next up previous contents
Next: The set of integers Up: The integers Previous: The integers   Contents


An equivalence relation on $ \mathbb{N}^2$

In the set $ \mathbb{N}^2$ of ordered pairs of natural numbers we define binary relation:

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

Proposition 6.2.1   The relation $ \sim$ is an equivalence relation in $ \mathbb{N}^2$.

Proof.      $ \qedsymbol$

On $ \mathbb{N}^2$ we define two operations, called addition and multiplication, by the following equations:

$\displaystyle \forall (a,b) \in \mathbb{N}^2, \; \forall (c,d) \in \mathbb{N}^2, \;$ $\displaystyle (a,b)+(c,d)=(a+c,b+d).$    
$\displaystyle \quad$ $\displaystyle (a,b)\cdot (c,d)= (ac+bd, ad+bc).$    

For example, $ (2,3)+(7,1)=(9,4)$ and $ (2,3) \cdot (7,1)= (17, 23)$. The multipication can be also denoted without any symbol, like $ (a,b)(c,d)$. Propositions 2.2, 2.3, 2.4 and 2.5 are very easy to prove; we leave the task to the reader.

Proposition 6.2.2 (Commutative law)   For any two elements $ (a,b)$ and $ (c,d)$ in $ \mathbb{N}^2$, the following hold:
(i)
$ (a,b)+(c,d)=(c,d)+(a,b)$.
(ii)
$ (a,b)(c,d)=(c,d)(a,b)$.

Proposition 6.2.3 (Associative law)   For any three elements $ (a,b)$,$ (c,d)$ and $ (e,f)$ in $ \mathbb{N}^2$, the following hold:
(i)
$ [(a,b)+(c,d)]+(e,f)=(a,b)+[(c,d)+(e,f)]$.
(ii)
$ [(a,b)(c,d)](e,f)=(a,b)[(c,d)(e,f)]$.

Proposition 6.2.4 (Neutral element)   For any $ (a,b) \in \mathbb{N}^2$ the following hold:
(i)
$ (a,b)+(0,0)=(0,0)+(a,b)=(a,b)$.
(ii)
$ (a,b)(1,0)=(1,0)(a,b)=(a,b)$.

Proposition 6.2.5 (Distributive law)   For any three elements $ (a,b)$,$ (c,d)$ and $ (e,f)$ in $ \mathbb{N}^2$, the following hold:
(i)
$ [(a,b)+(c,d)](e,f)=(a,b)(e,f)+(c,d)(e,f)$.
(ii)
$ (a,b)[(c,d)+(e,f)]=(a,b)(c,d)+(a,b)(e,f)$.

In Proposition 2.5 each equation is a consequence of the other beacuse of Proposition 2.2. The addition and the multiplication in $ \mathbb{N}^2$ are compatible with the equivalence relation:

Proposition 6.2.6   For any four elements $ (a,b)$, $ (a',b')$, $ (c,d)$ and $ (c',d')$ in $ \mathbb{N}^2$, the following properties hold:
  1. \begin{displaymath}\begin{cases}
(a,b) \sim (a',b') \\  (c,d) \sim (c',d')
\end{cases}
\Longrightarrow
(a,b)+(c,d) \sim (a',b')+(c',d').\end{displaymath}
  2. \begin{displaymath}\begin{cases}
(a,b) \sim (a',b') \\  (c,d) \sim (c',d')
\end{cases}
\Longrightarrow
(a,b)(c,d) \sim (a',b')(c',d').\end{displaymath}

Proof. We prove the first property; we have:

\begin{displaymath}\begin{cases}
 (a,b) \sim (a',b') \\  (c,d) \sim (c',d')
 \en...
...ghtarrow 
 \begin{cases}
 a+b'=a'+b \\  c+d'=c'+d
 \end{cases}.\end{displaymath}    

By addition we have $ a+b'+c+d'=a'+b+c'+d$, i.e. $ (a+c)+(b'+d')=(a'+b')+(c+d)$ and this equivalent to $ (a,b)+(c,d) \sim (a',b')+(c',d').$ Because of the commutativity of the multiplication, we can show only the following property instead of the second equation above:

$\displaystyle (a,b) \sim (a',b') \Longrightarrow (a,b)(c,d) \sim (a',b')(c,d).$    

We leave the task to the reader. $ \qedsymbol$


next up previous contents
Next: The set of integers Up: The integers Previous: The integers   Contents
root 2002-06-10