# Can we modify the Peano axioms like this? [closed]

I am wondering if the following modifications of the Peano axioms result in a set of axioms equivalent to the Peano axioms, in the sense that any set of numbers satisfies these modified axioms if and only if they satisfy the Peano axioms:

1. Delete the axiom of induction, namely "Let $$P(n)$$ be any property pertaining to a natural number $$n$$. Suppose that $$P(0)$$ is true, and suppose that whenever $$P(n)$$ is true, $$P(S(n)$$ is also true. Then $$P(n)$$ is true for every natural number $$n$$."

2. Add the following axiom: "$$n$$ is a natural number only if it can be expressed as the sum of 2 natural numbers"

3. (This may or may not be a modification) Define addition by defining $$0 + 0 = 0$$, and if $$n+m = x$$, then $$S(n) + m = n+S(m) = S(x)$$

where $$S(n)$$ is the successor of $$n$$.

I think this works because:

1. We can guarantee that for any 2 numbers $$n,m$$ which are in a number system satisfying these axioms, that the addition of $$n+m$$ is defined- as because both $$n$$ and $$m$$ are, by the axiom in the second modification above, the result of addition, they are the result of repeated applications of the successor operation on $$0$$; therefore they are in the scope of the recursion contained in the addition definition.

2. The principle of induction still holds: because every number in a system satisfying these axioms can be obtained by repeatedly applying the successor operations on $$0$$, due to how addition is defined for them as above, $$P(0)$$ and $$P(n) \implies P(S(n))$$ implies that for all $$n, P(n)$$.

Is this reasoning correct?

• Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Mathematics Meta, or in Mathematics Chat. Comments continuing discussion may be removed. Commented May 24 at 0:35

Consider the set of ordered pairs of natural numbers i.e. $$\{(x,y)\mid (x,y) \in \mathbb{N}, x \leq y\}$$. Let $$(x,y) + (a,b) = (x+a,y+b)$$ and $$(x,y) \times (a,b) = (x\times a, y \times b)$$. Let the succesor function be defined such that $$S'((a,b)) = (a,S(b))$$. Let $$(x,y) \leq (a,b)$$ be $$x < a \vee (x = a\wedge y \leq b)$$. Then the set is a totally ordered set that satisfies your axioms but does not satisfy the induction axiom.

Edit: For a counterexample that satisfies the Robinson arithmetic, see this answer

• what is $S'$ here? Commented May 20 at 1:15
• @PrincessMia it is the successor function that I chose for the given set I defined, just to differentiate it from the successor function of natural numbers. Commented May 20 at 1:35
• @CameronBuie oh that's my mistake, I meant to say $(x,y) + (a,b) = (x+a,y+b)$ and $(x,y) \times (a,b) = (x\times a, y \times b)$ Commented May 20 at 4:02
• @PrincessMia The point of my addition is that is it is a solution to your axioms but isn't adequate to what you want. That's why I said it "satisfies your axioms but does not satisfy the induction axiom". Your "definition" simply isn't strong enough to rule out numbers that violate induction itself. Induction axiom is simply a lot stronger than you think. Commented May 21 at 17:00
• @PrincessMia here's a post that's related to what you might want. In short, induction is independent from other axioms of the Peano axioms, or ${PA}^-$. So it doesn't matter how you try to redefine addition or multiplication via axioms. math.stackexchange.com/questions/331733/… Commented May 21 at 17:17

Consider the set $$\Bbb H:=\left\{\frac12 n\mid n\in \Bbb N\right\},$$ where $$\Bbb N$$ is the usual set of natural numbers, so that $$\Bbb H$$ is a subset of the rational numbers that is closed under rational addition. Let $$S:\Bbb H\to\Bbb H$$ be given by $$S(x):=x+1,$$ where $$+$$ is the typical addition of rational numbers. This setup satisfies the following axioms:

• $$0\in\Bbb H$$
• $$\forall x\in\Bbb H, S(x)\in\Bbb H$$
• $$\forall x\in\Bbb H,S(x)\neq 0$$
• $$\forall x\in\Bbb H,\forall y\in \Bbb H,\bigl[S(x)=S(y)\implies x=y\bigr]$$

These are the Peano axioms except for the Induction Axiom. However, it does not satisfy the Induction axiom, as $$\Bbb N$$ is a proper subset which contains $$0$$ and is closed under the successor operation.

On the other hand, it satisfies the following axiom: $$\forall x\in\Bbb H,\exists y\in H:\exists z\in H:x=y+z.$$ This is true because $$0+x=x$$ for all $$x\in\Bbb H,$$ and (together with closure under addition) means that the axiom you're wanting to replace Induction with has been satisfied, so your replacement axiom won't do the job.

Added: In response to your edit, note that $$\Bbb H$$ also satisfies the following properties:

• $$0+0=0$$
• $$\forall x\in\Bbb H,\forall y\in\Bbb H,S(x)+y=S(x+y)$$
• $$\forall x\in\Bbb H,\forall y\in\Bbb H,S(x+y)=x+S(y)$$
• Thanks, is it not the case that because one 'modification' above was defining addition to result in only things which could be produced from repeated applications of the successor function to $0$, if we stipulate that every natural number must be the result of addition, it follows that the non-integer elements of $\mathbb{H}$ cannot actually be the result of addition, and fail the axiom with which I wanted to replace the axiom of induction? Commented May 20 at 4:17
• @PrincessMia: there is nothing in your axiom that limits the naturals to things which you can reach from $0$ by successor. In addition, your axiom 2 is second order logic, which PA avoids. The usual second order arithmetic induction is much simpler-it basically says all the naturals can be reached by successor from $0$. Second order logic does not have the nice proof properties that first order logic has, but this is a complicated subject. Commented May 20 at 4:52
• @RossMillikan can you tell me where exactly the error is in the following reasoning: 1. every result of addition can be reached from $0$ by a successor- as we start off with $0+0 = 0$, and every other result of addition, $n+m$, is the successor of a 'previous' version of addition- it is either $S(S^{-1}(n) + m)$ or $S(n+ S^{-1}(m))$. So there is ultimately a chain of successors from $0$ to $n+m$. 2. Because every natural number is the result of addition by axiom 2, every natural number can be reached from $0$ by a successor. Commented May 20 at 4:58
• @PrincessMia: nothing you have said implies there are not naturals not reachable from $0$. The example with the half integers is a good one. They satisfy your axiom 2 using the usual addition function. You are hoping the only if will rule out numbers that are not the usual naturals, but we get to build a model with whatever we want as long as it satisfies the axioms. So $0+\frac 12=\frac 12$ and I have shown that $\frac 12$ is the sum of two naturals. Commented May 20 at 5:05
• @RossMillikan isn't addition undefined for $1 \over 2$ though, according to the definition I have given above (using the successor function given in the post to which we are replying?) Commented May 20 at 5:11