The relation $(a,b)S(c,d) \iff a-b=c-d$ such that $(a,b),(c,d) \in \mathbb{N} \times \mathbb{N}$
Need to find equivalent set for ($6,6)$ and $(2,5)$
I found for $[(6,6)] = \{(c,d) \in\mathbb{N} \times \mathbb N \mid c=d \}$
and for $[(2,5)] = \{(c,d) \in\mathbb{N} \times \mathbb N \mid c=d+3 \}$
I think I'm correct so far.. now I need to find the collection of equivalence classes $\mathbb{N} \times \mathbb{N} /S$
I found that the relation condition works when for example in $(2,5)$ its equivalence set all the difference between $5$ and $2$ that for every $c = d+3$ belongs, $(4,7), (10,13)$ i don't know how to put that perspective in mathematic words
First, a given pair $(a,b)$ is equivalent to $(0,b-a)$ if $a\ge b$ and to $(a-b,0)$ if $a>b$, so the pairs with $a$ or $b$ zero form a full representative set.
The equivalence class of $(0,n)$ is $\{(a,a+n):a\in\Bbb N\}$, and similarly the equivalence class of $(n,0)$ is $\{(a+n,a):a\in\Bbb N\}$.
Note that this is the construction of the set of integer numbers $\Bbb Z$ based on the natural numbers, where a pair $(a,b)$ represents the difference $b-a$, so that $[(0,n)]$ represents the natural $n$ and $[(n,0)]$ represents the negative number $-n$ if $n>0$.