Is Peano's successor function the same as $+1$?

Question

I heard various times by watching YouTube videos (I don't remember which, exactly) that we can show after the definition of the addition in Peano's arithmetic that the successor operation is in definitive the $+1$ function. But I have some doubts about the exactitude of it. Here are my points (I speak about Peano's arithmetic only):

-1) the successor function is defined before the addition, and it is independent from it. The addition definition relies on the successor function. So, defining the successors based on addition looks to me like something circular.

-2) The successor function can work only with positive natural numbers, and the $+1$ can work with all categories of numbers like negatives and imaginary. For example, if $+1$ is the successor function, so, the successor of $-1$ would be $0$ because $-1+1=0$, but there is an axiom that states that $0$ is not the successor of anything.

So, I'm interested in the average point of view about that. Is the successor function the same as $+1$? If yes, what is the proof?

Thank you.

-Edit: -1)My first point was answered clearly by someone who delete he's answer. It was that the +1 is the generalisation of the successor function so they are a bit different but +1 include the successor function inside of it. -2)The second point is answered in the answer of fleablood. It is that the $0$ is not the successor of any positive natural numbers but can be the successor of a negative number like $-1$

I'm satifyed with theses answers and I don't have other points till now so I will mark this question as answered.

Thank you for all your contributions.

The successor function precedes all other operations on numbers! the map $n \mapsto n+1$ for natural numbers $n$ is, by definition, the successor function. The Peano Axioms ensure that this map works how we expect $+1$ to work. Good question! — JMM, Oct 16 '24 at 19:29
@JMM If $n \mapsto n+1$ with an addition in it define the successor. And the successor preced all other opperations like addition. If you are agree with it: You are saying me a circular thing. — lazare, Oct 16 '24 at 19:35
No no, I mean you start with the successor function. That is immutable and foundational. The concept of $+1$ is then defined in terms of the successor function. — JMM, Oct 16 '24 at 19:37
Why did someone downvote this question? It is totally valid. Stop discouraging new users! — JMM, Oct 16 '24 at 19:40
Closely related: https://math.stackexchange.com/questions/4795154/why-define-addition-with-successor — MJD, Oct 16 '24 at 21:01
The required addition function can be constructed (i.e. proven to exist) from Peano's Axioms as defined here: http://mathworld.wolfram.com/PeanosAxioms.html — Dan Christensen, Oct 17 '24 at 15:23

score 4 · Accepted Answer · edited Oct 22 '24 at 02:11

Defining $s(n)$ as $n+1$ is certainly circular! But if axiomatically declare "Every $n$ has a successor $s(n)$" without a definition but purely axiomatically that is perfectly fine. Then later we define $n+1$ as $s(n)$ (not the other way around) that is just fine.
The axiom actually states that $0$ is not the successor of any natural number. $0$ can, and will be, the successor of things that are not natural numbers. But before we can declare $s(-1)=0$ we must define the negative numbers. We do that after establishing everything about the natural numbers first.

SPECULATIVE OPINION FOLLOWS. for natural numbers, we start with an axiomatic successor function. We use that to define $+1$. We then somehow come up with a way to define $m+n; m,n\in \mathbb N$.

Then we extend the natural numbers to fit into a bigger set ($\mathbb Z$ or $\mathbb Q$ or $\mathbb R$) and extend the definition of addition.

Now at this point negative numbers do not have any meaningful definition of successor and maybe we don't need one.

But if we want to define $s(z): z\in \mathbb R$ we may do it this way

$s(z) = \begin{cases}\text{if }z\in \mathbb N, s(z)\text{ is the axiomatic successor function previously axiomized}\\\text{if }z \not \in\mathbb N, s(z)=z+1\end{cases}$

The is not circular because the order of definitions is in this order: successor functions of the naturals, addition of the naturals, the bigger set than the naturals, addition on the bigger set, successor function on the bigger set.

That's purely linear and no circles are involved.

If I recall correctly, in Peano's original work he wrote "$+1$" rather than something like a more modern $s(n)$, which as you basically pointed out can lead to confusion when trying to rigorously define addition. But I am definitely in favor of what you've laid out here. — Mark S., Oct 16 '24 at 20:38
@MarkS. Oh, probably. I suppose the shortest answer is "Ultimately, yes, the successor is $+1$. But we come to that by derivation; not circular definitions" — fleablood, Oct 17 '24 at 02:03
The speculative opinion is spot on and highlights an important point. We have many concepts in math that can be defined in different equivalent ways. When you develop the structure it is important to make sure you define one and prove the rest in a noncircular manner. After we do that we know all these equivalent formulations and use whichever is convenient. — Ross Millikan, Oct 22 '24 at 02:22

score 0 · Answer 2 · answered Oct 16 '24 at 22:46

The purpose of the Peano axioms is to provide a minimal amount of structure to define the natural numbers. The successor function is not, on its own, defined in terms of addition, it's just a unary operator on the set. It's only in combination with the axioms that describe how it interacts with addition that it becomes the same as adding one.

score 0 · Answer 3 · answered Oct 20 '24 at 20:28

You are right that if we define $s$ in terms of $+$, and $+$ in terms of $s$, we are in trouble.

But that is not what the typical Peano Arithmetic Axioms do.

Typically the first two axioms of Peano Arithmetic are:

$\forall x \ s(x) \neq 0$

$\forall x \forall y (s(x) = s(y) \to x = y)$

These two axioms 'define' the $s$ function in so far as that any domain that satisfies these two axioms will have to include a infinite number of objects that stand in a successive $s$ chain without any cycling back ... so the same signature as the natural number line.

Note that $+$ does not occur in these axioms!

The next two axioms 'define' addition:

$\forall x \ x+0=x$

$\forall x \forall y \ x+s(y) = s(x+y)$

So yes, here we see a reference back to $s$. Also, I use 'define', since without some kind of inductive axiom, there are domains that satisfy these two axioms but where the interpretation of the $+$ is not analogous to how the $+$ works in normal natural number arithmetic.

Anyway, with these axioms, we can now prove that $\forall x \ x + s(0) = s(x)$, i.e. that '$x+1=s(x)$':

As such, we can say that the $s$ and '$+1$' functions are Peano Arithmetically equivalent.

Is Peano's successor function the same as $+1$?

3 Answers3