Please vote to close this question. It's really dumb as when I was reading the Peano axioms, axiom 8 didn't register. Don't waste your time reading this question.... I also cannot delete it (I have tried but it won't let me).

$1$ is defined as $S(0).$

An essentially equivalent question to the one I am asking is, "Why isn't $0\neq 1$ a Peano axiom?"

If we accept Wikipedia's $9$ Peano axioms, as well as the two axioms of addition, how do we show then that $0\neq 1?$ Or do we need to accept the multiplication axioms as well in order to show that $0\neq 1?$ Or is there another axiom I am not seeing that you need, in order to show that $0\neq 1?$

Basically, as far as I can tell, none of the axioms tell us that we cannot have $0=1=2=3=\ldots.$

I think that what I am referring to is (essentially?) the group (semigroup? ring?) $\langle \{0\}, + \rangle\ $ along with $=$ defined as an equivalence relation, and an axiom that enables substitution.

$$$$ $0 \neq 1$ not provable in axiomatic arithmetic?

I looked here, but I'm not sure the question is the same as my one and I don't understand the answers. I also don't think the answers relate to wikipedia's version of the Peano axioms, but maybe I am wrong?

  • $\begingroup$ Don't we also have to assume a set theory as well? In ZFC for instance we know that $0\neq 1$ because $x\neq \{x\}$ by the axiom of regularity. $\endgroup$
    – JMoravitz
    Sep 21, 2022 at 11:48
  • $\begingroup$ @JMoravitz I am not familiar with the set theory formulation of Peano's axioms. Just wikpedia's "standard" approach. I know set theory approach is probably more formal or more correct or whatever, but I am not familiar with it. $\endgroup$ Sep 21, 2022 at 11:56
  • $\begingroup$ I don’t get all the downvotes. I tried deleting this as soon as the answers came, but you’re not allowed to… Maybe I will ask this as a question on meta $\endgroup$ Sep 22, 2022 at 17:13
  • 2
    $\begingroup$ I’m voting to close this question because I don't want to continue to receive downvotes. $\endgroup$ Sep 22, 2022 at 22:23
  • $\begingroup$ After thinking about your meta post for a few days, I decided to upvote this. I believe it is a natural Question to ask about where $0\neq 1$ is implied by the Peano axioms, in spite of it being a central consideration. You showed some effort in stating the problem by bringing up a trivial model $0=1=2=...$, so with my upvote you now have net positive effect on your reputation. Not that you should be worried. The two succinct Answers are also worth keeping! $\endgroup$
    – hardmath
    Sep 24, 2022 at 17:34

2 Answers 2


Axiom 8 in the Wikipedia list of axioms says that $0$ is not the successor of any natural number. Since $1 = S(0)$, we have $1 \neq 0$.

  • $\begingroup$ Somehow I missed this. Wow. $\endgroup$ Sep 21, 2022 at 11:50

You only need two of the axioms:

  1. Axiom 1: $0$ is a natural number.
  2. Axiom 8: For every natural number $n$, $S(n)=0$ is false.

In particular, the statement of axiom $8$ this is also true for $n=0$ (because, from Axiom 1, we know $n$ is a natural number). Therefore, $S(0)=0$ is false.

  • $\begingroup$ Yes. My brain isn't working properly for me to miss this... $\endgroup$ Sep 21, 2022 at 11:49

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