# Peano's Axioms: Mathematical Philosophy

In Peano Axioms, why is it necessary to define number and successor. Does not using them imply that we know what they mean? Or could they have just as easily been any two arbitrary terms which are not associated with “numbers” say widget and descendent respectively? Where by having the following:

• 0 is a widget
• The descendent of every widget is a widget

Given the axioms, are we to assume that we are at the starting point in which we don’t know anything about numbers (analogous to the discovery of an element) or, do we already know how numbers work are the axioms simply describe their behavior?

Assuming we don’t know what is meant by successor (because it has to be defined), why isn’t it necessary to include in the axioms that every number has one successor

• The axioms are a starting point. The Peano Axioms are one way to "define" numbers, if we want to look at the foundations of mathematics. – Akiva Weinberger Mar 23 '15 at 19:16
• Using your widgets and descendants: That system is isomorphic (basically, "the same thing") with the usual Peano Axioms. I should add, by the way, that the axioms allow us to write really formal proofs. – Akiva Weinberger Mar 23 '15 at 19:17
• From what I understand, in PA you don't "define" numbers, you just assume they "are" and PA specifies what properties they have. – Wojowu Mar 23 '15 at 19:18
• In the wikipedia edition of the axioms, axiom number 6 does say that every number has a successor. – Arthur Mar 23 '15 at 19:19
• Re the specific question in your final paragraph, the Peano axioms do in fact include the axiom that every number has exactly one successor. – WillO Mar 24 '15 at 3:20

I suggest you to read Peano's paper (1889). You can see the English translation in :

Among the signs [i.e. notions] of arithmetic, those that can be expressed by other signs of arithmetic together with the signs of logic represent the ideas that we can define.

Thus, I have defined all signs except the four that are contained in the explanations of §1 [The primitive, i.e. undefined, arithmetic notions are : "number", "one", "successor", and "is equal to"].

If, as I think, these cannot be reduced any further, it is not possible to define the ideas expressed by them through ideas assumed to be known previously.

Propositions that are deduced from others by the operations of logic are theorems; propositions that are not thus deduced I have called axioms. There are nine [four regarding "equality" and five "arithmetical"] of these axioms (§1), and they express the fundamental properties of the signs that lack definition [emphasis mine].

Thus, omitting "is equal to", that today we prefer to class among the logical notions, the primitive notions of "number", "one", "successor" are undefined.

The arithmetical axioms are :

$1 \in \mathbb N$ : $1$ is a number

$a \in \mathbb N \to a+1 \in \mathbb N$ : the successor of any number is a number

$a,b \in \mathbb N \to (a=b \leftrightarrow a+1 = b+1)$ : two numbers are equal iff their successors are equal

$a \in \mathbb N \to \lnot (a + 1 = 1)$ : $1$ is not the successor of any number

and the induction axiom.

One can make either choice: interpreting the Peano axioms as (1) assertions about actual numbers of which we have some prior knowledge, or (2) a list of axioms for a logical theory, in which it's as good to call an element a widget as a number. In case (2) we have to explain why in the world we picked these particular axioms, and the explanation is that the axioms abstract certain properties of real-world objects we repeatedly observe. On the other hand many people have been suspicious about the kind of assertion in (1) that we can have direct, pre-logical knowledge of the natural numbers, in which case a more formal approach such as (2) is called for.

• If you start by denying that we have any pre-logical knowledge of the natural numbers, you're going to have a difficult time making sense out of axioms like every number has exactly one successor''. – WillO Mar 24 '15 at 3:22
• @WillO to defend a position to which I do not adhere: Perhaps one could have prior knowledge of some particular numbers without claiming any knowledge about an arbitrary number, or about the collection of all numbers. – Kevin Arlin Mar 24 '15 at 8:03
• @KevinCarlson: To attack your defence of a position to which you do not adhere: We could have only prior knowledge of finitely many 'numbers', whatever they may be, and have to resort to philosophical induction or mathematical induction to infer the axiom of induction. That's circular. =) – user21820 Apr 22 '15 at 13:59
• @user21820 Of course one proposes the Peano axioms via inductive reasoning. There's nothing circular about using real-world understanding of induction to write down a mathematical axiom modeling it. – Kevin Arlin Apr 22 '15 at 17:46
• @KevinCarlson: Yep. My comment was merely to emphasize the fact that we cannot arrive at the induction axiom without any pre-logical knowledge of induction itself and the 'fact' that natural numbers are closed under the successor operation. (For induction we don't need distinct successors.) – user21820 Apr 23 '15 at 2:26

When it comes down to it, you can't have a mathematical foundation without some sort of acceptance of the natural numbers as self evident knowledge. Almost all mathematics is carried out with a foundation in first order logic, most foundations being first order theories of some sort. However, if you accept first order logic, you are implicitly accepting the natural numbers. The structure of the natural numbers is embedded in first order logic in many ways, I give an example here.

Take for example the statement ∃x,y∀z( y∈x & (z∈x->y=z) ). This means that there there exists an element x such that there is exactly one element y such that y∈x. Note that this statement does not rely on any first order theory, this is simply a statement in general first order logic. Then there is a statement in first order logic (again, the language in general, not an axiomatized theory in it), that captures the notion of '1'. Likewise, ∃x,y,z∀w( y∈x & z∈x & y≠z & (w∈x->w=y V w=z)) captures the notion of '2' (there are exactly 2 elements in x). This can be done for all natural numbers (0 can be represented by ∃x∀y(y∉x)).

Addition can be seen in general first order logic as (taking 1+2 as an example):

∃x,y,z[ ∃a∀b(a∈y&(b∈y->a=b)) & ∃a,b∀c(a∈z&b∈z&a≠b&(c∈z->c=a V c=b)) & (a)~(a∈y&a∈z) & (a)(a∈x <-> a∈y V a∈z ) ]

This says there is exactly one element in y (in the sense that there is exactly one a such that a∈y), and there are exactly 2 elements in z. Furthermore, y and z are disjoint, and x is their union. Then x contains 1+2 elements. First order logic permits a proof that x consequently has 3 elements in the same sense as described earlier.

Multiplication is more complicated, so I'll just outline it. To represent k = m*n, assert x has m elements, and assert each of these has n elements, and that the elements of x are disjoint. Then assert that y has precisely those elements that are contained in elements of x. Then y has k=m*n elements. Again, first order logic allows a proof that y has k elements in the direct sense given earlier.

In this manner, first order logic exhibits all the complexity of the integers under the operations '<', '+', and 'x' ('<' can be encoded similarly). In fact, a lot more complexity than this is contained in first order logic, without any axiomatized thoery. The analogs of the Peano axioms associated with the above discussion cannot be proven from general first order logic, as quantification over formulas is not allowed, but very precise and formal outlines can be given that immediately produce proofs of every expressible instance of the peano axioms, and so the axioms are, to the extent they are expressible, available.

The point of all this is that if you accept first order logic at all, you have already accepted a system which has a structure more complex than the structure of the natural numbers under the basic operations.

So, although you can say that at the start of your mathematical development you know nothing of the natural numbers, such a perspective contradicts the acceptance of the very language in which the Peano Axioms will be written. So, the reality is, in one way or another, we accept the natural numbers as self evident knowledge. Anyone who says otherwise and who supposes any first order theory embodies self evident truth is just accepting the integers as self evident without realizing it.

• Wait a minute. First-order logic need not have $\in$. You still can get the properties of naturals, but not induction. – user21820 Apr 22 '15 at 13:57
• I do not mean to imply any special meaning of $\in$ other than that it's a binary relation. You need a binary relation for PA, $\in$ is just my choice of notation for an arbitrary binary relation. The axioms of first order logic are not as fundamental as the rules of induction, in that you can prove any consequence of any axiom set in the base language, in the form of $A\rightarrow B$, where $A$ is a finite subset of the Axioms and $B$ is the consequence. So once you define first order logic with a binary relation, you have the machinery of every theory, whether you want it or not. – jack Apr 22 '15 at 15:57
• But PA excluding induction can be encoded without any binary predicate, since just a unary function will do, and the encoding will simply be a statement of PA minus induction, so what is your point? And induction cannot be stated as an axiom in first-order logic. – user21820 Apr 22 '15 at 16:21
• The stuff I said could be done with a unary function, just define $x\in y$ to be $f(x)=y$. So anything with one binary relation or one unary function includes the machinery I described. As for induction, as I noted it cannot be encoded in this way, although any particular instance of an inductive formula can be proven. This is enough to demonstrate the integers are encoded in first order logic in the way I said. In particular, it is clear that full induction is true. So the integers are there, and accepting first order logic then acting hesitant about the integers is not a consistent position. – jack Apr 22 '15 at 16:59

Many properties of the natural numbers were known for centuries before Peano. Dedekind and Peano, in the late 19th century set out to identify their essential properties from which, it was hoped, all others could be derived. So successful were they in this regard, that, for all practical purposes, Peano's Axioms have come to define the natural numbers. From these axioms, along with the axioms of set theory, we can construct the addition, multiplication and exponentiation functions. Likewise, we can construct the integers, rational, real and complex numbers along with the arithmetic operations associated with these number systems.

The Peano Axioms are all really quite intuitive. Here I present a modern, set-theoretic version.

We start with a set $\mathbb{N}$.

1. $0$ is a number: $$0\in\mathbb{N}$$

2. For every number there exists a unique successor, i.e. we have function $S$ such that: $$\forall x\in \mathbb{N}: S(x)\in \mathbb{N}$$

3. If two numbers $x$ and $y$ have the same successor, then $x=y$: $$\forall x,y \in\mathbb{N}: [S(x)=S(y)\implies x=y]$$

4. No numbers have a successor of $0$: $$\forall x\in\mathbb{N}:S(x)\ne 0$$

5. Let $P$ be any subset of $\mathbb{N}$. Suppose $0\in P$. Suppose further that if $k\in P$ then $S(k)\in P$. Then every number is in set $P$:

$$\forall P\subset \mathbb{N}:[[0\in P \land \forall k\in P: S(k)\in P]\implies \forall x\in \mathbb{N}: x\in P]$$

• The Peano axioms can be quite non-intuitive depending on where you stand: as you said, what the axioms describe are (maybe) the essential characteristics of the natural numbers, meaning they're all we would need to learn (almost?) everything regarding to these numbers. However, if we start from the absolute ground-up, assuming no prior knowledge about the natural numbers, these axioms may seem really insuficient to construct them. – YoTengoUnLCD Mar 24 '15 at 4:21
• @YoTengoUnLCD If you want to construct the natural numbers from another set, you can start with any Dedekind-infinite set $X$. Being Dedekind-infinite, we would have a function $f:X\to X$ such that $f$ is injective, but not surjective. With $f$ not surjective, we would then have $x_0\in X$ such that $\forall x\in X:f(x)\ne x_0$. From $X$, we can extract a subset $N$ such that $(N,f,x_0)$ would satisfy the Peano axioms, with $f$ being the successor function and $x_0$ the first element in $N$. $N$ would just be the smallest subset of $X$ such that $x_0\in N$ and $\forall x\in N:f(x)\in N$. – Dan Christensen Mar 24 '15 at 4:40
• For a formal development using my own ZFC-like axioms of set theory, see dcproof.com/ProofByInduction.html – Dan Christensen Mar 24 '15 at 4:48
• @YoTengoUnLCD Also (maybe this was what you were getting at), the structure determined by these axioms is unique, i.e. if both $(N,S,0)$ and $(N',S',0')$ satisfy the Peano axioms, then both structures will be essentially identical. We could map $0\leftrightarrow 0'$, $S(0)\leftrightarrow S'(0')$, $S(S(0))\leftrightarrow S'(S'(0'))$, and so on. They would be of the same structure and the same size -- only the names will differ. – Dan Christensen Mar 24 '15 at 19:24
• @YoTengoUnLCD You may find helpful my posting "What is a number again?" at my math blog dcproof.wordpress.com – Dan Christensen Mar 25 '15 at 4:36

Erase from your mind all comprehension of numbers and start (almost) from the philosophical starting gate. Here we give the 'widget' answer, creating the natural numbers.

So grappling with the concept of infinity, we come up with an abstract machine as depicted here:

The machine takes in a ruler and extends it by attaching the input to an extension with a rivet (both supplied by the machine).

If we consider the collection of all rulers, the machine will map each one to another one, but the $$\mathsf |$$, the initial 'seed', will not be seen as an output.

Ok, this thought experiment gives us confidence that we can 'get abstract', using the set theory framework.

Definition: A Dedekind Push Entlang Maschine is a tuple $$(f, X, n_0)$$ with
$$f: X \to X$$ an injective function such that $$n_0 \in X$$ but $$n_0 \notin f(X)$$.

Axiom: There exist a Dedekind Push Entlang Maschine.

Here comes the amazing aspect of this - buried inside a Dedekind Push Entlang Maschine is a 'copy' of the Create Rulers Machine.

Proposition 1: Let $$(f, X, n_0)$$ be any Dedekind Push Entlang Maschine. Then there exist a subset $$N$$ of $$X$$ with the properties that

$$\tag a \text{The restriction } (f, N, n_0) \text{ is another Dedekind Push Entlang Maschine}$$

$$\tag b \text{If } M \subset N \text{ and } (f, M, n_0) \text{ is a Dedekind Push Entlang Maschine, then } M = N$$

Proof
Just let $$N$$ be the intersection of all subsets $$L$$ contained in $$X$$ and defining a restricted machine, $$(f, L, n_0)$$. $$\blacksquare$$

For this minimal set $$N$$ we will use the symbol $$\sigma$$ for the restricted function $$f$$.

The function $$\sigma: N \to N$$ satisfies

$$\tag 1 \sigma \text{ is an injection}$$
$$\tag 2 n_0 \in N \text{ but } n_0 \notin \sigma(N)$$
$$\tag 3 \text{If } K \subset N \text{ and } n_0 \in K \text{ and } \sigma(K) \subset K \text{ then } K = N$$

Observe that $$\text{(3)}$$ is just stating that $$N$$ is the minimal set that can be extracted from $$X$$ to create a $$(\sigma, N, n_0)$$ Dedekind Push Entlang Maschine. But this is also the familiar induction principle.

If we continue studying this object, we will see that we can define addition and multiplication (see Appendix I of Serge Lang's Undergraduate Algebra) on $$N$$ and that $$\text{(1)}$$, $$\text{(2)}$$ and $$\text{(3)}$$ make any two such abstract creations isomorphic.

So we have it - the natural numbers $$\mathbb N$$!

The bonus of this abstract approach is we don't have to describe adding together two ruler pieces (uh, if both pieces greater than $$\mathsf |$$, drill-out that rivet and ...).