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If $(a,b) \sim (A,B)$ and $(c,d) \sim (C,D)$, where all pairs are whole numbers,

Prove that: $(a,b)*(c,d)\sim(A,B)*(C,D)$

Relation defined on the following:

$(a,b)\sim (A,B) \iff a+B=A+b$

$(c,d)\sim(C,D) \iff c+D=C+d$

For this I am assume we have to work a little backwards to show the proof, any tips on how to go about this.

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

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    $\begingroup$ You forgot to define the relation $\sim\ \ $ $\endgroup$ – Bill Dubuque Dec 6 '16 at 2:54
  • $\begingroup$ Could you define your symbols a bit more? $~$ is a general binary relation and it makes me nervous to interpret $*$ willy-nilly in that case. $\endgroup$ – Kitter Catter Dec 6 '16 at 2:55
  • $\begingroup$ @BillDubuque that is what I pointed out to my professor, but as he explained we can assume the old material we learned in the sense that (a,b)~(c,d)=(a+c, b+d) and (a,b)*(c,d)=(ac+bd, ad+bc) is all we covered in this class....So I am stuck trying prove the above question. Any idea on what I should do thanks! $\endgroup$ – user385853 Dec 6 '16 at 2:56
  • $\begingroup$ @KitterCatter * as in multiply on my exam question its a dot slightly higher then the middle, the assumption I made along with my class mates is that he meant multiply. $\endgroup$ – user385853 Dec 6 '16 at 2:59
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    $\begingroup$ You need to add that context to your question since it is impossible infer otherwise. $\endgroup$ – Bill Dubuque Dec 6 '16 at 3:05
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Notice:

$(a,b)~(A,B)⟹ a+B=b+A ⟹ B-A=b-a$ This ending will call **

$(c,d)~(C,D)⟹ c+D=d+C ⟹ c-d=C-D$ This ending will call **

Use ** so that we get: $(C-D)(B-A)=(c-d)(b-a)$

Foil:

$BC+AD-AC-BD=bc-ad-ac+ad$

Rearrange the terms to get: $ac+bd+AD+BC=ad+bc+AC+BD$

Which is $(ac+bd,ad+bc)~(AC+BD,AD+BC)$

$(a,b)*(c,d)~(A,B)(C,D)$

As we needed to prove.

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If you already know that $\sim$ is equivalence relation, then it suffices to show that $$(a,b) \sim (A,B) \implies (a,b)\ast(c,d) \sim (A,B)\ast(c,d). \tag{1}$$

Once you have this, you can get also $(a,b)\ast(c,d)\sim(A,B)\ast(c,d)$ using transitivity of $\sim$ and commutativity of $\ast$.


So to prove (1) we assume that $(a,b)\sim(A,B)$, i.e., that $a+B=A+b$. And we try to say something about $(ac+bd,ad+bc)$ and $(Ac+Bd,Ad+Bc)$.

We get $$(ac+bd) + (Ad+Bc) = (a+B)c+(A+b)d = (A+b)c+(a+B)d = (Ac+Bd)+(ad+bc)$$ which is exactly $$(ac+bd,ad+bc)\sim(Ac+Bd,Ad+Bc).$$


It is worth mentioning that this is equivalence relation is often used to construct integers from natural numbers. Which is a reason why it is better to avoid using subtraction in the proof - we assume that only operation which was defined so far.

We can embed cancellative commutative semigroup into a group using a similar approach.

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    $\begingroup$ Beat me to finishing it up. I like the second link thanks for a good read! $\endgroup$ – OLE Dec 9 '16 at 18:16
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You only need to prove $ac+bd-AC-BD=ad+bc-AD-BC$, so you only need to prove $a(c-d)+A(D-C) +b(d-c)+B(C-D)=0$, left is equal to $(a-b)(c-d)+(A-B)(D-C)$ Clearly it is zero, since $A-a=B-b, C-c=D-d$

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    $\begingroup$ mind explaining how you derive the first equation you have gotten. $\endgroup$ – OLE Dec 9 '16 at 2:59

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