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Within set theory, having the natural numbers $\mathbb{N}$ built as the minimal inductive set with the corresponding additive and multiplicative operations defined, integers $\mathbb{Z}$ can be set as equivalence classes of parallel diagonals of $\mathbb{N}\times\mathbb{N}$, which contain a copy of the natural numbers. See Set Theoretic Definition of Numbers .

Is there any alternative definition of the set $\mathbb{Z}$, starting from $\mathbb{N}$ already defined as usual, such that $$\mathbb{N}\subset\mathbb{Z}$$ as sets, preserving the sum and product operations?

Dr Potato
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4 Answers4

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Yes, one can construct a model of $\mathbb{Z}$ which contains $\mathbb{N}$. The following construction does not use equivalence classes or embedding technic to make $\mathbb{N}$ subset $\mathbb{Z}$. It rather extends a particular model of natural numbers.

Introduction. Let $(\mathbb{N},\mathrel{+} ,\cdot , \leq)$ be the system of Von Neumann natural numbers.

A natural number $m$ is a particular set with $m$ elements $m = \{0,...,m-1\}$ for $m>0$ and $0=\{\}$.

Subtraction and division can be defined for some pairs of natural numbers.

For $m,n \in \mathbb{N}, m - n$ is the natural number $d$, if there is any, such that $m = n + d.$

For $m,n \in \mathbb{N}, n>0, m\div n$ is the natural number $q$, if there is any, such that $m=n⋅q$.

In what follows I will use the fact that $n\subset\mathbb{N}$ and carry set-theoretical operations on natural numbers.

Construction of Integers

Definition 1. Let $n\in \mathbb{N}$ be a natural number. An opposite number $\overline{n}$ is a subset of $\mathbb{N}$ defined as: $$\overline{n}:=\begin{cases}0&\text{if } n=0\\ \mathbb{N}\setminus n &\text{if } n\neq 0.\end{cases}$$ The set of all opposite numbers is denoted by $\overline{\mathbb{N}}=\{\overline{n}|n\in\mathbb{N}\}$.

Intuition 1. An opposite number $\overline{n}$ is a particular set with $n$ elements being missing. Intuitively if we are missing $n$ elements and we receive $n$ then we do not miss anything and therefore we have nothing. This justifies our definition of $\overline{0}=0$.

Definition 2. We define define the set $\mathbb{Z}$ of integers as $ \mathbb{Z}=\mathbb{N}\cup \overline{\mathbb{N}}$.

We extend the domain of our definition of $\overline{a}:=\mathbb{N}\setminus a$ to all $a\in\overline{\mathbb{N}}\setminus\{0\}$.

Definition 3.

We define projection functions $$\mathsf{proj}_0: \mathbb{Z} \to \mathbb{N},a \mapsto a_0:= \begin{cases} a, & \text{if } a \in\mathbb{N} \\[2ex] 0, & \text{if } a \in\overline{\mathbb{N}} \end{cases} $$ $$\mathsf{proj}_1: \mathbb{Z} \to \mathbb{N},a \mapsto a_1:= \begin{cases} \overline{a}, & \text{if } \overline{a} \in\mathbb{N} \\[2ex] 0, & \text{if } \overline{a} \in\overline{\mathbb{N}} \end{cases} $$ Definition 4. We define the balance function as follows $$ \mathsf{bal}: \mathbb{N}\times\overline{\mathbb{N}}\to\mathbb{Z}, (m,\overline{n})\mapsto (m-\mathrm{min}\{m,n\})\cup(\overline{n-\mathrm{min}\{m,n\}}). $$

The balance function is well-defined as either $m-\mathrm{min}\{m,n\}=0$ or $\overline{n-\mathrm{min}\{m,n\}}=0$.

Intuition 2. For a natural number $m$ and an opposite number $\overline{n}$ we find a balance between $m$ and $\overline{n}$. If we are missing $n$ elements and we receive $m$ then we have $m-n$ elements if $m<n$, we don’t have or don’t miss any elements if $m=n$ and finally we miss $n-m$ elements if $m>n$.

Definition 5. We define,$+_\mathbb{Z}, \cdot_\mathbb{Z}$binary operations and $\leq_\mathbb{Z}$ an order on $\mathbb{Z}$ as follows:

$$+_\mathbb{Z} :\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}, (a,b)\mapsto \mathsf{bal}(a_0+b_0,\overline{a_1+b_1})$$

$$\cdot_\mathbb{Z}:\mathbb{Z}\times \mathbb{Z} \to \mathbb{Z}, (a,b)↦(a_0\cdot b_0+ a_1\cdot b_1 )\cup(\overline{ a_0\cdot b_1+ a_1\cdot b_0}) $$

$$ a \leq_\mathbb{Z} b :\Longleftrightarrow a_0 + b_1\leq a_1 + b_0.$$

Again as either $a_0\cdot b_0 + a_1\cdot b_1=0$ or $\overline{a_0\cdot b_1+ a_1\cdot b_0}=0$ and the binary operation $\cdot_\mathbb{Z}$ is well-defined.

Proposition 3. The binary operations and the order on $\mathbb{Z}$ restricted to natural numbers are the same as the binary operations and the order on $\mathbb{N}$.

As a bonus I present a construction of the rational numbers in the same spirit.

Construction of Rationals

Definition 6. Let $m, n\in \mathbb{N}$ and $n>0$. A ratio of $m : n$ is a subset of $\mathbb{N}$ defined as follows:

$$m:n=(m+n)\div\mathrm{gcd}\{m,n\}\setminus\{m \div\mathrm{gcd}\{m,n\}\}.$$

The set of ratios is the set $\mathbb{L}:=\{m:n|m, n \in \mathbb{N} \text{ and } n\neq 0\}.$ (Ancient and Modern λόγος (lógos) ‘ratio’.)

Intuition 3. We represent natural numbers $m,n$ as intervals $[0; m), [0; n)$ then the ratio of $[0; m) : [0; n)$ is the same as ratio $[0; m), [m; m+n)$. A ratio is a partion of $[0,m+n)$ which we represent by removing the point $m$ from $[0; m+n)$, i.e. $[0; m+n)\setminus \{m\}=[0; m)\cup(m; m+n)$.

Proposition 4. For coprime natural numbers $m, n \in \mathbb{N}$ and $n>0$ $$m : n := m\cup ((m+n)\setminus (m+1)).$$

For all natural numbers $m \in \mathbb{N}$: $$m : 1 = m.$$

We can now define addition, multiplication and the order on $\mathbb{L}$.

Definition 7. We define $$+_\mathbb{L} :\mathbb{L} \times \mathbb{L} \to \mathbb{L}, (a,a')\mapsto (m\cdot n'+m'\cdot n ) : (n\cdot n' )$$

$$\cdot_\mathbb{L} :\mathbb{L}\times \mathbb{L} \to \mathbb{L}, (a,a')\mapsto (m\cdot m'):(n\cdot n' ) $$

$$ a \leq_\mathbb{L} a' :\Longleftrightarrow m\cdot n'\leq m'\cdot n .$$

Definition 8. Let $a=m:n, a'=m':n'$ be ratios with $a\leq_\mathbb{L}a'$. We define subtraction $$a'-_\mathbb{L}a:=(m'\cdot n-m\cdot n' ) : (n'\cdot n ). $$

The order condition on the ratios is what is needed for the subtraction of natural numbers to be well-defined.

Proposition 5. The set of natural numbers is subset of $\mathbb{L}$ and operations of addition, multiplication, subtractions and the linear order on $\mathbb{L}$ extend those on $\mathbb{N}$.

Definition 9. Let $a\in \mathbb{L}$ be a ratio. An opposite ratio $\overline{a}$ is a subset of $\mathbb{N}$ defined as: $$\overline{a}:=\begin{cases}0&\text{if } a=0\\ \mathbb{N}\setminus n &\text{if } a\neq 0.\end{cases}$$ The set of all opposite ratios is denoted by $\overline{\mathbb{L}}=\{\overline{a}|a\in\mathbb{L}\}$.

Definition 10. We define the set $\mathbb{Q}$ of rational numbers as

$$ \mathbb{Q}=\mathbb{L}\cup \overline{\mathbb{L}}.$$

We repeat Defintions 3, 4, 5 subsituting $\mathbb{N}$ with $\mathbb{L}$, $\mathbb{Z}$ with $\mathbb{Q}$, and use operations defined on the set of ratios, rather than on the set of natural numbers.

Proposition 6. $$\mathbb{N}\subset\mathbb{Z}\subset\mathbb{Q}.$$

The binary operations and the order on $\mathbb{Q}$ restricted to natural numbers are the same as the binary operations and the order on $\mathbb{N}$.

The binary operations and the order on $\mathbb{Q}$ restricted to integers are the same as the binary operations and the order on $\mathbb{Z}$.

Definition 11. We define the fraction function as follows

$$\mathsf{frac}:\mathbb{Z}\times \mathbb{Z} \to \mathbb{Q},$$ $$ (a,b)↦\frac{a}{b}:=(a_0\cdot b_0+ a_1\cdot b_1 ):(b_0^2+b_1^2)\cup\overline{(a_0\cdot b_1+ a_1\cdot b_0):(b_0^2+b_1^2)} $$

Again as either $a_0\cdot b_0 + a_1\cdot b_1=0$ or $\overline{a_0\cdot b_1+ a_1\cdot b_0}=0$ so the fraction function is well-defined.

I don't know how this construction can be extended any further but there are various unique representations of reals as continuous fraction so I think it's a possibility that we can construct real numbers as subsets of $\mathbb{N}$.

As a bonus 2 I recap below Dedekind constructions on $\mathbb{L}$. The following variant is due to Holmes, page 95. Elementary Set Theory with a Universal Set http://math.boisestate.edu/~holmes/holmes/head.pdf.

Construction of Reals

Definition 11. A magnitude $x$ is a proper initial segment of $\mathbb{L}$ with no greatest element.

The set of magnitudes is the set $$\mathbb{M}:=\{x\subset\mathbb{L}| x \neq \mathbb{L}, \text{for all } a\in \mathbb{L}, a\in x \Longleftrightarrow \text{ for some } b\in x: a<b \}.$$

We define addition, multiplication and the order on $\mathbb{M}$.

Definition 12. We define $$+_\mathbb{M} :\mathbb{M} \times \mathbb{M} \to \mathbb{M}, (x,y)\mapsto \{a+ b|a\in x, b\in y\}$$

$$\cdot_\mathbb{M} :\mathbb{M}\times \mathbb{M} \to \mathbb{M}, (x,y)\mapsto \{a\cdot b|a\in x, b\in y\} $$

$$ x \leq_\mathbb{M} y :\Leftrightarrow x\subset y.$$

Definition 13. Let $x, y$ be magnitudes with $x\leq_\mathbb{L}y$. We define substruction $$y-_\mathbb{M}x:=\{b-a|b\in y \text{ and } a\notin x\}. $$

Definition 14. Let $x\in \mathbb{M}$ be a magnitude. An opposite magnitude $\overline{x}$ is a subset of $\mathbb{L}$ defined as: $$\overline{x}:=\begin{cases}0&\text{if } x=0\\ \mathbb{L}\setminus x &\text{if } x\neq 0.\end{cases}$$

The set of all opposite magnitudes is denoted by $\overline{\mathbb{M}}=\{\overline{x}|x\in\mathbb{M}\}$.

Definition 15. We define the set $\mathbb{R}$ of real numbers as

$$ \mathbb{R}=\mathbb{M}\cup \overline{\mathbb{M}}.$$

From this moment we carry forward the same as for $\mathbb{Z}$ and $\mathbb{Q}$ and again we re-use defintions 3, 4, 5 with analogous changes.

Question.

We can construct a model of $\mathbb{Z}$, $\mathbb{Q}$ where all integers, rationals are subsets of $\mathbb{N}$.

Can we construct a model of $\mathbb{R}$ where all reals are subsets $\mathbb{N}$?

Update 1 I've looked at the constructions of real numbers via continued fractions and I think the answer is one can code real numbers as subsets.

G. J. Rieger. A new approach to the real numbers (motivated by continued fractions). AOh. Brauceig. Wis. Ge, 33:205–217, 1982.

A. Knopfmacher and J. Knopfmacher. Two constructions of the real numbers via alternating series. International Journal of Mathematics and Mathematical Sciences, 12(3):603–613, 1989.

Definition 16 Let $a:N\to\mathbb{N}$ be a sequence of natural numbers where $N\in \mathbb{N}$ or $N= \mathbb{N}$ such that $a_{N-1}>1, N\in\mathbb{N}$. We define recurslively a sequence $ q:N \to\mathbb{N}$, $q_0=1$, $q_1=a_1$, $q_n=a_n\cdot q_{n-1}+q_{n-2}$ for $ n\geq2$. A continued ratio is a subset of $\mathbb{N}$ defined as follows

$$ \lambda(a)=a_0\cup \bigcup_{n\in N\setminus 1} \{a_0+q_n\}$$

Intuition 4 For a continued fraction we have:

$$ a_0+\underset{n\in N\setminus 1 }{\LARGE\mathbb{K}}\frac{1}{a_n} = a_0 +\sum_{n\in N\setminus 1}\frac{(-1)^n}{q_n\cdot q_{n-1}} $$

and the set $\lambda(a)$ captures all details of the sequence $a$.

The challenge is to explicitly define the arithmetic of continued fraction (analogous to Definition 13, 14, 15) and then re-use Definitions 3, 4, 5 to complete the construction.

http://mathworld.wolfram.com/RegularContinuedFraction.html

Update 2 I've added the extension of the overline operation for completeness sake. A more detailed version of this note is available here.

A Dz
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  • Nice! The exception $\overline{0}=0$ is somehow arbitrary, don't you think? Anyway... where did you get this idea? is it yours or you got it from somewhere else? – Dr Potato Jan 03 '20 at 02:56
  • Yes, I know it is some what arbitrary and I have experimented with defining $\overline{0}=\mathbb{N}$. It has some nice intuition that if you don't miss any element then you basically have them all. So alternatively you can define $\mathbb{Z} :=\mathbb{N}\oplus\overline{\mathbb{N}}$ it captures the intuition of having and missing elements, then one needs to again define an injection from $\mathsf{j}:\mathbb{N}\to\mathbb{Z}$ which would identify $n$ with ${n, \mathbb{N}}$. I've been inspired by the constructions of numbers in Holmes and posts in this question. – A Dz Jan 03 '20 at 11:41
  • As for $\mathbb{Z}$ I’ve seen a construction of it as $\mathbb{N}\times{0}\cup{0}\times\mathbb{N}$, and my own version it’s not far of it. It just avoids ordered pairs. The construction of $\mathbb{Q}$ is of my own invention, I have not seen it anywhere. Again it was inpired by the construction of rationals and reals in Holmes and the challenge to use only subsets of $\mathbb{N}$. – A Dz Jan 03 '20 at 11:47
  • I meant to redefine $\mathbb{Z}$ as disjoint sum not a direct sum, so more like that $\mathbb{Z} :=\left{ { m, \mathbb{N}},|, m\in \mathbb{N} \right}\cup \left{ { 0, \overline{n}},|, n\in \mathbb{N} \right}$. – A Dz Jan 03 '20 at 13:54
  • It can be built on both ways. In some sense, the integers have a copy of the naturals but as sets they are not exactly the same. Although if we see it as an inductive set, then they must be contained and the natural numbers are therefore built in distinct ways. – Dr Potato Jul 14 '24 at 06:46
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Let $\Bbb Z^-= (\Bbb N\setminus \{0\})\times \{M\}$ where $M$ is some (any) set such that $\Bbb Z^-$ is disjoint from $\Bbb N.$ And let $\Bbb Z=\Bbb N \cup \Bbb Z^-.$ Extend the operations $+$ and $\times$ from $\Bbb N$ to $\Bbb Z$ as follows:

$0+z=z+0=z$ and $0\times z=z\times 0=0$ for all $z\in \Bbb Z^-$.

If $x=(m, M)$ and $y=(n,M)$ belong to $\Bbb Z^-$ then $x+y=(m+n,M)$ and $x\times y=m\times n.$

If $x=(m,M)\in \Bbb Z^-$ and $0\ne n\in \Bbb N$ then $x\times n=n\times x=(m\times n, M).$

If $x=(m,M)\in \Bbb Z^-$ and $n\in \Bbb N$ then

$\quad (i)\;\; x+n=n+x= n-m$ if $n\ge m$

$\quad (ii)\; x+n=n+x=(m-n,M)$ if $n<m.$

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Yes (and the same argument goes for the other number sets).

Any construction of $\mathbb{Z}$ comes equipped with an embedding $i : \mathbb{N} \hookrightarrow \mathbb{Z}$, by letting $i(n)$ be '$n$ considered as an integer'.

Now define $$\mathbb{Z}' = \mathbb{N} \cup (\mathbb{Z} \setminus i[\mathbb{N}])$$ where $i[\mathbb{N}] = \{ i(n) \mid n \in \mathbb{N} \}$ is the image of $i : \mathbb{N} \hookrightarrow \mathbb{Z}$.

Evidently $\mathbb{N} \subseteq \mathbb{Z}'$ and the arithmetic operations of $\mathbb{N}$ are preserved in $\mathbb{Z}'$. And indeed, $\mathbb{Z}'$ is a perfectly good 'set of integers', since there is an easy-to-define bijection $\mathbb{Z}' \to \mathbb{Z}$ given by $$n \mapsto \begin{cases} i(n) & \text{if } n \in \mathbb{N} \\ n & \text{if } n \in \mathbb{Z} \setminus i[\mathbb{N}] \end{cases}$$

[Slight caveat: if your construction of $\mathbb{Z}$ already contains some natural numbers for whatever reason, replace $\mathbb{Z} \setminus i[\mathbb{N}]$ by an isomorphic set that contains no natural numbers, such as $(\mathbb{Z} \setminus i[\mathbb{N}]) \times \{ 0 \}$.]

  • I'm not sure the answers the question, since you're starting with "any construction of $\Bbb{Z}$", and the question asks about starting with $\Bbb{N}$. – Sort of Damocles Dec 04 '18 at 20:04
  • @dbx: It does answer the question. The construction of $\mathbb{Z}$ in the question takes $\mathbb{Z}$ to be the quotient of $\mathbb{N} \times \mathbb{N}$ by the equivalence relation $(a,b) \sim (c,d) \Leftrightarrow a+d=b+c$, and then the inclusion $i : \mathbb{N} \hookrightarrow \mathbb{Z}$ is given by $i(n) = [(0,n)]$ for all $n \in \mathbb{N}$. This is one example of a construction of $\mathbb{Z}$; my answer is saying that it doesn't matter what construction you pick. (So, in particular, my answer applies to the construction mentioned in the question.) – Clive Newstead Dec 04 '18 at 21:00
  • Although it answers the question, I am looking for something more elegant. – Dr Potato Dec 04 '18 at 22:22
  • @DrPotato: If you want to construct $\mathbb{Z}$ from $\mathbb{N}$ (with $\mathbb{N}$ as a subset), all you really need is a way of encoding $-n$ for $n \in \mathbb{N}$. (This is assuming $0 \in \mathbb{N}$, of course.) So you could define $-n = (n,0)$ for $n>0$, and then $\mathbb{Z} = \mathbb{N} \cup (\mathbb{N}^+ \times { 0 })$. I don't know if this qualifies as 'elegant' to you, but I'm not sure how much elegance you can expect to find in a set theoretic construction of $\mathbb{Z}$ from $\mathbb{N}$. – Clive Newstead Dec 05 '18 at 02:37
  • The question asks for more than just a way to encode $-n$. I interpret the line: "preserves sum and product operations" as follows: they are looking for a way to define $\Bbb{Z}$ so that the operations on the sets correspond directly to the operations on the elements. In the example given, we have $n \sim (\ldots, n-1,n,n+1,\ldots)$ and $m \sim (\ldots, m-1, m, m+1, \ldots)$, and $n+m \sim (\ldots, (n-1)+(m-1), (n+m), (n+1, m+1), \ldots))$, for example -- the operation of $+$ in $\Bbb{Z}$ recognizably corresponds to the operation of $+$ in $\Bbb{N}$. – Sort of Damocles Dec 05 '18 at 15:11
  • @dbx: I don't think I want to keep arguing about this, but I will entertain you one more time. The question wants a construction of $\mathbb{Z}$ that has $\mathbb{N}$ as a subset —if $\mathbb{N} \subseteq \mathbb{Z}$, and the 'integer $n$' is the same as the 'natural number $n$', then the arithmetic operations of $\mathbb{N}$ are preserved in $\mathbb{Z}$. This is why I said all you need to do is encode $-n$ somehow, if you're requiring that $\mathbb{N}$ be a subset of $\mathbb{Z}$. (continued...) – Clive Newstead Dec 05 '18 at 15:19
  • @dbx: I'm not sure what you mean by $n \sim (\dots,n-1,n,n+1,\dots)$ but I'm sure you're mistaken. The element of $\mathbb{Z}$ corresponding to $n \in \mathbb{N}$ in the construction in the question is the equivalence class ${ (a,a+n) \mid a \in \mathbb{N} }$, so e.g. $(0,n)\sim(1,n+1)\sim\cdots$. In particular $(0,n)$ is in a unique equivalence class for each $n \in \mathbb{N}$, and so we can define $[(0,n)]+[(0,m)] = [(0,n+m)]$. Is this what you meant? This is presumably what is meant by 'preservation of arithmetic operations', but it doesn't give $\mathbb{N}$ as a subset of $\mathbb{Z}$. – Clive Newstead Dec 05 '18 at 15:21
  • I'm basing my comments on the description: "Parallel diagonals of $\Bbb{N}\times\Bbb{N}$, each containing a copy of the natural numbers." As you noted, comments are not really for this sort of back-and-forth, so I'll bow out since the asker knows better than I do what they mean anyway. – Sort of Damocles Dec 05 '18 at 15:26
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    @dbx: Just to close this up, 'parallel diagonals' means sets of the form ${ (a,a+n) \mid a \in \mathbb{N} }$. The individual elements of each diagonal don't correspond to the different natural numbers; rather, each diagonal itself refers to a single natural number—the diagonal ${ (a,a+n) \mid a \in \mathbb{N} }$ corresponds with the natural number $n$. – Clive Newstead Dec 05 '18 at 16:22
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In this section we construct a commutative group. The second section develops the theory that uses this group in its presentation of the integers.

In set theory our starting point is the set of natural numbers $\Bbb N = \{0,1,2,\dots\}$ under addition $\text{+}$. Recall that any natural number $m \gt 0$ has a unique predecessor denoted by $m - 1$.

Define the set

$$\quad \Bbb Z = \{ (x,y) \in \Bbb N \times \Bbb N \; | \; y = 0 \text{ or } x = 0\}$$

Define the bijective transformation $\sigma: \Bbb Z \to \Bbb Z$ by $$ \sigma(x,y) = \left\{\begin{array}{lr} (x+1,y)\, \;\;\;\text{ |} & \text{for } y = 0\\ (x,y-1) \,\;\;\; \text{ |} & \text{for } y \gt 0 \end{array}\right\} $$

Recall that if $n \in \Bbb N$ the composition of $\sigma$ with itself $n$ times is denoted by $\sigma^n$, where $\sigma^0 = id_{\Bbb Z}$.

Let $\tau$ be the inverse mapping of $\sigma$.

If $\beta$ is in the group $\mathcal Z$ of bijective transformations generated by $\sigma$ then

$$ [\exists!\, m \in \Bbb N \text{ such that } \beta = \sigma^m] \;\text{ XOR } \;[\exists!\, m \in \Bbb N \text{ such that } m \gt 0 \land \beta = \tau^m ]$$

In the next section we show how the commutative group $(\mathcal Z,\circ)$ can be regarded as
the set of all integers.

If the reader wants to skip the next section, they should note that the mapping

$$ m \mapsto \sigma^m$$

is an injective morphism of $(\Bbb N,+)$ into $(\mathcal Z,\circ)$.


To supply the arguments constructing the integers, you'll need the following prerequisites:

A basic knowledge of elementary algebra and the following result,

Theorem 1: Let $X$ be a set with $x_0 \in X$ and $\psi: X \to X$ be any function. Then there exist one and only one function $\rho: \Bbb N \to X$ satisfyings

$\tag 1 \rho(0) = x_0$ $\tag 2 \forall n \in \Bbb N, \; \rho(n+1) = \psi(\rho(n))$

See, for example, the wikipedia article recursive definition.

Using theorem 1 and induction, it can be demonstrated that if $f: X \to X$ is any function then

$\tag 3 f \mapsto f^n, \; n \in \Bbb N$

is well-defined, where $f^0 = id_X$ and $f^{n+1} = f^{n} \circ f$ and that

$\tag 4 f^{m+n} = f^{m} \circ f^{n} \; \forall m,n \in \Bbb N$

If $f: X \to X$ is a bijection let $h$ denote the inverse of $f$. For $a \in X$, define

$\tag 5 X_a = \{x \in X \, \colon \, \exists k \in \Bbb N \text{ such that } [\,x = f^k(a)\,] \lor [\,x = h^k(a)\,]\}$

A simple argument show that $f(X_a) = X_a$ and $h(X_a) = X_a$
(when $X_a$ is finite, $f_{|X_a}$ correspond to a cyclic permutation).

Theorem 2: There exist a set $\Bbb Z$ with $\Bbb N \subset \Bbb Z$ and a bijective transformations $\sigma$ on $\Bbb Z$ such that

$\tag 6 \sigma(n) = n + 1 \text{ for } n \in \Bbb N$ $\tag 7 \text{If } [\,S \subset \Bbb Z\,] \land [\,S \ne \emptyset \,]\land [\,\sigma(S) = S \,]\text{ Then } S = \Bbb Z$

If $\tau$ is the inverse of $\sigma$, the group $\mathcal Z$ generated by $\sigma$ and $\tau$ is commutative.
Also, given any $m \in \Bbb Z$ there exist one and only one $\gamma \in \mathcal Z$ such that $\gamma(0) = m$.
In this way, the group structure on $(\mathcal Z,\circ)$ can be bijectively transported to the set $\Bbb Z$ creating the additive group $(\Bbb Z,+)$ and having the property that $(\mathbb{N},+) \hookrightarrow (\mathbb{Z},+)$ is a morphism.
Finally, the set $\Bbb Z$ is unique up to a bijective correspondence, in alignment with a (natural) commutative diagram.

We leave it to the interested reader to 'lift multiplication' from $\Bbb N$ to $\Bbb Z$.

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