Constructing the Integers from the Naturals

Question

I'm watching a video right now about the construction of the Integers from the Naturals.

The way to do so was to define the equivalence relation

$$(a,b)\text{ is equivalent to }(c,d)\text{ if }a+d=c+b$$

It was then said that "with a little algebra, this is equivalent to $a-b = c-d$ and thus the ordered pair $(a,b)$ represents the integer $a-b$.

Now, we are in the process of constructing the integers, which includes the negative naturals. My question is, how can we just say that $a+d=c+b$ means $a-b=c-d$ if we haven't constructed the negative naturals yet?

Answer 1

What we want is to construct the additive structure of $\mathbb{Z}$ from that of $\mathbb{N}$. We consider the set $Z=\{(m,n):m,n\in\mathbb{N}\}$ and define the binary operation (addition) by $(m_1,n_1)+_{Z}(m_2,n_2)=(m_1+_{\mathbb{N}}m_2,n_1+_{\mathbb{N}}n_2)$, where $+_{\mathbb{N}}$ is given by the additive structure of $\mathbb{N}$. With tedious work, we can show that $Z$ with $+_{Z}$ is just isomorphic to what we want, that is the additive structure of $\mathbb{Z}$. That "with some little algebra" means $(m,n)=(m,0)+_{Z}(0,n)$ as in the isomorphism from ${Z}$ to $\mathbb{Z}$, $(m,0)$ and $(0,n)$ are mapped to $m$ and $-n$ respectively. Hence, we see $(m,n)$ as "$m+(-n)$" or "$m-n$".