Having done a lot of hard work putting the natural numbers $\mathbb{N}$ on a sound footing the next step, to build $\mathbb{Z}$, is rather more straightforward. The most direct method is perhaps to define $\mathbb{Z}$ as a disjoint union of two copies of $\mathbb{N}$, one copy being $\left\{0,1,2,3,\dots \right\}$ and the other copy being $\left\{-1,-2,-3,\dots \right\}$. A slightly more sophisticated method is to use equivalence relations.

Definition.

Let $\sim $ be the following relation defined on $\mathbb{N}\times \mathbb{N}$: we define $({x}_{1},{y}_{1})\sim ({x}_{2},{y}_{2})$ to mean ${x}_{1}+{y}_{2}={x}_{2}+{y}_{1}$.

Proposition.

$\sim $ in an equivalence relation on $\mathbb{N}\times \mathbb{N}$.

**Proof.**

Exercise.

Definition.

We write $[a,b]$ for the equivalence class of $(a,b)\in \mathbb{N}\times \mathbb{N}$. The set $\mathbb{Z}$ is the set $\mathbb{N}\times \mathbb{N}/\sim $ of equivalence classes.

Proposition.

The function $i$ defined by $i\left(n\right)=[n,0]$ is a one-to-one function mapping $\mathbb{N}$ into $\mathbb{Z}$.

**Proof.**

Let $n,m\in \mathbb{N}$ with $i\left(n\right)=i\left(m\right)$. Then $[n,0]=[m,0]$ hence $(n,0)\sim (m,0)$ hence $n+0=m+0$ hence $n=m$.

Definition.

We identify each $n\in \mathbb{Z}$ with its image $i\left(n\right)$ under the map $i$. In particular, $0$ is the element $[0,0]\in \mathbb{Z}$, and $1$ is $[1,0]\in \mathbb{Z}$.

Definition.

We define addition, multiplication and order relations on $\mathbb{Z}$ by

- $[{x}_{1},{y}_{1}]+[{x}_{2},{y}_{2}]=[{x}_{1}+{x}_{2},{y}_{1}+{y}_{2}]$
- $[{x}_{1},{y}_{1}]\xb7[{x}_{2},{y}_{2}]=[{x}_{1}{x}_{2}+{y}_{1}{y}_{2},{x}_{1}{y}_{2}+{y}_{1}{x}_{2}]$
- $[{x}_{1},{y}_{1}]<[{x}_{2},{y}_{2}]\leftrightarrow {x}_{1}+{y}_{2}<{x}_{2}+{y}_{1}$

Proposition.

The operations $+,\xb7$ and relation $\u2a7d$ on $\mathbb{Z}$ are well-defined, i.e., the definitions above do not depend on the particular choice of representatives $({x}_{1},{y}_{1}),({x}_{2},{y}_{2})$.

**Proof.**

Exercise.

Proposition.

The embedding $i:\mathbb{N}\to \mathbb{Z}$ is a homomorphism respecting $+,\xb7$ and $\u2a7d$:

- $i(n+m)=i\left(n\right)+i\left(m\right)$;
- $i(n\xb7m)=i\left(n\right)\xb7i\left(m\right)$; and
- $n<m\leftrightarrow i\left(n\right)<i\left(m\right)$;

for all $n,m\in \mathbb{N}$.

**Proof.**

Exercise.

Having constructed the set of integers and its main operations, we need to prove that it satisfies the key axioms we expect. We list here the key axioms and in a few cases sketch the proof that the set satisfies the axioms.

Firstly, $\mathbb{Z}$, with the addition and multiplication operations forms a nice algebraic structure. Division is not available but almost everything else you'd hope works doe so. In particular the integers forms a commutative ring with $1$.

Theorem.

The set $\mathbb{Z}$, with the addition and multiplication operations, is a commutative ring with $1$. I.e.,

- $(\mathbb{Z},+)$ is an abelian group with identity element $0$.
- $(\mathbb{Z}\setminus \left\{0\right\},\xb7)$ is an abelian semigroup with identity element $1$. (This means that the commutative and associative laws hold, and also $\forall x\in \mathbb{Z}(1\xb7x=x=x\xb71)$, but that multiplicative inverses don't necessarily exist.)
- The distibutivity law holds.

**Proof.**

Exercise.

The ring of integers has a further nice property concerning multiplication.

Theorem.

The ring of integers is an integral domain. That is, it has no zero divisors, i.e, satisfies the additional axiom

$$\forall x\in \mathbb{Z}\forall y\in \mathbb{Z}(x\xb7y=0\Rightarrow (x=0\vee y=0\left)\right)$$
**Proof.**

Exercise.

So far we haven't yet mentioned the order structure $\u2a7d$ on the integers. This is easily remedied, and the axioms look very similar to those for Archimedean ordered fields.

As for $\mathbb{N}$, etc., we can work with $x<y$ or with the relation $x\u2a7dy$ defined to mean $x<y\vee x=y$. We shall work with $<$ here.

Theorem.

$<$ defined by $x\u2a7dy\wedge x\ne y$ is a linear order on $\mathbb{Z}$, i.e., satisfies

- Transitivity: $\forall x\in \mathbb{Z}\forall y\in \mathbb{Z}\forall z\in \mathbb{Z}\left(\right(xy\wedge yz)\Rightarrow xz)$
- Irreflexivity: $\forall x\in \mathbb{Z}\left(\neg xx\right)$
- Linearity: $\forall x\in \mathbb{Z}\forall y\in \mathbb{Z}(xy\vee x=y\vee yx)$

**Proof.**

Exercise.

We also need to add axioms stating the relationship between the order and the arithmetic operations.

Theorem.

The integers with $<$ forms an ordered ring, i.e., also satisfies the following axioms:

- One is positive: $0<1$.
- Addition respects $<$: $\forall x\in \mathbb{Z}\forall y\in \mathbb{Z}\forall z\in \mathbb{Z}(xy\Rightarrow x+zy+z)$
- Multiplication respects $<$: $\forall x\in \mathbb{Z}\forall y\in \mathbb{Z}\forall z\in \mathbb{Z}\left(\right(0z\wedge xy)\Rightarrow x\xb7zy\xb7z)$

In addition, this order is discrete, that is, there are no elements between $x$ and $x+1$:

- $\forall x\in \mathbb{Z}\neg \exists y\in \mathbb{Z}(xy\wedge yx+1)$

**Proof.**

Exercise.

We have been presenting all the properties as they most naturally arise. However, our axiomatisation is not optimal in the sense that some axioms could be omitted since they follow from others. A case in point is where the new axioms for the order automatically implies the no-zero-divisor property.

Exercise.

Show that any ordered ring is an integral domain.

The se of integers forms an Archimedean structure in the same sense as in Archimedean ordered fields. The direct translation of this axiom from the earlier web page is perfectly correct, though looks somewhat silly when one recalls the correct picture. It says that no single integer $z\in \mathbb{Z}$ is greater than all natural numbers. Phrased in a slightly more meaningful form for the integers, this becomes

Theorem.

No integer $z\in \mathbb{Z}$ is greater than all natural numbers. In other words, the set of nonnegative integers $\{z\in \mathbb{Z}:x\u2a7e0\}$ is exactly $\mathbb{N}$.

**Proof.**

Exercise.

By the least number principle in $\mathbb{N}$, this can be rephrased in an equivalent but rather more powerful form.

Theorem.

Let $P\left(z\right)$ hold for some integer $z\in \mathbb{Z}$ and suppose that there is $k\in \mathbb{Z}$ such that $\forall w\in \mathbb{Z}(wk\Rightarrow \neg P(w\left)\right)$. Then there is a least ${z}_{0}$ in $\mathbb{Z}$ such that $P\left({z}_{0}\right)$.

**Proof.**

Use the least number principle on the property $Q\left(n\right)$ of natural numbers where $Q\left(n\right)$ says that $P(n+k)$ holds.

For an example where the conditions of the last result are not met and where the least number principle fails, consider the property $P\left(n\right)$ that says that $n<0\wedge -n\text{is prime}$.

We now have all the information we need about the integers. The next task is to combine these to make an Archimedean field, the field of rationals, containing the integers.