Show that, for the following definition of $\mathbb{Z}$ and $\cdot$, that $\cdot$ is well-defined. I'm reading Real Numbers and Real Analysis by Ethan D. Bloch (2011), and in the textbook one definition for $\mathbb{Z}$ is as the set of equivalence classes of $\mathbb{N} \times \mathbb{N}$ under the relation $(a,b) \text{~} (c,d)$ iff $a + d = b + c$. Then, $[(a,b)] \cdot [(c,d)] = [(ac + bd, ad + bc)]$. The exercise is to prove that $\cdot$ is well-defined. You may assume that the following is true:


*

*Peano's Postulates

*$\mathbb{N}$ is an abelian semigroup under addition

*$\mathbb{N}$ is an abelian monoid under multiplication

*Multiplication in $\mathbb{N}$ is distributive.

*$\mathbb{N}$ is a well-ordered set under the ordering $a \leq b$ if and only if $\exists p \in \mathbb{N}$ such that $a + p = b$ or $a = b$


Also, the exercise gives the following hint:

Use the fact that if $a, b \in \mathbb{N}$, then $a + a = b + b$ if and only if $a = b$

What I've worked out so far:

To show that $\cdot$ is well-defined, we must show that, regardless of the representation of the equivalence classes, the resulting equivalence class from the binary operation is the same. In other words, suppose $[(a,b)] = [(e,f)]$ and $[(c,d)] = [(g,h)]$. Then show that
  $[(a,b)] \cdot [(c,d)] = [(e,f)] \cdot [(g,h)]$.
We also know that for $x$ and $y$ which are elements of a set, $[x] = [y]$ if and only if $ x \text{ ~ } y$. Therefore $[(a,b)] = [(e,f)] \implies (a,b) \text{ ~ } (e,f) \implies a + f = b + e$, and similarily
  $c + h = d + g$
To show that $\cdot$ is well-defined, we must show that
  $(ac + bd, ad + bc) \text{~} (eg + fh, eh + fg)$. Therefore, we must
  show that $(ac + bd) + (eh + fg) = (ad + bc) + (eg + fh)$

Any help would be appreciated.
 A: I came up with an answer, although I'll leave the question open because I'd like to see an answer which uses the hint provided. The motivation of this answer was backwards work, and the cancellation property.
We want to show that
$(ac + bd) + (eh + fg) = (ad + bc) + (eg + fh)$.
Well, since this won't hurt, let's add try adding $ah$ on both sides and see what happens. By the distributive property, we know that $ac + ah = a(c+h)$, and that's one of the sums seen in our equations, so it's neat to see it, right?
$(a(c+h) + bd) + (eh + fg) = (ad + bc) + (eg + ah + fh)$
$= (ad + bc) + (eg + h(a + f))$ 
Oh hey, that's neat! $a + f$ is another one of our summands. But we don't have a multiple of $h$ on the left side. Hmm. $a + f = b + e$, so maybe if we add $hb$ to both expressions, something else will happen!
The resulting expression on the left of the equality is now $(a(c + h) + bd) + (h(b + e) + fg)$ The resulting express on the right is now $(ad + b(c + h)) + (eg + h(a+f))$
Oh cool, $c + h$ again! Since $c + h = d + g$ ... we could try adding $ag$ to both sides of the equality!
$L.H.S = (a(c + h) + bd) + (h(b + e) + g(f + a))$
$R.H.S. =(a(d + g) + b(c + h)) + (eg + h(a + f))$
Ok. let's add $bg$ to both sides and I think we'll be done.
$L.H.S = (a(c + h) + b(d + g)) + (h(b + e) + g(f + a))$
$R.H.S. =(a(d + g) + b(c + h)) + (h(a + f) + g(b + e))$
Arranged like this, it is very easy to see that $L.H.S = R. H.S$.
Now, just read this thought process backwards (which works since $a + c = b + c$ if and only if $a = b$ for $a, b, c \in \mathbb{N}$), fill in a few details, and we're done.
