TITLE: I don't understand this definition of the integers.
QUESTION [6 upvotes]: Definition: The set of integers  is $\mathbb{Z}:=\frac{\mathbb{N} \times \mathbb{N}}{R}=\{[(m,n)]:(m,n)\in \mathbb{N}\times \mathbb{N}\}$.
I understand this is the set of all the equivalence classes of $\mathbb{N}\times \mathbb{N}$ using the equivalence relation $R$, but who are $m$ and $n$? place holders for what?
Introducing, the relation of equivalence:
$(m,n)R(p,q)\Leftrightarrow m + q = n + p.$
I didn't see this before!

REPLY [3 votes]: Formally you can't define a subtraction on $\mathbb{N}$, since an expression like $2-3$ wouldn't make any sense in $\mathbb{N}$. Nevertheless  you can still define the subtraction implicitly via the addition by rewriting the relation
$$b-x = a \Leftrightarrow a+x = b.$$
With the equation on the right-hand side $x$ is uniquely determined by the pair $(a,b) \in \mathbb{N} \times \mathbb{N}$, but choosing a different pair $(a',b')$ could result in the same $x$ as before, for example for $(a',b')=(a+1,b+1)$ we get
$$a+x = b \Leftrightarrow (a+1)+x = (b+1).$$
You now want to identify all pairs $(a,b)$, which define the same $x$, so your goal is 
$$(a,b) \sim (a',b') :\Leftrightarrow (a,b) \text{ and } (a',b') \text{ define the same } x.$$
This is exactly done by the equivalence relation you mentioned above and the set of all solutions to the problem $a+x=b$ for $a,b \in \mathbb{N}$ is obviously $\mathbb{Z}$.