hcju/proofwikiqa huggingface dataset:
From the definition of the Beta function: {{begin-eqn}} {{eqn | l = \map \Beta {z_1, z_2}       | r = \frac {\map \Gamma {z_1} \, \map \Gamma {z_2} } {\map \Gamma {z_1 + z_2} }        | c =  }} {{eqn | r = \int_0^1 u^{z_1 - 1} \paren {1 - u}^{z_2 - 1} \rd u       | c = Equivalence of Definitions of Beta Function }} {{end-eqn}} Letting $z_1 = z_2 = z$ gives: {{begin-eqn}} {{eqn | l = \frac {\map \Gamma z \, \map \Gamma z} {\map \Gamma {2 z} }       | r = \int_0^1 u^{z - 1} \paren {1 - u}^{z - 1} \rd u       | c =  }} {{eqn | r = \frac 1 2 \int_{-1}^1 \paren {\frac {1 + x} 2 }^{z - 1} \paren {\frac {1 - x} 2}^{z - 1} \rd x       | c = Substitute $u = \dfrac {1 + x} 2$ }} {{eqn | r = \frac 1 {2^{2 z - 1} } \int_{-1}^1 \paren {1 - x^2}^{z - 1} \rd x       | c =  }} {{eqn | ll= \leadsto       | n = 1       | l = 2^{2 z - 1} \map \Gamma z \, \map \Gamma z       | r = 2 \map \Gamma {2 z} \int_0^1 \paren {1 - x^2}^{z - 1} \rd x       | c = by Definite Integral of Even Function }} {{end-eqn}} Now substituting $u = x^2$ into the Beta function: :$\displaystyle \map \Beta {z_1, z_2} = \int_0^1 x^{2 z_1 - 2} \paren {1 - x^2}^{z_2 - 1} 2 x \rd x$ Letting $z_1 = \dfrac 1 2$ and $z_2 = z$ gives: :$(2): \quad \displaystyle \map \Beta {\frac 1 2, z} = 2 \int_0^1 \paren {1 - x^2}^{z - 1} \rd x$ Combining results $(1)$ and $(2)$: {{begin-eqn}} {{eqn | l = 2^{2 z - 1} \map \Gamma z \, \map \Gamma z        | r = \map \Gamma {2 z} \map \Beta {\frac 1 2, z}       | c = }} {{eqn | r = \map \Gamma {2 z} \frac {\map \Gamma {\frac 1 2} \, \map \Gamma z} {\map \Gamma {\frac 1 2 + z} }       | c =  }} {{eqn | ll= \leadsto       | l = \map \Gamma z \, \map \Gamma {z + \dfrac 1 2}        | r = 2^{1 - 2 z} \map \Gamma {\frac 1 2} \, \map \Gamma {2 z} }} {{end-eqn}} From Gamma Function of One Half: :$\map \Gamma {\dfrac 1 2} = \sqrt \pi$ It follows that: :$\map \Gamma z \, \map \Gamma {z + \dfrac 1 2} = 2^{1 - 2 z} \sqrt \pi \, \map \Gamma {2 z}$ {{qed}}
From Gauss Multiplication Formula: :$\displaystyle \prod_{k \mathop = 0}^{n - 1} \Gamma \left({z + \frac k n}\right) = \left({2 \pi}\right)^{\left({n - 1}\right) / 2} n^{1/2 - n z} \Gamma \left({n z}\right)$ Substituting $n = 2$ yields: {{begin-eqn}} {{eqn | l = \Gamma \left({z}\right) \Gamma \left({z + \frac 1 2}\right)       | r = \left({2 \pi}\right)^{1 / 2} 2^{1/2 - 2 z} \Gamma \left({2 z}\right)       | c =  }} {{eqn | r = 2^{1 - 2 z} \sqrt \pi \, \Gamma \left({2 z}\right)       | c =  }} {{end-eqn}} {{qed}}
From Stirling's Formula for Gamma Function: :$\ln \Gamma \left({z}\right) = \left({z - \dfrac 1 2}\right) \ln z - z + \dfrac {\ln 2 \pi} 2 + \mathcal O \left({z^{-1}}\right)$ Taking the derivative: :$(1): \quad \dfrac {\Gamma' \left({z}\right)} {\Gamma \left({z}\right)} = \ln z - \dfrac 1 {2 z} + \dfrac {\mathrm d} {\mathrm dz} \mathcal O \left({z^{-1}}\right)$ Since there exists $c \left({\epsilon}\right) > 0$ such that: :$- \dfrac c {\left\vert{z^{-1} }\right\vert} < \left\vert{\mathcal O \left({z^{-1} }\right)}\right\vert < \dfrac c{\left\vert{z^{-1} }\right\vert}, \quad \left\vert{z}\right\vert > 1$ {{explain|The above statement needs to be explained and justified.}} it follows directly that the third term in $(1)$ is $\mathcal O \left({z^{-1}}\right)$. {{qed}} Category:Gamma Function Category:Reciprocals 0l44q7x1pqd6cadsxsgxeptvnavacjq
This is immediate from, say, the Euler form of $\Gamma$ and the fact that complex conjugation preserves products and sums. Category:Gamma Function Category:Complex Conjugates 6vkdpw0tos2m2r2lw5vlcxtn3rmio06
From the definition of integer combination (or just because it's obvious): :$\Bbb S \subseteq \Z$ and clearly: :$\Bbb S \ne \O$ as $a 0 + b 0 = 0 \in \Bbb S$. Let $w_1, w_2 \in \Bbb S$: :$w_1 = a x_1 + b y_1, w_2 = a x_2 + b y_2$ Then $w_1 + \paren {-w_2} = a \paren {x_1 - x_2} + b \paren {y_1 - y_2} \in \Bbb S$ as $x_1 - x_2 \in \Z$ and $y_1 - y_2 \in \Z$. Let $r \in \Z$. Then: :$w_1 \times r = a x_1 r + b y_1 r = a \paren {x_1 r} + b \paren {y_1 r} \in \Bbb S$ :$r \times w_1 = r a x_1 + r b y_1 = \paren {r x_1} a + \paren {r y_1} b \in \Bbb S$ The result follows from Test for Ideal. {{qed}}
By Intersection of Ring Ideals is Ideal we have that $\ideal k = \ideal m \cap \ideal n$ is an ideal of $\Z$. By Ring of Integers is Principal Ideal Domain we have that $\ideal m$, $\ideal n$ and $\ideal k$ are all necessarily principal ideals. By Subrings of Integers are Sets of Integer Multiples we have that: :$\ideal m = m \Z, \ideal n = n \Z$ Thus: :$\ideal k = \set {x \in \Z: n \divides x \land m \divides x}$ The result follows by LCM iff Divides All Common Multiples. {{qed}}
=== Necessary Condition === Suppose $m = \lcm \set {a, b}$. Then from LCM Divides Common Multiple: : $a \divides n \land b \divides n \implies m \divides n$ {{qed|lemma}} === Sufficient Condition === Now suppose $a \divides m \land b \divides m$, and $m$ also divides any $n$ that $a$ and $b$ also divide. From $a \divides m \land b \divides m$, we see that $m$ is a common multiple of $a$ and $b$. From $a \divides n \land b \divides n$, we see that $n$ is also a common multiple of $a$ and $b$. Also, we have that $m \divides n$. From Absolute Value of Integer is not less than Divisors, we see that (in the domain of $\Z_{>0}$) $m \divides n \implies m \le n$. Thus, whatever $m$ may be, it is no larger than $n$. Therefore, $m$ must be the least of all the common multiples of $a$ and $b$. {{qed}} Category:Lowest Common Multiple ojgfu5gmd7pmtf8ef7xaj8tmtiinj36
Let $m = \lcm \set {a, b}$. Then $a \divides m$ and $b \divides m$ by definition. Suppose $n$ is some other common multiple of $a$ and $b$ such that $m \nmid n$ ($m$ does not divide $n$). Then from the Division Theorem: :$n = k m + r$ for some integer $k$ and with $0 < r < m$. Then since $r = n - k m$, using $a \divides n$ and $a \divides m$: : $a \divides r$ Similarly: : $b \divides r$ Then $r$ is a common multiple of $a$ and $b$. But we have that $r < m$. This contradicts the fact that $m$ is the lowest common multiple of $a$ and $b$. So, by contradiction, it follows that $m \divides n$. {{qed}} {{Euclid Note|35|VII|}}
By Sum of Ideals is Ideal we have that $\ideal d = \ideal m + \ideal n$ is an ideal of $\Z$. By Ring of Integers is Principal Ideal Domain we have that $\ideal m$, $\ideal n$ and $\ideal d$ are all necessarily principal ideals. By Subrings of Integers are Sets of Integer Multiples we have that: :$\ideal m = m \Z, \ideal n = n \Z$ Thus: :$\ideal d = \ideal m + \ideal n = \set {x \in \Z: \exists a, b \in \Z: x = a m + b n}$ That is, $\ideal d$ is the set of all integer combinations of $m$ and $n$. The result follows by Bézout's Lemma. {{qed}}
Consider the additive group $\struct {R, +}$. From Trivial Quotient Group is Quotient Group: :$\struct {R, +} / \set {0_R} \cong \struct {R, +}$ {{finish}}
We have that Ring Homomorphism of Addition is Group Homomorphism. The result follows from Group Homomorphism Preserves Inverses. {{qed}}
By definition of the Gamma function: :$\displaystyle \Gamma \left({\frac s 2}\right) = \int_0^\infty t^{s / 2 - 1} e^{-t} \rd t$ First substitute $t = \pi n^2 x$ to give: :$\displaystyle \pi^{-s / 2} \Gamma \left({\frac s 2}\right) n^{-s} = \int_0^\infty x^{s / 2 - 1} e^{-\pi n^2 x} \rd x$ By definition, on $\operatorname{Re} \left({s}\right) > 1$: :$\displaystyle \zeta \left({s}\right) = \sum_{n \mathop = 1}^\infty n^{-s}$ Therefore summing over $n \ge 1$: :$\displaystyle \pi^{-s / 2} \Gamma \left({\frac s 2}\right) \zeta \left({s}\right) = \int_0^\infty x^{s / 2 - 1} \omega \left({x}\right) \rd x$ where: :$\displaystyle \omega \left({x}\right) = \sum_{n \mathop = 1}^\infty e^{-\pi n^2 x}$ Next the integral is split at $x = 1$ as follows: {{begin-eqn}} {{eqn | l = \pi^{-s / 2}\Gamma \left({\frac s 2}\right) \zeta \left({s}\right)       | r = \int_0^1 x^{s / 2 - 1} \omega \left({x}\right) \rd x + \int_1^\infty x^{s / 2 - 1} \omega \left({x}\right) \rd x       | c = }} {{eqn | r = \int_1^\infty x^{-s / 2 - 1} \omega \left({\frac 1 x}\right) \rd x + \int_1^\infty x^{s / 2 - 1} \omega \left({x}\right) \rd x       | c = substituting $x \mapsto \dfrac 1 x$ in the first integral. }} {{end-eqn}} Recall the Jacobi theta function: {{begin-eqn}} {{eqn | l = \theta \left({x}\right)       | r = \sum_{n \mathop \in \Z} e^{- \pi n^2 x}       | c =  }} {{eqn | r = 2 \omega \left({x}\right) + 1       | c =  }} {{end-eqn}} Since $e^{-x^2}$ is a fixed point of the Fourier transform, by Properties of the Fourier Transform: :$\mathcal F \left[{e^{-\pi t^2 x} }\right] \left({u}\right) = x^{-1 / 2} e^{-\pi u^2 / x}$ where $\mathcal F$ denotes the Fourier transform. Therefore, by the Poisson Summation Formula: :$\theta \left({x}\right) = \dfrac 1 {\sqrt x} \theta \left({\dfrac 1 x}\right)$ whence: :$\omega \left({\dfrac 1 x}\right) = -\dfrac 1 2 + \dfrac 1 2 \sqrt x + \sqrt x \omega \left({x}\right)$ Therefore: {{begin-eqn}} {{eqn | l = \int_1^\infty x^{-s / 2 - 1} \omega \left({\frac 1 x}\right) \rd x       | r = \int_1^\infty x^{-s / 2 - 1} \left({-\frac 1 2 + \frac 1 2 \sqrt x + \sqrt x \omega \left({x}\right)}\right) \rd x       | c =  }} {{eqn | r = - \frac 1 s + \frac 1 {s - 1} + \int_1^\infty x^{-\left({s + 1}\right)/2} \omega \left({x}\right) \rd x       | c =  }} {{end-eqn}} So: :$\displaystyle \pi^{-s / 2} \Gamma \left({\frac s 2}\right) \zeta \left({s}\right) = -\frac 1 {s \left({1 - s}\right)} + \int_1^\infty \left[{x^{s / 2} + x^{\left({1 - s}\right) / 2} }\right] \omega \left({x}\right) \frac {\rd x} x$ as required. {{qed}}
We have Numbers of Type Integer a plus b root 2 form Subdomain of Reals. First we note that: :$\forall x \in \Z \sqbrk {\sqrt 2}: \map \phi x \in \Z \sqbrk {\sqrt 2}$ === Proof of Bijection === Let $\map \phi {a + b \sqrt 2} = \map \phi {a' + b' \sqrt 2}$. Then: :$a - b \sqrt 2 = a' - b' \sqrt 2$ and so: :$a + b \sqrt 2 = a' + b' \sqrt 2$ So $\phi$ is injective. Now let $y = c + d \sqrt 2 \in \Z \sqbrk {\sqrt 2}$. We have that: {{begin-eqn}} {{eqn | l = \map \phi {c + \paren {-d} \sqrt 2}       | r = c - \paren {-d} \sqrt 2       | c =  }} {{eqn | r = c + d \sqrt 2       | c =  }} {{end-eqn}} Hence: :$\forall y \in \Z \sqbrk {\sqrt 2}: \exists x \in \Z \sqbrk {\sqrt 2}: \map \phi x = y$ and so $\phi$ is surjective. So $\phi$ is a bijection. === Proof of Morphism === Now consider $x = a + b \sqrt 2, y = c + d \sqrt 2$. {{begin-eqn}} {{eqn | l = \map \phi x + \map \phi y       | r = \map \phi {a + b \sqrt 2} + \map \phi {c + d \sqrt 2}       | c =  }} {{eqn | r = \paren {a - b \sqrt 2} + \paren {c - d \sqrt 2}       | c = Definition of $\phi$ }} {{eqn | r = \paren {a + c} - \paren {b + d} \sqrt 2       | c =  }} {{eqn | r = \map \phi {\paren {a + c} + \paren {b + d} \sqrt 2}       | c = Definition of $\phi$ }} {{eqn | r = \map \phi {x + y}       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \map \phi x \, \map \phi y       | r = \map \phi {a + b \sqrt 2} \, \map \phi {c + d \sqrt 2}       | c =  }} {{eqn | r = \paren {a - b \sqrt 2} \paren {c - d \sqrt 2}       | c = Definition of $\phi$ }} {{eqn | r = \paren {a c + 2 b d} - \paren {b c + a d} \sqrt 2       | c =  }} {{eqn | r = \map \phi {\paren {a c + 2 b d} + \paren {b c + a d} \sqrt 2}       | c = Definition of $\phi$ }} {{eqn | r = \map \phi {x y}       | c =  }} {{end-eqn}} So $\phi$ is a ring homomorphism which is also a bijection, whose image equals its domain. Hence the result by definition of ring automorphism. {{qed}}
Let $a, b \in \Z$. Then: {{begin-eqn}} {{eqn | l = \map \phi {a + b}       | r = \eqclass {a + b} m       | c = Definition of $\phi$ }} {{eqn | r = \eqclass a m +_m \eqclass b m       | c = {{Defof|Modulo Addition}} }} {{eqn | r = \map \phi a +_m \map \phi b       | c = Definition of $\phi$ }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map \phi {a \times b}       | r = \eqclass {a \times b} m       | c = Definition of $\phi$ }} {{eqn | r = \eqclass a m \times_m \eqclass b m       | c = {{Defof|Modulo Multiplication}} }} {{eqn | r = \map \phi a \times_m \map \phi b       | c = Definition of $\phi$ }} {{end-eqn}} Hence $\phi$ is a ring homomorphism. Now let $\eqclass a m \in \Z_m$. By definition of residue class modulo $m$: :$\eqclass a m = \set {x \in \Z: \exists k \in \Z: z = a + k m}$ Setting $k = 0$: :$\map \phi a = \eqclass a m$ and so: :$\map {\phi^{-1} } {\eqclass a m} \ne \O$ Thus $\phi$ is a surjection. Now setting $k = 1$, for example, we have that: :$\map \phi {a + m} = \eqclass a m$ and so: :$\map \phi a = \map \phi {a + m}$. So $\phi$ is specifically not an injection. It follows by definition that $\phi$ is a ring epimorphism, but specifically not a ring monomorphism. Next we note that: :$\forall x \in \Z: \map \phi x \in \Z_m$ and so: :$\Img \phi = \Z_m$ Finally, we have that the kernel of $\phi$ is: :$\map \ker \phi = \set {x \in \Z: \map \phi x = \eqclass 0 m}$ Let $\map \phi x = \eqclass 0 m$ Then $x = 0 + k m$ for some $k \in \Z$. That is, $x \in m \Z$ and so: :$\map \ker \phi \subseteq m \Z$ Now let $x \in m \Z$. Then: :$\exists k \in \Z: x = 0 + k m$ and so by definition: :$\map \phi x = \eqclass 0 m$ So: :$m \Z \subseteq \map \ker \phi$ Hence: :$\map \ker \phi = m \Z$ {{qed}}
An element of $F \sqbrk X$ whose degree is $0$ is merely an element of $F$. But note that $0_F$, considered as an element of $F \sqbrk X$, has a degree which is not defined, so the null polynomial is seen to be excluded. Any element $a$ of $F$ has an inverse $1_F / a$. So all the elements of $F \sqbrk X$ whose degree is $0$ are units of $F$ and hence of $F \sqbrk X$. Now suppose $\map a X \in F \sqbrk X$ is a unit of $F \sqbrk X$. Then $\map a X \, \map q X = 1_F$ for some $\map q X \in F \sqbrk X$. Neither $\map a X$ nor $\map q X$ can be null. So by Properties of Degree $\map \deg {\map a X} + \map \deg {\map q X} = 0$. So $\map \deg {\map a X} = 0$. {{qed}}
For $a + b \sqrt 2$ to be a unit of $\struct {\Z \sqbrk {\sqrt 2}, +, \times}$, we require that: :$\exists c, d \in \Z: \paren {a + b \sqrt 2} \paren {c + d \sqrt 2} = 1$ In Numbers of Type Integer a plus b root 2 are Not a Field it is shown that the product inverse of $\paren {a + b \sqrt 2}$ is $\dfrac a {a^2 - 2 b^2} + \dfrac {b \sqrt 2} {a^2 - 2 b^2}$. So if $a^2 - 2 b^2 = \pm 1$ it follows that $c$ and $d$ are integers. Hence the result. {{qed}}
From Gaussian Integer Units are 4th Roots of Unity: : $\left\{{1, i, -1, -i}\right\}$ constitutes the set of the $4$th roots of unity. The result follows from Roots of Unity under Multiplication form Cyclic Group. {{qed}}
By definition of the imaginary unit $i$: {{begin-eqn}} {{eqn | l = i^2       | r = -1 }} {{eqn | l = i^3       | r = -i }} {{eqn | l = i^4       | r = 1 }} {{end-eqn}} thus demonstrating that $U_\C$ is generated by $i$. Thus $\left({U_\C, \times}\right)$ is by definition a cyclic group of order $4$. {{qed}}
From Units of Gaussian Integers, $U_\C$ is the set of units of the ring of Gaussian integers. From Group of Units is Group, $\left({U_\C, \times}\right)$ forms a group. It remains to note that: {{begin-eqn}} {{eqn | l = i^2       | r = -1 }} {{eqn | l = i^3       | r = -i }} {{eqn | l = i^4       | r = 1 }} {{end-eqn}} thus demonstrating that $U_\C$ is  cyclic. {{qed}}
Let $a + b i$ be a unit of $\struct {\Z \sqbrk i, +, \times}$. Then $a$ and $b$ are not both $0$ as then $a + b i$ would be the zero of $\struct {\Z \sqbrk i, +, \times}$. Then: :$\exists c, d \in \Z: \paren {a + b i} \paren {c + d i} = 1 + 0 i$ This leads (after algebra) to: :$c = \dfrac a {a^2 + b^2}, d = \dfrac {-b} {a^2 + b^2}$ Let $a^2 + b^2 = n$. We have that $n \in \Z, n > 0$. If $c$ and $d$ are integers, then $a$ and $b$ must both be divisible by $n$. Let $a = n x, b = n y$. Then: :$n^2 x^2 + n^2 y^2 = n$ and so: :$n \paren {x^2 + y^2} = 1$ Thus $n = a^2 + b^2 = 1$ and so as $a, b \in \Z$ we have: :$a^2 = 1, b^2 = 0$ or: :$a^2 = 0, b^2 = 1$ from which the result follows. {{qed}}
Let $\alpha = a + b i$ be a unit of $\struct {\Z \sqbrk i, +, \times}$. Then by definition of unit: :$\exists\beta = c + d i \in \Z \sqbrk i: \alpha \beta = 1$ Let $\cmod \alpha$ denote the modulus of $\alpha$. Then: {{begin-eqn}} {{eqn | l = \cmod \alpha^2 \cdot \cmod \beta^2       | r = \cmod {\alpha \beta}^2       | c = Modulus of Product }} {{eqn | r = \cmod 1^2       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} By Divisors of One: :$\cmod a^2 = 1$ or $-1$ Since $\cmod \alpha$ and $\cmod \beta$ are positive integers: :$\cmod \alpha^2 = a^2 + b^2 = 1$ and so either: :$\cmod a = 1$ and $\cmod b = 0$ or: :$\cmod b = 1$ and $\cmod a = 0$. Hence the set of units of $\struct {\Z \sqbrk i, +, \times}$ is $\set {\pm 1, \pm i}$. {{qed}}
From Integers form Ordered Integral Domain we have that $\struct {\Z, +, \times, \le}$ forms an ordered integral domain. For all $a \in \Z$, the absolute value of $a$ is defined as: :$\size a = \begin{cases} a & : 0 \le a \\ -a & : a < 0 \end{cases}$ By Product of Absolute Values on Ordered Integral Domain we have: :$\forall a, b \in \Z: \size a \cdot \size b = \size {a b}$ Since $\size b > 0$ it follows from Relation Induced by Strict Positivity Property is Compatible with Multiplication that $\size a \cdot \size b \ge \size a$. The second criterion: :For any $a, b \in R$, $b \ne 0$, there exist $q, r \in R$ with $\map \nu r < \map \nu b$, or $r = 0$ such that: ::$ a = q \circ b + r$ follows from the Division Theorem. {{qed}}
We have by definition that $\Z \sqbrk i \subseteq \C$. Let $a, b \in \Z \sqbrk i$. We have from Modulus of Product that $\cmod a \cdot \cmod b = \cmod {a b}$. From Complex Modulus is Non-Negative: :$\forall a \in \C: \cmod a \ge 0$ and: :$\cmod a = 0 \iff a = 0$ Let $a = x + i y$. Suppose $a \in \Z \sqbrk i \ne 0$. Then $x \ne 0$ or $y \ne 0$ and $x^2 \ge 1$ or $y^2 \ge 1$. So: :$\cmod a \ge 1$ Similarly, $b \in \Z \sqbrk i, b \ne 0 \implies \cmod b \ge 1$. Thus it follows that $\cmod {a b} \ge \cmod a$ and so $\nu$ is a Euclidean valuation on $\Z \left[{i}\right]$. {{finish|definition of Euclidean valuation requires codomain $\N$, consider modulus squared instead}} Now, consider $x, y \in \Z \sqbrk i$. We want to find $q, r \in \Z \sqbrk i$ such that $x = q y + r$. Note that this means we want $r = y \paren {\dfrac x y - q}$ where $\dfrac x y$ is complex but not necessarily Gaussian. We extend $\nu$ to the complex numbers and define $\nu: \C \to \C$ as: :$\forall z \in \C: \map \nu z = \cmod z$ Then we have: :$\map \nu r = \map \nu y \cdot \map \nu {\dfrac x y - q}$ Thus if $\map \nu {\dfrac x y - q} < 1$ we have: :$\map \nu r < \map \nu y$ Consider the point $P = \dfrac x y$ as a point on the complex plane. Let $Q$ lie at a point representing the Gaussian integer $q$ which lies closest to $P$. The distance $P Q$ is at most half the length of a diagonal of a unit square in the complex plane. Thus: :$\map \nu {\dfrac x y - q} = \cmod {\dfrac x y - q} \le \dfrac {\sqrt 2} 2 = \dfrac 1 {\sqrt 2} < 1$ This element $q$ and the element $r$, where $\map \nu r < \map \nu y$, are the required values. Thus $\Z \sqbrk i$ is a Euclidean domain. {{qed}} {{Improve}}
We have by definition that $\Z \sqbrk i \subseteq \C$. Let $a, b \in \Z \sqbrk i$. We have from Modulus of Product that: :$\cmod a \cdot \cmod b = \cmod {a b}$ From Complex Modulus is Norm we have that: :$\forall a \in \C: \cmod a \ge 0$ :$\cmod a = 0 \iff a = 0$ Suppose $a = x + i y \in \Z \sqbrk i$ and $a \ne 0$. Then either $x \ne 0$ or $y \ne 0$, so either $x^2 \ge 1$ or $y^2 \ge 1$. So $\cmod a^2 \ge 1$. If also $b \in \Z \sqbrk i, b \ne 0$, then: :$\map \nu {a b} = \cmod {a b}^2 \ge \cmod a^2 = \map \nu a$ Now, consider $x, y \in Z \sqbrk i$ with $y \ne 0$. We want to find $q, r \in Z \sqbrk i$ such that $x = q y + r$ and $\map \nu r < \map \nu y$. Note that this means we want $r = y \paren {\dfrac x y - q}$ where $\dfrac x y$ is complex but not necessarily Gaussian. Consider the complex number $p = \dfrac x y = p_r + i p_i$. We extend $\nu$ to the complex numbers and define $\nu: \C \to \C$ as: :$\forall z \in \C: \map \nu z = \cmod z^2$ Let $q = q_r + i q_i$ be the Gaussian integer such that: :$\map \nu {p - q} = \cmod {p - q}^2 = \paren {p_r - q_r}^2 + \paren {p_i - q_i}^2$ is minimal. That is, $q_r$ is the nearest integer to $p_r$ and $q_i$ is the nearest integer to $p_i$. A given real number can be at a distance at most $1/2$ from an integer, so it follows that: :$(1): \quad \map \nu {p - q} \le \paren {\dfrac 1 2}^2 + \paren {\dfrac 1 2}^2 = \dfrac 1 2$ Now by Complex Modulus is Norm, for any two complex numbers $z_1, z_2$ we have: :$\map \nu {z_1 z_2} = \map \nu {z_1} \, \map \nu {z_2}$ Thus we obtain: {{begin-eqn}} {{eqn | l = \map \nu {y \paren {p - q} }       | r = \map \nu y \, \map \nu {p - q} }} {{eqn | r = \dfrac {\map \nu y} 2       | o = \le       | c = by $(1)$ }} {{eqn | r = \map \nu y       | o = <       | c = since $\map \nu y \ne 0$ }} {{end-eqn}} On the other hand: {{begin-eqn}} {{eqn | l = \map \nu {y \paren {p - q} }       | r = \map \nu {y \paren {\frac x y - q} } }} {{eqn | r = \map \nu {x - y q} }} {{end-eqn}} So letting $r = x - y q$ we have $\map \nu r < \map \nu y$. Moreover we trivially have $x = q y + r$. Thus $\Z \sqbrk i$ is a Euclidean domain. {{qed}}
Let the leading coefficient of: * $f \left({X}\right)$ be $a_n$ * $g \left({X}\right)$ be $b_n$. Then: * $f \left({X}\right) = a_n X^n + \cdots + a_0$ * $g \left({X}\right) = b_n X^n + \cdots + b_0$ Consider the leading coefficient of the product $f \left({X}\right) g \left({X}\right)$: call it $c$. From the definition of polynomial addition and polynomial multiplication:  * $f \left({X}\right) g \left({X}\right) = c X^{n+m} + \cdots + a_0 b_0$ Clearly the highest term of $f \left({X}\right) g \left({X}\right)$ can have an index no higher than $n+m$. Hence the result: :$\deg \left({f g}\right) \not > \deg \left({f}\right) + \deg \left({g}\right)$ Next, note that the general ring with unity $(R, +, \circ)$ may have proper zero divisors. Therefore it is possible that $X^{n+m}$ may equal $0_R$. If that is the case, then the highest term will have an index definitely less than $n+m$. That is, in that particular case: :$\deg \left({f g}\right) < \deg \left({f}\right) + \deg \left({g}\right)$ Thus, for a general ring with unity $(R, +, \circ)$: :$\deg \left({f g}\right) \le \deg \left({f}\right) + \deg \left({g}\right)$ {{qed}}
First we associate to $f = a_0 + a_1 X + \cdots + a_n X^n \in R \sqbrk X$ a formal vector $x_f = \tuple {a_0, a_1, \ldots, a_n, 0_R, \ldots} \in R^\infty$. Let $x_f^i \in R$ denote the element at the $i$th position.  Then: :$\displaystyle \map \deg f = \sup \set {i \in \N : x_f^i \ne 0_R}$ The sum $+$ in the polynomial ring $R \sqbrk X$ gives rise to the following identity in $R^\infty$: :$x_{f + g}^i = x_f^i + x_g^i$ Next, let $f, g \in R \sqbrk X$, and let $d = \max \set {\map \deg f, \map \deg g}$. Then $x_f^i = 0_R = x_g^i$ for all $i > d$, so we have: :$\displaystyle \map \deg {f + g} = \sup \set {i \in \N : x_{f + g}^i \ne 0_R} = \sup \set {i \in \N : x_f^i + x_g^i \ne 0_R} \le d$ {{qed}}
From Degree of Product of Polynomials over Integral Domain, we have: :$\map \deg {f g} = \map \deg f + \map \deg g$ But $\map \deg g \ge 0$ by definition of degree, as $g$ is not null. Hence the result. {{qed}}
From Degree of Product of Polynomials over Integral Domain not Less than Degree of Factors we have that: : $\forall a, b \in F \sqbrk X, a \ne 0_F, b \ne 0_F: \map \deg {a b} \ge \map \deg a$ where $\map \deg a$ denotes the degree of $a$. From Division Theorem for Polynomial Forms over Field: :$\forall a, b \in F \sqbrk X, b \ne 0_F: \exists q, r \in F \sqbrk X: a = q b + r$ where $\map \deg r < \map \deg b$ (or $r = 0$), and $q$ and $r$ are unique. So $\deg$ is a Euclidean valuation on $F \sqbrk X$. Hence the result. {{qed}}