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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!

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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.

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

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  • $\begingroup$ But I do not have the definition for subtraction. $\endgroup$ – socrates Dec 10 '16 at 3:10
  • $\begingroup$ @socrates I thought you said you were working with whole numbers? So it is just subtraction for whole numbers, just as the other operations are defined in terms of addition and multiplication of whole numbers. $\endgroup$ – Bram28 Dec 10 '16 at 3:12
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    $\begingroup$ I think he means natural numbers and is trying to construct the integers. $\endgroup$ – Zelos Malum Dec 10 '16 at 4:22
  • $\begingroup$ @ZelosMalum Oh! OK, that makes sense. I took 'whole numbers' to be integers. OK, then your approach is much better! $\endgroup$ – Bram28 Dec 10 '16 at 12:43

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