If we want fully understand numbers and number sets, we must learn how to construct each sets of numbers.

## Construction of number systems: Natural numbers

## Peano Axioms

The set of natural numbers is axiomatic introduced by Italian mathematician Giuseppe Peano. These axioms are known as Peano’s axioms.

**Peano’s Axioms **for **natural numbers**:

$\mathbb{N}$ is a set with the following properties:

(1) $\mathbb{N}$ has a distinguished element which we call $’1’$

(2) There exists $s : \mathbb{N} \rightarrow\mathbb{N}$

(3) $s$ is one-to-one function (injective function)

(4) There does not exists an element $x \in \mathbb{N}$ such that $s(x)=1$ (not surjective function)

(5) If $M \subset \mathbb{N}$ such that: (Principal of Induction)

(a)$1 \in S$,

(b) if $x \in M \Longrightarrow s(x) \in S$,

then set $M=\mathbb{N}$.

A set $\mathbb{N}$ is the set of natural numbers and its element are natural numbers. A function $s$ is called a **successor function**.

**Lemma 1**. If $x \in \mathbb{N}$ and $x \neq 1$, then there exists $y \in \mathbb{N}$ such that $s(y) = x$.

One of the main features of a set $\mathbb{N}$ is that each element in it, except number $1$, has an immediate predecessor and each element has an immediate follower. There is the smallest natural number, number $1$, however, there is no the largest natural number.

Now, we can define operation of addition $+$ and multiplaying $\cdot$ by recursive rules.

**Addition of natural numbers **

There is a unique function $+: \mathbb{N} \times \mathbb{N} \to \mathbb{N}$, $(x,y) \mapsto x + y$, with the following properties:

$(i)$ $x+1 = s(x), \forall x \in \mathbb{N}$,

$(ii)$ $x + s(y) = s(x+y) , \forall x, y \in \mathbb{N}$.

**Theorem 1**. (*Associativity*) For all three natural numbers $x, y$ and $z$ the following is valid:

$$(x+y) + z = x + (y +z).$$

*Proof*.

Let $M \subseteq \mathbb{N}$ be a set of all natural numbers $z$ for which is valid $(x+y) + z = x + (y +z)$, that is:

$$M := \{z \in \mathbb{N}: \forall x, y \in \mathbb{N}, (x + y) + z = x + (y + z) \}.$$

We must prove that $M = \mathbb{N}$. Proof we conduct by using the principle of induction by $z$.

Firstly, we must show that $1 \in M$, that is $ (x + y) + 1 = x + (y + 1) $:

$$(x+y) + 1 $$

$$ s(x +y) $$

$$ x + s(y) $$

$$ x + (y +1).$$

It follows that $1 \in M$.

Now, suppose that $ z \in M$, that is $(x+y) + z = x + (y + z)$. We must show that $s(z)$ is also in the set $M$, that is

$$(x + y) + s(z) = x + (y + s(z)).$$

We have

$$( x + y) + s(z) $$

$$= s ((x+y) + z) $$

$$= s(x +( y + z)) $$

$$ = x + s( y + z) $$

$$= x + (y + s(z)) .$$

Therefore, $s(z) \in M$.

Since $1 \in M$ and from the assumption that $ z \in M$ follows $s(z) \in M$, by the principle of induction we conclude $M = \mathbb{N}$. The statement is true for all natural numbers $x, y$ and $z$.

**Theorem 2**. (*Commutativity*) For any two natural numbers $x$ and $y$ the following is valid:

$$x + y = y + x.$$

For proof the commutativity of natural numbers we need to impose two minor lemma.

**Lemma 2**. For any two natural numbers $x$ and $y$ is valid:

$$s(x) + y = s ( x+ y).$$

**Lemma 3**. For all natural number $x$ is valid:

$$ 1 + x = s(x).$$

*Proof*. ( **Theorem 2.**)

Proof we conduct by using the principle of induction by $y$. We define a set $M$ as:

$$M:= \{y \in \mathbb{N}: \forall x \in \mathbb{N}, x + y = y + x \}.$$

By the 5. Peano’s axiom we need to prove:

(1) $1 \in M$:

$$x + 1 $$

$$= s(x) $$

$$= 1 + x $$

Therefore, $1 \in M$.

(2) Suppose that $ x + y = y +x$. Then $s(y) \in M$, that is $x + s(y) = s(y) + x$:

$$x + s(y) $$

$$= s(x + y) $$

$$ = s( y + x) $$

$$ = s(y) + x .$$

We have proven that $s(y) \in M$.

Since $1 \in M$ and from the assumption that $ y \in M$ follows $s(y) \in M$, by the principle of induction we conclude $M = \mathbb{N}$. The statement is true for all natural numbers $x$ and $y$.

**Multiplication of natural numbers**

There is a unique function $\cdot: \mathbb{N} \times \mathbb{N} \to \mathbb{N}$, $(x,y) \mapsto x \cdot y$, with the following properties:

$(i)$ $\forall x \in \mathbb{N}$, $x \cdot 1=x$,

$(ii)$ $\forall x, y \in \mathbb{N}$, $x \cdot s(y)=(x \cdot y )+x$.

**The properties of multiplication of natural numbers**.

**Associativity**. For all three natural numbers $x, y$ and $z$ is valid:

$$(x \cdot y) \cdot z = x \cdot (y \cdot z).$$

**Commutativity**. For all two natural numbers $x$ and $y$ is valid:

$$ x \cdot y = y \cdot x.$$

**Distributive law**. For all natural numbers $x, y$ and $z$ is valid:

$$x \cdot ( y + z) = x \cdot y + x \cdot z.$$

**Ordering on** $\mathbb{N}$

**Definition**. Let $x, y \in \mathbb{N}$. We say that $x < y$ if there exists a $ z \in \mathbb{N}$ such that x + z = y.

## Construction of number systems: Integers

Now, we can construct integers. Of course, we will use natural number to construct integers.

Consider the set $S= \mathbb{N} \times \mathbb{N}$, and relations $(a,b)\sim (c,d)$ if is valid $a+b=c+d$. Let the set $\mathbb{Z}$ be equivalence classes under this relation. Now we can define on set $\mathbb{Z}$ like:

If sets $A$ and $B$ $\in \mathbb{Z}$, then sets $A$ and $B$ are non-empty subsets of set $S$ and thus we may pick elements $(a,b)$ form set $A$, and elements (c,d) from set $B$.

Now we can define an operation (*addition*) denoted by $\oplus$.

$A\oplus B=[(a+c, b+c)]$

Now, we want to define *multiplication of integers* denoted by $\sim$ like:

$A\sim B=[(a \cdot c+b \cdot d, a \cdot d+b \cdot c)]$.

For natural numbers and integers we can write down corresponding properties.

- Associativity of addition: $a+(b+c)=(a+b)+c$
- Commutativity od addition: $a+b=b+a$
- Additive inverse: $a+(-a)=(-a)+a=0$
- Existing of neutral element for addition: $a+0=0+a=a$
- Distributivity: $a(b+c)=a \cdot b+a \cdot c$
- Associative of multiplication: $a \cdot (b \cdot c)=(a \cdot b) \cdot c$
- Commutativity of multiplication: $a \cdot b=b \cdot a$
- Existing of neutral element for multiplication: $a \cdot 1=1 \cdot a=a$