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-1300 | Linearly Independent Subset also Independent in Generated Subspace | Let $G$ be a finitely generated $K$-vector space.
Let $S$ be a linearly independent subset of $G$.
Let $M$ be the subspace of $G$ generated by $S$.
If $M \ne G$, then $\forall b \in G: b \notin M$, the set $S \cup \set b$ is linearly independent. | Suppose that:
:$\ds \sum_{k \mathop = 1}^n \lambda_k x_k + \lambda b = 0$
where $\sequence {x_n}$ is a sequence of distinct vectors of $S$.
If $\lambda \ne 0$, then $\ds b = -\lambda^{-1} \paren {\sum_{k \mathop = 1}^n \lambda_k x_k} \in M$ which contradicts the definition of $b$.
Hence $\lambda = 0$, and so:
:$\ds \su... | Let $G$ be a [[Definition:Finitely Generated Module|finitely generated]] [[Definition:Vector Space|$K$-vector space]].
Let $S$ be a [[Definition:Linearly Independent Set|linearly independent subset]] of $G$.
Let $M$ be the [[Definition:Vector Subspace|subspace]] of $G$ [[Definition:Generator of Module|generated]] by ... | Suppose that:
:$\ds \sum_{k \mathop = 1}^n \lambda_k x_k + \lambda b = 0$
where $\sequence {x_n}$ is a [[Definition:Sequence of Distinct Terms|sequence of distinct vectors]] of $S$.
If $\lambda \ne 0$, then $\ds b = -\lambda^{-1} \paren {\sum_{k \mathop = 1}^n \lambda_k x_k} \in M$ which contradicts the definition of ... | Linearly Independent Subset also Independent in Generated Subspace | https://proofwiki.org/wiki/Linearly_Independent_Subset_also_Independent_in_Generated_Subspace | https://proofwiki.org/wiki/Linearly_Independent_Subset_also_Independent_in_Generated_Subspace | [
"Linear Independence"
] | [
"Definition:Finitely Generated Module",
"Definition:Vector Space",
"Definition:Linearly Independent/Set",
"Definition:Vector Subspace",
"Definition:Generator of Module",
"Definition:Linearly Independent/Set"
] | [
"Definition:Sequence of Distinct Terms",
"Definition:Linearly Independent/Set"
] |
proofwiki-1301 | Sufficient Conditions for Basis of Finite Dimensional Vector Space | Let $K$ be a division ring.
Let $n \ge 0$ be a natural number.
Let $E$ be an $n$-dimensional vector space over $K$.
Let $B \subseteq E$ be a subset such that $\card B = n$.
{{TFAE}}
:$(1): \quad$ $B$ is a basis of $E$.
:$(2): \quad$ $B$ is linearly independent.
:$(3): \quad$ $B$ is a generator for $E$. | === 1 implies 2 and 3 ===
Let $B$ be a basis of $E$.
Then conditions $(2)$ and $(3)$ follow directly by the definition of basis.
{{qed|lemma}} | Let $K$ be a [[Definition:Division Ring|division ring]].
Let $n \ge 0$ be a [[Definition:Natural Number|natural number]].
Let $E$ be an [[Definition:Dimension of Vector Space|$n$-dimensional]] [[Definition:Vector Space|vector space]] over $K$.
Let $B \subseteq E$ be a [[Definition:Subset|subset]] such that $\card B ... | === 1 implies 2 and 3 ===
Let $B$ be a [[Definition:Basis of Vector Space|basis]] of $E$.
Then conditions $(2)$ and $(3)$ follow directly by the definition of [[Definition:Basis of Vector Space|basis]].
{{qed|lemma}} | Sufficient Conditions for Basis of Finite Dimensional Vector Space | https://proofwiki.org/wiki/Sufficient_Conditions_for_Basis_of_Finite_Dimensional_Vector_Space | https://proofwiki.org/wiki/Sufficient_Conditions_for_Basis_of_Finite_Dimensional_Vector_Space | [
"Sufficient Conditions for Basis of Finite Dimensional Vector Space",
"Generators of Vector Spaces",
"Bases of Vector Spaces",
"Finite Dimensional Vector Spaces"
] | [
"Definition:Division Ring",
"Definition:Natural Numbers",
"Definition:Dimension of Vector Space",
"Definition:Vector Space",
"Definition:Subset",
"Definition:Basis of Vector Space",
"Definition:Linearly Independent/Set",
"Definition:Generator of Module"
] | [
"Definition:Basis of Vector Space",
"Definition:Basis of Vector Space",
"Definition:Basis of Vector Space"
] |
proofwiki-1302 | Dimension of Proper Subspace is Less Than its Superspace | Let $G$ be a vector space whose dimension is $n$.
Let $H$ be a subspace of $G$.
Then $H$ is finite dimensional, and $\map \dim H \le \map \dim G$.
{{refactor|level = medium|This page contains two results. A separate page for $\map \dim H \le \map \dim G$ is needed.<br/>Go to it.}}
If $H$ is a proper subspace of $G$, th... | Let $H$ be a subspace of $G$.
Every linearly independent subset of the vector space $H$ is a linearly independent subset of the vector space $G$.
Therefore, it has no more than $n$ elements by Size of Linearly Independent Subset is at Most Size of Finite Generator.
So the set of all natural numbers $k$ such that $H$ ha... | Let $G$ be a [[Definition:Vector Space|vector space]] whose [[Definition:Dimension of Vector Space|dimension]] is $n$.
Let $H$ be a [[Definition:Vector Subspace|subspace]] of $G$.
Then $H$ is [[Definition:Finite Dimensional Vector Space|finite dimensional]], and $\map \dim H \le \map \dim G$.
{{refactor|level = med... | Let $H$ be a [[Definition:Vector Subspace|subspace]] of $G$.
Every [[Definition:Linearly Independent Set|linearly independent subset]] of the [[Definition:Vector Space|vector space]] $H$ is a [[Definition:Linearly Independent Set|linearly independent subset]] of the [[Definition:Vector Space|vector space]] $G$.
There... | Dimension of Proper Subspace is Less Than its Superspace | https://proofwiki.org/wiki/Dimension_of_Proper_Subspace_is_Less_Than_its_Superspace | https://proofwiki.org/wiki/Dimension_of_Proper_Subspace_is_Less_Than_its_Superspace | [
"Dimension of Proper Subspace"
] | [
"Definition:Vector Space",
"Definition:Dimension of Vector Space",
"Definition:Vector Subspace",
"Definition:Dimension of Vector Space/Finite",
"Definition:Vector Subspace/Proper Subspace"
] | [
"Definition:Vector Subspace",
"Definition:Linearly Independent/Set",
"Definition:Vector Space",
"Definition:Linearly Independent/Set",
"Definition:Vector Space",
"Size of Linearly Independent Subset is at Most Size of Finite Generator",
"Definition:Natural Numbers",
"Definition:Linearly Independent/Se... |
proofwiki-1303 | Grassmann's Identity | Let $K$ be a division ring.
Let $\struct {G, +_G, \circ}_K$ be a $K$-vector space.
Let $M$ and $N$ be finite-dimensional subspaces of $G$.
Then the sum $M + N$ and intersection $M \cap N$ are finite-dimensional, and:
:$\map \dim {M + N} + \map \dim {M \cap N} = \map \dim M + \map \dim N$ | === Outline ===
Starting from a basis of $M \cap N$, we complete it to a basis of $M$ and one of $N$.
We then verify that the union of these basis is a basis of $M + N$.
=== Proof ===
First, suppose $M \subseteq N$ or $N \subseteq M$.
Then the assertion is clear.
Assume that $M \cap N$ is a proper subspace of both $M$ ... | Let $K$ be a [[Definition:Division Ring|division ring]].
Let $\struct {G, +_G, \circ}_K$ be a [[Definition:Vector Space|$K$-vector space]].
Let $M$ and $N$ be [[Definition:Finite Dimensional Vector Space|finite-dimensional]] [[Definition:Vector Subspace|subspaces]] of $G$.
Then the [[Definition:Setwise Addition|sum... | === Outline ===
Starting from a [[Definition:Basis of Vector Space|basis]] of $M \cap N$, we complete it to a [[Definition:Basis of Vector Space|basis]] of $M$ and one of $N$.
We then verify that the [[Definition:Set Union|union]] of these [[Definition:Basis of Vector Space|basis]] is a [[Definition:Basis of Vector S... | Grassmann's Identity/Proof 1 | https://proofwiki.org/wiki/Grassmann's_Identity | https://proofwiki.org/wiki/Grassmann's_Identity/Proof_1 | [
"Linear Algebra",
"Grassmann's Identity"
] | [
"Definition:Division Ring",
"Definition:Vector Space",
"Definition:Dimension of Vector Space/Finite",
"Definition:Vector Subspace",
"Definition:Subset Product",
"Definition:Intersection",
"Definition:Dimension of Vector Space/Finite"
] | [
"Definition:Basis of Vector Space",
"Definition:Basis of Vector Space",
"Definition:Set Union",
"Definition:Basis of Vector Space",
"Definition:Basis of Vector Space",
"Definition:Vector Subspace/Proper Subspace",
"Definition:Basis of Vector Space",
"Dimension of Proper Subspace is Less Than its Super... |
proofwiki-1304 | Grassmann's Identity | Let $K$ be a division ring.
Let $\struct {G, +_G, \circ}_K$ be a $K$-vector space.
Let $M$ and $N$ be finite-dimensional subspaces of $G$.
Then the sum $M + N$ and intersection $M \cap N$ are finite-dimensional, and:
:$\map \dim {M + N} + \map \dim {M \cap N} = \map \dim M + \map \dim N$ | By the second isomorphism theorem:
:$\dfrac {M + N} M \equiv \dfrac N {M \cap N}$
The result follows.
{{qed}} | Let $K$ be a [[Definition:Division Ring|division ring]].
Let $\struct {G, +_G, \circ}_K$ be a [[Definition:Vector Space|$K$-vector space]].
Let $M$ and $N$ be [[Definition:Finite Dimensional Vector Space|finite-dimensional]] [[Definition:Vector Subspace|subspaces]] of $G$.
Then the [[Definition:Setwise Addition|sum... | By the [[Second Isomorphism Theorem/Vector Spaces|second isomorphism theorem]]:
:$\dfrac {M + N} M \equiv \dfrac N {M \cap N}$
The result follows.
{{qed}} | Grassmann's Identity/Proof 2 | https://proofwiki.org/wiki/Grassmann's_Identity | https://proofwiki.org/wiki/Grassmann's_Identity/Proof_2 | [
"Linear Algebra",
"Grassmann's Identity"
] | [
"Definition:Division Ring",
"Definition:Vector Space",
"Definition:Dimension of Vector Space/Finite",
"Definition:Vector Subspace",
"Definition:Subset Product",
"Definition:Intersection",
"Definition:Dimension of Vector Space/Finite"
] | [
"Second Isomorphism Theorem/Vector Spaces"
] |
proofwiki-1305 | Rank Plus Nullity Theorem | Let $G$ be an $n$-dimensional vector space.
Let $H$ be a vector space.
Let $\phi: G \to H$ be a linear transformation.
Let $\map \rho \phi$ and $\map \nu \phi$ be the rank and nullity respectively of $\phi$.
Then the image of $\phi$ is finite-dimensional, and:
:$\map \rho \phi + \map \nu \phi = n$
By definition of rank... | If $\phi = 0$ then the assertion is clear.
Let $\phi$ be a non-zero linear transformation.
By Dimension of Proper Subspace is Less Than its Superspace and Generator of Vector Space Contains Basis, there is an ordered basis $\sequence {a_n}$ of $G$ such that:
:$\exists r \in \N_n: \set {a_k: r + 1 \le k \le n}$ is a bas... | Let $G$ be an [[Definition:Dimension of Vector Space|$n$-dimensional]] [[Definition:Vector Space|vector space]].
Let $H$ be a [[Definition:Vector Space|vector space]].
Let $\phi: G \to H$ be a [[Definition:Linear Transformation on Vector Space|linear transformation]].
Let $\map \rho \phi$ and $\map \nu \phi$ be the ... | If $\phi = 0$ then the assertion is clear.
Let $\phi$ be a non-zero [[Definition:Linear Transformation on Vector Space|linear transformation]].
By [[Dimension of Proper Subspace is Less Than its Superspace]] and [[Generator of Vector Space Contains Basis]], there is an [[Definition:Ordered Basis|ordered basis]] $\se... | Rank Plus Nullity Theorem | https://proofwiki.org/wiki/Rank_Plus_Nullity_Theorem | https://proofwiki.org/wiki/Rank_Plus_Nullity_Theorem | [
"Linear Algebra",
"Linear Transformations",
"Named Theorems"
] | [
"Definition:Dimension of Vector Space",
"Definition:Vector Space",
"Definition:Vector Space",
"Definition:Linear Transformation/Vector Space",
"Definition:Rank/Linear Transformation",
"Definition:Nullity/Linear Transformation",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Dimension of ... | [
"Definition:Linear Transformation/Vector Space",
"Dimension of Proper Subspace is Less Than its Superspace",
"Generator of Vector Space Contains Basis",
"Definition:Ordered Basis",
"Unique Linear Transformation Between Vector Spaces"
] |
proofwiki-1306 | Linear Transformation of Vector Space Monomorphism | Let $G$ and $H$ be a $K$-vector space.
Let $\phi: G \to H$ be a linear transformation.
Then $\phi$ is a monomorphism {{iff}} for every linearly independent sequence $\sequence {a_n}$ of vectors of $G$, $\sequence {\map \phi {a_n} }$ is a linearly independent sequence of vectors of $H$. | Suppose $\phi$ is a monomorphism.
Let $\sequence {a_n}$ be a linearly independent sequence.
Let:
:$\ds \sum_{k \mathop = 1}^n \lambda_k \map \phi {a_k} = 0$
Then:
:$\ds \map \phi {\sum_{k \mathop = 1}^n \lambda_k a_k} = 0$
So {{hypothesis}}:
:$\ds \sum_{k \mathop = 1}^n \lambda_k a_k = 0$
Hence:
:$\forall k \in \closed... | Let $G$ and $H$ be a [[Definition:Vector Space|$K$-vector space]].
Let $\phi: G \to H$ be a [[Definition:Linear Transformation on Vector Space|linear transformation]].
Then $\phi$ is a [[Definition:Vector Space Monomorphism|monomorphism]] {{iff}} for every [[Definition:Linearly Independent Sequence|linearly independ... | Suppose $\phi$ is a [[Definition:Vector Space Monomorphism|monomorphism]].
Let $\sequence {a_n}$ be a [[Definition:Linearly Independent Sequence|linearly independent sequence]].
Let:
:$\ds \sum_{k \mathop = 1}^n \lambda_k \map \phi {a_k} = 0$
Then:
:$\ds \map \phi {\sum_{k \mathop = 1}^n \lambda_k a_k} = 0$
So {{hy... | Linear Transformation of Vector Space Monomorphism | https://proofwiki.org/wiki/Linear_Transformation_of_Vector_Space_Monomorphism | https://proofwiki.org/wiki/Linear_Transformation_of_Vector_Space_Monomorphism | [
"Linear Transformations"
] | [
"Definition:Vector Space",
"Definition:Linear Transformation/Vector Space",
"Definition:Vector Space Monomorphism",
"Definition:Linearly Independent/Sequence",
"Definition:Vector/Linear Algebra",
"Definition:Linearly Independent/Sequence",
"Definition:Vector/Linear Algebra"
] | [
"Definition:Vector Space Monomorphism",
"Definition:Linearly Independent/Sequence",
"Definition:Linearly Independent/Sequence",
"Quotient Theorem for Group Epimorphisms",
"Definition:Isomorphism (Abstract Algebra)/R-Algebraic Structure Isomorphism/Vector Space Isomorphism",
"Definition:Vector Space Monomo... |
proofwiki-1307 | Linear Transformation of Vector Space Equivalent Statements | Let $G$ and $H$ be $n$-dimensional vector spaces.
Let $\phi: G \to H$ be a linear transformation.
{{TFAE}}
:$(1): \quad \phi$ is an isomorphism.
:$(2): \quad \phi$ is a monomorphism.
:$(3): \quad \phi$ is an epimorphism.
:$(4): \quad \map \phi B$ is a basis of $H$ for every basis $B$ of $G$.
:$(5): \quad \map \phi B$ i... | $(1)$ implies $(2)$ by definition.
$(2)$ implies $(4)$ by Linear Transformation of Vector Space Monomorphism and Results concerning Generators and Bases of Vector Spaces.
$(4)$ implies $(5)$ by basic logic.
Suppose $\map \phi B$ is a basis of $H$.
Then the image of $\phi$ is a subspace of $H$ generating $H$ and hence i... | Let $G$ and $H$ be [[Definition:Dimension (Linear Algebra)|$n$-dimensional]] [[Definition:Vector Space|vector spaces]].
Let $\phi: G \to H$ be a [[Definition:Linear Transformation|linear transformation]].
{{TFAE}}
:$(1): \quad \phi$ is an [[Definition:Vector Space Isomorphism|isomorphism]].
:$(2): \quad \phi$ is a ... | $(1)$ implies $(2)$ by definition.
$(2)$ implies $(4)$ by [[Linear Transformation of Vector Space Monomorphism]] and [[Results concerning Generators and Bases of Vector Spaces]].
$(4)$ implies $(5)$ by basic logic.
Suppose $\map \phi B$ is a [[Definition:Basis (Linear Algebra)|basis]] of $H$.
Then the [[Definition... | Linear Transformation of Vector Space Equivalent Statements | https://proofwiki.org/wiki/Linear_Transformation_of_Vector_Space_Equivalent_Statements | https://proofwiki.org/wiki/Linear_Transformation_of_Vector_Space_Equivalent_Statements | [
"Linear Algebra"
] | [
"Definition:Dimension (Linear Algebra)",
"Definition:Vector Space",
"Definition:Linear Transformation",
"Definition:Isomorphism (Abstract Algebra)/R-Algebraic Structure Isomorphism/Vector Space Isomorphism",
"Definition:Vector Space Monomorphism",
"Definition:R-Algebraic Structure Epimorphism",
"Definit... | [
"Linear Transformation of Vector Space Monomorphism",
"Results concerning Generators and Bases of Vector Spaces",
"Definition:Basis (Linear Algebra)",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Vector Subspace",
"Definition:Generator of Vector Space",
"Definition:Injection",
"Definit... |
proofwiki-1308 | Equivalent Statements for Vector Subspace Dimension One Less | Let $K$ be a field.
Let $M$ be a subspace of the $n$-dimensional vector space $K^n$.
The following statements are equivalent:
{{begin-itemize}}
{{item|(1):|$\map \dim M {{=}} n - 1$}}
{{item|(2):|$M$ is the kernel of a nonzero linear form}}
{{item|(3):|There exists a sequence $\sequence {\alpha_n} $ of scalars, not all... | Let $M^\circ$ be the annihilator of $M$.
Let $N = M^{\circ}$.
By Results Concerning Annihilator of Vector Subspace, $N$ is one-dimensional and $M = \map {J^{-1} } {N^\circ}$.
Let $\phi \in N: \phi \ne 0$.
Then $N$ is the set of all scalar multiples of $\phi$.
Because:
:$\map {J^{-1} } {N^\circ} = \set {x \in K^n: \fora... | Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $M$ be a [[Definition:Vector Subspace|subspace]] of the [[Definition:Dimension of Vector Space|$n$-dimensional]] [[Definition:Vector Space|vector space $K^n$]].
The following statements are equivalent:
{{begin-itemize}}
{{item|(1):|$\map \dim M {{=}} n ... | Let $M^\circ$ be the [[Definition:Annihilator on Algebraic Dual|annihilator]] of $M$.
Let $N = M^{\circ}$.
By [[Results Concerning Annihilator of Vector Subspace]], $N$ is [[Definition:Dimension of Vector Space|one-dimensional]] and $M = \map {J^{-1} } {N^\circ}$.
Let $\phi \in N: \phi \ne 0$.
Then $N$ is the set o... | Equivalent Statements for Vector Subspace Dimension One Less | https://proofwiki.org/wiki/Equivalent_Statements_for_Vector_Subspace_Dimension_One_Less | https://proofwiki.org/wiki/Equivalent_Statements_for_Vector_Subspace_Dimension_One_Less | [
"Linear Algebra"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Vector Subspace",
"Definition:Dimension of Vector Space",
"Definition:Vector Space",
"Definition:Kernel of Linear Transformation",
"Definition:Linear Form (Linear Algebra)",
"Definition:Sequence",
"Definition:Scalar/Vector Space"
] | [
"Definition:Annihilator on Algebraic Dual",
"Results Concerning Annihilator of Vector Subspace",
"Definition:Dimension of Vector Space",
"Definition:Scalar Multiplication/Vector Space",
"Definition:Kernel of Linear Transformation",
"Rank Plus Nullity Theorem",
"Definition:Scalar/Vector Space",
"Defini... |
proofwiki-1309 | Rank and Nullity of Transpose | Let $G$ and $H$ be $n$-dimensional vector spaces over a field.
Let $\map \LL {G, H}$ be the set of all linear transformations from $G$ to $H$.
Let $u \in \map \LL {G, H}$.
Let $u^\intercal$ be the transpose of $u$.
Then:
:$u$ and $u^\intercal$ have the same rank and nullity | From Rank Plus Nullity Theorem and Results Concerning Annihilator of Vector Subspace:
{{begin-eqn}}
{{eqn | l = \map \dim {\map {u^\intercal} {H^*} }
| r = n - \map \dim {\map \ker {u^\intercal} }
| c =
}}
{{eqn | r = n - \map \dim {\paren {\map u G}^\circ}
| c =
}}
{{eqn | r = \map \dim {\map u G}
... | Let $G$ and $H$ be [[Definition:Dimension of Vector Space|$n$-dimensional]] [[Definition:Vector Space|vector spaces]] over a [[Definition:Field (Abstract Algebra)|field]].
Let $\map \LL {G, H}$ be [[Definition:Set of All Linear Transformations|the set of all linear transformations]] from $G$ to $H$.
Let $u \in \map \... | From [[Rank Plus Nullity Theorem]] and [[Results Concerning Annihilator of Vector Subspace]]:
{{begin-eqn}}
{{eqn | l = \map \dim {\map {u^\intercal} {H^*} }
| r = n - \map \dim {\map \ker {u^\intercal} }
| c =
}}
{{eqn | r = n - \map \dim {\paren {\map u G}^\circ}
| c =
}}
{{eqn | r = \map \dim {\... | Rank and Nullity of Transpose | https://proofwiki.org/wiki/Rank_and_Nullity_of_Transpose | https://proofwiki.org/wiki/Rank_and_Nullity_of_Transpose | [
"Linear Algebra"
] | [
"Definition:Dimension of Vector Space",
"Definition:Vector Space",
"Definition:Field (Abstract Algebra)",
"Definition:Set of All Linear Transformations",
"Definition:Transpose of Linear Transformation",
"Definition:Rank/Linear Transformation",
"Definition:Nullity/Linear Transformation"
] | [
"Rank Plus Nullity Theorem",
"Results Concerning Annihilator of Vector Subspace",
"Definition:Rank/Linear Transformation",
"Definition:Nullity/Linear Transformation"
] |
proofwiki-1310 | Linear Operator on the Plane | Let $\phi$ be a linear operator on the real vector space of two dimensions $\R^2$.
Then $\phi$ is completely determined by an ordered tuple of $4$ real numbers. | Let $\phi$ be a linear operator on $\R^2$.
Let $\alpha_{11}, \alpha_{12}, \alpha_{21}, \alpha_{22} \in \R$ be the real numbers which satisfy the equations:
{{begin-eqn}}
{{eqn | l = \map \phi {e_1}
| r = \alpha_{11} e_1 + \alpha_{21} e_2
| c =
}}
{{eqn | l = \map \phi {e_2}
| r = \alpha_{12} e_1 + \a... | Let $\phi$ be a [[Definition:Linear Operator|linear operator]] on the [[Definition:Real Vector Space|real vector space]] of [[Definition:Dimension of Vector Space|two dimensions]] $\R^2$.
Then $\phi$ is completely determined by an [[Definition:Ordered Tuple|ordered tuple]] of $4$ [[Definition:Real Number|real numbers... | Let $\phi$ be a [[Definition:Linear Operator|linear operator]] on $\R^2$.
Let $\alpha_{11}, \alpha_{12}, \alpha_{21}, \alpha_{22} \in \R$ be the [[Definition:Real Number|real numbers]] which satisfy the equations:
{{begin-eqn}}
{{eqn | l = \map \phi {e_1}
| r = \alpha_{11} e_1 + \alpha_{21} e_2
| c =
}}
... | Linear Operator on the Plane | https://proofwiki.org/wiki/Linear_Operator_on_the_Plane | https://proofwiki.org/wiki/Linear_Operator_on_the_Plane | [
"Linear Operators",
"Analytic Geometry"
] | [
"Definition:Linear Operator",
"Definition:Real Vector Space",
"Definition:Dimension of Vector Space",
"Definition:Ordered Tuple",
"Definition:Real Number"
] | [
"Definition:Linear Operator",
"Definition:Real Number",
"Definition:Standard Ordered Basis",
"Definition:Linear Operator",
"Condition for Linear Transformation",
"Definition:Ordered Tuple",
"Definition:Real Number"
] |
proofwiki-1311 | Similarity Mapping is Linear Operator | Let $G$ be a vector space over a field $\struct {K, +, \times}$.
Let $\beta \in K$.
Then the similarity $s_\beta: G \to G$ defined as:
:$\forall \mathbf x \in G: \map {s_\beta} {\mathbf x} = \beta \mathbf x$
is a linear operator on $G$. | To prove that $s_\beta$ is a linear operator it is sufficient to demonstrate that:
:$(1): \quad \forall \mathbf x, \mathbf y \in G: \map {s_\beta} {\mathbf x + \mathbf y} = \map {s_\beta} {\mathbf x} + \map {s_\beta} {\mathbf y}$
:$(2): \quad \forall \mathbf x \in G: \forall \lambda \in K: \map {s_\beta} {\lambda \math... | Let $G$ be a [[Definition:Vector Space|vector space]] over a [[Definition:Field (Abstract Algebra)|field]] $\struct {K, +, \times}$.
Let $\beta \in K$.
Then the [[Definition:Similarity Mapping|similarity]] $s_\beta: G \to G$ defined as:
:$\forall \mathbf x \in G: \map {s_\beta} {\mathbf x} = \beta \mathbf x$
is a [[... | To prove that $s_\beta$ is a [[Definition:Linear Operator|linear operator]] it is sufficient to demonstrate that:
:$(1): \quad \forall \mathbf x, \mathbf y \in G: \map {s_\beta} {\mathbf x + \mathbf y} = \map {s_\beta} {\mathbf x} + \map {s_\beta} {\mathbf y}$
:$(2): \quad \forall \mathbf x \in G: \forall \lambda \in K... | Similarity Mapping is Linear Operator | https://proofwiki.org/wiki/Similarity_Mapping_is_Linear_Operator | https://proofwiki.org/wiki/Similarity_Mapping_is_Linear_Operator | [
"Similarity Mappings",
"Linear Operators"
] | [
"Definition:Vector Space",
"Definition:Field (Abstract Algebra)",
"Definition:Similarity Mapping",
"Definition:Linear Operator"
] | [
"Definition:Linear Operator"
] |
proofwiki-1312 | Cantor-Dedekind Hypothesis | The points on an infinite straight line are in one-to-one correspondence with the set $\R$ of real numbers.
Hence the set of all points on an infinite straight line and $\R$ are equinumerous. | === Step 1 ===
We will show that there exists a mapping from the infinite straight line $L$ to the set of real numbers $\R$.
Let us establish a relation $h$ between points on $L$ and elements of $\R$.
We allow the Axiom of Choice to set up a choice function to allow the points of $L$ to be selected systematically.
Pick... | The [[Definition:Point|points]] on an [[Definition:Infinite Straight Line|infinite straight line]] are in [[Definition:Bijection|one-to-one correspondence]] with the set $\R$ of [[Definition:Real Number|real numbers]].
Hence the [[Definition:Set|set]] of all [[Definition:Point|points]] on an [[Definition:Infinite Str... | === Step 1 ===
We will show that there exists a [[Definition:Mapping|mapping]] from the [[Definition:Infinite Straight Line|infinite straight line]] $L$ to the [[Definition:Set|set]] of [[Definition:Real Number|real numbers]] $\R$.
Let us establish a [[Definition:Relation|relation]] $h$ between [[Definition:Point|poi... | Cantor-Dedekind Hypothesis | https://proofwiki.org/wiki/Cantor-Dedekind_Hypothesis | https://proofwiki.org/wiki/Cantor-Dedekind_Hypothesis | [
"Analytic Geometry",
"Euclidean Geometry"
] | [
"Definition:Point",
"Definition:Line/Infinite Straight Line",
"Definition:Bijection",
"Definition:Real Number",
"Definition:Set",
"Definition:Point",
"Definition:Line/Infinite Straight Line",
"Definition:Set Equivalence"
] | [
"Definition:Mapping",
"Definition:Line/Infinite Straight Line",
"Definition:Set",
"Definition:Real Number",
"Definition:Relation",
"Definition:Point",
"Axiom:Axiom of Choice",
"Definition:Choice Function",
"Definition:Point",
"Definition:Coordinate System/Origin",
"Definition:Zero (Number)",
"... |
proofwiki-1313 | Rotation of Plane about Origin is Linear Operator | Let $r_\alpha$ be the rotation of the plane about the origin through an angle of $\alpha$.
That is, let $r_\alpha: \R^2 \to \R^2$ be the mapping defined as:
:$\forall x \in \R^2: \map {r_\alpha} x = \text { the point into which a rotation of $\alpha$ carries $x$}$
Then $r_\alpha$ is a linear operator. | Let $P = \tuple {\lambda_1, \lambda_2}$ be an arbitrary point in $\R^2$.
From Equations defining Plane Rotation:
:$\map {r_\alpha} P = \tuple {\lambda_1 \cos \alpha - \lambda_2 \sin \alpha, \lambda_1 \sin \alpha + \lambda_2 \cos \alpha}$
This demonstrates that $r_\alpha$ can be expressed as an ordered tuple of $4$ real... | Let $r_\alpha$ be the [[Definition:Plane Rotation|rotation]] of [[Definition:The Plane|the plane]] about the [[Definition:Origin|origin]] through an [[Definition:Angle|angle]] of $\alpha$.
That is, let $r_\alpha: \R^2 \to \R^2$ be the [[Definition:Mapping|mapping]] defined as:
:$\forall x \in \R^2: \map {r_\alpha} x =... | Let $P = \tuple {\lambda_1, \lambda_2}$ be an arbitrary [[Definition:Point|point]] in $\R^2$.
From [[Equations defining Plane Rotation/Cartesian|Equations defining Plane Rotation]]:
:$\map {r_\alpha} P = \tuple {\lambda_1 \cos \alpha - \lambda_2 \sin \alpha, \lambda_1 \sin \alpha + \lambda_2 \cos \alpha}$
This demo... | Rotation of Plane about Origin is Linear Operator | https://proofwiki.org/wiki/Rotation_of_Plane_about_Origin_is_Linear_Operator | https://proofwiki.org/wiki/Rotation_of_Plane_about_Origin_is_Linear_Operator | [
"Geometric Rotations",
"Linear Operators"
] | [
"Definition:Rotation (Geometry)/Plane",
"Definition:Plane Surface/The Plane",
"Definition:Coordinate System/Origin",
"Definition:Angle",
"Definition:Mapping",
"Definition:Linear Operator"
] | [
"Definition:Point",
"Equations defining Plane Rotation/Cartesian",
"Definition:Ordered Tuple",
"Definition:Real Number",
"Linear Operator on the Plane"
] |
proofwiki-1314 | Reflection of Plane in Line through Origin is Linear Operator | Let $M$ be a straight line in the plane $\R^2$ passing through the origin.
Let $s_M$ be the '''reflection''' of $\R^2$ in $M$.
Then $s_M$ is a linear operator for every straight line $M$ through the origin. | Let the angle between $M$ and the $x$-axis be $\alpha$.
To prove that $s_M$ is a '''linear operator''' it is sufficient to demonstrate that:
:$(1): \quad \forall P_1, P_2 \in \R^2: \map {s_M} {P_1 + P_2} = \map {s_M} {P_1} + \map {s_M} {P_2}$
:$(2): \quad \forall \lambda \in \R: \map {s_M} {\lambda P_1} = \lambda \map ... | Let $M$ be a [[Definition:Infinite Line|straight line]] in [[Definition:The Plane|the plane]] $\R^2$ passing through the [[Definition:Origin|origin]].
Let $s_M$ be the '''[[Definition:Plane Reflection|reflection]]''' of $\R^2$ in $M$.
Then $s_M$ is a [[Definition:Linear Operator|linear operator]] for every [[Definit... | Let the [[Definition:Plane Angle|angle]] between $M$ and the [[Definition:X-Axis|$x$-axis]] be $\alpha$.
To prove that $s_M$ is a '''[[Definition:Linear Operator|linear operator]]''' it is sufficient to demonstrate that:
:$(1): \quad \forall P_1, P_2 \in \R^2: \map {s_M} {P_1 + P_2} = \map {s_M} {P_1} + \map {s_M} {P... | Reflection of Plane in Line through Origin is Linear Operator | https://proofwiki.org/wiki/Reflection_of_Plane_in_Line_through_Origin_is_Linear_Operator | https://proofwiki.org/wiki/Reflection_of_Plane_in_Line_through_Origin_is_Linear_Operator | [
"Geometric Reflections",
"Linear Operators"
] | [
"Definition:Line/Infinite",
"Definition:Plane Surface/The Plane",
"Definition:Coordinate System/Origin",
"Definition:Reflection (Geometry)/Plane",
"Definition:Linear Operator",
"Definition:Line/Straight Line",
"Definition:Coordinate System/Origin"
] | [
"Definition:Angle",
"Definition:Axis/X-Axis",
"Definition:Linear Operator",
"Definition:Point",
"Definition:Plane Surface/The Plane",
"Equations defining Plane Reflection/Cartesian",
"Equations defining Plane Reflection/Cartesian",
"Equations defining Plane Reflection/Cartesian",
"Equations defining... |
proofwiki-1315 | Projection in Plane between Lines passing through Origin is Linear Operator | Let $M$ and $N$ be distinct straight lines through the plane through the origin.
Let $\pr_{M, N}$ be the projection on $M$ along $N$.
Then $\pr_{M, N}$ is a linear operator. | Let the angle between $M$ and the $x$-axis be $\theta$.
Let the angle between $N$ and the $x$-axis be $\phi$.
Let $P = \tuple {x, y}$ be an arbitrary point in the plane.
Then from Equations defining Projection in Plane:
:$\map {\pr_{M, N} } P = \begin {cases}
\tuple {0, y - x \tan \phi} & : \theta = \dfrac \pi 2 \\
\t... | Let $M$ and $N$ be distinct [[Definition:Straight Line|straight lines]] through [[Definition:The Plane|the plane]] through the [[Definition:Origin|origin]].
Let $\pr_{M, N}$ be the [[Definition:Projection in Plane|projection on $M$ along $N$]].
Then $\pr_{M, N}$ is a [[Definition:Linear Operator|linear operator]]. | Let the [[Definition:Plane Angle|angle]] between $M$ and the [[Definition:X-Axis|$x$-axis]] be $\theta$.
Let the [[Definition:Plane Angle|angle]] between $N$ and the [[Definition:X-Axis|$x$-axis]] be $\phi$.
Let $P = \tuple {x, y}$ be an arbitrary [[Definition:Point|point]] in [[Definition:The Plane|the plane]].
Th... | Projection in Plane between Lines passing through Origin is Linear Operator | https://proofwiki.org/wiki/Projection_in_Plane_between_Lines_passing_through_Origin_is_Linear_Operator | https://proofwiki.org/wiki/Projection_in_Plane_between_Lines_passing_through_Origin_is_Linear_Operator | [
"Linear Operators",
"Geometric Projections"
] | [
"Definition:Line/Straight Line",
"Definition:Plane Surface/The Plane",
"Definition:Coordinate System/Origin",
"Definition:Projection (Geometry)/Plane",
"Definition:Linear Operator"
] | [
"Definition:Angle",
"Definition:Axis/X-Axis",
"Definition:Angle",
"Definition:Axis/X-Axis",
"Definition:Point",
"Definition:Plane Surface/The Plane",
"Equations defining Projection in Plane/Cartesian",
"Definition:Ordered Tuple",
"Definition:Real Number",
"Linear Operator on the Plane"
] |
proofwiki-1316 | Condition for Straight Lines in Plane to be Parallel/General Equation | Let $L: \alpha_1 x + \alpha_2 y = \beta$ be a straight line in $\R^2$.
Then the straight line $L'$ is parallel to $L$ {{iff}} there is a $\beta' \in \R^2$ such that:
:$L' = \set {\tuple {x, y} \in \R^2: \alpha_1 x + \alpha_2 y = \beta'}$ | === Necessary Condition ===
When $L' = L$, the claim is trivial.
Let $L' \ne L$ be described by the equation:
:$\alpha'_1 x + \alpha'_2 y = \beta'$
{{WLOG}}, let $\alpha'_1 \ne 0$ (the case $\alpha'_2 \ne 0$ is similar).
Then for $\tuple {x, y} \in L'$ to hold, one needs:
{{begin-eqn}}
{{eqn | l = \alpha'_1 x + \alpha'... | Let $L: \alpha_1 x + \alpha_2 y = \beta$ be a [[Definition:Straight Line|straight line]] in $\R^2$.
Then the [[Definition:Straight Line|straight line]] $L'$ is [[Definition:Parallel Lines|parallel]] to $L$ {{iff}} there is a $\beta' \in \R^2$ such that:
:$L' = \set {\tuple {x, y} \in \R^2: \alpha_1 x + \alpha_2 y = ... | === Necessary Condition ===
When $L' = L$, the claim is trivial.
Let $L' \ne L$ be [[Equation of Straight Line in Plane|described by]] the equation:
:$\alpha'_1 x + \alpha'_2 y = \beta'$
{{WLOG}}, let $\alpha'_1 \ne 0$ (the case $\alpha'_2 \ne 0$ is similar).
Then for $\tuple {x, y} \in L'$ to hold, one needs:
{... | Condition for Straight Lines in Plane to be Parallel/General Equation | https://proofwiki.org/wiki/Condition_for_Straight_Lines_in_Plane_to_be_Parallel/General_Equation | https://proofwiki.org/wiki/Condition_for_Straight_Lines_in_Plane_to_be_Parallel/General_Equation | [
"Condition for Straight Lines in Plane to be Parallel"
] | [
"Definition:Line/Straight Line",
"Definition:Line/Straight Line",
"Definition:Parallel (Geometry)/Lines"
] | [
"Equation of Straight Line in Plane",
"Definition:Parallel (Geometry)/Lines",
"Equation of Straight Line in Plane",
"Definition:Parallel (Geometry)/Lines",
"Definition:Parallel (Geometry)/Lines"
] |
proofwiki-1317 | Condition for Planes to be Parallel | Let $P: \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma$ be a plane in $\R^3$.
Then the plane $P'$ is parallel to $P$ {{iff}} there is a $\gamma' \in \R$ such that:
:$P' = \set {\tuple {x_1, x_2, x_3} \in \R^3 : \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma'}$ | === Necessary Condition ===
We have that $P$ and $P'$ are parallel.
If $P = P'$, then $\gamma' = \gamma$ and we are done.
Let $P \neq P'$.
Let $\mathbf x$ be a point in $P$.
Construct a line $L$ through $\mathbf x$ perpendicular to $P'$.
Let $L_{P'}$ be a line in $P'$ that intersects $L$.
Let $L_P$ be a line in $P'$ th... | Let $P: \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma$ be a [[Equation of Plane|plane]] in $\R^3$.
Then the plane $P'$ is [[Definition:Parallel Planes|parallel]] to $P$ {{iff}} there is a $\gamma' \in \R$ such that:
:$P' = \set {\tuple {x_1, x_2, x_3} \in \R^3 : \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gam... | === Necessary Condition ===
We have that $P$ and $P'$ are [[Definition:Parallel Planes|parallel]].
If $P = P'$, then $\gamma' = \gamma$ and we are done.
Let $P \neq P'$.
Let $\mathbf x$ be a [[Definition:Point|point]] in $P$.
[[Construction of Straight Line Perpendicular to Plane from point not on Plane|Construct... | Condition for Planes to be Parallel | https://proofwiki.org/wiki/Condition_for_Planes_to_be_Parallel | https://proofwiki.org/wiki/Condition_for_Planes_to_be_Parallel | [
"Parallel Planes",
"Planes",
"Solid Analytic Geometry"
] | [
"Equation of Plane",
"Definition:Parallel (Geometry)/Planes"
] | [
"Definition:Parallel (Geometry)/Planes",
"Definition:Point",
"Construction of Straight Line Perpendicular to Plane from point not on Plane",
"Definition:Right Angle/Perpendicular/Plane",
"Definition:Line",
"Definition:Intersection (Geometry)",
"Definition:Line",
"Definition:Intersection (Geometry)",
... |
proofwiki-1318 | Lines are Subspaces of Plane | The one-dimensional subspaces of $\R^2$ are precisely the homogeneous lines of plane analytic geometry. | Follows directly from Equivalent Statements for Vector Subspace Dimension One Less.
{{qed}}
{{explain|Explain how}} | The [[Definition:Dimension of Vector Space|one-dimensional]] [[Definition:Vector Subspace|subspaces]] of $\R^2$ are precisely the [[Definition:Homogeneous (Analytic Geometry)|homogeneous lines]] of [[Definition:Plane Analytic Geometry|plane analytic geometry]]. | Follows directly from [[Equivalent Statements for Vector Subspace Dimension One Less]].
{{qed}}
{{explain|Explain how}} | Lines are Subspaces of Plane | https://proofwiki.org/wiki/Lines_are_Subspaces_of_Plane | https://proofwiki.org/wiki/Lines_are_Subspaces_of_Plane | [
"Linear Algebra",
"Plane Analytic Geometry"
] | [
"Definition:Dimension of Vector Space",
"Definition:Vector Subspace",
"Definition:Homogeneous (Analytic Geometry)",
"Definition:Analytic Geometry/Plane"
] | [
"Equivalent Statements for Vector Subspace Dimension One Less"
] |
proofwiki-1319 | Planes are Subspaces of Space | The two-dimensional subspaces of $\R^3$ are precisely the homogeneous planes of solid analytic geometry. | Follows directly from Equivalent Statements for Vector Subspace Dimension One Less.
{{qed}}
{{explain|How?}} | The [[Definition:Dimension of Vector Space|two-dimensional]] [[Definition:Vector Subspace|subspaces]] of $\R^3$ are precisely the [[Definition:Homogeneous (Analytic Geometry)|homogeneous planes]] of [[Definition:Solid Analytic Geometry|solid analytic geometry]]. | Follows directly from [[Equivalent Statements for Vector Subspace Dimension One Less]].
{{qed}}
{{explain|How?}} | Planes are Subspaces of Space | https://proofwiki.org/wiki/Planes_are_Subspaces_of_Space | https://proofwiki.org/wiki/Planes_are_Subspaces_of_Space | [
"Linear Algebra",
"Solid Analytic Geometry"
] | [
"Definition:Dimension of Vector Space",
"Definition:Vector Subspace",
"Definition:Homogeneous (Analytic Geometry)",
"Definition:Analytic Geometry/Solid"
] | [
"Equivalent Statements for Vector Subspace Dimension One Less"
] |
proofwiki-1320 | Matrix Space Semigroup under Hadamard Product | Let $\map {\MM_S} {m, n}$ be the matrix space over a semigroup $\struct {S, \cdot}$.
Then the algebraic structure $\struct {\map {\MM_S} {m, n}, \circ}$, where $\circ$ is the Hadamard product, is also a semigroup.
If $\struct {S, \cdot}$ is a commutative semigroup then so is $\struct {\map {\MM_S} {m, n}, \circ}$.
If $... | $\struct {S, \cdot}$ is a semigroup and is therefore closed and associative.
As $\struct {S, \cdot}$ is closed, then so is $\struct {\map {\MM_S} {m, n}, \circ}$ from Closure of Hadamard Product.
As $\struct {S, \cdot}$ is associative, then so is $\struct {\map {\MM_S} {m, n}, \circ}$ from Associativity of Hadamard Pro... | Let $\map {\MM_S} {m, n}$ be the [[Definition:Matrix Space|matrix space]] over a [[Definition:Semigroup|semigroup]] $\struct {S, \cdot}$.
Then the [[Definition:Algebraic Structure with One Operation|algebraic structure]] $\struct {\map {\MM_S} {m, n}, \circ}$, where $\circ$ is the [[Definition:Hadamard Product|Hadama... | $\struct {S, \cdot}$ is a [[Definition:Semigroup|semigroup]] and is therefore [[Definition:Closed Algebraic Structure|closed]] and [[Definition:Associative Algebraic Structure|associative]].
As $\struct {S, \cdot}$ is [[Definition:Closed Algebraic Structure|closed]], then so is $\struct {\map {\MM_S} {m, n}, \circ}$ f... | Matrix Space Semigroup under Hadamard Product | https://proofwiki.org/wiki/Matrix_Space_Semigroup_under_Hadamard_Product | https://proofwiki.org/wiki/Matrix_Space_Semigroup_under_Hadamard_Product | [
"Hadamard Product"
] | [
"Definition:Matrix Space",
"Definition:Semigroup",
"Definition:Algebraic Structure/One Operation",
"Definition:Hadamard Product",
"Definition:Semigroup",
"Definition:Commutative Semigroup",
"Definition:Monoid"
] | [
"Definition:Semigroup",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Semigroup",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Closure of Hadamard Product",
"Definition:Semigroup",
"Associativity of Hadamard Product",
"Definition:Semigroup",
"Definition:Co... |
proofwiki-1321 | Hadamard Product over Group forms Group | Let $\struct {G, \cdot}$ be a group whose identity is $e$.
Let $\map {\MM_G} {m, n}$ be a $m \times n$ matrix space over $\struct {G, \cdot}$.
Then $\struct {\map {\MM_G} {m, n}, \circ}$, where $\circ$ is Hadamard product, is also a group. | As $\struct {G, \cdot}$, being a group, is a monoid, it follows from Matrix Space Semigroup under Hadamard Product that $\struct {\map {\MM_G} {m, n}, \circ}$ is also a monoid.
As $\struct {G, \cdot}$ is a group, it follows from Negative Matrix is Inverse for Hadamard Product that all elements of $\struct {\map {\MM_G}... | Let $\struct {G, \cdot}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $\map {\MM_G} {m, n}$ be a [[Definition:Matrix Space|$m \times n$ matrix space]] over $\struct {G, \cdot}$.
Then $\struct {\map {\MM_G} {m, n}, \circ}$, where $\circ$ is [[Definition:Hadamard Product|Ha... | As $\struct {G, \cdot}$, being a [[Definition:Group|group]], is a [[Definition:Monoid|monoid]], it follows from [[Matrix Space Semigroup under Hadamard Product]] that $\struct {\map {\MM_G} {m, n}, \circ}$ is also a [[Definition:Monoid|monoid]].
As $\struct {G, \cdot}$ is a [[Definition:Group|group]], it follows from ... | Hadamard Product over Group forms Group | https://proofwiki.org/wiki/Hadamard_Product_over_Group_forms_Group | https://proofwiki.org/wiki/Hadamard_Product_over_Group_forms_Group | [
"Hadamard Product",
"Examples of Groups"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Matrix Space",
"Definition:Hadamard Product",
"Definition:Group"
] | [
"Definition:Group",
"Definition:Monoid",
"Matrix Space Semigroup under Hadamard Product",
"Definition:Monoid",
"Definition:Group",
"Negative Matrix is Inverse for Hadamard Product",
"Definition:Element",
"Definition:Inverse (Abstract Algebra)/Inverse"
] |
proofwiki-1322 | Matrix Multiplication is Associative | Matrix multiplication (conventional) is associative. | Let $\mathbf A = \sqbrk a_{m n}, \mathbf B = \sqbrk b_{n p}, \mathbf C = \sqbrk c_{p q}$ be matrices.
From inspection of the subscripts, we can see that both $\paren {\mathbf A \mathbf B} \mathbf C$ and $\mathbf A \paren {\mathbf B \mathbf C}$ are defined:
$\mathbf A$ has $n$ columns and $\mathbf B$ has $n$ rows, while... | [[Definition:Matrix Product (Conventional)|Matrix multiplication (conventional)]] is [[Definition:Associative Operation|associative]]. | Let $\mathbf A = \sqbrk a_{m n}, \mathbf B = \sqbrk b_{n p}, \mathbf C = \sqbrk c_{p q}$ be [[Definition:Matrix|matrices]].
From inspection of the subscripts, we can see that both $\paren {\mathbf A \mathbf B} \mathbf C$ and $\mathbf A \paren {\mathbf B \mathbf C}$ are defined:
$\mathbf A$ has $n$ [[Definition:Colum... | Matrix Multiplication is Associative | https://proofwiki.org/wiki/Matrix_Multiplication_is_Associative | https://proofwiki.org/wiki/Matrix_Multiplication_is_Associative | [
"Conventional Matrix Multiplication",
"Examples of Associative Operations"
] | [
"Definition:Matrix Product (Conventional)",
"Definition:Associative Operation"
] | [
"Definition:Matrix",
"Definition:Matrix/Column",
"Definition:Matrix/Row",
"Definition:Matrix/Column",
"Definition:Matrix/Row"
] |
proofwiki-1323 | Set of Linear Transformations is Isomorphic to Matrix Space | Let $R$ be a ring with unity.
Let $F$, $G$ and $H$ be free $R$-modules of finite dimension $p, n, m > 0$ respectively.
Let $\sequence {a_p}$, $\sequence {b_n}$ and $\sequence {c_m}$ be ordered bases
Let $\map {\LL_R} {G, H}$ denote the set of all linear transformations from $G$ to $H$.
Let $\map {\MM_R} {m, n}$ be the... | Let $u, v \in \map {\LL_R} {G, H}$ such that:
:$\map M u = \map M v$
We have that the matrix of $u$ relative to $\sequence {b_n}$ and $\sequence {c_m}$ is defined as the $m \times n$ matrix $\sqbrk \alpha_{m n}$ where:
:$\ds \forall \tuple {i, j} \in \closedint 1 m \times \closedint 1 n: \map u {b_j} = \sum_{i \mathop ... | Let $R$ be a [[Definition:Ring with Unity|ring with unity]].
Let $F$, $G$ and $H$ be [[Definition:Free Module over Ring|free $R$-modules]] of [[Definition:Finite Dimensional Free Module|finite dimension]] $p, n, m > 0$ respectively.
Let $\sequence {a_p}$, $\sequence {b_n}$ and $\sequence {c_m}$ be [[Definition:Ordere... | Let $u, v \in \map {\LL_R} {G, H}$ such that:
:$\map M u = \map M v$
We have that the [[Definition:Relative Matrix of Linear Transformation|matrix of $u$ relative to $\sequence {b_n}$ and $\sequence {c_m}$]] is defined as the [[Definition:Matrix|$m \times n$ matrix]] $\sqbrk \alpha_{m n}$ where:
:$\ds \forall \tuple ... | Set of Linear Transformations is Isomorphic to Matrix Space | https://proofwiki.org/wiki/Set_of_Linear_Transformations_is_Isomorphic_to_Matrix_Space | https://proofwiki.org/wiki/Set_of_Linear_Transformations_is_Isomorphic_to_Matrix_Space | [
"Linear Transformations",
"Matrix Algebra",
"Set of Linear Transformations is Isomorphic to Matrix Space"
] | [
"Definition:Ring with Unity",
"Definition:Free Module over Ring",
"Definition:Dimension of Module/Finite",
"Definition:Ordered Basis",
"Definition:Set of All Linear Transformations",
"Definition:Matrix Space",
"Definition:Relative Matrix of Linear Transformation",
"Definition:Isomorphism (Abstract Alg... | [
"Definition:Relative Matrix of Linear Transformation",
"Definition:Matrix",
"Definition:Injection"
] |
proofwiki-1324 | Matrix Multiplication Distributes over Matrix Addition | Matrix multiplication (conventional) is distributive over matrix entrywise addition. | Let $\mathbf A = \sqbrk a_{m n}, \mathbf B = \sqbrk b_{n p}, \mathbf C = \sqbrk c_{n p}$ be matrices over a ring $\struct {R, +, \circ}$.
Consider $\mathbf A \paren {\mathbf B + \mathbf C}$.
Let $\mathbf R = \sqbrk r_{n p} = \mathbf B + \mathbf C, \mathbf S = \sqbrk s_{m p} = \mathbf A \paren {\mathbf B + \mathbf C}$.
... | [[Definition:Matrix Product (Conventional)|Matrix multiplication (conventional)]] is [[Definition:Distributive Operation|distributive]] over [[Definition:Matrix Entrywise Addition|matrix entrywise addition]]. | Let $\mathbf A = \sqbrk a_{m n}, \mathbf B = \sqbrk b_{n p}, \mathbf C = \sqbrk c_{n p}$ be [[Definition:Matrix|matrices]] over a [[Definition:Ring (Abstract Algebra)|ring]] $\struct {R, +, \circ}$.
Consider $\mathbf A \paren {\mathbf B + \mathbf C}$.
Let $\mathbf R = \sqbrk r_{n p} = \mathbf B + \mathbf C, \mathbf S... | Matrix Multiplication Distributes over Matrix Addition | https://proofwiki.org/wiki/Matrix_Multiplication_Distributes_over_Matrix_Addition | https://proofwiki.org/wiki/Matrix_Multiplication_Distributes_over_Matrix_Addition | [
"Conventional Matrix Multiplication",
"Matrix Entrywise Addition",
"Examples of Distributive Operations"
] | [
"Definition:Matrix Product (Conventional)",
"Definition:Distributive Operation",
"Definition:Matrix Entrywise Addition"
] | [
"Definition:Matrix",
"Definition:Ring (Abstract Algebra)"
] |
proofwiki-1325 | Unit Matrix is Unity of Ring of Square Matrices | Let $R$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\struct {\map {\MM_R} n, +, \times}$ denote the ring of square matrices of order $n$ over $R$.
The unit matrix over $R$:
:$\mathbf I_n = \begin {pmatrix} 1_R & 0_R & 0_R & \cdots & 0_R \... | In Unit Matrix is Identity for Matrix Multiplication, it is demonstrated that:
:$\forall \mathbf A \in \map {\MM_R} n: \mathbf A \mathbf I_n = \mathbf A = \mathbf I_n \mathbf A$
Hence the result, by definition of identity element
{{qed}} | Let $R$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
Let $\struct {\map {\MM_R} n, +, \times}$ denote the [[Definiti... | In [[Unit Matrix is Identity for Matrix Multiplication]], it is demonstrated that:
:$\forall \mathbf A \in \map {\MM_R} n: \mathbf A \mathbf I_n = \mathbf A = \mathbf I_n \mathbf A$
Hence the result, by definition of [[Definition:Identity Element|identity element]]
{{qed}} | Unit Matrix is Unity of Ring of Square Matrices | https://proofwiki.org/wiki/Unit_Matrix_is_Unity_of_Ring_of_Square_Matrices | https://proofwiki.org/wiki/Unit_Matrix_is_Unity_of_Ring_of_Square_Matrices | [
"Rings of Square Matrices",
"Unit Matrices"
] | [
"Definition:Ring with Unity",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Strictly Positive/Integer",
"Definition:Ring of Square Matrices",
"Definition:Unit Matrix",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Unit Matrix is Identity for Matrix Multiplication",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] |
proofwiki-1326 | Matrix Multiplication over Order n Square Matrices is Closed | Let $\struct {R, +, \circ}$ be a ring.
Let $\map {\MM_R} n$ be a $n \times n$ matrix space over $R$.
Then matrix multiplication (conventional) over $\map {\MM_R} n$ is closed. | From the definition of matrix multiplication, the product of two matrices is another matrix.
The order of an $m \times n$ multiplied by an $n \times p$ matrix is $m \times p$.
The entries of that product matrix are elements of the ring over which the matrix is formed.
Thus an $n \times n$ matrix over $R$ multiplied by ... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\map {\MM_R} n$ be a [[Definition:Matrix Space|$n \times n$ matrix space]] over $R$.
Then [[Definition:Matrix Product (Conventional)|matrix multiplication (conventional)]] over $\map {\MM_R} n$ is [[Definition:Closure (Abstract Algebra... | From the definition of [[Definition:Matrix Product (Conventional)|matrix multiplication]], the product of two [[Definition:Matrix|matrices]] is another [[Definition:Matrix|matrix]].
The [[Definition:Order of Matrix|order]] of an $m \times n$ [[Definition:Matrix Product (Conventional)|multiplied]] by an $n \times p$ [[... | Matrix Multiplication over Order n Square Matrices is Closed | https://proofwiki.org/wiki/Matrix_Multiplication_over_Order_n_Square_Matrices_is_Closed | https://proofwiki.org/wiki/Matrix_Multiplication_over_Order_n_Square_Matrices_is_Closed | [
"Conventional Matrix Multiplication",
"Algebraic Closure"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Matrix Space",
"Definition:Matrix Product (Conventional)",
"Definition:Closure (Abstract Algebra)"
] | [
"Definition:Matrix Product (Conventional)",
"Definition:Matrix",
"Definition:Matrix",
"Definition:Matrix/Order",
"Definition:Matrix Product (Conventional)",
"Definition:Matrix",
"Definition:Matrix/Element",
"Definition:Matrix Product (Conventional)",
"Definition:Matrix",
"Definition:Element",
"D... |
proofwiki-1327 | Square Matrices over Real Numbers under Multiplication form Monoid | Let $\map {\MM_\R} n$ be a $n \times n$ matrix space over the set of real numbers $\R$.
Then the set of all $n \times n$ real matrices $\map {\MM_\R} n$ under matrix multiplication (conventional) forms a monoid. | :Matrix Multiplication over Order n Square Matrices is Closed.
:Matrix Multiplication is Associative.
:The Unit Matrix is Unity of Ring of Square Matrices.
{{qed}} | Let $\map {\MM_\R} n$ be a [[Definition:Matrix Space|$n \times n$ matrix space]] over the [[Definition:Real Number|set of real numbers $\R$]].
Then the set of all $n \times n$ real matrices $\map {\MM_\R} n$ under [[Definition:Matrix Product (Conventional)|matrix multiplication (conventional)]] forms a [[Definition:M... | :[[Matrix Multiplication over Order n Square Matrices is Closed]].
:[[Matrix Multiplication is Associative]].
:The [[Unit Matrix is Unity of Ring of Square Matrices]].
{{qed}} | Square Matrices over Real Numbers under Multiplication form Monoid | https://proofwiki.org/wiki/Square_Matrices_over_Real_Numbers_under_Multiplication_form_Monoid | https://proofwiki.org/wiki/Square_Matrices_over_Real_Numbers_under_Multiplication_form_Monoid | [
"Matrix Algebra",
"Examples of Monoids"
] | [
"Definition:Matrix Space",
"Definition:Real Number",
"Definition:Matrix Product (Conventional)",
"Definition:Monoid"
] | [
"Matrix Multiplication over Order n Square Matrices is Closed",
"Matrix Multiplication is Associative",
"Unit Matrix is Unity of Ring of Square Matrices"
] |
proofwiki-1328 | Ring of Square Matrices over Commutative Ring with Unity | Let $R$ be a commutative ring with unity.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\struct {\map {\MM_R} n, +, \times}$ denote the ring of square matrices of order $n$ over $R$.
Then $\struct {\map {\MM_R} n, +, \times}$ is a ring with unity.
However, for $n \ge 2$, $\struct {\map {\MM_R} n, +, \times... | From Ring of Square Matrices over Ring with Unity we have that $\struct {\map {\MM_R} n, +, \times}$ is a ring with unity.
However, Matrix Multiplication is not Commutative.
Hence $\struct {\map {\MM_R} n, +, \times}$ is not a commutative ring for $n \ge 2$.
For $n = 1$ we have that:
{{begin-eqn}}
{{eqn | q = \forall \... | Let $R$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
Let $\struct {\map {\MM_R} n, +, \times}$ denote the [[Definition:Ring of Square Matrices|ring of square matrices of order $n$ over $R$]]... | From [[Ring of Square Matrices over Ring with Unity]] we have that $\struct {\map {\MM_R} n, +, \times}$ is a [[Definition:Ring with Unity|ring with unity]].
However, [[Matrix Multiplication is not Commutative]].
Hence $\struct {\map {\MM_R} n, +, \times}$ is not a [[Definition:Commutative Ring|commutative ring]] for... | Ring of Square Matrices over Commutative Ring with Unity | https://proofwiki.org/wiki/Ring_of_Square_Matrices_over_Commutative_Ring_with_Unity | https://proofwiki.org/wiki/Ring_of_Square_Matrices_over_Commutative_Ring_with_Unity | [
"Commutative Algebra",
"Rings of Square Matrices",
"Examples of Rings with Unity"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Strictly Positive/Integer",
"Definition:Ring of Square Matrices",
"Definition:Ring with Unity",
"Definition:Commutative Ring"
] | [
"Ring of Square Matrices over Ring with Unity",
"Definition:Ring with Unity",
"Matrix Multiplication is not Commutative",
"Definition:Commutative Ring",
"Definition:Commutative Ring",
"Definition:Commutative Ring"
] |
proofwiki-1329 | Change of Basis Matrix is Nonsingular | Let $R$ be a ring with unity.
Let $M$ be a free $R$-module of finite dimension $n > 0$.
Let $\AA$ and $\BB$ be ordered bases of $M$.
Let $\mathbf P$ be the change of basis matrix from $\AA$ to $\BB$.
Then $\mathbf P$ is nonsingular, and its inverse $\mathbf P^{-1}$ is the change of basis matrix from $\BB$ to $\AA$. | From Product of Change of Basis Matrices and Change of Basis Matrix Between Equal Bases:
:$\sqbrk {I_M; \AA, \BB} \sqbrk {I_M; \BB, \AA} = \sqbrk {I_M; \AA, \AA} = I_n$
:$\sqbrk {I_M; \BB, \AA} \sqbrk {I_M; \AA, \BB} = \sqbrk {I_M; \BB, \BB} = I_n$
Hence the result.
{{qed}} | Let $R$ be a [[Definition:Ring with Unity|ring with unity]].
Let $M$ be a [[Definition:Free Module over Ring|free $R$-module]] of [[Definition:Dimension (Linear Algebra)|finite dimension]] $n > 0$.
Let $\AA$ and $\BB$ be [[Definition:Ordered Basis|ordered bases]] of $M$.
Let $\mathbf P$ be the [[Definition:Change of... | From [[Product of Change of Basis Matrices]] and [[Change of Basis Matrix Between Equal Bases]]:
:$\sqbrk {I_M; \AA, \BB} \sqbrk {I_M; \BB, \AA} = \sqbrk {I_M; \AA, \AA} = I_n$
:$\sqbrk {I_M; \BB, \AA} \sqbrk {I_M; \AA, \BB} = \sqbrk {I_M; \BB, \BB} = I_n$
Hence the result.
{{qed}} | Change of Basis Matrix is Nonsingular | https://proofwiki.org/wiki/Change_of_Basis_Matrix_is_Nonsingular | https://proofwiki.org/wiki/Change_of_Basis_Matrix_is_Nonsingular | [
"Nonsingular Matrices",
"Change of Basis"
] | [
"Definition:Ring with Unity",
"Definition:Free Module over Ring",
"Definition:Dimension (Linear Algebra)",
"Definition:Ordered Basis",
"Definition:Change of Basis Matrix",
"Definition:Nonsingular Matrix",
"Definition:Inverse Matrix",
"Definition:Change of Basis Matrix"
] | [
"Product of Change of Basis Matrices",
"Change of Basis Matrix from Basis to Itself is Identity"
] |
proofwiki-1330 | Invertible Matrix Corresponds with Change of Basis | Let $R$ be a commutative ring with unity.
Let $G$ be an $n$-dimensional unitary $R$-module.
Let $\sequence {a_n}$ be an ordered basis of $G$.
Let $\mathbf P = \sqbrk \alpha_n$ be a square matrix of order $n$ over $R$.
Let $\ds \forall j \in \closedint 1 n: b_j = \sum_{i \mathop = 1}^n \alpha_{i j} a_i$.
Then $\sequence... | From Change of Basis Matrix is Nonsingular, if $\sequence {b_n}$ is an ordered basis of $G$ then $\mathbf P$ is nonsingular.
Now let $\mathbf P$ be nonsingular.
Then by {{Corollary|Set of Linear Transformations is Isomorphic to Matrix Space}}, there is an automorphism $u$ of $G$ which satisfies $\mathbf P = \sqbrk {u; ... | Let $R$ be a [[Definition:Commutative Ring|commutative ring]] [[Definition:Ring with Unity|with unity]].
Let $G$ be an [[Definition:Dimension (Linear Algebra)|$n$-dimensional]] [[Definition:Unitary Module|unitary $R$-module]].
Let $\sequence {a_n}$ be an [[Definition:Ordered Basis|ordered basis]] of $G$.
Let $\mathb... | From [[Change of Basis Matrix is Nonsingular]], if $\sequence {b_n}$ is an [[Definition:Ordered Basis|ordered basis]] of $G$ then $\mathbf P$ is [[Definition:Nonsingular Matrix|nonsingular]].
Now let $\mathbf P$ be [[Definition:Nonsingular Matrix|nonsingular]].
Then by {{Corollary|Set of Linear Transformations is Is... | Invertible Matrix Corresponds with Change of Basis | https://proofwiki.org/wiki/Invertible_Matrix_Corresponds_with_Change_of_Basis | https://proofwiki.org/wiki/Invertible_Matrix_Corresponds_with_Change_of_Basis | [
"Linear Algebra",
"Matrix Algebra"
] | [
"Definition:Commutative Ring",
"Definition:Ring with Unity",
"Definition:Dimension (Linear Algebra)",
"Definition:Unitary Module over Ring",
"Definition:Ordered Basis",
"Definition:Matrix/Square Matrix",
"Definition:Ordered Basis",
"Definition:Nonsingular Matrix"
] | [
"Change of Basis Matrix is Nonsingular",
"Definition:Ordered Basis",
"Definition:Nonsingular Matrix",
"Definition:Nonsingular Matrix",
"Definition:Module Automorphism",
"Definition:Ordered Basis"
] |
proofwiki-1331 | Matrix Equivalence is Equivalence Relation | Matrix equivalence is an equivalence relation. | Checking in turn each of the criteria for equivalence: | [[Definition:Equivalent Matrices|Matrix equivalence]] is an [[Definition:Equivalence Relation|equivalence relation]]. | Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]: | Matrix Equivalence is Equivalence Relation | https://proofwiki.org/wiki/Matrix_Equivalence_is_Equivalence_Relation | https://proofwiki.org/wiki/Matrix_Equivalence_is_Equivalence_Relation | [
"Matrix Algebra",
"Examples of Equivalence Relations"
] | [
"Definition:Matrix Equivalence",
"Definition:Equivalence Relation"
] | [
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] |
proofwiki-1332 | Matrix Similarity is Equivalence Relation | Matrix similarity is an equivalence relation. | Follows directly from Matrix Equivalence is Equivalence Relation.
{{qed}} | [[Definition:Matrix Similarity|Matrix similarity]] is an [[Definition:Equivalence Relation|equivalence relation]]. | Follows directly from [[Matrix Equivalence is Equivalence Relation]].
{{qed}} | Matrix Similarity is Equivalence Relation/Proof 1 | https://proofwiki.org/wiki/Matrix_Similarity_is_Equivalence_Relation | https://proofwiki.org/wiki/Matrix_Similarity_is_Equivalence_Relation/Proof_1 | [
"Matrix Algebra",
"Examples of Equivalence Relations",
"Matrix Similarity is Equivalence Relation"
] | [
"Definition:Matrix Similarity",
"Definition:Equivalence Relation"
] | [
"Matrix Equivalence is Equivalence Relation"
] |
proofwiki-1333 | Matrix Similarity is Equivalence Relation | Matrix similarity is an equivalence relation. | Checking in turn each of the criteria for equivalence:
=== Reflexive ===
$\mathbf A = \mathbf{I_n}^{-1} \mathbf A \mathbf{I_n}$ trivially, for all order $n$ square matrices $\mathbf A$.
So matrix similarity is reflexive.
{{qed|lemma}}
=== Symmetric ===
Let $\mathbf B = \mathbf P^{-1} \mathbf A \mathbf P$.
As $\mathbf P... | [[Definition:Matrix Similarity|Matrix similarity]] is an [[Definition:Equivalence Relation|equivalence relation]]. | Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]:
=== Reflexive ===
$\mathbf A = \mathbf{I_n}^{-1} \mathbf A \mathbf{I_n}$ trivially, for all order $n$ [[Definition:Square Matrix|square matrices]] $\mathbf A$.
So [[Definition:Matrix Similarity|matrix similarity]] is [[Defini... | Matrix Similarity is Equivalence Relation/Proof 2 | https://proofwiki.org/wiki/Matrix_Similarity_is_Equivalence_Relation | https://proofwiki.org/wiki/Matrix_Similarity_is_Equivalence_Relation/Proof_2 | [
"Matrix Algebra",
"Examples of Equivalence Relations",
"Matrix Similarity is Equivalence Relation"
] | [
"Definition:Matrix Similarity",
"Definition:Equivalence Relation"
] | [
"Definition:Equivalence Relation",
"Definition:Matrix/Square Matrix",
"Definition:Matrix Similarity",
"Definition:Reflexive Relation",
"Definition:Nonsingular Matrix",
"Definition:Matrix Similarity",
"Definition:Symmetric Relation",
"Inverse of Matrix Product",
"Product of Matrices is Nonsingular if... |
proofwiki-1334 | Similar Matrices are Equivalent | If two square matrices over a ring with unity $R$ are similar, then they are equivalent.
That is:
:every equivalence class for the similarity relation on $\map {\MM_R} n$ is contained in an equivalence class for the relation of matrix equivalence.
where $\map {\MM_R} n$ denotes the $n \times n$ matrix space over $R$. | If $\mathbf A \sim \mathbf B$ then $\mathbf B = \mathbf P^{-1} \mathbf A \mathbf P$.
Let $\mathbf Q = \mathbf P$.
Then $\mathbf A$ are equivalent to $\mathbf B$, as:
:$\mathbf B = \mathbf Q^{-1} \mathbf A \mathbf P$
{{qed}} | If two [[Definition:Square Matrix|square matrices]] over a [[Definition:Ring with Unity|ring with unity]] $R$ are [[Definition:Matrix Similarity|similar]], then they are [[Definition:Matrix Equivalence|equivalent]].
That is:
:every [[Definition:Equivalence Class|equivalence class]] for the [[Definition:Matrix Similar... | If $\mathbf A \sim \mathbf B$ then $\mathbf B = \mathbf P^{-1} \mathbf A \mathbf P$.
Let $\mathbf Q = \mathbf P$.
Then $\mathbf A$ are [[Definition:Matrix Equivalence|equivalent]] to $\mathbf B$, as:
:$\mathbf B = \mathbf Q^{-1} \mathbf A \mathbf P$
{{qed}} | Similar Matrices are Equivalent | https://proofwiki.org/wiki/Similar_Matrices_are_Equivalent | https://proofwiki.org/wiki/Similar_Matrices_are_Equivalent | [
"Matrix Algebra"
] | [
"Definition:Matrix/Square Matrix",
"Definition:Ring with Unity",
"Definition:Matrix Similarity",
"Definition:Matrix Equivalence",
"Definition:Equivalence Class",
"Definition:Matrix Similarity",
"Definition:Equivalence Class",
"Definition:Matrix Equivalence",
"Definition:Matrix Space"
] | [
"Definition:Matrix Equivalence"
] |
proofwiki-1335 | Equivalent Matrices have Equal Rank | Let $\mathbf A$ and $\mathbf B$ be $m \times n$ matrices over a field $K$.
Let $\map \phi {\mathbf A}$ denote the rank of $\mathbf A$.
Let $\mathbf A \equiv \mathbf B$ denote that $\mathbf A$ and $\mathbf B$ are matrix equivalent.
Then:
:$\mathbf A \equiv \mathbf B$
{{iff}}:
:$\map \phi {\mathbf A} = \map \phi {\mathbf... | Let $\mathbf A$ and $\mathbf B$ be $m \times n$ matrices over a field $K$ such that $\mathbf A \equiv \mathbf B$.
Let $S$ and $T$ be vector spaces of dimensions $n$ and $m$ over $K$.
Let $\mathbf A$ be the matrix of a linear transformation $u: S \to T$ relative to the ordered bases $\sequence {a_n}$ of $S$ and $\sequen... | Let $\mathbf A$ and $\mathbf B$ be [[Definition:Matrix|$m \times n$ matrices]] over a [[Definition:Field (Abstract Algebra)|field]] $K$.
Let $\map \phi {\mathbf A}$ denote the [[Definition:Rank of Matrix|rank]] of $\mathbf A$.
Let $\mathbf A \equiv \mathbf B$ denote that $\mathbf A$ and $\mathbf B$ are [[Definition:M... | Let $\mathbf A$ and $\mathbf B$ be [[Definition:Matrix|$m \times n$ matrices]] over a [[Definition:Field (Abstract Algebra)|field]] $K$ such that $\mathbf A \equiv \mathbf B$.
Let $S$ and $T$ be [[Definition:Vector Space|vector spaces]] of [[Definition:Dimension of Vector Space|dimensions]] $n$ and $m$ over $K$.
Let ... | Equivalent Matrices have Equal Rank | https://proofwiki.org/wiki/Equivalent_Matrices_have_Equal_Rank | https://proofwiki.org/wiki/Equivalent_Matrices_have_Equal_Rank | [
"Matrix Equivalence",
"Rank of Matrix"
] | [
"Definition:Matrix",
"Definition:Field (Abstract Algebra)",
"Definition:Rank/Matrix",
"Definition:Matrix Equivalence"
] | [
"Definition:Matrix",
"Definition:Field (Abstract Algebra)",
"Definition:Vector Space",
"Definition:Dimension of Vector Space",
"Definition:Relative Matrix of Linear Transformation",
"Definition:Ordered Basis",
"Definition:Isomorphism (Abstract Algebra)/R-Algebraic Structure Isomorphism/Vector Space Isom... |
proofwiki-1336 | Number of Matrix Equivalence Classes | Let $K$ be a field.
Let $\map {\MM_K} {m, n}$ be the $m \times n$ matrix space over $K$.
Let $\mathbf A$ be an $m \times n$ matrix of rank $r$ over $K$.
Then:
:<nowiki>$\mathbf A \equiv \begin{cases}
\sqbrk {0_K}_{m n} & : r = 0 \\
& \\
\begin{bmatrix}
\mathbf I_r & \bszero \\
\bszero & \bszero
\end{bmatrix} & : 0... | Follows from Equivalent Matrices have Equal Rank.
{{qed}} | Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $\map {\MM_K} {m, n}$ be the [[Definition:Matrix Space|$m \times n$ matrix space]] over $K$.
Let $\mathbf A$ be an [[Definition:Matrix|$m \times n$ matrix]] of [[Definition:Rank of Matrix|rank]] $r$ over $K$.
Then:
:<nowiki>$\mathbf A \equiv \begin{ca... | Follows from [[Equivalent Matrices have Equal Rank]].
{{qed}} | Number of Matrix Equivalence Classes | https://proofwiki.org/wiki/Number_of_Matrix_Equivalence_Classes | https://proofwiki.org/wiki/Number_of_Matrix_Equivalence_Classes | [
"Matrix Equivalence"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Matrix Space",
"Definition:Matrix",
"Definition:Rank/Matrix",
"Definition:Equivalence Class",
"Definition:Matrix Equivalence",
"Definition:Zero Matrix"
] | [
"Equivalent Matrices have Equal Rank"
] |
proofwiki-1337 | Transpose of Matrix Product | Let $\mathbf A$ and $\mathbf B$ be matrices over a commutative ring such that $\mathbf A \mathbf B$ is defined.
Then $\mathbf B^\intercal \mathbf A^\intercal$ is defined, and:
:$\paren {\mathbf A \mathbf B}^\intercal = \mathbf B^\intercal \mathbf A^\intercal$
where $\mathbf X^\intercal$ is the transpose of $\mathbf X$. | Let $\mathbf A = \sqbrk a_{m n}$, $\mathbf B = \sqbrk b_{n p}$
Let $\mathbf A \mathbf B = \sqbrk c_{m p}$.
Then from the definition of matrix product:
:$\ds \forall i \in \closedint 1 m, j \in \closedint 1 p: c_{i j} = \sum_{k \mathop = 1}^n a_{i k} \circ b_{k j}$
So, let $\paren {\mathbf A \mathbf B}^\intercal = \sqbr... | Let $\mathbf A$ and $\mathbf B$ be [[Definition:Matrix|matrices]] over a [[Definition:Commutative Ring|commutative ring]] such that $\mathbf A \mathbf B$ is [[Definition:Matrix Product (Conventional)|defined]].
Then $\mathbf B^\intercal \mathbf A^\intercal$ is [[Definition:Matrix Product (Conventional)|defined]], and... | Let $\mathbf A = \sqbrk a_{m n}$, $\mathbf B = \sqbrk b_{n p}$
Let $\mathbf A \mathbf B = \sqbrk c_{m p}$.
Then from the definition of [[Definition:Matrix Product (Conventional)|matrix product]]:
:$\ds \forall i \in \closedint 1 m, j \in \closedint 1 p: c_{i j} = \sum_{k \mathop = 1}^n a_{i k} \circ b_{k j}$
So, le... | Transpose of Matrix Product | https://proofwiki.org/wiki/Transpose_of_Matrix_Product | https://proofwiki.org/wiki/Transpose_of_Matrix_Product | [
"Transposes of Matrices",
"Conventional Matrix Multiplication"
] | [
"Definition:Matrix",
"Definition:Commutative Ring",
"Definition:Matrix Product (Conventional)",
"Definition:Matrix Product (Conventional)",
"Definition:Transpose of Matrix"
] | [
"Definition:Matrix Product (Conventional)",
"Definition:Matrix/Underlying Structure",
"Definition:Commutative Ring"
] |
proofwiki-1338 | Rank is Dimension of Subspace | Let $K$ be a field.
Let $\mathbf A$ be an $m \times n$ matrix over $K$.
Then the rank of $\mathbf A$ is the dimension of the subspace of $K^n$ generated by the rows of $\mathbf A$. | Let $u: K^n \to K^m$ be the linear transformation such that $\mathbf A$ is the matrix of $u$ relative to the standard ordered bases of $K^n$ and $K^m$.
Let $\map \rho {\mathbf A}$ be the rank of $\mathbf A$.
Let $\mathbf A^\intercal$ be the transpose of $\mathbf A$.
Similar notations on $u$ denote the rank and transpos... | Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $\mathbf A$ be an [[Definition:Matrix|$m \times n$ matrix]] over $K$.
Then the [[Definition:Rank of Matrix|rank]] of $\mathbf A$ is the [[Definition:Dimension of Vector Space|dimension]] of the [[Definition:Vector Subspace|subspace]] of $K^n$ [[Definiti... | Let $u: K^n \to K^m$ be the [[Definition:Linear Transformation|linear transformation]] such that $\mathbf A$ is the [[Definition:Relative Matrix of Linear Transformation|matrix of $u$ relative to]] the [[Definition:Standard Ordered Basis|standard ordered bases]] of $K^n$ and $K^m$.
Let $\map \rho {\mathbf A}$ be the [... | Rank is Dimension of Subspace | https://proofwiki.org/wiki/Rank_is_Dimension_of_Subspace | https://proofwiki.org/wiki/Rank_is_Dimension_of_Subspace | [
"Rank of Matrix"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Matrix",
"Definition:Rank/Matrix",
"Definition:Dimension of Vector Space",
"Definition:Vector Subspace",
"Definition:Generator of Module",
"Definition:Row Matrix"
] | [
"Definition:Linear Transformation",
"Definition:Relative Matrix of Linear Transformation",
"Definition:Standard Ordered Basis",
"Definition:Rank/Matrix",
"Definition:Transpose of Matrix",
"Definition:Rank/Linear Transformation",
"Definition:Transpose of Linear Transformation",
"Rank and Nullity of Tra... |
proofwiki-1339 | General Linear Group is Group | Let $K$ be a field.
Let $\GL {n, K}$ be the general linear group of order $n$ over $K$.
Then $\GL {n, K}$ is a group. | Taking the group axioms in turn: | Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $\GL {n, K}$ be the [[Definition:General Linear Group|general linear group]] of [[Definition:Order of Square Matrix|order $n$]] over $K$.
Then $\GL {n, K}$ is a [[Definition:Group|group]]. | Taking the [[Axiom:Group Axioms|group axioms]] in turn: | General Linear Group is Group | https://proofwiki.org/wiki/General_Linear_Group_is_Group | https://proofwiki.org/wiki/General_Linear_Group_is_Group | [
"General Linear Group"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:General Linear Group",
"Definition:Matrix/Square Matrix/Order",
"Definition:Group"
] | [
"Axiom:Group Axioms"
] |
proofwiki-1340 | Transpose of Row Matrix is Column Matrix | Let $\mathbf x = \sqbrk x_{1 n} = \begin {bmatrix} x_1 & x_2 & \cdots & x_n \end {bmatrix}$ be a row matrix.
Then $\mathbf x^\intercal$, the transpose of $\mathbf x$, is a column matrix:
:$\begin {bmatrix} x_1 & x_2 & \cdots & x_n \end{bmatrix}^\intercal = \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix}$ | Self-evident.
{{Qed}} | Let $\mathbf x = \sqbrk x_{1 n} = \begin {bmatrix} x_1 & x_2 & \cdots & x_n \end {bmatrix}$ be a [[Definition:Row Matrix|row matrix]].
Then $\mathbf x^\intercal$, the [[Definition:Transpose of Matrix|transpose]] of $\mathbf x$, is a [[Definition:Column Matrix|column matrix]]:
:$\begin {bmatrix} x_1 & x_2 & \cdots & ... | Self-evident.
{{Qed}} | Transpose of Row Matrix is Column Matrix | https://proofwiki.org/wiki/Transpose_of_Row_Matrix_is_Column_Matrix | https://proofwiki.org/wiki/Transpose_of_Row_Matrix_is_Column_Matrix | [
"Transposes of Matrices",
"Row Matrices",
"column Matrices"
] | [
"Definition:Row Matrix",
"Definition:Transpose of Matrix",
"Definition:Column Matrix"
] | [] |
proofwiki-1341 | Transpose of Transpose of Matrix | Let $\mathbf A$ be a matrix.
Let $\mathbf A^\intercal$ be the transpose of $\mathbf A$.
Then:
:$\paren {\mathbf A^\intercal}^\intercal = \mathbf A$ | Follows directly from the definition of the transpose of a matrix.
{{Qed}} | Let $\mathbf A$ be a [[Definition:Matrix|matrix]].
Let $\mathbf A^\intercal$ be the [[Definition:Transpose of Matrix|transpose]] of $\mathbf A$.
Then:
:$\paren {\mathbf A^\intercal}^\intercal = \mathbf A$ | Follows directly from the definition of the [[Definition:Transpose of Matrix|transpose of a matrix]].
{{Qed}} | Transpose of Transpose of Matrix | https://proofwiki.org/wiki/Transpose_of_Transpose_of_Matrix | https://proofwiki.org/wiki/Transpose_of_Transpose_of_Matrix | [
"Transposes of Matrices"
] | [
"Definition:Matrix",
"Definition:Transpose of Matrix"
] | [
"Definition:Transpose of Matrix"
] |
proofwiki-1342 | Solution to Simultaneous Linear Equations | Let $\ds \forall i \in \closedint 1 m: \sum _{j \mathop = 1}^n {\alpha_{i j} x_j} = \beta_i$ be a system of simultaneous linear equations
where all of $\alpha_1, \ldots, a_n, x_1, \ldots x_n, \beta_i, \ldots, \beta_m$ are elements of a field $K$.
Then $x = \tuple {x_1, x_2, \ldots, x_n}$ is a solution of this system {{... | We can see the truth of this by writing them out in full.
:$\ds \sum_{j \mathop = 1}^n {\alpha_{i j} x_j} = \beta_i$
can be written as:
{{begin-eqn}}
{{eqn | l = \alpha_{1 1} x_1 + \alpha_{1 2} x_2 + \ldots + \alpha_{1 n} x_n
| r = \beta_1
| c =
}}
{{eqn | l = \alpha_{2 1} x_1 + \alpha_{2 2} x_2 + \ldots +... | Let $\ds \forall i \in \closedint 1 m: \sum _{j \mathop = 1}^n {\alpha_{i j} x_j} = \beta_i$ be a system of [[Definition:Simultaneous Linear Equations|simultaneous linear equations]]
where all of $\alpha_1, \ldots, a_n, x_1, \ldots x_n, \beta_i, \ldots, \beta_m$ are elements of a [[Definition:Field (Abstract Algebra)|... | We can see the truth of this by writing them out in full.
:$\ds \sum_{j \mathop = 1}^n {\alpha_{i j} x_j} = \beta_i$
can be written as:
{{begin-eqn}}
{{eqn | l = \alpha_{1 1} x_1 + \alpha_{1 2} x_2 + \ldots + \alpha_{1 n} x_n
| r = \beta_1
| c =
}}
{{eqn | l = \alpha_{2 1} x_1 + \alpha_{2 2} x_2 + \ldot... | Solution to Simultaneous Linear Equations | https://proofwiki.org/wiki/Solution_to_Simultaneous_Linear_Equations | https://proofwiki.org/wiki/Solution_to_Simultaneous_Linear_Equations | [
"Simultaneous Linear Equations"
] | [
"Definition:Simultaneous Equations/Linear Equations",
"Definition:Field (Abstract Algebra)",
"Definition:Simultaneous Equations/Solution",
"Definition:Matrix"
] | [
"Definition:Simultaneous Equations/Linear Equations",
"Definition:Matrix Space"
] |
proofwiki-1343 | Infinite Cyclic Group is Isomorphic to Integers | Let $G$ be an infinite cyclic group.
Then $G$ is isomorphic to the additive group of integers: $G \cong \struct {\Z, +}$. | From the definition of an infinite cyclic group, we have:
:$G = \gen a = \set {a^k: k \in \Z}$
Let us define the mapping:
:$\phi: \Z \to G: \map \phi k = a^k$.
We now show that $\phi$ is an isomorphism.
From Mapping from Additive Group of Integers to Powers of Group Element is Homomorphism, $\phi$ is a homomorphism.
No... | Let $G$ be an [[Definition:Infinite Cyclic Group|infinite cyclic group]].
Then $G$ is [[Definition:Group Isomorphism|isomorphic]] to the [[Definition:Additive Group of Integers|additive group of integers]]: $G \cong \struct {\Z, +}$. | From the definition of an [[Definition:Infinite Cyclic Group|infinite cyclic group]], we have:
:$G = \gen a = \set {a^k: k \in \Z}$
Let us define the [[Definition:Mapping|mapping]]:
:$\phi: \Z \to G: \map \phi k = a^k$.
We now show that $\phi$ is an [[Definition:Group Isomorphism|isomorphism]].
From [[Mapping from ... | Infinite Cyclic Group is Isomorphic to Integers | https://proofwiki.org/wiki/Infinite_Cyclic_Group_is_Isomorphic_to_Integers | https://proofwiki.org/wiki/Infinite_Cyclic_Group_is_Isomorphic_to_Integers | [
"Infinite Cyclic Group",
"Additive Group of Integers",
"Examples of Group Isomorphisms"
] | [
"Definition:Infinite Cyclic Group",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Additive Group of Integers"
] | [
"Definition:Infinite Cyclic Group",
"Definition:Mapping",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Mapping from Additive Group of Integers to Powers of Group Element is Homomorphism",
"Definition:Group Homomorphism",
"Definition:Surjection",
"Definition:Cyclic Group",
"Definitio... |
proofwiki-1344 | Subgroup of Infinite Cyclic Group is Infinite Cyclic Group | Let $G = \gen a$ be an infinite cyclic group generated by $a$, whose identity is $e$.
Let $g \in G, g \ne e: \exists k \in \Z, k \ne 0: g = a^k$.
Let $H = \gen g$.
Then $H \le G$ and $H \cong G$.
Thus, all non-trivial subgroups of an infinite cyclic group are themselves infinite cyclic groups.
A subgroup of $G = \gen a... | The fact that $H \le G$ follows from the definition of subgroup generator.
By Infinite Cyclic Group is Isomorphic to Integers:
:$G \cong \struct {\Z, +}$
Now we show that $H$ is of infinite order.
Suppose $\exists h \in H, h \ne e: \exists r \in \Z, r > 0: h^r = e$.
But:
:$h \in H \implies \exists s \in \Z, s > 0: h = ... | Let $G = \gen a$ be an [[Definition:Infinite Cyclic Group|infinite cyclic group]] [[Definition:Generator of Cyclic Group|generated]] by $a$, whose [[Definition:Identity Element|identity]] is $e$.
Let $g \in G, g \ne e: \exists k \in \Z, k \ne 0: g = a^k$.
Let $H = \gen g$.
Then $H \le G$ and $H \cong G$.
Thus, al... | The fact that $H \le G$ follows from the definition of [[Definition:Generator of Subgroup|subgroup generator]].
By [[Infinite Cyclic Group is Isomorphic to Integers]]:
:$G \cong \struct {\Z, +}$
Now we show that $H$ is of [[Definition:Infinite Group|infinite order]].
Suppose $\exists h \in H, h \ne e: \exists r \in... | Subgroup of Infinite Cyclic Group is Infinite Cyclic Group | https://proofwiki.org/wiki/Subgroup_of_Infinite_Cyclic_Group_is_Infinite_Cyclic_Group | https://proofwiki.org/wiki/Subgroup_of_Infinite_Cyclic_Group_is_Infinite_Cyclic_Group | [
"Subgroups",
"Infinite Cyclic Group"
] | [
"Definition:Infinite Cyclic Group",
"Definition:Cyclic Group/Generator",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Non-Trivial Subgroup",
"Definition:Subgroup",
"Definition:Infinite Cyclic Group",
"Definition:Infinite Cyclic Group",
"Definition:Subgroup",
"Subgroups of... | [
"Definition:Generator of Subgroup",
"Infinite Cyclic Group is Isomorphic to Integers",
"Definition:Infinite Group",
"Definition:Order of Group Element/Finite",
"Definition:Finite Group",
"Definition:Infinite Group",
"Subgroup of Cyclic Group is Cyclic",
"Definition:Cyclic Group",
"Definition:Cyclic ... |
proofwiki-1345 | Quotient Group of Infinite Cyclic Group by Subgroup | Let $C_n$ be the cyclic group of order $n$.
Then:
:$C_n \cong \dfrac {\struct {\Z, +} } {\struct {n \Z, +} } = \dfrac \Z {n \Z}$
where:
:$\Z$ is the additive group of integers
:$n \Z$ is the additive group of integer multiples
:$\Z / n \Z$ is the quotient group of $\Z$ by $n \Z$.
Thus, every cyclic group is isomorphic ... | Let $C_n = \gen {a: a^n = e_{C_n} }$, that is, let $a$ be a generator of $C_n$.
Let us define $\phi: \struct {\Z, +} \to C_n$ such that:
:$\forall k \in \Z: \map \phi k = a^k$
Then from the First Isomorphism Theorem:
:$\Img \phi = C_n = \struct {\Z, +} / \map \ker \phi$
We now need to show that $\map \ker \phi = n \Z$.... | Let $C_n$ be the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Group|order $n$]].
Then:
:$C_n \cong \dfrac {\struct {\Z, +} } {\struct {n \Z, +} } = \dfrac \Z {n \Z}$
where:
:$\Z$ is the [[Definition:Additive Group of Integers|additive group of integers]]
:$n \Z$ is the [[Definition:Additive Group... | Let $C_n = \gen {a: a^n = e_{C_n} }$, that is, let $a$ be a [[Definition:Generator of Cyclic Group|generator]] of $C_n$.
Let us define $\phi: \struct {\Z, +} \to C_n$ such that:
:$\forall k \in \Z: \map \phi k = a^k$
Then from the [[First Isomorphism Theorem for Groups|First Isomorphism Theorem]]:
:$\Img \phi = C_n =... | Quotient Group of Infinite Cyclic Group by Subgroup | https://proofwiki.org/wiki/Quotient_Group_of_Infinite_Cyclic_Group_by_Subgroup | https://proofwiki.org/wiki/Quotient_Group_of_Infinite_Cyclic_Group_by_Subgroup | [
"Infinite Cyclic Group",
"Quotient Groups",
"Additive Groups of Integer Multiples",
"Additive Group of Integers"
] | [
"Definition:Cyclic Group",
"Definition:Order of Structure",
"Definition:Additive Group of Integers",
"Definition:Additive Group of Integer Multiples",
"Definition:Quotient Group",
"Definition:Cyclic Group",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] | [
"Definition:Cyclic Group/Generator",
"First Isomorphism Theorem/Groups"
] |
proofwiki-1346 | Property of Being an Ideal is not Transitive | Let $J_1$ be an ideal of a ring $R$.
Let $J_2$ be an ideal of $J_1$.
Then $J_2$ need not necessarily be an ideal of $R$. | Let $R = \Q \sqbrk X$ be the ring of polynomials in one variable $X$ over $\Q$.
Let:
:$J_1 = \set {a_0 + a_1 X + \cdots + a_n X^n \in R : a_0 = a_1 = 0}$
and
:$J_2 = \set {a_0 + a_1 X + \cdots + a_n X^n \in R : a_0 = a_1 = a_3 = 0}$
First let us show that $J_1$ is an ideal of $R$.
We establish the properties of the ide... | Let $J_1$ be an [[Definition:Ideal of Ring|ideal]] of a [[Definition:Ring (Abstract Algebra)|ring]] $R$.
Let $J_2$ be an [[Definition:Ideal of Ring|ideal]] of $J_1$.
Then $J_2$ need not necessarily be an [[Definition:Ideal of Ring|ideal]] of $R$. | Let $R = \Q \sqbrk X$ be the [[Definition:Polynomial Ring in One Variable|ring of polynomials in one variable]] $X$ over $\Q$.
Let:
:$J_1 = \set {a_0 + a_1 X + \cdots + a_n X^n \in R : a_0 = a_1 = 0}$
and
:$J_2 = \set {a_0 + a_1 X + \cdots + a_n X^n \in R : a_0 = a_1 = a_3 = 0}$
First let us show that $J_1$ is an id... | Property of Being an Ideal is not Transitive | https://proofwiki.org/wiki/Property_of_Being_an_Ideal_is_not_Transitive | https://proofwiki.org/wiki/Property_of_Being_an_Ideal_is_not_Transitive | [
"Ideal Theory"
] | [
"Definition:Ideal of Ring",
"Definition:Ring (Abstract Algebra)",
"Definition:Ideal of Ring",
"Definition:Ideal of Ring"
] | [
"Definition:Polynomial Ring",
"Test for Ideal",
"Definition:Polynomial Addition/Polynomial Forms",
"Definition:Multiplication of Polynomials",
"Definition:Polynomial Addition/Polynomial Forms",
"Definition:Multiplication of Polynomials",
"Category:Ideal Theory"
] |
proofwiki-1347 | Ideals Containing Ideal Form Lattice | Let $J$ be an ideal of a ring $R$.
Let $\mathbb L_J$ be the set of all ideals of $R$ which contain $J$.
Then the ordered set $\struct {\mathbb L_J, \subseteq}$ is a lattice. | Let $b_1, b_2 \in \mathbb L_J$.
Then from Set of Ideals forms Complete Lattice:
:$(1): \quad b_1 + b_2 \in \mathbb L_J$ and is the supremum of $\set {b_1, b_2}$
:$(2): \quad b_1 \cap b_2 \in \mathbb L_J$ and is the infimum of $\set {b_1, b_2}$
Thus $\struct {\mathbb L_J, \subseteq}$ is a lattice.
{{qed}} | Let $J$ be an [[Definition:Ideal of Ring|ideal]] of a [[Definition:Ring (Abstract Algebra)|ring]] $R$.
Let $\mathbb L_J$ be the set of all [[Definition:Ideal of Ring|ideals]] of $R$ which contain $J$.
Then the [[Definition:Ordered Set|ordered set]] $\struct {\mathbb L_J, \subseteq}$ is a [[Definition:Lattice (Order ... | Let $b_1, b_2 \in \mathbb L_J$.
Then from [[Set of Ideals forms Complete Lattice]]:
:$(1): \quad b_1 + b_2 \in \mathbb L_J$ and is the [[Definition:Supremum of Set|supremum]] of $\set {b_1, b_2}$
:$(2): \quad b_1 \cap b_2 \in \mathbb L_J$ and is the [[Definition:Infimum of Set|infimum]] of $\set {b_1, b_2}$
Thus $\... | Ideals Containing Ideal Form Lattice | https://proofwiki.org/wiki/Ideals_Containing_Ideal_Form_Lattice | https://proofwiki.org/wiki/Ideals_Containing_Ideal_Form_Lattice | [
"Ideal Theory",
"Lattice Theory"
] | [
"Definition:Ideal of Ring",
"Definition:Ring (Abstract Algebra)",
"Definition:Ideal of Ring",
"Definition:Ordered Set",
"Definition:Lattice (Order Theory)"
] | [
"Set of Ideals forms Complete Lattice",
"Definition:Supremum of Set",
"Definition:Infimum of Set",
"Definition:Lattice (Order Theory)"
] |
proofwiki-1348 | Ideals Containing Ideal Isomorphic to Quotient Ring | Let $J$ be an ideal of a ring $R$.
Let $\mathbb L_J$ be the set of all ideals of $R$ which contain $J$.
Let the ordered set $\struct {\map {\mathbb L} {R / J}, \subseteq}$ be the set of all ideals of $R / J$.
Let the mapping $\Phi_J: \struct {\mathbb L_J, \subseteq} \to \struct {\map {\mathbb L} {R / J}, \subseteq}$ be... | Let $b \in \mathbb L_J$.
From the way $\mathbb L_J$ is defined:
:$J \subseteq b$
Thus by Preimage of Image of Subring under Ring Homomorphism:
:$\map {q_J^{-1} } {\map {q_J} b} = b + J = b$
Let $c$ be an ideal of $R / J$.
Then, by Image of Preimage of Subring under Ring Epimorphism:
:$\map {q_J} {\map {q_J^{-1} } c} = ... | Let $J$ be an [[Definition:Ideal of Ring|ideal]] of a [[Definition:Ring (Abstract Algebra)|ring]] $R$.
Let $\mathbb L_J$ be the [[Definition:Set|set]] of all [[Definition:Ideal of Ring|ideals]] of $R$ which contain $J$.
Let the [[Definition:Ordered Set|ordered set]] $\struct {\map {\mathbb L} {R / J}, \subseteq}$ be ... | Let $b \in \mathbb L_J$.
From the way $\mathbb L_J$ is defined:
:$J \subseteq b$
Thus by [[Preimage of Image of Subring under Ring Homomorphism]]:
:$\map {q_J^{-1} } {\map {q_J} b} = b + J = b$
Let $c$ be an ideal of $R / J$.
Then, by [[Image of Preimage of Subring under Ring Epimorphism]]:
:$\map {q_J} {\map {q_J^... | Ideals Containing Ideal Isomorphic to Quotient Ring | https://proofwiki.org/wiki/Ideals_Containing_Ideal_Isomorphic_to_Quotient_Ring | https://proofwiki.org/wiki/Ideals_Containing_Ideal_Isomorphic_to_Quotient_Ring | [
"Ideal Theory",
"Quotient Rings"
] | [
"Definition:Ideal of Ring",
"Definition:Ring (Abstract Algebra)",
"Definition:Set",
"Definition:Ideal of Ring",
"Definition:Ordered Set",
"Definition:Set",
"Definition:Ideal of Ring",
"Definition:Mapping",
"Definition:Quotient Epimorphism",
"Definition:Quotient Ring",
"Definition:Isomorphism (Ab... | [
"Preimage of Image of Subring under Ring Homomorphism",
"Image of Preimage of Subring under Ring Epimorphism",
"Bijection iff Left and Right Inverse",
"Definition:Bijection",
"Definition:Isomorphism (Abstract Algebra)/Ordered Structure Isomorphism",
"Image of Subset under Mapping is Subset of Image",
"D... |
proofwiki-1349 | Ring of Integers is Principal Ideal Domain | The integers $\Z$ form a principal ideal domain. | Let $J$ be an ideal of $\Z$.
Then $J$ is a subring of $\Z$, and so $\left({J, +}\right)$ is a subgroup of $\left({\Z, +}\right)$.
But by Integers under Addition form Infinite Cyclic Group, the group $\left({\Z, +}\right)$ is cyclic, generated by $1$.
Thus by Subgroup of Cyclic Group is Cyclic, $\left({J, +}\right)$ is ... | The [[Definition:Integer|integers]] $\Z$ form a [[Definition:Principal Ideal Domain|principal ideal domain]]. | Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $\Z$.
Then $J$ is a [[Definition:Subring|subring]] of $\Z$, and so $\left({J, +}\right)$ is a [[Definition:Subgroup|subgroup]] of $\left({\Z, +}\right)$.
But by [[Integers under Addition form Infinite Cyclic Group]], the group $\left({\Z, +}\right)$ is [[Definition:... | Ring of Integers is Principal Ideal Domain/Proof 1 | https://proofwiki.org/wiki/Ring_of_Integers_is_Principal_Ideal_Domain | https://proofwiki.org/wiki/Ring_of_Integers_is_Principal_Ideal_Domain/Proof_1 | [
"Integers",
"Examples of Principal Ideal Domains",
"Ring of Integers is Principal Ideal Domain"
] | [
"Definition:Integer",
"Definition:Principal Ideal Domain"
] | [
"Definition:Ideal of Ring",
"Definition:Subring",
"Definition:Subgroup",
"Integers under Addition form Infinite Cyclic Group",
"Definition:Cyclic Group",
"Subgroup of Cyclic Group is Cyclic",
"Definition:Principal Ideal of Ring",
"Definition:Principal Ideal of Ring"
] |
proofwiki-1350 | Ring of Integers is Principal Ideal Domain | The integers $\Z$ form a principal ideal domain. | We have that Integers are Euclidean Domain.
Then we have that Euclidean Domain is Principal Ideal Domain.
Hence the result.
{{qed}} | The [[Definition:Integer|integers]] $\Z$ form a [[Definition:Principal Ideal Domain|principal ideal domain]]. | We have that [[Integers are Euclidean Domain]].
Then we have that [[Euclidean Domain is Principal Ideal Domain]].
Hence the result.
{{qed}} | Ring of Integers is Principal Ideal Domain/Proof 2 | https://proofwiki.org/wiki/Ring_of_Integers_is_Principal_Ideal_Domain | https://proofwiki.org/wiki/Ring_of_Integers_is_Principal_Ideal_Domain/Proof_2 | [
"Integers",
"Examples of Principal Ideal Domains",
"Ring of Integers is Principal Ideal Domain"
] | [
"Definition:Integer",
"Definition:Principal Ideal Domain"
] | [
"Integers are Euclidean Domain",
"Euclidean Domain is Principal Ideal Domain"
] |
proofwiki-1351 | Ring of Integers is Principal Ideal Domain | The integers $\Z$ form a principal ideal domain. | Let $U$ be an arbitrary ideal of $\Z$.
Let $c$ be a non-zero element of $U$.
Then both $c$ and $-c$ belong to $\ideal a$ and one of them is positive.
Thus $U$ contains strictly positive elements.
Let $b$ be the smallest strictly positive element of $U$.
By the Set of Integers Bounded Below by Integer has Smallest Eleme... | The [[Definition:Integer|integers]] $\Z$ form a [[Definition:Principal Ideal Domain|principal ideal domain]]. | Let $U$ be an arbitrary [[Definition:Ideal of Ring|ideal]] of $\Z$.
Let $c$ be a non-[[Definition:Ring Zero|zero]] [[Definition:Element|element]] of $U$.
Then both $c$ and $-c$ belong to $\ideal a$ and one of them is [[Definition:Positive Integer|positive]].
Thus $U$ contains [[Definition:Strictly Positive Integer|s... | Ring of Integers is Principal Ideal Domain/Proof 3 | https://proofwiki.org/wiki/Ring_of_Integers_is_Principal_Ideal_Domain | https://proofwiki.org/wiki/Ring_of_Integers_is_Principal_Ideal_Domain/Proof_3 | [
"Integers",
"Examples of Principal Ideal Domains",
"Ring of Integers is Principal Ideal Domain"
] | [
"Definition:Integer",
"Definition:Principal Ideal Domain"
] | [
"Definition:Ideal of Ring",
"Definition:Ring Zero",
"Definition:Element",
"Definition:Positive/Integer",
"Definition:Strictly Positive/Integer",
"Definition:Smallest Element",
"Definition:Strictly Positive/Integer",
"Set of Integers Bounded Below by Integer has Smallest Element",
"Definition:Generat... |
proofwiki-1352 | Principal Ideals of Integers | Let $J$ be a non-zero ideal of $\Z$.
Then $J = \ideal b$ where $b$ is the smallest strictly positive integer belonging to $J$. | It follows from Ring of Integers is Principal Ideal Domain that $J$ is a principal ideal.
Let $c \in J, c \ne 0$.
Then $-c \in J$ and by Natural Numbers are Non-Negative Integers, exactly one of them is strictly positive.
Thus $J$ ''does'' actually contain strictly positive elements, so that's a start.
Let $b$ be the s... | Let $J$ be a [[Definition:Non-Null Ideal|non-zero]] [[Definition:Ideal of Ring|ideal]] of $\Z$.
Then $J = \ideal b$ where $b$ is the smallest [[Definition:Strictly Positive|strictly positive]] [[Definition:Integer|integer]] belonging to $J$. | It follows from [[Ring of Integers is Principal Ideal Domain]] that $J$ is a [[Definition:Principal Ideal of Ring|principal ideal]].
Let $c \in J, c \ne 0$.
Then $-c \in J$ and by [[Natural Numbers are Non-Negative Integers]], exactly one of them is strictly positive.
Thus $J$ ''does'' actually contain strictly pos... | Principal Ideals of Integers | https://proofwiki.org/wiki/Principal_Ideals_of_Integers | https://proofwiki.org/wiki/Principal_Ideals_of_Integers | [
"Ideal Theory"
] | [
"Definition:Non-Null Ideal",
"Definition:Ideal of Ring",
"Definition:Strictly Positive",
"Definition:Integer"
] | [
"Ring of Integers is Principal Ideal Domain",
"Definition:Principal Ideal of Ring",
"Natural Numbers are Non-Negative Integers",
"Natural Numbers are Non-Negative Integers",
"Well-Ordering Principle",
"Definition:Principal Ideal of Ring",
"Division Theorem"
] |
proofwiki-1353 | Natural Numbers Set Equivalent to Ideals of Integers | Let $S$ be the set of all ideals of $\Z$.
Let the mapping $\psi: \N \to S$ be defined as:
:$\forall b \in \N: \map \psi b = \ideal b$
where $\ideal b$ is the principal ideal of $\Z$ generated by $b$.
Then $\psi$ is a bijection. | First we show that $\psi$ is injective.
Suppose $0 < b < c$.
Then $b \in \ideal b$, but $b \notin \ideal c$, because from Principal Ideals of Integers, $c$ is the smallest positive integer in $\ideal c$.
Thus $\ideal b \ne \ideal c$.
It is also apparent that $b > 0 \implies \ideal b \ne \ideal 0$ as $\ideal 0 = \set 0$... | Let $S$ be the set of all [[Definition:Ideal of Ring|ideals]] of $\Z$.
Let the [[Definition:Mapping|mapping]] $\psi: \N \to S$ be defined as:
:$\forall b \in \N: \map \psi b = \ideal b$
where $\ideal b$ is the [[Definition:Principal Ideal of Ring|principal ideal]] of $\Z$ generated by $b$.
Then $\psi$ is a [[Definit... | First we show that $\psi$ is [[Definition:Injection|injective]].
Suppose $0 < b < c$.
Then $b \in \ideal b$, but $b \notin \ideal c$, because from [[Principal Ideals of Integers]], $c$ is the smallest [[Definition:Positive Integer|positive integer]] in $\ideal c$.
Thus $\ideal b \ne \ideal c$.
It is also apparent t... | Natural Numbers Set Equivalent to Ideals of Integers | https://proofwiki.org/wiki/Natural_Numbers_Set_Equivalent_to_Ideals_of_Integers | https://proofwiki.org/wiki/Natural_Numbers_Set_Equivalent_to_Ideals_of_Integers | [
"Ideal Theory",
"Integers",
"Natural Numbers"
] | [
"Definition:Ideal of Ring",
"Definition:Mapping",
"Definition:Principal Ideal of Ring",
"Definition:Bijection"
] | [
"Definition:Injection",
"Principal Ideals of Integers",
"Definition:Positive/Integer",
"Definition:Injection",
"Principal Ideals of Integers",
"Definition:Principal Ideal of Ring"
] |
proofwiki-1354 | Quotient Epimorphism from Integers by Principal Ideal | Let $m$ be a strictly positive integer.
Let $\ideal m$ be the principal ideal of $\Z$ generated by $m$.
The restriction to $\N_m$ of the quotient epimorphism $q_m$ from the ring $\struct {\Z, +, \times}$ onto $\struct {\Z, +, \times} / \ideal m$ is an isomorphism from the ring $\struct {\N_m, +_m, \times_m}$ of integer... | Let $x, y \in \N_m$.
By the Division Theorem:
{{begin-eqn}}
{{eqn | q = \exists q, r \in \Z
| l = x + y
| r = m q + r
| c = for $0 \le r < m$
}}
{{eqn | q = \exists p, s \in \Z
| l = x y
| r = m p + s
| c = for $0 \le s < m$
}}
{{end-eqn}}
Then $x +_m y = r$ and $x \times_m y = s$, s... | Let $m$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
Let $\ideal m$ be the [[Definition:Principal Ideal of Ring|principal ideal]] of $\Z$ generated by $m$.
The [[Definition:Restriction of Mapping|restriction]] to $\N_m$ of the [[Definition:Quotient Epimorphism|quotient epimorphism]] $q_m$... | Let $x, y \in \N_m$.
By the [[Division Theorem]]:
{{begin-eqn}}
{{eqn | q = \exists q, r \in \Z
| l = x + y
| r = m q + r
| c = for $0 \le r < m$
}}
{{eqn | q = \exists p, s \in \Z
| l = x y
| r = m p + s
| c = for $0 \le s < m$
}}
{{end-eqn}}
Then $x +_m y = r$ and $x \times_m y... | Quotient Epimorphism from Integers by Principal Ideal | https://proofwiki.org/wiki/Quotient_Epimorphism_from_Integers_by_Principal_Ideal | https://proofwiki.org/wiki/Quotient_Epimorphism_from_Integers_by_Principal_Ideal | [
"Modulo Arithmetic",
"Ideal Theory",
"Quotient Epimorphisms"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Principal Ideal of Ring",
"Definition:Restriction/Mapping",
"Definition:Quotient Epimorphism",
"Definition:Ring (Abstract Algebra)",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Ring of Integers Modulo m",
"Definition:Quotient Ring",
"D... | [
"Division Theorem",
"Definition:Restriction/Mapping",
"Definition:Ring Homomorphism",
"Definition:Restriction/Mapping",
"Definition:Surjection",
"Definition:Kernel of Ring Homomorphism",
"Definition:Restriction/Mapping",
"Quotient Theorem for Group Epimorphisms",
"Definition:Restriction/Mapping",
... |
proofwiki-1355 | Integer Divisor is Equivalent to Subset of Ideal | Let $\Z$ be the set of all integers.
Let $\Z_{>0}$ be the set of strictly positive integers.
Let $m \in \Z_{>0}$ and let $n \in \Z$.
Let $\ideal m$ be the principal ideal of $\Z$ generated by $m$.
Then:
:$m \divides n \iff \ideal n \subseteq \ideal m$ | The ring of integers is a principal ideal domain.
The result follows directly from Principal Ideals in Integral Domain.
{{qed}} | Let $\Z$ be the set of all [[Definition:Integer|integers]].
Let $\Z_{>0}$ be the set of [[Definition:Strictly Positive Integer|strictly positive integers]].
Let $m \in \Z_{>0}$ and let $n \in \Z$.
Let $\ideal m$ be the [[Definition:Principal Ideal of Ring|principal ideal]] of $\Z$ generated by $m$.
Then:
:$m \divi... | The [[Ring of Integers is Principal Ideal Domain|ring of integers is a principal ideal domain]].
The result follows directly from [[Principal Ideals in Integral Domain]].
{{qed}} | Integer Divisor is Equivalent to Subset of Ideal | https://proofwiki.org/wiki/Integer_Divisor_is_Equivalent_to_Subset_of_Ideal | https://proofwiki.org/wiki/Integer_Divisor_is_Equivalent_to_Subset_of_Ideal | [
"Integers",
"Ideal Theory"
] | [
"Definition:Integer",
"Definition:Strictly Positive/Integer",
"Definition:Principal Ideal of Ring"
] | [
"Ring of Integers is Principal Ideal Domain",
"Principal Ideals in Integral Domain"
] |
proofwiki-1356 | Principal Ideal Domain is Unique Factorization Domain | Every principal ideal domain is a unique factorization domain. | Let $R$ be a principal ideal domain whose zero is $0_R$.
Let $a \in R$ be an arbitrary element of $R$ which is neither $0_R$ nor a unit of $R$.
By Element of Principal Ideal Domain is Finite Product of Irreducible Elements, $a$ has a factorization into finitely many irreducible elements.
Note that this uses the Axiom o... | Every [[Definition:Principal Ideal Domain|principal ideal domain]] is a [[Definition:Unique Factorization Domain|unique factorization domain]]. | Let $R$ be a [[Definition:Principal Ideal Domain|principal ideal domain]] whose [[Definition:Ring Zero|zero]] is $0_R$.
Let $a \in R$ be an [[Definition:Arbitrary|arbitrary]] [[Definition:Element|element]] of $R$ which is neither $0_R$ nor a [[Definition:Unit of Ring|unit]] of $R$.
By [[Element of Principal Ideal Dom... | Principal Ideal Domain is Unique Factorization Domain | https://proofwiki.org/wiki/Principal_Ideal_Domain_is_Unique_Factorization_Domain | https://proofwiki.org/wiki/Principal_Ideal_Domain_is_Unique_Factorization_Domain | [
"Integral Domains",
"Principal Ideal Domains",
"Unique Factorization Domains",
"Ideal Theory",
"Factorization"
] | [
"Definition:Principal Ideal Domain",
"Definition:Unique Factorization Domain"
] | [
"Definition:Principal Ideal Domain",
"Definition:Ring Zero",
"Definition:Arbitrary",
"Definition:Element",
"Definition:Unit of Ring",
"Element of Principal Ideal Domain is Finite Product of Irreducible Elements/Proof 2",
"Definition:Divisor (Algebra)/Factorization",
"Definition:Finite Set",
"Definit... |
proofwiki-1357 | Maximal Ideal iff Quotient Ring is Field | Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $J$ be an ideal of $R$.
The following are equivalent:
:$(1): \quad$ $J$ is a maximal ideal.
:$(2): \quad$ The quotient ring $R / J$ is a field. | === Maximal Ideal implies Quotient Ring is Field ===
{{:Maximal Ideal iff Quotient Ring is Field/Proof 1/Maximal Ideal implies Quotient Ring is Field}}{{qed|lemma}}
=== Quotient Ring is Field implies Ideal is Maximal ===
{{:Maximal Ideal iff Quotient Ring is Field/Proof 1/Quotient Ring is Field implies Ideal is Maximal... | 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 $1_R$.
Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
The following are [[Definition:Logically Equivalent|equ... | === [[Maximal Ideal iff Quotient Ring is Field/Proof 1/Maximal Ideal implies Quotient Ring is Field|Maximal Ideal implies Quotient Ring is Field]] ===
{{:Maximal Ideal iff Quotient Ring is Field/Proof 1/Maximal Ideal implies Quotient Ring is Field}}{{qed|lemma}}
=== [[Maximal Ideal iff Quotient Ring is Field/Proof 1/... | Maximal Ideal iff Quotient Ring is Field/Proof 1 | https://proofwiki.org/wiki/Maximal_Ideal_iff_Quotient_Ring_is_Field | https://proofwiki.org/wiki/Maximal_Ideal_iff_Quotient_Ring_is_Field/Proof_1 | [
"Quotient Rings",
"Maximal Ideal iff Quotient Ring is Field",
"Maximal Ideals of Rings",
"Field Theory"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Ideal of Ring",
"Definition:Logical Equivalence",
"Definition:Maximal Ideal of Ring",
"Definition:Quotient Ring",
"Definition:Field (Abstract Algebra)"
] | [
"Maximal Ideal iff Quotient Ring is Field/Proof 1/Maximal Ideal implies Quotient Ring is Field",
"Maximal Ideal iff Quotient Ring is Field/Proof 1/Quotient Ring is Field implies Ideal is Maximal"
] |
proofwiki-1358 | Maximal Ideal iff Quotient Ring is Field | Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $J$ be an ideal of $R$.
The following are equivalent:
:$(1): \quad$ $J$ is a maximal ideal.
:$(2): \quad$ The quotient ring $R / J$ is a field. | Let $\mathbb L_J$ be the set of all ideals of $R$ which contain $J$.
Let the ordered set $\struct {\map {\mathbb L} {R / J}, \subseteq}$ be the set of all ideals of $R / J$.
Let the mapping $\Phi_J: \struct {\mathbb L_J, \subseteq} \to \struct {\map {\mathbb L} {R / J}, \subseteq}$ be defined as:
:$\forall a \in \mathb... | 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 $1_R$.
Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
The following are [[Definition:Logically Equivalent|equ... | Let $\mathbb L_J$ be the set of all [[Definition:Ideal of Ring|ideals]] of $R$ which contain $J$.
Let the [[Definition:Ordered Set|ordered set]] $\struct {\map {\mathbb L} {R / J}, \subseteq}$ be the [[Definition:Set|set]] of all [[Definition:Ideal of Ring|ideals]] of $R / J$.
Let the [[Definition:Mapping|mapping]] ... | Maximal Ideal iff Quotient Ring is Field/Proof 2 | https://proofwiki.org/wiki/Maximal_Ideal_iff_Quotient_Ring_is_Field | https://proofwiki.org/wiki/Maximal_Ideal_iff_Quotient_Ring_is_Field/Proof_2 | [
"Quotient Rings",
"Maximal Ideal iff Quotient Ring is Field",
"Maximal Ideals of Rings",
"Field Theory"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Ideal of Ring",
"Definition:Logical Equivalence",
"Definition:Maximal Ideal of Ring",
"Definition:Quotient Ring",
"Definition:Field (Abstract Algebra)"
] | [
"Definition:Ideal of Ring",
"Definition:Ordered Set",
"Definition:Set",
"Definition:Ideal of Ring",
"Definition:Mapping",
"Definition:Quotient Epimorphism/Ring",
"Ideals Containing Ideal Isomorphic to Quotient Ring",
"Definition:Isomorphism (Abstract Algebra)",
"Quotient Ring Defined by Ring Itself ... |
proofwiki-1359 | Maximal Ideal iff Quotient Ring is Field | Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $J$ be an ideal of $R$.
The following are equivalent:
:$(1): \quad$ $J$ is a maximal ideal.
:$(2): \quad$ The quotient ring $R / J$ is a field. | Let $J$ be a maximal ideal.
We have by definition of quotient ring that $J$ is the zero element of $R / J$.
Let $A \in R / J$ be a non-zero element of $R / J$.
Let $x \in A$.
Since $A \ne J$, we have that $x \notin J$.
Let the ideal $K = J + A$ of $R$ be formed.
This contains all the elements of the form $j + r a$, wit... | 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 $1_R$.
Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
The following are [[Definition:Logically Equivalent|equ... | Let $J$ be a [[Definition:Maximal Ideal of Ring|maximal ideal]].
We have by definition of [[Definition:Quotient Ring|quotient ring]] that $J$ is the [[Definition:Ring Zero|zero element]] of $R / J$.
Let $A \in R / J$ be a non-[[Definition:Ring Zero|zero element]] of $R / J$.
Let $x \in A$.
Since $A \ne J$, we have ... | Maximal Ideal iff Quotient Ring is Field/Proof 3 | https://proofwiki.org/wiki/Maximal_Ideal_iff_Quotient_Ring_is_Field | https://proofwiki.org/wiki/Maximal_Ideal_iff_Quotient_Ring_is_Field/Proof_3 | [
"Quotient Rings",
"Maximal Ideal iff Quotient Ring is Field",
"Maximal Ideals of Rings",
"Field Theory"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Ideal of Ring",
"Definition:Logical Equivalence",
"Definition:Maximal Ideal of Ring",
"Definition:Quotient Ring",
"Definition:Field (Abstract Algebra)"
] | [
"Definition:Maximal Ideal of Ring",
"Definition:Quotient Ring",
"Definition:Ring Zero",
"Definition:Ring Zero",
"Definition:Ideal of Ring",
"Definition:Element",
"Definition:Maximal Ideal of Ring",
"Definition:Product Inverse",
"Definition:Ring Zero",
"Definition:Product Inverse",
"Definition:Fi... |
proofwiki-1360 | Prime Number iff Generates Principal Maximal Ideal | Let $\Z_{>0}$ be the set of strictly positive integers.
Let $p \in \Z_{>0}$.
Let $\ideal p$ be the principal ideal of $\Z$ generated by $p$.
Then $p$ is prime {{iff}} $\ideal p$ is a maximal ideal of $\Z$. | First, note that $\Z$ is a principal ideal domain, so all ideals are principal. | Let $\Z_{>0}$ be the set of [[Definition:Strictly Positive Integer|strictly positive integers]].
Let $p \in \Z_{>0}$.
Let $\ideal p$ be the [[Definition:Principal Ideal of Ring|principal ideal]] of $\Z$ generated by $p$.
Then $p$ is [[Definition:Prime Number|prime]] {{iff}} $\ideal p$ is a [[Definition:Maximal Ide... | First, note that [[Ring of Integers is Principal Ideal Domain|$\Z$ is a principal ideal domain]], so all ideals are principal. | Prime Number iff Generates Principal Maximal Ideal | https://proofwiki.org/wiki/Prime_Number_iff_Generates_Principal_Maximal_Ideal | https://proofwiki.org/wiki/Prime_Number_iff_Generates_Principal_Maximal_Ideal | [
"Ideal Theory"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Principal Ideal of Ring",
"Definition:Prime Number",
"Definition:Maximal Ideal of Ring"
] | [
"Ring of Integers is Principal Ideal Domain"
] |
proofwiki-1361 | Integral Domain of Prime Order is Field | Let $\struct {\Z_p, +_p, \times_p}$ be the ring of integers modulo $p$.
The following statements are equivalent:
:$(1): \quad p$ is a prime.
:$(2): \quad \struct {\Z_p, +_p, \times_p}$ is an integral domain.
:$(3): \quad \struct {\Z_p, +_p, \times_p}$ is a field. | By Prime Number iff Generates Principal Maximal Ideal and Maximal Ideal iff Quotient Ring is Field, $(1)$ implies $(3)$, and from Field is Integral Domain, $(3)$ implies $(2)$.
By definition of integral domain, $\Z_p$ is an integral domain {{iff}} it has no proper zero divisors.
That is, {{iff}} $\struct {\Z_p^*, \time... | Let $\struct {\Z_p, +_p, \times_p}$ be the [[Definition:Ring of Integers Modulo m|ring of integers modulo $p$]].
The following statements are equivalent:
:$(1): \quad p$ is a [[Definition:Prime Number|prime]].
:$(2): \quad \struct {\Z_p, +_p, \times_p}$ is an [[Definition:Integral Domain|integral domain]].
:$(3): \qua... | By [[Prime Number iff Generates Principal Maximal Ideal]] and [[Maximal Ideal iff Quotient Ring is Field]], $(1)$ implies $(3)$, and from [[Field is Integral Domain]], $(3)$ implies $(2)$.
By definition of [[Definition:Integral Domain/Definition 2|integral domain]], $\Z_p$ is an [[Definition:Integral Domain|integral ... | Integral Domain of Prime Order is Field | https://proofwiki.org/wiki/Integral_Domain_of_Prime_Order_is_Field | https://proofwiki.org/wiki/Integral_Domain_of_Prime_Order_is_Field | [
"Integral Domains",
"Galois Fields"
] | [
"Definition:Ring of Integers Modulo m",
"Definition:Prime Number",
"Definition:Integral Domain",
"Definition:Field (Abstract Algebra)"
] | [
"Prime Number iff Generates Principal Maximal Ideal",
"Maximal Ideal iff Quotient Ring is Field",
"Field is Integral Domain",
"Definition:Integral Domain/Definition 2",
"Definition:Integral Domain",
"Definition:Proper Zero Divisor",
"Definition:Semigroup",
"Definition:Principal Ideal of Ring",
"Quot... |
proofwiki-1362 | Quotient Ring of Integers and Zero | Let $\struct {\Z, +, \times}$ be the integral domain of integers.
Let $\ideal 0$ be the principal ideal of $\struct {\Z, +, \times}$ generated by $0$.
The quotient ring $\struct {\Z / \ideal 0, +, \times}$ is isomorphic to $\struct {\Z, +, \times}$. | {{begin-eqn}}
{{eqn | l = \ideal 0
| r = \set {\sum^n_{i \mathop = 1} r_i \times 0 \times s_i: n \in \N; r_i, s_i \in \Z}
| c = {{Defof|Principal Ideal of Ring}}
}}
{{eqn | r = \set {\sum^n_{i \mathop = 1} 0: n \in \N}
| c = $0$ is the zero under integer multiplication
}}
{{eqn | r = \set 0
| c ... | Let $\struct {\Z, +, \times}$ be the [[Integers form Integral Domain|integral domain of integers]].
Let $\ideal 0$ be the [[Definition:Principal Ideal of Ring|principal ideal of $\struct {\Z, +, \times}$ generated by $0$]].
The [[Definition:Quotient Ring|quotient ring]] $\struct {\Z / \ideal 0, +, \times}$ is [[Defi... | {{begin-eqn}}
{{eqn | l = \ideal 0
| r = \set {\sum^n_{i \mathop = 1} r_i \times 0 \times s_i: n \in \N; r_i, s_i \in \Z}
| c = {{Defof|Principal Ideal of Ring}}
}}
{{eqn | r = \set {\sum^n_{i \mathop = 1} 0: n \in \N}
| c = $0$ is the [[Integer Multiplication has Zero|zero]] under [[Definition:Intege... | Quotient Ring of Integers and Zero | https://proofwiki.org/wiki/Quotient_Ring_of_Integers_and_Zero | https://proofwiki.org/wiki/Quotient_Ring_of_Integers_and_Zero | [
"Quotient Rings",
"Integers"
] | [
"Integers form Integral Domain",
"Definition:Principal Ideal of Ring",
"Definition:Quotient Ring",
"Definition:Isomorphism (Abstract Algebra)/Ring Isomorphism"
] | [
"Integer Multiplication has Zero",
"Definition:Multiplication/Integers",
"Integer Addition Identity is Zero",
"Category:Quotient Rings",
"Category:Integers"
] |
proofwiki-1363 | Quotient Ring of Integers and Principal Ideal from Unity | Let $\struct {\Z, +, \times}$ be the integral domain of integers.
Let $\ideal 1$ be the principal ideal of $\struct {\Z, +, \times}$ generated by $1$.
The quotient ring $\struct {\Z, +, \times} / \ideal 1$ is isomorphic to the null ring. | {{ProofWanted|Integers form Integral Domain and Element in Integral Domain is Unit iff Principal Ideal is Whole Domain}}
Category:Ideal Theory
Category:Quotient Rings
Category:Integers
my5ivmidti5dm6u697af8prw2n7jmhk | Let $\struct {\Z, +, \times}$ be the [[Integers form Integral Domain|integral domain of integers]].
Let $\ideal 1$ be the [[Definition:Principal Ideal of Ring|principal ideal]] of $\struct {\Z, +, \times}$ [[Definition:Generated Ideal of Ring|generated]] by $1$.
The [[Definition:Quotient Ring|quotient ring]] $\struc... | {{ProofWanted|[[Integers form Integral Domain]] and [[Element in Integral Domain is Unit iff Principal Ideal is Whole Domain]]}}
[[Category:Ideal Theory]]
[[Category:Quotient Rings]]
[[Category:Integers]]
my5ivmidti5dm6u697af8prw2n7jmhk | Quotient Ring of Integers and Principal Ideal from Unity | https://proofwiki.org/wiki/Quotient_Ring_of_Integers_and_Principal_Ideal_from_Unity | https://proofwiki.org/wiki/Quotient_Ring_of_Integers_and_Principal_Ideal_from_Unity | [
"Ideal Theory",
"Quotient Rings",
"Integers"
] | [
"Integers form Integral Domain",
"Definition:Principal Ideal of Ring",
"Definition:Generated Ideal of Ring",
"Definition:Quotient Ring",
"Definition:Isomorphism (Abstract Algebra)/Ring Isomorphism",
"Definition:Null Ring"
] | [
"Integers form Integral Domain",
"Element in Integral Domain is Unit iff Principal Ideal is Whole Domain",
"Category:Ideal Theory",
"Category:Quotient Rings",
"Category:Integers"
] |
proofwiki-1364 | Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal | Let $\struct {D, +, \circ}$ be a principal ideal domain.
Let $\ideal p$ be the principal ideal of $D$ generated by $p$.
Then $p$ is irreducible {{iff}} $\ideal p$ is a maximal ideal of $D$. | === Necessary Condition ===
{{:Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal/Forward Implication}} | Let $\struct {D, +, \circ}$ be a [[Definition:Principal Ideal Domain|principal ideal domain]].
Let $\ideal p$ be the [[Definition:Principal Ideal of Ring|principal ideal of $D$ generated by $p$]].
Then $p$ is [[Definition:Irreducible Element of Ring|irreducible]] {{iff}} $\ideal p$ is a [[Definition:Maximal Ideal of... | === [[Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal/Forward Implication|Necessary Condition]] ===
{{:Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal/Forward Implication}} | Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal | https://proofwiki.org/wiki/Principal_Ideal_of_Principal_Ideal_Domain_is_of_Irreducible_Element_iff_Maximal | https://proofwiki.org/wiki/Principal_Ideal_of_Principal_Ideal_Domain_is_of_Irreducible_Element_iff_Maximal | [
"Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal",
"Principal Ideals of Rings",
"Maximal Ideals of Rings",
"Principal Ideal Domains",
"Factorization"
] | [
"Definition:Principal Ideal Domain",
"Definition:Principal Ideal of Ring",
"Definition:Irreducible Element of Ring",
"Definition:Maximal Ideal of Ring"
] | [
"Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal/Forward Implication"
] |
proofwiki-1365 | Subring Generated by Unity of Ring with Unity | Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let the mapping $g: \Z \to R$ be defined as $\forall n \in \Z: \map g n = n 1_R$, where $n 1_R$ the $n$th power of $1_R$.
Let $\ideal x$ be the principal ideal of $\struct {R, +, \circ}$ generated by $x$.
Then $g$ is an epimo... | By the Index Law for Sum of Indices and Integral Multiple of Ring Element, we have:
:$\paren {n 1_R} \paren {m 1_R} = n \paren {m 1_R} = \paren {n m} 1_R$
Thus $g$ is an epimorphism from $\Z$ onto $S$.
{{AimForCont}} $R$ has no proper zero divisors.
By Kernel of Ring Epimorphism is Ideal, the kernel of $g$ is an ideal ... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
Let the mapping $g: \Z \to R$ be defined as $\forall n \in \Z: \map g n = n 1_R$, where $n 1_R$ the [[Definition:Power of Element|$n$th pow... | By the [[Index Laws for Monoids/Sum of Indices|Index Law for Sum of Indices]] and [[Integral Multiple of Ring Element]], we have:
:$\paren {n 1_R} \paren {m 1_R} = n \paren {m 1_R} = \paren {n m} 1_R$
Thus $g$ is an [[Definition:Ring Epimorphism|epimorphism]] from $\Z$ onto $S$.
{{AimForCont}} $R$ has no [[Definitio... | Subring Generated by Unity of Ring with Unity | https://proofwiki.org/wiki/Subring_Generated_by_Unity_of_Ring_with_Unity | https://proofwiki.org/wiki/Subring_Generated_by_Unity_of_Ring_with_Unity | [
"Ideal Theory"
] | [
"Definition:Ring with Unity",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Power of Element",
"Definition:Principal Ideal of Ring",
"Definition:Ring Epimorphism",
"Definition:Subring",
"Definition:Generated Subring",
"Definition:Proper Zero Divisor",
"Definition:... | [
"Index Laws for Monoids/Sum of Indices",
"Integral Multiple of Ring Element",
"Definition:Ring Epimorphism",
"Definition:Proper Zero Divisor",
"Kernel of Ring Epimorphism is Ideal",
"Definition:Kernel of Ring Homomorphism",
"Ring of Integers is Principal Ideal Domain",
"Definition:Kernel of Ring Homom... |
proofwiki-1366 | Null Ring iff Characteristic is One | The only ring whose characteristic is $1$ is the null ring. | From Null Ring iff Zero and Unity Coincide, $1_R \ne 0_R$ except when $R = \left\{{0_R}\right\}$.
{{qed}}
Category:Ring Theory
6oqldbeeezzek7hrj6fbojap6vj42gf | The only [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Characteristic of Ring|characteristic]] is $1$ is the [[Definition:Null Ring|null ring]]. | From [[Null Ring iff Zero and Unity Coincide]], $1_R \ne 0_R$ except when $R = \left\{{0_R}\right\}$.
{{qed}}
[[Category:Ring Theory]]
6oqldbeeezzek7hrj6fbojap6vj42gf | Null Ring iff Characteristic is One | https://proofwiki.org/wiki/Null_Ring_iff_Characteristic_is_One | https://proofwiki.org/wiki/Null_Ring_iff_Characteristic_is_One | [
"Ring Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Characteristic of Ring",
"Definition:Null Ring"
] | [
"Null Ring iff Zero and Unity Coincide",
"Category:Ring Theory"
] |
proofwiki-1367 | Characteristic of Finite Ring with No Zero Divisors | Let $\struct {R, +, \circ}$ be a finite ring with unity with no proper zero divisors whose zero is $0_R$ and whose unity is $1_R$.
Let $n \ne 0$ be the characteristic of $R$.
Then:
:$(1): \quad n$ must be a prime number
:$(2): \quad n$ is the order of all non-zero elements in $\struct {R, +}$.
It follows that $\struct ... | Follows directly from Subring Generated by Unity of Ring with Unity.
{{qed}} | Let $\struct {R, +, \circ}$ be a [[Definition:Finite Ring|finite]] [[Definition:Ring with Unity|ring with unity]] with no [[Definition:Proper Zero Divisor|proper zero divisors]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
Let $n \ne 0$ be the [[Definition:Charact... | Follows directly from [[Subring Generated by Unity of Ring with Unity]].
{{qed}} | Characteristic of Finite Ring with No Zero Divisors/Proof 1 | https://proofwiki.org/wiki/Characteristic_of_Finite_Ring_with_No_Zero_Divisors | https://proofwiki.org/wiki/Characteristic_of_Finite_Ring_with_No_Zero_Divisors/Proof_1 | [
"Finite Rings",
"Rings with Unity",
"Characteristics of Rings",
"Characteristic of Finite Ring with No Zero Divisors"
] | [
"Definition:Finite Ring",
"Definition:Ring with Unity",
"Definition:Proper Zero Divisor",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Characteristic of Ring",
"Definition:Prime Number",
"Definition:Order of Group Element",
"Definition:Cyclic Group",
"Definition:... | [
"Subring Generated by Unity of Ring with Unity"
] |
proofwiki-1368 | Characteristic of Finite Ring with No Zero Divisors | Let $\struct {R, +, \circ}$ be a finite ring with unity with no proper zero divisors whose zero is $0_R$ and whose unity is $1_R$.
Let $n \ne 0$ be the characteristic of $R$.
Then:
:$(1): \quad n$ must be a prime number
:$(2): \quad n$ is the order of all non-zero elements in $\struct {R, +}$.
It follows that $\struct ... | Suppose $\Char R = n$ where $n$ is composite.
Let $n = r s$, where $r, s \in \Z, r > 1, s > 1$.
First note that:
{{begin-eqn}}
{{eqn | l = \paren {r \cdot 1_R} \circ \paren {s \cdot 1_R}
| r = \paren {r s} \paren {1_R \circ 1_R}
| c = Integral Multiple of Ring Element
}}
{{eqn | r = \paren {r s} 1_R
|... | Let $\struct {R, +, \circ}$ be a [[Definition:Finite Ring|finite]] [[Definition:Ring with Unity|ring with unity]] with no [[Definition:Proper Zero Divisor|proper zero divisors]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
Let $n \ne 0$ be the [[Definition:Charact... | Suppose $\Char R = n$ where $n$ is [[Definition:Composite Number|composite]].
Let $n = r s$, where $r, s \in \Z, r > 1, s > 1$.
First note that:
{{begin-eqn}}
{{eqn | l = \paren {r \cdot 1_R} \circ \paren {s \cdot 1_R}
| r = \paren {r s} \paren {1_R \circ 1_R}
| c = [[Integral Multiple of Ring Element/Ge... | Characteristic of Finite Ring with No Zero Divisors/Proof 2 | https://proofwiki.org/wiki/Characteristic_of_Finite_Ring_with_No_Zero_Divisors | https://proofwiki.org/wiki/Characteristic_of_Finite_Ring_with_No_Zero_Divisors/Proof_2 | [
"Finite Rings",
"Rings with Unity",
"Characteristics of Rings",
"Characteristic of Finite Ring with No Zero Divisors"
] | [
"Definition:Finite Ring",
"Definition:Ring with Unity",
"Definition:Proper Zero Divisor",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Characteristic of Ring",
"Definition:Prime Number",
"Definition:Order of Group Element",
"Definition:Cyclic Group",
"Definition:... | [
"Definition:Composite Number",
"Integral Multiple of Ring Element/General Result",
"Definition:Proper Zero Divisor",
"Definition:Prime Number",
"Characteristic times Ring Element is Ring Zero",
"Element to Power of Multiple of Order is Identity",
"Definition:Prime Number",
"Null Ring iff Characteristi... |
proofwiki-1369 | Integral Domain with Characteristic Zero | In an integral domain with characteristic zero, every non-zero element has infinite order under ring addition. | Let $\struct {D, +, \circ}$ be an integral domain, whose zero is $0_D$ and whose unity is $1_D$, such that $\Char D = 0$.
Let $x \in D, x \ne 0_D$.
Then:
{{begin-eqn}}
{{eqn | q = \forall n \in \Z_{>0}
| l = n \cdot x
| r = n \cdot \paren {x \circ 1_D}
| c =
}}
{{eqn | r = \paren {n \circ 1_D} \cdot ... | In an [[Definition:Integral Domain|integral domain]] with [[Definition:Characteristic of Ring|characteristic zero]], every non-zero element has [[Definition:Order of Group Element|infinite order]] under [[Definition:Ring Addition|ring addition]]. | Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]], whose [[Definition:Ring Zero|zero]] is $0_D$ and whose [[Definition:Unity of Ring|unity]] is $1_D$, such that $\Char D = 0$.
Let $x \in D, x \ne 0_D$.
Then:
{{begin-eqn}}
{{eqn | q = \forall n \in \Z_{>0}
| l = n \cdot x
| ... | Integral Domain with Characteristic Zero | https://proofwiki.org/wiki/Integral_Domain_with_Characteristic_Zero | https://proofwiki.org/wiki/Integral_Domain_with_Characteristic_Zero | [
"Integral Domains"
] | [
"Definition:Integral Domain",
"Definition:Characteristic of Ring",
"Definition:Order of Group Element",
"Definition:Ring (Abstract Algebra)/Addition"
] | [
"Definition:Integral Domain",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Integral Multiple of Ring Element",
"Definition:Order of Group Element"
] |
proofwiki-1370 | Characteristic of Field is Zero or Prime | Let $F$ be a field.
Then the characteristic of $F$ is either zero or a prime number. | From the definition, a field is a ring with no zero divisors.
So by Characteristic of Finite Ring with No Zero Divisors, if $\Char F \ne 0$ then it is prime.
{{Qed}} | Let $F$ be a [[Definition:Field (Abstract Algebra)|field]].
Then the [[Definition:Characteristic of Field|characteristic]] of $F$ is either [[Definition:Zero (Number)|zero]] or a [[Definition:Prime Number|prime number]]. | From the definition, a [[Definition:Field (Abstract Algebra)|field]] is a [[Definition:Ring (Abstract Algebra)|ring]] with no [[Definition:Zero Divisor of Ring|zero divisors]].
So by [[Characteristic of Finite Ring with No Zero Divisors]], if $\Char F \ne 0$ then it is [[Definition:Prime Number|prime]].
{{Qed}} | Characteristic of Field is Zero or Prime/Proof 1 | https://proofwiki.org/wiki/Characteristic_of_Field_is_Zero_or_Prime | https://proofwiki.org/wiki/Characteristic_of_Field_is_Zero_or_Prime/Proof_1 | [
"Characteristic of Field is Zero or Prime",
"Characteristics of Fields"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Characteristic of Field",
"Definition:Zero (Number)",
"Definition:Prime Number"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Ring (Abstract Algebra)",
"Definition:Zero Divisor/Ring",
"Characteristic of Finite Ring with No Zero Divisors",
"Definition:Prime Number"
] |
proofwiki-1371 | Characteristic of Field is Zero or Prime | Let $F$ be a field.
Then the characteristic of $F$ is either zero or a prime number. | From the definition, a field is an instance of an integral domain.
The result follows from Characteristic of Integral Domain is Zero or Prime.
{{Qed}} | Let $F$ be a [[Definition:Field (Abstract Algebra)|field]].
Then the [[Definition:Characteristic of Field|characteristic]] of $F$ is either [[Definition:Zero (Number)|zero]] or a [[Definition:Prime Number|prime number]]. | From the definition, a [[Definition:Field (Abstract Algebra)|field]] is an instance of an [[Definition:Integral Domain|integral domain]].
The result follows from [[Characteristic of Integral Domain is Zero or Prime]].
{{Qed}} | Characteristic of Field is Zero or Prime/Proof 2 | https://proofwiki.org/wiki/Characteristic_of_Field_is_Zero_or_Prime | https://proofwiki.org/wiki/Characteristic_of_Field_is_Zero_or_Prime/Proof_2 | [
"Characteristic of Field is Zero or Prime",
"Characteristics of Fields"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Characteristic of Field",
"Definition:Zero (Number)",
"Definition:Prime Number"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Integral Domain",
"Characteristic of Integral Domain is Zero or Prime"
] |
proofwiki-1372 | Field of Characteristic Zero has Unique Prime Subfield | Let $F$ be a field, whose zero is $0_F$ and whose unity is $1_F$, with characteristic zero.
Then there exists a unique $P \subseteq F$ such that:
:$(1): \quad P$ is a subfield of $F$
:$(2): \quad P$ is isomorphic to the field of rational numbers $\struct {\Q, +, \times}$.
That is, $P \cong \Q$ is a unique minimal subfi... | Follows directly from:
:Subring Generated by Unity of Ring with Unity
:Quotient Theorem for Monomorphisms
{{qed}} | Let $F$ be a [[Definition:Field (Abstract Algebra)|field]], whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$, with [[Definition:Characteristic of Field|characteristic]] zero.
Then there exists a unique $P \subseteq F$ such that:
:$(1): \quad P$ is a [[Definition:Sub... | Follows directly from:
:[[Subring Generated by Unity of Ring with Unity]]
:[[Quotient Theorem for Monomorphisms]]
{{qed}} | Field of Characteristic Zero has Unique Prime Subfield/Proof 1 | https://proofwiki.org/wiki/Field_of_Characteristic_Zero_has_Unique_Prime_Subfield | https://proofwiki.org/wiki/Field_of_Characteristic_Zero_has_Unique_Prime_Subfield/Proof_1 | [
"Characteristics of Fields",
"Subfields",
"Field of Characteristic Zero has Unique Prime Subfield"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Multiplicative Identity",
"Definition:Characteristic of Field",
"Definition:Subfield",
"Definition:Isomorphism (Abstract Algebra)/Field Isomorphism",
"Definition:Field of Rational Numbers",
"Definition:Subfield",
"Definition... | [
"Subring Generated by Unity of Ring with Unity",
"Quotient Theorem for Monomorphisms"
] |
proofwiki-1373 | Field of Characteristic Zero has Unique Prime Subfield | Let $F$ be a field, whose zero is $0_F$ and whose unity is $1_F$, with characteristic zero.
Then there exists a unique $P \subseteq F$ such that:
:$(1): \quad P$ is a subfield of $F$
:$(2): \quad P$ is isomorphic to the field of rational numbers $\struct {\Q, +, \times}$.
That is, $P \cong \Q$ is a unique minimal subfi... | Let $\struct {F, +, \circ}$ be a field such that $\Char F = 0$.
Let $P$ be a prime subfield of $F$.
From Field has Prime Subfield, this has been shown to exist.
As $P$ is a subfield of $F$, we apply Zero and Unity of Subfield and see that the unity of $P$ is $1_F$.
As $P$ is closed:
:$\forall m \in \Z: m \cdot 1_F \in ... | Let $F$ be a [[Definition:Field (Abstract Algebra)|field]], whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$, with [[Definition:Characteristic of Field|characteristic]] zero.
Then there exists a unique $P \subseteq F$ such that:
:$(1): \quad P$ is a [[Definition:Sub... | Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]] such that $\Char F = 0$.
Let $P$ be a [[Definition:Prime Subfield|prime subfield]] of $F$.
From [[Field has Prime Subfield]], this has been shown to exist.
As $P$ is a [[Definition:Subfield|subfield]] of $F$, we apply [[Zero and Unity of... | Field of Characteristic Zero has Unique Prime Subfield/Proof 2 | https://proofwiki.org/wiki/Field_of_Characteristic_Zero_has_Unique_Prime_Subfield | https://proofwiki.org/wiki/Field_of_Characteristic_Zero_has_Unique_Prime_Subfield/Proof_2 | [
"Characteristics of Fields",
"Subfields",
"Field of Characteristic Zero has Unique Prime Subfield"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Multiplicative Identity",
"Definition:Characteristic of Field",
"Definition:Subfield",
"Definition:Isomorphism (Abstract Algebra)/Field Isomorphism",
"Definition:Field of Rational Numbers",
"Definition:Subfield",
"Definition... | [
"Definition:Field (Abstract Algebra)",
"Definition:Prime Subfield",
"Field has Prime Subfield",
"Definition:Subfield",
"Zero and Unity of Subfield",
"Definition:Multiplicative Identity",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Field Zero",
"Definition:Division Produc... |
proofwiki-1374 | Field of Prime Characteristic has Unique Prime Subfield | Let $F$ be a field whose characteristic is $p$.
Then there exists a unique $P \subseteq F$ such that:
:$(1): \quad P$ is a subfield of $F$
:$(2): \quad P \cong \Z_p$.
That is, $P \cong \Z_p$ is a unique minimal subfield of $F$, and all other subfields of $F$ contain $P$.
This field $P$ is called the prime subfield of $... | Let $\struct {F, +, \times}$ be a field whose unity is $1_F$ such that $\Char F = p$.
Let $P$ be a prime subfield of $F$.
From Field has Prime Subfield, this has been shown to exist.
We can consistently define a mapping $\phi: \Z_p \to F$ by:
:$\forall n \in \Z_p: \map \phi {\eqclass n p} = n \cdot 1_F$
Suppose $a, b \... | Let $F$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Characteristic of Field|characteristic]] is $p$.
Then there exists a [[Definition:Unique|unique]] $P \subseteq F$ such that:
:$(1): \quad P$ is a [[Definition:Subfield|subfield]] of $F$
:$(2): \quad P \cong \Z_p$.
That is, $P \cong \Z_p$ i... | Let $\struct {F, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Unity of Field|unity]] is $1_F$ such that $\Char F = p$.
Let $P$ be a [[Definition:Prime Subfield|prime subfield]] of $F$.
From [[Field has Prime Subfield]], this has been shown to exist.
We can consistently define a ... | Field of Prime Characteristic has Unique Prime Subfield | https://proofwiki.org/wiki/Field_of_Prime_Characteristic_has_Unique_Prime_Subfield | https://proofwiki.org/wiki/Field_of_Prime_Characteristic_has_Unique_Prime_Subfield | [
"Prime Fields"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Characteristic of Field",
"Definition:Unique",
"Definition:Subfield",
"Definition:Subfield",
"Definition:Subfield",
"Definition:Prime Subfield"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Multiplicative Identity",
"Definition:Prime Subfield",
"Field has Prime Subfield",
"Definition:Mapping",
"Definition:Well-Defined/Mapping",
"Integral Multiple Distributes over Ring Addition",
"Definition:Ring Homomorphism",
"Ring Homomorphism from F... |
proofwiki-1375 | Intersection of All Division Subrings is Prime Subfield | Let $\struct {K, +, \circ}$ be a division ring.
Let $P$ be the intersection of the set of all division subrings of $K$.
Then $P$ is the prime subfield of $K$. | By Intersection of Division Subrings is Division Subring, the intersection $P$ of the set of all division subrings of $K$ is a division ring.
Let $\map Z K$ be the center of $K$.
From Center of Ring is Commutative Subring, $\map Z K$ is a commutative subring of $K$.
Therefore $\map Z K$ is a commutative division ring
T... | Let $\struct {K, +, \circ}$ be a [[Definition:Division Ring|division ring]].
Let $P$ be the [[Definition:Set Intersection|intersection]] of the [[Definition:Set|set]] of all [[Definition:Division Subring|division subrings]] of $K$.
Then $P$ is the [[Definition:Prime Subfield|prime subfield]] of $K$. | By [[Intersection of Division Subrings is Division Subring]], the [[Definition:Set Intersection|intersection]] $P$ of the [[Definition:Set|set]] of all [[Definition:Division Subring|division subrings]] of $K$ is a [[Definition:Division Ring|division ring]].
Let $\map Z K$ be the [[Definition:Center of Ring|center]] of... | Intersection of All Division Subrings is Prime Subfield | https://proofwiki.org/wiki/Intersection_of_All_Division_Subrings_is_Prime_Subfield | https://proofwiki.org/wiki/Intersection_of_All_Division_Subrings_is_Prime_Subfield | [
"Subfields",
"Division Subrings"
] | [
"Definition:Division Ring",
"Definition:Set Intersection",
"Definition:Set",
"Definition:Division Subring",
"Definition:Prime Subfield"
] | [
"Intersection of Division Subrings is Division Subring",
"Definition:Set Intersection",
"Definition:Set",
"Definition:Division Subring",
"Definition:Division Ring",
"Definition:Center (Abstract Algebra)/Ring",
"Center of Ring is Commutative Subring",
"Definition:Commutative Ring",
"Definition:Subrin... |
proofwiki-1376 | Characteristic of Ordered Integral Domain is Zero | Let $\struct {D, +, \circ}$ be an ordered integral domain whose zero is $0_D$ and whose unity is $1_D$.
Let $\Char D$ be the characteristic of $D$.
Then $\Char D = 0$.
Let $g: \Z \to D$ be the mapping defined as:
:$\forall n \in \Z: \map g n = n \cdot 1_D$
where $n \cdot 1_D$ is defined as the $n$th power of $1_D$.
The... | By Properties of Ordered Ring $(5)$:
:$\forall n \in \Z_{>0}: n \cdot 1_D > 0$
Thus:
:$\forall p > 0: \Char D \ne p$
Hence:
:$\Char D = 0$
and so $g$ is a monomorphism from $\Z$ into $D$.
Also, if $m < p$, then $p - m \in \Z_+$, so:
:$p \cdot 1_D - m \cdot 1_D > 0_D$
Hence:
:$\map g m < \map g p$
Thus by Monomorphism f... | Let $\struct {D, +, \circ}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$ and whose [[Definition:Unity of Ring|unity]] is $1_D$.
Let $\Char D$ be the [[Definition:Characteristic of Ring|characteristic of $D$]].
Then $\Char D = 0$.
Let $g: \Z \to D... | By [[Properties of Ordered Ring]] $(5)$:
:$\forall n \in \Z_{>0}: n \cdot 1_D > 0$
Thus:
:$\forall p > 0: \Char D \ne p$
Hence:
:$\Char D = 0$
and so $g$ is a [[Definition:Ring Monomorphism|monomorphism]] from $\Z$ into $D$.
Also, if $m < p$, then $p - m \in \Z_+$, so:
:$p \cdot 1_D - m \cdot 1_D > 0_D$
Hence:
:$... | Characteristic of Ordered Integral Domain is Zero | https://proofwiki.org/wiki/Characteristic_of_Ordered_Integral_Domain_is_Zero | https://proofwiki.org/wiki/Characteristic_of_Ordered_Integral_Domain_is_Zero | [
"Ordered Integral Domains"
] | [
"Definition:Ordered Integral Domain",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Characteristic of Ring",
"Definition:Mapping",
"Definition:Power of Element",
"Definition:Ring Monomorphism",
"Definition:Ordered Ring",
"Definition:Ordered Ring"
] | [
"Properties of Ordered Ring",
"Definition:Ring Monomorphism",
"Monomorphism from Total Ordering",
"Definition:Ring Monomorphism",
"Definition:Ordered Integral Domain"
] |
proofwiki-1377 | Monomorphism from Rational Numbers to Totally Ordered Field | Let $\struct {F, +, \circ, \le}$ be a totally ordered field.
There is one and only one (ring) monomorphism from the totally ordered field $\Q$ onto $F$.
Its image is the prime subfield of $F$. | Follows from:
:Characteristic of Ordered Integral Domain is Zero
:Order Embedding between Quotient Fields is Unique.
{{Qed}} | Let $\struct {F, +, \circ, \le}$ be a [[Definition:Totally Ordered Field|totally ordered field]].
There is one and only one [[Definition:Ring Monomorphism|(ring) monomorphism]] from the [[Definition:Totally Ordered Field|totally ordered field]] $\Q$ onto $F$.
Its [[Definition:Image of Mapping|image]] is the [[Defini... | Follows from:
:[[Characteristic of Ordered Integral Domain is Zero]]
:[[Order Embedding between Quotient Fields is Unique]].
{{Qed}} | Monomorphism from Rational Numbers to Totally Ordered Field | https://proofwiki.org/wiki/Monomorphism_from_Rational_Numbers_to_Totally_Ordered_Field | https://proofwiki.org/wiki/Monomorphism_from_Rational_Numbers_to_Totally_Ordered_Field | [
"Totally Ordered Fields",
"Ring Monomorphisms"
] | [
"Definition:Totally Ordered Field",
"Definition:Ring Monomorphism",
"Definition:Totally Ordered Field",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Prime Subfield"
] | [
"Characteristic of Ordered Integral Domain is Zero",
"Order Embedding between Quotient Fields is Unique"
] |
proofwiki-1378 | Set of Polynomials over Integral Domain is Subring | Let $\struct {R, +, \circ}$ be a commutative ring.
Let $\struct {D, +, \circ}$ be an integral subdomain of $R$.
Then $\forall x \in R$, the set $D \sqbrk x$ of polynomials in $x$ over $D$ is a subring of $R$. | By application of the Subring Test:
As $D$ is an integral domain, it has a unity $1_D$ and so $x = 1_D x$.
Hence $x \in D \sqbrk x$ and so $D \sqbrk x \ne \O$.
Let $p, q \in D \sqbrk x$.
Then let:
:$\ds p = \sum_{k \mathop = 0}^m a_k \circ x^k, q = \sum_{k \mathop = 0}^n b_k \circ x^k$
Thus:
:$\ds -q = -\sum_{k \mathop... | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative Ring|commutative ring]].
Let $\struct {D, +, \circ}$ be an [[Definition:Subdomain|integral subdomain]] of $R$.
Then $\forall x \in R$, the [[Definition:Set|set]] $D \sqbrk x$ of [[Definition:Polynomial in Ring Element|polynomials]] in $x$ over $D$ is a [[Defi... | By application of the [[Subring Test]]:
As $D$ is an [[Definition:Integral Domain|integral domain]], it has a [[Definition:Unity of Ring|unity]] $1_D$ and so $x = 1_D x$.
Hence $x \in D \sqbrk x$ and so $D \sqbrk x \ne \O$.
Let $p, q \in D \sqbrk x$.
Then let:
:$\ds p = \sum_{k \mathop = 0}^m a_k \circ x^k, q = \... | Set of Polynomials over Integral Domain is Subring | https://proofwiki.org/wiki/Set_of_Polynomials_over_Integral_Domain_is_Subring | https://proofwiki.org/wiki/Set_of_Polynomials_over_Integral_Domain_is_Subring | [
"Polynomial Theory",
"Subrings"
] | [
"Definition:Commutative Ring",
"Definition:Subdomain",
"Definition:Set",
"Definition:Polynomial in Ring Element",
"Definition:Subring"
] | [
"Subring Test",
"Definition:Integral Domain",
"Definition:Unity (Abstract Algebra)/Ring",
"Polynomials Closed under Addition",
"Polynomials Closed under Ring Product",
"Subring Test"
] |
proofwiki-1379 | Polynomials Closed under Ring Product | Let $\struct {R, +, \circ}$ be a commutative ring.
Let $\ds f = \sum_{k \mathop \in Z} a_k \mathbf X^k$, $\ds g = \sum_{k \in Z} b_k \mathbf X^k$ be polynomials in the indeterminates $\set {X_j: j \in J}$ over $R$, where $Z$ is the set of all multiindices indexed by $\set {X_j: j \in J}$.
Define the product
:$\ds f \ot... | {{handwaving}}
It is immediate that $f \otimes g$ is a map from the free commutative monoid to $R$, so we need only prove that $f \otimes g$ is nonzero on finitely many $\mathbf X^k$, $k \in Z$.
Suppose that for some $k \in Z$:
:$\ds \sum_{\substack {p + q = k \\ p, q \mathop \in Z}} a_p b_q \ne 0$
Therefore if $c_k \n... | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative Ring|commutative ring]].
Let $\ds f = \sum_{k \mathop \in Z} a_k \mathbf X^k$, $\ds g = \sum_{k \in Z} b_k \mathbf X^k$ be [[Definition:Polynomial|polynomials]] in the [[Definition:Indeterminate (Polynomial Theory)|indeterminates]] $\set {X_j: j \in J}$ over $R... | {{handwaving}}
It is immediate that $f \otimes g$ is a map from the [[Definition:Free Commutative Monoid|free commutative monoid]] to $R$, so we need only prove that $f \otimes g$ is nonzero on finitely many $\mathbf X^k$, $k \in Z$.
Suppose that for some $k \in Z$:
:$\ds \sum_{\substack {p + q = k \\ p, q \mathop \... | Polynomials Closed under Ring Product | https://proofwiki.org/wiki/Polynomials_Closed_under_Ring_Product | https://proofwiki.org/wiki/Polynomials_Closed_under_Ring_Product | [
"Polynomial Theory"
] | [
"Definition:Commutative Ring",
"Definition:Polynomial",
"Definition:Polynomial Ring/Indeterminate",
"Definition:Multiindex"
] | [
"Definition:Free Commutative Monoid",
"Category:Polynomial Theory"
] |
proofwiki-1380 | Unique Representation in Polynomial Forms | Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $\struct {D, +, \circ}$ be an integral subdomain of $R$.
Let $X \in R$ be transcendental over $D$.
Let $D \sqbrk X$ be the ring of polynomials in $X$ over $D$.
Then each non-zero member of $D \left[{X}\right]$... | Suppose $f \in D \sqbrk X \setminus \set {0_R}$ has more than one way of being expressed in the above form.
Then you would be able to subtract one from the other and get a polynomial in $D \sqbrk X$ equal to zero.
As $f$ is transcendental, the result follows.
{{qed}} | 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 $1_R$.
Let $\struct {D, +, \circ}$ be an [[Definition:Subdomain|integral subdomain]] of $R$.
Let $X \in R$ be [[Definiti... | Suppose $f \in D \sqbrk X \setminus \set {0_R}$ has more than one way of being expressed in the above form.
Then you would be able to subtract one from the other and get a polynomial in $D \sqbrk X$ equal to zero.
As $f$ is [[Definition:Transcendental over Integral Domain|transcendental]], the result follows.
{{qed}} | Unique Representation in Polynomial Forms | https://proofwiki.org/wiki/Unique_Representation_in_Polynomial_Forms | https://proofwiki.org/wiki/Unique_Representation_in_Polynomial_Forms | [
"Polynomial Theory"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Subdomain",
"Definition:Transcendental (Abstract Algebra)/Ring",
"Definition:Ring of Polynomials in Ring Element"
] | [
"Definition:Transcendental (Abstract Algebra)/Ring"
] |
proofwiki-1381 | Ring of Polynomial Forms is Integral Domain | Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\struct {D, +, \circ}$ be an integral subdomain of $R$.
Let $X \in R$ be transcendental over $D$.
Let $D \sqbrk X$ be the ring of polynomials in $X$ over $D$.
Then $D \sqbrk X$ is an integral domain. | By Ring of Polynomial Forms is Commutative Ring with Unity we know that $D \sqbrk X$ is a commutative ring with unity.
Let neither $\ds \map f X = \sum_{k \mathop = 0}^n a_k x^k$ nor $\ds \map g X = \sum_{k \mathop = 0}^m b_k X^k$ be the null polynomial.
Then their leading coefficients $a_n$ and $b_m$ are non-zero.
The... | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative Ring with Unity|commutative ring with unity]].
Let $\struct {D, +, \circ}$ be an [[Definition:Subdomain|integral subdomain]] of $R$.
Let $X \in R$ be [[Definition:Transcendental over Integral Domain|transcendental over $D$]].
Let $D \sqbrk X$ be the [[Definit... | By [[Ring of Polynomial Forms is Commutative Ring with Unity]] we know that $D \sqbrk X$ is a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
Let neither $\ds \map f X = \sum_{k \mathop = 0}^n a_k x^k$ nor $\ds \map g X = \sum_{k \mathop = 0}^m b_k X^k$ be the [[Definition:Null Polynomial over... | Ring of Polynomial Forms is Integral Domain | https://proofwiki.org/wiki/Ring_of_Polynomial_Forms_is_Integral_Domain | https://proofwiki.org/wiki/Ring_of_Polynomial_Forms_is_Integral_Domain | [
"Polynomial Rings",
"Integral Domains"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Subdomain",
"Definition:Transcendental (Abstract Algebra)/Ring",
"Definition:Ring of Polynomials in Ring Element",
"Definition:Integral Domain"
] | [
"Ring of Polynomial Forms is Commutative Ring with Unity",
"Definition:Commutative and Unitary Ring",
"Definition:Null Polynomial/Ring",
"Definition:Leading Coefficient of Polynomial",
"Definition:Ring Zero",
"Definition:Integral Domain",
"Definition:Ring (Abstract Algebra)/Product",
"Definition:Multi... |
proofwiki-1382 | Rings of Polynomials in Ring Elements are Isomorphic | Let $R_1, R_2$ be commutative rings with unity.
Let $D$ be an integral subdomain of both $R_1$ and $R_2$.
Let $X_1, X_2 \in R$ be transcendental over $D$.
Let $D \sqbrk {X_1}, D \sqbrk {X_2}$ be the rings of polynomials in $X_1$ and $X_2$ over $D$.
Then $D \sqbrk {X_1}$ is isomorphic to $D \sqbrk {X_2}$. | First it is shown that the mapping $\phi: D \sqbrk {X_1} \to D \sqbrk {X_2}$ given by:
:$\ds \map \phi {\sum_{k \mathop = 0}^n a_k \circ X_1^k} = \sum_{k \mathop = 0}^n a_k \circ X_2^k$
is a bijection.
Let $p, q \in \phi: D \sqbrk {X_1}$.
Suppose $\map \phi p = \map \phi q$.
Then the coefficients of $\map \phi p$ and $... | Let $R_1, R_2$ be [[Definition:Commutative Ring with Unity|commutative rings with unity]].
Let $D$ be an [[Definition:Subdomain|integral subdomain]] of both $R_1$ and $R_2$.
Let $X_1, X_2 \in R$ be [[Definition:Transcendental over Integral Domain|transcendental over $D$]].
Let $D \sqbrk {X_1}, D \sqbrk {X_2}$ be the... | First it is shown that the [[Definition:Mapping|mapping]] $\phi: D \sqbrk {X_1} \to D \sqbrk {X_2}$ given by:
:$\ds \map \phi {\sum_{k \mathop = 0}^n a_k \circ X_1^k} = \sum_{k \mathop = 0}^n a_k \circ X_2^k$
is a [[Definition:Bijection|bijection]].
Let $p, q \in \phi: D \sqbrk {X_1}$.
Suppose $\map \phi p = \map ... | Rings of Polynomials in Ring Elements are Isomorphic | https://proofwiki.org/wiki/Rings_of_Polynomials_in_Ring_Elements_are_Isomorphic | https://proofwiki.org/wiki/Rings_of_Polynomials_in_Ring_Elements_are_Isomorphic | [
"Polynomial Theory",
"Ring Isomorphisms"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Subdomain",
"Definition:Transcendental (Abstract Algebra)/Ring",
"Definition:Ring of Polynomials in Ring Element",
"Definition:Isomorphism (Abstract Algebra)/Ring Isomorphism"
] | [
"Definition:Mapping",
"Definition:Bijection",
"Definition:Coefficient of Polynomial",
"Definition:Injection",
"Definition:Mapping",
"Definition:Injection",
"Injection is Bijection iff Inverse is Injection",
"Definition:Bijection",
"Definition:Ring Homomorphism",
"Definition:Multiplication of Polyn... |
proofwiki-1383 | Injection is Bijection iff Inverse is Injection | Let $\phi: S \to T$ be an injection.
Then $\phi$ is a bijection {{iff}} its inverse $\phi^{-1}$ is also an injection. | === Necessary Condition ===
Let $\phi$ be a bijection.
Then from Bijection iff Inverse is Bijection, its inverse $\phi^{-1}$ is also a bijection and therefore by definition an injection.
{{qed|lemma}} | Let $\phi: S \to T$ be an [[Definition:Injection|injection]].
Then $\phi$ is a [[Definition:Bijection|bijection]] {{iff}} its [[Definition:Inverse of Mapping|inverse]] $\phi^{-1}$ is also an [[Definition:Injection|injection]]. | === Necessary Condition ===
Let $\phi$ be a [[Definition:Bijection|bijection]].
Then from [[Bijection iff Inverse is Bijection]], its inverse $\phi^{-1}$ is also a [[Definition:Bijection|bijection]] and therefore by definition an [[Definition:Injection|injection]].
{{qed|lemma}} | Injection is Bijection iff Inverse is Injection | https://proofwiki.org/wiki/Injection_is_Bijection_iff_Inverse_is_Injection | https://proofwiki.org/wiki/Injection_is_Bijection_iff_Inverse_is_Injection | [
"Injections",
"Bijections"
] | [
"Definition:Injection",
"Definition:Bijection",
"Definition:Inverse of Mapping",
"Definition:Injection"
] | [
"Definition:Bijection",
"Inverse of Bijection is Bijection",
"Definition:Bijection",
"Definition:Injection",
"Definition:Injection",
"Definition:Injection"
] |
proofwiki-1384 | Division Theorem for Polynomial Forms over Field | Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $X$ be transcendental over $F$.
Let $F \sqbrk X$ be the ring of polynomials in $X$ over $F$.
Let $d$ be an element of $F \sqbrk X$ of degree $n \ge 1$.
Then $\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$... | From the equation $0_F = 0_F \circ d + 0_F$, the theorem is true for the trivial case $f = 0_F$.
So, if there is a counterexample to be found, it will have a degree.
{{AimForCont}} there exists at least one counterexample.
By a version of the Well-Ordering Principle, we can assign a number $m$ to the lowest degree poss... | Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$.
Let $X$ be [[Definition:Transcendental over Field|transcendental over $F$]].
Let $F \sqbrk X$ be the [[Definition:Ring of Polynomials in ... | From the equation $0_F = 0_F \circ d + 0_F$, the theorem is true for the trivial case $f = 0_F$.
So, if there is a [[Definition:Counterexample|counterexample]] to be found, it will have a [[Definition:Degree of Polynomial over Field|degree]].
{{AimForCont}} there exists at least one [[Definition:Counterexample|count... | Division Theorem for Polynomial Forms over Field/Proof 1 | https://proofwiki.org/wiki/Division_Theorem_for_Polynomial_Forms_over_Field | https://proofwiki.org/wiki/Division_Theorem_for_Polynomial_Forms_over_Field/Proof_1 | [
"Field Theory",
"Polynomial Theory",
"Division Theorem for Polynomial Forms over Field",
"Division Theorem"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Multiplicative Identity",
"Definition:Transcendental (Abstract Algebra)/Field Extension/Element",
"Definition:Ring of Polynomials in Ring Element",
"Definition:Element",
"Definition:Degree of Polynomial/Field",
"Definition:Deg... | [
"Definition:Counterexample",
"Definition:Degree of Polynomial/Field",
"Definition:Counterexample",
"Well-Ordering Principle",
"Definition:Degree of Polynomial/Field",
"Definition:Counterexample",
"Definition:Counterexample",
"Definition:Degree of Polynomial/Field",
"Definition:Counterexample",
"De... |
proofwiki-1385 | Division Theorem for Polynomial Forms over Field | Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $X$ be transcendental over $F$.
Let $F \sqbrk X$ be the ring of polynomials in $X$ over $F$.
Let $d$ be an element of $F \sqbrk X$ of degree $n \ge 1$.
Then $\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$... | Suppose $\map \deg f < \map \deg d$.
Then we take $\map q X = 0$ and $\map r X = \map f X$ and the result holds.
Otherwise, $\map \deg f \ge \map \deg d$.
Let:
:$\map f X = a_0 + a_1 X + a_2 x^2 + \cdots + a_m X^m$
:$\map d X = b_0 + b_1 X + b_2 x^2 + \cdots + b_n X^n$
We can subtract from $f$ a suitable multiple of $d... | Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$.
Let $X$ be [[Definition:Transcendental over Field|transcendental over $F$]].
Let $F \sqbrk X$ be the [[Definition:Ring of Polynomials in ... | Suppose $\map \deg f < \map \deg d$.
Then we take $\map q X = 0$ and $\map r X = \map f X$ and the result holds.
Otherwise, $\map \deg f \ge \map \deg d$.
Let:
:$\map f X = a_0 + a_1 X + a_2 x^2 + \cdots + a_m X^m$
:$\map d X = b_0 + b_1 X + b_2 x^2 + \cdots + b_n X^n$
We can subtract from $f$ a suitable multiple... | Division Theorem for Polynomial Forms over Field/Proof 2 | https://proofwiki.org/wiki/Division_Theorem_for_Polynomial_Forms_over_Field | https://proofwiki.org/wiki/Division_Theorem_for_Polynomial_Forms_over_Field/Proof_2 | [
"Field Theory",
"Polynomial Theory",
"Division Theorem for Polynomial Forms over Field",
"Division Theorem"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Multiplicative Identity",
"Definition:Transcendental (Abstract Algebra)/Field Extension/Element",
"Definition:Ring of Polynomials in Ring Element",
"Definition:Element",
"Definition:Degree of Polynomial/Field",
"Definition:Deg... | [
"Definition:Degree of Polynomial/Field",
"Definition:Degree of Polynomial/Field",
"Definition:Degree of Polynomial/Field",
"Second Principle of Mathematical Induction"
] |
proofwiki-1386 | Division Theorem for Polynomial Forms over Field | Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $X$ be transcendental over $F$.
Let $F \sqbrk X$ be the ring of polynomials in $X$ over $F$.
Let $d$ be an element of $F \sqbrk X$ of degree $n \ge 1$.
Then $\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$... | Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ provided that $\map \deg f < n$
=== Basis for the Induction ===
$\map P 0$ is the statement that $q$ and $r$ exist when $f = 0$.
This is shown trivially to be true... | Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$.
Let $X$ be [[Definition:Transcendental over Field|transcendental over $F$]].
Let $F \sqbrk X$ be the [[Definition:Ring of Polynomials in ... | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ provided that $\map \deg f < n$
=== Basis for the Induction ===
$\map P 0$ is the statement ... | Division Theorem for Polynomial Forms over Field/Proof 3 | https://proofwiki.org/wiki/Division_Theorem_for_Polynomial_Forms_over_Field | https://proofwiki.org/wiki/Division_Theorem_for_Polynomial_Forms_over_Field/Proof_3 | [
"Field Theory",
"Polynomial Theory",
"Division Theorem for Polynomial Forms over Field",
"Division Theorem"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Multiplicative Identity",
"Definition:Transcendental (Abstract Algebra)/Field Extension/Element",
"Definition:Ring of Polynomials in Ring Element",
"Definition:Element",
"Definition:Degree of Polynomial/Field",
"Definition:Deg... | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Definition:Coefficient of Polynomial",
"Division Theorem for Polynomial Forms over Field/Proof 3",
"Degree of Sum of Polynomials",
"... |
proofwiki-1387 | Polynomial Forms over Field form Principal Ideal Domain | Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $X$ be transcendental over $F$.
Let $F \sqbrk X$ be the ring of polynomials in $X$ over $F$.
Then $F \sqbrk X$ is a principal ideal domain. | For any $d \in F \sqbrk X$, let $\ideal d$ denote the principal ideal of $F \sqbrk X$ generated by $d$.
Let $J$ be any ideal of $F \sqbrk X$. What we need to prove is that $J$ is a principal ideal.
Let us first distinguish the following two cases for $J$:
:If $J = \set {0_F}$, then by Zero Element Generates Null Ideal ... | Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$.
Let $X$ be [[Definition:Transcendental over Field|transcendental over $F$]].
Let $F \sqbrk X$ be the [[Definition:Ring of Polynomials in ... | For any $d \in F \sqbrk X$, let $\ideal d$ denote the [[Definition:Principal Ideal of Ring|principal ideal of $F \sqbrk X$ generated by $d$]].
Let $J$ be any [[Definition:Ideal of Ring|ideal]] of $F \sqbrk X$. What we need to prove is that $J$ is a [[Definition:Principal Ideal of Ring|principal ideal]].
Let us first... | Polynomial Forms over Field form Principal Ideal Domain/Proof 1 | https://proofwiki.org/wiki/Polynomial_Forms_over_Field_form_Principal_Ideal_Domain | https://proofwiki.org/wiki/Polynomial_Forms_over_Field_form_Principal_Ideal_Domain/Proof_1 | [
"Field Theory",
"Polynomial Theory",
"Principal Ideal Domains",
"Polynomial Forms over Field form Principal Ideal Domain"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Multiplicative Identity",
"Definition:Transcendental (Abstract Algebra)/Field Extension/Element",
"Definition:Ring of Polynomials in Ring Element",
"Definition:Principal Ideal Domain"
] | [
"Definition:Principal Ideal of Ring",
"Definition:Ideal of Ring",
"Definition:Principal Ideal of Ring",
"Definition:Distinct/Singular",
"Zero Element Generates Null Ideal",
"Definition:Principal Ideal of Ring",
"Ideal of Unit is Whole Ring/Corollary",
"Definition:Principal Ideal of Ring",
"Definitio... |
proofwiki-1388 | Polynomial Forms over Field form Principal Ideal Domain | Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $X$ be transcendental over $F$.
Let $F \sqbrk X$ be the ring of polynomials in $X$ over $F$.
Then $F \sqbrk X$ is a principal ideal domain. | We have that Polynomial Forms over Field is Euclidean Domain.
We also have that Euclidean Domain is Principal Ideal Domain.
Hence the result.
{{qed}} | Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$.
Let $X$ be [[Definition:Transcendental over Field|transcendental over $F$]].
Let $F \sqbrk X$ be the [[Definition:Ring of Polynomials in ... | We have that [[Polynomial Forms over Field is Euclidean Domain]].
We also have that [[Euclidean Domain is Principal Ideal Domain]].
Hence the result.
{{qed}} | Polynomial Forms over Field form Principal Ideal Domain/Proof 2 | https://proofwiki.org/wiki/Polynomial_Forms_over_Field_form_Principal_Ideal_Domain | https://proofwiki.org/wiki/Polynomial_Forms_over_Field_form_Principal_Ideal_Domain/Proof_2 | [
"Field Theory",
"Polynomial Theory",
"Principal Ideal Domains",
"Polynomial Forms over Field form Principal Ideal Domain"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Multiplicative Identity",
"Definition:Transcendental (Abstract Algebra)/Field Extension/Element",
"Definition:Ring of Polynomials in Ring Element",
"Definition:Principal Ideal Domain"
] | [
"Polynomial Forms over Field is Euclidean Domain",
"Euclidean Domain is Principal Ideal Domain"
] |
proofwiki-1389 | Equal Consecutive Prime Number Gaps are Multiples of Six | If you list the gaps between consecutive primes greater than $5$:
:$2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, \ldots$
you will notice that consecutive gaps that are equal are of the form $6 x$.
This is ''always'' the case.
{{OEIS|A001223}} | Suppose there were two consecutive gaps between $3$ consecutive prime numbers that were equal, but not divisible by $6$.
Then the difference is $2 k$ where $k$ is not divisible by $3$.
Therefore the (supposed) prime numbers will be:
:$p, p + 2 k, p + 4 k$
But then $p + 4 k$ is congruent modulo $3$ to $p + k$.
That make... | If you list the gaps between consecutive [[Definition:Prime Number|primes]] greater than $5$:
:$2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, \ldots$
you will notice that consecutive gaps that are equal are of the form $6 x$.
This is ''always'' the case.
{{OEIS|A001223}} | Suppose there were two consecutive gaps between $3$ consecutive [[Definition:Prime Number|prime numbers]] that were equal, but not [[Definition:Divisor of Integer|divisible]] by $6$.
Then the difference is $2 k$ where $k$ is not [[Definition:Divisor of Integer|divisible]] by $3$.
Therefore the (supposed) [[Definition... | Equal Consecutive Prime Number Gaps are Multiples of Six | https://proofwiki.org/wiki/Equal_Consecutive_Prime_Number_Gaps_are_Multiples_of_Six | https://proofwiki.org/wiki/Equal_Consecutive_Prime_Number_Gaps_are_Multiples_of_Six | [
"Prime Numbers"
] | [
"Definition:Prime Number"
] | [
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Prime Number",
"Definition:Congruence (Number Theory)/Integers",
"Definition:Congruence (Number Theory)/Integers",
"Definition:Divisor (Algebra)/Integer",
"Definition:Prime Number",
... |
proofwiki-1390 | Standard Discrete Metric is Metric | The standard discrete metric is a metric. | Let $d: S \times S \to \R$ denote the standard discrete metric on the underlying set $S$ of some space $\struct {S, d}$.
By definition:
:$\forall x, y \in S: \map d {x, y} = \begin {cases} 0 & : x = y \\ 1 & : x \ne y \end {cases}$ | The [[Definition:Standard Discrete Metric|standard discrete metric]] is a [[Definition:Metric|metric]]. | Let $d: S \times S \to \R$ denote the [[Definition:Standard Discrete Metric|standard discrete metric]] on the [[Definition:Underlying Set of Metric Space|underlying set]] $S$ of some [[Definition:Metric Space|space]] $\struct {S, d}$.
By definition:
:$\forall x, y \in S: \map d {x, y} = \begin {cases} 0 & : x = y \\ ... | Standard Discrete Metric is Metric | https://proofwiki.org/wiki/Standard_Discrete_Metric_is_Metric | https://proofwiki.org/wiki/Standard_Discrete_Metric_is_Metric | [
"Standard Discrete Metric"
] | [
"Definition:Standard Discrete Metric",
"Definition:Metric Space/Metric"
] | [
"Definition:Standard Discrete Metric",
"Definition:Underlying Set/Metric Space",
"Definition:Metric Space"
] |
proofwiki-1391 | Derivative of Constant | Let $\map {f_c} x$ be the constant function on $\R$, where $c \in \R$.
Then:
:$\map { {f_c}'} x = 0$ | The function $f_c: \R \to \R$ is defined as:
:$\forall x \in \R: \map {f_c} x = c$
Thus:
{{begin-eqn}}
{{eqn | l = \map { {f_c}'} x
| r = \lim_{\delta x \mathop \to 0} \frac {\map {f_c} {x + \delta x} - \map {f_c} x} {\delta x}
| c = {{Defof|Differentiation}}
}}
{{eqn | r = \lim_{\delta x \mathop \to 0} \fr... | Let $\map {f_c} x$ be the [[Definition:Constant Mapping|constant function]] on $\R$, where $c \in \R$.
Then:
:$\map { {f_c}'} x = 0$ | The function $f_c: \R \to \R$ is defined as:
:$\forall x \in \R: \map {f_c} x = c$
Thus:
{{begin-eqn}}
{{eqn | l = \map { {f_c}'} x
| r = \lim_{\delta x \mathop \to 0} \frac {\map {f_c} {x + \delta x} - \map {f_c} x} {\delta x}
| c = {{Defof|Differentiation}}
}}
{{eqn | r = \lim_{\delta x \mathop \to 0} \... | Derivative of Constant | https://proofwiki.org/wiki/Derivative_of_Constant | https://proofwiki.org/wiki/Derivative_of_Constant | [
"Derivatives",
"Constant Mappings"
] | [
"Definition:Constant Mapping"
] | [] |
proofwiki-1392 | Product Rule for Derivatives | Let $\map f x, \map j x, \map k x$ be real functions defined on the open interval $I$.
Let $\xi \in I$ be a point in $I$ at which both $j$ and $k$ are differentiable.
Let $\map f x = \map j x \map k x$.
Then:
:$\map {f'} \xi = \map j \xi \map {k'} \xi + \map {j'} \xi \map k \xi$
It follows from the definition of deriva... | First we note that from Differentiable Function is Continuous, $j$ is continuous at $\xi$.
Hence:
:$(1): \quad \map j {\xi + h} \to \map j \xi$ as $h \to 0$
So:
{{begin-eqn}}
{{eqn | l = \map {f'} \xi
| r = \lim_{h \mathop \to 0} \frac {\map f {\xi + h} - \map f \xi} h
| c = {{Defof|Derivative}}
}}
{{eqn | ... | Let $\map f x, \map j x, \map k x$ be [[Definition:Real Function|real functions]] defined on the [[Definition:Open Real Interval|open interval]] $I$.
Let $\xi \in I$ be a point in $I$ at which both $j$ and $k$ are [[Definition:Differentiable Real Function at Point|differentiable]].
Let $\map f x = \map j x \map k x$... | First we note that from [[Differentiable Function is Continuous]], $j$ is [[Definition:Continuous Real Function|continuous]] at $\xi$.
Hence:
:$(1): \quad \map j {\xi + h} \to \map j \xi$ as $h \to 0$
So:
{{begin-eqn}}
{{eqn | l = \map {f'} \xi
| r = \lim_{h \mathop \to 0} \frac {\map f {\xi + h} - \map f \xi}... | Product Rule for Derivatives/Proof | https://proofwiki.org/wiki/Product_Rule_for_Derivatives | https://proofwiki.org/wiki/Product_Rule_for_Derivatives/Proof | [
"Product Rule for Derivatives",
"Derivatives",
"Differential Calculus"
] | [
"Definition:Real Function",
"Definition:Real Interval/Open",
"Definition:Differentiable Mapping/Real Function/Point",
"Definition:Derivative/Real Function/Derivative on Interval",
"Definition:Differentiable Mapping/Real Function/Interval",
"Definition:Real Interval/Open"
] | [
"Differentiable Function is Continuous",
"Definition:Continuous Real Function",
"Definition:Fraction/Numerator",
"Combination Theorem for Limits of Functions/Real/Sum Rule",
"Combination Theorem for Limits of Functions/Real/Product Rule"
] |
proofwiki-1393 | Derivative of Composite Function | Let $I, J$ be open real intervals.
Let $g : I \to J$ and $f : J \to \R$ be real functions.
Let $h : I \to \R$ be the real function defined as:
:$\forall x \in I: \map h x = \map {f \circ g} x = \map f {\map g x}$
Then, for each $x_0 \in I$ such that:
:$g$ is differentiable at $x_0$
:$f$ is differentiable at $\map g {x_... | Some sources, introducing the Derivative of Composite Function at elementary level, provide the following non-rigorous argument:
:''If $z$ is a function of $y$ where $y$ itself is some function of $x$,''
::''it is obvious that:''
:::$\dfrac {\delta z} {\delta x} = \dfrac {\delta z} {\delta y} \cdot \dfrac {\delta y} {\... | Let $I, J$ be [[Definition:Open Real Interval|open real intervals]].
Let $g : I \to J$ and $f : J \to \R$ be [[Definition:Real Function|real functions]].
Let $h : I \to \R$ be the [[Definition:Real Function|real function]] defined as:
:$\forall x \in I: \map h x = \map {f \circ g} x = \map f {\map g x}$
Then, for ... | Some sources, introducing the [[Derivative of Composite Function]] at elementary level, provide the following non-rigorous argument:
:''If $z$ is a [[Definition:Real Function|function]] of $y$ where $y$ itself is some [[Definition:Real Function|function]] of $x$,''
::''it is obvious that:''
:::$\dfrac {\delta z} {\del... | Derivative of Composite Function/Informal Proof | https://proofwiki.org/wiki/Derivative_of_Composite_Function | https://proofwiki.org/wiki/Derivative_of_Composite_Function/Informal_Proof | [
"Derivative of Composite Function",
"Differential Calculus"
] | [
"Definition:Real Interval/Open",
"Definition:Real Function",
"Definition:Real Function",
"Definition:Differentiable Mapping/Real Function/Point",
"Definition:Differentiable Mapping/Real Function/Point",
"Definition:Differentiable Mapping/Real Function/Point",
"Definition:Derivative/Real Function"
] | [
"Derivative of Composite Function",
"Definition:Real Function",
"Definition:Real Function",
"Definition:Finite",
"Definition:Arithmetic",
"Definition:Limit of Real Function",
"Definition:Derivative",
"Definition:Fraction"
] |
proofwiki-1394 | Derivative of Composite Function | Let $I, J$ be open real intervals.
Let $g : I \to J$ and $f : J \to \R$ be real functions.
Let $h : I \to \R$ be the real function defined as:
:$\forall x \in I: \map h x = \map {f \circ g} x = \map f {\map g x}$
Then, for each $x_0 \in I$ such that:
:$g$ is differentiable at $x_0$
:$f$ is differentiable at $\map g {x_... | Let $f, g, x_0$ satisfy the conditions of the theorem.
Define $g^* : I \to \R$ as:
:$\map {g^*} x = \begin{cases} \dfrac {\map g x - \map g {x_0}} {x - x_0} & : x \ne x_0 \\ \map {g'} {x_0} & : x = x_0 \end{cases}$
Then, for every $x \in I$:
:$\paren 1: \quad \map g x - \map g {x_0} = \paren {x - x_0} \map {g^*} x$
for... | Let $I, J$ be [[Definition:Open Real Interval|open real intervals]].
Let $g : I \to J$ and $f : J \to \R$ be [[Definition:Real Function|real functions]].
Let $h : I \to \R$ be the [[Definition:Real Function|real function]] defined as:
:$\forall x \in I: \map h x = \map {f \circ g} x = \map f {\map g x}$
Then, for ... | Let $f, g, x_0$ satisfy the conditions of the theorem.
Define $g^* : I \to \R$ as:
:$\map {g^*} x = \begin{cases} \dfrac {\map g x - \map g {x_0}} {x - x_0} & : x \ne x_0 \\ \map {g'} {x_0} & : x = x_0 \end{cases}$
Then, for every $x \in I$:
:$\paren 1: \quad \map g x - \map g {x_0} = \paren {x - x_0} \map {g^*} x$
f... | Derivative of Composite Function/Proof 1 | https://proofwiki.org/wiki/Derivative_of_Composite_Function | https://proofwiki.org/wiki/Derivative_of_Composite_Function/Proof_1 | [
"Derivative of Composite Function",
"Differential Calculus"
] | [
"Definition:Real Interval/Open",
"Definition:Real Function",
"Definition:Real Function",
"Definition:Differentiable Mapping/Real Function/Point",
"Definition:Differentiable Mapping/Real Function/Point",
"Definition:Differentiable Mapping/Real Function/Point",
"Definition:Derivative/Real Function"
] | [
"Definition:Derivative/Real Function/Derivative at Point",
"Definition:Continuous Real Function/Point",
"Definition:Continuous Real Function/Point",
"Combination Theorem for Limits of Functions/Real/Product Rule",
"Limit of Composite Function",
"Differentiable Function is Continuous",
"Definition:Deriva... |
proofwiki-1395 | Derivative of Composite Function | Let $I, J$ be open real intervals.
Let $g : I \to J$ and $f : J \to \R$ be real functions.
Let $h : I \to \R$ be the real function defined as:
:$\forall x \in I: \map h x = \map {f \circ g} x = \map f {\map g x}$
Then, for each $x_0 \in I$ such that:
:$g$ is differentiable at $x_0$
:$f$ is differentiable at $\map g {x_... | Let $\map g x = y$, and let:
{{begin-eqn}}
{{eqn | l = \map g {x + \delta x}
| r = y + \delta y
| c =
}}
{{eqn | ll= \leadsto
| l = \delta y
| r = \map g {x + \delta x} - \map g x
| c =
}}
{{end-eqn}}
Thus:
:$\delta y \to 0$ as $\delta x \to 0$
and:
:$(1): \quad \dfrac {\delta y} {\delta... | Let $I, J$ be [[Definition:Open Real Interval|open real intervals]].
Let $g : I \to J$ and $f : J \to \R$ be [[Definition:Real Function|real functions]].
Let $h : I \to \R$ be the [[Definition:Real Function|real function]] defined as:
:$\forall x \in I: \map h x = \map {f \circ g} x = \map f {\map g x}$
Then, for ... | Let $\map g x = y$, and let:
{{begin-eqn}}
{{eqn | l = \map g {x + \delta x}
| r = y + \delta y
| c =
}}
{{eqn | ll= \leadsto
| l = \delta y
| r = \map g {x + \delta x} - \map g x
| c =
}}
{{end-eqn}}
Thus:
:$\delta y \to 0$ as $\delta x \to 0$
and:
:$(1): \quad \dfrac {\delta y} {\de... | Derivative of Composite Function/Proof 2 | https://proofwiki.org/wiki/Derivative_of_Composite_Function | https://proofwiki.org/wiki/Derivative_of_Composite_Function/Proof_2 | [
"Derivative of Composite Function",
"Differential Calculus"
] | [
"Definition:Real Interval/Open",
"Definition:Real Function",
"Definition:Real Function",
"Definition:Differentiable Mapping/Real Function/Point",
"Definition:Differentiable Mapping/Real Function/Point",
"Definition:Differentiable Mapping/Real Function/Point",
"Definition:Derivative/Real Function"
] | [
"Proof by Cases"
] |
proofwiki-1396 | Derivative of Inverse Function | Let $I = \closedint a b$ and $J = \closedint c d$ be closed real intervals.
Let $I^o = \openint a b$ and $J^o = \openint c d$ be the corresponding open real intervals.
Let $f: I \to J$ be a real function which is continuous on $I$ and differentiable on $I^o$ such that $J = f \sqbrk I$.
Let either:
:$\forall x \in I^o: ... | From Derivative of Monotone Function, it follows that $f$ is either:
:strictly increasing on $I$ (if $\forall x \in I^o: D \, \map f x > 0$)
or:
:strictly decreasing on $I$ (if $\forall x \in I^o: D \, \map f x < 0$).
Therefore from Inverse of Strictly Monotone Function it follows that $f^{-1}: J \to I$ exists.
As $f$ ... | Let $I = \closedint a b$ and $J = \closedint c d$ be [[Definition:Closed Real Interval|closed real intervals]].
Let $I^o = \openint a b$ and $J^o = \openint c d$ be the corresponding [[Definition:Open Real Interval|open real intervals]].
Let $f: I \to J$ be a [[Definition:Real Function|real function]] which is [[Def... | From [[Derivative of Monotone Function]], it follows that $f$ is either:
:[[Definition:Strictly Increasing Real Function|strictly increasing]] on $I$ (if $\forall x \in I^o: D \, \map f x > 0$)
or:
:[[Definition:Strictly Decreasing Real Function|strictly decreasing]] on $I$ (if $\forall x \in I^o: D \, \map f x < 0$).
... | Derivative of Inverse Function | https://proofwiki.org/wiki/Derivative_of_Inverse_Function | https://proofwiki.org/wiki/Derivative_of_Inverse_Function | [
"Derivative of Inverse Function",
"Differential Calculus"
] | [
"Definition:Real Interval/Closed",
"Definition:Real Interval/Open",
"Definition:Real Function",
"Definition:Continuous Real Function/Interval",
"Definition:Differentiable Mapping/Real Function/Interval",
"Definition:Continuous Real Function/Interval",
"Definition:Differentiable Mapping/Real Function/Int... | [
"Derivative of Monotone Function",
"Definition:Strictly Increasing/Real Function",
"Definition:Strictly Decreasing/Real Function",
"Inverse of Strictly Monotone Function",
"Definition:Continuous Real Function/Interval",
"Image of Real Interval under Continuous Real Function is Real Interval",
"Definitio... |
proofwiki-1397 | Upper Darboux Sum Never Smaller than Lower Darboux Sum | Let $\closedint a b$ be a closed interval of the set $\R$ of real numbers.
Let $P = \set {x_0, x_1, x_2, \ldots, x_{n - 1}, x_n}$ be a finite subdivision of $\closedint a b$.
Let $f: \R \to \R$ be a real function.
Let $f$ be bounded on $\closedint a b$.
Let $\map L P$ be the lower Darboux sum of $\map f x$ on $\closedi... | For all $\nu \in 1, 2, \ldots, n$, let $\closedint {x_{\nu - 1} } {x_\nu}$ be a closed subinterval of $\closedint a b$.
As $f$ is bounded on $\closedint a b$, it is bounded on $\closedint {x_{\nu - 1} } {x_\nu}$.
So, let $m_\nu$ be the infimum and $M_\nu$ be the supremum of $\map f x$ on the interval $\closedint {x_{\n... | Let $\closedint a b$ be a [[Definition:Closed Real Interval|closed interval]] of the set $\R$ of [[Definition:Real Number|real numbers]].
Let $P = \set {x_0, x_1, x_2, \ldots, x_{n - 1}, x_n}$ be a [[Definition:Finite Subdivision|finite subdivision]] of $\closedint a b$.
Let $f: \R \to \R$ be a [[Definition:Real Func... | For all $\nu \in 1, 2, \ldots, n$, let $\closedint {x_{\nu - 1} } {x_\nu}$ be a [[Definition:Closed Real Interval|closed subinterval]] of $\closedint a b$.
As $f$ is [[Definition:Bounded Real-Valued Function|bounded]] on $\closedint a b$, it is [[Definition:Bounded Real-Valued Function|bounded]] on $\closedint {x_{\n... | Upper Darboux Sum Never Smaller than Lower Darboux Sum | https://proofwiki.org/wiki/Upper_Darboux_Sum_Never_Smaller_than_Lower_Darboux_Sum | https://proofwiki.org/wiki/Upper_Darboux_Sum_Never_Smaller_than_Lower_Darboux_Sum | [
"Real Analysis"
] | [
"Definition:Real Interval/Closed",
"Definition:Real Number",
"Definition:Subdivision of Interval/Finite",
"Definition:Real Function",
"Definition:Bounded Mapping/Real-Valued",
"Definition:Lower Darboux Sum",
"Definition:Upper Darboux Sum"
] | [
"Definition:Real Interval/Closed",
"Definition:Bounded Mapping/Real-Valued",
"Definition:Bounded Mapping/Real-Valued",
"Definition:Infimum of Mapping/Real-Valued Function",
"Definition:Supremum of Mapping/Real-Valued Function"
] |
proofwiki-1398 | Dedekind's Theorem | Let $\tuple {L, R}$ be a Dedekind cut of the set of real numbers $\R$.
Then there exists a unique real number which is a producer of $\tuple {L, R}$. | Suppose $P$ and $Q$ are two properties which are mutually exclusive.
Suppose that one of either of $P$ and $Q$ are possessed by every $x \in \R$.
Suppose that any number having $P$ is less than any which have $Q$.
Let us call the numbers with $P$ the ''left hand set'' $L$, and the ones with $Q$ the ''right hand set'' $... | Let $\tuple {L, R}$ be a [[Definition:Dedekind Cut|Dedekind cut]] of the set of [[Definition:Real Number|real numbers]] $\R$.
Then there exists a [[Definition:Unique|unique]] [[Definition:Real Number|real number]] which is a [[Definition:Producer of Dedekind Cut|producer]] of $\tuple {L, R}$. | Suppose $P$ and $Q$ are two [[Definition:Property|properties]] which are mutually exclusive.
Suppose that one of either of $P$ and $Q$ are possessed by every $x \in \R$.
Suppose that any number having $P$ is less than any which have $Q$.
Let us call the numbers with $P$ the ''left hand set'' $L$, and the ones with $... | Dedekind's Theorem/Proof 1 | https://proofwiki.org/wiki/Dedekind's_Theorem | https://proofwiki.org/wiki/Dedekind's_Theorem/Proof_1 | [
"Real Analysis",
"Dedekind Cuts",
"Dedekind's Theorem"
] | [
"Definition:Dedekind Cut",
"Definition:Real Number",
"Definition:Unique",
"Definition:Real Number",
"Definition:Producer of Dedekind Cut"
] | [
"Definition:Property",
"Definition:Subset",
"Definition:Rational Number",
"Definition:Dedekind Cut",
"Definition:Rational Number",
"Rational Numbers are Densely Ordered"
] |
proofwiki-1399 | Dedekind's Theorem | Let $\tuple {L, R}$ be a Dedekind cut of the set of real numbers $\R$.
Then there exists a unique real number which is a producer of $\tuple {L, R}$. | === Proof of Uniqueness ===
{{AimForCont}} both $\alpha$ and $\beta$ produce $\tuple {L, R}$.
By the Trichotomy Law for Real Numbers either $\beta < \alpha$ or $\alpha < \beta$.
Suppose that $\beta < \alpha$.
From Real Numbers are Densely Ordered, there exists at least one real number $c$ such that $\beta < c$ and $c <... | Let $\tuple {L, R}$ be a [[Definition:Dedekind Cut|Dedekind cut]] of the set of [[Definition:Real Number|real numbers]] $\R$.
Then there exists a [[Definition:Unique|unique]] [[Definition:Real Number|real number]] which is a [[Definition:Producer of Dedekind Cut|producer]] of $\tuple {L, R}$. | === Proof of Uniqueness ===
{{AimForCont}} both $\alpha$ and $\beta$ [[Definition:Producer of Dedekind Cut|produce]] $\tuple {L, R}$.
By the [[Trichotomy Law for Real Numbers]] either $\beta < \alpha$ or $\alpha < \beta$.
Suppose that $\beta < \alpha$.
From [[Real Numbers are Densely Ordered]], there exists at leas... | Dedekind's Theorem/Proof 2 | https://proofwiki.org/wiki/Dedekind's_Theorem | https://proofwiki.org/wiki/Dedekind's_Theorem/Proof_2 | [
"Real Analysis",
"Dedekind Cuts",
"Dedekind's Theorem"
] | [
"Definition:Dedekind Cut",
"Definition:Real Number",
"Definition:Unique",
"Definition:Real Number",
"Definition:Producer of Dedekind Cut"
] | [
"Definition:Producer of Dedekind Cut",
"Trichotomy Law for Real Numbers",
"Real Numbers are Densely Ordered",
"Definition:Real Number",
"Definition:Dedekind Cut",
"Definition:Set Partition",
"Definition:Disjoint Sets",
"Definition:Contradiction",
"Definition:Contradiction",
"Definition:Unique",
... |
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