Applying Peano Axioms to Subsets of Natural Numbers

Question

Concise Question

Is it possible to apply the Peano axioms to subsets of the natural numbers and thus define arithmetic operations on those subsets strictly in terms of those subsets? If so, is it also possible to create a correspondence between those arithmetic operations and the natural numbers' operations?

Wordy Question

My knowledge of these subjects is largely informal, so I welcome corrections of the errors I am about to make.

My informal understanding of Peano axioms is similar to my software engineering experience: If I can define a suitable initial value and successor function (satisfying the contract that is the Peano axioms), then I can meaningfully talk about arithmetic on the resulting set.

Could an appropriate (initial, successor) be defined for subsets of ℕ? Consider, for example, the set of squared natural numbers $S = \{n : n, n^2 ∈ ℕ\}$. Its zero is 0, conceptually the same 0 as the initial value for ℕ. Its successor function... is beyond me. I don't know if it exists — and the implications if it doesn't. And if it does exist, I would expect it to have some correspondence with ℕ's successor function since $S ⊂ ℕ$; does such a correspondence really exist?

The only hint I can conceive regarding the successor is that it must not contradict ℕ's addition operator. Or can it contradict ℕ's $+$? Addition of elements of $S$ is not closed — at least, according to ℕ's $+$ — although it is closed in ℕ.

More generally, it seems I am asking if every ordered set could be said to have Peano arithmetic, and the relationship between such arithmetics and their supersets' artihmetics.

References For Others Like Me

• What Does It Mean to Be a Number? (The Peano Axioms) | Infinite Series
• How multiplication is really defined in Peano arithmetic | Devlin's Angle
• For fun: Church numerals

Answer 1

According to my understanding of Peano's axioms, it actually defines a non self-crossing "chain" that starts with an "initial element"(not necessary the well-known $0$ in the ordinary natural number set) and a successor function $S(n)$ that connects the chain.

For example, you can define a very "weird" "natural number set" in the form of $"3\rightarrow 4\rightarrow 0\rightarrow 1\rightarrow 2\rightarrow 5\rightarrow6\rightarrow ..."$ Here the $"0"$ is $4$, and $S(n)$ is represented using the $"\rightarrow"$.

Once you have the two core concepts, you can define "addition" and "multiplication" accordingly using $S(n)$, which can be very weird and not obeying the real-life additions and multiplications on natural numbers.

In your example, you can simply set $"0"$ to be the real-world $0$, and set $S(n^2)=(n+1)^2$. Then the "chain" will look like $0\rightarrow 1\rightarrow 4\rightarrow 9\rightarrow ...$. Then an "addition" (denoted as "$+'$") may look like $a^2+'b^2=(a+b)^2$, and a "multiplication" (denoted as "$\times'$") may look like $a^2\times' b^2=(ab)^2$. As you can see, the defined "addition" and "multiplications" are not true under the well-known addition and multiplication defined in natural numbers.

Answer 2

Sure. There are many (I am guessing uncountably many) ways to do this.

For example, take as the domain the subset of numbers $1,2,3, ...$ (i.e. exclude $0$), and now have $1$ act as if it is the number $0$, have $2$ act as if it were $1$, etc. Thus, for example, what under the standard interpretation would be a proof of $1+1=2$ now becomes a proof of $2 +' 2 = 3$, where $+'$ obviously can no longer be the normal addition function, but is something else. Likewise, what would normally be considered a proof of $2*3=6$ would now be seen as a proof that $3 *' 4 = 7$, where again $*'$ is no longer the normal multiplication function.

To do this, interpret the logical constant symbol $\mathbf{0}$ as the mathematical number $1$, and interpret the function symbols $\mathbf{s}$, $\mathbf{+}$, and $\mathbf{*}$ as the functions $s'$, $+'$, and $*'$ defined as follows (where $+$, $-$, and $*$ are the normal mathematical functions of addition, subtraction, and multiplication):

$s'(x) = x+1$

$x +' y = (x-1) + (y-1) + 1$

$x *' y = (x-1) * (y-1) + 1$

... but you see the problem: these functions $+'$ and $*'$ are 'weird functions that we hardly ever will have a use for in real life or doing practical mathematics. Hence, the theorems I can prove for the domain of positive integers using Peano Arithmetic are correspondingly uninteresting.