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-4200 | Gelfond-Schneider Theorem | Let $\alpha$ and $\beta$ be algebraic numbers (possibly complex) such that $\alpha \notin \set {0, 1}$.
Let $\beta$ be irrational, that is:
:$\beta \notin \Q$
This includes $\beta$ being imaginary or complex.
Then any value of $\alpha^\beta$ is transcendental. | Let $\alpha$ be an algebraic number such that $\alpha \ne 0$ and $\alpha \ne 1$.
Let $\beta$ be an algebraic number such that some value of $\alpha^\beta$ is algebraic.
The result will follow if we can show that $\beta \in \Q$.
We treat only the special case that $\alpha, \beta \in \R$ and $\alpha > 0$, assuming that $... | Let $\alpha$ and $\beta$ be [[Definition:Algebraic Number|algebraic numbers]] (possibly [[Definition:Complex Number|complex]]) such that $\alpha \notin \set {0, 1}$.
Let $\beta$ be [[Definition:Irrational Number|irrational]], that is:
:$\beta \notin \Q$
This includes $\beta$ being [[Definition:Wholly Imaginary|imagina... | Let $\alpha$ be an [[Definition:Algebraic Number|algebraic number]] such that $\alpha \ne 0$ and $\alpha \ne 1$.
Let $\beta$ be an [[Definition:Algebraic Number|algebraic number]] such that some value of $\alpha^\beta$ is [[Definition:Algebraic Number|algebraic]].
The result will follow if we can show that $\beta \in... | Gelfond-Schneider Theorem | https://proofwiki.org/wiki/Gelfond-Schneider_Theorem | https://proofwiki.org/wiki/Gelfond-Schneider_Theorem | [
"Gelfond-Schneider Theorem",
"Transcendental Number Theory",
"Analysis"
] | [
"Definition:Algebraic Number",
"Definition:Complex Number",
"Definition:Irrational Number",
"Definition:Complex Number/Wholly Imaginary",
"Definition:Complex Number",
"Definition:Transcendental Number"
] | [
"Definition:Algebraic Number",
"Definition:Algebraic Number",
"Definition:Algebraic Number",
"Definition:Algebraic Number",
"Definition:Real Number",
"Definition:Image (Set Theory)/Mapping/Element",
"Definition:Algebraic Number",
"Definition:Integer",
"Definition:Sufficiently Large",
"Definition:S... |
proofwiki-4201 | Associative Algebra has Multiplicative Inverses iff Unitary Division Algebra | Let $\struct {A_R, \oplus}$ be an associative algebra over the ring $A_R$.
Then:
:$\struct {A_R, \oplus}$ has a unique multiplicative inverse for every non-zero $a \in A_R$
{{iff}}:
:$\struct {A_R, \oplus}$ is a unitary division algebra. | Let $A = \struct {A_R, \oplus}$ be a unitary division algebra whose zero is $\mathbf 0_A$.
From the unitary nature of $A$ we have that $\oplus$ has a unit $1$ such that:
:$\forall a \in A_R, a \ne \mathbf 0_A: a \oplus 1 = a = 1 \oplus a$
From the division algebra nature of $A$ we have that $R$ is a field and that:
: $... | Let $\struct {A_R, \oplus}$ be an [[Definition:Associative Algebra|associative algebra]] over the [[Definition:Ring (Abstract Algebra)|ring]] $A_R$.
Then:
:$\struct {A_R, \oplus}$ has a unique [[Definition:Multiplicative Inverse|multiplicative inverse]] for every [[Definition:Zero Vector|non-zero]] $a \in A_R$
{{iff}... | Let $A = \struct {A_R, \oplus}$ be a [[Definition:Unitary Division Algebra|unitary division algebra]] whose [[Definition:Zero Vector|zero]] is $\mathbf 0_A$.
From the [[Definition:Unitary Algebra|unitary]] nature of $A$ we have that $\oplus$ has a [[Definition:Unit of Algebra|unit]] $1$ such that:
:$\forall a \in A_R,... | Associative Algebra has Multiplicative Inverses iff Unitary Division Algebra | https://proofwiki.org/wiki/Associative_Algebra_has_Multiplicative_Inverses_iff_Unitary_Division_Algebra | https://proofwiki.org/wiki/Associative_Algebra_has_Multiplicative_Inverses_iff_Unitary_Division_Algebra | [
"Associative Algebras",
"Multiplicative Inverses",
"Unitary Division Algebras"
] | [
"Definition:Associative Algebra",
"Definition:Ring (Abstract Algebra)",
"Definition:Multiplicative Inverse",
"Definition:Zero Vector",
"Definition:Unitary Division Algebra"
] | [
"Definition:Unitary Division Algebra",
"Definition:Zero Vector",
"Definition:Unital Algebra",
"Definition:Unit of Algebra",
"Definition:Division Algebra",
"Definition:Field (Abstract Algebra)",
"Definition:Associative Operation",
"Definition:Multiplicative Inverse",
"Definition:Multiplicative Invers... |
proofwiki-4202 | Norm of Unit of Normed Division Algebra | Let $\struct {A_F, \oplus}$ be a normed division algebra.
Let the unit of $\struct {A_F, \oplus}$ be $1_A$.
Then:
:$\norm {1_A} = 1$
where $\norm {1_A}$ denotes the norm of $1_A$. | By definition:
:$\forall a, b \in A_F: \norm {a \oplus b} = \norm a \norm b$
So:
:$\norm {1_A} = \norm {1_A \oplus 1_A} = \norm {1_A} \norm {1_A}$
So $\norm {1_A} \in \R$ is idempotent under real multiplication.
From Idempotent Elements of Ring with No Proper Zero Divisors, only $0 \in \R$ and $1 \in \R$ fit that bill.... | Let $\struct {A_F, \oplus}$ be a [[Definition:Normed Division Algebra|normed division algebra]].
Let the [[Definition:Unit of Algebra|unit]] of $\struct {A_F, \oplus}$ be $1_A$.
Then:
:$\norm {1_A} = 1$
where $\norm {1_A}$ denotes the [[Definition:Norm on Vector Space|norm]] of $1_A$. | By definition:
:$\forall a, b \in A_F: \norm {a \oplus b} = \norm a \norm b$
So:
:$\norm {1_A} = \norm {1_A \oplus 1_A} = \norm {1_A} \norm {1_A}$
So $\norm {1_A} \in \R$ is [[Definition:Idempotent Element|idempotent]] under [[Definition:Real Multiplication|real multiplication]].
From [[Idempotent Elements of Ring ... | Norm of Unit of Normed Division Algebra | https://proofwiki.org/wiki/Norm_of_Unit_of_Normed_Division_Algebra | https://proofwiki.org/wiki/Norm_of_Unit_of_Normed_Division_Algebra | [
"Normed Division Algebras",
"Norm Theory"
] | [
"Definition:Normed Division Algebra",
"Definition:Unit of Algebra",
"Definition:Norm/Vector Space"
] | [
"Definition:Idempotence/Element",
"Definition:Multiplication/Real Numbers",
"Idempotent Elements of Ring with No Proper Zero Divisors",
"Definition:Norm/Vector Space"
] |
proofwiki-4203 | Commutator on Algebra is Alternating Bilinear Mapping | Let $\struct {A_R, \oplus}$ be an algebra over a ring.
Then the commutator on $\struct {A_R, \oplus}$ is an alternating bilinear mapping:
:$\forall a, b \in A_R: \sqbrk {a, b} = -\sqbrk {b, a}$ | {{begin-eqn}}
{{eqn | l = \sqbrk {a, b}
| r = a \oplus b - b \oplus a
| c =
}}
{{eqn | r = -b \oplus a + a \oplus b
| c =
}}
{{eqn | r = -b \oplus a - \paren {-\paren {a \oplus b} }
| c =
}}
{{eqn | r = -\paren {b \oplus a - a \oplus b}
| c =
}}
{{eqn | r = -\sqbrk {b, a}
| c =
... | Let $\struct {A_R, \oplus}$ be an [[Definition:Algebra over Ring|algebra over a ring]].
Then the [[Definition:Commutator on Algebra|commutator]] on $\struct {A_R, \oplus}$ is an [[Definition:Alternating Bilinear Mapping|alternating bilinear mapping]]:
:$\forall a, b \in A_R: \sqbrk {a, b} = -\sqbrk {b, a}$ | {{begin-eqn}}
{{eqn | l = \sqbrk {a, b}
| r = a \oplus b - b \oplus a
| c =
}}
{{eqn | r = -b \oplus a + a \oplus b
| c =
}}
{{eqn | r = -b \oplus a - \paren {-\paren {a \oplus b} }
| c =
}}
{{eqn | r = -\paren {b \oplus a - a \oplus b}
| c =
}}
{{eqn | r = -\sqbrk {b, a}
| c =
... | Commutator on Algebra is Alternating Bilinear Mapping | https://proofwiki.org/wiki/Commutator_on_Algebra_is_Alternating_Bilinear_Mapping | https://proofwiki.org/wiki/Commutator_on_Algebra_is_Alternating_Bilinear_Mapping | [
"Algebras",
"Alternating Bilinear Mappings",
"Commutators"
] | [
"Definition:Algebra over Ring",
"Definition:Commutator/Algebra",
"Definition:Alternating Bilinear Mapping"
] | [] |
proofwiki-4204 | Quaternions Defined by Ordered Pairs | Consider the quaternions $\Bbb H$ as numbers in the form:
: $a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$
where:
: $a, b, c, d$ are real numbers;
: $\mathbf 1, \mathbf i, \mathbf j, \mathbf k$ are entities related to each other in the following way:
{{begin-eqn}}
{{eqn | l = \mathbf i \mathbf j = - \math... | First we identify the following:
{{begin-eqn}}
{{eqn | n = 1
| l = \mathbf 1
| r = \tuple {1, 0}
| c =
}}
{{eqn | n = 2
| l = \mathbf i
| r = \tuple {i, 0}
| c =
}}
{{eqn | n = 3
| l = \mathbf j
| r = \tuple {0, i}
| c =
}}
{{eqn | n = 4
| l = \mathbf k
... | Consider the [[Definition:Quaternion|quaternions]] $\Bbb H$ as numbers in the form:
: $a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$
where:
: $a, b, c, d$ are [[Definition:Real Number|real numbers]];
: $\mathbf 1, \mathbf i, \mathbf j, \mathbf k$ are entities related to each other in the following way:
{{beg... | First we identify the following:
{{begin-eqn}}
{{eqn | n = 1
| l = \mathbf 1
| r = \tuple {1, 0}
| c =
}}
{{eqn | n = 2
| l = \mathbf i
| r = \tuple {i, 0}
| c =
}}
{{eqn | n = 3
| l = \mathbf j
| r = \tuple {0, i}
| c =
}}
{{eqn | n = 4
| l = \mathbf k
... | Quaternions Defined by Ordered Pairs | https://proofwiki.org/wiki/Quaternions_Defined_by_Ordered_Pairs | https://proofwiki.org/wiki/Quaternions_Defined_by_Ordered_Pairs | [
"Quaternions"
] | [
"Definition:Quaternion",
"Definition:Real Number",
"Definition:Quaternion",
"Definition:Ordered Pair",
"Definition:Complex Number",
"Definition:Quaternion/Multiplication",
"Definition:Complex Number",
"Definition:Complex Conjugate",
"Definition:Logical Equivalence"
] | [
"Quaternion Multiplication"
] |
proofwiki-4205 | Multiplicative Inverse in Nicely Normed Star-Algebra | Let $A = \struct {A_F, \oplus}$ be a nicely normed $*$-algebra whose conjugation is denoted $*$.
Let $a \in A$.
Then the multiplicative inverse of $a$ is given by:
:$a^{-1} = \dfrac {a^*} {\norm a^2}$
where:
:$a^*$ is the conjugate of $a$
:$\norm a$ is the norm of $a$. | For the result to hold, we need to show that $a \oplus \dfrac {a^*} {\norm a^2} = 1 = \dfrac {a^*} {\norm a^2} \oplus a$.
{{begin-eqn}}
{{eqn | o =
| r = a \oplus \dfrac {a^*} {\norm a^2}
| c =
}}
{{eqn | r = a \oplus a^* \cdot \dfrac 1 {\norm a^2}
| c =
}}
{{eqn | r = \norm a^2 \cdot \dfrac 1 {\no... | Let $A = \struct {A_F, \oplus}$ be a [[Definition:Nicely Normed Star-Algebra|nicely normed $*$-algebra]] whose [[Definition:Conjugation on Algebra|conjugation]] is denoted $*$.
Let $a \in A$.
Then the [[Definition:Multiplicative Inverse|multiplicative inverse]] of $a$ is given by:
:$a^{-1} = \dfrac {a^*} {\norm a^2}... | For the result to hold, we need to show that $a \oplus \dfrac {a^*} {\norm a^2} = 1 = \dfrac {a^*} {\norm a^2} \oplus a$.
{{begin-eqn}}
{{eqn | o =
| r = a \oplus \dfrac {a^*} {\norm a^2}
| c =
}}
{{eqn | r = a \oplus a^* \cdot \dfrac 1 {\norm a^2}
| c =
}}
{{eqn | r = \norm a^2 \cdot \dfrac 1 {\n... | Multiplicative Inverse in Nicely Normed Star-Algebra | https://proofwiki.org/wiki/Multiplicative_Inverse_in_Nicely_Normed_Star-Algebra | https://proofwiki.org/wiki/Multiplicative_Inverse_in_Nicely_Normed_Star-Algebra | [
"Nicely Normed Star-Algebras",
"Multiplicative Inverses"
] | [
"Definition:Nicely Normed Star-Algebra",
"Definition:Conjugation on Algebra",
"Definition:Multiplicative Inverse",
"Definition:Conjugation on Algebra/Conjugate",
"Definition:Norm/Vector Space"
] | [
"Definition:Associative Operation"
] |
proofwiki-4206 | Nicely Normed Alternative Algebra is Normed Division Algebra | $A = \struct {A_F, \oplus}$ be a nicely normed $*$-algebra which is also an alternative algebra.
Then $A$ is a normed division algebra. | Let $a, b \in A$.
Then all of $a, b, a^*, b^*$ can be generated by $\map \Im a$ and $\map \Im b$.
{{explain|Prove the above statement.}}
So as $A$ is an alternative algebra, it follows that $\oplus$ is associative for $a, b, a^*, b^*$.
So:
{{begin-eqn}}
{{eqn | l = \norm {a b}^2
| r = \paren {a \oplus b} \oplus \... | $A = \struct {A_F, \oplus}$ be a [[Definition:Nicely Normed Star-Algebra|nicely normed $*$-algebra]] which is also an [[Definition:Alternative Algebra|alternative algebra]].
Then $A$ is a [[Definition:Normed Division Algebra|normed division algebra]]. | Let $a, b \in A$.
Then all of $a, b, a^*, b^*$ can be generated by $\map \Im a$ and $\map \Im b$.
{{explain|Prove the above statement.}}
So as $A$ is an [[Definition:Alternative Algebra|alternative algebra]], it follows that $\oplus$ is [[Definition:Associative Operation|associative]] for $a, b, a^*, b^*$.
So:
{{b... | Nicely Normed Alternative Algebra is Normed Division Algebra | https://proofwiki.org/wiki/Nicely_Normed_Alternative_Algebra_is_Normed_Division_Algebra | https://proofwiki.org/wiki/Nicely_Normed_Alternative_Algebra_is_Normed_Division_Algebra | [
"Nicely Normed Star-Algebras",
"Alternative Algebras",
"Normed Division Algebras"
] | [
"Definition:Nicely Normed Star-Algebra",
"Definition:Alternative Algebra",
"Definition:Normed Division Algebra"
] | [
"Definition:Alternative Algebra",
"Definition:Associative Operation",
"Definition:Nicely Normed Star-Algebra",
"Definition:Associative Operation"
] |
proofwiki-4207 | Artin's Theorem on Alternative Algebras | Let $A = \struct {A_R, \oplus}$ be an algebra over the ring $R$ such that $A$ is ''not'' a boolean algebra.
Then $A$ is alternative {{iff}}:
:$\forall a, b \in A: \paren {a \oplus a} \oplus b = a \oplus \paren {a \oplus b}$
:$\forall a, b \in A: \paren {b \oplus a} \oplus a = b \oplus \paren {a \oplus a}$ | When $A$ is
So suppose that:
:$\forall a, b \in A: \paren {a \oplus a} \oplus b = a \oplus \paren {a \oplus b}$
:$\forall a, b \in A: \paren {b \oplus a} \oplus a = b \oplus \paren {a \oplus a}$
Then:
:$\sqbrk {a, a, b} = 0$
:$\sqbrk {b, a, a} = 0$
where $\sqbrk {a, a, b}$ denotes the associator of $a, b \in A_R$.
Now... | Let $A = \struct {A_R, \oplus}$ be an [[Definition:Algebra over Ring|algebra over the ring]] $R$ such that $A$ is ''not'' a [[Definition:Boolean Algebra|boolean algebra]].
Then $A$ is [[Definition:Alternative Algebra|alternative]] {{iff}}:
:$\forall a, b \in A: \paren {a \oplus a} \oplus b = a \oplus \paren {a \oplus... | When $A$ is
So suppose that:
:$\forall a, b \in A: \paren {a \oplus a} \oplus b = a \oplus \paren {a \oplus b}$
:$\forall a, b \in A: \paren {b \oplus a} \oplus a = b \oplus \paren {a \oplus a}$
Then:
:$\sqbrk {a, a, b} = 0$
:$\sqbrk {b, a, a} = 0$
where $\sqbrk {a, a, b}$ denotes the [[Definition:Associator|asso... | Artin's Theorem on Alternative Algebras | https://proofwiki.org/wiki/Artin's_Theorem_on_Alternative_Algebras | https://proofwiki.org/wiki/Artin's_Theorem_on_Alternative_Algebras | [
"Alternative Algebras"
] | [
"Definition:Algebra over Ring",
"Definition:Boolean Algebra",
"Definition:Alternative Algebra"
] | [
"Definition:Associator"
] |
proofwiki-4208 | Real Numbers form Algebra | The set of real numbers $\R$ forms an algebra over the field of real numbers.
This algebra is:
:$(1): \quad$ An associative algebra.
:$(2): \quad$ A commutative algebra.
:$(3): \quad$ A normed division algebra.
:$(4): \quad$ A nicely normed $*$-algebra whose $*$ operator is the identity mapping.
:$(5): \quad$ A real $*... | === Construction of Algebra ===
From Real Numbers form Field, $\struct {\R, +, \times}$ is a field.
Let this be expressed as $\struct {\R, +_\R, \times_\R}$ in order to call attention to the precise scope of the operators.
From Real Numbers form Vector Space, we have that $\struct {\R^1, +, \cdot}_\R$ is a vector space... | The [[Definition:Real Number|set of real numbers]] $\R$ forms an [[Definition:Algebra over Field|algebra]] over the [[Definition:Field of Real Numbers|field of real numbers]].
This [[Definition:Algebra over Field|algebra]] is:
:$(1): \quad$ An [[Definition:Associative Algebra|associative algebra]].
:$(2): \quad$ A [[D... | === Construction of Algebra ===
From [[Real Numbers form Field]], $\struct {\R, +, \times}$ is a [[Definition:Field (Abstract Algebra)|field]].
Let this be expressed as $\struct {\R, +_\R, \times_\R}$ in order to call attention to the precise scope of the operators.
From [[Real Numbers form Vector Space]], we have t... | Real Numbers form Algebra | https://proofwiki.org/wiki/Real_Numbers_form_Algebra | https://proofwiki.org/wiki/Real_Numbers_form_Algebra | [
"Real Numbers",
"Algebras"
] | [
"Definition:Real Number",
"Definition:Algebra over Field",
"Definition:Field of Real Numbers",
"Definition:Algebra over Field",
"Definition:Associative Algebra",
"Definition:Commutative Algebra (Abstract Algebra)",
"Definition:Normed Division Algebra",
"Definition:Nicely Normed Star-Algebra",
"Defin... | [
"Real Numbers form Field",
"Definition:Field (Abstract Algebra)",
"Real Numbers form Vector Space",
"Definition:Vector Space",
"Definition:Field (Abstract Algebra)",
"Definition:Abelian Group",
"Definition:Addition/Real Numbers",
"Real Numbers form Vector Space",
"Definition:Real Number",
"Definit... |
proofwiki-4209 | Real Numbers form Vector Space | The set of real numbers $\R$, with the operations of addition and multiplication, forms a vector space. | Let the field of real numbers be denoted $\struct {\R, +, \times}$.
From Real Vector Space is Vector Space, we have that $\struct {\R^n, +, \cdot}$ is a vector space, where:
:$\mathbf a + \mathbf b = \tuple {a_1 + b_1, a_2 + b_2, \ldots, a_n + b_n}$
:$\lambda \cdot \mathbf a = \tuple {\lambda \times a_1, \lambda \times... | The [[Definition:Real Number|set of real numbers]] $\R$, with the operations of [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]], forms a [[Definition:Vector Space|vector space]]. | Let the [[Definition:Field of Real Numbers|field of real numbers]] be denoted $\struct {\R, +, \times}$.
From [[Real Vector Space is Vector Space]], we have that $\struct {\R^n, +, \cdot}$ is a [[Definition:Vector Space|vector space]], where:
:$\mathbf a + \mathbf b = \tuple {a_1 + b_1, a_2 + b_2, \ldots, a_n + b_n}$... | Real Numbers form Vector Space | https://proofwiki.org/wiki/Real_Numbers_form_Vector_Space | https://proofwiki.org/wiki/Real_Numbers_form_Vector_Space | [
"Linear Algebra",
"Real Numbers",
"Examples of Vector Spaces"
] | [
"Definition:Real Number",
"Definition:Addition/Real Numbers",
"Definition:Multiplication/Real Numbers",
"Definition:Vector Space"
] | [
"Definition:Field of Real Numbers",
"Real Vector Space is Vector Space",
"Definition:Vector Space",
"Definition:Vector Space",
"Definition:Vector Space",
"Definition:Field of Real Numbers"
] |
proofwiki-4210 | Complex Numbers form Algebra | The set of complex numbers $\C$ forms an algebra over the field of real numbers.
This algebra is:
:$(1): \quad$ An associative algebra.
:$(2): \quad$ A commutative algebra.
:$(3): \quad$ A normed division algebra.
:$(4): \quad$ A nicely normed $*$-algebra.
However, $\C$ is not a real $*$-algebra. | The complex numbers $\C$ are formed by the Cayley-Dickson Construction from the real numbers $\R$.
From Real Numbers form Algebra, we have that $\R$ forms:
:$(1): \quad$ An associative algebra.
:$(2): \quad$ A commutative algebra.
:$(3): \quad$ A normed division algebra.
:$(4): \quad$ A nicely normed $*$-algebra whose ... | The [[Definition:Complex Number|set of complex numbers]] $\C$ forms an [[Definition:Algebra over Field|algebra]] over the [[Definition:Field of Real Numbers|field of real numbers]].
This [[Definition:Algebra over Field|algebra]] is:
:$(1): \quad$ An [[Definition:Associative Algebra|associative algebra]].
:$(2): \quad$... | The [[Definition:Complex Number|complex numbers]] $\C$ are formed by the [[Definition:Complex Number/Construction from Cayley-Dickson Construction|Cayley-Dickson Construction]] from the [[Definition:Real Number|real numbers]] $\R$.
From [[Real Numbers form Algebra]], we have that $\R$ forms:
:$(1): \quad$ An [[Definit... | Complex Numbers form Algebra | https://proofwiki.org/wiki/Complex_Numbers_form_Algebra | https://proofwiki.org/wiki/Complex_Numbers_form_Algebra | [
"Complex Numbers",
"Algebras"
] | [
"Definition:Complex Number",
"Definition:Algebra over Field",
"Definition:Field of Real Numbers",
"Definition:Algebra over Field",
"Definition:Associative Algebra",
"Definition:Commutative Algebra (Abstract Algebra)",
"Definition:Normed Division Algebra",
"Definition:Nicely Normed Star-Algebra",
"De... | [
"Definition:Complex Number",
"Definition:Complex Number/Construction from Cayley-Dickson Construction",
"Definition:Real Number",
"Real Numbers form Algebra",
"Definition:Associative Algebra",
"Definition:Commutative Algebra (Abstract Algebra)",
"Definition:Normed Division Algebra",
"Definition:Nicely... |
proofwiki-4211 | Characterization of Class Membership | Let $A$ and $B$ be classes.
Then:
:$\forall A, B: \paren {A \in B \iff \exists x: \paren {A = x \land x \in B} }$
where $x$ is specifically a set. | Let $V$ denote the universal class.
By Class is Subclass of Universal Class, $A \subseteq V$ and $B \subseteq V$.
By definition of universal class, every element of $V$ is a set.
Hence every element of $B$ is a set.
So if $A \in B$, then it follows that $A$ is itself a set.
Hence the result.
{{qed}} | Let $A$ and $B$ be [[Definition:Class (Class Theory)|classes]].
Then:
:$\forall A, B: \paren {A \in B \iff \exists x: \paren {A = x \land x \in B} }$
where $x$ is specifically a [[Definition:Set|set]]. | Let $V$ denote the [[Definition:Universal Class|universal class]].
By [[Class is Subclass of Universal Class]], $A \subseteq V$ and $B \subseteq V$.
By definition of [[Definition:Universal Class|universal class]], every [[Definition:Element of Class|element]] of $V$ is a [[Definition:Set|set]].
Hence every [[Definit... | Characterization of Class Membership | https://proofwiki.org/wiki/Characterization_of_Class_Membership | https://proofwiki.org/wiki/Characterization_of_Class_Membership | [
"Class Theory"
] | [
"Definition:Class (Class Theory)",
"Definition:Set"
] | [
"Definition:Universal Class",
"Class is Subclass of Universal Class",
"Definition:Universal Class",
"Definition:Element/Class",
"Definition:Set",
"Definition:Element/Class",
"Definition:Set",
"Definition:Set"
] |
proofwiki-4212 | Equality implies Substitution | Let $\map P x$ denote a Well-Formed Formula which contains $x$ as a free variable.
Then the following are tautologies:
:$\forall x: \paren {\map P x \iff \exists y: \paren {y = x \land \map P y} }$
:$\forall x: \paren {\map P x \iff \forall y: \paren {y = x \implies \map P y} }$
Note that when $y$ is substituted for $x... | {{questionable|Inelegant and unclear to the point of soliciting purging}}
By Universal Affirmative implies Particular Affirmative iff First Predicate is not Vacuous:
:$\paren {\exists y: y = x \land \forall y: \paren {y = x \implies \map P x} } \implies \exists y: \paren {y = x \land \map P x}$
Then:
{{begin-eqn}}
{{eq... | Let $\map P x$ denote a [[Definition:Well-Formed Formula|Well-Formed Formula]] which contains $x$ as a [[Definition:Free Variable|free variable]].
Then the following are [[Definition:Tautology|tautologies]]:
:$\forall x: \paren {\map P x \iff \exists y: \paren {y = x \land \map P y} }$
:$\forall x: \paren {\map P x \... | {{questionable|Inelegant and unclear to the point of soliciting purging}}
By [[Universal Affirmative implies Particular Affirmative iff First Predicate is not Vacuous]]:
:$\paren {\exists y: y = x \land \forall y: \paren {y = x \implies \map P x} } \implies \exists y: \paren {y = x \land \map P x}$
Then:
{{begin-eqn}... | Equality implies Substitution | https://proofwiki.org/wiki/Equality_implies_Substitution | https://proofwiki.org/wiki/Equality_implies_Substitution | [
"Predicate Logic"
] | [
"Definition:Well-Formed Formula",
"Definition:Free Variable",
"Definition:Tautology",
"Confusion of Bound Variables"
] | [
"Universal Affirmative implies Particular Affirmative iff First Predicate is not Vacuous",
"Equality is Reflexive",
"Existential Generalisation",
"Modus Ponendo Ponens",
"Substitutivity of Equality",
"Universal Generalisation",
"Hypothetical Syllogism",
"Substitutivity of Equality",
"Universal Gener... |
proofwiki-4213 | Fundamental Law of Universal Class | :$\forall x: x \in \Bbb U$
where:
:$\Bbb U$ denotes the universal class
:$x$ denotes a set. | From the definition of the universal class:
:$\Bbb U = \set {x: x = x}$
From this, it follows immediately that:
:$\forall x: \paren {x \in \Bbb U \iff x = x}$
From Equality is Reflexive, $x = x$ is a tautology.
Thus the asserted statement is also tautologous.
{{qed}} | :$\forall x: x \in \Bbb U$
where:
:$\Bbb U$ denotes the [[Definition:Universal Class|universal class]]
:$x$ denotes a [[Definition:Set|set]]. | From the definition of the [[Definition:Universal Class|universal class]]:
:$\Bbb U = \set {x: x = x}$
From this, it follows immediately that:
:$\forall x: \paren {x \in \Bbb U \iff x = x}$
From [[Equality is Reflexive]], $x = x$ is a [[Definition:Tautology|tautology]].
Thus the asserted statement is also tautolog... | Fundamental Law of Universal Class | https://proofwiki.org/wiki/Fundamental_Law_of_Universal_Class | https://proofwiki.org/wiki/Fundamental_Law_of_Universal_Class | [
"Universal Class"
] | [
"Definition:Universal Class",
"Definition:Set"
] | [
"Definition:Universal Class",
"Equality is Reflexive",
"Definition:Tautology"
] |
proofwiki-4214 | Cayley-Dickson Construction forms Star-Algebra | Let $A = \struct {A_F, \oplus}$ be a $*$-algebra.
Let $A' = \struct {A'_F, \oplus'} = \struct {A, \oplus}^2$ be the algebra formed from $A$ by the Cayley-Dickson construction.
Then $A'$ is also a $*$-algebra. | === Bilinearity of $\oplus'$ ===
First we need to show that $\oplus'$ is bilinear.
$(1): \quad$ Let $\tuple {a_1, b_1}, \tuple {a_2, b_2}, \tuple {c, d} \in A'$.
Then:
{{begin-eqn}}
{{eqn | o =
| r = \paren {\tuple {a_1, b_1} + \tuple {a_2, b_2} } \oplus' \tuple {c, d}
| c =
}}
{{eqn | r = \tuple {a_1 + a... | Let $A = \struct {A_F, \oplus}$ be a [[Definition:Star-Algebra|$*$-algebra]].
Let $A' = \struct {A'_F, \oplus'} = \struct {A, \oplus}^2$ be the [[Definition:Algebra over Field|algebra]] formed from $A$ by the [[Definition:Cayley-Dickson Construction|Cayley-Dickson construction]].
Then $A'$ is also a [[Definition:Sta... | === Bilinearity of $\oplus'$ ===
First we need to show that $\oplus'$ is [[Definition:Bilinear Mapping|bilinear]].
$(1): \quad$ Let $\tuple {a_1, b_1}, \tuple {a_2, b_2}, \tuple {c, d} \in A'$.
Then:
{{begin-eqn}}
{{eqn | o =
| r = \paren {\tuple {a_1, b_1} + \tuple {a_2, b_2} } \oplus' \tuple {c, d}
... | Cayley-Dickson Construction forms Star-Algebra | https://proofwiki.org/wiki/Cayley-Dickson_Construction_forms_Star-Algebra | https://proofwiki.org/wiki/Cayley-Dickson_Construction_forms_Star-Algebra | [
"Cayley-Dickson Construction",
"Star-Algebras"
] | [
"Definition:Star-Algebra",
"Definition:Algebra over Field",
"Definition:Cayley-Dickson Construction",
"Definition:Star-Algebra"
] | [
"Definition:Bilinear Mapping",
"Definition:Bilinear Mapping",
"Definition:Bilinear Mapping",
"Definition:Bilinear Mapping",
"Definition:Bilinear Mapping"
] |
proofwiki-4215 | Empty Set is Small | :$\O \in U$
where $U$ is the universal class. | {{begin-eqn}}
{{eqn | l = \exists x
| o = :
| r = \forall y: \paren {\neg \paren {y \in x} }
| c = {{axiom-link|the Empty Set|Set Theory}}
}}
{{eqn | ll= \leadsto
| l = \exists x
| o = :
| r = \forall y: \paren {y \in x \iff y \ne y}
| c = Equality is Reflexive
}}
{{eqn | ll= \... | :$\O \in U$
where $U$ is the [[Definition:Universal Class|universal class]]. | {{begin-eqn}}
{{eqn | l = \exists x
| o = :
| r = \forall y: \paren {\neg \paren {y \in x} }
| c = {{axiom-link|the Empty Set|Set Theory}}
}}
{{eqn | ll= \leadsto
| l = \exists x
| o = :
| r = \forall y: \paren {y \in x \iff y \ne y}
| c = [[Equality is Reflexive]]
}}
{{eqn | l... | Empty Set is Small | https://proofwiki.org/wiki/Empty_Set_is_Small | https://proofwiki.org/wiki/Empty_Set_is_Small | [
"Set Theory",
"Class Theory",
"Empty Set"
] | [
"Definition:Universal Class"
] | [
"Equality is Reflexive",
"Characterization of Class Membership",
"Fundamental Law of Universal Class"
] |
proofwiki-4216 | Algebra from Cayley-Dickson Construction is not Real Star-Algebra | Let $A$ be a $*$-algebra.
Let $A'$ be constructed from $A$ using the Cayley-Dickson construction.
Then $A'$ is not a real $*$-algebra. | Let the conjugation operator on $A$ be $*$.
{{AimForCont}} $A'$ is a real $*$-algebra whose conjugation operator is $*'$.
Then by definition:
:$\forall a \in A': {a^*}' = a$
Let $a = \tuple {x, y} \in A'$.
Then by definition of the Cayley-Dickson construction:
:$x, y \in A$
By definition of the conjugation operator:
:$... | Let $A$ be a [[Definition:Star-Algebra|$*$-algebra]].
Let $A'$ be constructed from $A$ using the [[Definition:Cayley-Dickson Construction|Cayley-Dickson construction]].
Then $A'$ is not a [[Definition:Real Star-Algebra|real $*$-algebra]]. | Let the [[Definition:Conjugation on Algebra|conjugation]] operator on $A$ be $*$.
{{AimForCont}} $A'$ is a [[Definition:Real Star-Algebra|real $*$-algebra]] whose [[Definition:Conjugation on Algebra|conjugation]] operator is $*'$.
Then by definition:
:$\forall a \in A': {a^*}' = a$
Let $a = \tuple {x, y} \in A'$.
... | Algebra from Cayley-Dickson Construction is not Real Star-Algebra | https://proofwiki.org/wiki/Algebra_from_Cayley-Dickson_Construction_is_not_Real_Star-Algebra | https://proofwiki.org/wiki/Algebra_from_Cayley-Dickson_Construction_is_not_Real_Star-Algebra | [
"Real Star-Algebras",
"Cayley-Dickson Construction"
] | [
"Definition:Star-Algebra",
"Definition:Cayley-Dickson Construction",
"Definition:Real Star-Algebra"
] | [
"Definition:Conjugation on Algebra",
"Definition:Real Star-Algebra",
"Definition:Conjugation on Algebra",
"Definition:Cayley-Dickson Construction",
"Definition:Conjugation on Algebra",
"Equality of Ordered Pairs",
"Definition:Star-Algebra",
"Definition:Unitary Division Algebra",
"Definition:Contradi... |
proofwiki-4217 | Cayley-Dickson Construction from Real Star-Algebra is Commutative | Let $A = \struct {A_F, \oplus}$ be a $*$-algebra.
Let $A' = \struct {A_F, \oplus'}$ be constructed from $A$ using the Cayley-Dickson construction.
Then $A$ is a real star-algebra {{iff}} $A'$ is a commutative algebra. | Let the conjugation operator on $A$ be $*$.
Let $\tuple {a, b}, \tuple {c, d} \in A'$.
Let $A$ be a real star-algebra.
{{begin-eqn}}
{{eqn | l = \tuple {a, b} \oplus' \tuple {c, d}
| r = \tuple {a \oplus c - d \oplus b^*, a^* \oplus d + c \oplus b}
| c =
}}
{{eqn | r = \tuple {a \oplus c - d \oplus b^*, a^... | Let $A = \struct {A_F, \oplus}$ be a [[Definition:Star-Algebra|$*$-algebra]].
Let $A' = \struct {A_F, \oplus'}$ be constructed from $A$ using the [[Definition:Cayley-Dickson Construction|Cayley-Dickson construction]].
Then $A$ is a [[Definition:Real Star-Algebra|real star-algebra]] {{iff}} $A'$ is a [[Definition:Com... | Let the [[Definition:Conjugation on Algebra|conjugation]] operator on $A$ be $*$.
Let $\tuple {a, b}, \tuple {c, d} \in A'$.
Let $A$ be a [[Definition:Real Star-Algebra|real star-algebra]].
{{begin-eqn}}
{{eqn | l = \tuple {a, b} \oplus' \tuple {c, d}
| r = \tuple {a \oplus c - d \oplus b^*, a^* \oplus d + c ... | Cayley-Dickson Construction from Real Star-Algebra is Commutative | https://proofwiki.org/wiki/Cayley-Dickson_Construction_from_Real_Star-Algebra_is_Commutative | https://proofwiki.org/wiki/Cayley-Dickson_Construction_from_Real_Star-Algebra_is_Commutative | [
"Cayley-Dickson Construction",
"Real Star-Algebras",
"Commutative Algebras"
] | [
"Definition:Star-Algebra",
"Definition:Cayley-Dickson Construction",
"Definition:Real Star-Algebra",
"Definition:Commutative Algebra (Abstract Algebra)"
] | [
"Definition:Conjugation on Algebra",
"Definition:Real Star-Algebra",
"Real Star-Algebra is Commutative",
"Real Addition is Commutative",
"Definition:Commutative Algebra (Abstract Algebra)",
"Definition:Commutative Algebra (Abstract Algebra)",
"Definition:Commutative Algebra (Abstract Algebra)",
"Defin... |
proofwiki-4218 | Cayley-Dickson Construction from Commutative Associative Algebra is Associative | Let $A = \struct {A_F, \oplus}$ be a $*$-algebra.
Let $A' = \struct {A_F, \oplus'}$ be constructed from $A$ using the Cayley-Dickson construction.
Then:
:$A'$ is an associative algebra
{{iff}}:
:$A$ is both a commutative algebra and an associative algebra. | Let the conjugation operator on $A$ be $*$.
Let $\tuple {a, b}, \tuple {c, d}, \tuple {e, f} \in A'$.
In order to streamline notation, let $\oplus$ and $\oplus'$ both be denoted by product notation:
:$a \oplus b =: a b$
:$x \oplus' y =: x y$
The context will make it clear which is meant.
Suppose $A$ is commutative and ... | Let $A = \struct {A_F, \oplus}$ be a [[Definition:Star-Algebra|$*$-algebra]].
Let $A' = \struct {A_F, \oplus'}$ be constructed from $A$ using the [[Definition:Cayley-Dickson Construction|Cayley-Dickson construction]].
Then:
:$A'$ is an [[Definition:Associative Algebra|associative algebra]]
{{iff}}:
:$A$ is both a [[... | Let the [[Definition:Conjugation on Algebra|conjugation]] operator on $A$ be $*$.
Let $\tuple {a, b}, \tuple {c, d}, \tuple {e, f} \in A'$.
In order to streamline notation, let $\oplus$ and $\oplus'$ both be denoted by [[Definition:Multiplicative Notation|product notation]]:
:$a \oplus b =: a b$
:$x \oplus' y =: x y... | Cayley-Dickson Construction from Commutative Associative Algebra is Associative | https://proofwiki.org/wiki/Cayley-Dickson_Construction_from_Commutative_Associative_Algebra_is_Associative | https://proofwiki.org/wiki/Cayley-Dickson_Construction_from_Commutative_Associative_Algebra_is_Associative | [
"Cayley-Dickson Construction",
"Associative Algebras",
"Commutative Algebras"
] | [
"Definition:Star-Algebra",
"Definition:Cayley-Dickson Construction",
"Definition:Associative Algebra",
"Definition:Commutative Algebra (Abstract Algebra)",
"Definition:Associative Algebra"
] | [
"Definition:Conjugation on Algebra",
"Definition:Multiplicative Notation",
"Definition:Commutative Algebra (Abstract Algebra)",
"Definition:Associative Algebra",
"Definition:Commutative Algebra (Abstract Algebra)",
"Definition:Associative Algebra",
"Definition:Associative Algebra",
"Definition:Commuta... |
proofwiki-4219 | Nicely Normed Cayley-Dickson Construction from Associative Algebra is Alternative | Let $A = \struct {A_F, \oplus}$ be a $*$-algebra.
Let $A' = \struct {A_F, \oplus'}$ be constructed from $A$ using the Cayley-Dickson construction.
Then $A'$ is a nicely normed alternative algebra {{iff}} $A$ is a nicely normed associative algebra. | Let the conjugation operator on $A$ be $*$.
Let $\tuple {a, b}, \tuple {c, d} \in A'$.
In order to streamline notation, let $\oplus$ and $\oplus'$ both be denoted by product notation:
{{begin-eqn}}
{{eqn | l = a \oplus b
| o = =:
| r = a b
}}
{{eqn | l = x \oplus' y
| o = =:
| r = x y
}}
{{end-e... | Let $A = \struct {A_F, \oplus}$ be a [[Definition:Star-Algebra|$*$-algebra]].
Let $A' = \struct {A_F, \oplus'}$ be constructed from $A$ using the [[Definition:Cayley-Dickson Construction|Cayley-Dickson construction]].
Then $A'$ is a [[Definition:Nicely Normed Star-Algebra|nicely normed]] [[Definition:Alternative Alg... | Let the [[Definition:Conjugation on Algebra|conjugation]] operator on $A$ be $*$.
Let $\tuple {a, b}, \tuple {c, d} \in A'$.
In order to streamline notation, let $\oplus$ and $\oplus'$ both be denoted by [[Definition:Multiplicative Notation|product notation]]:
{{begin-eqn}}
{{eqn | l = a \oplus b
| o = =:
... | Nicely Normed Cayley-Dickson Construction from Associative Algebra is Alternative | https://proofwiki.org/wiki/Nicely_Normed_Cayley-Dickson_Construction_from_Associative_Algebra_is_Alternative | https://proofwiki.org/wiki/Nicely_Normed_Cayley-Dickson_Construction_from_Associative_Algebra_is_Alternative | [
"Cayley-Dickson Construction",
"Associative Algebras",
"Nicely Normed Star-Algebras",
"Alternative Algebras"
] | [
"Definition:Star-Algebra",
"Definition:Cayley-Dickson Construction",
"Definition:Nicely Normed Star-Algebra",
"Definition:Alternative Algebra",
"Definition:Nicely Normed Star-Algebra",
"Definition:Associative Algebra"
] | [
"Definition:Conjugation on Algebra",
"Definition:Multiplicative Notation",
"Definition:Nicely Normed Star-Algebra",
"Definition:Associative Algebra",
"Definition:Associative Algebra",
"Definition:Nicely Normed Star-Algebra",
"Definition:Associative Algebra",
"Definition:Nicely Normed Star-Algebra",
... |
proofwiki-4220 | Element of Universe | Let $A$ be a class, which may be either a set or a proper class.
Then:
:$\forall A: \paren {A \in U \iff \exists x: x = A}$
where $U$ is the universal class.
That is, if $A$ is an element of $U$, then $A$ is a set. | {{begin-eqn}}
{{eqn | q = \forall A
| l = \leftparen {A \in U}
| o = \iff
| r = \rightparen {\exists x: \paren {x = A \land x \in U} }
| c = Characterization of Class Membership
}}
{{eqn | q = \forall x
| l = x \in U
| o =
| c = Fundamental Law of Universal Class
}}
{{eqn | ll=... | Let $A$ be a [[Definition:Class (Class Theory)|class]], which may be either a [[Definition:Set|set]] or a [[Definition:Proper Class|proper class]].
Then:
:$\forall A: \paren {A \in U \iff \exists x: x = A}$
where $U$ is the [[Definition:Universal Class|universal class]].
That is, if $A$ is an [[Definition:Element of... | {{begin-eqn}}
{{eqn | q = \forall A
| l = \leftparen {A \in U}
| o = \iff
| r = \rightparen {\exists x: \paren {x = A \land x \in U} }
| c = [[Characterization of Class Membership]]
}}
{{eqn | q = \forall x
| l = x \in U
| o =
| c = [[Fundamental Law of Universal Class]]
}}
{{e... | Element of Universe | https://proofwiki.org/wiki/Element_of_Universe | https://proofwiki.org/wiki/Element_of_Universe | [
"Class Theory"
] | [
"Definition:Class (Class Theory)",
"Definition:Set",
"Definition:Class (Class Theory)/Proper Class",
"Definition:Universal Class",
"Definition:Element/Class",
"Definition:Set"
] | [
"Characterization of Class Membership",
"Fundamental Law of Universal Class",
"Conjunction with Tautology"
] |
proofwiki-4221 | Unordered Pairs Exist | Let $A$ and $B$ be classes.
Then:
:$\forall A, B: \set {A, B} \in U$
where $U$ is the universal class. | {{begin-eqn}}
{{eqn | q = \forall A, B: \exists x: \forall y
| l = \leftparen {y \in x}
| o = \iff
| r = \rightparen {y = A \lor y = B}
| c = Axiom of Pairing
}}
{{eqn | ll= \leadsto
| q = \forall A, B: \exists x
| l = x
| r = \set {y: y = A \lor y = B}
| c = {{Defof|Set ... | Let $A$ and $B$ be [[Definition:Class (Class Theory)|classes]].
Then:
:$\forall A, B: \set {A, B} \in U$
where $U$ is the [[Definition:Universal Class|universal class]]. | {{begin-eqn}}
{{eqn | q = \forall A, B: \exists x: \forall y
| l = \leftparen {y \in x}
| o = \iff
| r = \rightparen {y = A \lor y = B}
| c = [[Axiom:Axiom of Pairing (Class Theory)|Axiom of Pairing]]
}}
{{eqn | ll= \leadsto
| q = \forall A, B: \exists x
| l = x
| r = \set {y: ... | Unordered Pairs Exist | https://proofwiki.org/wiki/Unordered_Pairs_Exist | https://proofwiki.org/wiki/Unordered_Pairs_Exist | [
"Class Theory",
"Doubletons"
] | [
"Definition:Class (Class Theory)",
"Definition:Universal Class"
] | [
"Axiom:Axiom of Pairing/Class Theory",
"Element of Universe"
] |
proofwiki-4222 | Conjunction with Tautology | :$p \land \top \dashv \vdash p$ | {{BeginTableau|p \land \top \vdash p}}
{{Premise|1|p \land \top}}
{{Simplification|2|1|p|1|1}}
{{EndTableau}}
{{qed|lemma}}
{{BeginTableau|p \vdash p \land \top}}
{{Premise|1|p}}
{{ExcludedMiddle|2|q \lor \neg q}}
{{ExcludedMiddle|3|\top}}
{{Conjunction|4|1|p \land \top|1|3}}
{{EndTableau|qed}} | :$p \land \top \dashv \vdash p$ | {{BeginTableau|p \land \top \vdash p}}
{{Premise|1|p \land \top}}
{{Simplification|2|1|p|1|1}}
{{EndTableau}}
{{qed|lemma}}
{{BeginTableau|p \vdash p \land \top}}
{{Premise|1|p}}
{{ExcludedMiddle|2|q \lor \neg q}}
{{ExcludedMiddle|3|\top}}
{{Conjunction|4|1|p \land \top|1|3}}
{{EndTableau|qed}} | Conjunction with Tautology/Proof 1 | https://proofwiki.org/wiki/Conjunction_with_Tautology | https://proofwiki.org/wiki/Conjunction_with_Tautology/Proof_1 | [
"Conjunction",
"Tautology",
"Conjunction with Tautology"
] | [] | [] |
proofwiki-4223 | Conjunction with Tautology | :$p \land \top \dashv \vdash p$ | We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations.
$\begin{array}{|c|ccc||c|} \hline
p & p & \land & \top & \top \\
\hline
\F & \F & \F & \T & \T \\
\T & \T & \T & \T & \T \\
\hline
\end{a... | :$p \land \top \dashv \vdash p$ | We apply the [[Method of Truth Tables]] to the proposition.
As can be seen by inspection, in each case, the [[Definition:Truth Value|truth values]] in the appropriate columns match for all [[Definition:Boolean Interpretation|boolean interpretations]].
$\begin{array}{|c|ccc||c|} \hline
p & p & \land & \top & \top \\
\... | Conjunction with Tautology/Proof by Truth Table | https://proofwiki.org/wiki/Conjunction_with_Tautology | https://proofwiki.org/wiki/Conjunction_with_Tautology/Proof_by_Truth_Table | [
"Conjunction",
"Tautology",
"Conjunction with Tautology"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Boolean Interpretation"
] |
proofwiki-4224 | Disjunction with Tautology | : $p \lor \top \dashv \vdash \top$ | {{BeginTableau|p \lor \top \vdash \top}}
{{Premise|1|p \lor \top}}
{{Assumption|2|\top}}
{{Assumption|3|p}}
{{Addition|4|3|p \lor \neg p|3|1}}
{{TableauLine
| n = 5
| pool = 3
| f = \top
| rlnk = Law of Excluded Middle/Proof Rule
| rtxt = Law of Excluded Middle
| dep = 4
| c = }}
{{ProofByCases|6|1|\top|1|2|2|3|5}}
{{E... | : $p \lor \top \dashv \vdash \top$ | {{BeginTableau|p \lor \top \vdash \top}}
{{Premise|1|p \lor \top}}
{{Assumption|2|\top}}
{{Assumption|3|p}}
{{Addition|4|3|p \lor \neg p|3|1}}
{{TableauLine
| n = 5
| pool = 3
| f = \top
| rlnk = Law of Excluded Middle/Proof Rule
| rtxt = Law of Excluded Middle
| dep = 4
| c = }}
{{ProofByCases|6|1|\top|1|2|2|3|5}}
{{E... | Disjunction with Tautology/Proof 1 | https://proofwiki.org/wiki/Disjunction_with_Tautology | https://proofwiki.org/wiki/Disjunction_with_Tautology/Proof_1 | [
"Disjunction",
"Tautology",
"Disjunction with Tautology"
] | [] | [] |
proofwiki-4225 | Disjunction with Tautology | : $p \lor \top \dashv \vdash \top$ | We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations.
$\begin{array}{|c|c||ccc|} \hline
p & \top & p & \lor & \top \\
\hline
F & T & F & T & T \\
T & T & T & T & T \\
\hline
\end{array}$
{{qe... | : $p \lor \top \dashv \vdash \top$ | We apply the [[Method of Truth Tables]] to the proposition.
As can be seen by inspection, in each case, the [[Definition:Truth Value|truth values]] in the appropriate columns match for all [[Definition:Boolean Interpretation|boolean interpretations]].
$\begin{array}{|c|c||ccc|} \hline
p & \top & p & \lor & \top \\
\... | Disjunction with Tautology/Proof 2 | https://proofwiki.org/wiki/Disjunction_with_Tautology | https://proofwiki.org/wiki/Disjunction_with_Tautology/Proof_2 | [
"Disjunction",
"Tautology",
"Disjunction with Tautology"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Boolean Interpretation"
] |
proofwiki-4226 | Conjunction with Contradiction | :$p \land \bot \dashv \vdash \bot$ | {{BeginTableau|p \land \bot \vdash \bot}}
{{Premise|1|p \land \bot}}
{{Simplification|2|1|\bot|1|2}}
{{EndTableau}}
{{qed|lemma}}
{{BeginTableau|\bot \vdash p \land \bot}}
{{Premise|1|\bot}}
{{Explosion|2|1|p \land \bot|1|From a bottom, we can prove what we like}}
{{EndTableau}}
{{qed}} | :$p \land \bot \dashv \vdash \bot$ | {{BeginTableau|p \land \bot \vdash \bot}}
{{Premise|1|p \land \bot}}
{{Simplification|2|1|\bot|1|2}}
{{EndTableau}}
{{qed|lemma}}
{{BeginTableau|\bot \vdash p \land \bot}}
{{Premise|1|\bot}}
{{Explosion|2|1|p \land \bot|1|From a bottom, we can prove what we like}}
{{EndTableau}}
{{qed}} | Conjunction with Contradiction/Proof 1 | https://proofwiki.org/wiki/Conjunction_with_Contradiction | https://proofwiki.org/wiki/Conjunction_with_Contradiction/Proof_1 | [
"Conjunction",
"Contradiction",
"Conjunction with Contradiction"
] | [] | [] |
proofwiki-4227 | Conjunction with Contradiction | :$p \land \bot \dashv \vdash \bot$ | We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations.
$\begin{array}{|c|ccc||c|ccc|} \hline
\bot & p & \land & \bot & p \\
\hline
F & F & F & F & F \\
F & T & F & F & T \\
\hline
\end{array}$... | :$p \land \bot \dashv \vdash \bot$ | We apply the [[Method of Truth Tables]] to the proposition.
As can be seen by inspection, in each case, the [[Definition:Truth Value|truth values]] in the appropriate columns match for all [[Definition:Boolean Interpretation|boolean interpretations]].
$\begin{array}{|c|ccc||c|ccc|} \hline
\bot & p & \land & \bot & p ... | Conjunction with Contradiction/Proof 2 | https://proofwiki.org/wiki/Conjunction_with_Contradiction | https://proofwiki.org/wiki/Conjunction_with_Contradiction/Proof_2 | [
"Conjunction",
"Contradiction",
"Conjunction with Contradiction"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Boolean Interpretation"
] |
proofwiki-4228 | Disjunction with Contradiction | :$p \lor \bot \dashv \vdash p$ | {{BeginTableau|p \lor \bot \vdash p}}
{{Premise|1|p \lor \bot}}
{{Assumption|2|p}}
{{Assumption|3|\bot}}
{{Explosion|4|3|p|3}}
{{ProofByCases|5|1|p|1|2|2|3|4}}
{{EndTableau|lemma}}
{{BeginTableau|p \vdash p \lor \bot}}
{{Premise|1|p}}
{{Addition|2|1|p \lor \bot|1|1}}
{{EndTableau|qed}} | :$p \lor \bot \dashv \vdash p$ | {{BeginTableau|p \lor \bot \vdash p}}
{{Premise|1|p \lor \bot}}
{{Assumption|2|p}}
{{Assumption|3|\bot}}
{{Explosion|4|3|p|3}}
{{ProofByCases|5|1|p|1|2|2|3|4}}
{{EndTableau|lemma}}
{{BeginTableau|p \vdash p \lor \bot}}
{{Premise|1|p}}
{{Addition|2|1|p \lor \bot|1|1}}
{{EndTableau|qed}} | Disjunction with Contradiction/Proof 1 | https://proofwiki.org/wiki/Disjunction_with_Contradiction | https://proofwiki.org/wiki/Disjunction_with_Contradiction/Proof_1 | [
"Disjunction",
"Contradiction",
"Disjunction with Contradiction"
] | [] | [] |
proofwiki-4229 | Disjunction with Contradiction | :$p \lor \bot \dashv \vdash p$ | We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations.
:<nowiki>$\begin{array}{|c|ccc|} \hline
p & p & \lor & \bot \\
\hline
\F & \F & \F & \F \\
\T & \T & \T & \F \\
\hline
\end{array}$</now... | :$p \lor \bot \dashv \vdash p$ | We apply the [[Method of Truth Tables]] to the proposition.
As can be seen by inspection, in each case, the [[Definition:Truth Value|truth values]] in the appropriate columns match for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin{array}{|c|ccc|} \hline
p & p & \lor & \bot \\
\h... | Disjunction with Contradiction/Proof by Truth Table | https://proofwiki.org/wiki/Disjunction_with_Contradiction | https://proofwiki.org/wiki/Disjunction_with_Contradiction/Proof_by_Truth_Table | [
"Disjunction",
"Contradiction",
"Disjunction with Contradiction"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Boolean Interpretation"
] |
proofwiki-4230 | Contradiction is Negation of Tautology | A contradiction implies and is implied by the negation of a tautology:
:$\bot \dashv \vdash \neg \top$
That is, a falsehood can not be true, and a non-truth is a falsehood. | {{BeginTableau|\bot \vdash \neg \top}}
{{Premise|1|\bot}}
{{Assumption|2|\top}}
{{Explosion|3|1|\neg \top|1|Any statement we want}}
{{Contradiction|4|1|\bot|2|3}}
{{Explosion|5|1|\neg \top|4}}
{{EndTableau|lemma}}
{{BeginTableau|\neg \top \vdash \bot}}
{{Premise|1|\neg \top}}
{{ExcludedMiddle|2|p \lor \neg p|From the L... | A [[Definition:Contradiction|contradiction]] implies and is implied by the [[Definition:Logical Not|negation]] of a [[Definition:Tautology|tautology]]:
:$\bot \dashv \vdash \neg \top$
That is, a falsehood can not be true, and a non-truth is a falsehood. | {{BeginTableau|\bot \vdash \neg \top}}
{{Premise|1|\bot}}
{{Assumption|2|\top}}
{{Explosion|3|1|\neg \top|1|Any statement we want}}
{{Contradiction|4|1|\bot|2|3}}
{{Explosion|5|1|\neg \top|4}}
{{EndTableau|lemma}}
{{BeginTableau|\neg \top \vdash \bot}}
{{Premise|1|\neg \top}}
{{ExcludedMiddle|2|p \lor \neg p|From the... | Contradiction is Negation of Tautology/Proof 1 | https://proofwiki.org/wiki/Contradiction_is_Negation_of_Tautology | https://proofwiki.org/wiki/Contradiction_is_Negation_of_Tautology/Proof_1 | [
"Contradiction",
"Tautology",
"Logical Negation",
"Contradiction is Negation of Tautology"
] | [
"Definition:Contradiction",
"Definition:Logical Not",
"Definition:Tautology"
] | [] |
proofwiki-4231 | Contradiction is Negation of Tautology | A contradiction implies and is implied by the negation of a tautology:
:$\bot \dashv \vdash \neg \top$
That is, a falsehood can not be true, and a non-truth is a falsehood. | Let $p$ be a propositional formula.
Let $v$ be any arbitrary boolean interpretation of $p$.
Then $\map v p = F \iff \map v {\neg p} = T$ by the definition of the logical not.
Since $v$ is arbitrary, $p$ is false in all interpretations {{iff}} $\neg p$ is true in all interpretations.
Hence:
:$\bot \dashv \vdash \neg \to... | A [[Definition:Contradiction|contradiction]] implies and is implied by the [[Definition:Logical Not|negation]] of a [[Definition:Tautology|tautology]]:
:$\bot \dashv \vdash \neg \top$
That is, a falsehood can not be true, and a non-truth is a falsehood. | Let $p$ be a [[Definition:Propositional Formula|propositional formula]].
Let $v$ be any arbitrary [[Definition:Boolean Interpretation|boolean interpretation]] of $p$.
Then $\map v p = F \iff \map v {\neg p} = T$ by the definition of the [[Definition:Logical Not|logical not]].
Since $v$ is arbitrary, $p$ is [[Defini... | Contradiction is Negation of Tautology/Proof 3 | https://proofwiki.org/wiki/Contradiction_is_Negation_of_Tautology | https://proofwiki.org/wiki/Contradiction_is_Negation_of_Tautology/Proof_3 | [
"Contradiction",
"Tautology",
"Logical Negation",
"Contradiction is Negation of Tautology"
] | [
"Definition:Contradiction",
"Definition:Logical Not",
"Definition:Tautology"
] | [
"Definition:Language of Propositional Logic/Formal Grammar/WFF",
"Definition:Boolean Interpretation",
"Definition:Logical Not",
"Definition:False",
"Definition:Boolean Interpretation",
"Definition:True",
"Definition:Boolean Interpretation"
] |
proofwiki-4232 | Contradiction is Negation of Tautology | A contradiction implies and is implied by the negation of a tautology:
:$\bot \dashv \vdash \neg \top$
That is, a falsehood can not be true, and a non-truth is a falsehood. | We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth values in the appropriate columns match.
$\begin{array}{|c||cc|} \hline
\top & \neg & \bot \\
\hline
\T & \T & \F \\
\hline
\end{array}$
{{qed}} | A [[Definition:Contradiction|contradiction]] implies and is implied by the [[Definition:Logical Not|negation]] of a [[Definition:Tautology|tautology]]:
:$\bot \dashv \vdash \neg \top$
That is, a falsehood can not be true, and a non-truth is a falsehood. | We apply the [[Method of Truth Tables]] to the proposition.
As can be seen by inspection, the [[Definition:Truth Value|truth values]] in the appropriate columns match.
$\begin{array}{|c||cc|} \hline
\top & \neg & \bot \\
\hline
\T & \T & \F \\
\hline
\end{array}$
{{qed}} | Contradiction is Negation of Tautology/Proof by Truth Table | https://proofwiki.org/wiki/Contradiction_is_Negation_of_Tautology | https://proofwiki.org/wiki/Contradiction_is_Negation_of_Tautology/Proof_by_Truth_Table | [
"Contradiction",
"Tautology",
"Logical Negation",
"Contradiction is Negation of Tautology"
] | [
"Definition:Contradiction",
"Definition:Logical Not",
"Definition:Tautology"
] | [
"Method of Truth Tables",
"Definition:Truth Value"
] |
proofwiki-4233 | Tautology is Negation of Contradiction | A tautology implies and is implied by the negation of a contradiction:
:$\top \dashv \vdash \neg \bot$
That is, a truth can not be false, and a non-falsehood must be a truth. | {{BeginTableau|\top \vdash \neg \bot}}
{{Premise|1|\top}}
{{Assumption|2|\bot|If a contradiction were assumed ...}}
{{Explosion|3|2|\neg \top|2}}
{{NonContradiction|4|1, 2|1|3}}
{{Contradiction|5|1|\neg \bot|2|4}}
{{EndTableau|lemma}}
{{BeginTableau|\neg \bot \vdash \top}}
{{Premise|1|\neg \bot}}
{{Assumption|2|\neg \t... | A [[Definition:Tautology|tautology]] implies and is implied by the [[Definition:Logical Not|negation]] of a [[Definition:Contradiction|contradiction]]:
:$\top \dashv \vdash \neg \bot$
That is, a truth can not be false, and a non-falsehood must be a truth. | {{BeginTableau|\top \vdash \neg \bot}}
{{Premise|1|\top}}
{{Assumption|2|\bot|If a contradiction were assumed ...}}
{{Explosion|3|2|\neg \top|2}}
{{NonContradiction|4|1, 2|1|3}}
{{Contradiction|5|1|\neg \bot|2|4}}
{{EndTableau|lemma}}
{{BeginTableau|\neg \bot \vdash \top}}
{{Premise|1|\neg \bot}}
{{Assumption|2|\neg ... | Tautology is Negation of Contradiction/Proof 1 | https://proofwiki.org/wiki/Tautology_is_Negation_of_Contradiction | https://proofwiki.org/wiki/Tautology_is_Negation_of_Contradiction/Proof_1 | [
"Tautology",
"Logical Negation",
"Contradiction",
"Tautology is Negation of Contradiction"
] | [
"Definition:Tautology",
"Definition:Logical Not",
"Definition:Contradiction"
] | [] |
proofwiki-4234 | Tautology is Negation of Contradiction | A tautology implies and is implied by the negation of a contradiction:
:$\top \dashv \vdash \neg \bot$
That is, a truth can not be false, and a non-falsehood must be a truth. | Let $p$ be a propositional formula.
Let $v$ be an arbitrary boolean interpretation of $p$.
Then:
:$\map v p = T \iff \map v {\neg p} = F$
by the definition of the logical not.
Since $v$ is arbitrary, $p$ is true in all interpretations {{iff}} $\neg p$ is false in all interpretations.
Hence:
:$\top \dashv \vdash \neg \... | A [[Definition:Tautology|tautology]] implies and is implied by the [[Definition:Logical Not|negation]] of a [[Definition:Contradiction|contradiction]]:
:$\top \dashv \vdash \neg \bot$
That is, a truth can not be false, and a non-falsehood must be a truth. | Let $p$ be a [[Definition:Propositional Formula|propositional formula]].
Let $v$ be an arbitrary [[Definition:Boolean Interpretation|boolean interpretation]] of $p$.
Then:
:$\map v p = T \iff \map v {\neg p} = F$
by the definition of the [[Definition:Logical Not|logical not]].
Since $v$ is arbitrary, $p$ is [[De... | Tautology is Negation of Contradiction/Proof 3 | https://proofwiki.org/wiki/Tautology_is_Negation_of_Contradiction | https://proofwiki.org/wiki/Tautology_is_Negation_of_Contradiction/Proof_3 | [
"Tautology",
"Logical Negation",
"Contradiction",
"Tautology is Negation of Contradiction"
] | [
"Definition:Tautology",
"Definition:Logical Not",
"Definition:Contradiction"
] | [
"Definition:Language of Propositional Logic/Formal Grammar/WFF",
"Definition:Boolean Interpretation",
"Definition:Logical Not",
"Definition:True",
"Definition:Boolean Interpretation",
"Definition:False",
"Definition:Boolean Interpretation"
] |
proofwiki-4235 | Tautology is Negation of Contradiction | A tautology implies and is implied by the negation of a contradiction:
:$\top \dashv \vdash \neg \bot$
That is, a truth can not be false, and a non-falsehood must be a truth. | We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth values in the appropriate columns match.
$\begin{array}{|c||cc|} \hline
\bot & \neg & \top \\
\hline
\F & \F & \T \\
\hline
\end{array}$
{{qed}} | A [[Definition:Tautology|tautology]] implies and is implied by the [[Definition:Logical Not|negation]] of a [[Definition:Contradiction|contradiction]]:
:$\top \dashv \vdash \neg \bot$
That is, a truth can not be false, and a non-falsehood must be a truth. | We apply the [[Method of Truth Tables]] to the proposition.
As can be seen by inspection, the [[Definition:Truth Value|truth values]] in the appropriate columns match.
$\begin{array}{|c||cc|} \hline
\bot & \neg & \top \\
\hline
\F & \F & \T \\
\hline
\end{array}$
{{qed}} | Tautology is Negation of Contradiction/Proof by Truth Table | https://proofwiki.org/wiki/Tautology_is_Negation_of_Contradiction | https://proofwiki.org/wiki/Tautology_is_Negation_of_Contradiction/Proof_by_Truth_Table | [
"Tautology",
"Logical Negation",
"Contradiction",
"Tautology is Negation of Contradiction"
] | [
"Definition:Tautology",
"Definition:Logical Not",
"Definition:Contradiction"
] | [
"Method of Truth Tables",
"Definition:Truth Value"
] |
proofwiki-4236 | Cayley-Dickson Construction from Nicely Normed Algebra is Nicely Normed | Let $A = \struct {A_F, \oplus}$ be a $*$-algebra.
Let $A' = \struct {A_F, \oplus'}$ be constructed from $A$ using the Cayley-Dickson construction.
Then $A'$ is a nicely normed algebra {{iff}} $A$ is also a nicely normed algebra. | Let the conjugation operator on $A$ be $*$.
Let $\tuple {a, b}, \tuple {c, d} \in A'$.
In order to streamline notation, let $\oplus$ and $\oplus'$ both be denoted by product notation:
:$a \oplus b =: a b$
:$x \oplus' y =: x y$
The context will make it clear which is meant.
Let $A$ be a nicely normed algebra.
Then:
{{be... | Let $A = \struct {A_F, \oplus}$ be a [[Definition:Star-Algebra|$*$-algebra]].
Let $A' = \struct {A_F, \oplus'}$ be constructed from $A$ using the [[Definition:Cayley-Dickson Construction|Cayley-Dickson construction]].
Then $A'$ is a [[Definition:Nicely Normed Star-Algebra|nicely normed algebra]] {{iff}} $A$ is also ... | Let the [[Definition:Conjugation on Algebra|conjugation]] operator on $A$ be $*$.
Let $\tuple {a, b}, \tuple {c, d} \in A'$.
In order to streamline notation, let $\oplus$ and $\oplus'$ both be denoted by [[Definition:Multiplicative Notation|product notation]]:
:$a \oplus b =: a b$
:$x \oplus' y =: x y$
The context ... | Cayley-Dickson Construction from Nicely Normed Algebra is Nicely Normed | https://proofwiki.org/wiki/Cayley-Dickson_Construction_from_Nicely_Normed_Algebra_is_Nicely_Normed | https://proofwiki.org/wiki/Cayley-Dickson_Construction_from_Nicely_Normed_Algebra_is_Nicely_Normed | [
"Cayley-Dickson Construction",
"Nicely Normed Star-Algebras"
] | [
"Definition:Star-Algebra",
"Definition:Cayley-Dickson Construction",
"Definition:Nicely Normed Star-Algebra",
"Definition:Nicely Normed Star-Algebra"
] | [
"Definition:Conjugation on Algebra",
"Definition:Multiplicative Notation",
"Definition:Nicely Normed Star-Algebra",
"Definition:Nicely Normed Star-Algebra",
"Definition:Real Element in Star-Algebra",
"Definition:Real Element in Star-Algebra",
"Definition:Nicely Normed Star-Algebra",
"Definition:Norm/V... |
proofwiki-4237 | Quaternions form Algebra | The set of quaternions $\Bbb H$ forms an algebra over the field of real numbers.
This algebra is:
:$(1): \quad$ An associative algebra, but '''not''' a commutative algebra.
:$(2): \quad$ A normed division algebra.
:$(3): \quad$ A nicely normed $*$-algebra. | The quaternions $\Bbb H$ are formed by the Cayley-Dickson Construction from the complex numbers $\C$.
From Complex Numbers form Algebra, we have that $\C$ forms:
:$(1): \quad$ An associative algebra
:$(2): \quad$ A commutative algebra
:$(3): \quad$ A normed division algebra
:$(4): \quad$ A nicely normed $*$-algebra.
Fr... | The [[Definition:Quaternion|set of quaternions]] $\Bbb H$ forms an [[Definition:Algebra over Field|algebra]] over the [[Definition:Field of Real Numbers|field of real numbers]].
This [[Definition:Algebra over Field|algebra]] is:
:$(1): \quad$ An [[Definition:Associative Algebra|associative algebra]], but '''not''' a [... | The [[Definition:Quaternion|quaternions]] $\Bbb H$ are formed by the [[Definition:Quaternion/Construction from Cayley-Dickson Construction|Cayley-Dickson Construction]] from the [[Definition:Complex Number|complex numbers]] $\C$.
From [[Complex Numbers form Algebra]], we have that $\C$ forms:
:$(1): \quad$ An [[Defini... | Quaternions form Algebra | https://proofwiki.org/wiki/Quaternions_form_Algebra | https://proofwiki.org/wiki/Quaternions_form_Algebra | [
"Quaternions",
"Algebras"
] | [
"Definition:Quaternion",
"Definition:Algebra over Field",
"Definition:Field of Real Numbers",
"Definition:Algebra over Field",
"Definition:Associative Algebra",
"Definition:Commutative Algebra (Abstract Algebra)",
"Definition:Normed Division Algebra",
"Definition:Nicely Normed Star-Algebra"
] | [
"Definition:Quaternion",
"Definition:Quaternion/Construction from Cayley-Dickson Construction",
"Definition:Complex Number",
"Complex Numbers form Algebra",
"Definition:Associative Algebra",
"Definition:Commutative Algebra (Abstract Algebra)",
"Definition:Normed Division Algebra",
"Definition:Nicely N... |
proofwiki-4238 | Octonions form Algebra | The set of octonions $\Bbb O$ forms an algebra over the field of real numbers.
This algebra is:
:$(1): \quad$ An alternative algebra, but '''not''' an associative algebra.
:$(2): \quad$ A normed division algebra.
:$(3): \quad$ A nicely normed $*$-algebra. | The octonions $\Bbb O$ are formed by the Cayley-Dickson construction from the quaternions $\Bbb H$.
From Quaternions form Algebra, we have that $\Bbb H$ forms:
:$(1): \quad$ An associative algebra
:$(2): \quad$ A normed division algebra
:$(3): \quad$ A nicely normed $*$-algebra.
From Cayley-Dickson Construction forms S... | The [[Definition:Octonion|set of octonions]] $\Bbb O$ forms an [[Definition:Algebra over Field|algebra]] over the [[Definition:Field of Real Numbers|field of real numbers]].
This [[Definition:Algebra over Field|algebra]] is:
:$(1): \quad$ An [[Definition:Alternative Algebra|alternative algebra]], but '''not''' an [[De... | The [[Definition:Octonion|octonions]] $\Bbb O$ are formed by the [[Definition:Cayley-Dickson Construction|Cayley-Dickson construction]] from the [[Definition:Quaternion|quaternions]] $\Bbb H$.
From [[Quaternions form Algebra]], we have that $\Bbb H$ forms:
:$(1): \quad$ An [[Definition:Associative Algebra|associative ... | Octonions form Algebra | https://proofwiki.org/wiki/Octonions_form_Algebra | https://proofwiki.org/wiki/Octonions_form_Algebra | [
"Octonions",
"Algebras"
] | [
"Definition:Octonion",
"Definition:Algebra over Field",
"Definition:Field of Real Numbers",
"Definition:Algebra over Field",
"Definition:Alternative Algebra",
"Definition:Associative Algebra",
"Definition:Normed Division Algebra",
"Definition:Nicely Normed Star-Algebra"
] | [
"Definition:Octonion",
"Definition:Cayley-Dickson Construction",
"Definition:Quaternion",
"Quaternions form Algebra",
"Definition:Associative Algebra",
"Definition:Normed Division Algebra",
"Definition:Nicely Normed Star-Algebra",
"Cayley-Dickson Construction forms Star-Algebra",
"Definition:Star-Al... |
proofwiki-4239 | Division Algebra has No Zero Divisors | Let $A = \struct {A_F, \oplus}$ be an algebra over a field $F$.
Then $A$ is a division algebra {{iff}} it has no zero divisors.
That is:
:$\forall a, b \in A_F: a \oplus b = \mathbf 0_A \implies a = \mathbf 0_A \lor b = \mathbf 0_A$
If the product of two elements of $A$ is zero, then at least one of those elements must... | Let $A$ be a division algebra, in the sense that:
:$\forall a, b \in A_F, b \ne \mathbf 0_A: \exists_1 x \in A_F, y \in A_F: a = b \oplus x, a = y \oplus b$
Suppose that:
:$\exists a, b \in A_F \setminus \set {\mathbf 0_A}: \mathbf 0_A = b \oplus a$
Then by definition of the zero vector we also have that $\mathbf 0_A =... | Let $A = \struct {A_F, \oplus}$ be an [[Definition:Algebra over Field|algebra over a field]] $F$.
Then $A$ is a [[Definition:Division Algebra|division algebra]] {{iff}} it has no [[Definition:Zero Divisor of Algebra|zero divisors]].
That is:
:$\forall a, b \in A_F: a \oplus b = \mathbf 0_A \implies a = \mathbf 0_A ... | Let $A$ be a [[Definition:Division Algebra|division algebra]], in the sense that:
:$\forall a, b \in A_F, b \ne \mathbf 0_A: \exists_1 x \in A_F, y \in A_F: a = b \oplus x, a = y \oplus b$
Suppose that:
:$\exists a, b \in A_F \setminus \set {\mathbf 0_A}: \mathbf 0_A = b \oplus a$
Then by definition of the [[Definiti... | Division Algebra has No Zero Divisors | https://proofwiki.org/wiki/Division_Algebra_has_No_Zero_Divisors | https://proofwiki.org/wiki/Division_Algebra_has_No_Zero_Divisors | [
"Division Algebras"
] | [
"Definition:Algebra over Field",
"Definition:Division Algebra",
"Definition:Zero Divisor/Algebra",
"Definition:Division Algebra"
] | [
"Definition:Division Algebra",
"Definition:Zero Vector",
"Definition:Zero Vector",
"Definition:Division Algebra",
"Definition:Bilinear Mapping",
"Category:Division Algebras"
] |
proofwiki-4240 | Normed Division Algebra is Unitary Division Algebra | Let $A = \struct {A_F, \oplus}$ be a normed divison algebra over a field $F$.
Let the unit of $A$ be $1_A$, and the zero of $A$ be $0_A$.
Then $A$ is a unitary division algebra.
Also:
:$\norm {1_A} = 1$
where $\norm {1_A}$ denotes the norm of $1_A$. | Let $A = \struct {A_F, \oplus}$ be a normed divison algebra as defined {{hypothesis}}.
The fact that $A$ is a unitary algebra is a consequence of the definition of normed divison algebra.
From the definition of a norm, we have that:
:$\forall a \in A: \norm a = 0 \iff a = 0_A$
So, let $a, b \in A \setminus \set {0_A}$.... | Let $A = \struct {A_F, \oplus}$ be a [[Definition:Normed Division Algebra|normed divison algebra]] over a [[Definition:Field (Abstract Algebra)|field]] $F$.
Let the [[Definition:Unit of Algebra|unit]] of $A$ be $1_A$, and the [[Definition:Zero Vector|zero]] of $A$ be $0_A$.
Then $A$ is a [[Definition:Unitary Divisio... | Let $A = \struct {A_F, \oplus}$ be a [[Definition:Normed Division Algebra|normed divison algebra]] as defined {{hypothesis}}.
The fact that $A$ is a [[Definition:Unitary Algebra|unitary algebra]] is a consequence of the definition of [[Definition:Normed Division Algebra|normed divison algebra]].
From the definition ... | Normed Division Algebra is Unitary Division Algebra | https://proofwiki.org/wiki/Normed_Division_Algebra_is_Unitary_Division_Algebra | https://proofwiki.org/wiki/Normed_Division_Algebra_is_Unitary_Division_Algebra | [
"Normed Division Algebras",
"Unitary Division Algebras"
] | [
"Definition:Normed Division Algebra",
"Definition:Field (Abstract Algebra)",
"Definition:Unit of Algebra",
"Definition:Zero Vector",
"Definition:Unitary Division Algebra",
"Definition:Norm/Vector Space"
] | [
"Definition:Normed Division Algebra",
"Definition:Unital Algebra",
"Definition:Normed Division Algebra",
"Definition:Norm/Vector Space",
"Definition:Division Algebra"
] |
proofwiki-4241 | Zermelo's Theorem (Set Theory) | Every set of cardinals is well-ordered with respect to $\le$. | Let $S_1$ and $S_2$ be sets which are not empty.
Suppose there exists an injection $f: S_1 \to S_2$ and another injection $g: S_2 \to S_1$.
Then by the Cantor-Bernstein-Schröder Theorem there exists a bijection between $S_1$ and $S_2$ and by definition $S_1$ is equivalent to $S_2$.
Let $\AA$ be the set of invertible ma... | Every set of [[Definition:Cardinal|cardinals]] is [[Definition:Well-Ordered Set|well-ordered]] with respect to $\le$. | Let $S_1$ and $S_2$ be [[Definition:Set|sets]] which are not [[Definition:Empty Set|empty]].
Suppose there exists an [[Definition:Injection|injection]] $f: S_1 \to S_2$ and another [[Definition:Injection|injection]] $g: S_2 \to S_1$.
Then by the [[Cantor-Bernstein-Schröder Theorem]] there exists a [[Definition:Biject... | Zermelo's Theorem (Set Theory) | https://proofwiki.org/wiki/Zermelo's_Theorem_(Set_Theory) | https://proofwiki.org/wiki/Zermelo's_Theorem_(Set_Theory) | [
"Set Theory"
] | [
"Definition:Cardinal",
"Definition:Well-Ordered Set"
] | [
"Definition:Set",
"Definition:Empty Set",
"Definition:Injection",
"Definition:Injection",
"Cantor-Bernstein-Schröder Theorem",
"Definition:Bijection",
"Definition:Set Equivalence",
"Definition:Inverse Mapping",
"Definition:Empty Set",
"Definition:Mapping",
"Definition:Inverse Mapping",
"Defini... |
proofwiki-4242 | Cardinals form Equivalence Classes | Let $\map \Card S$ denote the cardinal of the set $S$.
Then $\map \Card S$ induces an equivalence class which contains all sets which have the same cardinality as $S$. | Follows directly from:
:The definition of a cardinal as $S \sim T \iff \map \Card S = \map \Card T$
:Set Equivalence behaves like Equivalence Relation
:Relation Partitions Set iff Equivalence.
{{qed}}
Category:Cardinals
j21akpr62mrktbv4ttfsy07xrfrqfrt | Let $\map \Card S$ denote the [[Definition:Cardinal|cardinal]] of the set $S$.
Then $\map \Card S$ induces an [[Definition:Equivalence Class|equivalence class]] which contains all [[Definition:Set|sets]] which have the same [[Definition:Cardinality|cardinality]] as $S$. | Follows directly from:
:The definition of a [[Definition:Cardinal|cardinal]] as $S \sim T \iff \map \Card S = \map \Card T$
:[[Set Equivalence behaves like Equivalence Relation]]
:[[Relation Partitions Set iff Equivalence]].
{{qed}}
[[Category:Cardinals]]
j21akpr62mrktbv4ttfsy07xrfrqfrt | Cardinals form Equivalence Classes | https://proofwiki.org/wiki/Cardinals_form_Equivalence_Classes | https://proofwiki.org/wiki/Cardinals_form_Equivalence_Classes | [
"Cardinals"
] | [
"Definition:Cardinal",
"Definition:Equivalence Class",
"Definition:Set",
"Definition:Cardinality"
] | [
"Definition:Cardinal",
"Set Equivalence behaves like Equivalence Relation",
"Relation Partitions Set iff Equivalence",
"Category:Cardinals"
] |
proofwiki-4243 | Integers form Commutative Ring with Unity | The integers $\struct {\Z, +, \times}$ form a commutative ring with unity under addition and multiplication. | We have that:
:$\struct {\Z, +, \times}$ form a commutative ring.
:$\struct {\Z, +, \times}$ has a unity, and the unity is $1$.
{{Qed}} | The [[Definition:Integer|integers]] $\struct {\Z, +, \times}$ form a [[Definition:Commutative and Unitary Ring|commutative ring with unity]] under [[Definition:Integer Addition|addition]] and [[Definition:Integer Multiplication|multiplication]]. | We have that:
:[[Integers form Commutative Ring|$\struct {\Z, +, \times}$ form a commutative ring]].
:$\struct {\Z, +, \times}$ has a [[Definition:Unity of Ring|unity]], and the [[Integer Multiplication Identity is One|unity is $1$]].
{{Qed}} | Integers form Commutative Ring with Unity | https://proofwiki.org/wiki/Integers_form_Commutative_Ring_with_Unity | https://proofwiki.org/wiki/Integers_form_Commutative_Ring_with_Unity | [
"Integers",
"Commutative Algebra"
] | [
"Definition:Integer",
"Definition:Commutative and Unitary Ring",
"Definition:Addition/Integers",
"Definition:Multiplication/Integers"
] | [
"Integers form Commutative Ring",
"Definition:Unity (Abstract Algebra)/Ring",
"Integer Multiplication Identity is One"
] |
proofwiki-4244 | Ring of Square Matrices over Ring is Ring | Let $R$ be a ring.
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. | From Matrix Entrywise Addition forms Abelian Group we have that $\struct {\map {\MM_R} n, +}$ is an abelian group, because $\struct {R, +}$ is itself an abelian group.
Similarly, it is clear that $\struct {\map {\MM_R} n, \times}$ is a semigroup, as Matrix Multiplication over Order n Square Matrices is Closed and Matri... | Let $R$ be a [[Definition:Ring (Abstract Algebra)|ring]].
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 $R$.]]
Then $\struct {\map {\MM... | From [[Matrix Entrywise Addition forms Abelian Group]] we have that $\struct {\map {\MM_R} n, +}$ is an [[Definition:Abelian Group|abelian group]], because $\struct {R, +}$ is itself an [[Definition:Abelian Group|abelian group]].
Similarly, it is clear that $\struct {\map {\MM_R} n, \times}$ is a [[Definition:Semigrou... | Ring of Square Matrices over Ring is Ring | https://proofwiki.org/wiki/Ring_of_Square_Matrices_over_Ring_is_Ring | https://proofwiki.org/wiki/Ring_of_Square_Matrices_over_Ring_is_Ring | [
"Rings of Square Matrices"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Strictly Positive/Integer",
"Definition:Ring of Square Matrices",
"Definition:Ring (Abstract Algebra)"
] | [
"Matrix Entrywise Addition forms Abelian Group",
"Definition:Abelian Group",
"Definition:Abelian Group",
"Definition:Semigroup",
"Matrix Multiplication over Order n Square Matrices is Closed",
"Matrix Multiplication is Associative",
"Matrix Multiplication Distributes over Matrix Addition"
] |
proofwiki-4245 | Ring of Square Matrices over Ring with Unity | Let $R$ be a 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. | From Ring of Square Matrices over Ring is Ring we have that $\struct {\map {\MM_R} n, +, \times}$ is a ring.
As $R$ has a unity, the unit matrix can be formed.
The unity of $\struct {\map {\MM_R} n, +, \times}$ is this unit matrix.
{{qed}} | Let $R$ be a [[Definition:Ring with Unity|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 $R$.]]
Then $\struct {\map {... | From [[Ring of Square Matrices over Ring is Ring]] we have that $\struct {\map {\MM_R} n, +, \times}$ is a [[Definition:Ring (Abstract Algebra)|ring]].
As $R$ has a [[Definition:Unity of Ring|unity]], the [[Definition:Unit Matrix|unit matrix]] can be formed.
The [[Definition:Unity of Ring|unity]] of $\struct {\map {\... | Ring of Square Matrices over Ring with Unity | https://proofwiki.org/wiki/Ring_of_Square_Matrices_over_Ring_with_Unity | https://proofwiki.org/wiki/Ring_of_Square_Matrices_over_Ring_with_Unity | [
"Rings of Square Matrices",
"Rings with Unity"
] | [
"Definition:Ring with Unity",
"Definition:Strictly Positive/Integer",
"Definition:Ring of Square Matrices",
"Definition:Ring with Unity"
] | [
"Ring of Square Matrices over Ring is Ring",
"Definition:Ring (Abstract Algebra)",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Unit Matrix",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Unit Matrix"
] |
proofwiki-4246 | Ring of Square Matrices over Field is Ring with Unity | Let $F$ be a field.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\struct {\map {\MM_F} n, +, \times}$ denote the ring of square matrices of order $n$ over $F$.
Then $\struct {\map {\MM_F} n, +, \times}$ is a ring with unity, but is not a commutative ring. | We have by definition that a field is a division ring which is also commutative.
Hence $F$ is a commutative ring with unity.
So, from Ring of Square Matrices over Commutative Ring with Unity we have that $\struct {\map {\MM_F} n, +, \times}$ is a ring with unity.
From Matrix Multiplication is not Commutative, we have t... | Let $F$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
Let $\struct {\map {\MM_F} n, +, \times}$ denote the [[Definition:Ring of Square Matrices|ring of square matrices of order $n$ over $F$]].
Then $\struct {\map {\... | We have by definition that a [[Definition:Field (Abstract Algebra)|field]] is a [[Definition:Division Ring|division ring]] which is also [[Definition:Commutative Ring|commutative]].
Hence $F$ is a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
So, from [[Ring of Square Matrices over Commutat... | Ring of Square Matrices over Field is Ring with Unity | https://proofwiki.org/wiki/Ring_of_Square_Matrices_over_Field_is_Ring_with_Unity | https://proofwiki.org/wiki/Ring_of_Square_Matrices_over_Field_is_Ring_with_Unity | [
"Matrix Algebra",
"Examples of Rings with Unity"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Strictly Positive/Integer",
"Definition:Ring of Square Matrices",
"Definition:Ring with Unity",
"Definition:Commutative Ring"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Division Ring",
"Definition:Commutative Ring",
"Definition:Commutative and Unitary Ring",
"Ring of Square Matrices over Commutative Ring with Unity",
"Definition:Ring with Unity",
"Matrix Multiplication is not Commutative",
"Definition:Commutative Rin... |
proofwiki-4247 | Ring of Square Matrices over Real Numbers | 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, but is not a commutative ring. | Recall that Real Numbers form Field.
The result follows directly from Ring of Square Matrices over Field is Ring with Unity.
{{qed}} | 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 $\R$]].
Then $\struct {\map {\MM_\R} n, +, \times}$ is a [[Definition:Ring with Unity|rin... | Recall that [[Real Numbers form Field]].
The result follows directly from [[Ring of Square Matrices over Field is Ring with Unity]].
{{qed}} | Ring of Square Matrices over Real Numbers | https://proofwiki.org/wiki/Ring_of_Square_Matrices_over_Real_Numbers | https://proofwiki.org/wiki/Ring_of_Square_Matrices_over_Real_Numbers | [
"Matrix Algebra",
"Rings with Unity",
"Ring of Square Matrices over Real Numbers"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Ring of Square Matrices",
"Definition:Ring with Unity",
"Definition:Commutative Ring"
] | [
"Real Numbers form Field",
"Ring of Square Matrices over Field is Ring with Unity"
] |
proofwiki-4248 | Matrix Multiplication is not Commutative/Order 2 Square Matrices | Let $R$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $\map {\MM_R} 2$ denote the $2 \times 2$ matrix space over $R$.
The operation of (conventional) matrix multiplication is not commutative over $\map {\MM_R} 2$. | As $R$ is a ring with unity, we have that:
{{begin-eqn}}
{{eqn | l = 0_R
| o = \ne
| r = 1_R
}}
{{eqn | l = 0_R \times 0_R
| r = 0_R
| c =
}}
{{eqn | l = 0_R \times 1_R
| r = 0_R = 1_R \times 0_R
| c =
}}
{{eqn | l = 1_R \times 1_R
| r = 1_R
| c =
}}
{{end-eqn}}
Now le... | 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 $\map {\MM_R} 2$ denote the [[Definition:Matrix Space|$2 \times 2$ matrix space]] over $R$.
The operation of [[Definition:Matrix Product (Conventional)|(c... | As $R$ is a [[Definition:Ring with Unity|ring with unity]], we have that:
{{begin-eqn}}
{{eqn | l = 0_R
| o = \ne
| r = 1_R
}}
{{eqn | l = 0_R \times 0_R
| r = 0_R
| c =
}}
{{eqn | l = 0_R \times 1_R
| r = 0_R = 1_R \times 0_R
| c =
}}
{{eqn | l = 1_R \times 1_R
| r = 1_R
... | Matrix Multiplication is not Commutative/Order 2 Square Matrices | https://proofwiki.org/wiki/Matrix_Multiplication_is_not_Commutative/Order_2_Square_Matrices | https://proofwiki.org/wiki/Matrix_Multiplication_is_not_Commutative/Order_2_Square_Matrices | [
"Matrix Multiplication is not Commutative"
] | [
"Definition:Ring with Unity",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Matrix Space",
"Definition:Matrix Product (Conventional)",
"Definition:Commutative/Operation"
] | [
"Definition:Ring with Unity",
"Definition:Element",
"Definition:Commutative/Elements",
"Definition:Ring with Unity",
"Definition:Matrix Product (Conventional)",
"Definition:Commutative/Operation"
] |
proofwiki-4249 | Ring Zero is Unique | Let $\struct {R, +, \circ}$ be a ring.
Then the ring zero of $R$ is unique. | The ring zero is, by definition of a ring, the identity element of the additive group $\struct {R, +}$.
The result then follows from Identity of Group is Unique.
{{qed}} | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Then the [[Definition:Ring Zero|ring zero]] of $R$ is unique. | The [[Definition:Ring Zero|ring zero]] is, by definition of a [[Definition:Ring (Abstract Algebra)|ring]], the [[Definition:Identity Element|identity element]] of the [[Definition:Additive Group of Ring|additive group]] $\struct {R, +}$.
The result then follows from [[Identity of Group is Unique]].
{{qed}} | Ring Zero is Unique/Proof 1 | https://proofwiki.org/wiki/Ring_Zero_is_Unique | https://proofwiki.org/wiki/Ring_Zero_is_Unique/Proof_1 | [
"Ring Theory",
"Ring Zero is Unique"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Zero"
] | [
"Definition:Ring Zero",
"Definition:Ring (Abstract Algebra)",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Additive Group of Ring",
"Identity of Group is Unique"
] |
proofwiki-4250 | Ring Zero is Unique | Let $\struct {R, +, \circ}$ be a ring.
Then the ring zero of $R$ is unique. | From Ring Product with Zero we have that the ring zero of $R$ is indeed a zero element, as suggested by its name.
The result then follows from Zero Element is Unique.
{{qed}} | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Then the [[Definition:Ring Zero|ring zero]] of $R$ is unique. | From [[Ring Product with Zero]] we have that the [[Definition:Ring Zero|ring zero]] of $R$ is indeed a [[Definition:Zero Element|zero element]], as suggested by its name.
The result then follows from [[Zero Element is Unique]].
{{qed}} | Ring Zero is Unique/Proof 2 | https://proofwiki.org/wiki/Ring_Zero_is_Unique | https://proofwiki.org/wiki/Ring_Zero_is_Unique/Proof_2 | [
"Ring Theory",
"Ring Zero is Unique"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Zero"
] | [
"Ring Product with Zero",
"Definition:Ring Zero",
"Definition:Zero Element",
"Zero Element is Unique"
] |
proofwiki-4251 | Ring Zero is Unique | Let $\struct {R, +, \circ}$ be a ring.
Then the ring zero of $R$ is unique. | Suppose $0$ and $0'$ are both ring zeroes of $\struct {R, +, \circ}$.
Then by Ring Product with Zero:
:$0' \circ 0 = 0$ by dint of $0$ being a ring zero
:$0' \circ 0 = 0'$ by dint of $0'$ being a ring zero.
So $0 = 0' \circ 0 = 0'$.
So $0 = 0'$ and there is only one ring zero of $\struct {R, +, \circ}$ after all.
{{qed... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Then the [[Definition:Ring Zero|ring zero]] of $R$ is unique. | Suppose $0$ and $0'$ are both [[Definition:Ring Zero|ring zeroes]] of $\struct {R, +, \circ}$.
Then by [[Ring Product with Zero]]:
:$0' \circ 0 = 0$ by dint of $0$ being a [[Definition:Ring Zero|ring zero]]
:$0' \circ 0 = 0'$ by dint of $0'$ being a [[Definition:Ring Zero|ring zero]].
So $0 = 0' \circ 0 = 0'$.
So $0... | Ring Zero is Unique/Proof 3 | https://proofwiki.org/wiki/Ring_Zero_is_Unique | https://proofwiki.org/wiki/Ring_Zero_is_Unique/Proof_3 | [
"Ring Theory",
"Ring Zero is Unique"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Zero"
] | [
"Definition:Ring Zero",
"Ring Product with Zero",
"Definition:Ring Zero",
"Definition:Ring Zero",
"Definition:Ring Zero"
] |
proofwiki-4252 | Ring Negative is Unique | Let $\struct {R, +, \circ}$ be a ring.
Let $a \in R$.
Then the ring negative $-a$ of $a$ is unique. | The ring negative is, by definition of a ring, the inverse element of $a$ in the additive group $\struct {R, +}$.
The result then follows from Inverse in Group is Unique.
{{qed}} | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $a \in R$.
Then the [[Definition:Ring Negative|ring negative]] $-a$ of $a$ is unique. | The [[Definition:Ring Negative|ring negative]] is, by definition of a [[Definition:Ring (Abstract Algebra)|ring]], the [[Definition:Inverse Element|inverse element]] of $a$ in the [[Definition:Additive Group of Ring|additive group]] $\struct {R, +}$.
The result then follows from [[Inverse in Group is Unique]].
{{qed}} | Ring Negative is Unique | https://proofwiki.org/wiki/Ring_Negative_is_Unique | https://proofwiki.org/wiki/Ring_Negative_is_Unique | [
"Ring Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Negative"
] | [
"Definition:Ring Negative",
"Definition:Ring (Abstract Algebra)",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Additive Group of Ring",
"Inverse in Group is Unique"
] |
proofwiki-4253 | Negative of Ring Negative | Let $\struct {R, +, \circ}$ be a ring.
Let $a \in R$ and let $-a$ be the ring negative of $a$.
Then:
:$-\paren {-a} = a$ | The ring negative is, by definition of a ring, the inverse element of $a$ in the additive group $\struct {R, +}$.
The result then follows from Inverse of Group Inverse.
{{qed}} | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $a \in R$ and let $-a$ be the [[Definition:Ring Negative|ring negative]] of $a$.
Then:
:$-\paren {-a} = a$ | The [[Definition:Ring Negative|ring negative]] is, by definition of a [[Definition:Ring (Abstract Algebra)|ring]], the [[Definition:Inverse Element|inverse element]] of $a$ in the [[Definition:Additive Group of Ring|additive group]] $\struct {R, +}$.
The result then follows from [[Inverse of Group Inverse]].
{{qed}} | Negative of Ring Negative | https://proofwiki.org/wiki/Negative_of_Ring_Negative | https://proofwiki.org/wiki/Negative_of_Ring_Negative | [
"Ring Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Negative"
] | [
"Definition:Ring Negative",
"Definition:Ring (Abstract Algebra)",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Additive Group of Ring",
"Inverse of Group Inverse"
] |
proofwiki-4254 | Even Perfect Number is Triangular | All perfect numbers which are even are triangular. | Follows from:
: Even Perfect Number is Hexagonal
: Hexagonal Number is Triangular Number
{{qed}} | All [[Definition:Perfect Number|perfect numbers]] which are [[Definition:Even Integer|even]] are [[Definition:Triangular Number|triangular]]. | Follows from:
: [[Even Perfect Number is Hexagonal]]
: [[Hexagonal Number is Triangular Number]]
{{qed}} | Even Perfect Number is Triangular/Proof 2 | https://proofwiki.org/wiki/Even_Perfect_Number_is_Triangular | https://proofwiki.org/wiki/Even_Perfect_Number_is_Triangular/Proof_2 | [
"Even Perfect Number is Triangular",
"Euclidean Numbers",
"Triangular Numbers",
"Perfect Numbers"
] | [
"Definition:Perfect Number",
"Definition:Even Integer",
"Definition:Triangular Number"
] | [
"Even Perfect Number is Hexagonal",
"Hexagonal Number is Triangular Number"
] |
proofwiki-4255 | Field has no Proper Zero Divisors | Let $\struct {F, +, \times}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Then $\struct {F, +, \times}$ has no proper zero divisors.
That is:
:$a \times b = 0_F \implies a = 0_F \lor b = 0_F$ | {{begin-eqn}}
{{eqn | l = a
| o = \ne
| r = 0_F
| c =
}}
{{eqn | ll= \leadsto
| l = a^{-1} \times \paren {a \times b}
| r = a^{-1} \times 0_F
| c = $a^{-1}$ exists because $a \ne 0_F$
}}
{{eqn | ll= \leadsto
| l = \paren {a^{-1} \times a} \times b
| r = a^{-1} \times 0_F... | Let $\struct {F, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$.
Then $\struct {F, +, \times}$ has no [[Definition:Proper Zero Divisor|proper zero divisors]].
That is:
:$a \times b = 0_F \implies a = 0... | {{begin-eqn}}
{{eqn | l = a
| o = \ne
| r = 0_F
| c =
}}
{{eqn | ll= \leadsto
| l = a^{-1} \times \paren {a \times b}
| r = a^{-1} \times 0_F
| c = $a^{-1}$ exists because $a \ne 0_F$
}}
{{eqn | ll= \leadsto
| l = \paren {a^{-1} \times a} \times b
| r = a^{-1} \times 0_F... | Field has no Proper Zero Divisors/Proof 2 | https://proofwiki.org/wiki/Field_has_no_Proper_Zero_Divisors | https://proofwiki.org/wiki/Field_has_no_Proper_Zero_Divisors/Proof_2 | [
"Field Theory",
"Field has no Proper Zero Divisors"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Multiplicative Identity",
"Definition:Proper Zero Divisor"
] | [
"Field Product with Zero"
] |
proofwiki-4256 | Multiplicative Identity is Unique | Let $\struct {F, +, \times}$ be a field.
Then the multiplicative identity $1_F$ of $F$ is unique. | From the definition of multiplicative identity, $1_F$ is the identity element of the multiplicative group $\struct {F^*, \times}$.
The result follows from Identity of Group is Unique.
{{qed}} | Let $\struct {F, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]].
Then the [[Definition:Multiplicative Identity|multiplicative identity]] $1_F$ of $F$ is unique. | From the definition of [[Definition:Multiplicative Identity|multiplicative identity]], $1_F$ is the [[Definition:Identity Element|identity element]] of the [[Definition:Multiplicative Group|multiplicative group]] $\struct {F^*, \times}$.
The result follows from [[Identity of Group is Unique]].
{{qed}} | Multiplicative Identity is Unique | https://proofwiki.org/wiki/Multiplicative_Identity_is_Unique | https://proofwiki.org/wiki/Multiplicative_Identity_is_Unique | [
"Field Theory",
"Multiplication"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Multiplicative Identity"
] | [
"Definition:Multiplicative Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Multiplicative Group",
"Identity of Group is Unique"
] |
proofwiki-4257 | Multiplicative Inverse in Field is Unique | Let $\struct {F, +, \times}$ be a field whose zero is $0_F$.
Let $a \in F$ such that $a \ne 0_F$.
Then the multiplicative inverse $a^{-1}$ of $a$ is unique. | From the definition of multiplicative inverse, $a^{-1}$ is the inverse element of the multiplicative group $\struct {F^*, \times}$.
The result follows from Inverse in Group is Unique.
{{qed}} | Let $\struct {F, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$.
Let $a \in F$ such that $a \ne 0_F$.
Then the [[Definition:Multiplicative Inverse in Field|multiplicative inverse]] $a^{-1}$ of $a$ is [[Definition:Unique|unique]]. | From the definition of [[Definition:Multiplicative Inverse in Field|multiplicative inverse]], $a^{-1}$ is the [[Definition:Inverse Element|inverse element]] of the [[Definition:Multiplicative Group|multiplicative group]] $\struct {F^*, \times}$.
The result follows from [[Inverse in Group is Unique]].
{{qed}} | Multiplicative Inverse in Field is Unique/Proof 1 | https://proofwiki.org/wiki/Multiplicative_Inverse_in_Field_is_Unique | https://proofwiki.org/wiki/Multiplicative_Inverse_in_Field_is_Unique/Proof_1 | [
"Field Theory",
"Multiplicative Inverse in Field is Unique"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Multiplicative Inverse/Field",
"Definition:Unique"
] | [
"Definition:Multiplicative Inverse/Field",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Multiplicative Group",
"Inverse in Group is Unique"
] |
proofwiki-4258 | Multiplicative Inverse in Field is Unique | Let $\struct {F, +, \times}$ be a field whose zero is $0_F$.
Let $a \in F$ such that $a \ne 0_F$.
Then the multiplicative inverse $a^{-1}$ of $a$ is unique. | From the definition of a field as a division ring, every element of $F^*$ is a unit.
The result follows from Product Inverse in Ring is Unique.
{{qed}} | Let $\struct {F, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$.
Let $a \in F$ such that $a \ne 0_F$.
Then the [[Definition:Multiplicative Inverse in Field|multiplicative inverse]] $a^{-1}$ of $a$ is [[Definition:Unique|unique]]. | From the definition of a [[Definition:Field (Abstract Algebra)|field]] as a [[Definition:Division Ring|division ring]], every element of $F^*$ is a [[Definition:Unit of Ring|unit]].
The result follows from [[Product Inverse in Ring is Unique]].
{{qed}} | Multiplicative Inverse in Field is Unique/Proof 2 | https://proofwiki.org/wiki/Multiplicative_Inverse_in_Field_is_Unique | https://proofwiki.org/wiki/Multiplicative_Inverse_in_Field_is_Unique/Proof_2 | [
"Field Theory",
"Multiplicative Inverse in Field is Unique"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Multiplicative Inverse/Field",
"Definition:Unique"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Division Ring",
"Definition:Unit of Ring",
"Product Inverse in Ring is Unique"
] |
proofwiki-4259 | Inverse of Multiplicative Inverse | Let $\struct {F, +, \times}$ be a field whose zero is $0_F$.
Let $a \in F$ such that $a \ne 0_F$.
Let $a^{-1}$ be the multiplicative inverse of $a$.
Then $\paren {a^{-1} }^{-1} = a$. | {{begin-eqn}}
{{eqn | l = \paren {a^{-1} } \times a
| r = a \times \paren {a^{-1} }
| c = {{Field-axiom|M2}}
}}
{{eqn | r = 1_F
| c = {{Field-axiom|M4}}
}}
{{eqn | ll= \leadsto
| l = a
| r = \paren {a^{-1} }^{-1}
| c = {{Defof|Multiplicative Inverse in Field}}
}}
{{end-eqn}}
{{qed}} | Let $\struct {F, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$.
Let $a \in F$ such that $a \ne 0_F$.
Let $a^{-1}$ be the [[Definition:Multiplicative Inverse in Field|multiplicative inverse]] of $a$.
Then $\paren {a^{-1} }^{-1} = a$. | {{begin-eqn}}
{{eqn | l = \paren {a^{-1} } \times a
| r = a \times \paren {a^{-1} }
| c = {{Field-axiom|M2}}
}}
{{eqn | r = 1_F
| c = {{Field-axiom|M4}}
}}
{{eqn | ll= \leadsto
| l = a
| r = \paren {a^{-1} }^{-1}
| c = {{Defof|Multiplicative Inverse in Field}}
}}
{{end-eqn}}
{{qed}} | Inverse of Multiplicative Inverse/Proof 2 | https://proofwiki.org/wiki/Inverse_of_Multiplicative_Inverse | https://proofwiki.org/wiki/Inverse_of_Multiplicative_Inverse/Proof_2 | [
"Field Theory",
"Inverse of Multiplicative Inverse"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Multiplicative Inverse/Field"
] | [] |
proofwiki-4260 | Integral Multiple Distributes over Ring Addition | Let $\struct {R, +, \times}$ be a ring, or a field.
Let $a, b \in R$ and $m, n \in \Z$.
Then:
:$(1): \quad \paren {m + n} \cdot a = \paren {m \cdot a} + \paren {n \cdot a}$
:$(2): \quad m \cdot \paren {a + b} = \paren {m \cdot a} + \paren {m \cdot b}$
where $m \cdot a$ is as defined in integral multiple. | We have that the additive group $\struct {R, +}$ is an abelian group.
$(1): \quad \paren {m + n} \cdot a = \paren {m \cdot a} + \paren {n \cdot a}$:
This is an instance of Powers of Group Elements: Sum of Indices when expressed in additive notation:
:$\forall n, m \in \Z: \forall a \in R: m a + n a = \paren {m + n} a$
... | Let $\struct {R, +, \times}$ be a [[Definition:Ring (Abstract Algebra)|ring]], or a [[Definition:Field (Abstract Algebra)|field]].
Let $a, b \in R$ and $m, n \in \Z$.
Then:
:$(1): \quad \paren {m + n} \cdot a = \paren {m \cdot a} + \paren {n \cdot a}$
:$(2): \quad m \cdot \paren {a + b} = \paren {m \cdot a} + \paren... | We have that the [[Definition:Additive Group of Ring|additive group]] $\struct {R, +}$ is an [[Definition:Abelian Group|abelian group]].
$(1): \quad \paren {m + n} \cdot a = \paren {m \cdot a} + \paren {n \cdot a}$:
This is an instance of [[Powers of Group Elements/Sum of Indices|Powers of Group Elements: Sum of Ind... | Integral Multiple Distributes over Ring Addition | https://proofwiki.org/wiki/Integral_Multiple_Distributes_over_Ring_Addition | https://proofwiki.org/wiki/Integral_Multiple_Distributes_over_Ring_Addition | [
"Ring Theory",
"Field Theory",
"Examples of Distributive Operations"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Field (Abstract Algebra)",
"Definition:Integral Multiple"
] | [
"Definition:Additive Group of Ring",
"Definition:Abelian Group",
"Powers of Group Elements/Sum of Indices",
"Definition:Additive Notation",
"Power of Product in Abelian Group",
"Definition:Additive Notation"
] |
proofwiki-4261 | Integral Multiple of Integral Multiple | Let $\struct {F, +, \times}$ be a field.
Let $a \in F$ and $m, n \in \Z$.
Then:
:$\paren {m n} \cdot a = m \cdot \paren {n \cdot a}$
where $n \cdot a$ is as defined in integral multiple. | We have that $\struct {F^*, \times}$ is the multiplicative group of $\struct {F, +, \times}$.
Let $a \in F^*$, that is, $a \in F: a \ne 0_F$, where $0_F$ is the zero of $F$.
This is an instance of Powers of Group Elements when expressed in additive notation:
:$\forall m, n \in \Z: \paren {m n} a = m \paren {n a}$
{{qed... | Let $\struct {F, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $a \in F$ and $m, n \in \Z$.
Then:
:$\paren {m n} \cdot a = m \cdot \paren {n \cdot a}$
where $n \cdot a$ is as defined in [[Definition:Integral Multiple|integral multiple]]. | We have that $\struct {F^*, \times}$ is the [[Definition:Multiplicative Group|multiplicative group]] of $\struct {F, +, \times}$.
Let $a \in F^*$, that is, $a \in F: a \ne 0_F$, where $0_F$ is the [[Definition:Field Zero|zero]] of $F$.
This is an instance of [[Powers of Group Elements]] when expressed in [[Definitio... | Integral Multiple of Integral Multiple | https://proofwiki.org/wiki/Integral_Multiple_of_Integral_Multiple | https://proofwiki.org/wiki/Integral_Multiple_of_Integral_Multiple | [
"Field Theory"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Integral Multiple"
] | [
"Definition:Multiplicative Group",
"Definition:Field Zero",
"Powers of Group Elements",
"Definition:Additive Notation",
"Definition:Field Zero"
] |
proofwiki-4262 | Product of Integral Multiples | Let $\struct {F, +, \times}$ be a field.
Let $a, b \in F$ and $m, n \in \Z$.
Then:
:$\paren {m \cdot a} \times \paren {n \cdot b} = \paren {m n} \cdot \paren {a \times b}$
where $m \cdot a$ is as defined in Integral Multiple. | First let $m = 0$ or $n = 0$.
{{WLOG}}, let $m = 0$.
The case where $n = 0$ follows the same lines.
We have:
{{begin-eqn}}
{{eqn | q = \forall a, b \in R: \forall n \in \Z_{>0}
| l = \paren {m \cdot a} \times \paren {n \cdot b}
| r = \paren {0 \cdot a} \times \paren {n \cdot b}
| c = Definition of $m$... | Let $\struct {F, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $a, b \in F$ and $m, n \in \Z$.
Then:
:$\paren {m \cdot a} \times \paren {n \cdot b} = \paren {m n} \cdot \paren {a \times b}$
where $m \cdot a$ is as defined in [[Definition:Integral Multiple/Rings and Fields|Integral Multiple]]. | First let $m = 0$ or $n = 0$.
{{WLOG}}, let $m = 0$.
The case where $n = 0$ follows the same lines.
We have:
{{begin-eqn}}
{{eqn | q = \forall a, b \in R: \forall n \in \Z_{>0}
| l = \paren {m \cdot a} \times \paren {n \cdot b}
| r = \paren {0 \cdot a} \times \paren {n \cdot b}
| c = Definition of... | Product of Integral Multiples/Proof 2 | https://proofwiki.org/wiki/Product_of_Integral_Multiples | https://proofwiki.org/wiki/Product_of_Integral_Multiples/Proof_2 | [
"Field Theory",
"Product of Integral Multiples"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Integral Multiple/Rings and Fields"
] | [
"Definition:Ring Zero",
"General Distributivity Theorem",
"Powers of Group Elements/Product of Indices/Additive Notation",
"Powers of Group Elements"
] |
proofwiki-4263 | Annihilator of Ring Always Contains Zero | Let $\struct {R, +, \times}$ be a ring.
Let $\map {\mathrm {Ann} } R$ be the annihilator of $R$.
Then $0 \in \map {\mathrm {Ann} } R$. | We have by definition of integral multiple that:
:$\forall r \in R: 0 \cdot r = 0_R$
where $0_R$ is the zero of $R$.
Hence the result by definition of annihilator.
{{qed}}
Category:Ring Theory
sky2zptmfq5vwwjsru78klub6slkzlg | Let $\struct {R, +, \times}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $\map {\mathrm {Ann} } R$ be the [[Definition:Annihilator of Ring|annihilator]] of $R$.
Then $0 \in \map {\mathrm {Ann} } R$. | We have by definition of [[Definition:Integral Multiple|integral multiple]] that:
:$\forall r \in R: 0 \cdot r = 0_R$
where $0_R$ is the [[Definition:Ring Zero|zero]] of $R$.
Hence the result by definition of [[Definition:Annihilator of Ring|annihilator]].
{{qed}}
[[Category:Ring Theory]]
sky2zptmfq5vwwjsru78klub6slk... | Annihilator of Ring Always Contains Zero | https://proofwiki.org/wiki/Annihilator_of_Ring_Always_Contains_Zero | https://proofwiki.org/wiki/Annihilator_of_Ring_Always_Contains_Zero | [
"Ring Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Annihilator of Ring"
] | [
"Definition:Integral Multiple",
"Definition:Ring Zero",
"Definition:Annihilator of Ring",
"Category:Ring Theory"
] |
proofwiki-4264 | Non-Trivial Annihilator Contains Positive Integer | Let $\left({R, +, \times}\right)$ be a ring with unity.
Let $A = \operatorname{Ann} \left({R}\right)$ be the annihilator of $R$.
Let $a \in A$ such that $a \ne 0$.
Then $A$ contains at least one strictly positive integer. | Let the zero of $R$ be $0_R$ and the unity of $R$ be $1_R$.
First we note that:
{{begin-eqn}}
{{eqn | l = a
| o = \in
| r = \operatorname{Ann} \left({R}\right)
| c =
}}
{{eqn | ll= \leadsto
| l = a \cdot 1_R
| r = 0_R
| c =
}}
{{eqn | ll= \leadsto
| l = \left({-a}\right) \cdo... | Let $\left({R, +, \times}\right)$ be a [[Definition:Ring with Unity|ring with unity]].
Let $A = \operatorname{Ann} \left({R}\right)$ be the [[Definition:Annihilator of Ring|annihilator]] of $R$.
Let $a \in A$ such that $a \ne 0$.
Then $A$ contains at least one [[Definition:Strictly Positive Integer|strictly positiv... | Let the [[Definition:Ring Zero|zero]] of $R$ be $0_R$ and the [[Definition:Unity of Ring|unity]] of $R$ be $1_R$.
First we note that:
{{begin-eqn}}
{{eqn | l = a
| o = \in
| r = \operatorname{Ann} \left({R}\right)
| c =
}}
{{eqn | ll= \leadsto
| l = a \cdot 1_R
| r = 0_R
| c =
}}... | Non-Trivial Annihilator Contains Positive Integer | https://proofwiki.org/wiki/Non-Trivial_Annihilator_Contains_Positive_Integer | https://proofwiki.org/wiki/Non-Trivial_Annihilator_Contains_Positive_Integer | [
"Rings with Unity"
] | [
"Definition:Ring with Unity",
"Definition:Annihilator of Ring",
"Definition:Strictly Positive/Integer"
] | [
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Strictly Positive",
"Definition:Strictly Positive/Integer",
"Category:Rings with Unity"
] |
proofwiki-4265 | Area of Sector | Let $\CC = ABC$ be a circle whose center is $A$ and with radii $AB$ and $AC$.
Let $BAC$ be the sector of $\CC$ whose angle between $AB$ and $AC$ is $\theta$.
:300px
Then the area $\AA$ of sector $BAC$ is given by:
:$\AA = \dfrac {r^2 \theta} 2$
where:
:$r = AB$ is the length of the radius of the circle
:$\theta$ is mea... | Let $\CC$ be a circle of radius $r$ whose center $A$ is at the origin and with radii $AB$ and $AC$.
Let $B$ and $C$ be arbitrary points on the circumference of $\CC$.
Let $C$ be positioned at $\tuple {0, r}$ so that $AC$ coincides with the $y$-axis.
Let $BAC$ be the sector of $\CC$ whose angle between $AB$ and $AC$ is ... | Let $\CC = ABC$ be a [[Definition:Circle|circle]] whose [[Definition:Center of Circle|center]] is $A$ and with [[Definition:Radius of Circle|radii]] $AB$ and $AC$.
Let $BAC$ be the [[Definition:Sector of Circle|sector]] of $\CC$ whose [[Definition:Angle of Sector|angle]] between $AB$ and $AC$ is $\theta$.
:[[File:Sec... | Let $\CC$ be a [[Definition:Circle|circle]] of [[Definition:Radius of Circle|radius]] $r$ whose [[Definition:Center of Circle|center]] $A$ is at the [[Definition:Origin|origin]] and with [[Definition:Radius of Circle|radii]] $AB$ and $AC$.
Let $B$ and $C$ be arbitrary [[Definition:Point|points]] on the [[Definition:Ci... | Area of Sector/Proof 1 | https://proofwiki.org/wiki/Area_of_Sector | https://proofwiki.org/wiki/Area_of_Sector/Proof_1 | [
"Area of Sector",
"Sectors of Circles",
"Area Formulas"
] | [
"Definition:Circle",
"Definition:Circle/Center",
"Definition:Circle/Radius",
"Definition:Sector of Circle",
"Definition:Sector of Circle/Angle",
"File:Sector.png",
"Definition:Area",
"Definition:Sector of Circle",
"Definition:Linear Measure/Length",
"Definition:Circle/Radius",
"Definition:Circle... | [
"Definition:Circle",
"Definition:Circle/Radius",
"Definition:Circle/Center",
"Definition:Coordinate System/Origin",
"Definition:Circle/Radius",
"Definition:Point",
"Definition:Circumference",
"Definition:Axis/Y-Axis",
"Definition:Sector of Circle",
"Definition:Sector of Circle/Angle",
"File:Area... |
proofwiki-4266 | Area of Sector | Let $\CC = ABC$ be a circle whose center is $A$ and with radii $AB$ and $AC$.
Let $BAC$ be the sector of $\CC$ whose angle between $AB$ and $AC$ is $\theta$.
:300px
Then the area $\AA$ of sector $BAC$ is given by:
:$\AA = \dfrac {r^2 \theta} 2$
where:
:$r = AB$ is the length of the radius of the circle
:$\theta$ is mea... | {{questionable|To be replaced with something rigorous, based on calculus.}}
From Area of Circle, the area of $\CC$ is $\pi r^2$.
From Measurement of Full Angle, the angle within $\CC$ is $2 \pi$.
The fraction of the area of $\CC$ within the sector $BAC$ is therefore $\pi r^2 \times \dfrac \theta {2 \pi}$.
Hence the res... | Let $\CC = ABC$ be a [[Definition:Circle|circle]] whose [[Definition:Center of Circle|center]] is $A$ and with [[Definition:Radius of Circle|radii]] $AB$ and $AC$.
Let $BAC$ be the [[Definition:Sector of Circle|sector]] of $\CC$ whose [[Definition:Angle of Sector|angle]] between $AB$ and $AC$ is $\theta$.
:[[File:Sec... | {{questionable|To be replaced with something rigorous, based on calculus.}}
From [[Area of Circle]], the area of $\CC$ is $\pi r^2$.
From [[Measurement of Full Angle]], the angle within $\CC$ is $2 \pi$.
The fraction of the area of $\CC$ within the sector $BAC$ is therefore $\pi r^2 \times \dfrac \theta {2 \pi}$.
H... | Area of Sector/Proof 2 | https://proofwiki.org/wiki/Area_of_Sector | https://proofwiki.org/wiki/Area_of_Sector/Proof_2 | [
"Area of Sector",
"Sectors of Circles",
"Area Formulas"
] | [
"Definition:Circle",
"Definition:Circle/Center",
"Definition:Circle/Radius",
"Definition:Sector of Circle",
"Definition:Sector of Circle/Angle",
"File:Sector.png",
"Definition:Area",
"Definition:Sector of Circle",
"Definition:Linear Measure/Length",
"Definition:Circle/Radius",
"Definition:Circle... | [
"Area of Circle",
"Measurements of Common Angles/Full Angle"
] |
proofwiki-4267 | Rational Numbers form Prime Field | The field of rational numbers $\struct {\Q, +, \times}$ is a prime field.
That is, the only subset of $\Q$ which sustains both addition and multiplication are $\Q$ and $\set 0$, and vacuously $\O$. | Let $F$ be a subfield of $\struct {\Q, +, \times}$.
Then $F$ is by definition a field.
Thus by definition:
:$\struct {F, +}$ is an abelian group
and:
:$\struct {F^*, \times}$ is an abelian group, where $F^* = F \setminus \set 0$.
and $\times$ is distributive over $+$:
:$\forall a, b, c \in F: a \times \paren {b + c} = ... | The [[Definition:Field of Rational Numbers|field of rational numbers]] $\struct {\Q, +, \times}$ is a [[Definition:Prime Field|prime field]].
That is, the only [[Definition:Subset|subset]] of $\Q$ which sustains both [[Definition:Rational Addition|addition]] and [[Definition:Rational Multiplication|multiplication]] a... | Let $F$ be a [[Definition:Subfield|subfield]] of $\struct {\Q, +, \times}$.
Then $F$ is by definition a [[Definition:Field (Abstract Algebra)|field]].
Thus by definition:
:$\struct {F, +}$ is an [[Definition:Abelian Group|abelian group]]
and:
:$\struct {F^*, \times}$ is an [[Definition:Abelian Group|abelian group]], ... | Rational Numbers form Prime Field | https://proofwiki.org/wiki/Rational_Numbers_form_Prime_Field | https://proofwiki.org/wiki/Rational_Numbers_form_Prime_Field | [
"Rational Numbers",
"Prime Fields"
] | [
"Definition:Field of Rational Numbers",
"Definition:Prime Field",
"Definition:Subset",
"Definition:Addition/Rational Numbers",
"Definition:Multiplication/Rational Numbers",
"Definition:Vacuous Truth"
] | [
"Definition:Subfield",
"Definition:Field (Abstract Algebra)",
"Definition:Abelian Group",
"Definition:Abelian Group",
"Definition:Distributive Operation",
"Principle of Mathematical Induction",
"Definition:Positive/Integer",
"Definition:Negative/Integer",
"Subfield Test"
] |
proofwiki-4268 | Field of Integers Modulo Prime is Prime Field | Let $p$ be a prime number.
Let $\struct {\Z_p, +, \times}$ be the field of integers modulo $p$.
Then $\struct {\Z_p, +, \times}$ is a prime field. | If $\struct {F, +, \times}$ is a subfield of $\struct {\Z_p, +, \times}$, then $\struct {F, +}$ is a subgroup of $\struct {\Z_p, +}$.
But from Prime Group has no Proper Subgroups, $\struct {\Z_p, +}$ has no proper subgroup except the trivial group.
Hence $F = \Z_p$ and so follows the result.
{{qed}} | Let $p$ be a [[Definition:Prime Number|prime number]].
Let $\struct {\Z_p, +, \times}$ be the [[Ring of Integers Modulo Prime is Field|field of integers modulo $p$]].
Then $\struct {\Z_p, +, \times}$ is a [[Definition:Prime Field|prime field]]. | If $\struct {F, +, \times}$ is a [[Definition:Subfield|subfield]] of $\struct {\Z_p, +, \times}$, then $\struct {F, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\Z_p, +}$.
But from [[Prime Group has no Proper Subgroups]], $\struct {\Z_p, +}$ has no [[Definition:Proper Subgroup|proper subgroup]] except the [[... | Field of Integers Modulo Prime is Prime Field | https://proofwiki.org/wiki/Field_of_Integers_Modulo_Prime_is_Prime_Field | https://proofwiki.org/wiki/Field_of_Integers_Modulo_Prime_is_Prime_Field | [
"Modulo Arithmetic",
"Prime Fields",
"Field of Integers Modulo Prime"
] | [
"Definition:Prime Number",
"Ring of Integers Modulo Prime is Field",
"Definition:Prime Field"
] | [
"Definition:Subfield",
"Definition:Subgroup",
"Prime Group has no Proper Subgroups",
"Definition:Proper Subgroup",
"Definition:Trivial Group"
] |
proofwiki-4269 | Field has Prime Subfield | Let $\struct {F, +, \times}$ be a field.
Then $F$ has a subfield which is a prime field. | By definition of field, $F$ is a division ring where $\times$ is commutative.
Therefore all division subrings of $F$ are in fact subfields of $F$.
By Intersection of All Division Subrings is Prime Subfield, the intersection of all subfields of $F$ is a prime field which is a subfield of $F$.
{{qed}} | Let $\struct {F, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]].
Then $F$ has a [[Definition:Subfield|subfield]] which is a [[Definition:Prime Field|prime field]]. | By definition of [[Definition:Field (Abstract Algebra)|field]], $F$ is a [[Definition:Division Ring|division ring]] where $\times$ is [[Definition:Commutative Operation|commutative]].
Therefore all [[Definition:Division Subring|division subrings]] of $F$ are in fact [[Definition:Subfield|subfields]] of $F$.
By [[Inte... | Field has Prime Subfield | https://proofwiki.org/wiki/Field_has_Prime_Subfield | https://proofwiki.org/wiki/Field_has_Prime_Subfield | [
"Subfields"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Subfield",
"Definition:Prime Field"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Division Ring",
"Definition:Commutative/Operation",
"Definition:Division Subring",
"Definition:Subfield",
"Intersection of All Division Subrings is Prime Subfield",
"Definition:Set Intersection",
"Definition:Subfield",
"Definition:Prime Field",
"D... |
proofwiki-4270 | Constant Mapping to Identity is Homomorphism/Rings | Let $\struct {R_1, +_1, \circ_1}$ and $\struct {R_2, +_2, \circ_2}$ be rings with zeroes $0_1$ and $0_2$ respectively.
Let $\zeta$ be the zero homomorphism from $R_1$ to $R_2$, that is:
:$\forall x \in R_1: \map \zeta x = 0_2$
Then $\zeta$ is a ring homomorphism whose image is $\set {0_2}$ and whose kernel is $R_1$. | The additive groups of $\struct {R_1, +_1, \circ_1}$ and $\struct {R_2, +_2, \circ_2}$ are $\struct {R_1, +_1}$ and $\struct {R_2, +_2}$ respectively.
Their identities are $0_1$ and $0_2$ respectively.
Thus from the Constant Mapping to Group Identity is Homomorphism we have that $\zeta: \struct {R_1, +_1} \to \struct {... | Let $\struct {R_1, +_1, \circ_1}$ and $\struct {R_2, +_2, \circ_2}$ be [[Definition:Ring (Abstract Algebra)|rings]] with [[Definition:Ring Zero|zeroes]] $0_1$ and $0_2$ respectively.
Let $\zeta$ be the [[Definition:Zero Homomorphism|zero homomorphism]] from $R_1$ to $R_2$, that is:
:$\forall x \in R_1: \map \zeta x = ... | The [[Definition:Additive Group of Ring|additive groups]] of $\struct {R_1, +_1, \circ_1}$ and $\struct {R_2, +_2, \circ_2}$ are $\struct {R_1, +_1}$ and $\struct {R_2, +_2}$ respectively.
Their [[Definition:Identity Element|identities]] are $0_1$ and $0_2$ respectively.
Thus from the [[Constant Mapping to Group Iden... | Constant Mapping to Identity is Homomorphism/Rings | https://proofwiki.org/wiki/Constant_Mapping_to_Identity_is_Homomorphism/Rings | https://proofwiki.org/wiki/Constant_Mapping_to_Identity_is_Homomorphism/Rings | [
"Constant Mapping to Identity is Homomorphism",
"Ring Homomorphisms"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Zero",
"Definition:Zero Homomorphism",
"Definition:Ring Homomorphism",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Kernel of Ring Homomorphism"
] | [
"Definition:Additive Group of Ring",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Constant Mapping to Identity is Homomorphism/Groups",
"Definition:Group Homomorphism",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Kernel of Ring Homomorphism"
] |
proofwiki-4271 | Constant Mapping to Identity is Homomorphism/Groups | Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be groups with identities $e_1$ and $e_2$ respectively.
Let $\phi_e: \struct {G_1, \circ_1} \to \struct {G_2, \circ_2}$ be the constant mapping defined as:
:$\forall x \in G_1: \map {\phi_e} x = e_2$
Then $\phi_e$ is a group homomorphism whose image is $\set {e_... | Let $x, y \in G_1$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\phi_e} {x \circ_1 y}
| r = e_2
| c = as $x \circ_1 y \in G_1$
}}
{{eqn | r = \map {\phi_e} x \circ_2 \map {\phi_e} y
| c = as $\map {\phi_e} x = e_2$ and $\map {\phi_e} y = e_2$
}}
{{end-eqn}}
So $\phi_e$ is a group homomorphism.
The results a... | Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be [[Definition:Group|groups]] with [[Definition:Identity Element|identities]] $e_1$ and $e_2$ respectively.
Let $\phi_e: \struct {G_1, \circ_1} \to \struct {G_2, \circ_2}$ be the [[Definition:Constant Mapping|constant mapping]] defined as:
:$\forall x \in G_1:... | Let $x, y \in G_1$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\phi_e} {x \circ_1 y}
| r = e_2
| c = as $x \circ_1 y \in G_1$
}}
{{eqn | r = \map {\phi_e} x \circ_2 \map {\phi_e} y
| c = as $\map {\phi_e} x = e_2$ and $\map {\phi_e} y = e_2$
}}
{{end-eqn}}
So $\phi_e$ is a [[Definition:Group Homomorphi... | Constant Mapping to Identity is Homomorphism/Groups | https://proofwiki.org/wiki/Constant_Mapping_to_Identity_is_Homomorphism/Groups | https://proofwiki.org/wiki/Constant_Mapping_to_Identity_is_Homomorphism/Groups | [
"Constant Mapping to Identity is Homomorphism",
"Group Homomorphisms"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Constant Mapping",
"Definition:Group Homomorphism",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Kernel of Group Homomorphism"
] | [
"Definition:Group Homomorphism",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Kernel of Group Homomorphism"
] |
proofwiki-4272 | Identity Mapping is Automorphism/Groups | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Then $I_G: \struct {G, \circ} \to \struct {G, \circ}$ is a group automorphism.
Its kernel is $\set e$. | The main result Identity Mapping is Automorphism holds directly.
As $I_G$ is a bijection, the only element that maps to $e$ is $e$ itself.
Thus the kernel is $\set e$.
{{qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Then $I_G: \struct {G, \circ} \to \struct {G, \circ}$ is a [[Definition:Group Automorphism|group automorphism]].
Its [[Definition:Kernel of Group Homomorphism|kernel]] is $\set e$. | The main result [[Identity Mapping is Automorphism]] holds directly.
As $I_G$ is a [[Definition:Bijection|bijection]], the only element that maps to $e$ is $e$ itself.
Thus the [[Definition:Kernel of Group Homomorphism|kernel]] is $\set e$.
{{qed}} | Identity Mapping is Automorphism/Groups | https://proofwiki.org/wiki/Identity_Mapping_is_Automorphism/Groups | https://proofwiki.org/wiki/Identity_Mapping_is_Automorphism/Groups | [
"Group Automorphisms",
"Identity Mappings"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Group Automorphism",
"Definition:Kernel of Group Homomorphism"
] | [
"Identity Mapping is Automorphism",
"Definition:Bijection",
"Definition:Kernel of Group Homomorphism"
] |
proofwiki-4273 | Identity Mapping is Automorphism/Rings | Let $\struct {R, +, \circ}$ be a ring whose zero is $0$.
Then $I_R: \struct {R, +, \circ} \to \struct {R, +, \circ}$ is a ring automorphism.
Its kernel is $\set 0$. | The result Identity Mapping is Automorphism holds directly, for both $+$ and $\circ$.
As $I_R$ is a bijection, the only element that maps to $0$ is $0$ itself.
Thus the kernel is $\set 0$.
{{qed}} | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0$.
Then $I_R: \struct {R, +, \circ} \to \struct {R, +, \circ}$ is a [[Definition:Ring Automorphism|ring automorphism]].
Its [[Definition:Kernel of Ring Homomorphism|kernel]] is $\set 0$. | The result [[Identity Mapping is Automorphism]] holds directly, for both $+$ and $\circ$.
As $I_R$ is a [[Definition:Bijection|bijection]], the only element that maps to $0$ is $0$ itself.
Thus the [[Definition:Kernel of Ring Homomorphism|kernel]] is $\set 0$.
{{qed}} | Identity Mapping is Automorphism/Rings | https://proofwiki.org/wiki/Identity_Mapping_is_Automorphism/Rings | https://proofwiki.org/wiki/Identity_Mapping_is_Automorphism/Rings | [
"Ring Automorphisms",
"Identity Mappings"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Zero",
"Definition:Ring Automorphism",
"Definition:Kernel of Ring Homomorphism"
] | [
"Identity Mapping is Automorphism",
"Definition:Bijection",
"Definition:Kernel of Ring Homomorphism"
] |
proofwiki-4274 | Ring Monomorphism from Integers to Rationals | Let $\phi: \Z \to \Q$ be the mapping from the integers $\Z$ to the rational numbers $\Q$ defined as:
:$\forall x \in \Z: \map \phi x = \dfrac x 1$
Then $\phi$ is a (ring) monomorphism, but specifically not an epimorphism. | First note that:
:$\forall a, b \in \Z: a \ne b \implies \dfrac a 1 = \map \phi a \ne \map \phi b = \dfrac b 1$
and so clearly $\phi$ is an injection.
However, take for example $\dfrac 1 2$:
:$\not \exists a \in \Z: \map \phi a = \dfrac 1 2$
as $\dfrac 1 2 \notin \Img \phi$.
So $\phi$ is not a surjection.
Next let $a,... | Let $\phi: \Z \to \Q$ be the [[Definition:Mapping|mapping]] from the [[Definition:Integer|integers]] $\Z$ to the [[Definition:Rational Number|rational numbers]] $\Q$ defined as:
:$\forall x \in \Z: \map \phi x = \dfrac x 1$
Then $\phi$ is a [[Definition:Ring Monomorphism|(ring) monomorphism]], but specifically not an... | First note that:
:$\forall a, b \in \Z: a \ne b \implies \dfrac a 1 = \map \phi a \ne \map \phi b = \dfrac b 1$
and so clearly $\phi$ is an [[Definition:Injection|injection]].
However, take for example $\dfrac 1 2$:
:$\not \exists a \in \Z: \map \phi a = \dfrac 1 2$
as $\dfrac 1 2 \notin \Img \phi$.
So $\phi$ is not... | Ring Monomorphism from Integers to Rationals | https://proofwiki.org/wiki/Ring_Monomorphism_from_Integers_to_Rationals | https://proofwiki.org/wiki/Ring_Monomorphism_from_Integers_to_Rationals | [
"Ring Monomorphisms",
"Integers",
"Rational Numbers"
] | [
"Definition:Mapping",
"Definition:Integer",
"Definition:Rational Number",
"Definition:Ring Monomorphism",
"Definition:Ring Epimorphism"
] | [
"Definition:Injection",
"Definition:Surjection",
"Definition:Injection",
"Definition:Surjection",
"Definition:Ring Homomorphism",
"Definition:Ring Monomorphism",
"Definition:Ring Epimorphism"
] |
proofwiki-4275 | Field Homomorphism Preserves Unity | Let $\phi: \struct {F_1, +_1, \times_1} \to \struct {F_2, +_2, \times_2}$ be a field homomorphism.
Let:
:$1_{F_1}$ be the unity of $F_1$
:$1_{F_2}$ be the unity of $F_2$.
Then:
:$\map \phi {1_{F_1} } = 1_{F_2}$ | By definition, if $\struct {F_1, +_1, \times_1}$ and $\struct {F_2, +_2, \times_2}$ are fields then $\struct {F_1^*, \times_1}$ and $\struct {F_2^*, \times_2}$ are groups.
Again by definition:
:the unity of $\struct {F_1, +_1, \times_1}$ is the identity of $\struct {F_1^*, \times_1}$
:the unity of $\struct {F_2, +_2, \... | Let $\phi: \struct {F_1, +_1, \times_1} \to \struct {F_2, +_2, \times_2}$ be a [[Definition:Field Homomorphism|field homomorphism]].
Let:
:$1_{F_1}$ be the [[Definition:Unity of Field|unity]] of $F_1$
:$1_{F_2}$ be the [[Definition:Unity of Field|unity]] of $F_2$.
Then:
:$\map \phi {1_{F_1} } = 1_{F_2}$ | By definition, if $\struct {F_1, +_1, \times_1}$ and $\struct {F_2, +_2, \times_2}$ are [[Definition:Field (Abstract Algebra)|fields]] then $\struct {F_1^*, \times_1}$ and $\struct {F_2^*, \times_2}$ are [[Definition:Group|groups]].
Again by definition:
:the [[Definition:Unity of Field|unity]] of $\struct {F_1, +_1, \... | Field Homomorphism Preserves Unity | https://proofwiki.org/wiki/Field_Homomorphism_Preserves_Unity | https://proofwiki.org/wiki/Field_Homomorphism_Preserves_Unity | [
"Field Homomorphisms"
] | [
"Definition:Field Homomorphism",
"Definition:Multiplicative Identity",
"Definition:Multiplicative Identity"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Group",
"Definition:Multiplicative Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Multiplicative Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Group Homomorphism Preserves Identity"
] |
proofwiki-4276 | Field Homomorphism Preserves Product Inverses | Let $\phi: \struct {F_1, +_1, \times_1} \to \struct {F_2, +_2, \times_2}$ be a field homomorphism.
Then:
:$\forall x \in F_1^*: \map \phi {x^{-1} } = \map \phi x^{-1}$ | By definition, if $\struct {F_1, +_1, \times_1}$ and $\struct {F_2, +_2, \times_2}$ are fields then $\struct {F_1^*, \times_1}$ and $\struct {F_2^*, \times_2}$ are groups.
Again by definition:
:the product inverse of $x$ in $\struct {F_1, +_1, \times_1}$ for $\times_1$ is the product inverse of $x$ in $\struct {F_1^*, ... | Let $\phi: \struct {F_1, +_1, \times_1} \to \struct {F_2, +_2, \times_2}$ be a [[Definition:Field Homomorphism|field homomorphism]].
Then:
:$\forall x \in F_1^*: \map \phi {x^{-1} } = \map \phi x^{-1}$ | By definition, if $\struct {F_1, +_1, \times_1}$ and $\struct {F_2, +_2, \times_2}$ are [[Definition:Field (Abstract Algebra)|fields]] then $\struct {F_1^*, \times_1}$ and $\struct {F_2^*, \times_2}$ are [[Definition:Group|groups]].
Again by definition:
:the [[Definition:Product Inverse|product inverse]] of $x$ in $\s... | Field Homomorphism Preserves Product Inverses | https://proofwiki.org/wiki/Field_Homomorphism_Preserves_Product_Inverses | https://proofwiki.org/wiki/Field_Homomorphism_Preserves_Product_Inverses | [
"Field Homomorphisms"
] | [
"Definition:Field Homomorphism"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Group",
"Definition:Product Inverse",
"Definition:Product Inverse",
"Definition:Product Inverse",
"Definition:Product Inverse",
"Group Homomorphism Preserves Inverses"
] |
proofwiki-4277 | Ring Homomorphism from Field is Monomorphism or Zero Homomorphism | Let $\struct {F, +_F, \circ}$ be a field whose zero is $0_F$.
Let $\struct {S, +_S, *}$ be a ring whose zero is $0_S$.
Let $\phi: F \to S$ be a ring homomorphism.
Then either:
:$(1): \quad \phi$ is a monomorphism (that is, $\phi$ is injective)
or
:$(2): \quad \phi$ is the zero homomorphism (that is, $\forall a \in F: \... | We have by definition that a field is a division ring.
The result can be seen to be an application of Ring Homomorphism from Division Ring is Monomorphism or Zero Homomorphism.
{{qed}} | Let $\struct {F, +_F, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$.
Let $\struct {S, +_S, *}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_S$.
Let $\phi: F \to S$ be a [[Definition:Ring Homomorphism|ring homomorp... | We have by definition that a [[Definition:Field (Abstract Algebra)|field]] is a [[Definition:Division Ring|division ring]].
The result can be seen to be an application of [[Ring Homomorphism from Division Ring is Monomorphism or Zero Homomorphism]].
{{qed}} | Ring Homomorphism from Field is Monomorphism or Zero Homomorphism/Proof 1 | https://proofwiki.org/wiki/Ring_Homomorphism_from_Field_is_Monomorphism_or_Zero_Homomorphism | https://proofwiki.org/wiki/Ring_Homomorphism_from_Field_is_Monomorphism_or_Zero_Homomorphism/Proof_1 | [
"Ring Homomorphism from Field is Monomorphism or Zero Homomorphism",
"Field Theory",
"Monomorphisms (Abstract Algebra)"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Zero",
"Definition:Ring Homomorphism",
"Definition:Ring Monomorphism",
"Definition:Injection",
"Definition:Zero Homomorphism"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Division Ring",
"Ring Homomorphism from Division Ring is Monomorphism or Zero Homomorphism"
] |
proofwiki-4278 | Ring Homomorphism from Field is Monomorphism or Zero Homomorphism | Let $\struct {F, +_F, \circ}$ be a field whose zero is $0_F$.
Let $\struct {S, +_S, *}$ be a ring whose zero is $0_S$.
Let $\phi: F \to S$ be a ring homomorphism.
Then either:
:$(1): \quad \phi$ is a monomorphism (that is, $\phi$ is injective)
or
:$(2): \quad \phi$ is the zero homomorphism (that is, $\forall a \in F: \... | Let $\phi: F \to S$ be a ring homomorphism.
Suppose $\phi$ is not a monomorphism.
By definition, $\phi$ is not an injection.
So there must exist $a, b \in F: \map \phi a = \map \phi b$.
Let $k = a +_F \paren {-b}$.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi k
| r = \map \phi {a +_F \paren {-b} }
| c =
}}
{{... | Let $\struct {F, +_F, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$.
Let $\struct {S, +_S, *}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_S$.
Let $\phi: F \to S$ be a [[Definition:Ring Homomorphism|ring homomorp... | Let $\phi: F \to S$ be a [[Definition:Ring Homomorphism|ring homomorphism]].
Suppose $\phi$ is not a [[Definition:Ring Monomorphism|monomorphism]].
By definition, $\phi$ is not an [[Definition:Injection|injection]].
So there must exist $a, b \in F: \map \phi a = \map \phi b$.
Let $k = a +_F \paren {-b}$.
Then:
{{... | Ring Homomorphism from Field is Monomorphism or Zero Homomorphism/Proof 2 | https://proofwiki.org/wiki/Ring_Homomorphism_from_Field_is_Monomorphism_or_Zero_Homomorphism | https://proofwiki.org/wiki/Ring_Homomorphism_from_Field_is_Monomorphism_or_Zero_Homomorphism/Proof_2 | [
"Ring Homomorphism from Field is Monomorphism or Zero Homomorphism",
"Field Theory",
"Monomorphisms (Abstract Algebra)"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Zero",
"Definition:Ring Homomorphism",
"Definition:Ring Monomorphism",
"Definition:Injection",
"Definition:Zero Homomorphism"
] | [
"Definition:Ring Homomorphism",
"Definition:Ring Monomorphism",
"Definition:Injection",
"Definition:Product Inverse",
"Definition:Ring Monomorphism",
"Definition:Zero Homomorphism"
] |
proofwiki-4279 | Power to Characteristic of Field is Monomorphism | Let $F$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let the characteristic of $F$ be $p$ where $p \ne 0$.
Let $\phi: F \to F$ be the mapping on $F$ defined as:
:$\forall x \in F: \map \phi x = x^p$
Then $\phi$ is a (field) monomorphism. | Let $a, b \in F$.
First note that:
{{begin-eqn}}
{{eqn | q = \forall k: 0 < k < p
| l = \binom p k
| o = \equiv
| r = 0
| rr= \pmod p
| c = Binomial Coefficient of Prime
}}
{{eqn | ll= \leadsto
| l = \binom p k
| r = r p
| c = for some $r \in \Z$
}}
{{eqn | ll= \leadsto
... | 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 $1_F$.
Let the [[Definition:Characteristic of Field|characteristic]] of $F$ be $p$ where $p \ne 0$.
Let $\phi: F \to F$ be the [[Definition:Mapping|mapping]] on $F$... | Let $a, b \in F$.
First note that:
{{begin-eqn}}
{{eqn | q = \forall k: 0 < k < p
| l = \binom p k
| o = \equiv
| r = 0
| rr= \pmod p
| c = [[Binomial Coefficient of Prime]]
}}
{{eqn | ll= \leadsto
| l = \binom p k
| r = r p
| c = for some $r \in \Z$
}}
{{eqn | ll= \lea... | Power to Characteristic of Field is Monomorphism | https://proofwiki.org/wiki/Power_to_Characteristic_of_Field_is_Monomorphism | https://proofwiki.org/wiki/Power_to_Characteristic_of_Field_is_Monomorphism | [
"Characteristics of Fields",
"Field Monomorphisms"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Multiplicative Identity",
"Definition:Characteristic of Field",
"Definition:Mapping",
"Definition:Field Monomorphism"
] | [
"Binomial Coefficient of Prime",
"Characteristic of Field by Annihilator/Prime Characteristic",
"Binomial Theorem",
"Power of Product of Commutative Elements in Monoid",
"Definition:Field Homomorphism",
"Definition:Zero Homomorphism",
"Power of Identity is Identity",
"Ring Homomorphism from Field is M... |
proofwiki-4280 | Power to Characteristic Power of Field is Monomorphism | Let $F$ be a field whose characteristic is $p$ where $p \ne 0$.
Let $n \in \Z_{\ge 0}$ be any positive integer.
Let $\phi_n: F \to F$ be the mapping on $F$ defined as:
:$\forall x \in F: \map {\phi_n} x = x^{p^n}$
Then $\phi_n$ is a (field) monomorphism. | Proof by induction:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\phi_n$ is a (field) monomorphism.
$\map P 0$ is trivially true:
:$\map {\phi_0} x = x^{p^0} = x^1 = x$
and we see that $\phi_0$ is the identity automorphism.
This is not the zero homomorphism.
So from Ring Homomorphism from Field is M... | Let $F$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Characteristic of Field|characteristic]] is $p$ where $p \ne 0$.
Let $n \in \Z_{\ge 0}$ be any [[Definition:Positive Integer|positive integer]].
Let $\phi_n: F \to F$ be the [[Definition:Mapping|mapping]] on $F$ defined as:
:$\forall x \in ... | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\phi_n$ is a [[Definition:Field Monomorphism|(field) monomorphism]].
$\map P 0$ is trivially true:
:$\map {\phi_0} x = x^{p^0} = x^1 = x$
and we see that $\phi_0$ is ... | Power to Characteristic Power of Field is Monomorphism | https://proofwiki.org/wiki/Power_to_Characteristic_Power_of_Field_is_Monomorphism | https://proofwiki.org/wiki/Power_to_Characteristic_Power_of_Field_is_Monomorphism | [
"Characteristics of Fields",
"Field Monomorphisms"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Characteristic of Field",
"Definition:Positive/Integer",
"Definition:Mapping",
"Definition:Field Monomorphism"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Field Monomorphism",
"Definition:Identity Automorphism",
"Definition:Zero Homomorphism",
"Ring Homomorphism from Field is Monomorphism or Zero Homomorphism",
"Definition:Field Monomorphism",
"Definition:Field Monomorphism",
... |
proofwiki-4281 | Scalar Product with Identity | :$\lambda \ast e = 0_R \ast x = e$ | From {{Module-axiom|1}}, $y \to \lambda \ast y$ is an endomorphism of $\struct {G, +_G}$.
From {{Module-axiom|2}}, $\mu \to \mu \ast x$ is a homomorphism from $\struct {R, +_R}$ to $\struct {G, +_G}$.
The result follows from Homomorphism with Cancellable Codomain Preserves Identity.
{{qed}} | :$\lambda \ast e = 0_R \ast x = e$ | From {{Module-axiom|1}}, $y \to \lambda \ast y$ is an [[Definition:Endomorphism|endomorphism]] of $\struct {G, +_G}$.
From {{Module-axiom|2}}, $\mu \to \mu \ast x$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] from $\struct {R, +_R}$ to $\struct {G, +_G}$.
The result follows from [[Homomorphism wit... | Scalar Product with Identity | https://proofwiki.org/wiki/Scalar_Product_with_Identity | https://proofwiki.org/wiki/Scalar_Product_with_Identity | [
"Module Theory"
] | [] | [
"Definition:Endomorphism",
"Definition:Homomorphism (Abstract Algebra)",
"Homomorphism with Cancellable Codomain Preserves Identity"
] |
proofwiki-4282 | Scalar Product with Inverse | :$\lambda \ast \struct {-x} = \struct {-\lambda} \ast x = -\struct {\lambda \ast x}$ | From {{Module-axiom|1}}, $y \to \lambda \ast y$ is an endomorphism of $\struct {G, +_G}$.
From {{Module-axiom|2}}, $\mu \to \mu \ast x$ is a homomorphism from $\struct {R, +_R}$ to $\struct {G, +_G}$.
The result follows from Homomorphism with Identity Preserves Inverses.
{{qed}} | :$\lambda \ast \struct {-x} = \struct {-\lambda} \ast x = -\struct {\lambda \ast x}$ | From {{Module-axiom|1}}, $y \to \lambda \ast y$ is an [[Definition:Endomorphism|endomorphism]] of $\struct {G, +_G}$.
From {{Module-axiom|2}}, $\mu \to \mu \ast x$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] from $\struct {R, +_R}$ to $\struct {G, +_G}$.
The result follows from [[Homomorphism wit... | Scalar Product with Inverse | https://proofwiki.org/wiki/Scalar_Product_with_Inverse | https://proofwiki.org/wiki/Scalar_Product_with_Inverse | [
"Module Theory"
] | [] | [
"Definition:Endomorphism",
"Definition:Homomorphism (Abstract Algebra)",
"Homomorphism with Identity Preserves Inverses"
] |
proofwiki-4283 | Scalar Product with Sum | :$\ds \lambda \ast \paren {\sum_{k \mathop = 1}^m x_k} = \sum_{k \mathop = 1}^m \paren {\lambda \ast x_k}$ | This follows by induction from {{Module-axiom|1}}, as follows:
For all $m \in \N_{>0}$, let $\map P m$ be the proposition:
:$\ds \lambda \ast \paren {\sum_{k \mathop = 1}^m x_k} = \sum_{k \mathop = 1}^m \paren {\lambda \ast x_k}$ | :$\ds \lambda \ast \paren {\sum_{k \mathop = 1}^m x_k} = \sum_{k \mathop = 1}^m \paren {\lambda \ast x_k}$ | This follows by [[Principle of Mathematical Induction|induction]] from {{Module-axiom|1}}, as follows:
For all $m \in \N_{>0}$, let $\map P m$ be the [[Definition:Proposition|proposition]]:
:$\ds \lambda \ast \paren {\sum_{k \mathop = 1}^m x_k} = \sum_{k \mathop = 1}^m \paren {\lambda \ast x_k}$ | Scalar Product with Sum | https://proofwiki.org/wiki/Scalar_Product_with_Sum | https://proofwiki.org/wiki/Scalar_Product_with_Sum | [
"Module Theory"
] | [] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-4284 | Product with Sum of Scalar | :$\ds \paren {\sum_{k \mathop = 1}^m \lambda_k} \ast x = \sum_{k \mathop = 1}^m \paren {\lambda_k \ast x}$ | This follows by induction from {{Module-axiom|2}}, as follows.
For all $m \in \N_{>0}$, let $\map P m$ be the proposition:
:$\ds \paren {\sum_{k \mathop = 1}^m \lambda_k} \ast x = \sum_{k \mathop = 1}^m \paren {\lambda_k \ast x}$ | :$\ds \paren {\sum_{k \mathop = 1}^m \lambda_k} \ast x = \sum_{k \mathop = 1}^m \paren {\lambda_k \ast x}$ | This follows by [[Principle of Mathematical Induction|induction]] from {{Module-axiom|2}}, as follows.
For all $m \in \N_{>0}$, let $\map P m$ be the [[Definition:Proposition|proposition]]:
:$\ds \paren {\sum_{k \mathop = 1}^m \lambda_k} \ast x = \sum_{k \mathop = 1}^m \paren {\lambda_k \ast x}$ | Product with Sum of Scalar | https://proofwiki.org/wiki/Product_with_Sum_of_Scalar | https://proofwiki.org/wiki/Product_with_Sum_of_Scalar | [
"Module Theory"
] | [] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-4285 | Scalar Product with Product | :$\lambda \ast \paren {n \cdot x} = n \cdot \paren {\lambda \ast x} = \paren {n \cdot \lambda} \ast x$ | First let $n = 0$.
The assertion follows directly from Scalar Product with Identity.
Next, let $n > 0$.
The assertion follows directly from Scalar Product with Sum and Product with Sum of Scalar, by letting $m = n$ and making all the $\lambda$s and $x$s the same.
Finally, let $n < 0$.
The assertion follows from Scalar ... | :$\lambda \ast \paren {n \cdot x} = n \cdot \paren {\lambda \ast x} = \paren {n \cdot \lambda} \ast x$ | First let $n = 0$.
The assertion follows directly from [[Scalar Product with Identity]].
Next, let $n > 0$.
The assertion follows directly from [[Scalar Product with Sum]] and [[Product with Sum of Scalar]], by letting $m = n$ and making all the $\lambda$s and $x$s the same.
Finally, let $n < 0$.
The assertion f... | Scalar Product with Product | https://proofwiki.org/wiki/Scalar_Product_with_Product | https://proofwiki.org/wiki/Scalar_Product_with_Product | [
"Module Theory"
] | [] | [
"Scalar Product with Identity",
"Scalar Product with Sum",
"Product with Sum of Scalar",
"Scalar Product with Product",
"Scalar Product with Inverse",
"Index Laws for Monoids/Negative Index"
] |
proofwiki-4286 | Zero Vector is Linearly Dependent | Let $G$ be a group whose identity is $e$.
Let $R$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $\struct {G, +_G, \circ}_R$ be a unitary $R$-module.
Then the singleton set $\set e$ consisting of the zero vector is linearly dependent. | By Scalar Product with Identity we have:
:$\forall \lambda \in R: \lambda \circ e = e$
Hence the result by definition of linearly dependent.
{{qed}} | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
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 $\struct {G, +_G, \circ}_R$ be a [[Definition:Unitary Modul... | By [[Scalar Product with Identity]] we have:
:$\forall \lambda \in R: \lambda \circ e = e$
Hence the result by definition of [[Definition:Linearly Dependent Set|linearly dependent]].
{{qed}} | Zero Vector is Linearly Dependent | https://proofwiki.org/wiki/Zero_Vector_is_Linearly_Dependent | https://proofwiki.org/wiki/Zero_Vector_is_Linearly_Dependent | [
"Linear Dependence"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Ring with Unity",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Unitary Module over Ring",
"Definition:Singleton",
"Definition:Zero Vector",
"Definition:Linearly Dependent/S... | [
"Scalar Product with Identity",
"Definition:Linearly Dependent/Set"
] |
proofwiki-4287 | Expression of Vector as Linear Combination from Basis is Unique | Let $V$ be a vector space of dimension $n$.
Let $\BB = \set {\mathbf x_1, \mathbf x_2, \ldots, \mathbf x_n}$ be a basis for $V$.
Let $\mathbf x \in V$ be any vector of $V$.
Then $\mathbf x$ can be expressed as a unique linear combination of elements of $\BB$.
{{explain|The definition we have of linear combination doesn... | === Proof of Existence ===
By the definition of basis, $\BB$ is a spanning set.
Hence the result, by the definition of spanning set.
{{qed|lemma}} | Let $V$ be a [[Definition:Vector Space|vector space]] of [[Definition:Dimension (Linear Algebra)|dimension]] $n$.
Let $\BB = \set {\mathbf x_1, \mathbf x_2, \ldots, \mathbf x_n}$ be a [[Definition:Basis of Vector Space|basis]] for $V$.
Let $\mathbf x \in V$ be any [[Definition:Vector (Linear Algebra)|vector]] of $V$.... | === Proof of Existence ===
By the definition of [[Definition:Basis of Vector Space|basis]], $\BB$ is a [[Definition:Spanning Set of Vector Space|spanning set]].
Hence the result, by the definition of [[Definition:Spanning Set of Vector Space|spanning set]].
{{qed|lemma}} | Expression of Vector as Linear Combination from Basis is Unique | https://proofwiki.org/wiki/Expression_of_Vector_as_Linear_Combination_from_Basis_is_Unique | https://proofwiki.org/wiki/Expression_of_Vector_as_Linear_Combination_from_Basis_is_Unique | [
"Linear Algebra",
"Bases of Vector Spaces"
] | [
"Definition:Vector Space",
"Definition:Dimension (Linear Algebra)",
"Definition:Basis of Vector Space",
"Definition:Vector/Linear Algebra",
"Definition:Unique",
"Definition:Linear Combination"
] | [
"Definition:Basis of Vector Space",
"Definition:Generator of Vector Space",
"Definition:Generator of Vector Space",
"Definition:Basis of Vector Space"
] |
proofwiki-4288 | Generator of Vector Space Contains Basis | :$G$ contains a basis for $E$. | From:
:Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set
:Bases of Finitely Generated Vector Space have Equal Cardinality
and
:Sufficient Conditions for Basis of Finite Dimensional Vector Space
all we need to do is show that every infinite generator $S$ for $E$ contains a finite generator.... | :$G$ contains a [[Definition:Basis of Vector Space|basis]] for $E$. | From:
:[[Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set]]
:[[Bases of Finitely Generated Vector Space have Equal Cardinality]]
and
:[[Sufficient Conditions for Basis of Finite Dimensional Vector Space]]
all we need to do is show that every [[Definition:Infinite Set|infinite]] [[Definiti... | Generator of Vector Space Contains Basis | https://proofwiki.org/wiki/Generator_of_Vector_Space_Contains_Basis | https://proofwiki.org/wiki/Generator_of_Vector_Space_Contains_Basis | [
"Bases of Vector Spaces",
"Generators of Vector Spaces",
"Generator of Vector Space Contains Basis"
] | [
"Definition:Basis of Vector Space"
] | [
"Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set",
"Bases of Finitely Generated Vector Space have Equal Cardinality",
"Sufficient Conditions for Basis of Finite Dimensional Vector Space",
"Definition:Infinite Set",
"Definition:Generator of Vector Space",
"Definition:Finite ... |
proofwiki-4289 | Cardinality of Linearly Independent Set is No Greater than Dimension | :$H$ has at most $n$ elements. | Let $H$ be a linearly independent subset of $E$.
By definition of dimension of vector space, $E$ has a basis with exactly $n$ elements.
By Sufficient Conditions for Basis of Finite Dimensional Vector Space, $B$ is a generator for $E$.
Then by Size of Linearly Independent Subset is at Most Size of Finite Generator, $H$ ... | :$H$ has at most $n$ [[Definition:Element|elements]]. | Let $H$ be a [[Definition:Linearly Independent Set|linearly independent subset]] of $E$.
By definition of [[Definition:Dimension of Vector Space|dimension of vector space]], $E$ has a [[Definition:Basis of Vector Space|basis]] with exactly $n$ [[Definition:Element|elements]].
By [[Sufficient Conditions for Basis of F... | Cardinality of Linearly Independent Set is No Greater than Dimension | https://proofwiki.org/wiki/Cardinality_of_Linearly_Independent_Set_is_No_Greater_than_Dimension | https://proofwiki.org/wiki/Cardinality_of_Linearly_Independent_Set_is_No_Greater_than_Dimension | [
"Linear Independence"
] | [
"Definition:Element"
] | [
"Definition:Linearly Independent/Set",
"Definition:Dimension of Vector Space",
"Definition:Basis of Vector Space",
"Definition:Element",
"Sufficient Conditions for Basis of Finite Dimensional Vector Space",
"Definition:Generator of Module",
"Size of Linearly Independent Subset is at Most Size of Finite ... |
proofwiki-4290 | Cardinality of Generator of Vector Space is not Less than Dimension | Let $V$ be a vector space over a field $F$.
Let $\BB$ be a generator for $V$ containing $m$ elements.
Then:
:$\map {\dim_F} V \le m$
where $\map {\dim_F} V$ is the dimension of $V$. | Let $\BB = \set {x_1, x_2, \ldots, x_m}$ be a generator for $G$.
Let $\set {y_1, y_2, \ldots, y_n}$ be a subset of $G$ such that $n > m$.
As $\BB$ generates $G$, there exist $\alpha_{i j} \in F$ where $1 \le i \le m, 1 \le j \le n$ such that:
:$\ds \forall j: 1 \le j \le n: y_j = \sum_{i \mathop = 1}^m \alpha_{i j} x_i... | Let $V$ be a [[Definition:Vector Space|vector space]] over a [[Definition:Field (Abstract Algebra)|field]] $F$.
Let $\BB$ be a [[Definition:Generator of Vector Space|generator]] for $V$ containing $m$ [[Definition:Element|elements]].
Then:
:$\map {\dim_F} V \le m$
where $\map {\dim_F} V$ is the [[Definition:Dimensio... | Let $\BB = \set {x_1, x_2, \ldots, x_m}$ be a [[Definition:Generator of Vector Space|generator]] for $G$.
Let $\set {y_1, y_2, \ldots, y_n}$ be a [[Definition:Subset|subset]] of $G$ such that $n > m$.
As $\BB$ [[Definition:Generator of Module|generates]] $G$, there exist $\alpha_{i j} \in F$ where $1 \le i \le m, 1 ... | Cardinality of Generator of Vector Space is not Less than Dimension/Proof 1 | https://proofwiki.org/wiki/Cardinality_of_Generator_of_Vector_Space_is_not_Less_than_Dimension | https://proofwiki.org/wiki/Cardinality_of_Generator_of_Vector_Space_is_not_Less_than_Dimension/Proof_1 | [
"Cardinality of Generator of Vector Space is not Less than Dimension",
"Generators of Vector Spaces",
"Dimension of Vector Space"
] | [
"Definition:Vector Space",
"Definition:Field (Abstract Algebra)",
"Definition:Generator of Vector Space",
"Definition:Element",
"Definition:Dimension of Vector Space"
] | [
"Definition:Generator of Vector Space",
"Definition:Subset",
"Definition:Generator of Module",
"Homogeneous Simultaneous Linear Equations with More Unknowns than Equations",
"Definition:Linearly Dependent/Set",
"Definition:Dimension of Vector Space"
] |
proofwiki-4291 | Cardinality of Generator of Vector Space is not Less than Dimension | Let $V$ be a vector space over a field $F$.
Let $\BB$ be a generator for $V$ containing $m$ elements.
Then:
:$\map {\dim_F} V \le m$
where $\map {\dim_F} V$ is the dimension of $V$. | From Generator of Vector Space Contains Basis there exists a basis $B$ of $E$ such that $B \subseteq G$.
From Cardinality of Basis of Vector Space, $\card B = n$.
The result follows.
{{qed}} | Let $V$ be a [[Definition:Vector Space|vector space]] over a [[Definition:Field (Abstract Algebra)|field]] $F$.
Let $\BB$ be a [[Definition:Generator of Vector Space|generator]] for $V$ containing $m$ [[Definition:Element|elements]].
Then:
:$\map {\dim_F} V \le m$
where $\map {\dim_F} V$ is the [[Definition:Dimensio... | From [[Generator of Vector Space Contains Basis]] there exists a [[Definition:Basis of Vector Space|basis]] $B$ of $E$ such that $B \subseteq G$.
From [[Cardinality of Basis of Vector Space]], $\card B = n$.
The result follows.
{{qed}} | Cardinality of Generator of Vector Space is not Less than Dimension/Proof 2 | https://proofwiki.org/wiki/Cardinality_of_Generator_of_Vector_Space_is_not_Less_than_Dimension | https://proofwiki.org/wiki/Cardinality_of_Generator_of_Vector_Space_is_not_Less_than_Dimension/Proof_2 | [
"Cardinality of Generator of Vector Space is not Less than Dimension",
"Generators of Vector Spaces",
"Dimension of Vector Space"
] | [
"Definition:Vector Space",
"Definition:Field (Abstract Algebra)",
"Definition:Generator of Vector Space",
"Definition:Element",
"Definition:Dimension of Vector Space"
] | [
"Generator of Vector Space Contains Basis",
"Definition:Basis of Vector Space",
"Cardinality of Basis of Vector Space"
] |
proofwiki-4292 | Trichotomy Law for Real Numbers | The real numbers obey the '''trichotomy law'''.
That is, $\forall a, b \in \R$, exactly one of the following holds:
{{begin-axiom}}
{{axiom | n = 1
| lc= $a$ is greater than $b$:
| m = a > b
}}
{{axiom | n = 2
| lc= $a$ is equal to $b$:
| m = a = b
}}
{{axiom | n = 3
| lc= $a$ is... | This follows directly Real Numbers form Totally Ordered Field.
{{Qed}} | The [[Definition:Real Number|real numbers]] obey the '''[[Trichotomy Law (Ordering)|trichotomy law]]'''.
That is, $\forall a, b \in \R$, exactly one of the following holds:
{{begin-axiom}}
{{axiom | n = 1
| lc= $a$ is greater than $b$:
| m = a > b
}}
{{axiom | n = 2
| lc= $a$ is equal to $b$:
... | This follows directly [[Real Numbers form Totally Ordered Field]].
{{Qed}} | Trichotomy Law for Real Numbers/Proof 1 | https://proofwiki.org/wiki/Trichotomy_Law_for_Real_Numbers | https://proofwiki.org/wiki/Trichotomy_Law_for_Real_Numbers/Proof_1 | [
"Trichotomy Law for Real Numbers",
"Real Numbers",
"Inequalities",
"Trichotomy Law"
] | [
"Definition:Real Number",
"Trichotomy Law (Ordering)"
] | [
"Real Numbers form Totally Ordered Field"
] |
proofwiki-4293 | Trichotomy Law for Real Numbers | The real numbers obey the '''trichotomy law'''.
That is, $\forall a, b \in \R$, exactly one of the following holds:
{{begin-axiom}}
{{axiom | n = 1
| lc= $a$ is greater than $b$:
| m = a > b
}}
{{axiom | n = 2
| lc= $a$ is equal to $b$:
| m = a = b
}}
{{axiom | n = 3
| lc= $a$ is... | $\le$ is a total ordering on $\R$.
The trichotomy follows directly from Trichotomy Law.
{{Qed}} | The [[Definition:Real Number|real numbers]] obey the '''[[Trichotomy Law (Ordering)|trichotomy law]]'''.
That is, $\forall a, b \in \R$, exactly one of the following holds:
{{begin-axiom}}
{{axiom | n = 1
| lc= $a$ is greater than $b$:
| m = a > b
}}
{{axiom | n = 2
| lc= $a$ is equal to $b$:
... | $\le$ is a [[Definition:Total Ordering|total ordering]] on $\R$.
The trichotomy follows directly from [[Trichotomy Law (Ordering)|Trichotomy Law]].
{{Qed}} | Trichotomy Law for Real Numbers/Proof 2 | https://proofwiki.org/wiki/Trichotomy_Law_for_Real_Numbers | https://proofwiki.org/wiki/Trichotomy_Law_for_Real_Numbers/Proof_2 | [
"Trichotomy Law for Real Numbers",
"Real Numbers",
"Inequalities",
"Trichotomy Law"
] | [
"Definition:Real Number",
"Trichotomy Law (Ordering)"
] | [
"Definition:Total Ordering",
"Trichotomy Law (Ordering)"
] |
proofwiki-4294 | Real Number Ordering is Compatible with Addition | :$\forall a, b, c \in \R: a < b \implies a + c < b + c$ | From Real Numbers form Ordered Integral Domain, $\struct {\R, +, \times, \le}$ forms an ordered integral domain.
By definition of ordered integral domain, the usual ordering $\le$ is compatible with ring addition.
{{Qed}} | :$\forall a, b, c \in \R: a < b \implies a + c < b + c$ | From [[Real Numbers form Ordered Integral Domain]], $\struct {\R, +, \times, \le}$ forms an [[Definition:Ordered Integral Domain|ordered integral domain]].
By definition of [[Definition:Ordered Integral Domain|ordered integral domain]], the [[Definition:Usual Ordering|usual ordering]] $\le$ is [[Definition:Relation Co... | Real Number Ordering is Compatible with Addition | https://proofwiki.org/wiki/Real_Number_Ordering_is_Compatible_with_Addition | https://proofwiki.org/wiki/Real_Number_Ordering_is_Compatible_with_Addition | [
"Real Number Ordering is Compatible with Addition",
"Real Addition",
"Inequalities"
] | [] | [
"Real Numbers form Ordered Integral Domain",
"Definition:Ordered Integral Domain",
"Definition:Ordered Integral Domain",
"Definition:Usual Ordering",
"Definition:Relation Compatible with Operation",
"Definition:Ring (Abstract Algebra)/Addition"
] |
proofwiki-4295 | Real Number Inequalities can be Added | Let $a, b, c, d \in \R$ such that $a > b$ and $c > d$.
Then:
:$a + c > b + d$ | {{begin-eqn}}
{{eqn | l = a
| o = >
| r = b
| c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
| l = a + c
| o = >
| r = b + c
| c = Real Number Ordering is Compatible with Addition
}}
{{eqn | l = c
| o = >
| r = d
| c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
... | Let $a, b, c, d \in \R$ such that $a > b$ and $c > d$.
Then:
:$a + c > b + d$ | {{begin-eqn}}
{{eqn | l = a
| o = >
| r = b
| c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
| l = a + c
| o = >
| r = b + c
| c = [[Real Number Ordering is Compatible with Addition]]
}}
{{eqn | l = c
| o = >
| r = d
| c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
... | Real Number Inequalities can be Added/Proof 2 | https://proofwiki.org/wiki/Real_Number_Inequalities_can_be_Added | https://proofwiki.org/wiki/Real_Number_Inequalities_can_be_Added/Proof_2 | [
"Real Number Inequalities can be Added",
"Real Numbers",
"Addition",
"Inequalities"
] | [] | [
"Real Number Ordering is Compatible with Addition",
"Real Number Ordering is Compatible with Addition",
"Transitive Law"
] |
proofwiki-4296 | Positive Real Number Inequalities can be Multiplied | Let $a, b, c, d \in \R$ such that $a > b$ and $c > d$.
Let $b > 0$ and $d > 0$.
Then $a c > b d$.
If $b < 0$ or $d < 0$ the inequality does not hold. | Note that as $d > 0$ it follows that $c > d > 0$ and so $c > 0$.
{{begin-eqn}}
{{eqn | l = a
| o = >
| r = b
| c =
}}
{{eqn | ll= \leadsto
| l = a c
| o = >
| r = b c
| c = Real Number Ordering is Compatible with Multiplication, as $c > 0$
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | ... | Let $a, b, c, d \in \R$ such that $a > b$ and $c > d$.
Let $b > 0$ and $d > 0$.
Then $a c > b d$.
If $b < 0$ or $d < 0$ the inequality does not hold. | Note that as $d > 0$ it follows that $c > d > 0$ and so $c > 0$.
{{begin-eqn}}
{{eqn | l = a
| o = >
| r = b
| c =
}}
{{eqn | ll= \leadsto
| l = a c
| o = >
| r = b c
| c = [[Real Number Ordering is Compatible with Multiplication]], as $c > 0$
}}
{{end-eqn}}
{{begin-eqn}}
{... | Positive Real Number Inequalities can be Multiplied | https://proofwiki.org/wiki/Positive_Real_Number_Inequalities_can_be_Multiplied | https://proofwiki.org/wiki/Positive_Real_Number_Inequalities_can_be_Multiplied | [
"Positive Real Number Inequalities can be Multiplied",
"Real Numbers",
"Multiplication",
"Inequalities"
] | [] | [
"Real Number Ordering is Compatible with Multiplication",
"Real Number Ordering is Compatible with Multiplication",
"Trichotomy Law for Real Numbers"
] |
proofwiki-4297 | Positive Real Number Inequalities can be Multiplied | Let $a, b, c, d \in \R$ such that $a > b$ and $c > d$.
Let $b > 0$ and $d > 0$.
Then $a c > b d$.
If $b < 0$ or $d < 0$ the inequality does not hold. | Proof by Counterexample:
Let $a = c = -1, b = d = -2$.
Then $a c = 1$ but $b d = 2$.
{{qed}} | Let $a, b, c, d \in \R$ such that $a > b$ and $c > d$.
Let $b > 0$ and $d > 0$.
Then $a c > b d$.
If $b < 0$ or $d < 0$ the inequality does not hold. | [[Proof by Counterexample]]:
Let $a = c = -1, b = d = -2$.
Then $a c = 1$ but $b d = 2$.
{{qed}} | Positive Real Number Inequalities can be Multiplied/Disproof for Negative Parameters | https://proofwiki.org/wiki/Positive_Real_Number_Inequalities_can_be_Multiplied | https://proofwiki.org/wiki/Positive_Real_Number_Inequalities_can_be_Multiplied/Disproof_for_Negative_Parameters | [
"Positive Real Number Inequalities can be Multiplied",
"Real Numbers",
"Multiplication",
"Inequalities"
] | [] | [
"Proof by Counterexample"
] |
proofwiki-4298 | Real Ordering is not Compatible with Subtraction | Let $a, b, c, d \in R$ be real numbers such that $a > b$ and $c > d$.
Then it does not necessarily hold that:
:$a - c > b - d$
That is, the usual ordering is not compatible with subtraction. | Proof by Counterexample:
For example, set $a = 5, b = 3, c = 4, d = 1$
Then $a - c = 1$ while $b - d = 2$.
{{qed}} | Let $a, b, c, d \in R$ be [[Definition:Real Number|real numbers]] such that $a > b$ and $c > d$.
Then it does not necessarily hold that:
:$a - c > b - d$
That is, the [[Definition:Usual Ordering|usual ordering]] is not [[Definition:Relation Compatible with Operation|compatible]] with [[Definition:Real Subtraction|s... | [[Proof by Counterexample]]:
For example, set $a = 5, b = 3, c = 4, d = 1$
Then $a - c = 1$ while $b - d = 2$.
{{qed}} | Real Ordering is not Compatible with Subtraction | https://proofwiki.org/wiki/Real_Ordering_is_not_Compatible_with_Subtraction | https://proofwiki.org/wiki/Real_Ordering_is_not_Compatible_with_Subtraction | [
"Real Subtraction",
"Algebra"
] | [
"Definition:Real Number",
"Definition:Usual Ordering",
"Definition:Relation Compatible with Operation",
"Definition:Subtraction/Real Numbers"
] | [
"Proof by Counterexample"
] |
proofwiki-4299 | Real Ordering is not Compatible with Division | Let $a, b, c, d \in \R$ be real numbers such that $a > b$ and $c > d$.
Then it does not necessarily hold that:
:$\dfrac a c > \dfrac b d$
That is, the usual ordering is not compatible with division. | Proof by Counterexample:
For example, set $a = 5, b = 3, c = 4, d = 1$
Then $\dfrac a c = 1 \frac 1 4$ while $\dfrac b d = 3$.
{{qed}} | Let $a, b, c, d \in \R$ be [[Definition:Real Number|real numbers]] such that $a > b$ and $c > d$.
Then it does not necessarily hold that:
:$\dfrac a c > \dfrac b d$
That is, the [[Definition:Usual Ordering|usual ordering]] is not [[Definition:Relation Compatible with Operation|compatible]] with [[Definition:Real Di... | [[Proof by Counterexample]]:
For example, set $a = 5, b = 3, c = 4, d = 1$
Then $\dfrac a c = 1 \frac 1 4$ while $\dfrac b d = 3$.
{{qed}} | Real Ordering is not Compatible with Division | https://proofwiki.org/wiki/Real_Ordering_is_not_Compatible_with_Division | https://proofwiki.org/wiki/Real_Ordering_is_not_Compatible_with_Division | [
"Real Division",
"Algebra"
] | [
"Definition:Real Number",
"Definition:Usual Ordering",
"Definition:Relation Compatible with Operation",
"Definition:Division/Field/Real Numbers"
] | [
"Proof by Counterexample"
] |
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