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The set of integers

Definition 6.2.7   The quotient set $ \mathbb{N}^2 / \sim$ is denoted by $ \mathbb{Z}$ and its elements are called integers.

As usual, we denote by $ \overline{(a,b)}$ the equivalence class of the pair $ (a,b)$, i.e.

$\displaystyle \overline{(a,b)}=\overline{(c,d)}$    in $\displaystyle \mathbb{Z}
 \Longleftrightarrow 
 (a,b) \sim (c,d)$    in $\displaystyle \mathbb{N}^2.$    

In this set, we can define an addition and a multiplication, as follows:

Definition 6.2.8   For any two elements $ \overline{(a,b)}$ and $ \overline{(c,d)}$ in $ \mathbb{Z}$:
  1. $ \overline{(a,b)}+\overline{(c,d)}=\overline{(a,b)+(c,d)}$;
  2. $ \overline{(a,b)} \cdot \overline{(c,d)}=\overline{(a,b)(c,d)}$.

In other words, ``the sum of two classes is the class of the sum'' and ``the product of two classes is the class of the product''. By Proposition 2.6, these operations are well defined, i.e. the result does not depend on the choice of the representants of the equivalence classes. The description we gave until now of the integres does not fit the intuitive notion that everybody knows. Let us define a mapping $ \Phi: \; \mathbb{N}\longrightarrow \mathbb{Z}$ as follows:

$\displaystyle \forall a \in \mathbb{N}, \Phi (a) = \overline{(a,0)}.$    

Proposition 6.2.9       
  1. The mapping $ \Phi$ is an injection.
  2. For any two natural numbers $ a$ and $ b$, $ \Phi (a+b)=\Phi (a) + \Phi (b)$.
  3. For any two natural numbers $ a$ and $ b$, $ \Phi (ab)=\Phi (a) \; \Phi (b)$.

Proof.     
  1. Let $ a$ and $ b$ be two natural numbers such that $ \Phi (a)=\Phi (b)$. We have:

    $\displaystyle \Phi (a)=\Phi (b) \Longleftrightarrow \overline{(a,0)}=\overline{(b,0)}
 \Longleftrightarrow (a,0)\sim (b,0)
 \Longleftrightarrow a=b.$    

  2. For any two natural numbers $ a$ and $ b$, we have:

    $\displaystyle \Phi (a) + \Phi (b)=\overline{(a,0)}+\overline{(b,0)}
 =\overline{(a+b,0)}=\Phi (a+b).$    

  3. For any two natural numbers $ a$ and $ b$, we have:

    $\displaystyle \Phi (a) \cdot \Phi (b)=\overline{(a,0)} \cdot \overline{(b,0)}
 =\overline{(ab,0)}=\Phi (ab).$    

$ \qedsymbol$

As a consequence, we can identify the set $ \mathbb{N}$ of the natural numbers with a subset of the set $ \mathbb{Z}$ of integers. Now we can determine a useful partition of $ \mathbb{Z}$. Take $ (a,b) \in \mathbb{N}^2$.

Definition 6.2.10   Let $ a$ be a non zero natural number. An integer of the form $ \overline{(a,0)}$ is called a positive integer; we will denote it by $ +a$ or simply by $ a$. An integer of the form $ \overline{(0,a)}$ is called a negative integer;we will denote it by $ -a$. The set of non negative integers is denoted by $ \mathbb{Z}^+$ and the set of non positive integers is denoted $ \mathbb{Z}^-$.

Thus we have:

\begin{displaymath}\begin{cases}
 \mathbb{Z}^+ \cup \mathbb{Z}^-=\mathbb{Z}\\  \mathbb{Z}^+ \cap \mathbb{Z}^-= \{ 0 \}
 \end{cases}\end{displaymath}    

With the identification determined by the mapping $ \Phi$, we can write $ \mathbb{Z}^+=\mathbb{N}$. The sign rules are given by the following proposition:

Proposition 6.2.11   For any two natural numbers $ a$ and $ b$ we have:

$\displaystyle \begin{matrix}
 (+a)(+b)=+(ab)\\  (+a)(-b)=-(ab)\\  (-a)(+b)=-(ab)\\  (-a)(-b)=+(ab)
 \end{matrix}.$    

Proof.

$\displaystyle \begin{matrix}
 (+a)(+b)=\overline{(a,0)}\overline{(b,0)}=\overli...
...(-b)=\overline{(0,a)}\overline{(0,b)}=\overline{(ab,0)}=+(ab)\\  
 \end{matrix}$    

$ \qedsymbol$

An additional property is described in the following proposition:

Proposition 6.2.12   Let $ x$ be any integer. Then there exists a unique integer $ y$ such that $ x+y=y+x=0$.

The integer $ y$ will be called the opposite of $ x$ and will be denoted by $ -x$.

Proof. Take $ x=\overline{(a,b)}$ for some pair $ (a,b) \in \mathbb{N}^2$ and $ y= \overline{(b,a)}$. The w have:

$\displaystyle x+y=\overline{(a,b)}+\overline{(b,a)}=\overline{(a+b,b+a)}
 =\overline{(0,0)}=0.$    

The other equation is obtained now by the commutative law of the addition in $ \mathbb{Z}$. Assume now that an integer $ x=\overline{(a,b)}$ has two opposites $ y_1=\overline{(b_1,a_1)}$ and $ y_2=\overline{(b_2,a_2)}$. Then we have:

$\displaystyle \overline{(a,b)}+\overline{(b_1,a_1)}$ $\displaystyle =\overline{(a,b)}+\overline{(b_2,a_2)}$    
$\displaystyle \overline{(a+b_1,b+a_1)}$ $\displaystyle = \overline{(a+b_2,b+a_2)}$    
$\displaystyle (a+b_1,b+a_1)$ $\displaystyle \sim (a+b_2,b+a_2)$    
$\displaystyle a+b_1+b+a_2$ $\displaystyle = b+a_1+a+b_2$    
$\displaystyle b_1+a_2$ $\displaystyle = a_1+b_2$    by Prop 1.15    
$\displaystyle (b_1,a_1)$ $\displaystyle \sim (b_2,a_2)$    
$\displaystyle \overline{(b_1,a_1)}$ $\displaystyle = \overline{(b_2,a_2)}$    
$\displaystyle y_1$ $\displaystyle = y_2.$    

$ \qedsymbol$


next up previous contents
Next: Ordering Up: The integers Previous: An equivalence relation on   Contents
root 2002-06-10