Why is $ℤ$ defined as $ℤ := ℕ×ℕ \text{ / ~} $, where $ \text{~} := \{ [ (a,b),(c,d) ] | a+d = b+c \} $ 
Why is $ℤ$ defined as $ℤ := ℕ×ℕ \text{ / ~} $, where $ \text{~} := \{ [ (a,b),(c,d) ] | a+d = b+c \} $

We already defined $ℕ$ in class as $ ℕ:=∩ \{ I | I \text { is an inductive set} \} $; we also defined $+,·,<$ etc, and proved their properties.
Now, that we are defining $ℤ$ as mentioned in the title, I have a little trouble with this definition since our teacher is one of the oldest in our department, and sometimes forgets to write things down in his class notes; for example He didn't say what "/" means in $ℕ×ℕ \text{ / ~}$; I know $A$ \ $B$ is the old, well-know, difference of two sets, so I kinda feel like $ℕ×ℕ \text{ / ~}$ is just ~\ $ℕ×ℕ$, but even if that's the case, I don't know how possibly that would be the set of integers, let alone express $-4, 5, 2, -6$.
Could you tell me what \ stands for, why is $ℤ$ defined like this, and how to express a number with this definition, just like I would define $2$ as $\{∅,\{∅\}\}$ in $ℕ$.
Thanks in advance
 A: This is a way of constructing $\mathbb{Z}$ in terms of $\mathbb{N}$ (and set theory).
As to the meaning of $/$ in this context, it represents the “quotient modulo the equivalence relation $\sim$.”
Recall that if $X$ is a set and $\sim$ is an equivalence relation on $X$, then $\sim$ induces a partition of $X$: that is, a collection of subsets $\{B_i\}_{i\in I}$ such that:


*

*$B_i\subseteq X$ for each $i$;

*If $B_i\cap B_j\neq\varnothing$, then $B_i=B_j$ (or you can require that $B_i\cap B_j=\varnothing$ if $i\neq j$ instead).

*$\cup_{i\in I} B_i = X$.


Namely, you let $B_i$ be the equivalence classes of the equivalence relation $X$: for each $x\in X$, you can let $B_x = \{y\in X\mid y\sim x\}$ and these form a partition. The set of equivalence classes is denoted $X/\sim$.
(You probably saw a “Fundamental Theorem of Equivalence Relations”, that shows that an equivalence relation on $X$ is “the same” as a partition on $X$, meaning that a partition induces an equivalence relation, an equivalence relation induces a partition, and that these associations are inverses of each other).
An element of $\mathbb{Z}$ is then a collection of pairs of natural numbers: $[(a,b)] = \{(n,m)\in\mathbb{N}\times\mathbb{N} \mid (n,m)\sim(a,b)\}$.
So for example, one element of $\mathbb{Z}$ consists of all elements that are equivalent to $(1,0)$, namely
$$[(1,0)] = \{(1,0), (2,1), (3,2), (4,3), (5,4), \ldots, (n+1,n),\ldots\}.$$
To get an intuitive idea of what you are doing here, consider the set of equations of the form
$$a = x+b, \quad a,b\in\mathbb{N}.$$
Some of these equations have solutions in $\mathbb{N}$ (when $a\geq b$; or if your natural numbers do not include $0$, then when $a\gt b$). But some do not. We want to “extend” $\mathbb{N}$ to a larger set in which all these equations have solutions. So we identify each equation $a=x+b$ with the pair $(a,b)$, and we think of the pair $(a,b)$ as being “the solution to $a=x+b$.” However, different pairs yield the same solution: for example, $(a,b)$ is the same solutions as $(a+1,b+1)$, since if $a=x+b$, then $a+1 = x +(b+1)$. And more generally, $(a,b)$ represents the same solution as $(c,d)$ if and only if $a+d=b+c$. So you define an equivalence relation in which two pairs are “equivalent” precisely when they correspond to the same solution. And then you take the set of equivalence classes to get the larger set that includes all solutions to all of these equations. 
Some further work then establishes that this construction gives you something that essentially behaves like the $\mathbb{Z}$ we think we know and sometimes love, and so we declare this to be $\mathbb{Z}$.
CAVEAT: I don’t know how your book/notes motivate the construction. Instead of having $(a,b)$ “represent” the equation $a=x+b$, it’s possible that it is thought to represent the equation $a+x=b$; the idea is essentially the same, but you will eventually show that you have a “copy” of $\mathbb{N}$ sitting inside $\mathbb{Z}$, and whatever your motivation is will determine whether this is accomplished by sending $n$ to the class of $(n,0)$ or to the class of $(0,n)$. The construction is the same, just slightly different details.
A: If $A$ is a set and $\sim $ is an equivalence relation on $A$, then $A/{\sim}$ denotes the set of equivalence classes. 
Here $A=\Bbb N\times \Bbb N$ and one of the elements of $A/{\sim}$ would be $\{(0,2),(1,3),(2,4),(3,5),\ldots\}$, another $\{(3,0),(4,1),(5,2),\ldots\}$, and you will define addition in such a way that their sum is the equivalence class $\{(1,0),(2,1),(3,2),\ldots\}$. The additive inverse of $\{\ldots, (a,b),\ldots\}$ will turn out to be $\{\ldots,(b,a),\ldots\}$.
