id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-3200 | Distributivity is Preserved in Induced Structure | Let $\struct {T, \oplus, \otimes}$ be an algebraic structure, and let $S$ be a set.
Let $T^S$ denote the set of all mappings from $S$ to $T$.
Let $\struct {T^S, \oplus}$ be the structure on $T^S$ induced by $\oplus$.
Let $\struct {T^S, \otimes}$ be the structure on $T^S$ induced by $\otimes$.
If $\otimes$ is distributi... | Let $\struct {T, \oplus, \otimes}$ be an algebraic structure in which $\otimes$ distributes over $\oplus$.
Let $f, g, h \in T^S$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\paren {f \otimes \paren {g \oplus h} } } x
| r = \map f x \otimes \paren {\map g x \oplus \map h x}
| c = {{Defof|Pointwise Operation}}
}}... | Let $\struct {T, \oplus, \otimes}$ be an [[Definition:Algebraic Structure|algebraic structure]], and let $S$ be a [[Definition:Set|set]].
Let $T^S$ denote the [[Definition:Set of All Mappings|set of all mappings]] from $S$ to $T$.
Let $\struct {T^S, \oplus}$ be the structure on $T^S$ [[Definition:Induced Structure|in... | Let $\struct {T, \oplus, \otimes}$ be an [[Definition:Algebraic Structure|algebraic structure]] in which $\otimes$ distributes over $\oplus$.
Let $f, g, h \in T^S$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\paren {f \otimes \paren {g \oplus h} } } x
| r = \map f x \otimes \paren {\map g x \oplus \map h x}
... | Distributivity is Preserved in Induced Structure | https://proofwiki.org/wiki/Distributivity_is_Preserved_in_Induced_Structure | https://proofwiki.org/wiki/Distributivity_is_Preserved_in_Induced_Structure | [
"Pointwise Operations",
"Distributive Operations"
] | [
"Definition:Algebraic Structure",
"Definition:Set",
"Definition:Set of All Mappings",
"Definition:Pointwise Operation/Induced Structure",
"Definition:Pointwise Operation/Induced Structure",
"Definition:Distributive Operation",
"Definition:Pointwise Operation",
"Definition:Distributive Operation"
] | [
"Definition:Algebraic Structure",
"Definition:Distributive Operation/Left",
"Definition:Distributive Operation/Right",
"Category:Pointwise Operations",
"Category:Distributive Operations"
] |
proofwiki-3201 | Dirichlet Series Absolute Convergence Lemma | Let $\ds \map f s = \sum_{n \mathop = 1}^\infty \frac {a_n} {n^s}$ be a Dirichlet series.
Suppose that $f$ converges absolutely at $s_0 = \sigma_0 + i t_0 \in \C$.
Then $f$ converges absolutely at all points $s = \sigma + i t \in \C$ with $\sigma \ge \sigma_0$. | Suppose that $f$ converges absolutely at $\sigma_0 + i t_0$.
If $\sigma \ge \sigma_0$, then:
{{begin-eqn}}
{{eqn | l = \size {\frac {a_n} {n^s} }
| r = \frac {\size {a_n} } {n^\sigma}
}}
{{eqn | o = \le
| r = \frac {\size {a_n} } {n^{\sigma_0} }
}}
{{eqn | r = \size {\frac {a_n} { n^{s_0} } }
}}
{{end-eqn}}... | Let $\ds \map f s = \sum_{n \mathop = 1}^\infty \frac {a_n} {n^s}$ be a [[Definition:Dirichlet Series|Dirichlet series]].
Suppose that $f$ [[Definition:Absolutely Convergent Series|converges absolutely]] at $s_0 = \sigma_0 + i t_0 \in \C$.
Then $f$ [[Definition:Absolutely Convergent Series|converges absolutely]] at ... | Suppose that $f$ [[Definition:Absolutely Convergent Series|converges absolutely]] at $\sigma_0 + i t_0$.
If $\sigma \ge \sigma_0$, then:
{{begin-eqn}}
{{eqn | l = \size {\frac {a_n} {n^s} }
| r = \frac {\size {a_n} } {n^\sigma}
}}
{{eqn | o = \le
| r = \frac {\size {a_n} } {n^{\sigma_0} }
}}
{{eqn | r = \... | Dirichlet Series Absolute Convergence Lemma | https://proofwiki.org/wiki/Dirichlet_Series_Absolute_Convergence_Lemma | https://proofwiki.org/wiki/Dirichlet_Series_Absolute_Convergence_Lemma | [
"Dirichlet Series"
] | [
"Definition:Dirichlet Series",
"Definition:Absolutely Convergent Series",
"Definition:Absolutely Convergent Series"
] | [
"Definition:Absolutely Convergent Series",
"Definition:Absolutely Convergent Series",
"Definition:Absolutely Convergent Series"
] |
proofwiki-3202 | Existence of Abscissa of Absolute Convergence | Let $\ds \map f s = \sum_{n \mathop = 1}^\infty a_n n^{-s}$ be a Dirichlet series.
Let the series $\ds \sum_{n \mathop = 1}^\infty \size { a_n n^{-s} }$ not converge for all $s \in \C$, or diverge for all $s \in \C$.
Then there exists a real number $\sigma_a$ such that $\map f s$ converges absolutely for all $s = \sigm... | Let $S$ be the set of all complex numbers $s$ such that $\map f s$ converges absolutely.
By hypothesis, there is some $s_0 = \sigma_0 + it_0 \in \C$ such that $\map f {s_0}$ converges absolutely, so $S$ is not empty.
Moreover, $S$ is bounded below, for otherwise it follows from Dirichlet Series Absolute Convergence Lem... | Let $\ds \map f s = \sum_{n \mathop = 1}^\infty a_n n^{-s}$ be a [[Definition:Dirichlet Series|Dirichlet series]].
Let the series $\ds \sum_{n \mathop = 1}^\infty \size { a_n n^{-s} }$ not converge for all $s \in \C$, or diverge for all $s \in \C$.
Then there exists a real number $\sigma_a$ such that $\map f s$ [[De... | Let $S$ be the set of all complex numbers $s$ such that $\map f s$ [[Definition:Absolutely Convergent Series|converges absolutely]].
By hypothesis, there is some $s_0 = \sigma_0 + it_0 \in \C$ such that $\map f {s_0}$ converges absolutely, so $S$ is not [[Definition:Empty Set|empty]].
Moreover, $S$ is bounded below, ... | Existence of Abscissa of Absolute Convergence | https://proofwiki.org/wiki/Existence_of_Abscissa_of_Absolute_Convergence | https://proofwiki.org/wiki/Existence_of_Abscissa_of_Absolute_Convergence | [
"Dirichlet Series"
] | [
"Definition:Dirichlet Series",
"Definition:Absolutely Convergent Series"
] | [
"Definition:Absolutely Convergent Series",
"Definition:Empty Set",
"Dirichlet Series Absolute Convergence Lemma",
"Definition:Infimum of Set",
"Dirichlet Series Absolute Convergence Lemma"
] |
proofwiki-3203 | Dirichlet Series Convergence Lemma | Let $\ds \map f s = \sum_{n \mathop = 1}^\infty \frac {a_n} {n^s}$ be a Dirichlet series.
Let $\map f s$ converge at $s_0 = \sigma_0 + i t_0$.
Then $\map f s$ converge for all $s = \sigma + i t$ where $\sigma > \sigma_0$. | We begin with a lemma: | Let $\ds \map f s = \sum_{n \mathop = 1}^\infty \frac {a_n} {n^s}$ be a [[Definition:Dirichlet Series|Dirichlet series]].
Let $\map f s$ [[Definition:Convergent Series|converge]] at $s_0 = \sigma_0 + i t_0$.
Then $\map f s$ [[Definition:Convergent Series|converge]] for all $s = \sigma + i t$ where $\sigma > \sigma_0... | We begin with a [[Definition:Lemma|lemma]]: | Dirichlet Series Convergence Lemma | https://proofwiki.org/wiki/Dirichlet_Series_Convergence_Lemma | https://proofwiki.org/wiki/Dirichlet_Series_Convergence_Lemma | [
"Dirichlet Series"
] | [
"Definition:Dirichlet Series",
"Definition:Convergent Series",
"Definition:Convergent Series"
] | [
"Definition:Lemma"
] |
proofwiki-3204 | Existence of Abscissa of Convergence/General | Let $s = \sigma + i t$
Let $\ds \map f s = \sum_{n \mathop = 1}^\infty a_n e^{-\lambda_n s}$ be a general Dirichlet series.
Then there exists a extended real number, $\sigma_0$, such that
:$(1): \quad$ For $\sigma < \sigma_0$, $\map f s$ diverges
:$(2): \quad$ For $\sigma > \sigma_0$, $\map f s$ converges. | If there does not exist an $s_0$ such that $\map f {s_0}$ converges, then $\sigma_0 = \infty$ and the theorem is vacuously true
If there does exist such an $s_0$, let $\sigma_0$ be the infimum of the real part of all such $s_0$, where $\sigma_0= -\infty$ if the set is not bounded from below.
*It is clear that $\map f s... | Let $s = \sigma + i t$
Let $\ds \map f s = \sum_{n \mathop = 1}^\infty a_n e^{-\lambda_n s}$ be a [[Definition:General Dirichlet Series|general Dirichlet series]].
Then there exists a [[Definition:Extended Real Number Line|extended real number]], $\sigma_0$, such that
:$(1): \quad$ For $\sigma < \sigma_0$, $\map f s$... | If there does not exist an $s_0$ such that $\map f {s_0}$ converges, then $\sigma_0 = \infty$ and the theorem is [[Definition:Vacuous Truth|vacuously true]]
If there does exist such an $s_0$, let $\sigma_0$ be the [[Definition:Infimum of Subset of Real Numbers|infimum]] of the real part of all such $s_0$, where $\sigm... | Existence of Abscissa of Convergence/General | https://proofwiki.org/wiki/Existence_of_Abscissa_of_Convergence/General | https://proofwiki.org/wiki/Existence_of_Abscissa_of_Convergence/General | [
"General Dirichlet Series"
] | [
"Definition:General Dirichlet Series",
"Definition:Extended Real Number Line"
] | [
"Definition:Vacuous Truth",
"Definition:Infimum of Set/Real Numbers",
"Dirichlet Series Convergence Lemma",
"Category:General Dirichlet Series"
] |
proofwiki-3205 | Difference of Abscissae of Convergence | Let $\ds \map f s = \sum_{n \mathop = 1}^\infty a_n n^{-s}$ be a Dirichlet series.
Suppose that $\map f s$ has finite Abscissa of Convergence $\sigma_c$.
Then the Abscissa of Absolute Convergence $\sigma_a$ is finite, and:
:$0 \le \sigma_a - \sigma_c \le 1$ | It is trivial that $\sigma_a \ge \sigma_c$.
Suppose $s_0 = \sigma_0 + i t_0 \in \C$ such that $\map f {s_0}$ converges.
It is sufficient to show that $\map f s$ converges absolutely for all $s = \sigma + i t$ with $\sigma > \sigma_0 + 1$.
Pick an upper bound $M$ for the real numbers $\size {a_n n^{-s_0} }$.
Then for $s... | Let $\ds \map f s = \sum_{n \mathop = 1}^\infty a_n n^{-s}$ be a [[Definition:Dirichlet Series|Dirichlet series]].
Suppose that $\map f s$ has finite [[Abscissa of Convergence]] $\sigma_c$.
Then the [[Abscissa of Absolute Convergence]] $\sigma_a$ is finite, and:
:$0 \le \sigma_a - \sigma_c \le 1$ | It is trivial that $\sigma_a \ge \sigma_c$.
Suppose $s_0 = \sigma_0 + i t_0 \in \C$ such that $\map f {s_0}$ [[Definition:Convergent Series|converges]].
It is sufficient to show that $\map f s$ [[Definition:Absolutely Convergent Series|converges absolutely]] for all $s = \sigma + i t$ with $\sigma > \sigma_0 + 1$.
... | Difference of Abscissae of Convergence | https://proofwiki.org/wiki/Difference_of_Abscissae_of_Convergence | https://proofwiki.org/wiki/Difference_of_Abscissae_of_Convergence | [
"Dirichlet Series"
] | [
"Definition:Dirichlet Series",
"Existence of Abscissa of Convergence",
"Existence of Abscissa of Absolute Convergence"
] | [
"Definition:Convergent Series",
"Definition:Absolutely Convergent Series",
"Comparison Test"
] |
proofwiki-3206 | Completion Theorem (Metric Space) | Let $M = \struct {A, d}$ be a metric space.
Then there exists a completion $\tilde M = \struct {\tilde A, \tilde d}$ of $\struct {A, d}$.
Moreover, this completion is unique up to isometry.
That is, if $\struct {\hat A, \hat d}$ is another completion of $\struct {A, d}$, then there is a bijection $\tau: \tilde A \leftr... | 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:
:$\ds \sequence... | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
Then there exists a [[Definition:Completion of Metric Space|completion]] $\tilde M = \struct {\tilde A, \tilde d}$ of $\struct {A, d}$.
Moreover, this [[Definition:Completion of Metric Space|completion]] is unique up to [[Definition:Isometry (Me... | We construct the [[Definition:Completion of Metric Space|completion of a metric space]] as [[Definition:Equivalence Class|equivalence classes]] of the [[Definition:Set|set]] of [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequences]] in the [[Definition:Metric Space|space]] under a suitable [[Definition:Equivalen... | Completion Theorem (Metric Space) | https://proofwiki.org/wiki/Completion_Theorem_(Metric_Space) | https://proofwiki.org/wiki/Completion_Theorem_(Metric_Space) | [
"Completion Theorem (Metric Space)",
"Isometries (Metric Spaces)",
"Metric Spaces",
"Complete Metric Spaces",
"Completion Theorem"
] | [
"Definition:Metric Space",
"Definition:Completion (Metric Space)",
"Definition:Completion (Metric Space)",
"Definition:Isometry (Metric Spaces)",
"Definition:Completion (Metric Space)",
"Definition:Bijection",
"Definition:Restriction/Mapping",
"Definition:Identity Mapping",
"Definition:Metric Space/... | [
"Definition:Completion (Metric Space)",
"Definition:Equivalence Class",
"Definition:Set",
"Definition:Cauchy Sequence/Metric Space",
"Definition:Metric Space",
"Definition:Equivalence Relation",
"Definition:Metric Space",
"Definition:Set",
"Definition:Cauchy Sequence/Metric Space",
"Definition:Rel... |
proofwiki-3207 | Completeness Criterion (Metric Spaces) | Let $M = \struct {S, d}$ be a metric space.
Let $A \subseteq S$ be a dense subset.
Suppose that every Cauchy sequence in $A$ converges in $M$.
Then $M$ is complete. | Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $S$.
For each $n$ pick a Cauchy sequence $\sequence {y_{n, m} }_{m \mathop \in \N}$ in $A$ converging to $x_n$ like so:
:400pxCaption
Let $N \in \N$ be such that $\map d {x_{n_1}, x_{n_2} } < \epsilon / 3$ for all $n_1, n_2 > N$.
Let $M \in \N$ be such th... | Let $M = \struct {S, d}$ be a [[Definition:Metric Space|metric space]].
Let $A \subseteq S$ be a [[Definition:Everywhere Dense|dense]] [[Definition:Subset|subset]].
Suppose that every [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequence]] in $A$ [[Definition:Convergent Sequence (Metric Space)|converges]] in $M... | Let $\sequence {x_n}_{n \mathop \in \N}$ be a [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequence]] in $S$.
For each $n$ pick a [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequence]] $\sequence {y_{n, m} }_{m \mathop \in \N}$ in $A$ converging to $x_n$ like so:
:[[File:CompletenessCriterionProof.png|40... | Completeness Criterion (Metric Spaces)/Proof 1 | https://proofwiki.org/wiki/Completeness_Criterion_(Metric_Spaces) | https://proofwiki.org/wiki/Completeness_Criterion_(Metric_Spaces)/Proof_1 | [
"Metric Spaces",
"Complete Metric Spaces",
"Completeness Criterion (Metric Spaces)"
] | [
"Definition:Metric Space",
"Definition:Everywhere Dense",
"Definition:Subset",
"Definition:Cauchy Sequence/Metric Space",
"Definition:Convergent Sequence/Metric Space",
"Definition:Complete Metric Space"
] | [
"Definition:Cauchy Sequence/Metric Space",
"Definition:Cauchy Sequence/Metric Space",
"File:CompletenessCriterionProof.png",
"Triangle Inequality",
"Definition:Cauchy Sequence/Metric Space",
"Definition:Convergent Sequence/Metric Space",
"Definition:Limit of Sequence/Metric Space"
] |
proofwiki-3208 | Completeness Criterion (Metric Spaces) | Let $M = \struct {S, d}$ be a metric space.
Let $A \subseteq S$ be a dense subset.
Suppose that every Cauchy sequence in $A$ converges in $M$.
Then $M$ is 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\ran... | Let $M = \struct {S, d}$ be a [[Definition:Metric Space|metric space]].
Let $A \subseteq S$ be a [[Definition:Everywhere Dense|dense]] [[Definition:Subset|subset]].
Suppose that every [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequence]] in $A$ [[Definition:Convergent Sequence (Metric Space)|converges]] in $M... | Let $\left\langle{x_n}\right\rangle$ be a [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequence]] in $M$.
Since $A$ is [[Definition:Everywhere Dense|dense]], we can invoke the [[Axiom:Axiom of Countable Choice|Axiom of Countable Choice]] to choose for each $n$ some $y_n \in A$ which is within $1 / n$ of $x_n$.
... | Completeness Criterion (Metric Spaces)/Proof 2 | https://proofwiki.org/wiki/Completeness_Criterion_(Metric_Spaces) | https://proofwiki.org/wiki/Completeness_Criterion_(Metric_Spaces)/Proof_2 | [
"Metric Spaces",
"Complete Metric Spaces",
"Completeness Criterion (Metric Spaces)"
] | [
"Definition:Metric Space",
"Definition:Everywhere Dense",
"Definition:Subset",
"Definition:Cauchy Sequence/Metric Space",
"Definition:Convergent Sequence/Metric Space",
"Definition:Complete Metric Space"
] | [
"Definition:Cauchy Sequence/Metric Space",
"Definition:Everywhere Dense",
"Axiom:Axiom of Countable Choice",
"Definition:Cauchy Sequence/Metric Space",
"Definition:Cauchy Sequence/Metric Space",
"Triangle Inequality",
"Definition:Cauchy Sequence/Metric Space",
"Definition:Convergent Sequence/Metric Sp... |
proofwiki-3209 | Natural Number Addition is Closed | The operation of addition on the set of natural numbers $\N$ is closed:
:$\forall x, y \in \N: x + y \in \N$ | 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}} | The operation of [[Definition:Natural Number Addition|addition]] on the [[Definition:Set|set]] of [[Definition:Natural Numbers|natural numbers]] $\N$ is [[Definition:Closed Algebraic Structure|closed]]:
:$\forall x, y \in \N: x + y \in \N$ | Follows directly from [[Natural Numbers under Addition form Commutative Monoid]].
A [[Definition:Monoid|monoid]] by definition is a [[Definition:Semigroup|semigroup]].
Again by definition, the [[Definition:Binary Operation|operation]] in a [[Definition:Semigroup|semigroup]] is [[Definition:Closed Algebraic Structure|... | Natural Number Addition is Closed | https://proofwiki.org/wiki/Natural_Number_Addition_is_Closed | https://proofwiki.org/wiki/Natural_Number_Addition_is_Closed | [
"Natural Number Addition"
] | [
"Definition:Addition/Natural Numbers",
"Definition:Set",
"Definition:Natural Numbers",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] | [
"Natural Numbers under Addition form Commutative Monoid",
"Definition:Monoid",
"Definition:Semigroup",
"Definition:Operation/Binary Operation",
"Definition:Semigroup",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] |
proofwiki-3210 | Properties of Ordered Field | Let $\struct {F, +, \cdot}$ be an ordered field with unity $1$, zero $0$.
Denote the strict order by $<$ and the weak order by $\le$.
Let $\Char F$ denote the characteristic of $F$.
Then the following hold for all $x, y, z \in F$:
:$(1): \quad x < 0 \iff -x > 0$
:$(2): \quad x > y \iff x - y > 0$
:$(3): \quad x < y \if... | 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 dedu... | Let $\struct {F, +, \cdot}$ be an [[Definition:Ordered Field|ordered field]] with [[Definition:Unity of Field|unity]] $1$, [[Definition:Field Zero|zero]] $0$.
Denote the [[Definition:Strict Ordering|strict order]] by $<$ and the [[Definition:Weak Ordering|weak order]] by $\le$.
Let $\Char F$ denote the [[Definition:C... | By definition of [[Definition:Ordering|ordering]], the relation $\le$ is:
:[[Definition:Reflexive Relation|reflexive]]
:[[Definition:Transitive Relation|transitive]]
:[[Definition:Antisymmetric Relation|antisymmetric]]
and furthermore, every pair of elements is [[Definition:Comparable Elements|comparable]].
The [[Defi... | Properties of Ordered Field | https://proofwiki.org/wiki/Properties_of_Ordered_Field | https://proofwiki.org/wiki/Properties_of_Ordered_Field | [
"Ordered Fields"
] | [
"Definition:Ordered Field",
"Definition:Multiplicative Identity",
"Definition:Field Zero",
"Definition:Strict Ordering",
"Definition:Ordering",
"Definition:Characteristic of Field"
] | [
"Definition:Ordering",
"Definition:Reflexive Relation",
"Definition:Transitive Relation",
"Definition:Antisymmetric Relation",
"Definition:Comparable Elements",
"Definition:Ordering Compatible with Ring Structure",
"Category:Ordered Fields"
] |
proofwiki-3211 | Characterisation of Ordered Fields | Let $\struct {k, +, \cdot}$ be a field with unity $1$ and zero $0$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|There exists a total ordering $\le$ on $k$ such that $\struct {k, \le}$ is an ordered field}}
{{item|(2):|$-1$ cannot be written as a sum of squares of elements of $k$}}
{{item|(3):|$0$ cannot be written as a non-... | === $(2)$ iff $(3)$ ===
Suppose there exist $\set {x_i: i \in I}$ such that
:$\ds -1 = \sum_{i \mathop \in I} x_i^2$
Then
:$\ds 0 = 1^2 + \sum_{i \mathop \in I} x_i^2$
a non-empty sum of non-zero squared of $k$.
Conversely, suppose that there is a set $\set {x_i: i \in I} \ne \O$ with $x_i \ne 0$ for all $i \in I$ such... | Let $\struct {k, +, \cdot}$ be a [[Definition:Field (Abstract Algebra)|field]] with [[Definition:Unity of Field|unity]] $1$ and [[Definition:Field Zero|zero]] $0$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|There exists a [[Definition:Total Ordering|total ordering]] $\le$ on $k$ such that $\struct {k, \le}$ is an [[Defini... | === $(2)$ iff $(3)$ ===
Suppose there exist $\set {x_i: i \in I}$ such that
:$\ds -1 = \sum_{i \mathop \in I} x_i^2$
Then
:$\ds 0 = 1^2 + \sum_{i \mathop \in I} x_i^2$
a non-empty sum of non-zero squared of $k$.
Conversely, suppose that there is a set $\set {x_i: i \in I} \ne \O$ with $x_i \ne 0$ for all $i \in ... | Characterisation of Ordered Fields | https://proofwiki.org/wiki/Characterisation_of_Ordered_Fields | https://proofwiki.org/wiki/Characterisation_of_Ordered_Fields | [
"Field Theory",
"Ordered Fields"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Multiplicative Identity",
"Definition:Field Zero",
"Definition:Total Ordering",
"Definition:Ordered Field"
] | [] |
proofwiki-3212 | Hahn-Banach Theorem/Real Vector Space | Let $X$ be a vector space over $\R$.
Let $p : X \to \R$ be a sublinear functional.
Let $X_0$ be a linear subspace of $X$.
Let $f_0 : X_0 \to \R$ be a linear functional such that:
:$\map {f_0} x \le \map p x$ for each $x \in X_0$.
Then there exists a linear functional $f$ defined on the whole space $X$ which extends... | We first prove a lemma: | Let $X$ be a [[Definition:Vector Space|vector space]] over $\R$.
Let $p : X \to \R$ be a [[Definition:Sublinear Functional|sublinear functional]].
Let $X_0$ be a [[Definition:Linear Subspace|linear subspace]] of $X$.
Let $f_0 : X_0 \to \R$ be a [[Definition:Linear Functional|linear functional]] such that:
:$\map... | We first prove a lemma: | Hahn-Banach Theorem/Real Vector Space | https://proofwiki.org/wiki/Hahn-Banach_Theorem/Real_Vector_Space | https://proofwiki.org/wiki/Hahn-Banach_Theorem/Real_Vector_Space | [
"Hahn-Banach Theorem"
] | [
"Definition:Vector Space",
"Definition:Sublinear Functional",
"Definition:Linear Subspace",
"Definition:Linear Functional",
"Definition:Linear Functional",
"Definition:Vector Space",
"Definition:Extension of Mapping",
"Definition:Linear Functional"
] | [] |
proofwiki-3213 | Completion of Valued Field | Let $\struct {k, \norm {\,\cdot\,} }$ be a valued field.
Then there exists a completion $\struct {k', \norm {\,\cdot\,}'}$ of $\struct {k, \norm {\,\cdot\,} }$ such that $\struct {k', \norm {\,\cdot\,}'}$ is a valued field.
Furthermore, every completion of $\struct{k, \norm {\,\cdot\,} }$ is isometrically isomorphic to... | 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 I... | Let $\struct {k, \norm {\,\cdot\,} }$ be a [[Definition:Valued Field|valued field]].
Then there exists a [[Definition:Completion (Normed Division Ring)|completion]] $\struct {k', \norm {\,\cdot\,}'}$ of $\struct {k, \norm {\,\cdot\,} }$ such that $\struct {k', \norm {\,\cdot\,}'}$ is a [[Definition:Valued Field|valued... | By [[Completion of Normed Division Ring]] then $\struct {k, \norm {\, \cdot \,} }$ has a [[Definition:Completion (Normed Division Ring)|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 [[De... | Completion of Valued Field | https://proofwiki.org/wiki/Completion_of_Valued_Field | https://proofwiki.org/wiki/Completion_of_Valued_Field | [
"Valued Fields"
] | [
"Definition:Valued Field",
"Definition:Completion (Normed Division Ring)",
"Definition:Valued Field",
"Definition:Completion (Normed Division Ring)",
"Definition:Isometric Isomorphism/Normed Division Ring"
] | [
"Completion of Normed Division Ring",
"Definition:Completion (Normed Division Ring)",
"Normed Division Ring is Field iff Completion is Field",
"Definition:Field (Abstract Algebra)",
"Normed Division Ring Completions are Isometric and Isomorphic",
"Definition:Completion (Normed Division Ring)",
"Definiti... |
proofwiki-3214 | Hardy-Littlewood Circle Method | Let $\AA$ be a subset of the non-negative integers.
Let:
:$\ds \map T s = \sum_{a \mathop \in \AA} s^a$
be the generating function for $\AA$.
For $N \in \N$, let $\map {r_{\AA, \ell} } N$ be the number of solutions $\tuple {x_1, \ldots, x_\ell} \in \AA^\ell$ to the equation:
:$x_1 + \cdots + x_\ell = N$
Then:
:$\ds \fo... | We have:
:$\ds \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 ... | Let $\AA$ be a [[Definition:Subset|subset]] of the non-negative [[Definition:Integer|integers]].
Let:
:$\ds \map T s = \sum_{a \mathop \in \AA} s^a$
be the [[Definition:Generating Function|generating function]] for $\AA$.
For $N \in \N$, let $\map {r_{\AA, \ell} } N$ be the number of solutions $\tuple {x_1, \ldots,... | We have:
:$\ds \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 rec... | Hardy-Littlewood Circle Method | https://proofwiki.org/wiki/Hardy-Littlewood_Circle_Method | https://proofwiki.org/wiki/Hardy-Littlewood_Circle_Method | [
"Analytic Number Theory"
] | [
"Definition:Subset",
"Definition:Integer",
"Definition:Generating Function"
] | [
"Cauchy's Integral Formula/General Result",
"Definition:Complex Function",
"Definition:Holomorphic Function",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Path (Complex Plane)",
"Definition:Winding Number",
"Definition:Generating Function",
"Definition:Taylor Series",
"Definition:Circle",
... |
proofwiki-3215 | Vinogradov Circle Method | Let $\AA$ be a subset of the non-negative integers.
For $\alpha \in \R$, let:
:$\map e \alpha := \map \exp {2 \pi i \alpha}$
Let:
:$\ds \map {T_N} s = \sum_{\substack {a \mathop \in \AA \\ a \mathop \le N} } s^a$
be the truncated generating function for $\AA$.
{{explain|"truncated" generating function}}
Let:
:$\map {V_... | For $m \in \N$ let $\map {r_{\AA, \ell} } {m; N}$ be the number of solutions to $(1)$ with no $x_i$ exceeding $N$.
Then:
:$\forall m \le N: \map {r_{\AA, \ell} } {m; N} = \map {r_{\AA, \ell} } m$
and:
:$\forall m > \ell N: \map {r_{\AA, \ell} } {m; N} = 0$
Then we compute:
:$\ds \map {T_N} s^\ell = \sum_{m \mathop = 0}... | Let $\AA$ be a subset of the non-negative integers.
For $\alpha \in \R$, let:
:$\map e \alpha := \map \exp {2 \pi i \alpha}$
Let:
:$\ds \map {T_N} s = \sum_{\substack {a \mathop \in \AA \\ a \mathop \le N} } s^a$
be the truncated [[Definition:Generating Function|generating function]] for $\AA$.
{{explain|"truncate... | For $m \in \N$ let $\map {r_{\AA, \ell} } {m; N}$ be the number of solutions to $(1)$ with no $x_i$ exceeding $N$.
Then:
:$\forall m \le N: \map {r_{\AA, \ell} } {m; N} = \map {r_{\AA, \ell} } m$
and:
:$\forall m > \ell N: \map {r_{\AA, \ell} } {m; N} = 0$
Then we compute:
:$\ds \map {T_N} s^\ell = \sum_{m \mathop =... | Vinogradov Circle Method | https://proofwiki.org/wiki/Vinogradov_Circle_Method | https://proofwiki.org/wiki/Vinogradov_Circle_Method | [
"Analytic Number Theory"
] | [
"Definition:Generating Function",
"Definition:Generating Function"
] | [
"Exponentials Form Orthonormal Basis for L^2",
"Definition:Kronecker Delta",
"Category:Analytic Number Theory"
] |
proofwiki-3216 | Correspondence between Linear Group Actions and Linear Representations | Let $\struct {k, +, \cdot}$ be a field.
Let $V$ be a vector space over $k$ of finite dimension.
Let $\struct {G, *}$ be a finite group.
There is a one-to-one correspondence between linear group actions of $G$ on $V$ and linear representations of $G$ in $V$, as follows:
Let $\phi: G \times V \to V$ be a group action.
Le... | The first equivalence follows from Correspondence Between Group Actions and Permutation Representations. | Let $\struct {k, +, \cdot}$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $V$ be a [[Definition:Vector Space|vector space]] over $k$ of [[Definition:Finite Dimensional Free Module|finite]] [[Definition:Dimension (Linear Algebra)|dimension]].
Let $\struct {G, *}$ be a [[Definition:Finite Group|finite group]]... | The first equivalence follows from [[Correspondence Between Group Actions and Permutation Representations]]. | Correspondence between Linear Group Actions and Linear Representations | https://proofwiki.org/wiki/Correspondence_between_Linear_Group_Actions_and_Linear_Representations | https://proofwiki.org/wiki/Correspondence_between_Linear_Group_Actions_and_Linear_Representations | [
"Group Actions",
"Representation Theory"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Vector Space",
"Definition:Dimension of Module/Finite",
"Definition:Dimension (Linear Algebra)",
"Definition:Finite Group",
"Definition:Linear Group Action",
"Definition:Linear Representation",
"Definition:Group Action",
"Definition:Group Representa... | [
"Correspondence Between Group Actions and Permutation Representations"
] |
proofwiki-3217 | Set Difference with Disjoint Set | Let $S, T$ be sets.
Then:
:$S \cap T = \O \iff S \setminus T = S$
where:
:$S \cap T$ denotes set intersection
:$\O$ denotes the empty set
:$S \setminus T$ denotes set difference. | {{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 wit... | Let $S, T$ be [[Definition:Set|sets]].
Then:
:$S \cap T = \O \iff S \setminus T = S$
where:
:$S \cap T$ denotes [[Definition:Set Intersection|set intersection]]
:$\O$ denotes the [[Definition:Empty Set|empty set]]
:$S \setminus T$ denotes [[Definition:Set Difference|set difference]]. | {{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 = [[Intersecti... | Set Difference with Disjoint Set | https://proofwiki.org/wiki/Set_Difference_with_Disjoint_Set | https://proofwiki.org/wiki/Set_Difference_with_Disjoint_Set | [
"Set Difference",
"Disjoint Sets"
] | [
"Definition:Set",
"Definition:Set Intersection",
"Definition:Empty Set",
"Definition:Set Difference"
] | [
"Intersection with Complement is Empty iff Subset",
"Intersection with Subset is Subset",
"Set Difference as Intersection with Complement"
] |
proofwiki-3218 | Integer Subtraction is Closed | The set of integers is closed under subtraction:
:$\forall a, b \in \Z: a - b \in \Z$ | 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}} | The [[Definition:Set|set]] of [[Definition:Integer|integers]] is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Integer Subtraction|subtraction]]:
:$\forall a, b \in \Z: a - b \in \Z$ | From the definition of [[Definition:Integer Subtraction|subtraction]]:
:$a - b := a + \paren {-b}$
where $-b$ is the [[Definition:Inverse Element|inverse]] for [[Definition:Integer Addition|integer addition]].
From [[Integers under Addition form Abelian Group]], the [[Definition:Algebraic Structure|algebraic structure... | Integer Subtraction is Closed | https://proofwiki.org/wiki/Integer_Subtraction_is_Closed | https://proofwiki.org/wiki/Integer_Subtraction_is_Closed | [
"Integer Subtraction",
"Algebraic Closure"
] | [
"Definition:Set",
"Definition:Integer",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Subtraction/Integers"
] | [
"Definition:Subtraction/Integers",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Addition/Integers",
"Integers under Addition form Abelian Group",
"Definition:Algebraic Structure",
"Definition: Group",
"Definition:Subtraction/Integers",
"Definition:Closure (Abstract Algebra)/Algebraic S... |
proofwiki-3219 | Rational Subtraction is Closed | The set of rational numbers is closed under subtraction:
:$\forall a, b \in \Q: a - b \in \Q$ | 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 Infinite Abelian Group, $\struct {\Q, +}$ forms a group.
Thus:
:$\forall a, b \in \Q: a + \paren {-b} \in \Q$
Therefore rational number subtraction is closed.... | The [[Definition:Set|set]] of [[Definition:Rational Number|rational numbers]] is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Rational Subtraction|subtraction]]:
:$\forall a, b \in \Q: a - b \in \Q$ | From the definition of [[Definition:Rational Subtraction|subtraction]]:
:$a - b := a + \paren {-b}$
where $-b$ is the [[Definition:Inverse Element|inverse]] for [[Definition:Rational Addition|rational number addition]].
From [[Rational Numbers under Addition form Infinite Abelian Group]], $\struct {\Q, +}$ forms a [[D... | Rational Subtraction is Closed | https://proofwiki.org/wiki/Rational_Subtraction_is_Closed | https://proofwiki.org/wiki/Rational_Subtraction_is_Closed | [
"Rational Subtraction",
"Algebraic Closure"
] | [
"Definition:Set",
"Definition:Rational Number",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Subtraction/Rational Numbers"
] | [
"Definition:Subtraction/Rational Numbers",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Addition/Rational Numbers",
"Rational Numbers under Addition form Infinite Abelian Group",
"Definition:Group",
"Definition:Subtraction/Rational Numbers",
"Definition:Closure (Abstract Algebra)/Algebra... |
proofwiki-3220 | Real Subtraction is Closed | The set of real numbers is closed under subtraction:
:$\forall a, b \in \R: a - b \in \R$ | 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 Group:
:$\forall a, b \in \R: a + \paren {-b} \in \R$
Therefore real number subtraction is closed.
{{qed}} | The [[Definition:Set|set]] of [[Definition:Real Number|real numbers]] is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Real Subtraction|subtraction]]:
:$\forall a, b \in \R: a - b \in \R$ | From the definition of [[Definition:Real Subtraction|real subtraction]]:
:$a - b := a + \paren {-b}$
where $-b$ is the [[Definition:Inverse Element|inverse]] for [[Definition:Real Addition|real number addition]].
From [[Real Numbers under Addition form Group]]:
:$\forall a, b \in \R: a + \paren {-b} \in \R$
Therefore... | Real Subtraction is Closed | https://proofwiki.org/wiki/Real_Subtraction_is_Closed | https://proofwiki.org/wiki/Real_Subtraction_is_Closed | [
"Real Subtraction",
"Algebraic Closure"
] | [
"Definition:Set",
"Definition:Real Number",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Subtraction/Real Numbers"
] | [
"Definition:Subtraction/Real Numbers",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Addition/Real Numbers",
"Real Numbers under Addition form Group",
"Definition:Subtraction/Real Numbers",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] |
proofwiki-3221 | Complex Subtraction is Closed | The set of complex numbers is closed under subtraction:
:$\forall a, b \in \C: a - b \in \C$ | 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 Group, it follows that:
:$\forall a, b \in \C: a + \paren {-b} \in \C$
Therefore complex number subtraction is closed.
{{qed}} | The [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Complex Subtraction|subtraction]]:
:$\forall a, b \in \C: a - b \in \C$ | From the definition of [[Definition:Complex Subtraction|complex subtraction]]:
:$a - b := a + \paren {-b}$
where $-b$ is the [[Definition:Inverse Element|inverse]] for [[Definition:Complex Addition|complex number addition]].
From [[Complex Numbers under Addition form Group]], it follows that:
:$\forall a, b \in \C: a ... | Complex Subtraction is Closed | https://proofwiki.org/wiki/Complex_Subtraction_is_Closed | https://proofwiki.org/wiki/Complex_Subtraction_is_Closed | [
"Complex Subtraction"
] | [
"Definition:Set",
"Definition:Complex Number",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Subtraction/Complex Numbers"
] | [
"Definition:Subtraction/Complex Numbers",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Addition/Complex Numbers",
"Complex Numbers under Addition form Group",
"Definition:Subtraction/Complex Numbers",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] |
proofwiki-3222 | Strictly Positive Real Numbers are Closed under Division | The set $\R_{>0}$ of strictly positive real numbers is closed under division:
:$\forall a, b \in \R_{>0}: a \div b \in \R_{>0}$ | 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:... | The [[Definition:Set|set]] $\R_{>0}$ of [[Definition:Strictly Positive Real Number|strictly positive real numbers]] is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Real Division|division]]:
:$\forall a, b \in \R_{>0}: a \div b \in \R_{>0}$ | From the definition of [[Definition:Real Division|division]]:
:$a \div b := a \times \paren {\dfrac 1 b}$
where $\dfrac 1 b$ is the [[Definition:Inverse Element|inverse]] for [[Definition:Real Multiplication|real number multiplication]].
From [[Strictly Positive Real Numbers under Multiplication form Uncountable Abeli... | Strictly Positive Real Numbers are Closed under Division | https://proofwiki.org/wiki/Strictly_Positive_Real_Numbers_are_Closed_under_Division | https://proofwiki.org/wiki/Strictly_Positive_Real_Numbers_are_Closed_under_Division | [
"Real Division",
"Algebraic Closure"
] | [
"Definition:Set",
"Definition:Strictly Positive/Real Number",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Division/Field/Real Numbers"
] | [
"Definition:Division/Field/Real Numbers",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Multiplication/Real Numbers",
"Strictly Positive Real Numbers under Multiplication form Uncountable Abelian Group",
"Definition:Algebraic Structure",
"Definition:Group",
"Definition:Division/Field/Real... |
proofwiki-3223 | Rational Numbers form Integral Domain | The set of rational numbers $\Q$ forms an integral domain under addition and multiplication: $\struct {\Q, +, \times}$. | Recall that Rational Numbers form Field.
The result then follows directly from Field is Integral Domain.
{{qed}} | The [[Definition:Rational Number|set of rational numbers]] $\Q$ forms an [[Definition:Integral Domain|integral domain]] under [[Definition:Rational Addition|addition]] and [[Definition:Rational Multiplication|multiplication]]: $\struct {\Q, +, \times}$. | Recall that [[Rational Numbers form Field]].
The result then follows directly from [[Field is Integral Domain]].
{{qed}} | Rational Numbers form Integral Domain | https://proofwiki.org/wiki/Rational_Numbers_form_Integral_Domain | https://proofwiki.org/wiki/Rational_Numbers_form_Integral_Domain | [
"Examples of Integral Domains",
"Rational Numbers"
] | [
"Definition:Rational Number",
"Definition:Integral Domain",
"Definition:Addition/Rational Numbers",
"Definition:Multiplication/Rational Numbers"
] | [
"Rational Numbers form Field",
"Field is Integral Domain"
] |
proofwiki-3224 | Real Numbers form Integral Domain | The set of real numbers $\R$ forms an integral domain under addition and multiplication: $\struct {\R, +, \times}$. | Recall that Real Numbers form Field.
The result then follows directly from Field is Integral Domain.
{{qed}} | The [[Definition:Real Number|set of real numbers]] $\R$ forms an [[Definition:Integral Domain|integral domain]] under [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]]: $\struct {\R, +, \times}$. | Recall that [[Real Numbers form Field]].
The result then follows directly from [[Field is Integral Domain]].
{{qed}} | Real Numbers form Integral Domain | https://proofwiki.org/wiki/Real_Numbers_form_Integral_Domain | https://proofwiki.org/wiki/Real_Numbers_form_Integral_Domain | [
"Examples of Integral Domains",
"Real Numbers"
] | [
"Definition:Real Number",
"Definition:Integral Domain",
"Definition:Addition/Real Numbers",
"Definition:Multiplication/Real Numbers"
] | [
"Real Numbers form Field",
"Field is Integral Domain"
] |
proofwiki-3225 | Complex Numbers form Integral Domain | The set of complex numbers $\C$ forms an integral domain under addition and multiplication: $\struct {\C, +, \times}$. | Recall that Complex Numbers form Field.
The result then follows directly from Field is Integral Domain.
{{qed}} | The [[Definition:Complex Number|set of complex numbers]] $\C$ forms an [[Definition:Integral Domain|integral domain]] under [[Definition:Complex Addition|addition]] and [[Definition:Complex Multiplication|multiplication]]: $\struct {\C, +, \times}$. | Recall that [[Complex Numbers form Field]].
The result then follows directly from [[Field is Integral Domain]].
{{qed}} | Complex Numbers form Integral Domain | https://proofwiki.org/wiki/Complex_Numbers_form_Integral_Domain | https://proofwiki.org/wiki/Complex_Numbers_form_Integral_Domain | [
"Examples of Integral Domains",
"Complex Numbers"
] | [
"Definition:Complex Number",
"Definition:Integral Domain",
"Definition:Addition/Complex Numbers",
"Definition:Multiplication/Complex Numbers"
] | [
"Complex Numbers form Field",
"Field is Integral Domain"
] |
proofwiki-3226 | Gauss's Lemma on Unique Factorization Domains | Let $R$ be a unique factorization domain.
Then the ring of polynomials $R \sqbrk X$ is also a unique factorization domain. | {{ProofWanted|this possibly works}}
{{Namedfor|Carl Friedrich Gauss|cat = Gauss}}
Category:Factorization
Category:Unique Factorization Domains
nhk280w723paxiql05rzkp23k6sghaq | Let $R$ be a [[Definition:Unique Factorization Domain|unique factorization domain]].
Then the [[Definition:Ring of Polynomials|ring of polynomials]] $R \sqbrk X$ is also a [[Definition:Unique Factorization Domain|unique factorization domain]]. | {{ProofWanted|this possibly works}}
{{Namedfor|Carl Friedrich Gauss|cat = Gauss}}
[[Category:Factorization]]
[[Category:Unique Factorization Domains]]
nhk280w723paxiql05rzkp23k6sghaq | Gauss's Lemma on Unique Factorization Domains | https://proofwiki.org/wiki/Gauss's_Lemma_on_Unique_Factorization_Domains | https://proofwiki.org/wiki/Gauss's_Lemma_on_Unique_Factorization_Domains | [
"Factorization",
"Unique Factorization Domains"
] | [
"Definition:Unique Factorization Domain",
"Definition:Polynomial Ring",
"Definition:Unique Factorization Domain"
] | [
"Category:Factorization",
"Category:Unique Factorization Domains"
] |
proofwiki-3227 | Product with Inverse equals Identity iff Equality | Let $\struct {G, \circ}$ be a group whose identity element is $e$.
Then:
:$\forall a, b \in G: a \circ b^{-1} = e \iff a = b$ | 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... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity element]] is $e$.
Then:
:$\forall a, b \in G: a \circ b^{-1} = e \iff a = b$ | Using various properties of [[Definition:Group|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^... | Product with Inverse equals Identity iff Equality | https://proofwiki.org/wiki/Product_with_Inverse_equals_Identity_iff_Equality | https://proofwiki.org/wiki/Product_with_Inverse_equals_Identity_iff_Equality | [
"Group Theory"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Definition:Group",
"Cancellation Laws",
"Category:Group Theory"
] |
proofwiki-3228 | Gröbner Basis | Let $F$ be a finite set of polynomials.
{{explain|Explain what the underlying field of the polynomials is. Real numbers, abstract rings, general fields? Link to the appropriate definition for the case.}}
Let $\map {\operatorname {SP} } {f_1, f_2}$ be the $S$-polynomial of $f_1$ and $f_2$.
Let $g$ be a polynomial.
Let $... | === Necessary Condition ===
Let $F$ be a Gröbner basis.
Let $f_1, f_2 \in F$.
Then:
:$\map {\operatorname {SP} } {f_1, f_2} \in \map {\operatorname {Ideal} } F$
That is:
:$\map {\operatorname {SP} } {f_1, f_2} \equiv_F 0$
{{explain|Link to a result explaining the above.}}
By the relation between reduction and congruenc... | Let $F$ be a finite set of polynomials.
{{explain|Explain what the underlying field of the polynomials is. Real numbers, abstract rings, general fields? Link to the appropriate definition for the case.}}
Let $\map {\operatorname {SP} } {f_1, f_2}$ be the [[Definition:S-Polynomial|$S$-polynomial]] of $f_1$ and $f_2$.
... | === Necessary Condition ===
Let $F$ be a [[Definition:Gröbner Basis|Gröbner basis]].
Let $f_1, f_2 \in F$.
Then:
:$\map {\operatorname {SP} } {f_1, f_2} \in \map {\operatorname {Ideal} } F$
That is:
:$\map {\operatorname {SP} } {f_1, f_2} \equiv_F 0$
{{explain|Link to a result explaining the above.}}
By the relat... | Gröbner Basis | https://proofwiki.org/wiki/Gröbner_Basis | https://proofwiki.org/wiki/Gröbner_Basis | [] | [
"Definition:S-Polynomial",
"Definition:Reduced Form of Polynomial",
"Definition:Gröbner Basis"
] | [
"Definition:Gröbner Basis",
"Definition:Gröbner Basis",
"Definition:Church-Rosser Property",
"Definition:Normal Form Church-Rosser Property"
] |
proofwiki-3229 | Quadratic Integers over 2 form Subdomain of Reals | Let $\Z \sqbrk {\sqrt 2}$ denote the set of quadratic integers over $2$:
:$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$
That is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are integers.
Then the algebraic structure:
:$\struct {\Z \sqbrk {\sqrt 2}, +, \times}$
where $+$ and $\times$ are conve... | $\Z \sqbrk {\sqrt 2} \subseteq \R$ 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}
... | Let $\Z \sqbrk {\sqrt 2}$ denote the [[Definition:Set|set]] of [[Definition:Quadratic Integer|quadratic integers]] over $2$:
:$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$
That is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are [[Definition:Integer|integers]].
Then the [[Definition:Algebr... | $\Z \sqbrk {\sqrt 2} \subseteq \R$ where $\R$ is the set of [[Definition:Real Number|real numbers]], so we immediately have that [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]] are [[Definition:Well-Defined Operation|well-defined]].
=== Closure ===
Let $a_1 + b_1 \sqrt 2, ... | Quadratic Integers over 2 form Subdomain of Reals/Proof 1 | https://proofwiki.org/wiki/Quadratic_Integers_over_2_form_Subdomain_of_Reals | https://proofwiki.org/wiki/Quadratic_Integers_over_2_form_Subdomain_of_Reals/Proof_1 | [
"Examples of Integral Domains",
"Quadratic Integers",
"Quadratic Integers over 2 form Subdomain of Reals"
] | [
"Definition:Set",
"Definition:Algebraic Integer/Quadratic",
"Definition:Integer",
"Definition:Algebraic Structure/Two Operations",
"Definition:Addition/Real Numbers",
"Definition:Multiplication/Real Numbers",
"Definition:Real Number",
"Definition:Subdomain",
"Definition:Real Number"
] | [
"Definition:Real Number",
"Definition:Addition/Real Numbers",
"Definition:Multiplication/Real Numbers",
"Definition:Well-Defined/Operation",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Addition/Real Numbers",
"Definition:Multiplication/Real Numbers",
"Definition:Associativ... |
proofwiki-3230 | Quadratic Integers over 2 form Subdomain of Reals | Let $\Z \sqbrk {\sqrt 2}$ denote the set of quadratic integers over $2$:
:$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$
That is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are integers.
Then the algebraic structure:
:$\struct {\Z \sqbrk {\sqrt 2}, +, \times}$
where $+$ and $\times$ are conve... | 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 $a + b \sqrt 2$, wher... | Let $\Z \sqbrk {\sqrt 2}$ denote the [[Definition:Set|set]] of [[Definition:Quadratic Integer|quadratic integers]] over $2$:
:$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$
That is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are [[Definition:Integer|integers]].
Then the [[Definition:Algebr... | From [[Integers form Subdomain of Reals]], $\struct {\Z, +, \times}$ is an [[Definition:Subdomain|integral subdomain]] of the [[Definition:Real Number|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 b... | Quadratic Integers over 2 form Subdomain of Reals/Proof 2 | https://proofwiki.org/wiki/Quadratic_Integers_over_2_form_Subdomain_of_Reals | https://proofwiki.org/wiki/Quadratic_Integers_over_2_form_Subdomain_of_Reals/Proof_2 | [
"Examples of Integral Domains",
"Quadratic Integers",
"Quadratic Integers over 2 form Subdomain of Reals"
] | [
"Definition:Set",
"Definition:Algebraic Integer/Quadratic",
"Definition:Integer",
"Definition:Algebraic Structure/Two Operations",
"Definition:Addition/Real Numbers",
"Definition:Multiplication/Real Numbers",
"Definition:Real Number",
"Definition:Subdomain",
"Definition:Real Number"
] | [
"Integers form Subdomain of Reals",
"Definition:Subdomain",
"Definition:Real Number",
"Set of Polynomials over Integral Domain is Subring"
] |
proofwiki-3231 | Sufficient Conditions for Uncountability | Let $X$ be a set.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$X$ contains an uncountable subset}}
{{item|(2):|$X$ is uncountable}}
{{item|(3):|Every sequence of distinct points $\sequence {x_n}_{n \mathop \in \N}$ in $X$ omits at least one $x \in X$}}
{{item|(4):|There is no surjection $\N \twoheadrightarrow X$}}
{{item|(5... | {{Recall|Uncountable Set|uncountable set}}
:$X$ is uncountable {{iff}} there is no injection $X \hookrightarrow \N$. | Let $X$ be a [[Definition:Set|set]].
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$X$ contains an [[Definition:Uncountable Set|uncountable]] [[Definition:Subset|subset]]}}
{{item|(2):|$X$ is [[Definition:Uncountable Set|uncountable]]}}
{{item|(3):|Every [[Definition:Sequence of Distinct Terms|sequence of distinct points]] $... | {{Recall|Uncountable Set|uncountable set}}
:$X$ is [[Definition:Uncountable Set|uncountable]] {{iff}} there is no [[Definition:Injection|injection]] $X \hookrightarrow \N$. | Sufficient Conditions for Uncountability | https://proofwiki.org/wiki/Sufficient_Conditions_for_Uncountability | https://proofwiki.org/wiki/Sufficient_Conditions_for_Uncountability | [
"Uncountable Sets"
] | [
"Definition:Set",
"Definition:Uncountable/Set",
"Definition:Subset",
"Definition:Uncountable/Set",
"Definition:Sequence of Distinct Terms",
"Definition:Surjection",
"Definition:Infinite Set",
"Definition:Bijection",
"Continuum Hypothesis",
"Definition:Uncountable/Set",
"Definition:Extended Real ... | [
"Definition:Uncountable/Set",
"Definition:Injection",
"Definition:Injection",
"Definition:Uncountable/Set",
"Definition:Injection",
"Definition:Injection",
"Definition:Injection",
"Definition:Injection",
"Definition:Injection",
"Definition:Injection",
"Definition:Injection"
] |
proofwiki-3232 | Direct Product iff Nontrivial Idempotent | Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Then $R$ is the direct product of two non-trivial rings {{iff}} $R$ contains an idempotent element not equal to $0_R$ or $1_R$. | === Necessary Condition ===
Suppose $R_1$ and $R_2$ are non-trivial rings, such that $R = R_1 \times R_2$.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {1_{R_1}, 0_{R_2} }^2
| r = \tuple {1_{R_1}^2, 0_{R_2}^2}
| c =
}}
{{eqn | r = \tuple {1_{R_1}, 0_{R_2} }
| c =
}}
{{end-eqn}}
Thus $\tuple {1_{R_1}, 0_{R... | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
Then $R$ is the [[Definition:Ring Direct Product|direct product]] of two [[Definition:Non-Trivial Ring|non-trivia... | === Necessary Condition ===
Suppose $R_1$ and $R_2$ are [[Definition:Non-Trivial Ring|non-trivial rings]], such that $R = R_1 \times R_2$.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {1_{R_1}, 0_{R_2} }^2
| r = \tuple {1_{R_1}^2, 0_{R_2}^2}
| c =
}}
{{eqn | r = \tuple {1_{R_1}, 0_{R_2} }
| c =
}}
{{en... | Direct Product iff Nontrivial Idempotent | https://proofwiki.org/wiki/Direct_Product_iff_Nontrivial_Idempotent | https://proofwiki.org/wiki/Direct_Product_iff_Nontrivial_Idempotent | [
"Commutative Algebra"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Ring Direct Product",
"Definition:Non-Trivial Ring",
"Definition:Idempotence/Element"
] | [
"Definition:Non-Trivial Ring",
"Definition:Idempotence/Element",
"Definition:Idempotence/Element"
] |
proofwiki-3233 | Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive | Let $\struct {D, +, \times, \le}$ be an ordered integral domain whose zero is $0_D$.
Then:
:$\forall x \in D: x \ne 0_D \iff \map P {x \times x}$
where $\map P {x \times x}$ denotes that $x \times x$ has the (strict) positivity property.
That is, the square of any element of an ordered integral domain is (strictly) pos... | Let $x = 0_D$.
Then $x \times x = 0_D \times 0_D = 0_D$ by the properties of the ring zero.
Thus by definition of strict positivity property:
:$\neg \map P {0_D \times 0_D}$
{{qed|lemma}}
Now suppose $x \ne 0_D$.
One of two cases applies:
:$\map P x$
:$\neg \map P x$
Let $\map P x$.
Then by definition of (strict) posit... | Let $\struct {D, +, \times, \le}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$.
Then:
:$\forall x \in D: x \ne 0_D \iff \map P {x \times x}$
where $\map P {x \times x}$ denotes that $x \times x$ has the [[Definition:Strict Positivity Property|(strict... | Let $x = 0_D$.
Then $x \times x = 0_D \times 0_D = 0_D$ by the properties of the [[Definition:Ring Zero|ring zero]].
Thus by definition of [[Definition:Strict Positivity Property|strict positivity property]]:
:$\neg \map P {0_D \times 0_D}$
{{qed|lemma}}
Now suppose $x \ne 0_D$.
One of two cases applies:
:$\map P ... | Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive | https://proofwiki.org/wiki/Square_of_Non-Zero_Element_of_Ordered_Integral_Domain_is_Strictly_Positive | https://proofwiki.org/wiki/Square_of_Non-Zero_Element_of_Ordered_Integral_Domain_is_Strictly_Positive | [
"Ordered Integral Domains"
] | [
"Definition:Ordered Integral Domain",
"Definition:Ring Zero",
"Definition:Strict Positivity Property",
"Definition:Square/Mapping/Element",
"Definition:Element",
"Definition:Ordered Integral Domain",
"Definition:Strict Positivity Property",
"Definition:Element",
"Definition:Ring Zero"
] | [
"Definition:Ring Zero",
"Definition:Strict Positivity Property",
"Definition:Strict Positivity Property",
"Definition:Ordered Integral Domain/Trichotomy Law",
"Product of Ring Negatives"
] |
proofwiki-3234 | Open Sets in Real Number Line | Every non-empty open set $I \subseteq \R$ can be expressed as a countable union of pairwise disjoint open intervals.
If:
:$\ds I = \bigcup_{n \mathop \in \N} J_n$
:$\ds I = \bigcup_{n \mathop \in \N} K_n$
are two such expressions, then there exists a permutation $\sigma$ of $\N$ such that:
:$\forall n \in \N: J_n = K_{... | We know that $\R$ is a complete metric space.
Let $x \in I$.
By the definition of open sets in a metric space, there is an open interval $I_x$, included in $I$, that contains $x$.
Define the following:
{{begin-eqn}}
{{eqn | n = 1
| l = \map a x
| r = \inf \set {z: \openint z x \subseteq I}
}}
{{eqn | n = 2
... | Every [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Real Analysis)|open set]] $I \subseteq \R$ can be expressed as a [[Definition:Countable Union|countable union]] of [[Definition:Pairwise Disjoint|pairwise disjoint]] [[Definition:Open Real Interval|open intervals]].
If:
:$\ds I = \bigcup_{n \mathop \... | We know that [[Real Number Line is Complete Metric Space|$\R$ is a complete metric space]].
Let $x \in I$.
By the definition of [[Definition:Open Set (Metric Space)|open sets in a metric space]], there is an [[Definition:Open Real Interval|open interval]] $I_x$, [[Definition:Subset|included]] in $I$, that contains $x... | Open Sets in Real Number Line | https://proofwiki.org/wiki/Open_Sets_in_Real_Number_Line | https://proofwiki.org/wiki/Open_Sets_in_Real_Number_Line | [
"Open Sets",
"Real Number Line with Euclidean Metric"
] | [
"Definition:Non-Empty Set",
"Definition:Open Set/Real Analysis",
"Definition:Set Union/Countable Union",
"Definition:Pairwise Disjoint",
"Definition:Real Interval/Open",
"Definition:Permutation"
] | [
"Real Number Line is Complete Metric Space",
"Definition:Open Set/Metric Space",
"Definition:Real Interval/Open",
"Definition:Subset",
"Definition:Non-Empty Set",
"Definition:Real Interval/Open",
"Definition:Infimum of Set",
"Definition:Open Set/Real Analysis",
"Definition:Pairwise Disjoint",
"Def... |
proofwiki-3235 | Unity of Ordered Integral Domain is Strictly Positive | Let $\struct {D, +, \times \le}$ be an ordered integral domain whose unity is $1_D$.
Then:
:$\map P {1_D}$
where $P$ is the (strict) positivity property. | We have by definition of the unity that:
:$\forall a \in D: 1_D \times a = a = a \times 1_D$
This particularly applies to $1_D$ itself:
:$1_D = 1_D \times 1_D$
But then by Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive:
:$\map P {1_D \times 1_D} \implies \map P {1_D}$
{{qed}} | Let $\struct {D, +, \times \le}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]] whose [[Definition:Unity of Ring|unity]] is $1_D$.
Then:
:$\map P {1_D}$
where $P$ is the [[Definition:Strict Positivity Property|(strict) positivity property]]. | We have by definition of the [[Definition:Unity of Ring|unity]] that:
:$\forall a \in D: 1_D \times a = a = a \times 1_D$
This particularly applies to $1_D$ itself:
:$1_D = 1_D \times 1_D$
But then by [[Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive]]:
:$\map P {1_D \times 1_D} \implies \m... | Unity of Ordered Integral Domain is Strictly Positive | https://proofwiki.org/wiki/Unity_of_Ordered_Integral_Domain_is_Strictly_Positive | https://proofwiki.org/wiki/Unity_of_Ordered_Integral_Domain_is_Strictly_Positive | [
"Ordered Integral Domains"
] | [
"Definition:Ordered Integral Domain",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Strict Positivity Property"
] | [
"Definition:Unity (Abstract Algebra)/Ring",
"Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive"
] |
proofwiki-3236 | Integers form Ordered Integral Domain | The integers $\Z$ form an ordered integral domain under addition and multiplication. | From Integers form Integral Domain we have that $\struct {\Z, +, \times}$ forms an integral domain.
From Natural Numbers are Non-Negative Integers we have that the set $\N$ can be considered as a subset of the integers.
Then we have that $\struct {\N_{>0}, +, \times}$ is a (commutative) semiring.
So it follows by defin... | The [[Definition:Integer|integers]] $\Z$ form an [[Definition:Ordered Integral Domain|ordered integral domain]] under [[Definition:Integer Addition|addition]] and [[Definition:Integer Multiplication|multiplication]]. | From [[Integers form Integral Domain]] we have that $\struct {\Z, +, \times}$ forms an [[Definition:Integral Domain|integral domain]].
From [[Natural Numbers are Non-Negative Integers]] we have that the set $\N$ can be considered as a [[Definition:Subset|subset]] of the [[Definition:Integer|integers]].
Then we have t... | Integers form Ordered Integral Domain/Proof 1 | https://proofwiki.org/wiki/Integers_form_Ordered_Integral_Domain | https://proofwiki.org/wiki/Integers_form_Ordered_Integral_Domain/Proof_1 | [
"Integers",
"Examples of Ordered Integral Domains",
"Integers form Ordered Integral Domain"
] | [
"Definition:Integer",
"Definition:Ordered Integral Domain",
"Definition:Addition/Integers",
"Definition:Multiplication/Integers"
] | [
"Integers form Integral Domain",
"Definition:Integral Domain",
"Natural Numbers are Non-Negative Integers",
"Definition:Subset",
"Definition:Integer",
"Non-Zero Natural Numbers form Commutative Semiring",
"Definition:Semiring (Abstract Algebra)",
"Definition:Closure (Abstract Algebra)/Algebraic Struct... |
proofwiki-3237 | Integers form Ordered Integral Domain | The integers $\Z$ form an ordered integral domain under addition and multiplication. | From Integers form Integral Domain we have that $\struct {\Z, +, \times}$ forms an integral domain.
From Integers form Totally Ordered Ring, $\struct {\Z, +, \times}$ is a totally ordered ring.
Hence the result.
{{qed}} | The [[Definition:Integer|integers]] $\Z$ form an [[Definition:Ordered Integral Domain|ordered integral domain]] under [[Definition:Integer Addition|addition]] and [[Definition:Integer Multiplication|multiplication]]. | From [[Integers form Integral Domain]] we have that $\struct {\Z, +, \times}$ forms an [[Definition:Integral Domain|integral domain]].
From [[Integers form Totally Ordered Ring]], $\struct {\Z, +, \times}$ is a [[Definition:Totally Ordered Ring|totally ordered ring]].
Hence the result.
{{qed}} | Integers form Ordered Integral Domain/Proof 2 | https://proofwiki.org/wiki/Integers_form_Ordered_Integral_Domain | https://proofwiki.org/wiki/Integers_form_Ordered_Integral_Domain/Proof_2 | [
"Integers",
"Examples of Ordered Integral Domains",
"Integers form Ordered Integral Domain"
] | [
"Definition:Integer",
"Definition:Ordered Integral Domain",
"Definition:Addition/Integers",
"Definition:Multiplication/Integers"
] | [
"Integers form Integral Domain",
"Definition:Integral Domain",
"Integers form Totally Ordered Ring",
"Definition:Totally Ordered Ring"
] |
proofwiki-3238 | Rational Numbers form Ordered Integral Domain | The rational numbers $\Q$ form an ordered integral domain under addition and multiplication. | This follows directly from Rational Numbers form Totally Ordered Field:
The set of rational numbers $\Q$ forms an ordered field under addition and multiplication: $\struct {\Q, +, \times, \le}$.
An ordered field is also an ordered integral domain.
{{Qed}} | The [[Definition:Rational Number|rational numbers]] $\Q$ form an [[Definition:Ordered Integral Domain|ordered integral domain]] under [[Definition:Rational Addition|addition]] and [[Definition:Rational Multiplication|multiplication]]. | This follows directly from [[Rational Numbers form Totally Ordered Field]]:
The [[Definition:Rational Number|set of rational numbers]] $\Q$ forms an [[Definition:Ordered Field|ordered field]] under [[Definition:Rational Addition|addition]] and [[Definition:Rational Multiplication|multiplication]]: $\struct {\Q, +, \ti... | Rational Numbers form Ordered Integral Domain/Proof 1 | https://proofwiki.org/wiki/Rational_Numbers_form_Ordered_Integral_Domain | https://proofwiki.org/wiki/Rational_Numbers_form_Ordered_Integral_Domain/Proof_1 | [
"Rational Numbers",
"Examples of Ordered Integral Domains",
"Rational Numbers form Ordered Integral Domain"
] | [
"Definition:Rational Number",
"Definition:Ordered Integral Domain",
"Definition:Addition/Rational Numbers",
"Definition:Multiplication/Rational Numbers"
] | [
"Rational Numbers form Totally Ordered Field",
"Definition:Rational Number",
"Definition:Ordered Field",
"Definition:Addition/Rational Numbers",
"Definition:Multiplication/Rational Numbers",
"Definition:Ordered Field",
"Definition:Ordered Integral Domain"
] |
proofwiki-3239 | Real Numbers form Ordered Integral Domain | The set of real numbers $\R$ forms an ordered integral domain under addition and multiplication: $\struct {\R, +, \times, \le}$. | This follows directly from Real Numbers form Totally Ordered Field.
The set of real numbers $\R$ forms an ordered field under addition and multiplication: $\struct {\R, +, \times, \le}$.
An ordered field is also an ordered integral domain.
{{Qed}} | The [[Definition:Real Number|set of real numbers]] $\R$ forms an [[Definition:Ordered Integral Domain|ordered integral domain]] under [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]]: $\struct {\R, +, \times, \le}$. | This follows directly from [[Real Numbers form Totally Ordered Field]].
The [[Definition:Real Number|set of real numbers]] $\R$ forms an [[Definition:Ordered Field|ordered field]] under [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]]: $\struct {\R, +, \times, \le}$.
An [[De... | Real Numbers form Ordered Integral Domain/Proof 1 | https://proofwiki.org/wiki/Real_Numbers_form_Ordered_Integral_Domain | https://proofwiki.org/wiki/Real_Numbers_form_Ordered_Integral_Domain/Proof_1 | [
"Examples of Ordered Integral Domains",
"Real Numbers",
"Real Numbers form Ordered Integral Domain"
] | [
"Definition:Real Number",
"Definition:Ordered Integral Domain",
"Definition:Addition/Real Numbers",
"Definition:Multiplication/Real Numbers"
] | [
"Real Numbers form Totally Ordered Field",
"Definition:Real Number",
"Definition:Ordered Field",
"Definition:Addition/Real Numbers",
"Definition:Multiplication/Real Numbers",
"Definition:Ordered Field",
"Definition:Ordered Integral Domain"
] |
proofwiki-3240 | Real Numbers form Ordered Integral Domain | The set of real numbers $\R$ forms an ordered integral domain under addition and multiplication: $\struct {\R, +, \times, \le}$. | We have that the real numbers $\struct {\R, +, \times}$ form an integral domain.
It remains to specify a property $P$ on $\R$ such that:
:$(1): \quad \forall a, b \in \R: \map P a \land \map P b \implies \map P {a + b}$
:$(2): \quad \forall a, b \in \R: \map P a \land \map P b \implies \map P {a \times b}$
:$(3): \quad... | The [[Definition:Real Number|set of real numbers]] $\R$ forms an [[Definition:Ordered Integral Domain|ordered integral domain]] under [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]]: $\struct {\R, +, \times, \le}$. | We have that the [[Real Numbers form Integral Domain|real numbers $\struct {\R, +, \times}$ form an integral domain]].
It remains to specify a property $P$ on $\R$ such that:
:$(1): \quad \forall a, b \in \R: \map P a \land \map P b \implies \map P {a + b}$
:$(2): \quad \forall a, b \in \R: \map P a \land \map P b \... | Real Numbers form Ordered Integral Domain/Proof 2 | https://proofwiki.org/wiki/Real_Numbers_form_Ordered_Integral_Domain | https://proofwiki.org/wiki/Real_Numbers_form_Ordered_Integral_Domain/Proof_2 | [
"Examples of Ordered Integral Domains",
"Real Numbers",
"Real Numbers form Ordered Integral Domain"
] | [
"Definition:Real Number",
"Definition:Ordered Integral Domain",
"Definition:Addition/Real Numbers",
"Definition:Multiplication/Real Numbers"
] | [
"Real Numbers form Integral Domain",
"Rational Numbers form Ordered Integral Domain",
"Definition:Rational Number",
"Definition:Strict Positivity Property",
"Definition:Strict Positivity Property",
"Definition:Addition/Real Numbers",
"Definition:Multiplication/Real Numbers",
"Definition:Strict Positiv... |
proofwiki-3241 | Quadratic Integers over 2 form Ordered Integral Domain | Let $\Z \sqbrk {\sqrt 2}$ denote the set of quadratic integers over $2$:
:$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$
that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are integers.
Then the algebraic structure:
:$\struct {\Z \sqbrk {\sqrt 2}, +, \times}$
where $+$ and $\times$ are conve... | We have that Quadratic Integers over 2 form Subdomain of Reals.
We also have that such numbers are real.
The result follows from Real Numbers form Ordered Integral Domain.
{{explain|So is a subdomain of an ordered integral domain always an ordered integral domain? Does this need to be proved?}}
{{qed}} | Let $\Z \sqbrk {\sqrt 2}$ denote the [[Definition:Set|set]] of [[Definition:Quadratic Integer|quadratic integers]] over $2$:
:$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$
that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are [[Definition:Integer|integers]].
Then the [[Definition:Algebra... | We have that [[Quadratic Integers over 2 form Subdomain of Reals]].
We also have that such numbers are [[Definition:Real Number|real]].
The result follows from [[Real Numbers form Ordered Integral Domain]].
{{explain|So is a subdomain of an ordered integral domain always an ordered integral domain? Does this need to... | Quadratic Integers over 2 form Ordered Integral Domain | https://proofwiki.org/wiki/Quadratic_Integers_over_2_form_Ordered_Integral_Domain | https://proofwiki.org/wiki/Quadratic_Integers_over_2_form_Ordered_Integral_Domain | [
"Examples of Ordered Integral Domains",
"Quadratic Integers"
] | [
"Definition:Set",
"Definition:Algebraic Integer/Quadratic",
"Definition:Integer",
"Definition:Algebraic Structure",
"Definition:Addition/Real Numbers",
"Definition:Multiplication/Real Numbers",
"Definition:Real Number",
"Definition:Ordered Integral Domain"
] | [
"Quadratic Integers over 2 form Subdomain of Reals",
"Definition:Real Number",
"Real Numbers form Ordered Integral Domain"
] |
proofwiki-3242 | General Positivity Rule in Ordered Integral Domain | Let $\struct {D, +, \times}$ be an ordered integral domain, whose (strict) positivity property is denoted $P$.
Let $S \subset D$ be a subset of $D$ such that:
:$\forall s \in S: \map P x$
Then the following are true:
:$\ds \forall n \in \N_{>0}: \forall s_i \in S: \map P {\sum_{i \mathop = 1}^n s_i}$
:$\ds \forall n \i... | Trivially true for $n = 1$, and true by definition for $n = 2$.
Suppose it is true for all $n$ up to $n = k$ for some $k \in \N$.
Then:
:$\ds \forall s_i \in S: \map P {\sum_{i \mathop = 1}^k s_i}$
:$\ds \forall s_i \in S: \map P {\prod_{i \mathop = 1}^k s_i}$
Take any $\ds \sum_{i \mathop = 1}^{k + 1} s_i = \sum_{i \m... | Let $\struct {D, +, \times}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]], whose [[Definition:Strict Positivity Property|(strict) positivity property]] is denoted $P$.
Let $S \subset D$ be a [[Definition:Subset|subset]] of $D$ such that:
:$\forall s \in S: \map P x$
Then the following are tru... | Trivially true for $n = 1$, and true by definition for $n = 2$.
Suppose it is true for all $n$ up to $n = k$ for some $k \in \N$.
Then:
:$\ds \forall s_i \in S: \map P {\sum_{i \mathop = 1}^k s_i}$
:$\ds \forall s_i \in S: \map P {\prod_{i \mathop = 1}^k s_i}$
Take any $\ds \sum_{i \mathop = 1}^{k + 1} s_i = \sum... | General Positivity Rule in Ordered Integral Domain | https://proofwiki.org/wiki/General_Positivity_Rule_in_Ordered_Integral_Domain | https://proofwiki.org/wiki/General_Positivity_Rule_in_Ordered_Integral_Domain | [
"Ordered Integral Domains"
] | [
"Definition:Ordered Integral Domain",
"Definition:Strict Positivity Property",
"Definition:Subset",
"Definition:Ring (Abstract Algebra)/Addition",
"Definition:Ring (Abstract Algebra)/Product",
"Definition:Strict Positivity Property",
"Definition:Strict Positivity Property"
] | [
"Definition:Strict Positivity Property",
"Second Principle of Mathematical Induction"
] |
proofwiki-3243 | Finite Integral Domain cannot be Ordered | Let $\struct {D, +, \times}$ be an integral domain whose unity is $1_D$ and whose order is finite.
Then $D$ cannot be an ordered integral domain. | Let $\struct {D, +, \times}$ be an integral domain whose order is $n$.
{{AimForCont}} there exists a (strict) positivity property $P$ on $\struct {D, +, \times}$.
From Unity of Ordered Integral Domain is Strictly Positive, we have that $\map P {1_D}$.
Let $\struct {D, +}$ be the additive group of $\struct {D, +, \times... | Let $\struct {D, +, \times}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Unity of Ring|unity]] is $1_D$ and whose [[Definition:Order of Structure|order]] is [[Definition:Finite Order (Structure)|finite]].
Then $D$ cannot be an [[Definition:Ordered Integral Domain|ordered integral domain]]. | Let $\struct {D, +, \times}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Order of Structure|order]] is $n$.
{{AimForCont}} there exists a [[Definition:Strict Positivity Property|(strict) positivity property]] $P$ on $\struct {D, +, \times}$.
From [[Unity of Ordered Integral Domain is Stric... | Finite Integral Domain cannot be Ordered | https://proofwiki.org/wiki/Finite_Integral_Domain_cannot_be_Ordered | https://proofwiki.org/wiki/Finite_Integral_Domain_cannot_be_Ordered | [
"Ordered Integral Domains"
] | [
"Definition:Integral Domain",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Order of Structure",
"Definition:Order of Structure/Finite Structure",
"Definition:Ordered Integral Domain"
] | [
"Definition:Integral Domain",
"Definition:Order of Structure",
"Definition:Strict Positivity Property",
"Unity of Ordered Integral Domain is Strictly Positive",
"Definition:Additive Group of Ring",
"Order of Element Divides Order of Finite Group",
"Definition:Order of Group Element",
"Definition:Order... |
proofwiki-3244 | Ring of Integers Modulo m cannot be Ordered Integral Domain | Let $m \in \Z: m \ge 2$.
Let $\struct {\Z_m, +, \times}$ be the ring of integers modulo $m$.
Then $\struct {\Z_m, +, \times}$ cannot be an ordered integral domain. | First note that from Ring of Integers Modulo Prime is Integral Domain, $\struct {\Z_m, +, \times}$ is an integral domain only when $m$ is prime.
So for $m$ composite the result holds.
If $m$ is prime, and $\struct {\Z_m, +, \times}$ is therefore an integral domain, its order is finite.
The result follows from Finite I... | Let $m \in \Z: m \ge 2$.
Let $\struct {\Z_m, +, \times}$ be the [[Definition:Ring of Integers Modulo m|ring of integers modulo $m$]].
Then $\struct {\Z_m, +, \times}$ cannot be an [[Definition:Ordered Integral Domain|ordered integral domain]]. | First note that from [[Ring of Integers Modulo Prime is Integral Domain]], $\struct {\Z_m, +, \times}$ is an [[Definition:Integral Domain|integral domain]] only when $m$ is [[Definition:Prime Number|prime]].
So for $m$ [[Definition:Composite Number|composite]] the result holds.
If $m$ is [[Definition:Prime Number|pr... | Ring of Integers Modulo m cannot be Ordered Integral Domain | https://proofwiki.org/wiki/Ring_of_Integers_Modulo_m_cannot_be_Ordered_Integral_Domain | https://proofwiki.org/wiki/Ring_of_Integers_Modulo_m_cannot_be_Ordered_Integral_Domain | [
"Ring of Integers Modulo m",
"Ordered Integral Domains"
] | [
"Definition:Ring of Integers Modulo m",
"Definition:Ordered Integral Domain"
] | [
"Ring of Integers Modulo Prime is Integral Domain",
"Definition:Integral Domain",
"Definition:Prime Number",
"Definition:Composite Number",
"Definition:Prime Number",
"Definition:Integral Domain",
"Definition:Order of Structure",
"Definition:Order of Structure/Finite Structure",
"Finite Integral Dom... |
proofwiki-3245 | Strict Positivity Property induces Total Ordering | Let $\struct {D, +, \times}$ be an integral domain whose zero is $0_D$.
Let $D$ be endowed with a (strict) positivity property $P: D \to \set {\T, \F}$.
Then there exists a total ordering $\le$ on $\struct {D, +, \times}$ induced by $P$ which is compatible with the ring structure of $\struct {D, +, \times}$. | By definition of the strict positivity property:
{{:Definition:Strict Positivity Property}}
Let us define a relation $<$ on $D$ as:
:$\forall a, b \in D: a < b \iff \map P {-a + b}$
Setting $a = 0$:
:$\forall b \in D: 0 < b \iff \map P b$
demonstrating that (strictly) positive elements of $D$ are those which are greate... | Let $\struct {D, +, \times}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$.
Let $D$ be endowed with a [[Definition:Strict Positivity Property|(strict) positivity property]] $P: D \to \set {\T, \F}$.
Then there exists a [[Definition:Total Ordering|total ordering]] $... | By definition of the [[Definition:Strict Positivity Property|strict positivity property]]:
{{:Definition:Strict Positivity Property}}
Let us define a [[Definition:Relation|relation]] $<$ on $D$ as:
:$\forall a, b \in D: a < b \iff \map P {-a + b}$
Setting $a = 0$:
:$\forall b \in D: 0 < b \iff \map P b$
demonstratin... | Strict Positivity Property induces Total Ordering | https://proofwiki.org/wiki/Strict_Positivity_Property_induces_Total_Ordering | https://proofwiki.org/wiki/Strict_Positivity_Property_induces_Total_Ordering | [
"Ordered Integral Domains"
] | [
"Definition:Integral Domain",
"Definition:Ring Zero",
"Definition:Strict Positivity Property",
"Definition:Total Ordering",
"Definition:Total Ordering Induced by Strict Positivity Property",
"Definition:Ordering Compatible with Ring Structure"
] | [
"Definition:Strict Positivity Property",
"Definition:Relation",
"Definition:Strict Positivity Property",
"Definition:Ring Zero",
"Relation Induced by Strict Positivity Property is Compatible with Addition",
"Definition:Relation Compatible with Operation",
"Relation Induced by Strict Positivity Property ... |
proofwiki-3246 | Relation Induced by Strict Positivity Property is Transitive | Let $\struct {D, +, \times}$ be an ordered integral domain where $P$ is the (strict) positivity property.
Let the relation $<$ be defined on $D$ as:
:$\forall a, b \in D: a < b \iff \map P {-a + b}$
Then $<$ is a transitive relation. | Let $a < b$ and $b < c$.
Thus:
{{begin-eqn}}
{{eqn | o =
| r = a < b, b < c
| c =
}}
{{eqn | o = \leadsto
| r = \map P {-a + b}, \map P {-b + c}
| c = Definition of $<$
}}
{{eqn | o = \leadsto
| r = \map P {\paren {-a + b} + \paren {-b + c} }
| c = {{Defof|Strict Positivity Propert... | Let $\struct {D, +, \times}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]] where $P$ is the [[Definition:Strict Positivity Property|(strict) positivity property]].
Let the [[Definition:Relation|relation]] $<$ be defined on $D$ as:
:$\forall a, b \in D: a < b \iff \map P {-a + b}$
Then $<$ is ... | Let $a < b$ and $b < c$.
Thus:
{{begin-eqn}}
{{eqn | o =
| r = a < b, b < c
| c =
}}
{{eqn | o = \leadsto
| r = \map P {-a + b}, \map P {-b + c}
| c = Definition of $<$
}}
{{eqn | o = \leadsto
| r = \map P {\paren {-a + b} + \paren {-b + c} }
| c = {{Defof|Strict Positivity Prope... | Relation Induced by Strict Positivity Property is Transitive | https://proofwiki.org/wiki/Relation_Induced_by_Strict_Positivity_Property_is_Transitive | https://proofwiki.org/wiki/Relation_Induced_by_Strict_Positivity_Property_is_Transitive | [
"Ordered Integral Domains",
"Transitive Relations"
] | [
"Definition:Ordered Integral Domain",
"Definition:Strict Positivity Property",
"Definition:Relation",
"Definition:Transitive Relation"
] | [
"Definition:Transitive Relation"
] |
proofwiki-3247 | Relation Induced by Strict Positivity Property is Compatible with Addition | Let $\struct {D, +, \times}$ be an ordered integral domain where $P$ is the (strict) positivity property.
Let the relation $<$ be defined on $D$ as:
:$\forall a, b \in D: a < b \iff \map P {-a + b}$
Then $<$ is compatible with $+$, that is:
:$\forall x, y, z \in D: x < y \implies \paren {x + z} < \paren {y + z}$
:$\for... | Let $a < b$:
{{begin-eqn}}
{{eqn | o =
| r = a < b
| c =
}}
{{eqn | o = \leadsto
| r = \map P {-a + b}
| c = by definition of $<$
}}
{{eqn | o = \leadsto
| r = \map P {-a + b + \paren {-c} + c}
| c = for any $c \in D$
}}
{{eqn | o = \leadsto
| r = \map P {\paren {-a + \paren ... | Let $\struct {D, +, \times}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]] where $P$ is the [[Definition:Strict Positivity Property|(strict) positivity property]].
Let the [[Definition:Relation|relation]] $<$ be defined on $D$ as:
:$\forall a, b \in D: a < b \iff \map P {-a + b}$
Then $<$ is ... | Let $a < b$:
{{begin-eqn}}
{{eqn | o =
| r = a < b
| c =
}}
{{eqn | o = \leadsto
| r = \map P {-a + b}
| c = by definition of $<$
}}
{{eqn | o = \leadsto
| r = \map P {-a + b + \paren {-c} + c}
| c = for any $c \in D$
}}
{{eqn | o = \leadsto
| r = \map P {\paren {-a + \paren... | Relation Induced by Strict Positivity Property is Compatible with Addition | https://proofwiki.org/wiki/Relation_Induced_by_Strict_Positivity_Property_is_Compatible_with_Addition | https://proofwiki.org/wiki/Relation_Induced_by_Strict_Positivity_Property_is_Compatible_with_Addition | [
"Ordered Integral Domains"
] | [
"Definition:Ordered Integral Domain",
"Definition:Strict Positivity Property",
"Definition:Relation",
"Definition:Relation Compatible with Operation"
] | [
"Definition:Commutative/Operation",
"Definition:Relation Compatible with Operation"
] |
proofwiki-3248 | Relation Induced by Strict Positivity Property is Compatible with Multiplication | Let $\struct {D, +, \times}$ be an ordered integral domain where $P$ is the (strict) positivity property.
Let the relation $<$ be defined on $D$ as:
:$\forall a, b \in D: a < b \iff \map P {-a + b}$
Then $<$ is compatible with $\times$ in the following sense:
:$\forall x, y, z \in D: x < y, \map P z \implies \paren {z ... | If $x < y$ then $\map P {-x + y}$.
Hence:
{{begin-eqn}}
{{eqn | o =
| r = \map P {-x + y}, \, \map P z
| c =
}}
{{eqn | o = \leadsto
| r = \map P {z \times \paren {-x + y} }
| c = {{Defof|Strict Positivity Property}}
}}
{{eqn | o = \leadsto
| r = \map P {-z \times x + z \times y}
|... | Let $\struct {D, +, \times}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]] where $P$ is the [[Definition:Strict Positivity Property|(strict) positivity property]].
Let the [[Definition:Relation|relation]] $<$ be defined on $D$ as:
:$\forall a, b \in D: a < b \iff \map P {-a + b}$
Then $<$ is ... | If $x < y$ then $\map P {-x + y}$.
Hence:
{{begin-eqn}}
{{eqn | o =
| r = \map P {-x + y}, \, \map P z
| c =
}}
{{eqn | o = \leadsto
| r = \map P {z \times \paren {-x + y} }
| c = {{Defof|Strict Positivity Property}}
}}
{{eqn | o = \leadsto
| r = \map P {-z \times x + z \times y}
... | Relation Induced by Strict Positivity Property is Compatible with Multiplication | https://proofwiki.org/wiki/Relation_Induced_by_Strict_Positivity_Property_is_Compatible_with_Multiplication | https://proofwiki.org/wiki/Relation_Induced_by_Strict_Positivity_Property_is_Compatible_with_Multiplication | [
"Ordered Integral Domains"
] | [
"Definition:Ordered Integral Domain",
"Definition:Strict Positivity Property",
"Definition:Relation",
"Definition:Relation Compatible with Operation"
] | [
"Definition:Distributive Operation",
"Definition:Commutative/Operation",
"Definition:Integral Domain"
] |
proofwiki-3249 | Relation Induced by Strict Positivity Property is Trichotomy | Let $\struct {D, +, \times}$ be an ordered integral domain where $P$ is the (strict) positivity property.
Let the relation $<$ be defined on $D$ as:
:$\forall a, b \in D: a < b \iff \map P {-a + b}$
Then $\forall a, b \in D:$ exactly one of the following conditions applies:
:$a < b$
:$a = b$
:$a > b$
That is, $<$ is a ... | Take any $a, b \in D$ and consider $-a + b$.
From the trichotomy law of ordered integral domains, exactly one of the following applies:
:$(1): \quad \map P {-a + b}$
:$(2): \quad \map P {-\paren {-a + b} }$
:$(3): \quad -a + b = 0$
Taking each of these in turn and taking into account that $\struct {D, +}$ is the additi... | Let $\struct {D, +, \times}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]] where $P$ is the [[Definition:Strict Positivity Property|(strict) positivity property]].
Let the [[Definition:Relation|relation]] $<$ be defined on $D$ as:
:$\forall a, b \in D: a < b \iff \map P {-a + b}$
Then $\foral... | Take any $a, b \in D$ and consider $-a + b$.
From the [[Definition:Trichotomy Law (Integral Domain)|trichotomy law of ordered integral domains]], exactly one of the following applies:
:$(1): \quad \map P {-a + b}$
:$(2): \quad \map P {-\paren {-a + b} }$
:$(3): \quad -a + b = 0$
Taking each of these in turn and taki... | Relation Induced by Strict Positivity Property is Trichotomy | https://proofwiki.org/wiki/Relation_Induced_by_Strict_Positivity_Property_is_Trichotomy | https://proofwiki.org/wiki/Relation_Induced_by_Strict_Positivity_Property_is_Trichotomy | [
"Ordered Integral Domains"
] | [
"Definition:Ordered Integral Domain",
"Definition:Strict Positivity Property",
"Definition:Relation",
"Definition:Trichotomy"
] | [
"Definition:Ordered Integral Domain/Trichotomy Law",
"Definition:Additive Group of Ring"
] |
proofwiki-3250 | Relation Induced by Strict Positivity Property is Asymmetric and Antireflexive | Let $\struct {D, +, \times}$ be an ordered integral domain where $P$ is the (strict) positivity property.
Let the relation $<$ be defined on $D$ as:
:$\forall a, b \in D: a < b \iff \map P {-a + b}$
Then $<$ is asymmetric and antireflexive. | From the trichotomy law of ordered integral domains, for all $x \in D$ exactly one of the following applies:
:$(1): \quad \map P x$
:$(2): \quad \map P {-x}$
:$(3): \quad x = 0$
Let $a, b \in D: a < b$.
Then by definition:
:$\map P {-a + b}$
Suppose also that $b < a$.
Then by definition:
:$\map P {-b + a}$
and so:
:$\m... | Let $\struct {D, +, \times}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]] where $P$ is the [[Definition:Strict Positivity Property|(strict) positivity property]].
Let the [[Definition:Relation|relation]] $<$ be defined on $D$ as:
:$\forall a, b \in D: a < b \iff \map P {-a + b}$
Then $<$ is ... | From the [[Definition:Trichotomy Law (Integral Domain)|trichotomy law of ordered integral domains]], for all $x \in D$ exactly one of the following applies:
:$(1): \quad \map P x$
:$(2): \quad \map P {-x}$
:$(3): \quad x = 0$
Let $a, b \in D: a < b$.
Then by definition:
:$\map P {-a + b}$
Suppose also that $b < a$... | Relation Induced by Strict Positivity Property is Asymmetric and Antireflexive | https://proofwiki.org/wiki/Relation_Induced_by_Strict_Positivity_Property_is_Asymmetric_and_Antireflexive | https://proofwiki.org/wiki/Relation_Induced_by_Strict_Positivity_Property_is_Asymmetric_and_Antireflexive | [
"Ordered Integral Domains"
] | [
"Definition:Ordered Integral Domain",
"Definition:Strict Positivity Property",
"Definition:Relation",
"Definition:Asymmetric Relation",
"Definition:Antireflexive Relation"
] | [
"Definition:Ordered Integral Domain/Trichotomy Law",
"Definition:Ordered Integral Domain/Trichotomy Law",
"Definition:Asymmetric Relation",
"Definition:Ordered Integral Domain/Trichotomy Law",
"Definition:Antireflexive Relation",
"Category:Ordered Integral Domains"
] |
proofwiki-3251 | Absolute Value on Ordered Integral Domain is Strictly Positive except when Zero | Let $\struct {D, +, \times, \le}$ be an ordered integral domain.
For all $a \in D$, let $\size a$ denote the absolute value of $a$.
Then $\size a$ is (strictly) positive except when $a = 0$. | Let $P$ be the (strict) positivity property on $D$.
Let $<$ be the (strict) total ordering defined on $D$ as:
:$a < b \iff a \le b \land a \ne b$
From the trichotomy law, exactly one of three possibilities holds for any $ a \in D$:
$(1): \quad \map P a$:
In this case $0 < a$ and so $\size a = a$.
So:
:$\map P a \implie... | Let $\struct {D, +, \times, \le}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]].
For all $a \in D$, let $\size a$ denote the [[Definition:Absolute Value on Ordered Integral Domain|absolute value]] of $a$.
Then $\size a$ is [[Definition:Strictly Positive|(strictly) positive]] except when $a = 0... | Let $P$ be the [[Definition:Strict Positivity Property|(strict) positivity property]] on $D$.
Let $<$ be the [[Definition:Strict Total Ordering|(strict) total ordering]] defined on $D$ as:
:$a < b \iff a \le b \land a \ne b$
From the [[Definition:Trichotomy Law (Integral Domain)|trichotomy law]], exactly one of three... | Absolute Value on Ordered Integral Domain is Strictly Positive except when Zero | https://proofwiki.org/wiki/Absolute_Value_on_Ordered_Integral_Domain_is_Strictly_Positive_except_when_Zero | https://proofwiki.org/wiki/Absolute_Value_on_Ordered_Integral_Domain_is_Strictly_Positive_except_when_Zero | [
"Absolute Value Function"
] | [
"Definition:Ordered Integral Domain",
"Definition:Absolute Value/Ordered Integral Domain",
"Definition:Strictly Positive"
] | [
"Definition:Strict Positivity Property",
"Definition:Strict Total Ordering",
"Definition:Ordered Integral Domain/Trichotomy Law"
] |
proofwiki-3252 | Sum of Absolute Values on Ordered Integral Domain | Let $\struct {D, +, \times, \le}$ be an ordered integral domain.
For all $a \in D$, let $\size a$ denote the absolute value of $a$.
Then:
:$\size {a + b} \le \size a + \size b$ | Let $P$ be the (strict) positivity property on $D$.
Let $<$ be the (strict) total ordering defined on $D$ as:
:$a < b \iff a \le b \land a \ne b$
Let $N$ be the (strict) negativity property on $D$.
Let $a \in D$.
If $\map P a$ or $a = 0$ then $a \le \size a$.
If $\map N a$ then by Properties of Strict Negativity: $(1)$... | Let $\struct {D, +, \times, \le}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]].
For all $a \in D$, let $\size a$ denote the [[Definition:Absolute Value on Ordered Integral Domain|absolute value]] of $a$.
Then:
:$\size {a + b} \le \size a + \size b$ | Let $P$ be the [[Definition:Strict Positivity Property|(strict) positivity property]] on $D$.
Let $<$ be the [[Definition:Strict Total Ordering|(strict) total ordering]] defined on $D$ as:
:$a < b \iff a \le b \land a \ne b$
Let $N$ be the [[Definition:Strict Negativity Property|(strict) negativity property]] on $D$.... | Sum of Absolute Values on Ordered Integral Domain | https://proofwiki.org/wiki/Sum_of_Absolute_Values_on_Ordered_Integral_Domain | https://proofwiki.org/wiki/Sum_of_Absolute_Values_on_Ordered_Integral_Domain | [
"Absolute Value Function"
] | [
"Definition:Ordered Integral Domain",
"Definition:Absolute Value/Ordered Integral Domain"
] | [
"Definition:Strict Positivity Property",
"Definition:Strict Total Ordering",
"Definition:Strict Negativity Property",
"Properties of Strict Negativity",
"Definition:Absolute Value/Ordered Integral Domain",
"Definition:Transitive Relation",
"Definition:Relation Compatible with Operation"
] |
proofwiki-3253 | One Succeeds Zero in Well-Ordered Integral Domain | Let $\struct {D, +, \times, \le}$ be a well-ordered integral domain.
Let $0$ and $1$ be the zero and unity respectively of $D$.
Then $0$ is the immediate predecessor of $1$:
:$0 < 1$
:$\neg \exists a \in D: 0 < a < 1$ | {{AimForCont}} there exists an element $a \in D$ such that $0 < a < 1$.
Let us create the set $S$ of all such elements:
:$S = \set {x \in D: 0 < x < 1}$
We know $S$ is not empty because it has already been asserted that $a$ is in it.
And all the elements in $S$ are strictly positive by definition.
Because $D$ is well-o... | Let $\struct {D, +, \times, \le}$ be a [[Definition:Well-Ordered Integral Domain|well-ordered integral domain]].
Let $0$ and $1$ be the [[Definition:Ring Zero|zero]] and [[Definition:Unity of Ring|unity]] respectively of $D$.
Then $0$ is the [[Definition:Immediate Predecessor Element|immediate predecessor]] of $1$:
... | {{AimForCont}} there exists an [[Definition:Element|element]] $a \in D$ such that $0 < a < 1$.
Let us create the [[Definition:Set|set]] $S$ of all such [[Definition:Element|elements]]:
:$S = \set {x \in D: 0 < x < 1}$
We know $S$ is not [[Definition:Empty Set|empty]] because it has already been asserted that $a$ is i... | One Succeeds Zero in Well-Ordered Integral Domain | https://proofwiki.org/wiki/One_Succeeds_Zero_in_Well-Ordered_Integral_Domain | https://proofwiki.org/wiki/One_Succeeds_Zero_in_Well-Ordered_Integral_Domain | [
"Well-Ordered Integral Domains"
] | [
"Definition:Well-Ordered Integral Domain",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Immediate Predecessor Element"
] | [
"Definition:Element",
"Definition:Set",
"Definition:Element",
"Definition:Empty Set",
"Definition:Element",
"Definition:Strictly Positive",
"Definition:Well-Ordered Set",
"Definition:Minimal/Element",
"Square of Element Less than Unity in Ordered Integral Domain",
"Definition:Element",
"Definiti... |
proofwiki-3254 | Square of Element Less than Unity in Ordered Integral Domain | Let $\struct {D, +, \times, \le}$ be an ordered integral domain.
Let $x \in D$ such that $0 < x < 1$.
Then:
: $0 < x \times x < x$ | We have that $0 < x < 1$.
From Relation Induced by Strict Positivity Property is Compatible with Multiplication:
:$0 \times x < x \times x < 1 \times x$
Hence the result.
{{qed}}
Category:Ordered Integral Domains
payuibiqj3z0g90g32oa6vktwe0b6lr | Let $\struct {D, +, \times, \le}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]].
Let $x \in D$ such that $0 < x < 1$.
Then:
: $0 < x \times x < x$ | We have that $0 < x < 1$.
From [[Relation Induced by Strict Positivity Property is Compatible with Multiplication]]:
:$0 \times x < x \times x < 1 \times x$
Hence the result.
{{qed}}
[[Category:Ordered Integral Domains]]
payuibiqj3z0g90g32oa6vktwe0b6lr | Square of Element Less than Unity in Ordered Integral Domain | https://proofwiki.org/wiki/Square_of_Element_Less_than_Unity_in_Ordered_Integral_Domain | https://proofwiki.org/wiki/Square_of_Element_Less_than_Unity_in_Ordered_Integral_Domain | [
"Ordered Integral Domains"
] | [
"Definition:Ordered Integral Domain"
] | [
"Relation Induced by Strict Positivity Property is Compatible with Multiplication",
"Category:Ordered Integral Domains"
] |
proofwiki-3255 | Principle of Mathematical Induction/Well-Ordered Integral Domain | Let $\struct {D, +, \times, \le}$ be a well-ordered integral domain whose zero is $0_D$.
Let the unity of $D$ be $1_D$.
Let $S \subseteq D$ be such that:
:$1_D \in S$
:$a \in S \implies a + 1_D \in S$
Then:
:$D_{> 0_D} \subseteq S$
where $D_{> 0_D}$ denotes all the elements $d \in D$ such that $\map P d$.
That is, $D_{... | Let $\overline S$ be the set of all elements of $D_{> 0_D}$ that are not in $S$:
:$\overline S = D_{> 0_D} \setminus S$
{{AimForCont}} $\overline S$ is not empty.
Then as $D$ is well-ordered, it follows that $\overline S$ has a minimal element, which we will call $m$.
Then $m - 1 \notin \overline S$.
But $1 < m$ as $1 ... | Let $\struct {D, +, \times, \le}$ be a [[Definition:Well-Ordered Integral Domain|well-ordered integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$.
Let the [[Definition:Unity of Ring|unity]] of $D$ be $1_D$.
Let $S \subseteq D$ be such that:
:$1_D \in S$
:$a \in S \implies a + 1_D \in S$
Then:
:$D_{> 0_D... | Let $\overline S$ be the set of all elements of $D_{> 0_D}$ that are not in $S$:
:$\overline S = D_{> 0_D} \setminus S$
{{AimForCont}} $\overline S$ is not [[Definition:Empty Set|empty]].
Then as $D$ is [[Definition:Well-Ordered Integral Domain|well-ordered]], it follows that $\overline S$ has a [[Definition:Minimal ... | Principle of Mathematical Induction/Well-Ordered Integral Domain | https://proofwiki.org/wiki/Principle_of_Mathematical_Induction/Well-Ordered_Integral_Domain | https://proofwiki.org/wiki/Principle_of_Mathematical_Induction/Well-Ordered_Integral_Domain | [
"Well-Ordered Integral Domains",
"Principle of Mathematical Induction"
] | [
"Definition:Well-Ordered Integral Domain",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Element",
"Definition:Strictly Positive"
] | [
"Definition:Empty Set",
"Definition:Well-Ordered Integral Domain",
"Definition:Minimal/Element",
"One Succeeds Zero in Well-Ordered Integral Domain",
"Definition:Strictly Positive",
"Definition:Contradiction",
"Set Difference with Superset is Empty Set"
] |
proofwiki-3256 | Multiple of Divisor in Integral Domain Divides Multiple | Let $\struct {D, +, \times}$ be an integral domain.
Let $a, b, c \in D$.
Let $a \divides b$, where $\divides$ denotes divisibility.
Then $a \times c$ is a divisor of $b \times c$. | By definition, if $a \divides b$ then $\exists d \in D: a \times d = b$.
Then $\paren {a \times d} \times c = b \times c$, that is:
:$\paren {a \times c} \times d = b \times c$
which follows because $\times$ is commutative and associative in an integral domain.
Hence the result.
{{qed}} | Let $\struct {D, +, \times}$ be an [[Definition:Integral Domain|integral domain]].
Let $a, b, c \in D$.
Let $a \divides b$, where $\divides$ denotes [[Definition:Divisor of Ring Element|divisibility]].
Then $a \times c$ is a [[Definition:Divisor of Ring Element|divisor]] of $b \times c$. | By definition, if $a \divides b$ then $\exists d \in D: a \times d = b$.
Then $\paren {a \times d} \times c = b \times c$, that is:
:$\paren {a \times c} \times d = b \times c$
which follows because $\times$ is [[Definition:Commutative Operation|commutative]] and [[Definition:Associative Operation|associative]] in an ... | Multiple of Divisor in Integral Domain Divides Multiple | https://proofwiki.org/wiki/Multiple_of_Divisor_in_Integral_Domain_Divides_Multiple | https://proofwiki.org/wiki/Multiple_of_Divisor_in_Integral_Domain_Divides_Multiple | [
"Divisibility",
"Integral Domains"
] | [
"Definition:Integral Domain",
"Definition:Divisor (Algebra)/Ring with Unity",
"Definition:Divisor (Algebra)/Ring with Unity"
] | [
"Definition:Commutative/Operation",
"Definition:Associative Operation",
"Definition:Integral Domain"
] |
proofwiki-3257 | Integer Divisor Results/One Divides all Integers | {{begin-eqn}}
{{eqn | l = 1
| o = \divides
| r = n
}}
{{eqn | l = -1
| o = \divides
| r = n
}}
{{end-eqn}} | From Integers form Integral Domain, the concept '''divisibility''' is fully applicable to the integers.
Therefore this result follows directly from Unity Divides All Elements.
{{qed}} | {{begin-eqn}}
{{eqn | l = 1
| o = \divides
| r = n
}}
{{eqn | l = -1
| o = \divides
| r = n
}}
{{end-eqn}} | From [[Integers form Integral Domain]], the concept '''[[Definition:Divisor of Integer|divisibility]]''' is fully applicable to the [[Definition:Integer|integers]].
Therefore this result follows directly from [[Unity Divides All Elements]].
{{qed}} | Integer Divisor Results/One Divides all Integers | https://proofwiki.org/wiki/Integer_Divisor_Results/One_Divides_all_Integers | https://proofwiki.org/wiki/Integer_Divisor_Results/One_Divides_all_Integers | [
"Divisors"
] | [] | [
"Integers form Integral Domain",
"Definition:Divisor (Algebra)/Integer",
"Definition:Integer",
"Unity Divides All Elements"
] |
proofwiki-3258 | Integer Divisor Results/Integer Divides Itself | :$n \divides n$ | From Integer Multiplication Identity is One:
:$\forall n \in \Z: 1 \cdot n = n = n \cdot 1$
thus demonstrating that $n$ is a divisor of itself.
{{qed}} | :$n \divides n$ | From [[Integer Multiplication Identity is One]]:
:$\forall n \in \Z: 1 \cdot n = n = n \cdot 1$
thus demonstrating that $n$ is a [[Definition:Divisor of Integer|divisor]] of itself.
{{qed}} | Integer Divisor Results/Integer Divides Itself/Proof 1 | https://proofwiki.org/wiki/Integer_Divisor_Results/Integer_Divides_Itself | https://proofwiki.org/wiki/Integer_Divisor_Results/Integer_Divides_Itself/Proof_1 | [
"Divisors",
"Integer Divisor Results/Integer Divides Itself"
] | [] | [
"Integer Multiplication Identity is One",
"Definition:Divisor (Algebra)/Integer"
] |
proofwiki-3259 | Integer Divisor Results/Integer Divides Itself | :$n \divides n$ | As the set of integers form an integral domain, the concept '''divides''' is fully applicable to the integers.
Therefore this result follows directly from Element of Integral Domain is Divisor of Itself.
{{qed}} | :$n \divides n$ | As the [[Integers form Integral Domain|set of integers form an integral domain]], the concept '''[[Definition:Divisor of Ring Element|divides]]''' is fully applicable to the [[Definition:Integer|integers]].
Therefore this result follows directly from [[Element of Integral Domain is Divisor of Itself]].
{{qed}} | Integer Divisor Results/Integer Divides Itself/Proof 2 | https://proofwiki.org/wiki/Integer_Divisor_Results/Integer_Divides_Itself | https://proofwiki.org/wiki/Integer_Divisor_Results/Integer_Divides_Itself/Proof_2 | [
"Divisors",
"Integer Divisor Results/Integer Divides Itself"
] | [] | [
"Integers form Integral Domain",
"Definition:Divisor (Algebra)/Ring with Unity",
"Definition:Integer",
"Element of Integral Domain is Divisor of Itself"
] |
proofwiki-3260 | Integer Divisor Results/Integer Divides its Negative | {{begin-eqn}}
{{eqn | l = n
| o = \divides
| r = -n
}}
{{eqn | l = -n
| o = \divides
| r = n
}}
{{end-eqn}} | From Integers form Integral Domain, the integers are an integral domain.
Hence we can apply Product of Ring Negatives:
:$\forall n \in \Z: \exists -1 \in \Z: n = \paren {-1} \times \paren {-n}$
and Product with Ring Negative:
:$\forall n \in \Z: \exists -1 \in \Z: -n = \paren {-1} \times \paren n$
{{qed}} | {{begin-eqn}}
{{eqn | l = n
| o = \divides
| r = -n
}}
{{eqn | l = -n
| o = \divides
| r = n
}}
{{end-eqn}} | From [[Integers form Integral Domain]], the [[Definition:Integer|integers]] are an [[Definition:Integral Domain|integral domain]].
Hence we can apply [[Product of Ring Negatives]]:
:$\forall n \in \Z: \exists -1 \in \Z: n = \paren {-1} \times \paren {-n}$
and [[Product with Ring Negative]]:
:$\forall n \in \Z: \exis... | Integer Divisor Results/Integer Divides its Negative | https://proofwiki.org/wiki/Integer_Divisor_Results/Integer_Divides_its_Negative | https://proofwiki.org/wiki/Integer_Divisor_Results/Integer_Divides_its_Negative | [
"Divisors"
] | [] | [
"Integers form Integral Domain",
"Definition:Integer",
"Definition:Integral Domain",
"Product of Ring Negatives",
"Product with Ring Negative"
] |
proofwiki-3261 | Integer Divisor Results/Integer Divides its Absolute Value | {{begin-eqn}}
{{eqn | l = n
| o = \divides
| r = \size n
}}
{{eqn | l = \size n
| o = \divides
| r = n
}}
{{end-eqn}}
where:
:$\size n$ is the absolute value of $n$
:$\divides$ denotes divisibility. | Let $n > 0$.
Then $\size n = n$ and Integer Divides Itself applies.
Let $n = 0$.
Then Integer Divides Itself holds again.
Let $n < 0$.
Then $\size n = -n$ and Integer Divides its Negative applies.
{{qed}}
Category:Divisors
Category:Absolute Value Function
qwokq3o9x7dlgf3p8phr3bj8v8ig1wr | {{begin-eqn}}
{{eqn | l = n
| o = \divides
| r = \size n
}}
{{eqn | l = \size n
| o = \divides
| r = n
}}
{{end-eqn}}
where:
:$\size n$ is the [[Definition:Absolute Value|absolute value]] of $n$
:$\divides$ denotes [[Definition:Divisor of Integer|divisibility]]. | Let $n > 0$.
Then $\size n = n$ and [[Integer Divides Itself]] applies.
Let $n = 0$.
Then [[Integer Divides Itself]] holds again.
Let $n < 0$.
Then $\size n = -n$ and [[Integer Divides its Negative]] applies.
{{qed}}
[[Category:Divisors]]
[[Category:Absolute Value Function]]
qwokq3o9x7dlgf3p8phr3bj8v8ig1wr | Integer Divisor Results/Integer Divides its Absolute Value | https://proofwiki.org/wiki/Integer_Divisor_Results/Integer_Divides_its_Absolute_Value | https://proofwiki.org/wiki/Integer_Divisor_Results/Integer_Divides_its_Absolute_Value | [
"Divisors",
"Absolute Value Function"
] | [
"Definition:Absolute Value",
"Definition:Divisor (Algebra)/Integer"
] | [
"Integer Divisor Results/Integer Divides Itself",
"Integer Divisor Results/Integer Divides Itself",
"Integer Divisor Results/Integer Divides its Negative",
"Category:Divisors",
"Category:Absolute Value Function"
] |
proofwiki-3262 | Integer Divisor Results/Integer Divides Zero | :$n \divides 0$ | From Integers form Integral Domain, the concept '''divides''' is fully applicable to the integers.
Therefore this result follows directly from Element of Integral Domain Divides Zero.
{{qed}} | :$n \divides 0$ | From [[Integers form Integral Domain]], the concept '''[[Definition:Divisor of Ring Element|divides]]''' is fully applicable to the [[Definition:Integer|integers]].
Therefore this result follows directly from [[Element of Integral Domain Divides Zero]].
{{qed}} | Integer Divisor Results/Integer Divides Zero | https://proofwiki.org/wiki/Integer_Divisor_Results/Integer_Divides_Zero | https://proofwiki.org/wiki/Integer_Divisor_Results/Integer_Divides_Zero | [
"Divisors",
"Integer Divisor Results"
] | [] | [
"Integers form Integral Domain",
"Definition:Divisor (Algebra)/Ring with Unity",
"Definition:Integer",
"Element of Integral Domain Divides Zero"
] |
proofwiki-3263 | Integer Divisor Results/Divisors of Negative Values | :$m \divides n \iff -m \divides n \iff m \divides -n \iff -m \divides -n$ | Let $m \divides n$.
From Integer Divides its Negative, we have $-m \divides m$.
From Divisor Relation on Positive Integers is Partial Ordering it follows that $-m \divides n$.
From Integer Divides its Negative, we have $n \divides -n$.
From Divisor Relation on Positive Integers is Partial Ordering it follows that $m \d... | :$m \divides n \iff -m \divides n \iff m \divides -n \iff -m \divides -n$ | Let $m \divides n$.
From [[Integer Divides its Negative]], we have $-m \divides m$.
From [[Divisor Relation on Positive Integers is Partial Ordering]] it follows that $-m \divides n$.
From [[Integer Divides its Negative]], we have $n \divides -n$.
From [[Divisor Relation on Positive Integers is Partial Ordering]] ... | Integer Divisor Results/Divisors of Negative Values | https://proofwiki.org/wiki/Integer_Divisor_Results/Divisors_of_Negative_Values | https://proofwiki.org/wiki/Integer_Divisor_Results/Divisors_of_Negative_Values | [
"Divisors"
] | [] | [
"Integer Divisor Results/Integer Divides its Negative",
"Divisor Relation on Positive Integers is Partial Ordering",
"Integer Divisor Results/Integer Divides its Negative",
"Divisor Relation on Positive Integers is Partial Ordering",
"Integer Divisor Results/Integer Divides its Negative",
"Divisor Relatio... |
proofwiki-3264 | Divisors of One | The only divisors of $1$ are $1$ and $-1$.
That is:
:$a \divides 1 \iff a = \pm 1$ | === Necessary Condition ===
Let $a = \pm 1$.
From Integer Divides Itself we have that $1 \divides 1$.
From Integer Divides its Negative we have that $-1 \divides 1$.
{{qed|lemma}} | The only [[Definition:Divisor of Integer|divisors]] of $1$ are $1$ and $-1$.
That is:
:$a \divides 1 \iff a = \pm 1$ | === Necessary Condition ===
Let $a = \pm 1$.
From [[Integer Divides Itself]] we have that $1 \divides 1$.
From [[Integer Divides its Negative]] we have that $-1 \divides 1$.
{{qed|lemma}} | Divisors of One | https://proofwiki.org/wiki/Divisors_of_One | https://proofwiki.org/wiki/Divisors_of_One | [
"Divisors",
"1"
] | [
"Definition:Divisor (Algebra)/Integer"
] | [
"Integer Divisor Results/Integer Divides Itself",
"Integer Divisor Results/Integer Divides its Negative"
] |
proofwiki-3265 | Divisor Relation is Antisymmetric | ''Divides'' is a antisymmetric relation on $\Z_{>0}$, the set of positive integers.
That is:
:$\forall a, b \in \Z_{>0}: a \divides b \land b \divides a \implies a = b$ | Let $a, b \in \Z_{> 0}$ such that $a \divides b$ and $b \divides a$.
Then:
{{begin-eqn}}
{{eqn | l = a \divides b
| o = \implies
| r = \size a \le \size b
| c = Absolute Value of Integer is not less than Divisors
}}
{{eqn | l = b \divides a
| o = \implies
| r = \size b \le \size a
| ... | ''[[Definition:Divisor of Integer|Divides]]'' is a [[Definition:Antisymmetric Relation|antisymmetric relation]] on $\Z_{>0}$, the set of [[Definition:Positive Integer|positive integers]].
That is:
:$\forall a, b \in \Z_{>0}: a \divides b \land b \divides a \implies a = b$ | Let $a, b \in \Z_{> 0}$ such that $a \divides b$ and $b \divides a$.
Then:
{{begin-eqn}}
{{eqn | l = a \divides b
| o = \implies
| r = \size a \le \size b
| c = [[Absolute Value of Integer is not less than Divisors]]
}}
{{eqn | l = b \divides a
| o = \implies
| r = \size b \le \size a
... | Divisor Relation is Antisymmetric | https://proofwiki.org/wiki/Divisor_Relation_is_Antisymmetric | https://proofwiki.org/wiki/Divisor_Relation_is_Antisymmetric | [
"Divisors",
"Divisor Relation is Antisymmetric"
] | [
"Definition:Divisor (Algebra)/Integer",
"Definition:Antisymmetric Relation",
"Definition:Positive/Integer"
] | [
"Absolute Value of Integer is not less than Divisors",
"Absolute Value of Integer is not less than Divisors",
"Definition:Domain (Set Theory)/Relation",
"Definition:Positive/Integer"
] |
proofwiki-3266 | Divisor Relation is Antisymmetric | ''Divides'' is a antisymmetric relation on $\Z_{>0}$, the set of positive integers.
That is:
:$\forall a, b \in \Z_{>0}: a \divides b \land b \divides a \implies a = b$ | Let $a \divides b$ and $b \divides a$.
Then from Divisor Relation is Antisymmetric:
:$\size a = \size b$
The result follows from Integer Divides its Negative and Integer Divides its Absolute Value.
{{qed}} | ''[[Definition:Divisor of Integer|Divides]]'' is a [[Definition:Antisymmetric Relation|antisymmetric relation]] on $\Z_{>0}$, the set of [[Definition:Positive Integer|positive integers]].
That is:
:$\forall a, b \in \Z_{>0}: a \divides b \land b \divides a \implies a = b$ | Let $a \divides b$ and $b \divides a$.
Then from [[Divisor Relation is Antisymmetric]]:
:$\size a = \size b$
The result follows from [[Integer Divides its Negative]] and [[Integer Divides its Absolute Value]].
{{qed}} | Divisor Relation is Antisymmetric/Corollary/Proof 1 | https://proofwiki.org/wiki/Divisor_Relation_is_Antisymmetric | https://proofwiki.org/wiki/Divisor_Relation_is_Antisymmetric/Corollary/Proof_1 | [
"Divisors",
"Divisor Relation is Antisymmetric"
] | [
"Definition:Divisor (Algebra)/Integer",
"Definition:Antisymmetric Relation",
"Definition:Positive/Integer"
] | [
"Divisor Relation is Antisymmetric",
"Integer Divisor Results/Integer Divides its Negative",
"Integer Divisor Results/Integer Divides its Absolute Value"
] |
proofwiki-3267 | Divisor Relation is Antisymmetric | ''Divides'' is a antisymmetric relation on $\Z_{>0}$, the set of positive integers.
That is:
:$\forall a, b \in \Z_{>0}: a \divides b \land b \divides a \implies a = b$ | Let $a \divides b$ and $b \divides a$.
Then by definition of divisor:
:$\exists c, d \in \Z: a c = b, b d = a$
Thus:
{{begin-eqn}}
{{eqn | l = a c d
| r = a
| c =
}}
{{eqn | ll= \leadsto
| l = c d
| r = 1
| c =
}}
{{eqn | ll= \leadsto
| l = d
| r = \pm 1
| c = Divisors ... | ''[[Definition:Divisor of Integer|Divides]]'' is a [[Definition:Antisymmetric Relation|antisymmetric relation]] on $\Z_{>0}$, the set of [[Definition:Positive Integer|positive integers]].
That is:
:$\forall a, b \in \Z_{>0}: a \divides b \land b \divides a \implies a = b$ | Let $a \divides b$ and $b \divides a$.
Then by definition of [[Definition:Divisor of Integer|divisor]]:
:$\exists c, d \in \Z: a c = b, b d = a$
Thus:
{{begin-eqn}}
{{eqn | l = a c d
| r = a
| c =
}}
{{eqn | ll= \leadsto
| l = c d
| r = 1
| c =
}}
{{eqn | ll= \leadsto
| l = d
... | Divisor Relation is Antisymmetric/Corollary/Proof 2 | https://proofwiki.org/wiki/Divisor_Relation_is_Antisymmetric | https://proofwiki.org/wiki/Divisor_Relation_is_Antisymmetric/Corollary/Proof_2 | [
"Divisors",
"Divisor Relation is Antisymmetric"
] | [
"Definition:Divisor (Algebra)/Integer",
"Definition:Antisymmetric Relation",
"Definition:Positive/Integer"
] | [
"Definition:Divisor (Algebra)/Integer",
"Divisors of One"
] |
proofwiki-3268 | Real Function is Continuous at Point iff Oscillation is Zero | Let $f: D \to \R$ be a real function where $D \subseteq \R$.
Let $x$ be a point in $D$.
Let $N_x$ be the set of open subset neighborhoods of $x$.
Let $\map {\omega_f} x$ be the oscillation of $f$ at $x$:
:$\map {\omega_f} x = \ds \inf \set {\map {\omega_f} {I \cap D}: I \in N_x}$
where $\map {\omega_f} {I \cap D}$ deno... | === Necessary Condition ===
Let $\map {\omega_f} x = 0$.
Let $\epsilon > 0$.
Suppose that $\forall I: I \in N_x: \map {\omega_f} {I \cap D} \ge \epsilon$.
Then by definition of oscillation at point, $\map {\omega_f} x \ge \epsilon$.
This contradicts $\map {\omega_f} x = 0$.
From this contradiction we deduce that:
:$\e... | Let $f: D \to \R$ be a [[Definition:Real Function|real function]] where $D \subseteq \R$.
Let $x$ be a point in $D$.
Let $N_x$ be the [[Definition:Set|set]] of [[Definition:Neighborhood of Real Number|open subset neighborhoods]] of $x$.
Let $\map {\omega_f} x$ be the [[Definition:Oscillation/Real Space/Oscillation a... | === Necessary Condition ===
Let $\map {\omega_f} x = 0$.
Let $\epsilon > 0$.
Suppose that $\forall I: I \in N_x: \map {\omega_f} {I \cap D} \ge \epsilon$.
Then by definition of [[Definition:Oscillation/Real Space/Oscillation at Point/Infimum|oscillation at point]], $\map {\omega_f} x \ge \epsilon$.
This contradic... | Real Function is Continuous at Point iff Oscillation is Zero | https://proofwiki.org/wiki/Real_Function_is_Continuous_at_Point_iff_Oscillation_is_Zero | https://proofwiki.org/wiki/Real_Function_is_Continuous_at_Point_iff_Oscillation_is_Zero | [
"Real Analysis"
] | [
"Definition:Real Function",
"Definition:Set",
"Definition:Neighborhood (Real Analysis)/Open Subset",
"Definition:Oscillation/Real Space/Oscillation at Point/Infimum",
"Definition:Oscillation/Real Space/Oscillation on Set",
"Definition:Continuous Real Function"
] | [
"Definition:Oscillation/Real Space/Oscillation at Point/Infimum",
"Definition:Open Set/Real Analysis",
"Definition:Element",
"Definition:Neighborhood (Real Analysis)/Open Subset",
"Definition:Subset",
"Definition:Continuous Real Function",
"Definition:Continuous Real Function/Point",
"Definition:Conti... |
proofwiki-3269 | Quadratic Integers over 2 are Not a Field | Let $\Z \sqbrk {\sqrt 2}$ denote the set of quadratic integers over $2$:
:$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$
that is, all (real) numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are integers.
Then the algebraic structure:
:$\struct {\Z \sqbrk {\sqrt 2}, +, \times}$
where $+$ and $\times$ ar... | We start from the result Quadratic Integers over 2 form Subdomain of Reals.
We have from that result that $1 + 0 \sqrt 2$ is the unity.
Now consider a representative element $a + b \sqrt 2 \in \Z \sqbrk {\sqrt 2}$.
From Difference of Two Squares we have:
:$\paren {a + b \sqrt 2} \paren {a - b \sqrt 2} = a^2 - 2 b^2$
wh... | Let $\Z \sqbrk {\sqrt 2}$ denote the [[Definition:Set|set]] of [[Definition:Quadratic Integer|quadratic integers]] over $2$:
:$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$
that is, all [[Definition:Real Number|(real) numbers]] of the form $a + b \sqrt 2$ where $a$ and $b$ are [[Definition:Integer|integers]... | We start from the result [[Quadratic Integers over 2 form Subdomain of Reals]].
We have from that result that $1 + 0 \sqrt 2$ is the [[Definition:Unity of Ring|unity]].
Now consider a representative element $a + b \sqrt 2 \in \Z \sqbrk {\sqrt 2}$.
From [[Difference of Two Squares]] we have:
:$\paren {a + b \sqrt 2}... | Quadratic Integers over 2 are Not a Field | https://proofwiki.org/wiki/Quadratic_Integers_over_2_are_Not_a_Field | https://proofwiki.org/wiki/Quadratic_Integers_over_2_are_Not_a_Field | [
"Field Theory",
"Quadratic Integers"
] | [
"Definition:Set",
"Definition:Algebraic Integer/Quadratic",
"Definition:Real Number",
"Definition:Integer",
"Definition:Algebraic Structure",
"Definition:Addition/Real Numbers",
"Definition:Multiplication/Real Numbers",
"Definition:Real Number",
"Definition:Field (Abstract Algebra)"
] | [
"Quadratic Integers over 2 form Subdomain of Reals",
"Definition:Unity (Abstract Algebra)/Ring",
"Difference of Two Squares",
"Definition:Product Inverse",
"Definition:Integer",
"Definition:Product Inverse",
"Definition:Field (Abstract Algebra)"
] |
proofwiki-3270 | Real Numbers of Type Rational a plus b root 2 form Field | Let $\Q \sqbrk {\sqrt 2}$ denote the set:
:$\Q \sqbrk {\sqrt 2} := \set {x \in \R: x = a + b \sqrt 2: a, b \in \Q}$
that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are rational numbers.
Then the algebraic structure:
:$\struct {\Q \sqbrk {\sqrt 2}, +, \times}$
where $+$ and $\times$ are conventional a... | {{improve|Use subfield test}}
By definition, $\Q \sqbrk {\sqrt 2} \subseteq \R$ where $\R$ is the set of real numbers.
Thus $+$ and $\times$ on $\struct {\Q \sqbrk {\sqrt 2}, +, \times}$ are well-defined. | Let $\Q \sqbrk {\sqrt 2}$ denote the [[Definition:Set|set]]:
:$\Q \sqbrk {\sqrt 2} := \set {x \in \R: x = a + b \sqrt 2: a, b \in \Q}$
that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are [[Definition:Rational Number|rational numbers]].
Then the [[Definition:Algebraic Structure with Two Operations|a... | {{improve|Use subfield test}}
By definition, $\Q \sqbrk {\sqrt 2} \subseteq \R$ where $\R$ is the set of [[Definition:Real Number|real numbers]].
Thus $+$ and $\times$ on $\struct {\Q \sqbrk {\sqrt 2}, +, \times}$ are [[Definition:Well-Defined Operation|well-defined]]. | Real Numbers of Type Rational a plus b root 2 form Field | https://proofwiki.org/wiki/Real_Numbers_of_Type_Rational_a_plus_b_root_2_form_Field | https://proofwiki.org/wiki/Real_Numbers_of_Type_Rational_a_plus_b_root_2_form_Field | [
"Examples of Fields"
] | [
"Definition:Set",
"Definition:Rational Number",
"Definition:Algebraic Structure/Two Operations",
"Definition:Addition/Real Numbers",
"Definition:Multiplication/Real Numbers",
"Definition:Real Number",
"Definition:Field (Abstract Algebra)"
] | [
"Definition:Real Number",
"Definition:Well-Defined/Operation"
] |
proofwiki-3271 | Characteristic of Ring of Integers Modulo Prime | Let $\struct {\Z_p, +, \times}$ be the ring of integers modulo $p$, where $p$ is a prime number.
The characteristic of $\struct {\Z_p, +, \times}$ is $p$. | From Ring of Integers Modulo Prime is Field we have that $\struct {\Z_p, +, \times}$ is a field.
So Characteristic of Finite Ring with No Zero Divisors applies, and so the characteristic of $\struct {\Z_p, +, \times}$ is prime.
The result follows.
{{qed}} | Let $\struct {\Z_p, +, \times}$ be the [[Definition:Ring of Integers Modulo m|ring of integers modulo $p$]], where $p$ is a [[Definition:Prime Number|prime number]].
The [[Definition:Characteristic of Ring|characteristic]] of $\struct {\Z_p, +, \times}$ is $p$. | From [[Ring of Integers Modulo Prime is Field]] we have that $\struct {\Z_p, +, \times}$ is a [[Definition:Field (Abstract Algebra)|field]].
So [[Characteristic of Finite Ring with No Zero Divisors]] applies, and so the [[Definition:Characteristic of Ring|characteristic]] of $\struct {\Z_p, +, \times}$ is [[Definition... | Characteristic of Ring of Integers Modulo Prime | https://proofwiki.org/wiki/Characteristic_of_Ring_of_Integers_Modulo_Prime | https://proofwiki.org/wiki/Characteristic_of_Ring_of_Integers_Modulo_Prime | [
"Ring of Integers Modulo m",
"Finite Rings"
] | [
"Definition:Ring of Integers Modulo m",
"Definition:Prime Number",
"Definition:Characteristic of Ring"
] | [
"Ring of Integers Modulo Prime is Field",
"Definition:Field (Abstract Algebra)",
"Characteristic of Finite Ring with No Zero Divisors",
"Definition:Characteristic of Ring",
"Definition:Prime Number"
] |
proofwiki-3272 | Gamma Function Extends Factorial | :$\forall n \in \N: \map \Gamma {n + 1} = n!$ | For $n = 0$:
{{begin-eqn}}
{{eqn | l = \map \Gamma 1
| r = \int_0^\infty e^{-t} \rd t
| c = {{Defof|Integral Form of Gamma Function|Gamma Function}}
}}
{{eqn | r = \bigintlimits {-e^{-t} } 0 \infty
| c =
}}
{{eqn | r = 0 - \paren {-1}
}}
{{eqn | r = 1
}}
{{end-eqn}}
Then by Gamma Difference Equation:
... | :$\forall n \in \N: \map \Gamma {n + 1} = n!$ | For $n = 0$:
{{begin-eqn}}
{{eqn | l = \map \Gamma 1
| r = \int_0^\infty e^{-t} \rd t
| c = {{Defof|Integral Form of Gamma Function|Gamma Function}}
}}
{{eqn | r = \bigintlimits {-e^{-t} } 0 \infty
| c =
}}
{{eqn | r = 0 - \paren {-1}
}}
{{eqn | r = 1
}}
{{end-eqn}}
Then by [[Gamma Difference Equat... | Gamma Function Extends Factorial | https://proofwiki.org/wiki/Gamma_Function_Extends_Factorial | https://proofwiki.org/wiki/Gamma_Function_Extends_Factorial | [
"Gamma Function",
"Factorials"
] | [] | [
"Gamma Difference Equation"
] |
proofwiki-3273 | Uniform Limit of Analytic Functions is Analytic | Let $U$ be an open subset of $\C$.
Let $\sequence {f_n}_{n \mathop \in \N}$ be a sequence of analytic functions $f_n : U \to \C$.
Let $\sequence {f_n}$ converge locally uniformly to $f$ on $U$.
Then $f$ is analytic. | Let $z \in U$.
By definition of locally uniform convergence, there is an $\epsilon > 0$ so that:
:$\map {B_\epsilon} z \subset U$
and $f_n$ converges uniformly on $\map {B_\epsilon} z$.
Let $\gamma$ be any simple closed curve in $\map {B_\epsilon} z$.
Then $f_n \to f$ uniformly on $\gamma$, because $\gamma \subset \map... | Let $U$ be an [[Definition:Open Set (Complex Analysis)|open subset]] of $\C$.
Let $\sequence {f_n}_{n \mathop \in \N}$ be a [[Definition:Sequence|sequence]] of [[Definition:Analytic Complex Function|analytic functions]] $f_n : U \to \C$.
Let $\sequence {f_n}$ [[Definition:Locally Uniform Convergence|converge locally ... | Let $z \in U$.
By definition of [[Definition:Locally Uniform Convergence|locally uniform convergence]], there is an $\epsilon > 0$ so that:
:$\map {B_\epsilon} z \subset U$
and $f_n$ [[Definition:Uniform Convergence|converges uniformly]] on $\map {B_\epsilon} z$.
Let $\gamma$ be any simple [[Definition:Closed Contour... | Uniform Limit of Analytic Functions is Analytic | https://proofwiki.org/wiki/Uniform_Limit_of_Analytic_Functions_is_Analytic | https://proofwiki.org/wiki/Uniform_Limit_of_Analytic_Functions_is_Analytic | [
"Complex Analysis",
"Analytic Complex Functions",
"Uniform Convergence"
] | [
"Definition:Open Set/Complex Analysis",
"Definition:Sequence",
"Definition:Analytic Function/Complex Plane",
"Definition:Locally Uniform Convergence",
"Definition:Analytic Function"
] | [
"Definition:Locally Uniform Convergence",
"Definition:Uniform Convergence",
"Definition:Contour/Closed/Complex Plane",
"Definition:Uniform Convergence",
"Definite Integral of Limit of Uniformly Convergent Sequence of Integrable Functions",
"Definition:Analytic Function/Complex Plane",
"Cauchy-Goursat Th... |
proofwiki-3274 | Triangle Inequality for Integrals/Complex | Let $\closedint a b$ be a closed real interval.
Let $f: \closedint a b \to \C$ be a continuous complex function.
Then:
:$\ds \size {\int_a^b \map f t \rd t} \le \int_a^b \size {\map f t} \rd t$
where the first integral is a complex Riemann integral, and the second integral is a definite real integral. | Define:
:$z \in \C$ as the value of the complex Riemann integral:
::$z = \ds \int_a^b \map f t \rd t$
:$r \in \hointr 0 \to$ as the modulus of $z$
:$\theta \in \hointr 0 {2 \pi}$ as the argument of $z$.
From Modulus and Argument of Complex Exponential:
:$z = re^{i \theta}$
Then:
{{begin-eqn}}
{{eqn | l = r
| r = ... | Let $\closedint a b$ be a [[Definition:Closed Real Interval|closed real interval]].
Let $f: \closedint a b \to \C$ be a [[Definition:Continuous Complex Function|continuous]] [[Definition:Complex Function|complex function]].
Then:
:$\ds \size {\int_a^b \map f t \rd t} \le \int_a^b \size {\map f t} \rd t$
where the f... | Define:
:$z \in \C$ as the value of the [[Definition:Complex Riemann Integral|complex Riemann integral]]:
::$z = \ds \int_a^b \map f t \rd t$
:$r \in \hointr 0 \to$ as the [[Definition:Complex Modulus|modulus]] of $z$
:$\theta \in \hointr 0 {2 \pi}$ as the [[Definition:Argument of Complex Number|argument]] of $z$.
... | Triangle Inequality for Integrals/Complex | https://proofwiki.org/wiki/Triangle_Inequality_for_Integrals/Complex | https://proofwiki.org/wiki/Triangle_Inequality_for_Integrals/Complex | [
"Triangle Inequality for Integrals",
"Complex Analysis"
] | [
"Definition:Real Interval/Closed",
"Definition:Continuous Complex Function",
"Definition:Complex Function",
"Definition:Integrable Function/Complex",
"Definition:Definite Integral"
] | [
"Definition:Integrable Function/Complex",
"Definition:Complex Modulus",
"Definition:Argument of Complex Number",
"Modulus and Argument of Complex Exponential",
"Reciprocal of Complex Exponential",
"Linear Combination of Complex Integrals",
"Definition:Complex Number/Wholly Real",
"Triangle Inequality ... |
proofwiki-3275 | Estimation Lemma for Contour Integrals | Let $C$ be a contour.
Let $f: \Img C \to \C$ be a continuous complex function, where $\Img C$ denotes the image of $C$.
Then:
:$\ds \size {\int_C \map f z \rd z} \le \max_{z \mathop \in \Img C} \size {\map f z} \map L C$
where $\map L C$ denotes the length of $C$. | By definition of contour, $C$ is a concatenation of a finite sequence $C_1, \ldots, C_n$ of directed smooth curves.
Let $C_k$ be parameterized by the smooth path $\gamma_k: \closedint {a_k} {b_k} \to \C$ for all $k \in \set {1, \ldots, n}$.
Then:
{{begin-eqn}}
{{eqn | l = \size {\int_C \map f z \rd z}
| r = \size... | Let $C$ be a [[Definition:Contour (Complex Plane)|contour]].
Let $f: \Img C \to \C$ be a [[Definition:Continuous Complex Function|continuous complex function]], where $\Img C$ denotes the [[Definition:Image of Contour (Complex Plane)|image]] of $C$.
Then:
:$\ds \size {\int_C \map f z \rd z} \le \max_{z \mathop \in ... | By definition of [[Definition:Contour (Complex Plane)|contour]], $C$ is a [[Definition:Concatenation of Contours (Complex Plane)|concatenation]] of a [[Definition:Finite Sequence|finite sequence]] $C_1, \ldots, C_n$ of [[Definition:Directed Smooth Curve (Complex Plane)|directed smooth curves]].
Let $C_k$ be [[Definiti... | Estimation Lemma for Contour Integrals | https://proofwiki.org/wiki/Estimation_Lemma_for_Contour_Integrals | https://proofwiki.org/wiki/Estimation_Lemma_for_Contour_Integrals | [
"Complex Contour Integrals",
"Named Theorems",
"Triangle Inequality",
"Inequalities"
] | [
"Definition:Contour/Complex Plane",
"Definition:Continuous Complex Function",
"Definition:Contour/Image/Complex Plane",
"Definition:Contour/Length/Complex Plane"
] | [
"Definition:Contour/Complex Plane",
"Definition:Concatenation of Contours/Complex Plane",
"Definition:Finite Sequence",
"Definition:Directed Smooth Curve/Complex Plane",
"Definition:Directed Smooth Curve/Parameterization/Complex Plane",
"Definition:Smooth Path/Complex",
"Triangle Inequality/Complex Numb... |
proofwiki-3276 | Jensen's Formula | Let $S$ be an open subset of the complex plane containing the closed disk:
:$D_r = \set {z \in \C : \cmod z \le r}$
of radius $r$ about $0$.
Let $f: S \to \C$ be holomorphic on $S$.
Let $f$ have no zeroes on the circle $\cmod z = r$.
Let $\map f 0 \ne 0$.
Let $\rho_1, \ldots, \rho_n$ be the zeroes of $f$ in $D_r$, cou... | Write:
:$\map f z = \paren {z - \rho_1} \dotsm \paren {z - \rho_n} \map g z$
so $\map g z \ne 0$ for $z \in D_r$.
It is sufficient to check the equality for each factor of $f$ in this expansion.
First let:
:$\map h z = z - \rho_k$
for some $k \in \set {1, \ldots, n}$.
Making use of the substitution $u = r e^{i \theta} ... | Let $S$ be an [[Definition:Open Set (Complex Analysis)|open subset]] of the [[Definition:Complex Plane|complex plane]] containing the [[Definition:Closed Complex Disk|closed disk]]:
:$D_r = \set {z \in \C : \cmod z \le r}$
of [[Definition:Radius of Closed Ball|radius]] $r$ about $0$.
Let $f: S \to \C$ be [[Definition:... | Write:
:$\map f z = \paren {z - \rho_1} \dotsm \paren {z - \rho_n} \map g z$
so $\map g z \ne 0$ for $z \in D_r$.
It is sufficient to check the equality for each factor of $f$ in this expansion.
First let:
:$\map h z = z - \rho_k$
for some $k \in \set {1, \ldots, n}$.
Making use of the [[Integration by Substitution... | Jensen's Formula/Proof 1 | https://proofwiki.org/wiki/Jensen's_Formula | https://proofwiki.org/wiki/Jensen's_Formula/Proof_1 | [
"Complex Analysis",
"Jensen's Formula"
] | [
"Definition:Open Set/Complex Analysis",
"Definition:Complex Number/Complex Plane",
"Definition:Complex Disk/Closed",
"Definition:Closed Ball/Metric Space/Radius",
"Definition:Holomorphic Function",
"Definition:Root of Mapping",
"Definition:Circle",
"Definition:Root of Mapping",
"Definition:Multiplic... | [
"Integration by Substitution",
"Definition:Anticlockwise",
"Cauchy's Residue Theorem"
] |
proofwiki-3277 | Jensen's Formula | Let $S$ be an open subset of the complex plane containing the closed disk:
:$D_r = \set {z \in \C : \cmod z \le r}$
of radius $r$ about $0$.
Let $f: S \to \C$ be holomorphic on $S$.
Let $f$ have no zeroes on the circle $\cmod z = r$.
Let $\map f 0 \ne 0$.
Let $\rho_1, \ldots, \rho_n$ be the zeroes of $f$ in $D_r$, cou... | Write
:$\map f z = \dfrac {r^2 - z \overline {\rho_1} } {r \paren {z - \rho_1} } \cdots \dfrac {r^2 - z \overline {\rho_n} } {r \paren {z - \rho_n} } \map g z$
so $\map g z \ne 0$ for $z \in D_r$.
{{explain|Why can you write like this? Looks wrong, because the zeros do not correspond. For example, insert $z{{=}}\rho_1$... | Let $S$ be an [[Definition:Open Set (Complex Analysis)|open subset]] of the [[Definition:Complex Plane|complex plane]] containing the [[Definition:Closed Complex Disk|closed disk]]:
:$D_r = \set {z \in \C : \cmod z \le r}$
of [[Definition:Radius of Closed Ball|radius]] $r$ about $0$.
Let $f: S \to \C$ be [[Definition:... | Write
:$\map f z = \dfrac {r^2 - z \overline {\rho_1} } {r \paren {z - \rho_1} } \cdots \dfrac {r^2 - z \overline {\rho_n} } {r \paren {z - \rho_n} } \map g z$
so $\map g z \ne 0$ for $z \in D_r$.
{{explain|Why can you write like this? Looks wrong, because the zeros do not correspond. For example, insert $z{{=}}\rho_1... | Jensen's Formula/Proof 2 | https://proofwiki.org/wiki/Jensen's_Formula | https://proofwiki.org/wiki/Jensen's_Formula/Proof_2 | [
"Complex Analysis",
"Jensen's Formula"
] | [
"Definition:Open Set/Complex Analysis",
"Definition:Complex Number/Complex Plane",
"Definition:Complex Disk/Closed",
"Definition:Closed Ball/Metric Space/Radius",
"Definition:Holomorphic Function",
"Definition:Root of Mapping",
"Definition:Circle",
"Definition:Root of Mapping",
"Definition:Multiplic... | [
"Definition:Complex Conjugate",
"Definition:Holomorphic Function",
"Definition:Root of Mapping",
"Definition:Mean Value Property"
] |
proofwiki-3278 | Jensen's Formula | Let $S$ be an open subset of the complex plane containing the closed disk:
:$D_r = \set {z \in \C : \cmod z \le r}$
of radius $r$ about $0$.
Let $f: S \to \C$ be holomorphic on $S$.
Let $f$ have no zeroes on the circle $\cmod z = r$.
Let $\map f 0 \ne 0$.
Let $\rho_1, \ldots, \rho_n$ be the zeroes of $f$ in $D_r$, cou... | {{WLOG}}, assume that $r = 0$.
{{explain|I don't see why $r {{=}} 0$ is WLOG?}}
Write:
:$\map f z = \map {B_{\rho_1} } z \dotsm \map {B_{\rho_n} } z \map g z$
where $\map {B_\rho} z := \dfrac {z - \rho} {1 - z \overline \rho}$ is the Blaschke factor at $\rho$.
So:
:$\map g z$ is holomorphic and nonzero in $D_r$
and:
:... | Let $S$ be an [[Definition:Open Set (Complex Analysis)|open subset]] of the [[Definition:Complex Plane|complex plane]] containing the [[Definition:Closed Complex Disk|closed disk]]:
:$D_r = \set {z \in \C : \cmod z \le r}$
of [[Definition:Radius of Closed Ball|radius]] $r$ about $0$.
Let $f: S \to \C$ be [[Definition:... | {{WLOG}}, assume that $r = 0$.
{{explain|I don't see why $r {{=}} 0$ is WLOG?}}
Write:
:$\map f z = \map {B_{\rho_1} } z \dotsm \map {B_{\rho_n} } z \map g z$
where $\map {B_\rho} z := \dfrac {z - \rho} {1 - z \overline \rho}$ is the [[Definition:Blaschke Factor|Blaschke factor]] at $\rho$.
So:
:$\map g z$ is holo... | Jensen's Formula/Proof 3 | https://proofwiki.org/wiki/Jensen's_Formula | https://proofwiki.org/wiki/Jensen's_Formula/Proof_3 | [
"Complex Analysis",
"Jensen's Formula"
] | [
"Definition:Open Set/Complex Analysis",
"Definition:Complex Number/Complex Plane",
"Definition:Complex Disk/Closed",
"Definition:Closed Ball/Metric Space/Radius",
"Definition:Holomorphic Function",
"Definition:Root of Mapping",
"Definition:Circle",
"Definition:Root of Mapping",
"Definition:Multiplic... | [
"Definition:Blaschke Factor",
"Cauchy's Residue Theorem"
] |
proofwiki-3279 | Cauchy's Residue Theorem | Let $U$ be a simply connected open subset of the complex plane $\C$.
Let $a_1, a_2, \dots, a_n$ be finitely many points of $U$.
Let $f: U \to \C$ be analytic in $U \setminus \set {a_1, a_2, \dots, a_n}$.
Let $L$ be a contour in $\C$ oriented anticlockwise.
Let $\partial U_k$ denote the closed contour bounding $U_k$.
T... | Let $\set {U_1, \dotsc, U_n}$ be a set of open subsets of $U$ such that $a_i \in U_i$, and $a_i \notin U_j$ for $i \ne j$.
Let $U_i \cap U_j = \O$ for all $i \ne j$.
By Existence of Laurent Series, around each $a_k$ there is an expansion:
:$\ds \map f z = \sum_{j \mathop = -\infty}^\infty c_j \paren {z - a_k}^j$
conve... | Let $U$ be a [[Definition:Simply Connected|simply connected]] [[Definition:Open Set (Complex Analysis)|open]] [[Definition:Subset|subset]] of the [[Definition:Complex Number|complex plane]] $\C$.
Let $a_1, a_2, \dots, a_n$ be [[Definition:Finite Set|finitely many]] points of $U$.
Let $f: U \to \C$ be [[Definition:Ana... | Let $\set {U_1, \dotsc, U_n}$ be a [[Definition:Set|set]] of [[Definition:Open Set (Complex Analysis)|open]] [[Definition:Subset|subsets]] of $U$ such that $a_i \in U_i$, and $a_i \notin U_j$ for $i \ne j$.
Let $U_i \cap U_j = \O$ for all $i \ne j$.
By [[Existence of Laurent Series]], around each $a_k$ there is an ex... | Cauchy's Residue Theorem | https://proofwiki.org/wiki/Cauchy's_Residue_Theorem | https://proofwiki.org/wiki/Cauchy's_Residue_Theorem | [
"Cauchy's Residue Theorem",
"Complex Analysis"
] | [
"Definition:Simply Connected",
"Definition:Open Set/Complex Analysis",
"Definition:Subset",
"Definition:Complex Number",
"Definition:Finite Set",
"Definition:Analytic Function",
"Definition:Contour/Complex Plane",
"Definition:Anticlockwise",
"Definition:Contour/Closed/Complex Plane",
"Definition:R... | [
"Definition:Set",
"Definition:Open Set/Complex Analysis",
"Definition:Subset",
"Existence of Laurent Series",
"Definition:Convergent Series",
"Contour Integral of Concatenation of Contours",
"Definition:Order of Pole/Simple Pole",
"Definition:Subset",
"Definition:Holomorphic Function/Complex Plane",... |
proofwiki-3280 | Cauchy's Integral Formula | Let $D = \set {z \in \C: \cmod z \le r}$ be the closed disk of radius $r$ in $\C$.
Let $f: U \to \C$ be holomorphic on some open set containing $D$.
Then for each $a$ in the interior of $D$:
:$\ds \map f a = \frac 1 {2 \pi i} \oint_{\partial D} \frac {\map f z} {z - a} \rd z$
where $\partial D$ is the boundary of $D$, ... | Let $C$ be any arbitrary closed curve which defines a region $R$ where the function $\map f z$ is holomorphic on $R$.
Let $z_0$ be any point in the region $R$ such that:
:$\dfrac {\map f z} {z - z_0}$ is holomorphic on $R \setminus \set {z_0}$
We draw a circle $C_r$ with center at $z_0$ and radius $r$ such that $r \to ... | Let $D = \set {z \in \C: \cmod z \le r}$ be the [[Definition:Closed Complex Disk|closed disk]] of [[Definition:Radius of Disk|radius]] $r$ in $\C$.
Let $f: U \to \C$ be [[Definition:Holomorphic Function|holomorphic]] on some [[Definition:Open Set (Complex Analysis)|open set]] containing $D$.
Then for each $a$ in the... | Let $C$ be any arbitrary closed curve which defines a region $R$ where the function $\map f z$ is [[Definition:Holomorphic Function|holomorphic]] on $R$.
Let $z_0$ be any point in the region $R$ such that:
:$\dfrac {\map f z} {z - z_0}$ is [[Definition:Holomorphic Function|holomorphic]] on $R \setminus \set {z_0}$
W... | Cauchy's Integral Formula | https://proofwiki.org/wiki/Cauchy's_Integral_Formula | https://proofwiki.org/wiki/Cauchy's_Integral_Formula | [
"Complex Analysis",
"Cauchy's Integral Formula"
] | [
"Definition:Complex Disk/Closed",
"Definition:Disk/Radius",
"Definition:Holomorphic Function",
"Definition:Open Set/Complex Analysis",
"Definition:Interior (Complex Analysis)",
"Definition:Boundary (Geometry)",
"Definition:Anticlockwise"
] | [
"Definition:Holomorphic Function",
"Definition:Holomorphic Function",
"Cauchy-Goursat Theorem",
"Complex-Differentiable Function is Continuous",
"Definition:Continuous Complex Function"
] |
proofwiki-3281 | Cauchy's Integral Formula/General Result | Let $n \in \N$ be a natural number.
Then for each $a$ in the interior of $D$, the $n$-th derivative of $f$ at $a$ exists and can be written as:
:$\ds \map {f^\paren n} a = \frac {n!} {2 \pi i} \oint_{\partial D} \frac {\map f z} {\paren {z - a}^{n + 1} } \rd z$
where $\partial D$ is the boundary of $D$, and is traverse... | Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
:$\ds \map {f^\paren n} a = \frac {n!} {2 \pi i} \oint_{\partial D} \frac {\map f z} {\paren {z - a}^{n + 1} } \rd z$ | Let $n \in \N$ be a [[Definition:Natural Number|natural number]].
Then for each $a$ in the [[Definition:Interior (Topology)|interior]] of $D$, the $n$-th [[Definition:Higher Derivative|derivative]] of $f$ at $a$ exists and can be written as:
:$\ds \map {f^\paren n} a = \frac {n!} {2 \pi i} \oint_{\partial D} \frac {... | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \map {f^\paren n} a = \frac {n!} {2 \pi i} \oint_{\partial D} \frac {\map f z} {\paren {z - a}^{n + 1} } \rd z$ | Cauchy's Integral Formula/General Result | https://proofwiki.org/wiki/Cauchy's_Integral_Formula/General_Result | https://proofwiki.org/wiki/Cauchy's_Integral_Formula/General_Result | [
"Cauchy's Integral Formula",
"Proofs by Induction"
] | [
"Definition:Natural Numbers",
"Definition:Interior (Topology)",
"Definition:Derivative/Higher Derivatives/Higher Order",
"Definition:Boundary (Geometry)",
"Definition:Anticlockwise"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-3282 | Characteristic of Subfield of Complex Numbers is Zero | The characteristic of any subfield of the field of complex numbers is $0$. | {{AimForCont}} to the contrary.
Let $K$ be a subfield of $\C$ such that $\Char K = n$ where $n \in \N, n > 0$.
Then from Characteristic times Ring Element is Ring Zero:
:$\forall a \in K: n \cdot a = 0$
But as $K$ is a subfield of $\C$ it follows that $K \subseteq \C$ which means:
:$\exists a \in \C: n \cdot a = 0$
Thu... | The [[Definition:Characteristic of Field|characteristic]] of any [[Definition:Subfield|subfield]] of the [[Definition:Field of Complex Numbers|field of complex numbers]] is $0$. | {{AimForCont}} to the contrary.
Let $K$ be a [[Definition:Subfield|subfield]] of $\C$ such that $\Char K = n$ where $n \in \N, n > 0$.
Then from [[Characteristic times Ring Element is Ring Zero]]:
:$\forall a \in K: n \cdot a = 0$
But as $K$ is a [[Definition:Subfield|subfield]] of $\C$ it follows that $K \subseteq ... | Characteristic of Subfield of Complex Numbers is Zero | https://proofwiki.org/wiki/Characteristic_of_Subfield_of_Complex_Numbers_is_Zero | https://proofwiki.org/wiki/Characteristic_of_Subfield_of_Complex_Numbers_is_Zero | [
"Subfields",
"Characteristics of Fields",
"Complex Numbers"
] | [
"Definition:Characteristic of Field",
"Definition:Subfield",
"Definition:Field of Complex Numbers"
] | [
"Definition:Subfield",
"Characteristic times Ring Element is Ring Zero",
"Definition:Subfield",
"Definition:Characteristic of Field",
"Definition:Infinite Set",
"Definition:Infinite Set",
"Field has Algebraic Closure",
"Algebraically Closed Field is Infinite",
"Field has Characteristic of Zero iff e... |
proofwiki-3283 | Characteristic times Ring Element is Ring Zero | Let $\struct {R, +, \circ}$ be a ring with unity.
Let the zero of $R$ be $0_R$ and the unity of $R$ be $1_R$.
Let the characteristic of $R$ be $n$.
Then:
:$\forall a \in R: n \cdot a = 0_R$ | If $a = 0_R$ then $n \cdot a = 0_R$ is immediate.
So let $a \in R: a \ne 0_R$.
Then:
{{begin-eqn}}
{{eqn | l = n \cdot a
| r = n \cdot \paren {1_R \circ a}
| c = {{Defof|Unity of Ring}}
}}
{{eqn | r = \paren {n \cdot 1_R} \circ a
| c = Integral Multiple of Ring Element
}}
{{eqn | r = 0_R \circ a
... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]].
Let the [[Definition:Ring Zero|zero]] of $R$ be $0_R$ and the [[Definition:Unity of Ring|unity]] of $R$ be $1_R$.
Let the [[Definition:Characteristic of Ring|characteristic]] of $R$ be $n$.
Then:
:$\forall a \in R: n \cdot a = 0_R$ | If $a = 0_R$ then $n \cdot a = 0_R$ is immediate.
So let $a \in R: a \ne 0_R$.
Then:
{{begin-eqn}}
{{eqn | l = n \cdot a
| r = n \cdot \paren {1_R \circ a}
| c = {{Defof|Unity of Ring}}
}}
{{eqn | r = \paren {n \cdot 1_R} \circ a
| c = [[Integral Multiple of Ring Element]]
}}
{{eqn | r = 0_R \circ ... | Characteristic times Ring Element is Ring Zero | https://proofwiki.org/wiki/Characteristic_times_Ring_Element_is_Ring_Zero | https://proofwiki.org/wiki/Characteristic_times_Ring_Element_is_Ring_Zero | [
"Characteristics of Rings"
] | [
"Definition:Ring with Unity",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Characteristic of Ring"
] | [
"Integral Multiple of Ring Element"
] |
proofwiki-3284 | Characteristic of Finite Ring is Non-Zero | Let $\struct {R, +, \circ}$ be a finite ring with unity.
Then the characteristic of $R$ is not zero. | We have that $\struct {R, +, \circ}$ is finite, so its additive group $\struct {R, +}$ is likewise finite.
The result follows by Element of Finite Group is of Finite Order and the definition of characteristic.
{{qed}} | Let $\struct {R, +, \circ}$ be a [[Definition:Finite Ring|finite]] [[Definition:Ring with Unity|ring with unity]].
Then the [[Definition:Characteristic of Ring|characteristic]] of $R$ is not [[Definition:Zero (Number)|zero]]. | We have that $\struct {R, +, \circ}$ is [[Definition:Finite Ring|finite]], so its [[Definition:Additive Group of Ring|additive group]] $\struct {R, +}$ is likewise [[Definition:Finite Group|finite]].
The result follows by [[Element of Finite Group is of Finite Order]] and the definition of [[Definition:Characteristic ... | Characteristic of Finite Ring is Non-Zero | https://proofwiki.org/wiki/Characteristic_of_Finite_Ring_is_Non-Zero | https://proofwiki.org/wiki/Characteristic_of_Finite_Ring_is_Non-Zero | [
"Finite Rings"
] | [
"Definition:Finite Ring",
"Definition:Ring with Unity",
"Definition:Characteristic of Ring",
"Definition:Zero (Number)"
] | [
"Definition:Finite Ring",
"Definition:Additive Group of Ring",
"Definition:Finite Group",
"Element of Finite Group is of Finite Order",
"Definition:Characteristic of Ring"
] |
proofwiki-3285 | Field of Rational Functions is Field | Let $K$ be a field.
Let $K \sqbrk x$ be the integral domain of polynomial forms on $K$.
Let $\map K x$ be the field of rational functions on $K$.
Then $\map K x$ forms a field.
If the characteristic of $K$ is $p$, then the characteristic of $\map K x$ is non-zero. | {{ProofWanted|Reverted this to the version 9th May 2011 as it's mutated into something that makes less sense.}} | Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $K \sqbrk x$ be the [[Ring of Polynomial Forms is Integral Domain|integral domain of polynomial forms]] on $K$.
Let $\map K x$ be the [[Definition:Field of Rational Functions|field of rational functions]] on $K$.
Then $\map K x$ forms a [[Definition:Fi... | {{ProofWanted|Reverted this to the version 9th May 2011 as it's mutated into something that makes less sense.}} | Field of Rational Functions is Field | https://proofwiki.org/wiki/Field_of_Rational_Functions_is_Field | https://proofwiki.org/wiki/Field_of_Rational_Functions_is_Field | [
"Field Theory",
"Polynomial Theory"
] | [
"Definition:Field (Abstract Algebra)",
"Ring of Polynomial Forms is Integral Domain",
"Definition:Field of Rational Functions",
"Definition:Field (Abstract Algebra)",
"Definition:Characteristic of Field",
"Definition:Characteristic of Field",
"Definition:Zero (Number)"
] | [] |
proofwiki-3286 | Hadamard Factorization Theorem | Let $f: \C \to \C$ be an entire function of finite order $\omega$.
Let $0$ be a zero of $f$ of multiplicity $m \ge 0$.
Let $\sequence {a_n}$ be the sequence of non-zero zeroes of $f$, repeated according to multiplicity.
Then:
:$f$ has finite rank $p \le \omega$
and:
:there exists a polynomial $g$ of degree at most $\om... | By Convergence Exponent is Less Than Order, $f$ has finite exponent of convergence $\tau \le \omega$.
By Relation Between Rank and Exponent of Convergence, $f$ has finite rank $p \leq \omega$.
{{ProofWanted}} | Let $f: \C \to \C$ be an [[Definition:Entire Function|entire function]] of [[Definition:Order of Entire Function|finite order]] $\omega$.
Let $0$ be a [[Definition:Zero of Function|zero]] of $f$ of [[Definition:Multiplicity (Analytic Function)|multiplicity]] $m \ge 0$.
Let $\sequence {a_n}$ be the [[Definition:Sequen... | By [[Convergence Exponent is Less Than Order]], $f$ has finite [[Definition:Exponent of Convergence|exponent of convergence]] $\tau \le \omega$.
By [[Relation Between Rank and Exponent of Convergence]], $f$ has [[Definition:Rank of Entire Function|finite rank]] $p \leq \omega$.
{{ProofWanted}} | Hadamard Factorization Theorem | https://proofwiki.org/wiki/Hadamard_Factorization_Theorem | https://proofwiki.org/wiki/Hadamard_Factorization_Theorem | [
"Entire Functions",
"Infinite Products",
"Complex Analysis"
] | [
"Definition:Entire Function",
"Definition:Order of Entire Function",
"Definition:Root of Mapping",
"Definition:Multiplicity (Complex Analysis)",
"Definition:Sequence",
"Definition:Zero (Number)",
"Definition:Root of Mapping",
"Definition:Multiplicity (Complex Analysis)",
"Definition:Rank of Entire F... | [
"Exponent of Convergence is Less Than Order",
"Definition:Exponent of Convergence",
"Relation Between Rank and Exponent of Convergence",
"Definition:Rank of Entire Function"
] |
proofwiki-3287 | Approximation to Stirling's Formula for Gamma Function | Let $\Gamma: \R \to \R$ be the gamma function restricted to the real numbers.
Let $x \in \R_{>0}$.
Then:
:$\map \Gamma x \sim \sqrt {\dfrac {2 \pi} x} \paren {\dfrac x e}^x$
where $\sim$ denotes asymptotic equivalence. | {{Explain|How do these inequalities follow from Bohr-Mollerup Theorem, or perhaps Log of Gamma Function is Convex on Positive Reals?}}
From Bohr-Mollerup Theorem:
{{begin-eqn}}
{{eqn | n = 1
| l = \paren {y + n}^y n!
| r = \paren {y + n}^y \map \Gamma {n + 1}
| c = Gamma Function Extends Factorial
}}
... | Let $\Gamma: \R \to \R$ be the [[Definition:Gamma Function|gamma function]] [[Definition:Restriction of Mapping|restricted]] to the [[Definition:Real Numbers|real numbers]].
Let $x \in \R_{>0}$.
Then:
:$\map \Gamma x \sim \sqrt {\dfrac {2 \pi} x} \paren {\dfrac x e}^x$
where $\sim$ denotes [[Definition:Asymptotic Eq... | {{Explain|How do these inequalities follow from [[Bohr-Mollerup Theorem]], or perhaps [[Log of Gamma Function is Convex on Positive Reals]]?}}
From [[Bohr-Mollerup Theorem]]:
{{begin-eqn}}
{{eqn | n = 1
| l = \paren {y + n}^y n!
| r = \paren {y + n}^y \map \Gamma {n + 1}
| c = [[Gamma Function Exten... | Approximation to Stirling's Formula for Gamma Function | https://proofwiki.org/wiki/Approximation_to_Stirling's_Formula_for_Gamma_Function | https://proofwiki.org/wiki/Approximation_to_Stirling's_Formula_for_Gamma_Function | [
"Gamma Function",
"Stirling's Formula"
] | [
"Definition:Gamma Function",
"Definition:Restriction/Mapping",
"Definition:Real Number",
"Definition:Asymptotic Equality"
] | [
"Bohr-Mollerup Theorem",
"Log of Gamma Function is Convex on Positive Reals",
"Bohr-Mollerup Theorem",
"Gamma Function Extends Factorial",
"Definition:Natural Numbers",
"Stirling's Formula",
"Definition:Fraction/Numerator",
"Definition:Fraction/Denominator",
"Definition:Fraction/Numerator",
"Defin... |
proofwiki-3288 | Poisson Summation Formula | Let $f: \R \to \C$ be a Schwarz function.
Let $\hat f$ be its Fourier transform.
Then:
:$\ds \sum_{n \mathop \in \Z} \map f n = \sum_{m \mathop \in \Z} \map {\hat f} m$ | Let:
:$\ds \map F x = \sum_{n \mathop \in \Z} \map f {x + n}$
Then $\map F x$ is $1$-periodic (because of absolute convergence), and has Fourier coefficients:
{{begin-eqn}}
{{eqn | l = \hat F_k
| r = \int_0^1 \sum_{n \mathop \in \Z} \map f {x + n} e^{-2\pi i k x} \rd x
| c =
}}
{{eqn | r = \sum_{n \mathop ... | Let $f: \R \to \C$ be a [[Definition:Schwarz Function|Schwarz function]].
Let $\hat f$ be its [[Definition:Fourier Transform|Fourier transform]].
Then:
:$\ds \sum_{n \mathop \in \Z} \map f n = \sum_{m \mathop \in \Z} \map {\hat f} m$ | Let:
:$\ds \map F x = \sum_{n \mathop \in \Z} \map f {x + n}$
Then $\map F x$ is $1$-[[Definition:Periodic Function|periodic]] (because of absolute convergence), and has [[Definition:Fourier Coefficient|Fourier coefficients]]:
{{begin-eqn}}
{{eqn | l = \hat F_k
| r = \int_0^1 \sum_{n \mathop \in \Z} \map f {x... | Poisson Summation Formula | https://proofwiki.org/wiki/Poisson_Summation_Formula | https://proofwiki.org/wiki/Poisson_Summation_Formula | [
"Harmonic Analysis",
"Fourier Analysis"
] | [
"Definition:Schwarz Function",
"Definition:Fourier Transform"
] | [
"Definition:Periodic Function",
"Definition:Fourier Series/Fourier Coefficient",
"Definition:Schwarz Function",
"Definition:Uniform Convergence",
"Definition:Fourier Transform",
"Definition:Fourier Series"
] |
proofwiki-3289 | Harmonic Properties of Schwarz Functions | Let $f, g : \R \to \C$ be Schwarz functions.
Let $\hat f$, $\hat g$ be the Fourier transforms of $f$ and $g$ respectively.
Then:
:$(1): \quad \hat f$, $\hat g$ are Schwarz functions.
:$(2): \quad \map {\widehat {\paren {\hat f} } } x = \map f {-x}$ for all $x \in \R$.
:$(3): \quad$ If $f * g$ is the convolution of $f$ ... | {{ProofWanted}}
Category:Harmonic Analysis
rh6nke1c84ur75fg19blvabn2rzgntp | Let $f, g : \R \to \C$ be [[Definition:Schwarz Function|Schwarz functions]].
Let $\hat f$, $\hat g$ be the [[Definition:Fourier Transform|Fourier transforms]] of $f$ and $g$ respectively.
Then:
:$(1): \quad \hat f$, $\hat g$ are [[Definition:Schwarz Function|Schwarz functions]].
:$(2): \quad \map {\widehat {\paren ... | {{ProofWanted}}
[[Category:Harmonic Analysis]]
rh6nke1c84ur75fg19blvabn2rzgntp | Harmonic Properties of Schwarz Functions | https://proofwiki.org/wiki/Harmonic_Properties_of_Schwarz_Functions | https://proofwiki.org/wiki/Harmonic_Properties_of_Schwarz_Functions | [
"Harmonic Analysis"
] | [
"Definition:Schwarz Function",
"Definition:Fourier Transform",
"Definition:Schwarz Function",
"Definition:Convolution"
] | [
"Category:Harmonic Analysis"
] |
proofwiki-3290 | Euler's Reflection Formula | :$\forall z \notin \Z: \map \Gamma z \map \Gamma {1 - z} = \dfrac \pi {\map \sin {\pi z} }$ | We have the Weierstrass products:
:$\ds \map \sin {\pi z} = \pi z \prod_{n \mathop \ne 0} \paren {1 - \frac z n} \map \exp {\frac z n}$
From the Weierstrass form of the Gamma function:
:$\ds \frac 1 {\map \Gamma z} = z e^{\gamma z} \prod_{n \mathop = 1}^\infty \paren {1 + \frac z n} \map \exp {-\frac z n}$
from which:
... | :$\forall z \notin \Z: \map \Gamma z \map \Gamma {1 - z} = \dfrac \pi {\map \sin {\pi z} }$ | We have the [[Weierstrass Factorization Theorem|Weierstrass products]]:
:$\ds \map \sin {\pi z} = \pi z \prod_{n \mathop \ne 0} \paren {1 - \frac z n} \map \exp {\frac z n}$
From the [[Definition:Weierstrass Form of Gamma Function|Weierstrass form of the Gamma function]]:
:$\ds \frac 1 {\map \Gamma z} = z e^{\gamma... | Euler's Reflection Formula | https://proofwiki.org/wiki/Euler's_Reflection_Formula | https://proofwiki.org/wiki/Euler's_Reflection_Formula | [
"Euler's Reflection Formula",
"Gamma Function",
"Reflection Formulas"
] | [] | [
"Weierstrass Factorization Theorem",
"Definition:Gamma Function/Weierstrass Form",
"Euler Formula for Sine Function",
"Gamma Difference Equation"
] |
proofwiki-3291 | Legendre's Duplication Formula | :$\forall z \notin \set {-\dfrac n 2: n \in \N}: \map \Gamma z \map \Gamma {z + \dfrac 1 2} = 2^{1 - 2 z} \sqrt \pi \, \map \Gamma {2 z}$
where $\N$ denotes the natural numbers. | 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-... | :$\forall z \notin \set {-\dfrac n 2: n \in \N}: \map \Gamma z \map \Gamma {z + \dfrac 1 2} = 2^{1 - 2 z} \sqrt \pi \, \map \Gamma {2 z}$
where $\N$ denotes the [[Definition:Natural Number|natural numbers]]. | From the definition of the [[Definition:Beta Function|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 Defini... | Legendre's Duplication Formula/Proof 1 | https://proofwiki.org/wiki/Legendre's_Duplication_Formula | https://proofwiki.org/wiki/Legendre's_Duplication_Formula/Proof_1 | [
"Legendre's Duplication Formula",
"Gamma Function",
"Functional Equations"
] | [
"Definition:Natural Numbers"
] | [
"Definition:Beta Function",
"Equivalence of Definitions of Beta Function",
"Integration by Substitution",
"Definite Integral of Even Function",
"Definition:Beta Function",
"Gamma Function of One Half"
] |
proofwiki-3292 | Legendre's Duplication Formula | :$\forall z \notin \set {-\dfrac n 2: n \in \N}: \map \Gamma z \map \Gamma {z + \dfrac 1 2} = 2^{1 - 2 z} \sqrt \pi \, \map \Gamma {2 z}$
where $\N$ denotes the natural numbers. | From Gauss Multiplication Formula:
:$\ds \prod_{k \mathop = 0}^{n - 1} \map \Gamma {z + \frac k n} = \paren {2 \pi}^{\paren {n - 1} / 2} n^{1/2 - n z} \map \Gamma {n z}$
Substituting $n = 2$ yields:
{{begin-eqn}}
{{eqn | l = \map \Gamma z \map \Gamma {z + \frac 1 2}
| r = \paren {2 \pi}^{1 / 2} 2^{1/2 - 2 z} \map... | :$\forall z \notin \set {-\dfrac n 2: n \in \N}: \map \Gamma z \map \Gamma {z + \dfrac 1 2} = 2^{1 - 2 z} \sqrt \pi \, \map \Gamma {2 z}$
where $\N$ denotes the [[Definition:Natural Number|natural numbers]]. | From [[Gauss Multiplication Formula]]:
:$\ds \prod_{k \mathop = 0}^{n - 1} \map \Gamma {z + \frac k n} = \paren {2 \pi}^{\paren {n - 1} / 2} n^{1/2 - n z} \map \Gamma {n z}$
Substituting $n = 2$ yields:
{{begin-eqn}}
{{eqn | l = \map \Gamma z \map \Gamma {z + \frac 1 2}
| r = \paren {2 \pi}^{1 / 2} 2^{1/2 - 2 ... | Legendre's Duplication Formula/Proof 2 | https://proofwiki.org/wiki/Legendre's_Duplication_Formula | https://proofwiki.org/wiki/Legendre's_Duplication_Formula/Proof_2 | [
"Legendre's Duplication Formula",
"Gamma Function",
"Functional Equations"
] | [
"Definition:Natural Numbers"
] | [
"Gauss Multiplication Formula"
] |
proofwiki-3293 | Approximation to Reciprocal times Derivative of Gamma Function | For all $z \in \C$ such that $\cmod {\map \arg z} < \pi - \epsilon, \cmod z > 1$:
:$\dfrac {\map {\Gamma'} z} {\map \Gamma z} = \ln z + \map {\OO_\epsilon} {z^{-1} }$
where:
:$\map \OO {z^{-1} }$ denotes big-$\OO$ notation
:the implied constant depends on $\epsilon$. | From Logarithmic Approximation of Error Term of Stirling's Formula for Gamma Function:
:$\ln \map \Gamma z = \paren {z - \dfrac 1 2} \ln z - z + \dfrac {\ln 2 \pi} 2 + \map \OO {z^{-1} }$
Taking the derivative {{WRT|Differentiation}} $z$:
:$(1): \quad \dfrac {\map {\Gamma'} z} {\map \Gamma z} = \ln z - \dfrac 1 {2 z} +... | For all $z \in \C$ such that $\cmod {\map \arg z} < \pi - \epsilon, \cmod z > 1$:
:$\dfrac {\map {\Gamma'} z} {\map \Gamma z} = \ln z + \map {\OO_\epsilon} {z^{-1} }$
where:
:$\map \OO {z^{-1} }$ denotes [[Definition:Big-O Notation|big-$\OO$ notation]]
:the [[Definition:Implied Constant|implied constant]] depends on ... | From [[Logarithmic Approximation of Error Term of Stirling's Formula for Gamma Function]]:
:$\ln \map \Gamma z = \paren {z - \dfrac 1 2} \ln z - z + \dfrac {\ln 2 \pi} 2 + \map \OO {z^{-1} }$
Taking the [[Definition:Derivative|derivative]] {{WRT|Differentiation}} $z$:
:$(1): \quad \dfrac {\map {\Gamma'} z} {\map \Ga... | Approximation to Reciprocal times Derivative of Gamma Function | https://proofwiki.org/wiki/Approximation_to_Reciprocal_times_Derivative_of_Gamma_Function | https://proofwiki.org/wiki/Approximation_to_Reciprocal_times_Derivative_of_Gamma_Function | [
"Gamma Function",
"Reciprocals"
] | [
"Definition:Big-O Notation",
"Definition:Big-O Notation/Implied Constant"
] | [
"Logarithmic Approximation of Error Term of Stirling's Formula for Gamma Function",
"Definition:Derivative",
"Category:Gamma Function",
"Category:Reciprocals"
] |
proofwiki-3294 | Complex Conjugate of Gamma Function | :$\forall z \in \C \setminus \set {0, -1, -2, \ldots}: \map \Gamma {\overline z} = \overline {\map \Gamma z}$ | 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
bczi75z79zhoprjwzygznzihxjbx2de | :$\forall z \in \C \setminus \set {0, -1, -2, \ldots}: \map \Gamma {\overline z} = \overline {\map \Gamma z}$ | This is immediate from, say, the [[Definition:Euler Form of Gamma Function|Euler form]] of $\Gamma$ and the fact that [[Definition:Complex Conjugate|complex conjugation]] preserves products and sums.
[[Category:Gamma Function]]
[[Category:Complex Conjugates]]
bczi75z79zhoprjwzygznzihxjbx2de | Complex Conjugate of Gamma Function | https://proofwiki.org/wiki/Complex_Conjugate_of_Gamma_Function | https://proofwiki.org/wiki/Complex_Conjugate_of_Gamma_Function | [
"Gamma Function",
"Complex Conjugates"
] | [] | [
"Definition:Gamma Function/Euler Form",
"Definition:Complex Conjugate",
"Category:Gamma Function",
"Category:Complex Conjugates"
] |
proofwiki-3295 | Linear Combination of Integers is Ideal | Let $a, b$ be any integers.
Let $\Bbb S = \set {a x + b y: x, y \in \Z}$.
Then the algebraic structure:
:$\struct {\Bbb S, +, \times}$
is an ideal of $\Z$. | 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_... | Let $a, b$ be any [[Definition:Integer|integers]].
Let $\Bbb S = \set {a x + b y: x, y \in \Z}$.
Then the [[Definition:Algebraic Structure|algebraic structure]]:
:$\struct {\Bbb S, +, \times}$
is an [[Definition:Ideal of Ring|ideal]] of $\Z$. | From the definition of [[Definition:Integer Combination|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... | Linear Combination of Integers is Ideal | https://proofwiki.org/wiki/Linear_Combination_of_Integers_is_Ideal | https://proofwiki.org/wiki/Linear_Combination_of_Integers_is_Ideal | [
"Ideal Theory",
"Integers"
] | [
"Definition:Integer",
"Definition:Algebraic Structure",
"Definition:Ideal of Ring"
] | [
"Definition:Integer Combination",
"Test for Ideal"
] |
proofwiki-3296 | Intersection of Integer Ideals is Lowest Common Multiple | Let $\ideal m$ and $\ideal n$ be ideals of the integers $\Z$.
Let $\ideal k$ be the intersection of $\ideal m$ and $\ideal n$.
Then $k = \lcm \set {m, n}$. | 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:
:$\idea... | Let $\ideal m$ and $\ideal n$ be [[Definition:Ideal of Ring|ideals]] of the [[Definition:Integer|integers]] $\Z$.
Let $\ideal k$ be the [[Definition:Set Intersection|intersection]] of $\ideal m$ and $\ideal n$.
Then $k = \lcm \set {m, n}$. | By [[Intersection of Ring Ideals is Ideal]] we have that $\ideal k = \ideal m \cap \ideal n$ is an [[Definition:Ideal of Ring|ideal]] of $\Z$.
By [[Ring of Integers is Principal Ideal Domain]] we have that $\ideal m$, $\ideal n$ and $\ideal k$ are all necessarily [[Definition:Principal Ideal of Ring|principal ideals]]... | Intersection of Integer Ideals is Lowest Common Multiple | https://proofwiki.org/wiki/Intersection_of_Integer_Ideals_is_Lowest_Common_Multiple | https://proofwiki.org/wiki/Intersection_of_Integer_Ideals_is_Lowest_Common_Multiple | [
"Ideal Theory",
"Integers",
"Lowest Common Multiple"
] | [
"Definition:Ideal of Ring",
"Definition:Integer",
"Definition:Set Intersection"
] | [
"Intersection of Ring Ideals is Ideal",
"Definition:Ideal of Ring",
"Ring of Integers is Principal Ideal Domain",
"Definition:Principal Ideal of Ring",
"Subrings of Integers are Sets of Integer Multiples",
"LCM iff Divides All Common Multiples"
] |
proofwiki-3297 | LCM iff Divides All Common Multiples | Let $a, b \in \Z$ such that $a b \ne 0$.
Let $m \in \Z$ such that $m > 0$.
Then $m = \lcm \set {a, b}$ {{iff}}:
: $(1): \quad a \divides m \land b \divides m$
: $(2): \quad a \divides n \land b \divides n \implies m \divides n$
That is, in the set of positive integers, $m$ is the LCM of $a$ and $b$ {{iff}} $m$ is a com... | === 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}} | Let $a, b \in \Z$ such that $a b \ne 0$.
Let $m \in \Z$ such that $m > 0$.
Then $m = \lcm \set {a, b}$ {{iff}}:
: $(1): \quad a \divides m \land b \divides m$
: $(2): \quad a \divides n \land b \divides n \implies m \divides n$
That is, in the [[Definition:Set|set]] of [[Definition:Positive Integer|positive integ... | === 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}} | LCM iff Divides All Common Multiples | https://proofwiki.org/wiki/LCM_iff_Divides_All_Common_Multiples | https://proofwiki.org/wiki/LCM_iff_Divides_All_Common_Multiples | [
"Lowest Common Multiple"
] | [
"Definition:Set",
"Definition:Positive/Integer",
"Definition:Lowest Common Multiple/Integers",
"Definition:Common Multiple",
"Definition:Divisor (Algebra)/Integer",
"Definition:Common Multiple"
] | [
"LCM Divides Common Multiple"
] |
proofwiki-3298 | LCM Divides Common Multiple | Let $a, b \in \Z$ such that $a b \ne 0$.
Let $n$ be any common multiple of $a$ and $b$.
That is, let $n \in \Z: a \divides n, b \divides n$.
Then:
:$\lcm \set {a, b} \divides n$
where $\lcm \set {a, b}$ is the lowest common multiple of $a$ and $b$.
{{:Euclid:Proposition/VII/35}} | 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 \divide... | Let $a, b \in \Z$ such that $a b \ne 0$.
Let $n$ be any [[Definition:Common Multiple|common multiple]] of $a$ and $b$.
That is, let $n \in \Z: a \divides n, b \divides n$.
Then:
:$\lcm \set {a, b} \divides n$
where $\lcm \set {a, b}$ is the [[Definition:Lowest Common Multiple of Integers|lowest common multiple]] of... | Let $m = \lcm \set {a, b}$.
Then $a \divides m$ and $b \divides m$ [[Definition:Lowest Common Multiple of Integers|by definition]].
Suppose $n$ is some other [[Definition:Common Multiple|common multiple]] of $a$ and $b$ such that $m \nmid n$ ($m$ does not divide $n$).
Then from the [[Division Theorem]]:
:$n = k m + ... | LCM Divides Common Multiple | https://proofwiki.org/wiki/LCM_Divides_Common_Multiple | https://proofwiki.org/wiki/LCM_Divides_Common_Multiple | [
"Lowest Common Multiple"
] | [
"Definition:Common Multiple",
"Definition:Lowest Common Multiple/Integers"
] | [
"Definition:Lowest Common Multiple/Integers",
"Definition:Common Multiple",
"Division Theorem",
"Definition:Common Multiple",
"Proof by Contradiction",
"Definition:Lowest Common Multiple/Integers"
] |
proofwiki-3299 | Sum of Integer Ideals is Greatest Common Divisor | Let $\ideal m$ and $\ideal n$ be ideals of the integers $\Z$.
Let $\ideal d = \ideal m + \ideal n$.
Then $d = \gcd \set {m, n}$. | 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, \idea... | Let $\ideal m$ and $\ideal n$ be [[Definition:Ideal of Ring|ideals]] of the [[Definition:Integer|integers]] $\Z$.
Let $\ideal d = \ideal m + \ideal n$.
Then $d = \gcd \set {m, n}$. | By [[Sum of Ideals is Ideal]] we have that $\ideal d = \ideal m + \ideal n$ is an [[Definition:Ideal of Ring|ideal]] of $\Z$.
By [[Ring of Integers is Principal Ideal Domain]] we have that $\ideal m$, $\ideal n$ and $\ideal d$ are all necessarily [[Definition:Principal Ideal of Ring|principal ideals]].
By [[Subrings ... | Sum of Integer Ideals is Greatest Common Divisor | https://proofwiki.org/wiki/Sum_of_Integer_Ideals_is_Greatest_Common_Divisor | https://proofwiki.org/wiki/Sum_of_Integer_Ideals_is_Greatest_Common_Divisor | [
"Ideal Theory",
"Integers",
"Greatest Common Divisor"
] | [
"Definition:Ideal of Ring",
"Definition:Integer"
] | [
"Sum of Ideals is Ideal",
"Definition:Ideal of Ring",
"Ring of Integers is Principal Ideal Domain",
"Definition:Principal Ideal of Ring",
"Subrings of Integers are Sets of Integer Multiples",
"Definition:Integer Combination",
"Bézout's Identity"
] |
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