I repeat the Peano Axioms:

  1. Zero is a number.

  2. If a is a number, the successor of a is a number.

  3. zero is not the successor of a number.

  4. Two numbers of which the successors are equal are themselves equal.

  5. If a set S of numbers contains zero and also the successor of every number in S, then every number is in S.

Suppose to have two isomorph "copies" of the natural numbers $\mathbb{N}':=\{0',1',2'...\}$ and $\mathbb{N}'':=\{0'',1'',2''...\}$. Then the set $NUMBERS:=\mathbb{N}'\cup \mathbb{N}''$ with "Zero"$:=0'$ and the "natural" successor for each element in any of the two sets, seems to satisfy the axioms.

Yes, P5 is very strange now because, it says that when I start with a set which I know to contain at least $0'$ and every successor of the numbers in it, automatically contains $0''$ which is not a successor of any number.

If this way of reasoning is allowed we could also use a number of copies of $\mathbb{N}$ indicized by a continuous index so there will be TWO NOT ISOMORPHIC Peano sets.

Because this sounds very strange to me, It's possible that there is a problem in my argument. What do you think?

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    $\begingroup$ Actually, $\mathbb N'\cup\mathbb N''$ does not satisfy $5$. $\endgroup$ Jun 25, 2014 at 22:43
  • $\begingroup$ Basically stating "$0$ is a number" is a statement about a single constant. That is the nature of the (more formalized) statement. $\endgroup$ Jun 25, 2014 at 22:45
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    $\begingroup$ Related question about alternative models of the axioms you stated: math.stackexchange.com/questions/637693/… $\endgroup$
    – MJD
    Jun 25, 2014 at 23:02

4 Answers 4


Note that from the fifth axiom it follows that if $x\neq 0$ then $x=S(y)$ for some $y$.

Therefore if $x,y$ are both $0$'s (and are distinct) then at least one of them is a successor. Which means that one of them is not a $0$.

I think that the source of the confusion here is the fact that $0$ is seen as just any number. The truth is that $0$ is a constant in the formal language of arithmetic. And in a given structure there can only be one interpretation of a constant. Not two or more.

  • $\begingroup$ What you say is not explicit in the axiom. Couldn't be a (very common) interpretation? $\endgroup$
    – Benzio
    Jun 25, 2014 at 22:38
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    $\begingroup$ @Benzio Given a zero $0$, let $S$ be the set $\{x\mid x =0$ or $x$ is a successor$\}$. By P5, every number is in $S$, that is, every number not equal to $0$ is a successor, and in particular you cannot have more than one zero. Do you understand why P5 implies that $S$ contains all numbers? Do you agree with this? Otherwise you are just reading the axioms in a manner different from that which is intended. For instance, you may be reading P5 as saying that $S$ must contain "all zeros", not just the specific one I called $0$. This is not the intended reading of the axiom. $\endgroup$ Jun 25, 2014 at 22:43
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    $\begingroup$ @Benzio You seem to be very confused. Stating the axioms in the language of formal logic, and using formal logic rather than the informal way we argue in English may be needed here. Several books contain details of this formalized presentation, and it may be to your advantage to look for one, and read it carefully. If you are not familiar with formal logic, start with a book on the subject, before specializing to one covering (formal) Peano Arithmetic. I suggest Klenne's "Introduction to metamathematics", as more modern treatments may skip some of the formal details needed here. $\endgroup$ Jun 25, 2014 at 22:51
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    $\begingroup$ @Benzio: The axioms in some sense try to "define" the natural numbers by stating that they satisfy all these properties. Therefore, by the last axiom, whenever a collection of objects satisfy the property that "the object defined as 0 is inside and the collection is closed under S", then the collection is all that there is. As seen in the Andres' comment, the collection of objects which are either 0 or successor of some number satisfy this property. Therefore, the collection of objects that are "either 0 or successor of some number" is all the natural numbers. $\endgroup$
    – Burak
    Jun 25, 2014 at 22:59
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    $\begingroup$ You can prove that every number is either 0 or is a successor. So in particular any mythical $0'$ must either be $0$ or a successor. So there cannot be two distinct numbers $0$ and $0'$ neither of which is a successor. So it is axiom 5 that prevents the situation of this mythical $0'$. $\endgroup$ Jun 25, 2014 at 23:01

In logic, $0$ is a constant, not a property. That is, we can say "$x=0$" rather than "$x$ is a zero."

Even in second order logic, the statement of induction is $0\in S$. Not, "for all zeros $z, z\in S$."

You can, of course, interpret lots of language in lots of different ways. But mathematics uses "$0$ is a natural number" in a very specific way, and if you want to interpret that phrase differently, you are free to do so, but you are likely to confuse mathematicians and fail to communicate with them unless you are very explicit.

  • $\begingroup$ You can easily prove that the zero of any Peano set (defined in the obvious way) is unique. See my answer. $\endgroup$ Jun 26, 2014 at 4:52
  • $\begingroup$ Well, not if you take the obtuse interpretation of Peano that the OP was taking - that the axiom of induction means if you've proved it for all zeros... @DanChristensen $\endgroup$ Jun 26, 2014 at 5:10

I'm not fully sure I understand your question, but perhaps you will find this argument enlightening: consider the set

$$\mathbb{N}' = \{0', s(0'), s(s(0')), \ldots\}$$

which contains $0'$ and all its successors. (This is the same $\mathbb{N}'$ you defined in the question.) Hopefully it is clear that $\mathbb{N}'$ contains the successor of every element in $\mathbb{N}'$:

$$\forall n\in\mathbb{N}'\ s(n)\in\mathbb{N}'$$

So axiom 5 applies to $\mathbb{N}'$, which tells us that $\mathbb{N}'$ contains all numbers, or in other words, axiom 5 tells us that if $n$ is a number, $n\in\mathbb{N}'$. That in turn means anything not in $\mathbb{N}'$ is not a number.

Now, if you want, you can postulate the existence of another object, like $0''$, and even a set $\mathbb{N}''$ of that object and its successors. But you cannot claim that $0''$ (or any of its successors) is a number (as defined in the Peano axioms) without contradicting the result of the preceding argument, and thus contradicting axiom 5.

You can construct sets which contain numbers and other things, like $\mathbb{N}'\cup\mathbb{N}''$. This set contains every number, because every number is in $\mathbb{N}'$ and by the definition of the union operation, anything in $\mathbb{N}'$ is in the union of $\mathbb{N}'$ and any other set. It also contains a bunch of other things which are not numbers, namely the objects in the set $\mathbb{N}''$. Nothing says that every object in the set $S$ referenced in axiom 5 must be a number.



Suppose we have zeroes $0$ and $0'$ in a Peano set such that:

  1. $0$ is a number.

  2. $0'$ is a number

  3. If x is a number, the successor of x is a number.

  4. $0$ is not the successor of any number.

  5. $0'$ is not the successor of any number.

  6. Two numbers of which the successors are equal are themselves equal.

  7. If a set S of numbers contains $0$ and also the successor of every number in S, then every number is in S.

  8. If a set S of numbers contains $0'$ and also the successor of every number in S, then every number is in S.

Therefore $0=0'$, and there is a unique zero in every Peano set (defined in the obvious way).


We can easily prove by induction (7) that all non-$0$ numbers have a predecessor.

Suppose $0\neq 0'$. Therefore $0'$ must have a predecessor. But this contradicts (5). Therefore, we must have $0=0'$.


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