Variations in the successor function from Peano's axioms

Question

Concerning the successor function in Peano's axioms, what prevents me from defining it in the following way:

0 to 2, 2 to 1, 1 to 4, 4 to 3, 3 to 6, 6 to 5, 5 to 8, 8 to 7 ... and so on.

It seems this satisfies the axioms? Please help! Thank you :)

Answer 1

It works ...

We have to note that according to Peano's axioms the symbol $1$ is introduced as the name for the successor of $0$; thus, according to your approach, we have to use $1$ to denote the "real" number $2$ (the "new" successor of $0$).

Consider that Peano's axioms include the recursive axioms for sum and product.

For sum :

$\forall x(x+0=x)$

$\forall x \forall y (x+S(y)=S(x+y))$.

Thus, applying the second axiom with e.g. $x :=1$ and $y := 0$, we have that :

$1+S(0)=S(1)=4$,

because in the "reformed" sequence : $0,2,1,4,3,\ldots$, the successor of $1$ is $4$.

Consider now :

$2+S(0)=S(0)+S(0)$,

because $2$ is the "new" successor of $0$; thus :

$2+S(0)=S(0)+S(0)=S(S(0)+0)=S(S(0))=S(2)=1$,

because $1$ is the successor of $2$.

Thus, your proposed "reform" seems consistent, because the operation of "adding to $n$ the successor of $0$" still produces the successor of $n$.

Now, there is an easy way to avoid this apparent confusion : we can introduce new symbols :

$1^*$ for the successor of $0$, i.e. as a new "name" for $2$,

$2^*$ for the successor of the successor of $0$, i.e. for the successor of $2$, i.e. as a new "name" for $1$,

$3^*$ for the successor of the successor of the successor of $0$, i.e. for the successor of $1$, i.e. as a new "name" for $4$,

and so on.

In this way, we have relabeleld the "reformed" sequence : $0,2,1,4,3,\ldots$ with a sequence of new "names" : $0, 1^*, 2^*,3^*,\ldots$.

Using them in the above "computations", we get :

$2^*+S(0)=2^* + 1^*=3^*$

and it is correct, because $3^*$ is the new "name" for $4$.

In the same way :

$1^*+S(0)=1^*+1^*=2^*$

and again it is correct, because $2^*$ is the new "name" for $1$.

---

Conclusion : the proposed "reform" us useless.

According to Peano's axioms the "real" number $1$ has only one "relevant" property : to be the (unique) successor of $0$, i.e. the successor of the unique number without successor.

As long as we satisfy the two basic property of the successor function :

• $0$ has no successor

• no two different numbers can have the same successor

there are no "metaphysical" properties that can distinguish two numbers for each other if not their "relative position" with respect to $0$.

Answer 2

Only a single number is actually named in the Peano Axioms (either 0 or 1 depending on personal preference of the writer). There is nothing in them that prevents you from assigning goofy, non-standard names to the other numbers.

The second axiom, as usually presented, says that every number has a unique successor that is also a number, i.e. the successor relation is a function mapping $\mathbb{N}$ to itself.

$S: \mathbb{N}\to \mathbb{N}$

or equivalently

$\forall x\in \mathbb{N}:S(x)\in \mathbb{N}$

or equivalently

$\forall x\in \mathbb{N}: \exists y\in \mathbb{N}:S(x)=y$

Applying the latter statement for $x=0$ (assuming $0$ is the "first" number), we would then have:

$\exists y\in \mathbb{N}:S(0)=y$

Technically, we could then use the rule of existential specification to assign to $S(0)$ any name but $0$, say $S(0)=\text{Fluffy}$.

Similarly, we could assign to $S(\text{Fluffy})$ any name but but $0$ or Fluffy, say $S(\text{Fluffy})=\text{Bobo}$.

So, the first three numbers would then be: $0$, Fluffy and Bobo.

Answer 3

Given the standard successor function $S: n \mapsto n+1$ on $\Bbb N$, if $\beta$ is any bijection of $\Bbb N$ then

$\quad \displaystyle \beta^{-1} \circ S \circ \beta: \Bbb N \to \Bbb N $

satisfies the Peano axioms.

Let $\tau$ be defined by

$$ \tau(n) = \left\{\begin{array}{lr} 0\, \;\;\;\;\;\;\;\;\;\;\;\,\text{ |} & \text{for } n = 0\\ 2m-1 \,\;\;\; \text{ |} & \text{for } n =2m \text{ for } m \gt 0 \\ 2m+2 \,\;\;\; \text{ |} & \text{for } n =2m +1 \end{array}\right\} $$

The mapping $\tau$ is easily seen to be a bijection satisfying $\displaystyle \tau^{-1} = \tau$. Moreover the OP's candidate successor function is equal to

$\quad \displaystyle \tau \circ S \circ \tau$