Peano Axioms.

Let $\mathbb N \neq \emptyset$ be a set and $S:\mathbb N \to \mathbb N$ a function. The elements of $\mathbb N$ are the natural numbers. If $n \in \mathbb N$ then $S(n)$ is the successor of $n$. $\mathbb N$ and $S$ satisfy the following axioms:

    * $A_1$: $S$ is onte-to-one.
    * $A_2$: $\mathscr R(S) \neq \mathbb N$ i.e $S$ is not surjective.
    * $A_3$: If $u \notin \mathscr R(S)$ and $M \subseteq \mathbb N$ such that:
      $i)$ $u \in M$.
      $ii)$ If $n \in M$ then $S(n) \in M$.
    Then $M = \mathbb N.$


Theorem. $\exists! u \in \mathbb N: u \notin \mathscr R(S).$

We denote $u$ by $1$.

The problem.

Let $P = \mathbb Z$ and $S:P \to P$ defined by: $$ S(a) = \begin{cases} -a & \text{if $a > 0$} \\ 1 & \text{if $a=0$} \\ -(a-1) & \text{if $a<0$} \end{cases} $$

I already prove that $(P,S,0)$ satisfy axioms $A_1$ and $A_2$ but I have no idea how to prove $A_3$.

My try:

Let $A \subset P$ such that:

    1. $0 \in A$.
    2. If $a \in A$ then $S(a) \in A$.

I need to prove that $A = P$. In order to do that I define $B = P-A$ and trying to show that $B = \emptyset$. Lets prove this by contradiction.

If $B \neq \emptyset$ there is $a \in B$ then we have the following cases:

      If $a=0$ then $0 \in P-A \Rightarrow 0 \in P$ and $0 \notin A$. This is a contradiction by definition of A. Then $0 \notin B$.

I have no clue how to prove the other cases.


1 Answer 1


If $a\in B$ with $a>0$ we have $a\notin A$, let $b$ be such that $S(b)=a=-(b-1)$ and $b<0$. Here we have either $b\in A$ or $b\notin A$, clearly we cannot have $b\in A$ as that entails that $a\in A$ which we had already excluded. so $b\notin A$. We can continue this predeccesor process and will eventually reach that $S^n(0)=a$ for some $n$, but we already have that $0\in A$ and hence $S^n(0)=a\in A$ as well and we have a contradiction. This is similarly done for the negative case

  • $\begingroup$ How can you prove that this process ends with $S^n(0)=a$ for some $n$?. $\endgroup$ May 20, 2016 at 16:17
  • $\begingroup$ by virtue of how the function si defined, if $S(b)=a$ with $a>0$ then $a=-(b-1)=1-b$ for $b<0$ which gives us that $|a|=|b|+1$ and for $S(c)=b$ we have $b=-c$ and $|c|=|b|$ with $c>0$ at which we get the chain $|a|=|b|+1=|c|+1=|d|+2=|e|+2=\ldots$ which naturally stops, by virtue of your successor function, at $0$ $\endgroup$ May 20, 2016 at 16:21

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