In my syllabus for "Scientific Computing" it states the following idea.
Consider the case $a \leq b$, $a,b \in \mathbb{N}$. This is essentially the same as $a + x = b$, $x \in \mathbb{N}$. Therefore, the only solutions to this equation are when $a$ is less than or equal to $b$. However, when we require for there to be a solution we need to introduce the idea of pairs, i.e. $(a,b)$ that describes $x$. However, it is clear to see that $(a,b)$ is not the only way to represent $x$. We also have $(a+1,b+1)$ which describes the same number. Therefore, we introduce equivalence relations that comes directly from the equation $a+x=b$: $(a,b) \sim (c,d) \iff a+d=b+c$.
I'm just confused as to what this even means. How does $(a,b)$ describe $x$? If we have $1 \leq 5$, then the equation becomes $1+x=5 \Rightarrow x=4$. Then our equivalence relation becomes $(1,5)$ - so essentially the relation describes $x=b-a$? But then how does $(c,d)$ come into play?
From the Peano axioms you have constructed ${\mathbb N}=\{0,1,2,3,\ldots\}$ with $+$, $\times$, and $<\,$; and you have proven the usual rules of computing in this system. Now you want to enlarge ${\mathbb N}$ to the system ${\mathbb Z}$, which also contains the negative numbers, whereby the known rules are still valid.
A simple idea would be to create for each $n\in{\mathbb N}_{>0}$ its negative clone $-n$, and put these clones into the bag. But it has turned out that proving all the rules now is a hopeless undertaking, because of all the "cases". Instead one uses a certain algebraic trick, and your quoted text tries to indulge in this trick.
The idea is to represent all integer numbers, positive, $0$, and negative ones, by pairs of natural numbers. A pair $(a,b)\in{\mathbb N}^2$ then represents the solution $x$ to the equation $a+x=b$, i.e., the number that we usually call $b-a$. When $a\leq b$ then $b-a\in{\mathbb N}$ is already present in our world, and we would not need this pair-complication. But when $a> b$ then the pair $(a,b)$ is a new kind of thing, representing $a-b>0$ "unit steps backwards".
Introducing these pairs creates a new difficulty: The same "new number" $x$, being $\geq0$ or $<0$, is represented by infinitely many pairs, e.g., $3=(2,5)=(16,19)$. This means that an actual "new number" is an equivalence class of such pairs. Some work is then needed to install good operations $+$ and $\times$ on the set ${\mathbb Z}$ of these classes, whereby these operations have to give the same result as the already present operations on the "old numbers" $\geq0$.