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I'm using following definitions: On the set $\mathbb{N}\times\mathbb{N}$ we define an equivalence relation $\sim$ for all $(a,b)\in\mathbb{N}\times\mathbb{N}$ and $(c,d)\in\mathbb{N}\times\mathbb{N}$ as \begin{align*}\tag{I} (a,b)\sim (c,d):\Leftrightarrow a+d=b+c \end{align*} The set of integers $\mathbb{Z}$ is then defined as \begin{align*} \mathbb{Z}:=\{[(a,b)]_{\sim}\mid (a,b)\in\mathbb{N}\times\mathbb{N}\}, \end{align*} where $[(a,b)]_{\sim}$ denotes the equivalence class of $(a,b)$, which is defined as \begin{align*}\tag{II} [(a,b)]_{\sim}=\{(c,d)\in\mathbb{N}\times\mathbb{N}\mid (a,b)\sim (c,d)\}. \end{align*}

For convenience I will write $[(a,b)]$ instead of $[(a,b)]_{\sim}$.

Addition $+$ and multiplication $\cdot$ on the set $\mathbb{Z}$ are defined as \begin{align*}\tag{III} [(a,b)]+[(c,d)]:=[(a+c,b+d)] \end{align*} and \begin{align*}\tag{IV} [(a,b)]\cdot [(c,d)]:=[(ac+bd,ad+bc)]. \end{align*}

I've already proved that addition on $\mathbb{Z}$ is well-defined. However, I'm struggling to do the same for multiplication.

Let \begin{align*} [(a,b)]&=[(x,y)] \\ [(c,d)]&=[(z,w)]. \end{align*}

Then we have \begin{align*}\tag{Va} (a,b)\sim(x,y)&:\Leftrightarrow a+y=b+x \\ \tag{Vb} (c,d)\sim(z,w)&:\Leftrightarrow c+w=d+z. \end{align*}

We have to show that \begin{align*} [(a,b)]\cdot [(c,d)]=[(x,y)]\cdot [(z,w)] \end{align*} holds. Using $(IV)$ this turns into \begin{align*} [(ac+bd,ad+bc)]=[(xz+yw,xw+yz)] \end{align*} and by $(II)$ we then have \begin{align*} (ac+bd,ad+bc)\sim (xz+yw,xw+yz) \end{align*} Using $(I)$ this is equivalent to \begin{align*}\tag{VI} (ac+bd)+(xw+yz)=(ad+bc)+(xz+yw). \end{align*}

So I have to find a way to get from equations $(Va)$ and $(Vb)$ to $(VI)$.

PAT
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Remember, $x+y=x'+y$ is equivalent to $x=x'$ for natural numbers $x$, $x'\!$, and $y$. Thus, we have: \begin{align*} (ac+bd)+(xw+yz)&=(ad+bc)+(xz+yw)\\ \iff(ac+bd)+(xw+yz)+aw&=(ad+bc)+(xz+yw)+aw\\ \iff (ad+bd)+(xw+yz)+az&=(ad+bc)+(xz+bw)+xw\\ \iff bd+yz+az&=bc+xz+bw\\ \iff bd+xz+bz&=bd+xz+bz. \end{align*}

Here, the equivalences are:

  1. Adding $aw$.
  2. LHS: (Va) and distributivity, which gives $ac+aw=ad+az$. RHS: (Vb) and distributivity, which gives $aw+yw=bw+xw$.
  3. Cancelling $ad$ and $xw$.
  4. LHS: (Va) and distributy, which gives $az+yz=bz+xz$. RHS: (Vb) and distributivity, which gives $bc+bw=bd+bz$.
Kenta S
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  • How could you possibly come up with this acrobatic act? Anyways, the important thing is, it actually works out just fine, thanks. – PAT Aug 07 '21 at 15:27
  • Another approach is to fix $(a,b)$ and let $(c,d)\sim(z,w)$, then prove $[(a,b)]\cdot[(c,d)]=[(a,b)]\cdot[(z,w)]$. Then, do the same thing for the LHS. (i.e., just let each component vary individually.) This simplifies all the calculations. – Kenta S Aug 07 '21 at 15:31
  • You mean prove $[(a,b)]\cdot[(c,d)]=[(a,b)]\cdot[(z,w)]$ and then $[(a,b)]\cdot[(z,w)]=[(x,y)]\cdot[(z,w)]$? – PAT Aug 07 '21 at 15:38
  • Yes, that's exactly what I mean. (You could also start by observing that this multiplication operation commutes, so you only need to check one of those identities.) – Kenta S Aug 07 '21 at 15:41
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    sounds interesting.. and hopefully a bit more straightforward. Will definitely try to work this idea out. – PAT Aug 07 '21 at 15:50