Help proving inequality 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 $\delta(r) &lt; \frac12\delta(b)$? Any help appreciated. 

Thank you.

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

I don't know to get this.
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 $\frac12\delta(b) = \frac{11}{2}$.
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 $\frac12\delta(b) = \frac{11}{2}$.