Are there natural numbers that are not the descendant of 0?

Based on the Peano Axioms (wich are a way to correctly absolutely define the set of natural numbers – correct me if i’m wrong) it is possible to construct a set of symbols that doesn’t quite look the way i imagine the natural numbers:

If there is a circle of other symbols next to the infinite row of known natural numbers, doesn’t this also fit all the requirements?

So are there multiple unequal sets of natural numbers?




The axiom of induction fails for the structure above.
– mjqxxxx
Oct 20 at 18:10



The peano axioms aren’t so much definitions as… axioms. Definitions are new words for things that already exist. When working in the Peano axioms, one assumes there exists a structure satisfying those axioms… It’s different. More like axioms for set theory or Euclidean geometry.
– Noah Olander
2 days ago


3 Answers


The induction axiom ensures that N\Bbb N cannot contain a cycle like your a,b,ca,b,c cycle. It says that if

0∈A0\in A, and
for each n∈Nn\in\Bbb N, n∈An\in A implies that n+1∈An+1\in A,

then A=NA=\Bbb N. Take AA to be everything in your diagram except a,ba,b, and cc; this AA satisfies both of these requirements, yet it’s not the whole set shown in your diagram. Thus, the set in your diagram doesn’t satisfy the Peano axioms, and indeed they characterize N\Bbb N.

However, the induction axiom cannot be expressed in first-order logic, and there are structures other than N\Bbb N that satisfy the first-order counterpart of the Peano axioms, though they still don’t contain cycles. All of them are linearly ordered and consist of a copy of the standard N\Bbb N followed by copies of Z\Bbb Z (so that everything except 00 has a unique immediate predecessor). There are restrictions on how these copies of Z\Bbb Z can be ordered relative to one another. For instance, the only possibility for a countable non-standard model looks like N\Bbb N followed by Q×Z\Bbb Q\times\Bbb Z ordered lexicographically.



How does N\mathbb{N} followed by copies of Z\mathbb{Z} satisfy the induction axiom? It seems I should be able to take AA to be the copy of N\mathbb{N}.
– arkeet
Oct 20 at 18:22



@arkeet: I need to revise that slightly. It doesn’t satisfy the second-order induction axiom that I stated; it does satisfy the first-order axiom schema that replaces the induction axiom if you try to give a corresponding first-order axiomatization.
– Brian M. Scott
Oct 20 at 18:29

The induction axiom is:

If P(0)P(0) holds, and if P(n+1)P(n + 1) follows from P(n)P(n), then PP is true of all values.

If you let S={a,b,c}S = \{a, b, c\}, and P(n)=n∉SP(n) = n \not \in S, then the above axiom fails.

P(0)P(0) is true, P(n+1)P(n + 1) follows from P(n)P(n) (it hold for n∈Sn \in S due to vacuous implication), but P doesn’t hold for all numbers in your structure (it fails for n∈Sn \in S).

So the induction axiom will fail for any structure with non-natural numbers, just let SS be the set of numbers not in the natural number chain.

Going by wikipedia.

The first 8 axioms (particularly 8): For every natural number n, S(n) = 0 is false. That is, there is no natural number whose successor is 0.) seem to assure that the is the “number line” 0->1->2->3…. and that it is infinite and not circular. But they don’t say any thing about there not being any a->b->c->d…. or oom<->pah. which are completely distinct and separate from this set of number lines.

They do state that they must be distinct though. If S(S(S(….(a)))) = a then a ≠\ne S(S(…..(0))))). Or if b ∉\not \in 0->1->2->3…. then no predecessor or successor of b is either.

But 9) the induction axiom does imply there are not distinct threads.

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.

As K = 0->1->2->3 … is such a set we have N⊂K⊂N\mathbb N \subset K \subset \mathbb N and 0->1->2->3…. are precisely the natural numbers and nothing else are.

This is not just theoretical. 1/2 ->1 1/2 -> 2 1/2 -> 3 1/2 ⊄N\not \subset \mathbb N is just such a separate string that does exist but are not the natural numbers.

I like to think (and I’m probably abusing some theory and I could be wrong) that 1-8 define the smallest thing the natural numbers can be. And the induction axiom restricts it to the smallest it can be. I may be naively abusing math though.