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  "body": "\n## Proof\n\nFor a complex number $z = u + vi$, write $\\abs{z}^2 = u^2 + v^2$. Recall $\\abs{zw}^2 = \\abs{z}^2 \\abs{w}^2$, and $\\abs{\\gamma}^2 = N(\\gamma)$ for a [[Gaussian integer|0B268GVX95]] $\\gamma$, where $N$ is the [[norm|00HPT88ZX0]].\n\n1. Since $\\beta \\ne 0$, $N(\\beta) = \\beta\\bar\\beta$ is a positive integer. So $\\dfrac{\\alpha}{\\beta} = \\dfrac{\\alpha \\bar\\beta}{N(\\beta)} = s + ti$ with rational numbers $s$ and $t$.\n2. Choose integers $m$ and $n$ with $\\abs{s - m} \\le \\frac{1}{2}$ and $\\abs{t - n} \\le \\frac{1}{2}$, for example the nearest integers to $s$ and $t$.\n3. Put $q = m + ni$ and $r = \\alpha - q\\beta$. Both are Gaussian integers, since $\\mathbb{Z}[i]$ is closed under sums and products. And $\\alpha = q\\beta + r$.\n4. We have $r = \\beta \\left(\\dfrac{\\alpha}{\\beta} - q\\right) = \\beta \\big((s - m) + (t - n)i\\big)$.\n5. So $N(r) = N(\\beta) \\big((s - m)^2 + (t - n)^2\\big) \\le N(\\beta) \\left(\\frac{1}{4} + \\frac{1}{4}\\right) = \\frac{1}{2} N(\\beta)$.\n6. Since $N(\\beta) \u003e 0$, this gives $N(r) \u003c N(\\beta)$. This proves the division statement.\n7. The ring $\\mathbb{Z}[i]$ is a subring of the field $\\C$. A product of nonzero complex numbers is nonzero, so $\\mathbb{Z}[i]$ is an [[integral domain|09S3PFBEPM]].\n8. The norm takes nonzero Gaussian integers to positive integers, so it is a function $\\mathbb{Z}[i] \\setminus \\set{0} \\to \\N$.\n9. By steps 3 and 6, for all $\\alpha$ and $\\beta \\ne 0$ there are $q$, $r$ with $\\alpha = q\\beta + r$ and $r = 0$ or $N(r) \u003c N(\\beta)$. So $\\mathbb{Z}[i]$ with the norm is a [[Euclidean domain|0GJ2HRG95T]].\n\n## Idea\n\nDivide $\\alpha$ by $\\beta$ as complex numbers and round the quotient to the nearest lattice point $q$.\nThe rounding error is a complex number of absolute value at most $\\sqrt{2}/2$.\nSo the remainder $r = \\beta(\\alpha/\\beta - q)$ has norm at most half the norm of $\\beta$.\n",
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