1
$\begingroup$

I was reading through the Peano axioms here, and a question came up:

Can we define $S(0)=1$, and $S(1)=1$?

It seems to me (at least as it is stated) that it would satisfy all of the axioms listed. And I couldn't find any restrictions on the successor function, which would disallow this.

So would this be a valid Peano arithmetic or am I missing something here?

Thanks!

$\endgroup$
  • 2
    $\begingroup$ If $S(1)=1=S(0)$, then $0=1$ by Axiom 7 in the link. $\endgroup$ – logarithm May 13 at 13:28
  • $\begingroup$ This is still not a problem on its own, the contradiction arises with Axiom 8 which says that $0$ is not a successor, but $0=1$ implies $S(0)=1=0$. $\endgroup$ – logarithm May 13 at 13:37
  • $\begingroup$ @logarithm: Yes, if we're not considering $=$ to be built into the logic (but the Wikipedia article is equivocal about that). $\endgroup$ – Henning Makholm May 13 at 13:41
3
$\begingroup$

Your structure fails to satisfy the axiom that the Wikipedia article gives number 7:

  1. For all natural numbers $m$ and $n$, $m = n$ if and only if $S(m) = S(n)$. That is, $S$ is an injection.
$\endgroup$
  • $\begingroup$ Ah! I see. Didn't see the "only if" part. Thanks! $\endgroup$ – Valdeminas May 13 at 13:34
  • $\begingroup$ @Valdeminas: It's actually the "if" direction that sinks your structure. $\endgroup$ – Henning Makholm May 13 at 13:39
  • $\begingroup$ $\{0\}$ with $S(0)=0$ satisfies Axiom 7. It is Axiom 8, the one that fails. $\endgroup$ – logarithm May 13 at 13:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.