The Peano Axioms never explicitly state that the successor function is $n$ $+ 1$. Is this just taken by convention? It's as if we should already know what the natural numbers are and know that the successor of $0$ is $1$, and of $1$ is $2$.

It seems the Peano Axioms are describing an already known existing set, as opposing to constructing it.

Even with the principle of induction, it is never clear that the successor of a natural number $n$ should be this same number $n$ plus $1$

I would appreciate any clarification.

  • $\begingroup$ I'm pretty sure it would all work just as well if the successor function was $n+2$. Can you think of anything that would go wrong if it was, or any way you could tell the difference? $\endgroup$ Dec 27, 2022 at 0:30
  • $\begingroup$ no matter what your sucessor function is, if it satifiees the Peano axioms then $\{n_0, S(n_0), S(S(n_0)), S(S(S(n_0))), ....\}$ is your set of Natural Numbers. $\endgroup$
    – L F
    Dec 27, 2022 at 1:28
  • $\begingroup$ The short answer is that $S(n)=n+1$ is one of the consequences of the axioms (in particular, the axiom $x+S(y)=S(x+y)$ and the definition $1:=S(0)$). It is somewhat odd, and worth internalizing, that $1$ is defined out of other stuff and $S$ is taken as primitive. $\endgroup$ Dec 27, 2022 at 15:46

6 Answers 6


1 is defined as $S(0)$, where $S$ denotes the successor function. The relevant convention does not concern what is known about numbers, but only what we call them. Of course, our preconceptions about how natural numbers work greatly informs how the Peano Axioms were chosen. They didn't drop from the sky, after all. But now, given that we have articulated the axiomatic basis, we can adapt our conception of the natural numbers such that it reflects the axioms rather than our intuitions.

Indeed, that $S(n) = n + 1$ for all $n$ is not an axiom but something we must prove, like so: \begin{align} n + 1 &= n + S(0) = S(n+0) = S(n) \end{align} The proof only makes use of the axioms and the definition of addition.

  • $\begingroup$ Nice answer! But you should explicitly mention that $S$ denotes the successor function (although this may seem to be obvious). Also use mathjax. See math.stackexchange.com/help/notation. $\endgroup$ Dec 27, 2022 at 1:05
  • $\begingroup$ Thanks for the feedback! $\endgroup$ Dec 27, 2022 at 1:15
  • $\begingroup$ The addition function can actually be proven to exist. See my answer. $\endgroup$ Dec 28, 2022 at 22:35

The Peano Axioms are just that: a bunch of statements expressed in the language of first-order logic. And notice that they don't use the $1$ symbol, so you're right: how can we tell that $s(0)$ is supposed to be $1$?

But please note that in a way it is even worse than you think: why should the $0$ logic symbol even be the same as the mathematical $0$?! Again, as far as purely formal logic is concerned, it's all just a bunch of symbols. So for all we know, $0$ is referring to bananas, and $s$ is the function of putting something in a smoothie-maker, which would make $s(0)$ to be a banana-smoothie.

But of course, in creating the Peano Axioms, we didn't aim to be talking about bananas and smoothies. We intended the axioms to be about the natural numbers. Specifically, we use the $0$ symbol to represent the mathematical number $0$, and we used $s$ for the 'successor' function, and since $1$ is the successor of $0$, we can treat $s(0)$ as the number $1$.

So notice that we don't prove that $s(0)$ is $1$, but rather stipulate (define, if you want) that $1$ is the number that comes after $0$. That is, it is our intended interpretation that tells us that $s(0)$ is the same as $1$.

Interestingly, your question as to how we know that the successor of $n$ is $n+1$ actually is not a mere definition. Indeed, your question amounts to proving that $s(n) = n + s(0)$ ... and that is something we can do using the Peano Axioms that relate to addition:

$n + s(0) = s(n+0) = s(n)$

Hey, a first theorem we can derive from the Peano Axioms!

P.s. Using the Peano Axioms you can prove that $s(s(0)) \neq s(0)$. So I guess this means that the Peano Axioms can't be talking about banana-smoothies, since if you put a banana-smoothie into a smoothie-maker, you get back that very same banana-smoothie :P

  • $\begingroup$ The addition function can be proven to exist. See my answer. $\endgroup$ Dec 28, 2022 at 22:37
  • $\begingroup$ Sorry, I down-voted your answer by mistake and couldn't undo it. $\endgroup$ Dec 28, 2022 at 22:39
  • $\begingroup$ @DanChristensen No worrries! :) $\endgroup$
    – Bram28
    Dec 29, 2022 at 2:45

The axioms are sort of independent of the actual representation, whether you are adding a number like + 10 in decimal or + A in hex or + X in Roman numerals, it's all the same thing - just different definitions/representations of numbers that behave according to the axioms.

If you define $1$ as $S(0)$, the successor of natural number $0$, then $S(n) = n + 1$ follows from the definition of addition since $n + S(0) = S(n + 0) = S(n) = n + 1$. Then you can define $2$ as $S(1)$, the same as $S(S(0))$ (the successor of the successor of $0$), and so on.


Peano's Axioms as stated in the language of set theory define $(N,S,0)$ where $S$ is the successor function on $N$:

  1. $0\in N$
  2. $S: N \to N$
  3. $S$ is injective
  4. $\forall x: S(x)\neq 0$
  5. $\forall P\subset N:(0\in P \land \forall x\in P: S(x)\in P \implies P=N)$

Using them and the axioms of set theory, we can formally prove the existence of a binary function $Add$ such that:

  1. $\forall a,b \in N: Add(a,b) \in N$

  2. $\forall a\in N: Add(a,0)=a$

  3. $\forall a,b \in N: Add(a,S(b))=S(Add(a,b))$

(This can be used to "define" the addition function on $N$.)

Then we have $Add(x,1)= Add(x,S(0))=S(Add(x,0))=S(x)$ where $S(0)=1$.


For clarification/elucidation the OP should examine the $2^{nd}$ paragraph of the wikipedia posting on the subject:

The need to formalize arithmetic was not well appreciated until the work of Hermann Grassmann, who showed in the 1860s that many facts in arithmetic could be derived from more basic facts about the successor operation and induction.[1][2] In 1881, Charles Sanders Peirce provided an axiomatization of natural-number arithmetic.[3][4] In 1888, Richard Dedekind proposed another axiomatization of natural-number arithmetic, and in 1889, Peano published a simplified version of them as a collection of axioms in his book, The principles of arithmetic presented by a new method (Latin: Arithmetices principia, nova methodo exposita).


Wikipedia says the following:

Axioms 1, 6, 7, 8 define a unary representation of the intuitive notion of natural numbers: the number $1$ can be defined as $S(0)$, $2$ as $S(S(0))$, etc. However, considering the notion of natural numbers as being defined by these axioms, axioms 1, 6, 7, 8 do not imply that the successor function generates all the natural numbers different from $0$.

The intuitive notion that each natural number can be obtained by applying successor sufficiently often to zero requires an additional axiom, which is sometimes called the axiom of induction.

The axiom of induction is defined as

If $K$ is a set such that $0$ is in K, and for every natural number $n$, $n$ being in $K$ implies that $S(n)$ is in $K$, then $K$ contains every natural number.

I guess that the axiom of induction requires that $S(n) = n+1$. To see this, define $K$ as \begin{equation*} K = \{0, S(0), S(S(0)),\dots\} \subset \mathbb{N}. \end{equation*} The axiom of induction implies $K=\mathbb{N}$. So, if we assume $n<S(n)$, then it should be $S(n) = n+1$. I do not know whether the condition $n<S(n)$ is included in the definition of $S(n)$ or is derived from other axioms, and this is the reason I said 'guess'.


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