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Construction of number systems

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.

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

A\oplusB=[(a+c, b+c)]

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

A\simB=[(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

 

 

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