# 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?

-

7
All sets satisfying the Peano axioms are isomorphic. When we refer to the natural numbers, we are referring to this isomorphism class. – Alex Becker Apr 9 '11 at 1:15

1
How could one construct such an isomorphism? 1 should map to 1 and if x maps to y then x+1 should map to to y+1, but should I use induction to define it globally? – ashpool Apr 9 '11 at 1:20

10
The (first-order) version of the Peano axioms is not categorical, there are "non-standard" models. The question that ashpool asks is a deep and contentious one in the philosophy of mathematics. – André Nicolas Apr 9 '11 at 1:23
@ashpool: yes, you use induction on each side of the isomorphism. – Carl Mummert Apr 9 '11 at 2:33
@ashpool: Please include all the information on the body; your main question appeared only on the subject. – Arturo Magidin Apr 9 '11 at 4:24
show 3 more comments

## 6 Answers

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.

-

+1 I'm glad you posted this.  My answer was prompted by your early comment re: isomorphism!  I agree we can prove something in first-order ZFC about isomorphism, e.g. that a smallest infinite ordinal exists, but that this doesn't "fully capture the natural numbers."  It also seems that second-order semantics ought to be "more expressive" than first-order semantics, but I have doubts about the possible implication that $\mathbb{N}$ can be "fully captured" through some combination of second-order semantics and set theory. – hardmath Apr 13 '11 at 16:20
The main benefit of second-order semantics is that the Peano axioms are categorical in them, unlike in first-order semantics. Can you make your doubts more concrete, given that theorem? – Carl Mummert Apr 14 '11 at 0:14

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.

-

4
Nice detailed description of one set-theoretic definition. But in a sense, this pushes back the OP's question "What are the natural numbers, really?" to "What is the universe of sets, really?". And once we introduce ZFC, we are faced again with non-categoricity. – André Nicolas Apr 9 '11 at 1:39
Oof, perhaps I didn't fully understand what the OP was getting at then. I'm curious to see how this all fleshes out then. – yunone Apr 9 '11 at 1:43

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).

-

1
Peano's original axiomatization was a finite list of second-order axioms; the restatement as an infinite set of first-order axioms is a later development. In second-order logic, it can be proved that any two models are isomorphic, because the second-order axiomatization is more restrictive about which structures are models than the first-order axiomatization is. I've tried to address this in a separate answer. – Carl Mummert Apr 9 '11 at 12:09

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.

-

5
Best answer so far – Holowitz Apr 9 '11 at 10:06

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.

-

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.

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 61, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9427846670150757, "perplexity_flag": "head"}