How to prove the following statement about whole numbers using its definition?

Question

I would like to prove the following statement:

If $(a,b)$ ~ $(A,B)$ and $(c,d)$ ~ $(C,D)$, where all pairs are pairs of whole numbers, prove that $(a,b)*(c,d)$ ~ $(A,B)*(C,D)$.

Note: Whole numbers are defined as the following:

$(a,b)$ ~ $(c,d)$ iff $a+d=b+c$

$(a,b) + (c,d) = (a+c,b+d)$

$(a,b)*(c,d) = (ac+bd,ad+bc)$

My half-finished proof:

Given $(a,b)$ ~ $(A,B)$ and this means that $a+B = b+A$

Given $(c,d)$ ~ $(C,D)$ and this means that $c+D = d+C$

Prove $(a,b)*(c,d)$ ~ $(A,B)*(C,D)$ and this means that $(ac+bd,ad+bc)$ ~ $(AC+BD,AD+BC)$ which according to the definition is $((ac+bd)+(AD+BC))=((ad+bc)+(AC+BD))$

I do not know where to go from here. please help me!

Answer 1

First note that by transitivity we only need to show that $$(a,b)(c,d)=(A,B)(c,d)$$ as we can go $$(a,b)(c,d)=(A,B)(c,d)=(A,B)(C,D)$$ So we have then that $$ac+bd+Bc+Ad=(a+B)c+(b+A)d=(A+b)c+(B+a)d=Ac+bc+Bd+ad$$ at which the equality follows.

Answer 2

Given that $a + B = b + A$, you have that $b-a=B-A$.

Likewise, from $c+D=d+C$, we know that $d-c=D-C$.

So, $(d-c)(b-a)= (D-C)(B-A)$

So, $bd-bc-ad+ac=BD-BC-AD+AC$

And so, $ac + bd+AD+BC=ad+bc+AC+BD$