equation in Peano axioms indicates that the equation has no solution in the intuitive "natural numbers."
The resolution of Hilbert's Tenth Problem shows that there are udecidable problems of this sort.
A: I saw this issue discussed in Goldstern and Judah's The Incompleteness Phenomenon, but I'm sure there are other sources and I would welcome competing views.  They provide a description of all the countable models of the Peano axioms and prove that there is an uncountable number of countable models.
First order logic is very nice because you have the completeness and compactness theorems.  Unfortunately, the naturals are not categorical.  I suspect I am not alone in having learned a lot of number theory (arithmetic, prime numbers, unique factorization ...) before seeing any axioms for the naturals and maybe before learning the word "axiom".  It feels like the naturals should be unique.  To many people there should be "true arithmetic", the set of all true sentences in $\mathbb{N}$.  I haven't seen (but haven't looked hard) a claim that you can just ignore the issue like the analysts and push it off to the set theorists because $\mathbb{N}$ is absolute in ZFC.  The Handbook of Analysis and Its Foundations says that $\mathbb{R}$ is categorical only because of the second order logic least upper bound axiom, but people ignore the second order situation and get on with life.
A: Ravichandran's answer is right: the natural numbers are the numbers 0, 1, 2, 3, ... .  We can directly understand these numbers, based on our inductive definition of how to count in English or another natural language, even before we create an axiom system for them. 
The standard list of axioms that we use to characterize the natural numbers was stated by Peano. Several people have brought up first-order logic in answers here, but that isn't what you want to use to establish categoricity (I will come back to that). Dedekind was the first to prove that we can axiomatize the natural numbers in a way that all models of our axioms are isomorphic, using second-order logic with second-order semantics. 

Theorem (after Dedekind). Say that we have two structures $(A, 0_A, S_A)$ and $(B, 0_B,S_B)$ each of which is a model of Peano's axioms for the natural numbers with successor, and each of which has the property that every nonempy subset of the domain has a least element. Then there is a bijection $f\colon A \to B$ such that $f(0_A) = 0_B$ and for all $a \in A$ $f(S_A(a)) = S_B(f(a))$. 

The theorem is proved using the axiom of induction repeatedly, but the idea is completely transparent. Suppose that Sisyphus is given a new task. He will count out all the natural numbers in order in Greek, while a fellow tortured soul will count out all the natural numbers in order in English, at the same speed as Sisyphus. Just based on Ravichandran's assessment of what the natural numbers are, the map that sends each number spoken by Sisyphus to the number spoken by his companion at the same time is obviously an isomorphism between the "Greek natural numbers" and the "English natural numbers."
This argument cannot be captured in first-order logic, not even in first-order ZFC. But that isn't the fault of the natural numbers: first order logic can't give a categorical set of axioms for any infinite structure. Dedekind's proof can be cast in ZFC in the sense that it shows that the models of Peano's axioms in a certain model of ZFC are all isomorphic to each other in that model. Moreover, ZFC is sound in the sense that the things it proves about the natural numbers are correct. But as long as we use first-order semantics for ZFC, it can't fully capture the natural numbers. 
The key point of second order semantics is that "every nonempty subset" means every nonempty subset.  If a structure $(A, 0_A, S_A)$ satisfies a certain finite set of axioms that are all true in $\mathbb{N}$, then $\mathbb{N}$ can be identified with an initial segment of $A$. In this case, if $A - \mathbb{N}$ is nonempty it will not have a least element, and so $A$ will not satisfy the second-order Peano axioms. Again, this argument cannot be completely captured in first-order logic because $\mathbb{N}$ cannot be fully captured. 
Even if we think of Peano's axioms as only specifying an isomorphism class of structures, the situation is not like Noetherian rings, where there are many non-isomorphic examples. $\mathbb{N}$ is more like the finite field on two elements. The question of which model of the second-order Peano axioms is really the "natural numbers" is like the question of which isomorphic copy of $F_2$ is "really" $F_2$. It's a fine question for philosophers, but as mathematicians we have a perfectly good idea what $F_2$ is, up to isomorphism, and a good idea what $\mathbb{N}$ is, up to isomorphism. We also have axiom systems that let us prove things about these structures. $F_2$ is easier to apprehend because it's finite, but this only makes $\mathbb{N}$ more interesting. 
A: My understanding is that, in the constructive sense, the natural numbers are the smallest inductive set. By the Axiom of Infinity, there exists an inductive set, and the intersection of inductive sets is again an inductive set. 
To see this, recall that a set $A$ is inductive if $\emptyset\in A$, and for all $a$, if $a\in A$, then $a^+\in A$, where $a^+=a\cup\{a\}$ is the successor of $a$. 
Now let $T$ be a nonempty family of inductive sets. So $\cap T$ is also inductive, for clearly $\emptyset\in\cap T$, and if $a\in\cap T$, then $a\in A$ for all $A\in T$ since each $A$ is inductive, so $a^+\in A$ for all $A\in T$, so $a^+\in\cap T$. 
You can then define a natural number to be a set which belongs to every inductive set. To see that a set $\omega$ of such sets actually exists, consider $A$ to be an inductive set. Then let $$T=\{K\in\mathscr{P}(A)\ |\ K\textrm{ is inductive}\}$$ So $A\in T$, thus $T\neq\emptyset$. Let $\omega=\bigcap_{K\in T}K$. Indeed, $\omega$ consists exactly of the natural numbers under this definition. For let $n\in\omega$, and let $B$ be an inductive set. Then $A\cap B\in T$, so $n\in A\cap B$, and thus $n\in B$. Conversely, let $n$ be a natural number. Then $n\in K$ for all $K\in T$, so $n\in\omega$. So the usual idea of $\omega$ as the set of natural numbers makes sense.
So the natural numbers are precisely the intersection of all inductive sets, that is, elements which are in every inductive set. Also, this set is unique by Extensionality.
A: The use of the definite article before the term natural numbers is a convention adopted by mathematicians adhering to a platonist or realist philosophy of mathematics.  A uniqueness implied by such a convention is therefore not a mathematical theorem but rather a statement of adherence to a specific philosophical creed.  Not all mathematicians adhere to this, however; see e.g., this answer.
A: Each natural number is a set. A possibility would be:


*

*0 is the set that contains the empty set $\{\emptyset\}$.

*1 is the set that contains the empty set and 0 $\{\emptyset,\{\emptyset\}\}$

*2 is the set that contains the empty set and 1 $\{\emptyset,\{\emptyset,\{\emptyset\}\}\}$

*3 is the set that contains the empty set and 2 $\{\emptyset,\{\emptyset,\{\emptyset,\{\emptyset\}\}\}$


and so on. But keep in mind that the representation is not very important, because all correct representations (like the above) you can think of, are isomorphic.
