Constraction Of Number Systems - Free Math Worksheets

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 $(x + y) + 1 = x + (y + 1)$ :

$(x+y) + 1 \overset{\underset{\mathrm{(i)}}{}}{=} s(x +y) \overset{\underset{\mathrm{(ii)}}{}}{=} x + s(y) \overset{\underset{\mathrm{(i)}}{}}{=} x + (y +1).$

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

$\begin{aligned} ( x + y) + s(z) &= s ((x+y) + z) & \mbox{\{definition of addition (ii)\}} \\ &= s(x +( y + z)) & \mbox{\{by assumption\}} \\ & = x + s( y + z) & \mbox{\{definition of addition (ii)\}} \\ &= x + (y + s(z)) & \mbox{\{definition of addition (ii)\}}. \end{aligned}$

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) $1 \in M$ :

$\begin{aligned} x + 1 &= s(x) & \mbox{\{definition of addition (i) \}} \\ &= 1 + x & \mbox{\{Lemma 3.\}}. \end{aligned}$

Therefore, $1 \in M$ .

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

$\begin{aligned} x + s(y) &= s(x + y) & \mbox{\{definition of addition (ii) \}} \\ & = s( y + x) & \mbox{\{by assumption \}} \\ & = s(y) + x & \mbox{\{Lemma 2.\}}. \end{aligned}$

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 $B$ .