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$A=\mathbb{Z} \times \mathbb{Z}$ then for every $(a,b),(c,d) \in A$ there exists $(p,q),(r,s) \in A$ such that $$(a,b)= ((p,q)*(c,d))+(r,s) $$ with $r^2+s^2<c^2+d^2$

The operation $*$ and $+$ are defined as $$(a,b)+(c,d)=(a+c,b+d)$$ $$(a,b)*(c,d)=(ac-bd,bc+ad)$$

I think induction might be a way to prove the equality. But I am not sure. The hint given in the book also says to use induction but I am not sure how to use induction here as we are working on ordered pairs.

Jalil Ahmad
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    Exact duplicate of Prove that the Gaussian Integer's ring is a Euclidean domain – Parcly Taxel Jan 15 '21 at 01:52
  • @ParclyTaxel It might be but I am not familiar with the topics mentioned there. Also they are not part of my subject. – Jalil Ahmad Jan 15 '21 at 01:55
  • The hint that is given in the book also says to use induction. But I am stuck there. – Jalil Ahmad Jan 15 '21 at 01:56
  • You say induction might be a way to show existence. It's not. The duplicate is correct. – Parcly Taxel Jan 15 '21 at 02:00
  • Note that the statement is false for $c=d=0$. – Gerry Myerson Jan 15 '21 at 02:04
  • @GerryMyerson you might have missed the condition $r^2+s^2<c^2+d^2$ – Jalil Ahmad Jan 15 '21 at 02:07
  • Not at all. You're supposed to show $r,s$ exist with $r^2+s^2<c^2+d^2$. If $c=d=0$, then you're supposed to show $r,s$ exist with $r^2+s^2<0$. So what you are supposed to show is, in fact, false. – Gerry Myerson Jan 15 '21 at 02:10
  • @GerryMyerson I have checked again for any typos but this is exactly how it is asked in the book. Also it says to use induction in the book as I mentioned in the comments above. – Jalil Ahmad Jan 15 '21 at 02:14
  • I don't care what it says in the book; can you not see that it's wrong? They do make mistakes in books, you know. Fortunately, what's wrong here can be fixed. All you have to do is include in the hypotheses that $(c,d)$ is not $(0,0)$. As for induction, I would suggest that the first step is to calculate and simplify $(ac-bd)^2+(ad+bc)^2$. – Gerry Myerson Jan 15 '21 at 02:29
  • @GerryMyerson Got you point but could you please explain why do we have to calculate $(ac-bd)^2+(ad+bc)^2$ – Jalil Ahmad Jan 15 '21 at 02:39
  • You don't have to do anything. But when you come here to ask a math question, and someone who has been doing math for 45 years suggests you do a simple calculation, maybe you should carry out the suggestion first, and ask questions later. – Gerry Myerson Jan 15 '21 at 06:00
  • @GerryMyerson Sir I have tried and the result is $(a^2+b^2)(c^2+d^2)$ so what is the next step. I am not saying that what you are suggesting is not correct but actually I asked you because I did not understand logic behind the reason for this calculation hence could not continue. – Jalil Ahmad Jan 15 '21 at 06:42
  • OK. Now, you kow that given integers $a,b$, $b\ne0$, you can divide $a$ by $b$ and get an integer quotient $q$ and an integer remainder $r$ with $0\le r<|b|$, right? The remainder $r$ is less than the (absolute value of the) divisor $b$, because if it isn't then you've picked the wrong quotient $q$ – you could change $q$ and get a smaller remainder. So, this question is very similar. In effect, you are dividing $(a,b)$ by $(c,d)$ and getting a quotient $(p,q)$ and a remainder $(r,s)$, and you're trying to prove you can make the "absolute value" of $(r,s)$, which is $r^2+s^2$ (continued) – Gerry Myerson Jan 15 '21 at 11:36
  • (continued) less than the "absolute value" of the divisor $(c,d)$, which is $c^2+d^2$. Think about how the argument goes for ordinary division, and try to tweak it to go for this new kind of division. Notice that the $(a^2+b^2)(c^2+d^2)$ calculation shows that this really does act like an absolute value; the absolute value of the product is the product of the absolute values (just like $|xy|=|x||y|$). – Gerry Myerson Jan 15 '21 at 11:40
  • @GerryMyerson What I understand from you comments is to use the absolute values to show that the equation $$(a,b)= ((p,q)*(c,d))+(r,s) $$ is true for the condition $r^2+s^2<c^2+d^2$ – Jalil Ahmad Jan 15 '21 at 23:43
  • @GerryMyerson If we take absolute value on both sides then this gives $$|(a,b)|= |((p,q)(c,d))+(r,s)| \leq |((p,q)(c,d))|+|(r,s)| = (|(p,q)|*|(c,d)|)+|(r,s)| $$ which give us $a^2+b^2\leq (p^2+q^2)(c^2+d^2)+(r^2+s^2)$ and now using this result I somehow have to get to the condition $r^2+s^2 < c^2+d^2$ – Jalil Ahmad Jan 15 '21 at 23:45
  • You want to show that you can pick $p,q$ in such a way that $r^2+s^2<c^2+d^2$. So you want to show that if that inequality doesn't hold, then you can change $p,q$ to make $r^2+s&^2$ smaller. – Gerry Myerson Jan 16 '21 at 02:08
  • Typo there, sorry, should end "to make $r^2+s^2$ smaller." – Gerry Myerson Jan 17 '21 at 00:14

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