2
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3. $\delta$ is the standard Euclidean valuation on $\mathbb{Z}[i]$. For each of the following pairs $a,b \in \mathbb{Z}[i]$, find $q,r \in \mathbb{Z}[i]$ such that $a = qb + r$, where either $r = 0$ or $\delta(r) \leq \frac12\delta(b)$. Verify that $r$ satisfies this condition.

$a = 5$, $b= (2+3i)$

$\dfrac{5}{2+3i} = \dfrac{10}{13} - \dfrac{15}{13i}$

Take $q = 1-i$. Then $5 = (2+3i)(1-i)-1$, so $r = -i$.

How does one prove $\delta(r) < \frac12\delta(b)$? Any help appreciated.

Thank you.

Also, the solution in my book is $\delta(r) = 1 < \frac{11}{2} = \frac12\delta(b)$.

I don't know to get this.

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  • $\begingroup$ Change (correct) -1 for - i in 5 = (...... $\endgroup$ – Piquito Jun 6 '15 at 15:03
  • $\begingroup$ Change also (the ERRATA of the book) 11 by 13 $\endgroup$ – Piquito Jun 6 '15 at 15:07
-1
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I think that the standard Euclidean valuation on $\mathbb{Z}[i]$ is just defined as $\delta(x+yi) = x^2 + y^2$. If this is true, then

$\delta(r) = \delta(-i) = (-1)^2 = 1$

and

$\delta(b) = \delta(2 + 3i) = 2^2 + 3^2 = 13$, so $\frac12\delta(b) = \frac{13}{2}$.

So I'm not sure why the book says $\frac12\delta(b) = \frac{11}{2}$.

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  • $\begingroup$ so maybe the answer in my book is wrong? $\endgroup$ – italy Jun 6 '15 at 14:43
  • $\begingroup$ That is my suspicion. The good news is that $\delta(r) < \frac12\delta(b)$ either way. :) $\endgroup$ – Ken Jun 6 '15 at 14:53
  • $\begingroup$ Okay, thank you for your help. $\endgroup$ – italy Jun 6 '15 at 14:56
  • $\begingroup$ Oh, looks like there's an errata of the book; $\frac12\delta(b)$ is supposed to be $\frac{13}{2}$. $\endgroup$ – Ken Jun 6 '15 at 15:24

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