hcju/proofwikiqa huggingface dataset:
We construct the completion of a metric space as equivalence classes of the set of Cauchy sequences in the space under a suitable equivalence relation. Let $\struct {A, d}$ be a metric space. Let $\CC \sqbrk A$ denote the set of all Cauchy sequences in $A$. Define a relation $\sim$ on $\CC \sqbrk A$ by: :$\displaystyle \sequence {x_n} \sim \sequence {y_n} \iff \lim_{n \mathop \to \infty} \map d {x_n, y_n} = 0$ By Equivalence Relation on Cauchy Sequences, $\sim$ is an equivalence relation on $\CC \sqbrk A$. Denote the equivalence class of $\sequence {x_n} \in \CC \sqbrk A$ by $\sqbrk {x_n}$. Denote the set of equivalence classes under $\sim$ by $\tilde A$. By Relation Partitions Set iff Equivalence this is a partition of $\CC \sqbrk A$. That is, each $\sequence {x_n} \in \CC \sqbrk A$ lies in one and only one equivalence class under $\sim$. Define $\tilde d: \tilde A \to \R_{\ge 0}$ by: :$\displaystyle \map {\tilde d} {\sqbrk {x_n}, \sqbrk {y_n} } = \lim_{n \mathop \to \infty} \map d {x_n, y_n}$ === Lemma 1: $\tilde d$ is Well-Defined === {{:Completion Theorem (Metric Space)/Lemma 1}} We claim that $\struct {\tilde A, \tilde d}$ is a completion of $\struct {A, d}$. Therefore we must show that: :$\tilde d$ is a metric on $\tilde A$ :There exists an everywhere dense inclusion $\struct {A, d} \to \struct {\tilde A, \tilde d}$ preserving $d$. In addition the theorem claims that $\struct {\tilde A, \tilde d}$ is unique up to isometry. === Lemma 2: $\tilde d$ is a Metric === {{:Completion Theorem (Metric Space)/Lemma 2}} === Lemma 3: $\tilde A$ is a Completion of $A$ === {{:Completion Theorem (Metric Space)/Lemma 3}} === Lemma 4: Uniqueness of $\tilde A$ === {{:Completion Theorem (Metric Space)/Lemma 4}}{{qed}} {{ACC|Completion Theorem (Metric Space)/Lemma 3}}
Let $\left\langle{x_n}\right\rangle_{n \mathop \in \N}$ be a Cauchy sequence in $S$. For each $n$ pick a Cauchy sequence $\left\langle{y_{n, m}}\right\rangle_{m \mathop \in \N}$ in $A$ converging to $x_n$ like so: :400pxCaption Let $N \in \N$ be such that $d \left({x_{n_1}, x_{n_2}}\right) < \epsilon / 3$ for all $n_1, n_2 > N$. Let $M \in \N$ be such that $d \left({y_{n_i, m}, x_{n_i}}\right) < \epsilon / 3$ for all $m > M$ and all $n_1, n_2 > N$.  Let $m > M$. Let $n_1, n_2 > N$. We have: {{begin-eqn}} {{eqn | l = d \left({y_{n_1, m}, y_{n_2, m} }\right)       | o = \le       | r = d \left({y_{n_1, m}, x_{n_1} }\right) + d \left({x_{n_1}, x_{n_2} }\right) + d \left({x_{n_2}, y_{n_2, m} }\right)       | c = two applications of the Triangle Inequality }} {{eqn | o = <       | r = \epsilon }} {{end-eqn}} Therefore $\left\langle{y_{m, n}}\right\rangle_{n \in \N}$ is Cauchy in $A$ for $m > M$. So $\left\langle{y_{m, n}}\right\rangle_{n \mathop \in \N}$ converges to some limit $y_n \in S$. {{finish|Proof to complete}}
Let $\left\langle{x_n}\right\rangle$ be a Cauchy sequence in $M$. Since $A$ is dense, we can invoke the Axiom of Countable Choice to choose for each $n$ some $y_n \in A$ which is within $1 / n$ of $x_n$. We will show that $\left\langle{y_n}\right\rangle$ is Cauchy. Let $\epsilon > 0$. Since $\left\langle{x_n}\right\rangle$ is Cauchy, there exists some $N'$ such that: :$\forall n, m \ge N': d \left({x_n, x_m}\right) < \frac 1 3 \epsilon$ Let $N \in \N$ be greater than $N'$ and $3 \epsilon^{-1}$. Note that $n, m \ge N$ implies that both: : $(1): \quad n, m \ge N'$ : $(2): \quad \dfrac 1 n, \dfrac 1 m \le \dfrac 1 N < \dfrac 1 3 \epsilon$ Thus for $n, m \ge N$, by the Triangle Inequality and the above: {{begin-eqn}} {{eqn | l = d \left({y_n, y_m}\right)       | o = \le       | r = d \left({y_n, x_n}\right) + d \left({x_n, x_m}\right) + d \left({x_m, y_m}\right)       | c =  }} {{eqn | o = <       | r = \frac 1 n + \frac 1 3 \epsilon + \frac 1 m        | c =  }} {{eqn | o = <       | r = \frac 1 3 \epsilon + \frac 1 3 \epsilon + \frac 1 3 \epsilon       | c =  }} {{eqn | r = \epsilon       | c =  }} {{end-eqn}} Hence $\left\langle{y_n}\right\rangle$ is a Cauchy sequence in $A$. By assumption, it converges to some limit $y \in S$. Now, we verify that $\left\langle{x_n}\right\rangle$ converges to $y$ as well. Let $\epsilon > 0$. Since $\left\langle{y_n}\right\rangle$ converges, there exists some $N'$ such that: :$\forall n \ge N': d \left({y_n, y}\right) < \dfrac 1 2 \epsilon$ Let $N \in \N$ be greater than $N'$ and $2 \epsilon^{-1}$. Note that $n \ge N$ implies that both: : $(1): \quad n \ge N'$ : $(2): \quad \dfrac 1 n \le \dfrac 1 N < \dfrac 1 2 \epsilon$ Thus, for $n \ge N$, we have by the Triangle Inequality and above that: {{begin-eqn}} {{eqn | l = d \left({x_n, y}\right)       | o = \le       | r = d \left({x_n, y_n}\right) + d \left({y_n, y}\right)       | c =  }} {{eqn | o = <       | r = \frac 1 n + \frac 1 2 \epsilon       | c =  }} {{eqn | o = <       | r = \frac 1 2 \epsilon + \frac 1 2 \epsilon       | c =  }} {{eqn | r = \epsilon       | c =  }} {{end-eqn}} Hence $\left\langle{x_n}\right\rangle$ converges to $y$. {{qed}}
Follows directly from Natural Numbers under Addition form Commutative Monoid. A monoid by definition is a semigroup. Again by definition, the operation in a semigroup is closed. {{qed}}
By definition of ordering, the relation $\le$ is: :reflexive :transitive :antisymmetric and furthermore, every pair of elements is comparable. The order is compatible with $F$ in the sense that, for all $x, y, z, c \in F$: :$x < y \implies x + z < y + z$ :$c > 0,\ x < y \implies c x < c y$ The proof is by repeated deduction from these properties. * $(1): \quad x < 0 \iff -x > 0$: {{begin-eqn}} {{eqn | l = x       | o = <       | r = 0 }} {{eqn | ll= \leadsto       | l = x - x       | o = <       | r = 0 - x }} {{eqn | ll= \leadsto       | l = -x       | o = >       | r = 0 }} {{end-eqn}} Conversely: {{begin-eqn}} {{eqn | l = x       | o = >       | r = 0 }} {{eqn | ll= \leadsto       | l = x - x       | o = >       | r = 0 - x }} {{eqn | ll= \leadsto       | l = -x       | o = <       | r = 0 }} {{end-eqn}} * $(2): \quad x > y \iff x - y > 0$: {{begin-eqn}} {{eqn | l = x       | o = >       | r = y }} {{eqn | ll= \leadsto       | l = x - y       | o = >       | r = y - y = 0 }} {{end-eqn}} Conversely: {{begin-eqn}} {{eqn | l = x - y > 0       | o = >       | r = y }} {{eqn | ll= \leadsto       | l = x - y + y       | o = >       | r = y }} {{eqn | ll= \leadsto       | l = x       | o = >       | r = y }} {{end-eqn}} * $(3): \quad x < y \iff -x > -y$: {{begin-eqn}} {{eqn | l = x       | o = <       | r = y }} {{eqn | ll= \leadsto       | l = x - x - y       | o = <       | r = y - x - y }} {{eqn | ll= \leadsto       | l = -y       | o = <       | r = -x }} {{end-eqn}} Conversely: {{begin-eqn}} {{eqn | l = -y       | o = <       | r = -x }} {{eqn | ll= \leadsto       | l = -y + y + x       | o = <       | r = -x + y + x }} {{eqn | ll= \leadsto       | l = x       | o = <       | r = y }} {{end-eqn}} * $(4): \quad (z < 0) \land (x < y) \implies x z > y z$: By parts 1 and 3 above, if $z < 0$, $x < y$ then $-z > 0$ and $-x > -y$. Then: :$x z = \paren {-x} \paren {-z} > \paren {-y} \paren {-z} = y z$ * $(5): \quad x \ne 0 \implies x^2 > 0$: If $x > 0$, then: :$x^2 = x \cdot x > x \cdot 0 = 0$ If $x < 0$, then by 1, $-x > 0$, so: :$x^2 = \paren {-x} \cdot \paren {-x} > \paren {-x} \cdot 0 = 0$ * $(6): \quad 1 > 0$: This is immediate from $(5)$, noting that $1 = 1^2$ is a square. * $(7): \quad \Char k = 0$: By $(6)$, we have: :$0 < 1 < 1 + 1 < 1 + 1 + 1 < \cdots$ so $n \cdot 1 \ne 0$ for all $n \in \N$. * $(8): \quad x > y > 0 \iff y^{-1} > x^{-1} > 0$: First let $x > 0$, and suppose that $x^{-1} < 0$. Then by $(4)$: :$0 = 0 \cdot x^{-1} > x \cdot x^{-1} = 1$ which contradicts $(6)$, so $x^{-1} > 0$. Now let $x > y > 0$. Then: {{begin-eqn}} {{eqn | l = x^{-1} y^{-1} x       | o = >       | r = x^{-1} y^{-1} y > 0 }} {{eqn | ll= \leadsto       | l = y^{-1}       | o = >       | r = x^{-1} > 0 }} {{end-eqn}} The converse follows upon interchanging $x^{-1} \leftrightarrow x$ and $y^{-1} \leftrightarrow y$. {{qed}} Category:Ordered Fields jfttok3h9u1ed8f6bf74k928uc01v1b
Let a linear functional $g$ be called '''admissible''' {{iff}} :$\forall x \in \Dom g: \map g x \le \map p x$ A linear functional $h_1$ extends a linear functional $h_2$ {{iff}}: :$\Dom {h_2} \subseteq \Dom {h_1}$ and: :$\forall x \in \Dom {h_2}: \map {h_2} x = \map {h_1} x$ The proof consists of two steps: First, the set of '''admissible''' linear functionals that extend $f$ is inductive. Using Zorn's Lemma the existence of a maximal element is derived. Second, it is proved by contradiction that this functional is defined on the whole space $E$. {{finish}} {{Namedfor|Hans Hahn|name2 = Stefan Banach|cat = Hahn|cat2 = Banach}}
By Completion of Normed Division Ring then $\struct {k, \norm {\, \cdot \,} }$ has a normed division ring completion $\struct {k', \norm {\, \cdot \,}'}$ By Normed Division Ring is Field iff Completion is Field then $\struct {k', \norm {\, \cdot \,}'}$ is a field. By Normed Division Ring Completions are Isometric and Isomorphic then every completion of $\struct {k, \norm {\,\cdot\,} }$ is isometrically isomorphic to $\struct {k', \norm {\,\cdot\,}'}$. {{qed}}
We have: :$\displaystyle \map T s^\ell = \sum_{N \mathop = 0}^\infty \map {r_{\AA, \ell} } N s^N$ and: :$\map {\dfrac {\d^N} {\d s^N} } {\map T s^\ell} = N! \cdot \map {r_{\AA, \ell} } N + \map \OO s$ so: :$\map {r_{\AA, \ell} } N = \dfrac 1 {N!} \dfrac {\d^N} {\d s^N} \bigintlimits {\map T s^\ell} {s \mathop = 0}{}$ Now recall Cauchy's Integral Formula for Derivatives for a complex function $f$ holomorphic on a domain $D$, and a path $\gamma \subseteq D$ winding once around $a$: :$\displaystyle \valueat {\frac {\d^N} {\d s^N} \map f s} {s \mathop = a} = \frac {N!} {2 \pi i} \oint_\gamma \frac {\map f s} {\paren {s - a}^{N + 1} } \rd s$ Since $\map T s$ is defined by a generating function, $\map T s^\ell$ has a Taylor series about $s = 0$ which converges for all $\size s < 1$. Applying Cauchy's formula: {{begin-eqn}} {{eqn | l = \map {r_{\AA, \ell} } N       | r = \frac 1 {N!} \frac {\d^N} {\d s^N} \bigintlimits {\map T s^\ell} {s \mathop = 0} {} }} {{eqn | r = \frac 1 {2 \pi i} \oint_\gamma \frac {\map T s^\ell} {s^{N + 1} } \rd s }} {{end-eqn}} where $\gamma$ a circle about zero of radius $\rho$ for any $\rho < 1$. {{qed}} {{Namedfor|Godfrey Harold Hardy|name2 = John Edensor Littlewood|cat = Hardy|cat2 = Littlewood}} {{explain|Plenty of links to be included and explanations to be added.}} Category:Analytic Number Theory 3m6riycwzi0hkml8ggvt8kgztxgf2bn
For $m \in \N$ let $r_{\A, \ell} \left({m; N}\right)$ be the number of solutions to $(1)$ with no $x_i$ exceeding $N$. Then: :$\forall m \le N: r_{\A, \ell} \left({m; N}\right) = r_{\A, \ell} \left({m}\right)$ and: :$\forall m > \ell N: r_{\A, \ell} \left({m; N}\right) = 0$ Then we compute: :$\displaystyle T_N \left({s}\right)^\ell = \sum_{m \mathop = 0}^{\ell N} r_{\A, \ell} \left({m; N}\right) s^m$ and: :$(2): \quad \displaystyle V_N \left({\alpha}\right)^\ell =  \sum_{m \mathop = 0}^{\ell N} r_{\A, \ell} \left({m; N}\right) e \left({\alpha m}\right)$ Now it follows from Exponentials Form Orthonormal Basis for $\mathcal L^2$ that: :$\displaystyle \int_0^1 e \left({\alpha m}\right) e \left({-\alpha n}\right) \, \mathrm d \alpha = \delta_{m n}$ where $\delta_{m n}$ is the Kronecker delta. Therefore, we multiply $(2)$ by $e \left({-N \alpha}\right)$ and integrate: :$\displaystyle r_{\A, \ell} \left({N}\right) = r_{\A, \ell} \left({N; N}\right) = \int_0^1 \displaystyle V_N \left({\alpha}\right)^\ell e \left({-N \alpha}\right) \, \mathrm d \alpha $ {{qed}} {{Namedfor|Ivan Matveevich Vinogradov|cat = Vinogradov I M}} Category:Analytic Number Theory mo7rlqmgqxyozsgkmoi5hrxm6bywtv6
This is immediate from Product Form of Sum on Completely Multiplicative Function. {{namedfor|Leonhard Paul Euler|cat = Euler}} Category:Analytic Number Theory Category:Dirichlet Series p2ub82xp2f0y4q8r5tnwmokhanfez7x
From Euler Product: :$\displaystyle \sum_{n \mathop = 1}^\infty a_n n^{-z} = \prod_p \frac 1 {1 - a_p p^{-z} }$ {{iff}} $\displaystyle \sum_{n \mathop = 1}^\infty a_n n^{-z}$ is absolutely convergent. For all $n \in \Z_{\ge 1}$, let $a_n = 1$. From Convergence of P-Series: :$\displaystyle \sum_{n \mathop = 1}^\infty n^{-z}$ is absolutely convergent {{iff}}: :$\cmod z \ge 1$ It then follows that: :$\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^z} = \prod_p \frac 1 {1 - p^{-z} }$ {{qed}}
From Sum of Geometric Sequence: :$\dfrac 1 {1 - p^{-z} } = 1 + \dfrac 1 {p^z} + \dfrac 1 {p^{2 z} } + \cdots$ From Convergence of P-Series: :$\displaystyle \sum_{n \mathop = 1}^\infty n^{-z}$ is absolutely convergent {{iff}}: :$\cmod z \ge 1$ Thus: {{begin-eqn}} {{eqn | l = \sum_p \dfrac 1 {1 - p^{-z} }       | r = \sum_p \paren {1 + \dfrac 1 {p^z} + \dfrac 1 {p^{2 z} } + \dfrac 1 {p^{3 z} } + \cdots}       | c =  }} {{eqn | r = \paren {1 + \dfrac 1 {2^z} + \dfrac 1 {2^{2 z} } + \dfrac 1 {2^{3 z} } + \cdots}       | c =  }} {{eqn | o =        | ro= \times       | r = \paren {1 + \dfrac 1 {3^z} + \dfrac 1 {3^{2 z} } + \dfrac 1 {3^{3 z} } + \cdots}       | c =  }} {{eqn | o =        | ro= \times       | r = \paren {1 + \dfrac 1 {5^z} + \dfrac 1 {5^{2 z} } + \dfrac 1 {5^{3 z} } + \cdots}       | c =  }} {{eqn | o =        | ro= \times       | r = \cdots       | c =  }} {{eqn | r = 1 + \dfrac 1 {2^z} + \dfrac 1 {3^z} + \dfrac 1 {2^{2 z} } + \dfrac 1 {5^z}       | c =  }} {{eqn | o =        | ro= +       | r = \dfrac 1 {2^z 3^z} + \dfrac 1 {7^z} + \dfrac 1 {2^{3 z} } + \dfrac 1 {3^{2 z} }       | c =  }} {{eqn | o =        | ro= +       | r = \cdots       | c =  }} {{end-eqn}} The result follows from the Fundamental Theorem of Arithmetic. {{qed}} {{handwaving}}
{{begin-eqn}} {{eqn | l = S \cap T       | r = \O }} {{eqn | ll= \leadstoandfrom       | l = S       | o = \subseteq       | r = \map \complement T       | c = Intersection with Complement is Empty iff Subset }} {{eqn | ll= \leadstoandfrom       | l = S \cap \map \complement T       | r = S       | c = Intersection with Subset is Subset }} {{eqn | ll= \leadstoandfrom       | l = S \setminus T       | r = S       | c = Set Difference as Intersection with Complement }} {{end-eqn}} {{qed}}
From the definition of subtraction: :$a - b := a + \paren {-b}$ where $-b$ is the inverse for integer addition. From Integers under Addition form Abelian Group, the algebraic structure $\struct {\Z, +}$ is a group. Thus: :$\forall a, b \in \Z: a + \paren {-b} \in \Z$ Therefore integer subtraction is closed. {{qed}}
From the definition of subtraction: :$a - b := a + \paren {-b}$ where $-b$ is the inverse for rational number addition. From Rational Numbers under Addition form Abelian Group, $\struct {\Q, +}$ forms a group. Thus: :$\forall a, b \in \Q: a + \paren {-b} \in \Q$ Therefore rational number subtraction is closed. {{qed}}
From the definition of real subtraction: :$a - b := a + \paren {-b}$ where $-b$ is the inverse for real number addition. From Real Numbers under Addition form Abelian Group: :$\forall a, b \in \R: a + \paren {-b} \in \R$ Therefore real number subtraction is closed. {{qed}}
From the definition of complex subtraction: :$a - b := a + \paren {-b}$ where $-b$ is the inverse for complex number addition. From Complex Numbers under Addition form Abelian Group, it follows that: :$\forall a, b \in \C: a + \paren {-b} \in \C$ Therefore complex number subtraction is closed. {{qed}}
From the definition of division: :$a \div b := a \times \paren {\dfrac 1 b}$ where $\dfrac 1 b$ is the inverse for real number multiplication. From Strictly Positive Real Numbers under Multiplication form Uncountable Abelian Group, the algebraic structure $\struct {\R_{>0}, \times}$ forms a group. Thus it follows that: :$\forall a, b \in \R_{>0}: a \times \paren {\dfrac 1 b} \in \R$ Therefore real number division is closed in $\R_{>0}$. {{qed}}
Recall that Rational Numbers form Field. The result then follows directly from Field is Integral Domain. {{qed}}
Recall that Real Numbers form Field. The result then follows directly from Field is Integral Domain. {{qed}}
Recall that Complex Numbers form Field. The result then follows directly from Field is Integral Domain. {{qed}}
By definition, there is some polynomial $f \in K \left[{X}\right]$ such that $f \left({\alpha}\right) = 0$. Let $\mu$ be such a polynomial of minimal degree, and define: :$\mu_\alpha = \dfrac 1 a \mu $ where $a$ is the leading coefficient of $\mu$. Clearly this polynomial has the required properties, so we show uniqueness. Let $\mu'$ be another such polynomial. Then :$\mu_\alpha \left({\alpha}\right) - \mu' \left({\alpha}\right) = 0 - 0 = 0$ Since both polynomials are monic, $\mu_\alpha - \mu'$ has degree strictly less than $\mu_\alpha$. So we must have $\mu_\alpha - \mu' = 0$. Let $\mu_\alpha \divides f$. Then: :$f = g \mu_\alpha$ for some $g \in K \left[{X}\right]$ and: :$f \left({\alpha}\right) = 0 \cdot g \left({\alpha}\right) = 0$ Conversely, let $f \in K \left[{X}\right]$ be any polynomial such that $f \left({\alpha}\right) = 0$. By the Division Theorem for Polynomial Forms over Field, there exists $q, r \in K \left[{X}\right]$ such that: :$f = q \mu_\alpha + r$ and: :$\deg r < \deg \mu_\alpha$. Evaluating this expression at $\alpha$ we find that: :$f \left({\alpha}\right) = q \left({\alpha}\right) \mu_\alpha \left({\alpha}\right) + r \left({\alpha}\right) \implies r \left({\alpha}\right) = 0$ since $\mu_\alpha \left({\alpha}\right) = f \left({\alpha}\right) = 0$. But $\mu_\alpha$ has minimal degree among the non-zero polynomials that are zero at $\alpha$. Therefore as $\deg r < \deg \mu_\alpha$ we must have $r = 0$. Therefore: :$f = q \mu_\alpha$ That is, $\mu_\alpha$ divides $f$. {{qed}} Category:Field Extensions mg9uxte3mllfsldwer50b4uqk1lqt16
Since a UFD is Noetherian, and a Noetherian Domain is UFD if every irreducible element is prime, it is sufficient to prove that every irreducible element of $R \sqbrk X$ is prime. etc, {{ProofWanted|this possibly works}} {{Namedfor|Carl Friedrich Gauss|cat = Gauss}} Category:Factorization Category:Unique Factorization Domains bljlmlki1jz1fkjcafneorlm61c8zzm
Using various properties of groups: {{begin-eqn}} {{eqn | l = a \circ b^{-1}       | r = e       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {a \circ b^{-1} } \circ b       | r = e \circ b       | c = Cancellation Laws }} {{eqn | ll= \leadstoandfrom       | l = a \circ \paren {b^{-1} \circ b}       | r = e \circ b       | c = {{GroupAxiom|1}} }} {{eqn | ll= \leadstoandfrom       | l = a       | r = b       | c = {{GroupAxiom|2}} and {{GroupAxiom|3}} }} {{end-eqn}} {{qed}} Category:Group Theory hn59z6bi84g0y3d4d3ijnrryeortqfn
$\Z \sqbrk {\sqrt 2} \subseteq \R$ where $\R$ is the set of real numbers, so we immediately have that addition and multiplication are well-defined. === Closure === Let $a_1 + b_1 \sqrt 2, a_2 + b_2 \sqrt 2 \in \Z \sqbrk {\sqrt 2}$. Then: {{begin-eqn}} {{eqn | l = \paren {a_1 + b_1 \sqrt 2} + \paren {a_2 + b_2 \sqrt 2}       | r = \paren {a_1 + a_2} + \paren {b_1 + b_2} \sqrt 2       | c =  }} {{eqn | o = \in       | r = \Z \sqbrk {\sqrt 2}       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \paren {a_1 + b_1 \sqrt 2} \times \paren {a_2 + b_2 \sqrt 2}       | r = \paren {a_1 \times a_2 + 2 \times b_1 \times b_2} + \paren {a_1 \times b_2 + b_1 \times a_2} \sqrt 2       | c =  }} {{eqn | o = \in       | r = \Z \sqbrk {\sqrt 2}       | c =  }} {{end-eqn}} So both the operations $+$ and $\times$ are closed on $\Z \sqbrk {\sqrt 2}$. {{qed|lemma}} === Associativity === We have that addition and multiplication are associative on $\R$. Therefore it follows from Restriction of Associative Operation is Associative that they are also associative on $\Z \sqbrk {\sqrt 2}$. {{qed|lemma}} === Commutativity === We have that addition and multiplication are commutative on $\R$. Therefore it follows from Restriction of Commutative Operation is Commutative that they are also commutative on $\Z \sqbrk {\sqrt 2}$. {{qed|lemma}} === Identities === We have: {{begin-eqn}} {{eqn | l = \paren {a + b \sqrt 2} + \paren {0 + 0 \sqrt 2}       | r = \paren {a + 0} + \paren {b + 0} \sqrt 2       | c =  }} {{eqn | r = a + b \sqrt 2       | c =  }} {{end-eqn}} and similarly for $\paren {0 + 0 \sqrt 2} + \paren {a + b \sqrt 2}$. So $\paren {0 + 0 \sqrt 2}$ is the identity for $+$ on $\Z \sqbrk {\sqrt 2}$. Then: {{begin-eqn}} {{eqn | l = \paren {a + b \sqrt 2} \times \paren {1 + 0 \sqrt 2}       | r = \paren {a \times 1 + 2 \times b \times 0} + \paren {b \times 1 + a \times 0} \sqrt 2       | c =  }} {{eqn | r = a + b \sqrt 2       | c =  }} {{end-eqn}} and similarly for $\paren {a + b \sqrt 2} \times \paren {1 + 0 \sqrt 2}$. So $\paren {1 + 0 \sqrt 2}$ is the identity for $\times$ on $\Z \sqbrk {\sqrt 2}$. {{qed|lemma}} === Inverses === We have: {{begin-eqn}} {{eqn | l = \paren {a + b \sqrt 2} + \paren {-a + \paren {-b} \sqrt 2}       | r = \paren {a - a} + \paren {b - b} \sqrt 2       | c =  }} {{eqn | r = 0 + 0 \sqrt 2       | c =  }} {{end-eqn}} and similarly for $\paren {-a + \paren {-b} \sqrt 2} + \paren {a + b \sqrt 2}$. So $\paren {-a + \paren {-b} \sqrt 2}$ is the inverse of $\paren {a + b \sqrt 2}$ for $+$ on $\Z \sqbrk {\sqrt 2}$. We have no need to investigate inverses for $\times$ (which is convenient as $\times$ happens not to be closed for inverses on $\Z \sqbrk {\sqrt 2}$). {{qed|lemma}} === Distributivity === We have that Real Multiplication Distributes over Addition, so by Restriction of Operation Distributivity, $\times$ is distributive over $+$ on $\Z \sqbrk {\sqrt 2}$. {{qed|lemma}} === Divisors of Zero === By Real Numbers form Field, $\R$ is a field and so by Field is Integral Domain is itself an integral domain. Hence it has no proper zero divisors. {{qed|lemma}} The result follows by putting all the pieces together. {{qed}}
From Integers form Subdomain of Reals, $\struct {\Z, +, \times}$ is an integral subdomain of the real numbers $\R$. We have that $\sqrt 2 \in \R$. Every expression of the form: :$a_0 + a_1 \sqrt 2 + a_2 \paren {\sqrt 2}^2 + \cdots + a_n \paren {\sqrt 2}^n$ can be simplified to a number of the form $a + b \sqrt 2$, where $a, b \in \Z$. The result follows from Set of Polynomials over Integral Domain is Subring.