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I am just struggling a little bit understanding the brilliant answer found here, which is basically the proof that $\mathbb{Z}[i]$ is a euclidean domain. I do not quite understand (or I am not able to proove at the moment) the following section:

By a modified form of the division algorithm on the integers, $\exists q_1,q_2,r_1,r_2\in\Bbb Z$ such that $$ \begin{align}\alpha_1\beta_1+\alpha_2\beta_2&=N(b)q_1+r_1\\\alpha_1\beta_2-\alpha_2\beta_1&=N(b)q_2+r_2\end{align} $$ Where $-\frac12N(b)\le r_\ell\le\frac12N(b)$.

Where $a=\alpha_1+\alpha_2 i, b=\beta_1+\beta_2i$, $\alpha_1,\alpha_2,\beta_1,\beta_2\in\Bbb Z$ and $N$ is defined by $N(z) := z\overline{z}$ for $z \in \mathbb{C}$. I mean the $\frac{1}{2}$ makes sense , because any real number can be approximated by an integer to this precision. But somehow I do not quite see the formal way.

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TheGeekGreek
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Choose $q_1$ as the best integer approximation of $\frac{\alpha_1\beta_1+\alpha_2\beta_2}{N(b)}$. Choose $r_1 = \alpha_1\beta_1+\alpha_2\beta_2 - N(b)q_1$. Then $$\left|\frac{\alpha_1\beta_1+\alpha_2\beta_2}{N(b)}-q_1\right|\leq \frac{1}{2}$$ This can be multiplied by $N(b)$, because $N(b)$ is positive and real. You get $$\left|\alpha_1\beta_1+\alpha_2\beta_2-N(b)q_1\right|\leq \frac{1}{2}N(b)$$ Replacing $\alpha_1\beta_1+\alpha_2\beta_2 - N(b)q_1$ with $r_1$ results in $$\left|r_1\right|\leq \frac{1}{2}N(b)$$ which is equivalent with $$-\frac{1}{2}N(b)\leq r_1 \leq \frac{1}{2}N(b)$$

Reinhard Meier
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