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proofwiki-8300
Primes of form Power plus One
Let $q, n \in \Z_{>0}$ such that $q > 1$. Then $q^n + 1$ is prime only if: :$(1): \quad q$ is even and :$(2): \quad n$ is of the form $2^m$ for some positive integer $m$.
Note that if $q = 1$ then $q^n + 1 = 2$ which ''is'' prime. Hence the condition on $q$ in the statement of the theorem. So by hypothesis $q > 1$. Let $q$ be odd. Then by Two divides Power Plus One iff Odd, $q^n + 1$ is not prime. Let $q$ be even. Let $n$ be expressed in the form $r 2^m$ where $r$ is odd. Then $q^n + 1$...
Let $q, n \in \Z_{>0}$ such that $q > 1$. Then $q^n + 1$ is [[Definition:Prime Number|prime]] only if: :$(1): \quad q$ is [[Definition:Even Integer|even]] and :$(2): \quad n$ is of the form $2^m$ for some [[Definition:Positive Integer|positive integer]] $m$.
Note that if $q = 1$ then $q^n + 1 = 2$ which ''is'' [[Definition:Prime Number|prime]]. Hence the condition on $q$ in the statement of the theorem. So [[Definition:By Hypothesis|by hypothesis]] $q > 1$. Let $q$ be [[Definition:Odd Integer|odd]]. Then by [[Two divides Power Plus One iff Odd]], $q^n + 1$ is not [[De...
Primes of form Power plus One
https://proofwiki.org/wiki/Primes_of_form_Power_plus_One
https://proofwiki.org/wiki/Primes_of_form_Power_plus_One
[ "Number Theory" ]
[ "Definition:Prime Number", "Definition:Even Integer", "Definition:Positive/Integer" ]
[ "Definition:Prime Number", "Definition:By Hypothesis", "Definition:Odd Integer", "Two divides Power Plus One iff Odd", "Definition:Prime Number", "Definition:Even Integer", "Definition:Odd Integer", "Number Plus One divides Power Plus One iff Odd", "Definition:Divisor (Algebra)/Integer", "Definiti...
proofwiki-8301
Fermat Number whose Index is Sum of Integers
Let $F_n = 2^{\left({2^n}\right)} + 1$ be the $n$th Fermat number. Let $k \in \Z_{>0}$. Then: :$F_{n + k} - 1 = \left({F_n - 1}\right)^{2^k}$
By the definition of Fermat number {{begin-eqn}} {{eqn | l = F_{n + k} - 1 | r = 2^{2^{n + k} } | c = {{Defof|Fermat Number}} }} {{eqn | r = 2^{2^n 2^k} | c = }} {{eqn | r = \left({2^{2^n} }\right)^{2^k} | c = }} {{eqn | r = \left({F_n - 1}\right)^{2^k} | c = {{Defof|Fermat Number}} }} {...
Let $F_n = 2^{\left({2^n}\right)} + 1$ be the $n$th [[Definition:Fermat Number|Fermat number]]. Let $k \in \Z_{>0}$. Then: :$F_{n + k} - 1 = \left({F_n - 1}\right)^{2^k}$
By the definition of [[Definition:Fermat Number|Fermat number]] {{begin-eqn}} {{eqn | l = F_{n + k} - 1 | r = 2^{2^{n + k} } | c = {{Defof|Fermat Number}} }} {{eqn | r = 2^{2^n 2^k} | c = }} {{eqn | r = \left({2^{2^n} }\right)^{2^k} | c = }} {{eqn | r = \left({F_n - 1}\right)^{2^k} | c ...
Fermat Number whose Index is Sum of Integers
https://proofwiki.org/wiki/Fermat_Number_whose_Index_is_Sum_of_Integers
https://proofwiki.org/wiki/Fermat_Number_whose_Index_is_Sum_of_Integers
[ "Fermat Numbers" ]
[ "Definition:Fermat Number" ]
[ "Definition:Fermat Number" ]
proofwiki-8302
Number of Boolean Interpretations for Finite Set of Variables
Let $\PP_0$ be the vocabulary of language of propositional logic. Let $S \subseteq \PP_0$ be a finite set of $n$ letters from $\PP_0$. Then there are $2^n$ different partial boolean interpretations for $S$.
A partial boolean interpretation for $S$ is a mapping from $S$ to the Boolean domain $\set {T, F}$. By Cardinality of Set of All Mappings, the total number of mappings from $S$ to $T$ is: :$\card {T^S} = \card T^{\card S}$ The result follows directly. {{qed}}
Let $\PP_0$ be the [[Definition:Vocabulary of Propositional Logic|vocabulary]] of [[Definition:Language of Propositional Logic|language of propositional logic]]. Let $S \subseteq \PP_0$ be a [[Definition:Finite Set|finite set]] of $n$ [[Definition:Letter of Formal Language|letters]] from $\PP_0$. Then there are $2^n...
A [[Definition:Partial Boolean Interpretation|partial boolean interpretation]] for $S$ is a [[Definition:Mapping|mapping]] from $S$ to the [[Definition:Boolean Domain|Boolean domain]] $\set {T, F}$. By [[Cardinality of Set of All Mappings]], the total number of [[Definition:Mapping|mappings]] from $S$ to $T$ is: :$\c...
Number of Boolean Interpretations for Finite Set of Variables
https://proofwiki.org/wiki/Number_of_Boolean_Interpretations_for_Finite_Set_of_Variables
https://proofwiki.org/wiki/Number_of_Boolean_Interpretations_for_Finite_Set_of_Variables
[ "Boolean Interpretations" ]
[ "Definition:Language of Propositional Logic/Alphabet/Letter", "Definition:Language of Propositional Logic", "Definition:Finite Set", "Definition:Formal Language/Alphabet/Letter", "Definition:Boolean Interpretation" ]
[ "Definition:Boolean Interpretation", "Definition:Mapping", "Definition:Boolean Domain", "Cardinality of Set of All Mappings", "Definition:Mapping" ]
proofwiki-8303
Integers under Addition form Monoid
The set of integers under addition $\struct {\Z, +}$ forms a monoid.
Follows directly from Integers under Addition form Abelian Group. By definition, an abelian group is a group. Also by definition, a group is a monoid. Hence the result. {{qed}}
The [[Definition:Set|set]] of [[Definition:Integer|integers]] under [[Definition:Integer Addition|addition]] $\struct {\Z, +}$ forms a [[Definition:Monoid|monoid]].
Follows directly from [[Integers under Addition form Abelian Group]]. By definition, an [[Definition:Abelian Group|abelian group]] is a [[Definition:Group|group]]. Also by definition, a [[Definition:Group|group]] is a [[Definition:Monoid|monoid]]. Hence the result. {{qed}}
Integers under Addition form Monoid
https://proofwiki.org/wiki/Integers_under_Addition_form_Monoid
https://proofwiki.org/wiki/Integers_under_Addition_form_Monoid
[ "Integer Addition", "Examples of Monoids" ]
[ "Definition:Set", "Definition:Integer", "Definition:Addition/Integers", "Definition:Monoid" ]
[ "Integers under Addition form Abelian Group", "Definition:Abelian Group", "Definition:Group", "Definition:Group", "Definition:Monoid" ]
proofwiki-8304
Rational Numbers under Addition form Monoid
The set of rational numbers under addition $\struct {\Q, +}$ forms a monoid.
Follows directly from Rational Numbers under Addition form Infinite Abelian Group. By definition, an abelian group is a group. Also by definition, a group is a monoid. Hence the result. {{qed}}
The [[Definition:Set|set]] of [[Definition:Rational Number|rational numbers]] under [[Definition:Rational Addition|addition]] $\struct {\Q, +}$ forms a [[Definition:Monoid|monoid]].
Follows directly from [[Rational Numbers under Addition form Infinite Abelian Group]]. By definition, an [[Definition:Abelian Group|abelian group]] is a [[Definition:Group|group]]. Also by definition, a [[Definition:Group|group]] is a [[Definition:Monoid|monoid]]. Hence the result. {{qed}}
Rational Numbers under Addition form Monoid
https://proofwiki.org/wiki/Rational_Numbers_under_Addition_form_Monoid
https://proofwiki.org/wiki/Rational_Numbers_under_Addition_form_Monoid
[ "Rational Addition", "Examples of Monoids" ]
[ "Definition:Set", "Definition:Rational Number", "Definition:Addition/Rational Numbers", "Definition:Monoid" ]
[ "Rational Numbers under Addition form Infinite Abelian Group", "Definition:Abelian Group", "Definition:Group", "Definition:Group", "Definition:Monoid" ]
proofwiki-8305
Real Numbers form Field
The set of real numbers $\R$ forms a field under addition and multiplication: $\struct {\R, +, \times}$.
From Real Numbers under Addition form Infinite Abelian Group, we have that $\struct {\R, +}$ forms an abelian group. From Non-Zero Real Numbers under Multiplication form Abelian Group, we have that $\struct {\R_{\ne 0}, \times}$ forms an abelian group. Next we have that Real Multiplication Distributes over Addition. Th...
The [[Definition:Real Number|set of real numbers]] $\R$ forms a [[Definition:Field (Abstract Algebra)|field]] under [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]]: $\struct {\R, +, \times}$.
From [[Real Numbers under Addition form Infinite Abelian Group]], we have that $\struct {\R, +}$ forms an [[Definition:Abelian Group|abelian group]]. From [[Non-Zero Real Numbers under Multiplication form Abelian Group]], we have that $\struct {\R_{\ne 0}, \times}$ forms an [[Definition:Abelian Group|abelian group]]. ...
Real Numbers form Field
https://proofwiki.org/wiki/Real_Numbers_form_Field
https://proofwiki.org/wiki/Real_Numbers_form_Field
[ "Examples of Fields", "Real Numbers" ]
[ "Definition:Real Number", "Definition:Field (Abstract Algebra)", "Definition:Addition/Real Numbers", "Definition:Multiplication/Real Numbers" ]
[ "Real Numbers under Addition form Infinite Abelian Group", "Definition:Abelian Group", "Non-Zero Real Numbers under Multiplication form Abelian Group", "Definition:Abelian Group", "Real Multiplication Distributes over Addition", "Definition:Field (Abstract Algebra)" ]
proofwiki-8306
Rational Numbers form Field
Consider the algebraic structure $\struct {\Q, +, \times}$, where: :$\Q$ is the set of all rational numbers :$+$ is the operation of rational addition :$\times$ is the operation of rational multiplication. Then $\struct {\Q, +, \times}$ forms a field.
This is demonstrated in the formal definition of rational numbers. {{Qed}}
Consider the [[Definition:Algebraic Structure with Two Operations|algebraic structure]] $\struct {\Q, +, \times}$, where: :$\Q$ is the set of all [[Definition:Rational Number|rational numbers]] :$+$ is the operation of [[Definition:Rational Addition|rational addition]] :$\times$ is the operation of [[Definition:Ration...
This is demonstrated in the [[Definition:Rational Number/Formal Definition|formal definition of rational numbers]]. {{Qed}}
Rational Numbers form Field
https://proofwiki.org/wiki/Rational_Numbers_form_Field
https://proofwiki.org/wiki/Rational_Numbers_form_Field
[ "Examples of Fields", "Rational Numbers" ]
[ "Definition:Algebraic Structure/Two Operations", "Definition:Rational Number", "Definition:Addition/Rational Numbers", "Definition:Multiplication/Rational Numbers", "Definition:Field (Abstract Algebra)" ]
[ "Definition:Rational Number/Formal Definition" ]
proofwiki-8307
Real Numbers under Addition form Monoid
The set of real numbers under addition $\struct {\R, +}$ forms a monoid.
Taking the monoid axioms in turn:
The [[Definition:Set|set]] of [[Definition:Real Number|real numbers]] under [[Definition:Real Addition|addition]] $\struct {\R, +}$ forms a [[Definition:Monoid|monoid]].
Taking the [[Axiom:Monoid Axioms|monoid axioms]] in turn:
Real Numbers under Addition form Monoid
https://proofwiki.org/wiki/Real_Numbers_under_Addition_form_Monoid
https://proofwiki.org/wiki/Real_Numbers_under_Addition_form_Monoid
[ "Real Addition", "Examples of Monoids" ]
[ "Definition:Set", "Definition:Real Number", "Definition:Addition/Real Numbers", "Definition:Monoid" ]
[ "Axiom:Monoid Axioms" ]
proofwiki-8308
Integers Modulo m under Addition form Abelian Group
Let $\Z_m$ is the set of integers modulo $m$ Let $+_m$ be the operation of addition modulo $m$. Then the structure $\struct {\Z_m, +_m}$ is an abelian group.
From Integers Modulo m under Addition form Cyclic Group, $\struct {\Z_m, +_m}$ is a cyclic group. The result follows from Cyclic Group is Abelian. {{Qed}}
Let $\Z_m$ is the set of [[Definition:Integers Modulo m|integers modulo $m$]] Let $+_m$ be the [[Definition:Binary Operation|operation]] of [[Definition:Modulo Addition|addition modulo $m$]]. Then the [[Definition:Algebraic Structure with One Operation|structure]] $\struct {\Z_m, +_m}$ is an [[Definition:Abelian Grou...
From [[Integers Modulo m under Addition form Cyclic Group]], $\struct {\Z_m, +_m}$ is a [[Definition:Cyclic Group|cyclic group]]. The result follows from [[Cyclic Group is Abelian]]. {{Qed}}
Integers Modulo m under Addition form Abelian Group
https://proofwiki.org/wiki/Integers_Modulo_m_under_Addition_form_Abelian_Group
https://proofwiki.org/wiki/Integers_Modulo_m_under_Addition_form_Abelian_Group
[ "Additive Groups of Integers Modulo m", "Modulo Addition", "Examples of Abelian Groups" ]
[ "Definition:Integers Modulo m", "Definition:Operation/Binary Operation", "Definition:Modulo Addition", "Definition:Algebraic Structure/One Operation", "Definition:Abelian Group" ]
[ "Integers Modulo m under Addition form Cyclic Group", "Definition:Cyclic Group", "Cyclic Group is Abelian" ]
proofwiki-8309
Complex Numbers under Addition form Monoid
The set of complex numbers under addition $\left({\C, +}\right)$ forms a monoid.
Taking the monoid axioms in turn:
The [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] under [[Definition:Complex Addition|addition]] $\left({\C, +}\right)$ forms a [[Definition:Monoid|monoid]].
Taking the [[Axiom:Monoid Axioms|monoid axioms]] in turn:
Complex Numbers under Addition form Monoid
https://proofwiki.org/wiki/Complex_Numbers_under_Addition_form_Monoid
https://proofwiki.org/wiki/Complex_Numbers_under_Addition_form_Monoid
[ "Complex Addition", "Examples of Monoids" ]
[ "Definition:Set", "Definition:Complex Number", "Definition:Addition/Complex Numbers", "Definition:Monoid" ]
[ "Axiom:Monoid Axioms" ]
proofwiki-8310
Rational Numbers under Multiplication form Monoid
The set of rational numbers under multiplication $\struct {\Q, \times}$ forms a monoid.
Taking the monoid axioms in turn:
The [[Definition:Set|set]] of [[Definition:Rational Number|rational numbers]] under [[Definition:Rational Multiplication|multiplication]] $\struct {\Q, \times}$ forms a [[Definition:Monoid|monoid]].
Taking the [[Axiom:Monoid Axioms|monoid axioms]] in turn:
Rational Numbers under Multiplication form Monoid
https://proofwiki.org/wiki/Rational_Numbers_under_Multiplication_form_Monoid
https://proofwiki.org/wiki/Rational_Numbers_under_Multiplication_form_Monoid
[ "Rational Multiplication", "Examples of Monoids" ]
[ "Definition:Set", "Definition:Rational Number", "Definition:Multiplication/Rational Numbers", "Definition:Monoid" ]
[ "Axiom:Monoid Axioms" ]
proofwiki-8311
Real Numbers under Multiplication form Monoid
The set of real numbers under multiplication $\struct {\R, \times}$ forms a monoid.
Taking the monoid axioms in turn:
The [[Definition:Set|set]] of [[Definition:Real Number|real numbers]] under [[Definition:Real Multiplication|multiplication]] $\struct {\R, \times}$ forms a [[Definition:Monoid|monoid]].
Taking the [[Axiom:Monoid Axioms|monoid axioms]] in turn:
Real Numbers under Multiplication form Monoid
https://proofwiki.org/wiki/Real_Numbers_under_Multiplication_form_Monoid
https://proofwiki.org/wiki/Real_Numbers_under_Multiplication_form_Monoid
[ "Real Multiplication", "Examples of Monoids" ]
[ "Definition:Set", "Definition:Real Number", "Definition:Multiplication/Real Numbers", "Definition:Monoid" ]
[ "Axiom:Monoid Axioms" ]
proofwiki-8312
Complex Numbers under Multiplication form Monoid
The set of complex numbers under multiplication $\struct {\C, \times}$ forms a monoid.
Taking the monoid axioms in turn:
The [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] under [[Definition:Complex Multiplication|multiplication]] $\struct {\C, \times}$ forms a [[Definition:Monoid|monoid]].
Taking the [[Axiom:Monoid Axioms|monoid axioms]] in turn:
Complex Numbers under Multiplication form Monoid
https://proofwiki.org/wiki/Complex_Numbers_under_Multiplication_form_Monoid
https://proofwiki.org/wiki/Complex_Numbers_under_Multiplication_form_Monoid
[ "Complex Multiplication", "Examples of Monoids" ]
[ "Definition:Set", "Definition:Complex Number", "Definition:Multiplication/Complex Numbers", "Definition:Monoid" ]
[ "Axiom:Monoid Axioms" ]
proofwiki-8313
Non-Zero Natural Numbers under Multiplication form Commutative Monoid
Let $\N_{>0}$ be the set of natural numbers without zero, i.e. $\N_{>0} = \N \setminus \set 0$. The structure $\struct{\N_{>0}, \times}$ forms a commutative monoid.
From Non-Zero Natural Numbers under Multiplication form Commutative Semigroup, $\struct {\N_{>0}, \times}$ forms a commutative semigroup. From Identity Element of Natural Number Multiplication is One, $\struct {\N_{>0}, \times}$ has an identity element which is $1$. Hence the result, by definition of commutative monoid...
Let $\N_{>0}$ be the set of [[Definition:Natural Numbers|natural numbers]] without [[Definition:Zero (Number)|zero]], i.e. $\N_{>0} = \N \setminus \set 0$. The [[Definition:Algebraic Structure with One Operation|structure]] $\struct{\N_{>0}, \times}$ forms a [[Definition:Commutative Monoid|commutative monoid]].
From [[Non-Zero Natural Numbers under Multiplication form Commutative Semigroup]], $\struct {\N_{>0}, \times}$ forms a [[Definition:Commutative Semigroup|commutative semigroup]]. From [[Identity Element of Natural Number Multiplication is One]], $\struct {\N_{>0}, \times}$ has an [[Definition:Identity Element|identity...
Non-Zero Natural Numbers under Multiplication form Commutative Monoid
https://proofwiki.org/wiki/Non-Zero_Natural_Numbers_under_Multiplication_form_Commutative_Monoid
https://proofwiki.org/wiki/Non-Zero_Natural_Numbers_under_Multiplication_form_Commutative_Monoid
[ "Natural Number Multiplication", "Examples of Commutative Monoids" ]
[ "Definition:Natural Numbers", "Definition:Zero (Number)", "Definition:Algebraic Structure/One Operation", "Definition:Commutative Monoid" ]
[ "Non-Zero Natural Numbers under Multiplication form Commutative Semigroup", "Definition:Commutative Semigroup", "Identity Element of Natural Number Multiplication is One", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Commutative Monoid" ]
proofwiki-8314
Non-Zero Natural Numbers under Addition do not form Monoid
Let $\N_{>0}$ be the set of natural numbers without zero, i.e. $\N_{>0} = \N \setminus \set 0$. The structure $\struct {\N_{>0}, +}$ does ''not'' form a monoid.
From Natural Numbers under Addition form Commutative Monoid, $\struct {\N, +}$ forms a commutative monoid. From Natural Numbers Bounded Below under Addition form Commutative Semigroup, $\struct {\N_{>0}, +}$ forms a commutative semigroup. From Identity Element of Natural Number Addition is Zero, $0$ is the identity of ...
Let $\N_{>0}$ be the set of [[Definition:Natural Numbers|natural numbers]] without [[Definition:Zero (Number)|zero]], i.e. $\N_{>0} = \N \setminus \set 0$. The [[Definition:Algebraic Structure with One Operation|structure]] $\struct {\N_{>0}, +}$ does ''not'' form a [[Definition:Monoid|monoid]].
From [[Natural Numbers under Addition form Commutative Monoid]], $\struct {\N, +}$ forms a [[Definition:Commutative Monoid|commutative monoid]]. From [[Natural Numbers Bounded Below under Addition form Commutative Semigroup]], $\struct {\N_{>0}, +}$ forms a [[Definition:Commutative Semigroup|commutative semigroup]]. ...
Non-Zero Natural Numbers under Addition do not form Monoid
https://proofwiki.org/wiki/Non-Zero_Natural_Numbers_under_Addition_do_not_form_Monoid
https://proofwiki.org/wiki/Non-Zero_Natural_Numbers_under_Addition_do_not_form_Monoid
[ "Natural Number Addition", "Examples of Monoids" ]
[ "Definition:Natural Numbers", "Definition:Zero (Number)", "Definition:Algebraic Structure/One Operation", "Definition:Monoid" ]
[ "Natural Numbers under Addition form Commutative Monoid", "Definition:Commutative Monoid", "Natural Numbers Bounded Below under Addition form Commutative Semigroup", "Definition:Commutative Semigroup", "Identity Element of Natural Number Addition is Zero", "Definition:Identity (Abstract Algebra)/Two-Sided...
proofwiki-8315
Natural Numbers Bounded Below under Addition form Commutative Semigroup
Let $m \in \N$ where $\N$ is the set of natural numbers. Let $M \subseteq \N$ be defined as: :$M := \set {x \in \N: x \ge m}$ That is, $M$ is the set of all natural numbers greater than or equal to $m$. Then the algebraic structure $\struct {M, +}$ is a commutative semigroup.
We have that: :Natural Number Addition is Associative :Natural Number Addition is Commutative From Restriction of Associative Operation is Associative, $+$ is associative on $\struct {M, +}$. From Restriction of Commutative Operation is Commutative, $+$ is commutative on $\struct {M, +}$. It remains to be shown that $+...
Let $m \in \N$ where $\N$ is the [[Definition:Natural Numbers|set of natural numbers]]. Let $M \subseteq \N$ be defined as: :$M := \set {x \in \N: x \ge m}$ That is, $M$ is the set of all [[Definition:Natural Numbers|natural numbers]] greater than or equal to $m$. Then the [[Definition:Algebraic Structure with One Op...
We have that: :[[Natural Number Addition is Associative]] :[[Natural Number Addition is Commutative]] From [[Restriction of Associative Operation is Associative]], $+$ is [[Definition:Associative Operation|associative]] on $\struct {M, +}$. From [[Restriction of Commutative Operation is Commutative]], $+$ is [[Defini...
Natural Numbers Bounded Below under Addition form Commutative Semigroup
https://proofwiki.org/wiki/Natural_Numbers_Bounded_Below_under_Addition_form_Commutative_Semigroup
https://proofwiki.org/wiki/Natural_Numbers_Bounded_Below_under_Addition_form_Commutative_Semigroup
[ "Examples of Commutative Semigroups", "Natural Number Addition" ]
[ "Definition:Natural Numbers", "Definition:Natural Numbers", "Definition:Algebraic Structure/One Operation", "Definition:Commutative Semigroup" ]
[ "Natural Number Addition is Associative", "Natural Number Addition is Commutative", "Restriction of Associative Operation is Associative", "Definition:Associative Operation", "Restriction of Commutative Operation is Commutative", "Definition:Commutative/Operation", "Definition:Closure (Abstract Algebra)...
proofwiki-8316
Positive Real Numbers under Max Operation form Monoid
Let $\R_{\ge 0}$ be the set of positive (that is, non-negative) real numbers. Let $\max: \R_{\ge 0}^2 \to \R_{\ge 0}$ be the max operation on $\R_{\ge 0}$. Then $\struct {\R_{\ge 0}, \max}$ is a monoid whose identity is $0$.
From Real Numbers are Totally Ordered, $\R$ is a totally ordered set. From Max Operation on Toset forms Semigroup, $\struct {\R_{\ge 0}, \max}$ is a semigroup. By definition of $\R_{\ge 0}$: :$\forall x \in \R_{\ge 0}: 0 \le x$ Thus by definition of the max operation: :$\forall x \in \R_{\ge 0}: \map \max {0, x} = x = ...
Let $\R_{\ge 0}$ be the set of [[Definition:Positive Real Number|positive (that is, non-negative) real numbers]]. Let $\max: \R_{\ge 0}^2 \to \R_{\ge 0}$ be the [[Definition:Max Operation|max operation]] on $\R_{\ge 0}$. Then $\struct {\R_{\ge 0}, \max}$ is a [[Definition:Monoid|monoid]] whose [[Definition:Identity ...
From [[Real Numbers are Totally Ordered]], $\R$ is a [[Definition:Totally Ordered Set|totally ordered set]]. From [[Max Operation on Toset forms Semigroup]], $\struct {\R_{\ge 0}, \max}$ is a [[Definition:Semigroup|semigroup]]. By definition of $\R_{\ge 0}$: :$\forall x \in \R_{\ge 0}: 0 \le x$ Thus by definition of...
Positive Real Numbers under Max Operation form Monoid
https://proofwiki.org/wiki/Positive_Real_Numbers_under_Max_Operation_form_Monoid
https://proofwiki.org/wiki/Positive_Real_Numbers_under_Max_Operation_form_Monoid
[ "Real Numbers", "Max Operation", "Examples of Monoids" ]
[ "Definition:Positive/Real Number", "Definition:Max Operation", "Definition:Monoid", "Definition:Identity (Abstract Algebra)/Two-Sided Identity" ]
[ "Real Numbers are Totally Ordered", "Definition:Totally Ordered Set", "Max Operation on Toset forms Semigroup", "Definition:Semigroup", "Definition:Max Operation", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Monoid" ]
proofwiki-8317
Free Commutative Monoid on One Element is Isomorphic to Natural Numbers under Addition
Let $X = \set x$ be a singleton. Let $M$ be the free commutative monoid on $X$. Then $M$ is isomorphic to the additive monoid of natural numbers.
By definition, the free commutative monoid $M$ on $\set x$ is: :$M = \set {e, x, x^2, x^3, \ldots}$ where $e$ denotes the null sequence of elements of $X$. Let $\phi$ denote the mapping from $M$ to $\N$ as: :$\forall a \in M: \map \phi a = \begin{cases} 0 & : a = e \\ n & : a = x^n \end{cases}$ By definition of $\phi$:...
Let $X = \set x$ be a [[Definition:Singleton|singleton]]. Let $M$ be the [[Definition:Free Commutative Monoid|free commutative monoid]] on $X$. Then $M$ is [[Definition:Monoid Isomorphism|isomorphic]] to the [[Definition:Additive Monoid of Natural Numbers|additive monoid of natural numbers]].
By definition, the [[Definition:Free Commutative Monoid|free commutative monoid]] $M$ on $\set x$ is: :$M = \set {e, x, x^2, x^3, \ldots}$ where $e$ denotes the [[Definition:Null Sequence|null sequence]] of elements of $X$. Let $\phi$ denote the [[Definition:Mapping|mapping]] from $M$ to $\N$ as: :$\forall a \in M: \m...
Free Commutative Monoid on One Element is Isomorphic to Natural Numbers under Addition
https://proofwiki.org/wiki/Free_Commutative_Monoid_on_One_Element_is_Isomorphic_to_Natural_Numbers_under_Addition
https://proofwiki.org/wiki/Free_Commutative_Monoid_on_One_Element_is_Isomorphic_to_Natural_Numbers_under_Addition
[ "Natural Number Addition", "Free Monoids" ]
[ "Definition:Singleton", "Definition:Free Commutative Monoid", "Definition:Isomorphism (Abstract Algebra)/Monoid Isomorphism", "Definition:Additive Monoid of Natural Numbers" ]
[ "Definition:Free Commutative Monoid", "Definition:Null Sequence", "Definition:Mapping", "Definition:Injection", "Definition:Surjection", "Definition:Monoid Homomorphism", "Definition:Isomorphism (Abstract Algebra)/Monoid Isomorphism" ]
proofwiki-8318
Finite Monoid with Right Cancellable Operation is Group
Let $\struct {S, \circ}$ be a finite monoid. Let $\circ$ be a right cancellable operation. Then $\struct {S, \circ}$ is a group.
{{Group-axiom|0}}, {{Group-axiom|1}} and {{Group-axiom|2}} are satisfied by dint of $\struct {S, \circ}$ being a monoid. Recall the definition of right cancellable operation: :$\forall a, b, c \in S: a \circ c = b \circ c \implies a = b$ Let $\rho_c: S \to S$ be the right regular representation of $\struct {S, \circ}$ ...
Let $\struct {S, \circ}$ be a [[Definition:Finite Monoid|finite monoid]]. Let $\circ$ be a [[Definition:Right Cancellable Operation|right cancellable operation]]. Then $\struct {S, \circ}$ is a [[Definition:Group|group]].
{{Group-axiom|0}}, {{Group-axiom|1}} and {{Group-axiom|2}} are satisfied by dint of $\struct {S, \circ}$ being a [[Definition:Monoid|monoid]]. Recall the definition of [[Definition:Right Cancellable Operation|right cancellable operation]]: :$\forall a, b, c \in S: a \circ c = b \circ c \implies a = b$ Let $\rho_c: S ...
Finite Monoid with Right Cancellable Operation is Group
https://proofwiki.org/wiki/Finite_Monoid_with_Right_Cancellable_Operation_is_Group
https://proofwiki.org/wiki/Finite_Monoid_with_Right_Cancellable_Operation_is_Group
[ "Monoids", "Finite Groups" ]
[ "Definition:Finite Monoid", "Definition:Right Cancellable Operation", "Definition:Group" ]
[ "Definition:Monoid", "Definition:Right Cancellable Operation", "Definition:Regular Representations/Right Regular Representation", "Right Cancellable iff Right Regular Representation Injective", "Definition:Injection", "Right Regular Representation wrt Right Cancellable Element on Finite Semigroup is Bijec...
proofwiki-8319
Finite Monoid with Left Cancellable Operation is Group
Let $\struct {S, \circ}$ be a finite monoid. Let $\circ$ be a left cancellable operation. Then $\struct {S, \circ}$ is a group.
{{Group-axiom|0}}, {{Group-axiom|1}} and {{Group-axiom|2}} are satisfied by dint of $\paren {S, \circ}$ being a monoid. Recall the definition of left cancellable operation: :$\forall a, b, c \in S: c \circ a = c \circ b \implies a = b$ Let $\lambda_c: S \to S$ be the left regular representation of $\struct {S, \circ}$ ...
Let $\struct {S, \circ}$ be a [[Definition:Finite Monoid|finite monoid]]. Let $\circ$ be a [[Definition:Left Cancellable Operation|left cancellable operation]]. Then $\struct {S, \circ}$ is a [[Definition:Group|group]].
{{Group-axiom|0}}, {{Group-axiom|1}} and {{Group-axiom|2}} are satisfied by dint of $\paren {S, \circ}$ being a [[Definition:Monoid|monoid]]. Recall the definition of [[Definition:Left Cancellable Operation|left cancellable operation]]: :$\forall a, b, c \in S: c \circ a = c \circ b \implies a = b$ Let $\lambda_c: S ...
Finite Monoid with Left Cancellable Operation is Group
https://proofwiki.org/wiki/Finite_Monoid_with_Left_Cancellable_Operation_is_Group
https://proofwiki.org/wiki/Finite_Monoid_with_Left_Cancellable_Operation_is_Group
[ "Monoids", "Finite Groups" ]
[ "Definition:Finite Monoid", "Definition:Left Cancellable Operation", "Definition:Group" ]
[ "Definition:Monoid", "Definition:Left Cancellable Operation", "Definition:Regular Representations/Left Regular Representation", "Left Cancellable iff Left Regular Representation Injective", "Definition:Injection", "Left Regular Representation wrt Left Cancellable Element on Finite Semigroup is Bijection",...
proofwiki-8320
Multiplicative Group of Field is Abelian Group
Let $\struct {F, +, \times}$ be a field. Let $F^* := F \setminus \set 0$ be the set $F$ less its zero. The algebraic structure $\struct {F^*, \times}$ is an abelian group.
From the field axioms: {{begin-axiom}} {{axiom | n = \text M 0 | lc= Closure under product | q = \forall x, y \in F | m = x \circ y \in F }} {{axiom | n = \text M 1 | lc= Associativity of product | q = \forall x, y, z \in F | m = \paren {x \circ y} \circ z = x \circ \pare...
Let $\struct {F, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]]. Let $F^* := F \setminus \set 0$ be the set $F$ less its [[Definition:Field Zero|zero]]. The [[Definition:Algebraic Structure with One Operation|algebraic structure]] $\struct {F^*, \times}$ is an [[Definition:Abelian Group|abelian group...
From the [[Axiom:Field Axioms|field axioms]]: {{begin-axiom}} {{axiom | n = \text M 0 | lc= [[Definition:Closed Algebraic Structure|Closure]] under [[Definition:Field Product|product]] | q = \forall x, y \in F | m = x \circ y \in F }} {{axiom | n = \text M 1 | lc= [[Definition:Associati...
Multiplicative Group of Field is Abelian Group/Proof 1
https://proofwiki.org/wiki/Multiplicative_Group_of_Field_is_Abelian_Group
https://proofwiki.org/wiki/Multiplicative_Group_of_Field_is_Abelian_Group/Proof_1
[ "Field Theory", "Abelian Groups", "Multiplicative Group of Field is Abelian Group" ]
[ "Definition:Field (Abstract Algebra)", "Definition:Field Zero", "Definition:Algebraic Structure/One Operation", "Definition:Abelian Group" ]
[ "Axiom:Field Axioms", "Definition:Closure (Abstract Algebra)/Algebraic Structure", "Definition:Field (Abstract Algebra)/Product", "Definition:Associative Operation", "Definition:Field (Abstract Algebra)/Product", "Definition:Commutative/Operation", "Definition:Field (Abstract Algebra)/Product", "Defin...
proofwiki-8321
Multiplicative Group of Field is Abelian Group
Let $\struct {F, +, \times}$ be a field. Let $F^* := F \setminus \set 0$ be the set $F$ less its zero. The algebraic structure $\struct {F^*, \times}$ is an abelian group.
Recall that a field is a non-trivial commutative division ring. The result follows from Non-Zero Elements of Division Ring form Group. {{qed}}
Let $\struct {F, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]]. Let $F^* := F \setminus \set 0$ be the set $F$ less its [[Definition:Field Zero|zero]]. The [[Definition:Algebraic Structure with One Operation|algebraic structure]] $\struct {F^*, \times}$ is an [[Definition:Abelian Group|abelian group...
Recall that a [[Definition:Field (Abstract Algebra)|field]] is a [[Definition:Non-Trivial Ring|non-trivial]] [[Definition:Commutative Division Ring|commutative division ring]]. The result follows from [[Non-Zero Elements of Division Ring form Group]]. {{qed}}
Multiplicative Group of Field is Abelian Group/Proof 2
https://proofwiki.org/wiki/Multiplicative_Group_of_Field_is_Abelian_Group
https://proofwiki.org/wiki/Multiplicative_Group_of_Field_is_Abelian_Group/Proof_2
[ "Field Theory", "Abelian Groups", "Multiplicative Group of Field is Abelian Group" ]
[ "Definition:Field (Abstract Algebra)", "Definition:Field Zero", "Definition:Algebraic Structure/One Operation", "Definition:Abelian Group" ]
[ "Definition:Field (Abstract Algebra)", "Definition:Non-Trivial Ring", "Definition:Commutative Division Ring", "Non-Zero Elements of Division Ring form Group" ]
proofwiki-8322
One and Minus One form Subgroup of Multiplicative Group of Rational Numbers
Let $\struct {\Q_{\ne 0}, \times}$ be the multiplicative group of rational numbers. Let $S \subseteq \Q$ where $S = \set {1, -1}$. Then $\struct {S, \times}$ is a subgroup of $\struct {\Q_{\ne 0}, \times}$.
By hypothesis, $S$ is not empty. As $0 \notin S$, it follows that $S \subseteq \Q_{\ne 0}$. Recall that $-1 \times -1 = 1$ and also $1 \times 1 = 1$. Thus: :$\forall x \in S: x \times y^{-1} \in S$ The result follows from the One-Step Subgroup Test. {{qed}}
Let $\struct {\Q_{\ne 0}, \times}$ be the [[Definition:Multiplicative Group of Rational Numbers|multiplicative group of rational numbers]]. Let $S \subseteq \Q$ where $S = \set {1, -1}$. Then $\struct {S, \times}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\Q_{\ne 0}, \times}$.
[[Definition:By Hypothesis|By hypothesis]], $S$ is not [[Definition:Empty Set|empty]]. As $0 \notin S$, it follows that $S \subseteq \Q_{\ne 0}$. Recall that $-1 \times -1 = 1$ and also $1 \times 1 = 1$. Thus: :$\forall x \in S: x \times y^{-1} \in S$ The result follows from the [[One-Step Subgroup Test]]. {{qed}}
One and Minus One form Subgroup of Multiplicative Group of Rational Numbers
https://proofwiki.org/wiki/One_and_Minus_One_form_Subgroup_of_Multiplicative_Group_of_Rational_Numbers
https://proofwiki.org/wiki/One_and_Minus_One_form_Subgroup_of_Multiplicative_Group_of_Rational_Numbers
[ "Rational Numbers", "Examples of Subgroups" ]
[ "Definition:Multiplicative Group of Rational Numbers", "Definition:Subgroup" ]
[ "Definition:By Hypothesis", "Definition:Empty Set", "One-Step Subgroup Test" ]
proofwiki-8323
Symmetric Group is not Abelian
Let $S_n$ be the symmetric group of order $n$ where $n \ge 3$. Then $S_n$ is not abelian.
Let $\alpha \in S_n$ such that $\alpha$ is not the identity mapping. From Center of Symmetric Group is Trivial, $\alpha$ is not in the center $\map Z {S_n}$ of $S_n$. Thus $S_n \ne \map Z {S_n}$. The result follows by definition of abelian group. {{qed}}
Let $S_n$ be the [[Definition:Symmetric Group|symmetric group]] of order $n$ where $n \ge 3$. Then $S_n$ is not [[Definition:Abelian Group|abelian]].
Let $\alpha \in S_n$ such that $\alpha$ is not the [[Definition:Identity Mapping|identity mapping]]. From [[Center of Symmetric Group is Trivial]], $\alpha$ is not in the [[Definition:Center of Group|center]] $\map Z {S_n}$ of $S_n$. Thus $S_n \ne \map Z {S_n}$. The result follows by definition of [[Definition:Abeli...
Symmetric Group is not Abelian/Proof 1
https://proofwiki.org/wiki/Symmetric_Group_is_not_Abelian
https://proofwiki.org/wiki/Symmetric_Group_is_not_Abelian/Proof_1
[ "Symmetric Groups", "Symmetric Group is not Abelian" ]
[ "Definition:Symmetric Group", "Definition:Abelian Group" ]
[ "Definition:Identity Mapping", "Center of Symmetric Group is Trivial", "Definition:Center (Abstract Algebra)/Group", "Definition:Abelian Group" ]
proofwiki-8324
Symmetric Group is not Abelian
Let $S_n$ be the symmetric group of order $n$ where $n \ge 3$. Then $S_n$ is not abelian.
Let $a, b, c \in S$. Let $\alpha$ be the transposition on $S$ which exchanges $a$ and $b$. Let $\beta$ be the transposition on $S$ which exchanges $b$ and $c$. Then: :$\alpha \circ \beta$ maps $\tuple {a, b, c}$ to $\tuple {c, a, b}$ while: :$\beta \circ \alpha$ maps $\tuple {a, b, c}$ to $\tuple {b, c, a}$ Thus $\alph...
Let $S_n$ be the [[Definition:Symmetric Group|symmetric group]] of order $n$ where $n \ge 3$. Then $S_n$ is not [[Definition:Abelian Group|abelian]].
Let $a, b, c \in S$. Let $\alpha$ be the [[Definition:Transposition|transposition]] on $S$ which exchanges $a$ and $b$. Let $\beta$ be the [[Definition:Transposition|transposition]] on $S$ which exchanges $b$ and $c$. Then: :$\alpha \circ \beta$ maps $\tuple {a, b, c}$ to $\tuple {c, a, b}$ while: :$\beta \circ \al...
Symmetric Group is not Abelian/Proof 2
https://proofwiki.org/wiki/Symmetric_Group_is_not_Abelian
https://proofwiki.org/wiki/Symmetric_Group_is_not_Abelian/Proof_2
[ "Symmetric Groups", "Symmetric Group is not Abelian" ]
[ "Definition:Symmetric Group", "Definition:Abelian Group" ]
[ "Definition:Transposition", "Definition:Transposition", "Definition:Commutative/Elements", "Definition:Abelian Group" ]
proofwiki-8325
Möbius Function is Multiplicative/Corollary
Let $\gcd \set {m, n} > 1$. Then: :$\map \mu {m n} = 0$
Let $\gcd \set {m, n} = k$ where $k > 1$. Then $m = k r$ and $n = k s$ for some $r, s \in \Z$. Thus $m n = k^2 r s$. From Integer is Expressible as Product of Primes there exists $p \in \Z$ such that $p$ is prime and $p \divides k$. That is: :$\exists t \in \Z: k = p t$ and so: :$m n = p^2 t^2 r s$ That is: :$p^2 \divi...
Let $\gcd \set {m, n} > 1$. Then: :$\map \mu {m n} = 0$
Let $\gcd \set {m, n} = k$ where $k > 1$. Then $m = k r$ and $n = k s$ for some $r, s \in \Z$. Thus $m n = k^2 r s$. From [[Integer is Expressible as Product of Primes]] there exists $p \in \Z$ such that $p$ is [[Definition:Prime Number|prime]] and $p \divides k$. That is: :$\exists t \in \Z: k = p t$ and so: :$m ...
Möbius Function is Multiplicative/Corollary
https://proofwiki.org/wiki/Möbius_Function_is_Multiplicative/Corollary
https://proofwiki.org/wiki/Möbius_Function_is_Multiplicative/Corollary
[ "Multiplicative Functions", "Möbius Function" ]
[]
[ "Integer is Expressible as Product of Primes", "Definition:Prime Number", "Definition:Möbius Function", "Category:Multiplicative Functions", "Category:Möbius Function" ]
proofwiki-8326
Affine Group of One Dimension is Group
Let $\map {\operatorname {Af}_1} \R$ be the $1$-dimensional affine group on $\R$. Then $\map {\operatorname {Af}_1} \R$ is a group.
Taking the group axioms in turn: === {{Group-axiom|0|nolink}} === Let : :$a, c \in \R_{\ne 0} \land b, d \in \R$ Let: :$f_{ab}, f_{cd} \in \map {\operatorname {Af}_1} \R$ Then: {{begin-eqn}} {{eqn | l = \map {\paren {f_{ab} \circ f_{cd} } } x | r = \map {f_{ab} } {\map {f_{cd} } x} | c = {{Defof|Composition...
Let $\map {\operatorname {Af}_1} \R$ be the [[Definition:Affine Group of One Dimension|$1$-dimensional affine group on $\R$]]. Then $\map {\operatorname {Af}_1} \R$ is a [[Definition:Group|group]].
Taking the [[Axiom:Group Axioms|group axioms]] in turn: === {{Group-axiom|0|nolink}} === Let : :$a, c \in \R_{\ne 0} \land b, d \in \R$ Let: :$f_{ab}, f_{cd} \in \map {\operatorname {Af}_1} \R$ Then: {{begin-eqn}} {{eqn | l = \map {\paren {f_{ab} \circ f_{cd} } } x | r = \map {f_{ab} } {\map {f_{cd} } x} ...
Affine Group of One Dimension is Group/Proof 1
https://proofwiki.org/wiki/Affine_Group_of_One_Dimension_is_Group
https://proofwiki.org/wiki/Affine_Group_of_One_Dimension_is_Group/Proof_1
[ "Affine Group of One Dimension is Group", "Affine Groups" ]
[ "Definition:Affine Group of One Dimension", "Definition:Group" ]
[ "Axiom:Group Axioms", "Axiom:Field Axioms", "Definition:Closure (Abstract Algebra)/Algebraic Structure", "Composition of Mappings is Associative", "Definition:Associative Operation", "Identity of Affine Group of One Dimension", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Inverse in A...
proofwiki-8327
Affine Group of One Dimension is Group
Let $\map {\operatorname {Af}_1} \R$ be the $1$-dimensional affine group on $\R$. Then $\map {\operatorname {Af}_1} \R$ is a group.
The result follows from Affine Group of One Dimension as Semidirect Product and Semidirect Product of Groups is Group. {{qed}}
Let $\map {\operatorname {Af}_1} \R$ be the [[Definition:Affine Group of One Dimension|$1$-dimensional affine group on $\R$]]. Then $\map {\operatorname {Af}_1} \R$ is a [[Definition:Group|group]].
The result follows from [[Affine Group of One Dimension as Semidirect Product]] and [[Semidirect Product of Groups is Group]]. {{qed}}
Affine Group of One Dimension is Group/Proof 2
https://proofwiki.org/wiki/Affine_Group_of_One_Dimension_is_Group
https://proofwiki.org/wiki/Affine_Group_of_One_Dimension_is_Group/Proof_2
[ "Affine Group of One Dimension is Group", "Affine Groups" ]
[ "Definition:Affine Group of One Dimension", "Definition:Group" ]
[ "Affine Group of One Dimension as Semidirect Product", "Semidirect Product of Groups is Group" ]
proofwiki-8328
Identity of Affine Group of One Dimension
Let $\map {\mathrm {Af}_1} \R$ denote the $1$-dimensional affine group on $\R$. Then $\map {\mathrm {Af}_1} \R$ has $f_{1, 0}$ as an identity element.
Let $f_{a b} \in \map {\mathrm {Af}_1} \R$. Then: {{begin-eqn}} {{eqn | l = \map {\paren {f_{a b} \circ f_{1, 0} } } x | r = a \paren {1 x + 0} + b | c = }} {{eqn | r = a x + b | c = }} {{eqn | r = f_{a b} | c = }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map {\paren {f_{1, 0} \circ f_{a b} } }...
Let $\map {\mathrm {Af}_1} \R$ denote the [[Definition:Affine Group of One Dimension|$1$-dimensional affine group on $\R$]]. Then $\map {\mathrm {Af}_1} \R$ has $f_{1, 0}$ as an [[Definition:Identity Element|identity element]].
Let $f_{a b} \in \map {\mathrm {Af}_1} \R$. Then: {{begin-eqn}} {{eqn | l = \map {\paren {f_{a b} \circ f_{1, 0} } } x | r = a \paren {1 x + 0} + b | c = }} {{eqn | r = a x + b | c = }} {{eqn | r = f_{a b} | c = }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map {\paren {f_{1, 0} \circ f_{a b} ...
Identity of Affine Group of One Dimension
https://proofwiki.org/wiki/Identity_of_Affine_Group_of_One_Dimension
https://proofwiki.org/wiki/Identity_of_Affine_Group_of_One_Dimension
[ "Affine Groups" ]
[ "Definition:Affine Group of One Dimension", "Definition:Identity (Abstract Algebra)/Two-Sided Identity" ]
[ "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Category:Affine Groups" ]
proofwiki-8329
Inverse in Affine Group of One Dimension
Let $\map {\operatorname {Af}_1} \R$ denote the $1$-dimensional affine group on $\R$. Let $f_{a b} \in \map {\operatorname {Af}_1} \R$. Let $c = \dfrac 1 a$ and $d = \dfrac {-b} a$. Then $f_{c d} \in \map {\operatorname {Af}_1} \R$ is the inverse of $f_{a b}$.
{{begin-eqn}} {{eqn | l = y | r = a x + b | c = }} {{eqn | ll= \leadsto | l = y - b | r = a x | c = }} {{eqn | ll= \leadsto | l = \frac {y - b} a | r = x | c = }} {{eqn | ll= \leadsto | l = \frac y a + \frac {- b} a | r = x | c = }} {{end-eqn}} As $a...
Let $\map {\operatorname {Af}_1} \R$ denote the [[Definition:Affine Group of One Dimension|$1$-dimensional affine group on $\R$]]. Let $f_{a b} \in \map {\operatorname {Af}_1} \R$. Let $c = \dfrac 1 a$ and $d = \dfrac {-b} a$. Then $f_{c d} \in \map {\operatorname {Af}_1} \R$ is the [[Definition:Inverse Element|inv...
{{begin-eqn}} {{eqn | l = y | r = a x + b | c = }} {{eqn | ll= \leadsto | l = y - b | r = a x | c = }} {{eqn | ll= \leadsto | l = \frac {y - b} a | r = x | c = }} {{eqn | ll= \leadsto | l = \frac y a + \frac {- b} a | r = x | c = }} {{end-eqn}} As $...
Inverse in Affine Group of One Dimension
https://proofwiki.org/wiki/Inverse_in_Affine_Group_of_One_Dimension
https://proofwiki.org/wiki/Inverse_in_Affine_Group_of_One_Dimension
[ "Affine Groups" ]
[ "Definition:Affine Group of One Dimension", "Definition:Inverse (Abstract Algebra)/Inverse" ]
[]
proofwiki-8330
Commutation with Group Elements implies Commuation with Product with Inverse
Let $G$ be a group. Let $a, b, c \in G$ such that $a$ commutes with both $b$ and $c$. Then $a$ commutes with $b c^{-1}$.
{{begin-eqn}} {{eqn | l = a b c^{-1} | r = b a c^{-1} | c = as $a$ commutes with $b$ }} {{eqn | r = b c^{-1} a | c = Commutation with Inverse in Monoid }} {{end-eqn}} {{qed}}
Let $G$ be a [[Definition:Group|group]]. Let $a, b, c \in G$ such that $a$ [[Definition:Commute|commutes]] with both $b$ and $c$. Then $a$ [[Definition:Commute|commutes]] with $b c^{-1}$.
{{begin-eqn}} {{eqn | l = a b c^{-1} | r = b a c^{-1} | c = as $a$ [[Definition:Commute|commutes]] with $b$ }} {{eqn | r = b c^{-1} a | c = [[Commutation with Inverse in Monoid]] }} {{end-eqn}} {{qed}}
Commutation with Group Elements implies Commuation with Product with Inverse
https://proofwiki.org/wiki/Commutation_with_Group_Elements_implies_Commuation_with_Product_with_Inverse
https://proofwiki.org/wiki/Commutation_with_Group_Elements_implies_Commuation_with_Product_with_Inverse
[ "Group Theory", "Commutativity" ]
[ "Definition:Group", "Definition:Commutative/Elements", "Definition:Commutative/Elements" ]
[ "Definition:Commutative/Elements", "Commutation with Inverse in Monoid" ]
proofwiki-8331
Möbius Strip has Euler Characteristic Zero
Let $M$ be a Möbius Strip. Then: :$\map \chi M = 0$ where $\map \chi M$ denotes the Euler characteristic of the graph $M$.
Let the number of vertices, edges and faces of $M$ be $V$, $E$ and $F$ respectively. From Möbius Strip has no Vertices: :$V = 0$ From Möbius Strip has 1 Edge: :$E = 1$ From Möbius Strip has 1 Face: :$F = 1$ By definition of the Euler characteristic: {{begin-eqn}} {{eqn | l = \map \chi M | r = V - E + F | c ...
Let $M$ be a [[Definition:Möbius Strip|Möbius Strip]]. Then: :$\map \chi M = 0$ where $\map \chi M$ denotes the [[Definition:Euler Characteristic|Euler characteristic]] of the [[Definition:Graph (Graph Theory)|graph]] $M$.
Let the number of [[Definition:Vertex of Graph|vertices]], [[Definition:Edge of Graph|edges]] and [[Definition:Face of Graph|faces]] of $M$ be $V$, $E$ and $F$ respectively. From [[Möbius Strip has no Vertices]]: :$V = 0$ From [[Möbius Strip has 1 Edge]]: :$E = 1$ From [[Möbius Strip has 1 Face]]: :$F = 1$ By defin...
Möbius Strip has Euler Characteristic Zero
https://proofwiki.org/wiki/Möbius_Strip_has_Euler_Characteristic_Zero
https://proofwiki.org/wiki/Möbius_Strip_has_Euler_Characteristic_Zero
[ "Möbius Strip" ]
[ "Definition:Möbius Strip", "Definition:Euler Characteristic", "Definition:Graph (Graph Theory)" ]
[ "Definition:Graph (Graph Theory)/Vertex", "Definition:Graph (Graph Theory)/Edge", "Definition:Planar Graph/Face", "Möbius Strip has no Vertices", "Möbius Strip has 1 Edge", "Möbius Strip has 1 Face", "Definition:Euler Characteristic" ]
proofwiki-8332
Centralizer of Subset is Intersection of Centralizers of Elements
Let $\struct {G, \circ}$ be a group. Let $S \subseteq G$. Let $\map {C_G} S$ be the centralizer of $S$ in $G$. Then: :$\ds \map {C_G} S = \bigcap_{x \mathop \in S} \map {C_G} x$ where $\map {C_G} z$ is the centralizer of $x$ in $G$.
{{begin-eqn}} {{eqn | l = \map {C_G} S | r = \set {g \in G: \forall x \in S: g \circ x = x \circ g} | c = {{Defof|Centralizer of Group Subset}} }} {{eqn | r = \set {g \in G: \forall x \in S: g \in \map {C_G} x} | c = {{Defof|Centralizer of Group Element}} }} {{eqn | r = \bigcap_{x \mathop \in S} \map ...
Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. Let $S \subseteq G$. Let $\map {C_G} S$ be the [[Definition:Centralizer of Group Subset|centralizer of $S$ in $G$]]. Then: :$\ds \map {C_G} S = \bigcap_{x \mathop \in S} \map {C_G} x$ where $\map {C_G} z$ is the [[Definition:Centralizer of Group Element|cent...
{{begin-eqn}} {{eqn | l = \map {C_G} S | r = \set {g \in G: \forall x \in S: g \circ x = x \circ g} | c = {{Defof|Centralizer of Group Subset}} }} {{eqn | r = \set {g \in G: \forall x \in S: g \in \map {C_G} x} | c = {{Defof|Centralizer of Group Element}} }} {{eqn | r = \bigcap_{x \mathop \in S} \map ...
Centralizer of Subset is Intersection of Centralizers of Elements
https://proofwiki.org/wiki/Centralizer_of_Subset_is_Intersection_of_Centralizers_of_Elements
https://proofwiki.org/wiki/Centralizer_of_Subset_is_Intersection_of_Centralizers_of_Elements
[ "Centralizers", "Subsets" ]
[ "Definition:Group", "Definition:Centralizer/Group Subset", "Definition:Centralizer/Group Element" ]
[]
proofwiki-8333
Permutation Representation defines Group Action
Let $G$ be a group whose identity is $e$. Let $X$ be a set. Let $\map \Gamma X$ be the symmetric group of $X$. Let $\rho: G \to \map \Gamma X$ be a permutation representation, that is, a homomorphism. The mapping $\phi: G \times X \to X$ associated to $\rho$, defined by: :$\map \phi {g, x} = \map {\map \rho g} x$ is a ...
Let $g, h \in G$ and $x \in X$. We verify that $g * \paren {h * x} = \paren {g h} * x$: {{begin-eqn}} {{eqn | l = g * \paren {h * x} | r = \map {\map \rho g} {\map {\map \rho h} x} | c = Definition of $\phi$ }} {{eqn | r = \map {\paren {\map \rho g \circ \map \rho h} } x | c = {{Defof|Composition of M...
Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$. Let $X$ be a [[Definition:Set|set]]. Let $\map \Gamma X$ be the [[Definition:Symmetric Group|symmetric group]] of $X$. Let $\rho: G \to \map \Gamma X$ be a [[Definition:Permutation Representation|permutation representation...
Let $g, h \in G$ and $x \in X$. We verify that $g * \paren {h * x} = \paren {g h} * x$: {{begin-eqn}} {{eqn | l = g * \paren {h * x} | r = \map {\map \rho g} {\map {\map \rho h} x} | c = Definition of $\phi$ }} {{eqn | r = \map {\paren {\map \rho g \circ \map \rho h} } x | c = {{Defof|Composition of...
Permutation Representation defines Group Action
https://proofwiki.org/wiki/Permutation_Representation_defines_Group_Action
https://proofwiki.org/wiki/Permutation_Representation_defines_Group_Action
[ "Group Actions" ]
[ "Definition:Group", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Set", "Definition:Symmetric Group", "Definition:Group Representation/Permutation", "Definition:Group Homomorphism", "Definition:Mapping", "Definition:Group Action/Permutation Representation", "Definition:Gro...
[ "Definition:Group Homomorphism", "Definition:Identity Mapping", "Group Homomorphism Preserves Identity", "Set of all Self-Maps under Composition forms Monoid" ]
proofwiki-8334
Group Action of Symmetric Group
Let $\N_n$ denote the set $\set {1, 2, \ldots, n}$. Let $\struct {S_n, \circ}$ denote the symmetric group on $\N_n$. The mapping $*: S_n \times \N_n \to \N_n$ defined as: :$\forall \pi \in S_n, \forall n \in \N_n: \pi * n = \map \pi n$ is a group action.
The group action axioms are investigated in turn. Let $\pi, \rho \in S_n$ and $n \in \N_n$. Thus: {{begin-eqn}} {{eqn | l = \pi * \paren {\rho * n} | r = \pi * \map \rho n | c = Definition of $*$ }} {{eqn | r = \map \pi {\map \rho n} | c = Definition of $*$ }} {{eqn | r = \map {\paren {\pi \circ \rho}...
Let $\N_n$ denote the [[Definition:Set|set]] $\set {1, 2, \ldots, n}$. Let $\struct {S_n, \circ}$ denote the [[Definition:Symmetric Group on n Letters|symmetric group]] on $\N_n$. The [[Definition:Mapping|mapping]] $*: S_n \times \N_n \to \N_n$ defined as: :$\forall \pi \in S_n, \forall n \in \N_n: \pi * n = \map \p...
The [[Axiom:Group Action Axioms|group action axioms]] are investigated in turn. Let $\pi, \rho \in S_n$ and $n \in \N_n$. Thus: {{begin-eqn}} {{eqn | l = \pi * \paren {\rho * n} | r = \pi * \map \rho n | c = Definition of $*$ }} {{eqn | r = \map \pi {\map \rho n} | c = Definition of $*$ }} {{eqn |...
Group Action of Symmetric Group
https://proofwiki.org/wiki/Group_Action_of_Symmetric_Group
https://proofwiki.org/wiki/Group_Action_of_Symmetric_Group
[ "Symmetric Groups", "Group Action of Symmetric Group" ]
[ "Definition:Set", "Definition:Symmetric Group/n Letters", "Definition:Mapping", "Definition:Group Action" ]
[ "Axiom:Group Action Axioms", "Definition:Identity Mapping" ]
proofwiki-8335
Group Action of Symmetric Group Acts Transitively
Let $S$ be a set. Let $\struct {\map \Gamma S, \circ}$ be the symmetric group on $S$. Let $*: \map \Gamma S \times S \to S$ be the group action defined as: :$\forall \pi \in \map \Gamma S, \forall s \in S: \pi * s = \map \pi s$ Then $*$ is a transitive group action. In other words, $\struct {\map \Gamma S, \circ}$ acts...
By Group Action of Symmetric Group, $*: \map \Gamma S \times S \to S$ is indeed a group action Let $s, t \in S$. As $\map \Gamma S$ is the symmetric group on $S$, there exists a permutation $\pi \in \map \Gamma S$ such that: :$\map \pi s = t$ This holds for any $s, t \in S$. Thus: :$\forall t \in S: t \in \Orb s$ and s...
Let $S$ be a [[Definition:Set|set]]. Let $\struct {\map \Gamma S, \circ}$ be the [[Definition:Symmetric Group|symmetric group]] on $S$. Let $*: \map \Gamma S \times S \to S$ be the [[Definition:Group Action|group action]] defined as: :$\forall \pi \in \map \Gamma S, \forall s \in S: \pi * s = \map \pi s$ Then $*$ i...
By [[Group Action of Symmetric Group]], $*: \map \Gamma S \times S \to S$ is indeed a [[Definition:Group Action|group action]] Let $s, t \in S$. As $\map \Gamma S$ is the [[Definition:Symmetric Group|symmetric group]] on $S$, there exists a [[Definition:Permutation|permutation]] $\pi \in \map \Gamma S$ such that: :$\...
Group Action of Symmetric Group Acts Transitively
https://proofwiki.org/wiki/Group_Action_of_Symmetric_Group_Acts_Transitively
https://proofwiki.org/wiki/Group_Action_of_Symmetric_Group_Acts_Transitively
[ "Group Action of Symmetric Group", "Transitive Group Actions" ]
[ "Definition:Set", "Definition:Symmetric Group", "Definition:Group Action", "Definition:Transitive Group Action", "Definition:Transitive Group Action" ]
[ "Group Action of Symmetric Group", "Definition:Group Action", "Definition:Symmetric Group", "Definition:Permutation", "Definition:Orbit (Group Theory)", "Definition:Transitive Group Action" ]
proofwiki-8336
Trivial Group Action is Group Action
Let $\struct {G, \circ}$ be a group whose identity is $e$. Let $S$ be a non-empty set. {{explain|Why does $S$ have to be non-empty?}} Let $*: G \times S \to S$ be the trivial group action: :$\forall \tuple {g, s} \in G \times S: g * s = s$ Then $*$ is indeed a group action.
The group action axioms are investigated in turn. Let $g_1, g_2 \in G$ and $s \in S$. Thus: {{begin-eqn}} {{eqn | l = g_1 * \paren {g_2 * s} | r = g_1 * s | c = {{Defof|Trivial Group Action}} }} {{eqn | r = s | c = {{Defof|Trivial Group Action}} }} {{eqn | r = \paren {g_1 \circ g_2} * s | c = {{...
Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$. Let $S$ be a [[Definition:Non-Empty Set|non-empty set]]. {{explain|Why does $S$ have to be non-empty?}} Let $*: G \times S \to S$ be the [[Definition:Trivial Group Action|trivial group action]]: :$\forall ...
The [[Axiom:Group Action Axioms|group action axioms]] are investigated in turn. Let $g_1, g_2 \in G$ and $s \in S$. Thus: {{begin-eqn}} {{eqn | l = g_1 * \paren {g_2 * s} | r = g_1 * s | c = {{Defof|Trivial Group Action}} }} {{eqn | r = s | c = {{Defof|Trivial Group Action}} }} {{eqn | r = \paren ...
Trivial Group Action is Group Action
https://proofwiki.org/wiki/Trivial_Group_Action_is_Group_Action
https://proofwiki.org/wiki/Trivial_Group_Action_is_Group_Action
[ "Group Actions", "Trivial Group Action" ]
[ "Definition:Group", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Non-Empty Set", "Definition:Trivial Group Action", "Definition:Group Action" ]
[ "Axiom:Group Action Axioms", "Category:Group Actions", "Category:Trivial Group Action" ]
proofwiki-8337
Orbit of Trivial Group Action is Singleton
Let $\left({G, \circ}\right)$ be a group whose identity is $e$. Let $S$ be a set. Let $*: G \times S \to S$ be the trivial group action: :$\forall \left({g, s}\right) \in G \times S: g * s = s$ Let $s \in S$. Then the orbit of $s$ under $*$ is $\left\{{s}\right\}$.
By definition: :$\operatorname{Orb} \left({s}\right) = \left\{{t \in S: \exists g \in G: g * s = t}\right\}$ By definition of the trivial group action: :$\forall g \in G: g * s = s$ Hence the result. {{qed}}
Let $\left({G, \circ}\right)$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$. Let $S$ be a [[Definition:Set|set]]. Let $*: G \times S \to S$ be the [[Definition:Trivial Group Action|trivial group action]]: :$\forall \left({g, s}\right) \in G \times S: g * s = s$ Let $s \in S$. ...
By definition: :$\operatorname{Orb} \left({s}\right) = \left\{{t \in S: \exists g \in G: g * s = t}\right\}$ By definition of the [[Definition:Trivial Group Action|trivial group action]]: :$\forall g \in G: g * s = s$ Hence the result. {{qed}}
Orbit of Trivial Group Action is Singleton
https://proofwiki.org/wiki/Orbit_of_Trivial_Group_Action_is_Singleton
https://proofwiki.org/wiki/Orbit_of_Trivial_Group_Action_is_Singleton
[ "Trivial Group Action" ]
[ "Definition:Group", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Set", "Definition:Trivial Group Action", "Definition:Orbit (Group Theory)" ]
[ "Definition:Trivial Group Action" ]
proofwiki-8338
Right Regular Representation by Inverse is Group Action
Let $\struct {G, \circ}$ be a group whose identity is $e$. Let $*: G \times G \to G$ be the operation: :$\forall g, h \in G: g * h = \map {\rho_{g^{-1} } } h$ where $\rho_g$ is the right regular representation of $G$ with respect to $g$. Then $*$ is a group action.
The group action axioms are investigated in turn. Let $g, h, a \in G$. Thus: {{begin-eqn}} {{eqn | l = g * \paren {h * a} | r = g * \map {\rho_{h^{-1} } } a | c = Definition of $*$ }} {{eqn | r = g * \paren {a \circ h^{-1} } | c = {{Defof|Right Regular Representation}} }} {{eqn | r = \map {\rho_{g^{-1...
Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$. Let $*: G \times G \to G$ be the [[Definition:Binary Operation|operation]]: :$\forall g, h \in G: g * h = \map {\rho_{g^{-1} } } h$ where $\rho_g$ is the [[Definition:Right Regular Representation|right regu...
The [[Axiom:Group Action Axioms|group action axioms]] are investigated in turn. Let $g, h, a \in G$. Thus: {{begin-eqn}} {{eqn | l = g * \paren {h * a} | r = g * \map {\rho_{h^{-1} } } a | c = Definition of $*$ }} {{eqn | r = g * \paren {a \circ h^{-1} } | c = {{Defof|Right Regular Representation}...
Right Regular Representation by Inverse is Group Action
https://proofwiki.org/wiki/Right_Regular_Representation_by_Inverse_is_Group_Action
https://proofwiki.org/wiki/Right_Regular_Representation_by_Inverse_is_Group_Action
[ "Group Actions" ]
[ "Definition:Group", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Operation/Binary Operation", "Definition:Regular Representations/Right Regular Representation", "Definition:Group Action" ]
[ "Axiom:Group Action Axioms", "Inverse of Group Product", "Inverse of Identity Element is Itself" ]
proofwiki-8339
Left Regular Representation is Group Action
Let $\struct {G, \circ}$ be a group whose identity is $e$. Let $*: G \times G \to G$ be the operation: :$\forall g, h \in G: g * h = \map {\lambda_g} h$ where $\lambda_g$ is the left regular representation of $G$ with respect to $g$. Then $*$ is a group action.
The group action axioms are investigated in turn. Let $g, h, a \in G$. Thus: {{begin-eqn}} {{eqn | l = g * \paren {h * a} | r = g * \map {\lambda_h} a | c = Definition of $*$ }} {{eqn | r = g * \paren {h \circ a} | c = {{Defof|Left Regular Representation}} }} {{eqn | r = \map {\lambda_g} {h \circ a} ...
Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$. Let $*: G \times G \to G$ be the [[Definition:Binary Operation|operation]]: :$\forall g, h \in G: g * h = \map {\lambda_g} h$ where $\lambda_g$ is the [[Definition:Left Regular Representation|left regular r...
The [[Axiom:Group Action Axioms|group action axioms]] are investigated in turn. Let $g, h, a \in G$. Thus: {{begin-eqn}} {{eqn | l = g * \paren {h * a} | r = g * \map {\lambda_h} a | c = Definition of $*$ }} {{eqn | r = g * \paren {h \circ a} | c = {{Defof|Left Regular Representation}} }} {{eqn | ...
Left Regular Representation is Group Action
https://proofwiki.org/wiki/Left_Regular_Representation_is_Group_Action
https://proofwiki.org/wiki/Left_Regular_Representation_is_Group_Action
[ "Group Actions" ]
[ "Definition:Group", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Operation/Binary Operation", "Definition:Regular Representations/Left Regular Representation", "Definition:Group Action" ]
[ "Axiom:Group Action Axioms" ]
proofwiki-8340
Right Regular Representation by Inverse is Transitive Group Action
Let $\struct {G, \circ}$ be a group whose identity is $e$. Let $*: G \times G \to G$ be the group action: :$\forall g, h \in G: g * h = \map {\rho_{g^{-1} } } h$ where $\rho_g$ is the right regular representation of $G$ with respect to $g$. Then $*$ is a transitive group action.
Let $g, h \in G$. Then: {{begin-eqn}} {{eqn | q = \exists a \in G | l = h | r = a \circ g^{-1} | c = Group has Latin Square Property }} {{eqn | ll= \leadsto | l = h | r = \map {\rho_{g^{-1} } } a | c = {{Defof|Right Regular Representation}} }} {{eqn | ll= \leadsto | l = h ...
Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$. Let $*: G \times G \to G$ be the [[Definition:Group Action|group action]]: :$\forall g, h \in G: g * h = \map {\rho_{g^{-1} } } h$ where $\rho_g$ is the [[Definition:Right Regular Representation|right regul...
Let $g, h \in G$. Then: {{begin-eqn}} {{eqn | q = \exists a \in G | l = h | r = a \circ g^{-1} | c = [[Group has Latin Square Property]] }} {{eqn | ll= \leadsto | l = h | r = \map {\rho_{g^{-1} } } a | c = {{Defof|Right Regular Representation}} }} {{eqn | ll= \leadsto | l = h ...
Right Regular Representation by Inverse is Transitive Group Action
https://proofwiki.org/wiki/Right_Regular_Representation_by_Inverse_is_Transitive_Group_Action
https://proofwiki.org/wiki/Right_Regular_Representation_by_Inverse_is_Transitive_Group_Action
[ "Regular Representations", "Transitive Group Actions" ]
[ "Definition:Group", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Group Action", "Definition:Regular Representations/Right Regular Representation", "Definition:Transitive Group Action" ]
[ "Group has Latin Square Property", "Definition:Orbit (Group Theory)", "Definition:Transitive Group Action" ]
proofwiki-8341
Left Regular Representation is Transitive Group Action
Let $\struct {G, \circ}$ be a group whose identity is $e$. Let $*: G \times G \to G$ be the group action: :$\forall g, h \in G: g * h = \map {\lambda_g} h$ where $\lambda_g$ is the left regular representation of $G$ with respect to $g$. Then $*$ is a transitive group action.
Let $g, h \in G$. Then: {{begin-eqn}} {{eqn | q = \exists a \in G | l = h | r = g \circ a | c = Group has Latin Square Property }} {{eqn | ll= \leadsto | l = h | r = \map {\lambda_g} a | c = {{Defof|Left Regular Representation}} }} {{eqn | ll= \leadsto | l = h | r = g * a...
Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$. Let $*: G \times G \to G$ be the [[Definition:Group Action|group action]]: :$\forall g, h \in G: g * h = \map {\lambda_g} h$ where $\lambda_g$ is the [[Definition:Left Regular Representation|left regular re...
Let $g, h \in G$. Then: {{begin-eqn}} {{eqn | q = \exists a \in G | l = h | r = g \circ a | c = [[Group has Latin Square Property]] }} {{eqn | ll= \leadsto | l = h | r = \map {\lambda_g} a | c = {{Defof|Left Regular Representation}} }} {{eqn | ll= \leadsto | l = h | r = ...
Left Regular Representation is Transitive Group Action
https://proofwiki.org/wiki/Left_Regular_Representation_is_Transitive_Group_Action
https://proofwiki.org/wiki/Left_Regular_Representation_is_Transitive_Group_Action
[ "Transitive Group Actions", "Regular Representations" ]
[ "Definition:Group", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Group Action", "Definition:Regular Representations/Left Regular Representation", "Definition:Transitive Group Action" ]
[ "Group has Latin Square Property", "Definition:Orbit (Group Theory)", "Definition:Transitive Group Action" ]
proofwiki-8342
Conjugacy Action is not Transitive
Let $\struct {G, \circ}$ be a non-trivial group whose identity is $e$. Let $*: G \times G \to G$ be the conjugacy group action: :$\forall g, h \in G: g * h = g \circ h \circ g^{-1}$ Then $*$ is ''not'' a transitive group action.
Proof by Counterexample: For $G$ to be a transitive group action, the orbit of any element of $G$ needs to be the whole of $G$. Take $h = e$. Then: {{begin-eqn}} {{eqn | q = \forall g \in G | l = g * e | r = g \circ e \circ g^{-1} | c = {{Defof|Conjugacy Action}} }} {{eqn | r = g \circ g^{-1} | ...
Let $\struct {G, \circ}$ be a [[Definition:Non-Trivial Group|non-trivial]] [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$. Let $*: G \times G \to G$ be the [[Definition:Conjugacy Action|conjugacy group action]]: :$\forall g, h \in G: g * h = g \circ h \circ g^{-1}$ Then $*$ is ''not...
[[Proof by Counterexample]]: For $G$ to be a [[Definition:Transitive Group Action|transitive group action]], the [[Definition:Orbit (Group Theory)|orbit]] of any [[Definition:Element|element]] of $G$ needs to be the whole of $G$. Take $h = e$. Then: {{begin-eqn}} {{eqn | q = \forall g \in G | l = g * e |...
Conjugacy Action is not Transitive
https://proofwiki.org/wiki/Conjugacy_Action_is_not_Transitive
https://proofwiki.org/wiki/Conjugacy_Action_is_not_Transitive
[ "Conjugacy Action", "Transitive Group Actions" ]
[ "Definition:Non-Trivial Group", "Definition:Group", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Conjugacy Action", "Definition:Transitive Group Action" ]
[ "Proof by Counterexample", "Definition:Transitive Group Action", "Definition:Orbit (Group Theory)", "Definition:Element", "Definition:Orbit (Group Theory)", "Definition:Trivial Group", "Definition:Transitive Group Action" ]
proofwiki-8343
Conjugacy Action on Abelian Group is Trivial
Let $\struct {G, \circ}$ be an abelian group whose identity is $e$. Let $*: G \times G \to G$ be the conjugacy group action: :$\forall g, h \in G: g * h = g \circ h \circ g^{-1}$ Then $*$ is a trivial group action.
For $G$ to be a trivial group action, the orbit of any element of $G$ is a singleton containing only that element. Take $h \in G$. Then: {{begin-eqn}} {{eqn | q = \forall g \in G | l = g * h | r = g \circ h \circ g^{-1} | c = }} {{eqn | r = h \circ g \circ g^{-1} | c = {{Defof|Abelian Group}}: ...
Let $\struct {G, \circ}$ be an [[Definition:Abelian Group|abelian group]] whose [[Definition:Identity Element|identity]] is $e$. Let $*: G \times G \to G$ be the [[Definition:Conjugacy Action|conjugacy group action]]: :$\forall g, h \in G: g * h = g \circ h \circ g^{-1}$ Then $*$ is a [[Definition:Trivial Group Acti...
For $G$ to be a [[Definition:Trivial Group Action|trivial group action]], the [[Definition:Orbit (Group Theory)|orbit]] of any [[Definition:Element|element]] of $G$ is a [[Definition:Singleton|singleton]] containing only that [[Definition:Element|element]]. Take $h \in G$. Then: {{begin-eqn}} {{eqn | q = \forall g \i...
Conjugacy Action on Abelian Group is Trivial
https://proofwiki.org/wiki/Conjugacy_Action_on_Abelian_Group_is_Trivial
https://proofwiki.org/wiki/Conjugacy_Action_on_Abelian_Group_is_Trivial
[ "Conjugacy Action" ]
[ "Definition:Abelian Group", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Conjugacy Action", "Definition:Trivial Group Action" ]
[ "Definition:Trivial Group Action", "Definition:Orbit (Group Theory)", "Definition:Element", "Definition:Singleton", "Definition:Element", "Definition:Commutative/Elements", "Definition:Orbit (Group Theory)", "Definition:Trivial Group Action" ]
proofwiki-8344
Group Action on Subgroup by Left Regular Representation
Let $G$ be a group. Let $H$ be a subgroup of $G$. Let $*: H \times G \to G$ be the operation defined as: :$\forall h \in H: \forall g \in G: h * g = \map {\lambda_h} g$ where $\map {\lambda_h} g$ is the left regular representation of $g$ by $h$. Then $*$ is a group action.
The group action axioms are investigated in turn. Let $h_1, h_2 \in H$ and $g \in G$. Thus: {{begin-eqn}} {{eqn | l = h_1 * \paren {h_2 * g} | r = h_1 * \paren {\map {\lambda_{h_2} } g} | c = Definition of $*$ }} {{eqn | r = h_1 * \paren {h_2 \circ g} | c = {{Defof|Left Regular Representation}} }} {{e...
Let $G$ be a [[Definition:Group|group]]. Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$. Let $*: H \times G \to G$ be the [[Definition:Binary Operation|operation]] defined as: :$\forall h \in H: \forall g \in G: h * g = \map {\lambda_h} g$ where $\map {\lambda_h} g$ is the [[Definition:Left Regular Representat...
The [[Axiom:Group Action Axioms|group action axioms]] are investigated in turn. Let $h_1, h_2 \in H$ and $g \in G$. Thus: {{begin-eqn}} {{eqn | l = h_1 * \paren {h_2 * g} | r = h_1 * \paren {\map {\lambda_{h_2} } g} | c = Definition of $*$ }} {{eqn | r = h_1 * \paren {h_2 \circ g} | c = {{Defof|Le...
Group Action on Subgroup by Left Regular Representation
https://proofwiki.org/wiki/Group_Action_on_Subgroup_by_Left_Regular_Representation
https://proofwiki.org/wiki/Group_Action_on_Subgroup_by_Left_Regular_Representation
[ "Group Actions" ]
[ "Definition:Group", "Definition:Subgroup", "Definition:Operation/Binary Operation", "Definition:Regular Representations/Left Regular Representation", "Definition:Group Action" ]
[ "Axiom:Group Action Axioms", "Axiom:Group Action Axioms", "Definition:Group Action" ]
proofwiki-8345
Group Action on Subgroup by Right Regular Representation
Let $G$ be a group. Let $H$ be a subgroup of $G$. Let $*: H \times G \to G$ be the operation defined as: :$\forall \tuple {h, g} \in H \times G: h * g = \map {\rho_{h^{-1} } } g$ where $\map {\rho_{h^{-1} } } g$ is the right regular representation of $g$ by $h^{-1}$. Then $*$ is a group action.
The group action axioms are investigated in turn. Let $h_1, h_2 \in H$ and $g \in G$. Thus: {{begin-eqn}} {{eqn | l = h_1 * \paren {h_2 * g} | r = h_1 * \paren {\map {\rho_{h_2^{-1} } } g} | c = Definition of $*$ }} {{eqn | r = h_1 * \paren {g \circ h_2^{-1} } | c = {{Defof|Right Regular Representatio...
Let $G$ be a [[Definition:Group|group]]. Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$. Let $*: H \times G \to G$ be the [[Definition:Binary Operation|operation]] defined as: :$\forall \tuple {h, g} \in H \times G: h * g = \map {\rho_{h^{-1} } } g$ where $\map {\rho_{h^{-1} } } g$ is the [[Definition:Right Re...
The [[Axiom:Group Action Axioms|group action axioms]] are investigated in turn. Let $h_1, h_2 \in H$ and $g \in G$. Thus: {{begin-eqn}} {{eqn | l = h_1 * \paren {h_2 * g} | r = h_1 * \paren {\map {\rho_{h_2^{-1} } } g} | c = Definition of $*$ }} {{eqn | r = h_1 * \paren {g \circ h_2^{-1} } | c = {...
Group Action on Subgroup by Right Regular Representation
https://proofwiki.org/wiki/Group_Action_on_Subgroup_by_Right_Regular_Representation
https://proofwiki.org/wiki/Group_Action_on_Subgroup_by_Right_Regular_Representation
[ "Group Actions" ]
[ "Definition:Group", "Definition:Subgroup", "Definition:Operation/Binary Operation", "Definition:Regular Representations/Right Regular Representation", "Definition:Group Action" ]
[ "Axiom:Group Action Axioms", "Inverse of Group Product", "Inverse of Identity Element is Itself", "Axiom:Group Action Axioms", "Definition:Group Action" ]
proofwiki-8346
Group Action on Subgroup by Right Regular Representation is not Transitive
Let $G$ be a group. Let $H$ be a proper subgroup of $G$. Let $*: H \times G \to G$ be the group action defined as: :$\forall \tuple {h, g} \in H \times G: h * g = \map {\rho_{h^{-1} } } g$ where $\map {\rho_{h^{-1} } } g$ is the right regular representation of $g$ by $h^{-1}$. Then $*$ is not transitive.
From Group Action on Subgroup by Right Regular Representation it is established that $*$ is a group action. From Orbit of Group Action on Subgroup by Right Regular Representation is Right Coset: :$\forall x \in G: \Orb x = H x$ where $H x$ is the right coset of $H$ by $x$. From Right Coset Space forms Partition it is a...
Let $G$ be a [[Definition:Group|group]]. Let $H$ be a [[Definition:Proper Subgroup|proper subgroup]] of $G$. Let $*: H \times G \to G$ be the [[Definition:Group Action|group action]] defined as: :$\forall \tuple {h, g} \in H \times G: h * g = \map {\rho_{h^{-1} } } g$ where $\map {\rho_{h^{-1} } } g$ is the [[Defini...
From [[Group Action on Subgroup by Right Regular Representation]] it is established that $*$ is a [[Definition:Group Action|group action]]. From [[Orbit of Group Action on Subgroup by Right Regular Representation is Right Coset]]: :$\forall x \in G: \Orb x = H x$ where $H x$ is the [[Definition:Right Coset|right coset...
Group Action on Subgroup by Right Regular Representation is not Transitive
https://proofwiki.org/wiki/Group_Action_on_Subgroup_by_Right_Regular_Representation_is_not_Transitive
https://proofwiki.org/wiki/Group_Action_on_Subgroup_by_Right_Regular_Representation_is_not_Transitive
[ "Transitive Group Actions" ]
[ "Definition:Group", "Definition:Proper Subgroup", "Definition:Group Action", "Definition:Regular Representations/Right Regular Representation", "Definition:Transitive Group Action" ]
[ "Group Action on Subgroup by Right Regular Representation", "Definition:Group Action", "Orbit of Group Action on Subgroup by Right Regular Representation is Right Coset", "Definition:Coset/Right Coset", "Right Coset Space forms Partition", "Definition:Transitive Group Action" ]
proofwiki-8347
Orbit of Group Action on Subgroup by Right Regular Representation is Right Coset
Let $G$ be a group. Let $H$ be a proper subgroup of $G$. Let $*: H \times G \to G$ be the group action defined as: :$\forall \tuple {h, g} \in H \times G: h * g = \map {\rho_{h^{-1} } } g$ where $\map {\rho_{h^{-1} } } g$ is the right regular representation of $g$ by $h^{-1}$. Let $x \in G$. Then the orbit of $x$ under...
We are given $G$ is a group and $H$ is a proper subgroup of $G$, so: :$H < G$ We are given $*: H \times G \to G$ is the group action defined as: :$\forall \tuple {h, g} \in H \times G: h * g = \map {\rho_{h^{-1} } } g$ where $\map {\rho_{h^{-1} } } g$ is the right regular representation of $g$ by $h^{-1}$. We are also ...
Let $G$ be a [[Definition:Group|group]]. Let $H$ be a [[Definition:Proper Subgroup|proper subgroup]] of $G$. Let $*: H \times G \to G$ be the [[Definition:Group Action|group action]] defined as: :$\forall \tuple {h, g} \in H \times G: h * g = \map {\rho_{h^{-1} } } g$ where $\map {\rho_{h^{-1} } } g$ is the [[Definit...
We are [[Definition:Given|given]] $G$ is a [[Definition:Group|group]] and $H$ is a [[Definition:Proper Subgroup|proper subgroup]] of $G$, so: :$H < G$ We are [[Definition:Given|given]] $*: H \times G \to G$ is the [[Definition:Group Action|group action]] defined as: :$\forall \tuple {h, g} \in H \times G: h * g = \map...
Orbit of Group Action on Subgroup by Right Regular Representation is Right Coset
https://proofwiki.org/wiki/Orbit_of_Group_Action_on_Subgroup_by_Right_Regular_Representation_is_Right_Coset
https://proofwiki.org/wiki/Orbit_of_Group_Action_on_Subgroup_by_Right_Regular_Representation_is_Right_Coset
[ "Group Actions" ]
[ "Definition:Group", "Definition:Proper Subgroup", "Definition:Group Action", "Definition:Regular Representations/Right Regular Representation", "Definition:Orbit (Group Theory)", "Definition:Coset/Right Coset" ]
[ "Definition:Given", "Definition:Group", "Definition:Proper Subgroup", "Definition:Given", "Definition:Group Action", "Definition:Regular Representations/Right Regular Representation", "Definition:Given" ]
proofwiki-8348
Left Coset by Identity
Then: : $e H = H$ where $e H$ is the left coset of $H$ by $e$.
We have: {{begin-eqn}} {{eqn | l = e H | r = \set {y \in G: \exists h \in H: y = e h} | c = {{Defof|Left Coset}} of $H$ by $e$ }} {{eqn | r = \set {y \in G: \exists h \in H: y = h} | c = {{Defof|Identity Element}} }} {{eqn | r = \set {y \in G: y \in H} | c = }} {{eqn | r = H | c = }} {{e...
Then: : $e H = H$ where $e H$ is the [[Definition:Left Coset|left coset]] of $H$ by $e$.
We have: {{begin-eqn}} {{eqn | l = e H | r = \set {y \in G: \exists h \in H: y = e h} | c = {{Defof|Left Coset}} of $H$ by $e$ }} {{eqn | r = \set {y \in G: \exists h \in H: y = h} | c = {{Defof|Identity Element}} }} {{eqn | r = \set {y \in G: y \in H} | c = }} {{eqn | r = H | c = }} {{...
Left Coset by Identity
https://proofwiki.org/wiki/Left_Coset_by_Identity
https://proofwiki.org/wiki/Left_Coset_by_Identity
[ "Coset by Identity" ]
[ "Definition:Coset/Left Coset" ]
[]
proofwiki-8349
Right Coset by Identity
Then: : $H = H e$ where $H e$ is the right coset of $H$ by $e$.
We have: {{begin-eqn}} {{eqn | l = H e | r = \set {x \in G: \exists h \in H: x = h e} | c = {{Defof|Right Coset}} of $H$ by $e$ }} {{eqn | r = \set {x \in G: \exists h \in H: x = h} | c = {{Defof|Identity Element}} }} {{eqn | r = \set {x \in G: x \in H} | c = }} {{eqn | r = H | c = }} {{...
Then: : $H = H e$ where $H e$ is the [[Definition:Right Coset|right coset]] of $H$ by $e$.
We have: {{begin-eqn}} {{eqn | l = H e | r = \set {x \in G: \exists h \in H: x = h e} | c = {{Defof|Right Coset}} of $H$ by $e$ }} {{eqn | r = \set {x \in G: \exists h \in H: x = h} | c = {{Defof|Identity Element}} }} {{eqn | r = \set {x \in G: x \in H} | c = }} {{eqn | r = H | c = }} {...
Right Coset by Identity
https://proofwiki.org/wiki/Right_Coset_by_Identity
https://proofwiki.org/wiki/Right_Coset_by_Identity
[ "Coset by Identity" ]
[ "Definition:Coset/Right Coset" ]
[]
proofwiki-8350
Inversion Mapping is Permutation
Let $\struct {G, \circ}$ be a group. Let $\iota: G \to G$ be the inversion mapping on $G$. Then $\iota$ is a permutation on $G$.
The inversion mapping on $G$ is the mapping $\iota: G \to G$ defined by: :$\forall g \in G: \map \iota g = g^{-1}$ where $g^{-1}$ is the inverse or $g$. By Inversion Mapping is Involution, $\iota$ is an involution: :$\forall g \in G: \map \iota {\map \iota g} = g$ The result follows from Involution is Permutation. {{qe...
Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. Let $\iota: G \to G$ be the [[Definition:Inversion Mapping|inversion mapping]] on $G$. Then $\iota$ is a [[Definition:Permutation|permutation]] on $G$.
The [[Definition:Inversion Mapping|inversion mapping]] on $G$ is the [[Definition:Mapping|mapping]] $\iota: G \to G$ defined by: :$\forall g \in G: \map \iota g = g^{-1}$ where $g^{-1}$ is the [[Definition:Inverse Element|inverse]] or $g$. By [[Inversion Mapping is Involution]], $\iota$ is an [[Definition:Involution...
Inversion Mapping is Permutation/Proof 1
https://proofwiki.org/wiki/Inversion_Mapping_is_Permutation
https://proofwiki.org/wiki/Inversion_Mapping_is_Permutation/Proof_1
[ "Inversion Mapping is Permutation", "Inversion Mappings", "Permutations" ]
[ "Definition:Group", "Definition:Inversion Mapping", "Definition:Permutation" ]
[ "Definition:Inversion Mapping", "Definition:Mapping", "Definition:Inverse (Abstract Algebra)/Inverse", "Inversion Mapping is Involution", "Definition:Involution (Mapping)", "Involution is Permutation" ]
proofwiki-8351
Inversion Mapping is Permutation
Let $\struct {G, \circ}$ be a group. Let $\iota: G \to G$ be the inversion mapping on $G$. Then $\iota$ is a permutation on $G$.
=== Proof of Surjection === Let $a \in G$. By definition of $\iota$: :$\iota(a^{-1}) = \left({a^{-1}}\right)^{-1}$ By Inverse of Inverse: :$\left({a^{-1}}\right)^{-1} = a$ Hence $a$ has a preimage. Since $a$ was arbitrary, $\iota$ is a surjection. === Proof of Injection === Suppose for some $a, b \in G$ that: :$\iota \...
Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. Let $\iota: G \to G$ be the [[Definition:Inversion Mapping|inversion mapping]] on $G$. Then $\iota$ is a [[Definition:Permutation|permutation]] on $G$.
=== Proof of Surjection === Let $a \in G$. By definition of $\iota$: :$\iota(a^{-1}) = \left({a^{-1}}\right)^{-1}$ By [[Inverse of Inverse]]: :$\left({a^{-1}}\right)^{-1} = a$ Hence $a$ has a [[Definition:Preimage of Element under Mapping|preimage]]. Since $a$ was arbitrary, $\iota$ is a [[Definition:Surjection|...
Inversion Mapping is Permutation/Proof 2
https://proofwiki.org/wiki/Inversion_Mapping_is_Permutation
https://proofwiki.org/wiki/Inversion_Mapping_is_Permutation/Proof_2
[ "Inversion Mapping is Permutation", "Inversion Mappings", "Permutations" ]
[ "Definition:Group", "Definition:Inversion Mapping", "Definition:Permutation" ]
[ "Inverse of Inverse", "Definition:Preimage/Mapping/Element", "Definition:Surjection", "Inverse in Group is Unique", "Definition:Injection", "Definition:Bijection", "Definition:Bijection", "Definition:Set", "Definition:Permutation" ]
proofwiki-8352
Inversion Mapping is Mapping
Let $\struct {G, \circ}$ be a group. Let $\iota: G \to G$ be the inversion mapping on $G$. Then $\iota$ is indeed a mapping.
To show that $\iota$ is a mapping, it is sufficient to show that: :$\map \iota a \ne \map \iota b \implies a \ne b$: {{begin-eqn}} {{eqn | l = \map \iota a | o = \ne | r = \map \iota b | c = {{hypothesis}} }} {{eqn | ll= \leadsto | l = a^{-1} | o = \ne | r = b^{-1} | c = Defini...
Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. Let $\iota: G \to G$ be the [[Definition:Inversion Mapping|inversion mapping]] on $G$. Then $\iota$ is indeed a [[Definition:Mapping|mapping]].
To show that $\iota$ is a [[Definition:Mapping|mapping]], it is sufficient to show that: :$\map \iota a \ne \map \iota b \implies a \ne b$: {{begin-eqn}} {{eqn | l = \map \iota a | o = \ne | r = \map \iota b | c = {{hypothesis}} }} {{eqn | ll= \leadsto | l = a^{-1} | o = \ne | r = ...
Inversion Mapping is Mapping
https://proofwiki.org/wiki/Inversion_Mapping_is_Mapping
https://proofwiki.org/wiki/Inversion_Mapping_is_Mapping
[ "Inversion Mappings" ]
[ "Definition:Group", "Definition:Inversion Mapping", "Definition:Mapping" ]
[ "Definition:Mapping", "Cancellation Laws", "Category:Inversion Mappings" ]
proofwiki-8353
Cartesian Product of Group Actions
Let $\struct {G, \circ}$ be a group. Let $S$ and $T$ be sets. Let $*_S: G \times S \to S$ and $*_T: G \times T \to T$ be group actions. Then the operation $*: G \times \paren {S \times T} \to S \times T$ defined as: :$\forall \tuple {g, \tuple {s, t} } \in G \times \paren {S \times T}: g * \tuple {s, t} = \tuple {g *_S...
The group action axioms are investigated in turn. Let $g, h \in G$ and $s, t \in S$. Thus: {{begin-eqn}} {{eqn | l = g * \tuple {h * \tuple {s, t} } | r = g * \tuple {h *_S s, h *_T t} | c = Definition of $*$ }} {{eqn | r = \tuple {g *_S \tuple {h *_S s}, g *_T \tuple {h *_T t} } | c = Definition of $...
Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. Let $S$ and $T$ be [[Definition:Set|sets]]. Let $*_S: G \times S \to S$ and $*_T: G \times T \to T$ be [[Definition:Group Action|group actions]]. Then the [[Definition:Operation|operation]] $*: G \times \paren {S \times T} \to S \times T$ defined as: :$\fora...
The [[Axiom:Group Action Axioms|group action axioms]] are investigated in turn. Let $g, h \in G$ and $s, t \in S$. Thus: {{begin-eqn}} {{eqn | l = g * \tuple {h * \tuple {s, t} } | r = g * \tuple {h *_S s, h *_T t} | c = Definition of $*$ }} {{eqn | r = \tuple {g *_S \tuple {h *_S s}, g *_T \tuple {h *_...
Cartesian Product of Group Actions
https://proofwiki.org/wiki/Cartesian_Product_of_Group_Actions
https://proofwiki.org/wiki/Cartesian_Product_of_Group_Actions
[ "Group Actions", "Cartesian Product" ]
[ "Definition:Group", "Definition:Set", "Definition:Group Action", "Definition:Operation", "Definition:Group Action" ]
[ "Axiom:Group Action Axioms", "Axiom:Group Action Axioms", "Definition:Group Action" ]
proofwiki-8354
Stabilizer of Cartesian Product of Group Actions
Let $\struct {G, \circ}$ be a group. Let $S$ and $T$ be sets. Let $*_S: G \times S \to S$ and $*_T: G \times T \to T$ be group actions. Let the group action $*: G \times \paren {S \times T} \to S \times T$ be defined as: :$\forall \tuple {g, \tuple {s, t} } \in G \times \paren {S \times T}: g * \tuple {s, t} = \tuple {...
By definition, the stabilizer of an element $x$ of $S$ is defined as: :$\Stab x := \set {g \in G: g * x = x}$ where $*$ denotes the group action. So: {{begin-eqn}} {{eqn | l = \Stab {s, t} | r = \set {g \in G: g * \tuple {s, t} = \tuple {s, t} } | c = {{Defof|Stabilizer}} }} {{eqn | r = \set {g \in G: \tupl...
Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. Let $S$ and $T$ be [[Definition:Set|sets]]. Let $*_S: G \times S \to S$ and $*_T: G \times T \to T$ be [[Definition:Group Action|group actions]]. Let the [[Definition:Group Action|group action]] $*: G \times \paren {S \times T} \to S \times T$ be defined as:...
By definition, the [[Definition:Stabilizer|stabilizer]] of an [[Definition:Element|element]] $x$ of $S$ is defined as: :$\Stab x := \set {g \in G: g * x = x}$ where $*$ denotes the [[Definition:Group Action|group action]]. So: {{begin-eqn}} {{eqn | l = \Stab {s, t} | r = \set {g \in G: g * \tuple {s, t} = \tupl...
Stabilizer of Cartesian Product of Group Actions
https://proofwiki.org/wiki/Stabilizer_of_Cartesian_Product_of_Group_Actions
https://proofwiki.org/wiki/Stabilizer_of_Cartesian_Product_of_Group_Actions
[ "Group Actions", "Cartesian Product", "Stabilizers" ]
[ "Definition:Group", "Definition:Set", "Definition:Group Action", "Definition:Group Action", "Definition:Stabilizer", "Definition:Stabilizer" ]
[ "Definition:Stabilizer", "Definition:Element", "Definition:Group Action" ]
proofwiki-8355
Index in Subgroup
Let $G$ be a group. Let $H, K$ be subgroups of finite index of $G$. Then: :$\index H {H \cap K} \le \index G K$ where $\index G K$ denotes the index of $K$ in $G$. Equality happens {{iff}} $G = H K$.
We list out all the left cosets of $H \cap K$ in $H$: :$H / \paren {H \cap K} = \set {h_n \paren {H \cap K}: h_n \in H, n \in I}$ where $I$ is some finite indexing set. For each pair $h_i, h_j \in H \subseteq G$, where $i \ne j$: :$h_i^{-1} h_j \notin H \cap K \quad$ Cosets are Equal iff Product with Inverse in Subgrou...
Let $G$ be a [[Definition:Group|group]]. Let $H, K$ be [[Definition:Subgroup|subgroups]] of [[Definition:Finite Index|finite index]] of $G$. Then: :$\index H {H \cap K} \le \index G K$ where $\index G K$ denotes the [[Definition:Index of Subgroup|index]] of $K$ in $G$. Equality happens {{iff}} $G = H K$.
We list out all the [[Definition:Left Coset|left cosets]] of $H \cap K$ in $H$: :$H / \paren {H \cap K} = \set {h_n \paren {H \cap K}: h_n \in H, n \in I}$ where $I$ is some [[Definition:Finite Set|finite]] indexing set. For each pair $h_i, h_j \in H \subseteq G$, where $i \ne j$: :$h_i^{-1} h_j \notin H \cap K \qua...
Index in Subgroup
https://proofwiki.org/wiki/Index_in_Subgroup
https://proofwiki.org/wiki/Index_in_Subgroup
[ "Subgroups", "Index of Subgroups" ]
[ "Definition:Group", "Definition:Subgroup", "Definition:Index of Subgroup/Finite", "Definition:Index of Subgroup" ]
[ "Definition:Coset/Left Coset", "Definition:Finite Set", "Cosets are Equal iff Product with Inverse in Subgroup", "Definition:Subgroup", "Cosets are Equal iff Product with Inverse in Subgroup", "Definition:Coset/Left Coset", "Definition:Coset/Left Coset", "Definition:Coset/Left Coset", "Definition:Co...
proofwiki-8356
Finite Cyclic Group has Euler Phi Generators
Let $C_n$ be a (finite) cyclic group of order $n$. Then $C_n$ has $\map \phi n$ generators, where $\map \phi n$ denotes the Euler $\phi$ function.
From List of Elements in Finite Cyclic Group, the elements of $G$ are: :$\set {g^k: g \in G, 0 \le k < n}$ From Power of Generator of Cyclic Group is Generator iff Power is Coprime with Order, $g^k$ generates $G$ {{iff}} $k \perp n$. The result follows by definition of the Euler $\phi$ function. {{qed}}
Let $C_n$ be a [[Definition:Finite Cyclic Group|(finite) cyclic group]] of [[Definition:Order of Structure|order $n$]]. Then $C_n$ has $\map \phi n$ [[Definition:Generator of Cyclic Group|generators]], where $\map \phi n$ denotes the [[Definition:Euler Phi Function|Euler $\phi$ function]].
From [[List of Elements in Finite Cyclic Group]], the [[Definition:Element|elements]] of $G$ are: :$\set {g^k: g \in G, 0 \le k < n}$ From [[Power of Generator of Cyclic Group is Generator iff Power is Coprime with Order]], $g^k$ [[Definition:Generator of Cyclic Group|generates]] $G$ {{iff}} $k \perp n$. The result f...
Finite Cyclic Group has Euler Phi Generators
https://proofwiki.org/wiki/Finite_Cyclic_Group_has_Euler_Phi_Generators
https://proofwiki.org/wiki/Finite_Cyclic_Group_has_Euler_Phi_Generators
[ "Finite Cyclic Groups" ]
[ "Definition:Finite Cyclic Group", "Definition:Order of Structure", "Definition:Cyclic Group/Generator", "Definition:Euler Phi Function" ]
[ "List of Elements in Finite Cyclic Group", "Definition:Element", "Power of Generator of Cyclic Group is Generator iff Power is Coprime with Order", "Definition:Cyclic Group/Generator", "Definition:Euler Phi Function" ]
proofwiki-8357
Composition of Left Regular Representations
:$\lambda_x \circ \lambda_y = \lambda_{x * y}$
Let $z \in S$. {{begin-eqn}} {{eqn | l = \map {\lambda_x \circ \lambda_y} z | r = \map {\lambda_x} {\map {\lambda_y} z} | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \map {\lambda_x} {y * z} | c = {{Defof|Left Regular Representation}} }} {{eqn | r = x * \paren {y * z} | c = {{Defof|Left...
:$\lambda_x \circ \lambda_y = \lambda_{x * y}$
Let $z \in S$. {{begin-eqn}} {{eqn | l = \map {\lambda_x \circ \lambda_y} z | r = \map {\lambda_x} {\map {\lambda_y} z} | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \map {\lambda_x} {y * z} | c = {{Defof|Left Regular Representation}} }} {{eqn | r = x * \paren {y * z} | c = {{Defof|Lef...
Composition of Left Regular Representations
https://proofwiki.org/wiki/Composition_of_Left_Regular_Representations
https://proofwiki.org/wiki/Composition_of_Left_Regular_Representations
[ "Semigroups", "Regular Representations" ]
[]
[]
proofwiki-8358
Composition of Right Regular Representations
:$\rho_x \circ \rho_y = \rho_{y * x}$
Let $z \in S$. {{begin-eqn}} {{eqn | l = \map {\paren {\rho_x \circ \rho_y} } z | r = \map {\rho_x} {\map {\rho_y} z} | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \map {\rho_x} {z * y} | c = {{Defof|Right Regular Representation}} }} {{eqn | r = \paren {z * y} * x | c = {{Defof|Right Re...
:$\rho_x \circ \rho_y = \rho_{y * x}$
Let $z \in S$. {{begin-eqn}} {{eqn | l = \map {\paren {\rho_x \circ \rho_y} } z | r = \map {\rho_x} {\map {\rho_y} z} | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \map {\rho_x} {z * y} | c = {{Defof|Right Regular Representation}} }} {{eqn | r = \paren {z * y} * x | c = {{Defof|Right R...
Composition of Right Regular Representations
https://proofwiki.org/wiki/Composition_of_Right_Regular_Representations
https://proofwiki.org/wiki/Composition_of_Right_Regular_Representations
[ "Semigroups", "Regular Representations" ]
[]
[]
proofwiki-8359
Composition of Left Regular Representation with Right
:$\lambda_x \circ \rho_y = \rho_y \circ \lambda_x$
Let $z \in S$. {{begin-eqn}} {{eqn | l = \map {\paren {\lambda_x \circ \rho_y} } z | r = \map {\lambda_x} {\map {\rho_y} z} | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \map {\lambda_x} {z * y} | c = {{Defof|Right Regular Representation}} }} {{eqn | r = x * \paren {z * y} | c = {{Defof...
:$\lambda_x \circ \rho_y = \rho_y \circ \lambda_x$
Let $z \in S$. {{begin-eqn}} {{eqn | l = \map {\paren {\lambda_x \circ \rho_y} } z | r = \map {\lambda_x} {\map {\rho_y} z} | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \map {\lambda_x} {z * y} | c = {{Defof|Right Regular Representation}} }} {{eqn | r = x * \paren {z * y} | c = {{Defo...
Composition of Left Regular Representation with Right
https://proofwiki.org/wiki/Composition_of_Left_Regular_Representation_with_Right
https://proofwiki.org/wiki/Composition_of_Left_Regular_Representation_with_Right
[ "Semigroups", "Regular Representations" ]
[]
[]
proofwiki-8360
Extendability Theorem for Derivatives Continuous on Open Intervals
Let $f$ be a continuous real function defined on an interval $\closedint a b$ where $a < b$. Then $f$ is continuously differentiable on $\closedint a b$ {{iff}}: :$f$ is continuously differentiable on $\openint a b$ and: :$\ds \lim_{x \mathop \to a^+} \map {f'} x$ and $\ds \lim_{x \mathop \to b^-} \map {f'} x$ exist.
=== Necessary Condition === Suppose that $f$ is continuously differentiable on $\closedint a b$. We need to show that: :$f$ is continuously differentiable on $\openint a b$ and: :$\ds \lim_{x \mathop \to a^+} \map {f'} x$ and $\ds \lim_{x \mathop \to b^-} \map {f'} x$ exist. $f$ is continuously differentiable on $\open...
Let $f$ be a [[Definition:Continuous Real Function|continuous real function]] defined on an [[Definition:Real Interval|interval]] $\closedint a b$ where $a < b$. Then $f$ is [[Definition:Continuously Differentiable Real Function|continuously differentiable]] on $\closedint a b$ {{iff}}: :$f$ is [[Definition:Continuou...
=== Necessary Condition === Suppose that $f$ is [[Definition:Continuously Differentiable Real Function|continuously differentiable]] on $\closedint a b$. We need to show that: :$f$ is [[Definition:Continuously Differentiable Real Function|continuously differentiable]] on $\openint a b$ and: :$\ds \lim_{x \mathop \...
Extendability Theorem for Derivatives Continuous on Open Intervals
https://proofwiki.org/wiki/Extendability_Theorem_for_Derivatives_Continuous_on_Open_Intervals
https://proofwiki.org/wiki/Extendability_Theorem_for_Derivatives_Continuous_on_Open_Intervals
[ "Differential Calculus" ]
[ "Definition:Continuous Real Function", "Definition:Real Interval", "Definition:Continuously Differentiable/Real Function", "Definition:Continuously Differentiable/Real Function" ]
[ "Definition:Continuously Differentiable/Real Function", "Definition:Continuously Differentiable/Real Function", "Definition:Continuously Differentiable/Real Function", "Definition:Continuously Differentiable/Real Function", "Definition:Subset", "Definition:Differentiable Mapping/Real Function", "Definit...
proofwiki-8361
Universal Affirmative and Universal Negative are Contrary iff First Predicate is not Vacuous
Consider the categorical statements: {{begin-axiom}} {{axiom | ll = \mathbf A: | lc= The universal affirmative: | q = \forall x | m = \map S x \implies \map P x }} {{axiom | ll= \mathbf E: | lc= The universal negative: | q = \forall x | m = \map S x \implies \neg \map P x...
=== Sufficient Condition === Let $\exists x: \map S x$. Suppose $\mathbf A$ and $\mathbf E$ are both true. As $\mathbf A$ is true, then by Modus Ponendo Ponens: :$\map P x$ As $\mathbf E$ is true, then by Modus Ponendo Ponens: :$\neg \map P x$ It follows by Proof by Contradiction that $\mathbf A$ and $\mathbf E$ are no...
Consider the [[Definition:Categorical Statement|categorical statements]]: {{begin-axiom}} {{axiom | ll = \mathbf A: | lc= The [[Definition:Universal Affirmative|universal affirmative]]: | q = \forall x | m = \map S x \implies \map P x }} {{axiom | ll= \mathbf E: | lc= The [[Definition:U...
=== Sufficient Condition === Let $\exists x: \map S x$. Suppose $\mathbf A$ and $\mathbf E$ are both [[Definition:True|true]]. As $\mathbf A$ is [[Definition:True|true]], then by [[Modus Ponendo Ponens]]: :$\map P x$ As $\mathbf E$ is [[Definition:True|true]], then by [[Modus Ponendo Ponens]]: :$\neg \map P x$ It ...
Universal Affirmative and Universal Negative are Contrary iff First Predicate is not Vacuous
https://proofwiki.org/wiki/Universal_Affirmative_and_Universal_Negative_are_Contrary_iff_First_Predicate_is_not_Vacuous
https://proofwiki.org/wiki/Universal_Affirmative_and_Universal_Negative_are_Contrary_iff_First_Predicate_is_not_Vacuous
[ "Categorical Statements" ]
[ "Definition:Categorical Statement", "Definition:Universal Affirmative", "Definition:Universal Negative", "Definition:Contrary Statements", "Definition:Symbolic Logic", "Definition:Predicate Logic" ]
[ "Definition:True", "Definition:True", "Modus Ponendo Ponens", "Definition:True", "Modus Ponendo Ponens", "Proof by Contradiction", "Definition:True", "Definition:Contrary Statements", "Definition:Contrary Statements", "Definition:True", "Definition:True", "Definition:Contrary Statements", "P...
proofwiki-8362
Particular Affirmative and Particular Negative are Subcontrary iff First Predicate is not Vacuous
Consider the categorical statements: {{begin-axiom}} {{axiom | ll = \mathbf I: | lc= The particular affirmative: | q = \exists x | m = \map S x \land \map P x }} {{axiom | ll= \mathbf O: | lc= The particular negative: | q = \exists x | m = \map S x \land \neg \map P x }} ...
=== Sufficient Condition === Let $\exists x: \map S x$. Suppose $\mathbf I$ and $\mathbf O$ are both false. As $\mathbf I$ is false, then by the Rule of Conjunction: :$\neg \map P x$ As $\mathbf O$ is false, then by the Rule of Conjunction: :$\neg \neg \map P x$ and so by Double Negation: :$\map P x$ It follows by Proo...
Consider the [[Definition:Categorical Statement|categorical statements]]: {{begin-axiom}} {{axiom | ll = \mathbf I: | lc= The [[Definition:Particular Affirmative|particular affirmative]]: | q = \exists x | m = \map S x \land \map P x }} {{axiom | ll= \mathbf O: | lc= The [[Definition:Pa...
=== Sufficient Condition === Let $\exists x: \map S x$. Suppose $\mathbf I$ and $\mathbf O$ are both [[Definition:False|false]]. As $\mathbf I$ is [[Definition:False|false]], then by the [[Rule of Conjunction/Proof Rule|Rule of Conjunction]]: :$\neg \map P x$ As $\mathbf O$ is [[Definition:False|false]], then by th...
Particular Affirmative and Particular Negative are Subcontrary iff First Predicate is not Vacuous
https://proofwiki.org/wiki/Particular_Affirmative_and_Particular_Negative_are_Subcontrary_iff_First_Predicate_is_not_Vacuous
https://proofwiki.org/wiki/Particular_Affirmative_and_Particular_Negative_are_Subcontrary_iff_First_Predicate_is_not_Vacuous
[ "Categorical Statements" ]
[ "Definition:Categorical Statement", "Definition:Particular Affirmative", "Definition:Particular Negative", "Definition:Subcontrary Statements", "Definition:Symbolic Logic", "Definition:Predicate Logic" ]
[ "Definition:False", "Definition:False", "Rule of Conjunction/Proof Rule", "Definition:False", "Rule of Conjunction/Proof Rule", "Double Negation", "Proof by Contradiction", "Definition:False", "Definition:Subcontrary Statements", "Definition:Subcontrary Statements", "Definition:False", "Defini...
proofwiki-8363
Universal Affirmative and Particular Negative are Contradictory
Consider the categorical statements: {{begin-axiom}} {{axiom | ll = \mathbf A: | lc= The universal affirmative: | q = \forall x | m = \map S x \implies \map P x }} {{axiom | ll= \mathbf O: | lc= The particular negative: | q = \exists x | m = \map S x \land \neg \map P x }...
{{begin-eqn}} {{eqn | o = | r = \mathbf A | c = }} {{eqn | ll= \leadsto | q = \forall x | o = | r = \map S x \implies \map P x | c = Definition of $\mathbf A$ }} {{eqn | ll= \leadsto | q = \forall x | o = | r = \neg \paren {\map S x \land \neg \map P x} | ...
Consider the [[Definition:Categorical Statement|categorical statements]]: {{begin-axiom}} {{axiom | ll = \mathbf A: | lc= The [[Definition:Universal Affirmative|universal affirmative]]: | q = \forall x | m = \map S x \implies \map P x }} {{axiom | ll= \mathbf O: | lc= The [[Definition:P...
{{begin-eqn}} {{eqn | o = | r = \mathbf A | c = }} {{eqn | ll= \leadsto | q = \forall x | o = | r = \map S x \implies \map P x | c = Definition of $\mathbf A$ }} {{eqn | ll= \leadsto | q = \forall x | o = | r = \neg \paren {\map S x \land \neg \map P x} | ...
Universal Affirmative and Particular Negative are Contradictory
https://proofwiki.org/wiki/Universal_Affirmative_and_Particular_Negative_are_Contradictory
https://proofwiki.org/wiki/Universal_Affirmative_and_Particular_Negative_are_Contradictory
[ "Categorical Statements" ]
[ "Definition:Categorical Statement", "Definition:Universal Affirmative", "Definition:Particular Negative", "Definition:Contradictory/Statements", "Definition:Symbolic Logic", "Definition:Predicate Logic" ]
[ "Conditional is Equivalent to Negation of Conjunction with Negative", "De Morgan's Laws (Predicate Logic)/Denial of Existence", "Conjunction with Negative is Equivalent to Negation of Conditional", "De Morgan's Laws (Predicate Logic)/Denial of Universality", "Definition:Contradictory/Statements" ]
proofwiki-8364
Particular Affirmative and Universal Negative are Contradictory
Consider the categorical statements: {{begin-axiom}} {{axiom | ll = \mathbf I: | lc= The particular affirmative: | q = \exists x | m = \map S x \land \map P x }} {{axiom | ll= \mathbf E: | lc= The universal negative: | q = \forall x | m = \map S x \implies \neg \map P x }...
{{begin-eqn}} {{eqn | o = | r = \mathbf E | c = }} {{eqn | ll= \therefore | l = \forall x: | o = | r = \map S x \implies \neg \map P x | c = Definition of $\mathbf E$ }} {{eqn | ll= \therefore | l = \forall x: | o = | r = \neg \paren {\map S x \land \map P x} ...
Consider the [[Definition:Categorical Statement|categorical statements]]: {{begin-axiom}} {{axiom | ll = \mathbf I: | lc= The [[Definition:Particular Affirmative|particular affirmative]]: | q = \exists x | m = \map S x \land \map P x }} {{axiom | ll= \mathbf E: | lc= The [[Definition:Un...
{{begin-eqn}} {{eqn | o = | r = \mathbf E | c = }} {{eqn | ll= \therefore | l = \forall x: | o = | r = \map S x \implies \neg \map P x | c = Definition of $\mathbf E$ }} {{eqn | ll= \therefore | l = \forall x: | o = | r = \neg \paren {\map S x \land \map P x} ...
Particular Affirmative and Universal Negative are Contradictory
https://proofwiki.org/wiki/Particular_Affirmative_and_Universal_Negative_are_Contradictory
https://proofwiki.org/wiki/Particular_Affirmative_and_Universal_Negative_are_Contradictory
[ "Categorical Statements" ]
[ "Definition:Categorical Statement", "Definition:Particular Affirmative", "Definition:Universal Negative", "Definition:Contradictory/Statements", "Definition:Symbolic Logic", "Definition:Predicate Logic" ]
[ "Modus Ponendo Tollens/Variant", "De Morgan's Laws (Predicate Logic)/Denial of Existence", "Conjunction is Equivalent to Negation of Conditional of Negative", "De Morgan's Laws (Predicate Logic)/Denial of Universality", "Definition:Contradictory/Statements" ]
proofwiki-8365
Universal Affirmative implies Particular Affirmative iff First Predicate is not Vacuous
Consider the categorical statements: {{begin-axiom}} {{axiom | ll = \map {\mathbf A} {S, P}: | lc= The universal affirmative: | q = \forall x | m = \map S x \implies \map P x }} {{axiom | ll= \map {\mathbf I} {S, P}: | lc= The particular affirmative: | q = \exists x | m =...
=== Sufficient Condition === Let $\exists x: \map S x$. Let $\map {\mathbf A} {S, P}$ be true. As $\map {\mathbf A} {S, P}$ is true, then by Modus Ponendo Ponens: :$\map P x$ From the Rule of Conjunction: :$\map S x \land \map P x$ Thus $\map {\mathbf I} {S, P}$ holds. So by the Rule of Implication: :$\map {\mathbf A} ...
Consider the [[Definition:Categorical Statement|categorical statements]]: {{begin-axiom}} {{axiom | ll = \map {\mathbf A} {S, P}: | lc= The [[Definition:Universal Affirmative|universal affirmative]]: | q = \forall x | m = \map S x \implies \map P x }} {{axiom | ll= \map {\mathbf I} {S, P}: ...
=== Sufficient Condition === Let $\exists x: \map S x$. Let $\map {\mathbf A} {S, P}$ be [[Definition:True|true]]. As $\map {\mathbf A} {S, P}$ is [[Definition:True|true]], then by [[Modus Ponendo Ponens]]: :$\map P x$ From the [[Rule of Conjunction/Proof Rule|Rule of Conjunction]]: :$\map S x \land \map P x$ Thus...
Universal Affirmative implies Particular Affirmative iff First Predicate is not Vacuous
https://proofwiki.org/wiki/Universal_Affirmative_implies_Particular_Affirmative_iff_First_Predicate_is_not_Vacuous
https://proofwiki.org/wiki/Universal_Affirmative_implies_Particular_Affirmative_iff_First_Predicate_is_not_Vacuous
[ "Categorical Statements" ]
[ "Definition:Categorical Statement", "Definition:Universal Affirmative", "Definition:Particular Affirmative", "Definition:Symbolic Logic", "Definition:Predicate Logic" ]
[ "Definition:True", "Definition:True", "Modus Ponendo Ponens", "Rule of Conjunction/Proof Rule", "Rule of Implication", "Definition:True", "Rule of Conjunction/Proof Rule", "Definition:True" ]
proofwiki-8366
Universal Negative implies Particular Negative iff First Predicate is not Vacuous
Consider the categorical statements: {{begin-axiom}} {{axiom | ll = \map {\mathbf E} {S, P}: | lc= The universal negative: | q = \forall x | m = \map S x \implies \neg \map P x }} {{axiom | ll= \map {\mathbf O} {S, P}: | lc= The particular negative: | q = \exists x | m = ...
=== Sufficient Condition === Let $\exists x: \map S x$. Let $\map {\mathbf E} {S, P}$ be true. As $\map {\mathbf E} {S, P}$ is true, then by Modus Ponendo Ponens: :$\neg \map P x$ From the Rule of Conjunction: :$\map S x \land \neg \map P x$ Thus $\map {\mathbf O} {S, P}$ holds. So by the Rule of Implication: :$\map {\...
Consider the [[Definition:Categorical Statement|categorical statements]]: {{begin-axiom}} {{axiom | ll = \map {\mathbf E} {S, P}: | lc= The [[Definition:Universal Negative|universal negative]]: | q = \forall x | m = \map S x \implies \neg \map P x }} {{axiom | ll= \map {\mathbf O} {S, P}: ...
=== Sufficient Condition === Let $\exists x: \map S x$. Let $\map {\mathbf E} {S, P}$ be [[Definition:True|true]]. As $\map {\mathbf E} {S, P}$ is [[Definition:True|true]], then by [[Modus Ponendo Ponens]]: :$\neg \map P x$ From the [[Rule of Conjunction/Proof Rule|Rule of Conjunction]]: :$\map S x \land \neg \map ...
Universal Negative implies Particular Negative iff First Predicate is not Vacuous
https://proofwiki.org/wiki/Universal_Negative_implies_Particular_Negative_iff_First_Predicate_is_not_Vacuous
https://proofwiki.org/wiki/Universal_Negative_implies_Particular_Negative_iff_First_Predicate_is_not_Vacuous
[ "Categorical Statements" ]
[ "Definition:Categorical Statement", "Definition:Universal Negative", "Definition:Particular Negative", "Definition:Symbolic Logic", "Definition:Predicate Logic" ]
[ "Definition:True", "Definition:True", "Modus Ponendo Ponens", "Rule of Conjunction/Proof Rule", "Rule of Implication", "Definition:True", "Rule of Conjunction/Proof Rule", "Definition:True" ]
proofwiki-8367
Socrates is Mortal
:$(1): \quad$ ''All humans are mortal.'' :$(2): \quad$ ''{{AuthorRef|Socrates}} is human.'' :$(3): \quad$ ''Therefore {{AuthorRef|Socrates}} is mortal.''
Let $x$ be an object variable from the universe of '''rational beings'''. Let $\map H x$ denote the propositional function ''$x$ is '''human'''''. Let $\map M x$ denote the propositional function ''$x$ is '''mortal'''''. Let $S$ be a proper name that denotes {{AuthorRef|Socrates}}. The argument can then be expressed as...
:$(1): \quad$ ''All humans are mortal.'' :$(2): \quad$ ''{{AuthorRef|Socrates}} is human.'' :$(3): \quad$ ''Therefore {{AuthorRef|Socrates}} is mortal.''
Let $x$ be an [[Definition:Object Variable|object variable]] from the [[Definition:Universe of Discourse|universe]] of '''rational beings'''. Let $\map H x$ denote the [[Definition:Propositional Function|propositional function]] ''$x$ is '''human'''''. Let $\map M x$ denote the [[Definition:Propositional Function|pro...
Socrates is Mortal
https://proofwiki.org/wiki/Socrates_is_Mortal
https://proofwiki.org/wiki/Socrates_is_Mortal
[ "Socrates is Mortal", "Logic", "Classic Problems" ]
[]
[ "Definition:Variable", "Definition:Universe of Discourse", "Definition:Propositional Function", "Definition:Propositional Function", "Definition:Proper Name", "Universal Instantiation", "Modus Ponendo Ponens" ]
proofwiki-8368
Right-Hand Differentiable Function is Right-Continuous
Let $f$ be a real function defined on an interval $I$. Let $a$ be a point in $I$ where $f$ is right-hand differentiable. Then $f$ is right-continuous at $a$.
By hypothesis, $\map {f'_+} a$ exists. First we note that $a$ cannot be the right hand end point of $I$ because values in $I$ greater than $a$ need to exist for $\map {f'_+} a$ to exist. We form the following expression: :$\ds \lim_{x \mathop \to a^+} \paren {\map f x - \map f a}$ We need to show that it is defined and...
Let $f$ be a [[Definition:Real Function|real function]] defined on an [[Definition:Real Interval|interval]] $I$. Let $a$ be a point in $I$ where $f$ is [[Definition:Right-Hand Derivative|right-hand differentiable]]. Then $f$ is [[Definition:Right-Continuous at Point|right-continuous]] at $a$.
By hypothesis, $\map {f'_+} a$ exists. First we note that $a$ cannot be the right hand [[Definition:Endpoint of Real Interval|end point]] of $I$ because values in $I$ greater than $a$ need to exist for $\map {f'_+} a$ to exist. We form the following expression: :$\ds \lim_{x \mathop \to a^+} \paren {\map f x - \map...
Right-Hand Differentiable Function is Right-Continuous
https://proofwiki.org/wiki/Right-Hand_Differentiable_Function_is_Right-Continuous
https://proofwiki.org/wiki/Right-Hand_Differentiable_Function_is_Right-Continuous
[ "Continuous Real Functions", "Differentiable Real Functions" ]
[ "Definition:Real Function", "Definition:Real Interval", "Definition:Right-Hand Derivative", "Definition:Continuous Real Function/Right-Continuous" ]
[ "Definition:Real Interval/Endpoints", "Definition:Fraction/Denominator", "Combination Theorem for Limits of Functions/Real/Product Rule", "Combination Theorem for Limits of Functions/Real/Sum Rule", "Definition:Continuous Real Function/Right-Continuous", "Category:Continuous Real Functions", "Category:D...
proofwiki-8369
Left-Hand Differentiable Function is Left-Continuous
Let $f$ be a real function defined on an interval $I$. Let $a$ be a point in $I$ where $f$ is left-hand differentiable. Then $f$ is left-continuous at $a$.
By hypothesis, $\map {f'_-} a$ exists. First we note that $a$ cannot be the left-hand end point of $I$ because values in $I$ less than $a$ need to exist for $\map {f'_-} a$ to exist. We form the following expression: :$\ds \lim_{x \mathop \to a^-} \paren {\map f x - \map f a}$ We need to show that it is defined and to ...
Let $f$ be a [[Definition:Real Function|real function]] defined on an [[Definition:Real Interval|interval]] $I$. Let $a$ be a point in $I$ where $f$ is [[Definition:Left-Hand Derivative|left-hand differentiable]]. Then $f$ is [[Definition:Left-Continuous at Point|left-continuous]] at $a$.
By hypothesis, $\map {f'_-} a$ exists. First we note that $a$ cannot be the left-hand [[Definition:Endpoint of Real Interval|end point]] of $I$ because values in $I$ less than $a$ need to exist for $\map {f'_-} a$ to exist. We form the following expression: :$\ds \lim_{x \mathop \to a^-} \paren {\map f x - \map f a...
Left-Hand Differentiable Function is Left-Continuous
https://proofwiki.org/wiki/Left-Hand_Differentiable_Function_is_Left-Continuous
https://proofwiki.org/wiki/Left-Hand_Differentiable_Function_is_Left-Continuous
[ "Continuous Real Functions", "Differentiable Real Functions" ]
[ "Definition:Real Function", "Definition:Real Interval", "Definition:Left-Hand Derivative", "Definition:Continuous Real Function/Left-Continuous" ]
[ "Definition:Real Interval/Endpoints", "Definition:Fraction/Denominator", "Combination Theorem for Limits of Functions/Real/Product Rule", "Combination Theorem for Limits of Functions/Real/Sum Rule", "Definition:Continuous Real Function/Left-Continuous", "Category:Continuous Real Functions", "Category:Di...
proofwiki-8370
Left-Hand and Right-Hand Differentiable Function is Continuous
Let $f$ be a real function defined on an interval $I$. Let $a$ be a point in $I$ where $f$ is left- and right-hand differentiable. Then $f$ is continuous at $a$.
By Left-Hand Differentiable Function is Left-Continuous, $f$ is left-continuous at $a$. By Right-Hand Differentiable Function is Right-Continuous, $f$ is right-continuous at $a$. By Continuous at Point iff Left-Continuous and Right-Continuous, $f$ is continuous at $a$. {{qed}} Category:Continuous Real Functions Categor...
Let $f$ be a [[Definition:Real Function|real function]] defined on an [[Definition:Real Interval|interval]] $I$. Let $a$ be a point in $I$ where $f$ is [[Definition:Left-Hand Derivative|left-]] and [[Definition:Right-Hand Derivative|right-hand differentiable]]. Then $f$ is [[Definition:Continuous Real Function at Po...
By [[Left-Hand Differentiable Function is Left-Continuous]], $f$ is [[Definition:Left-Continuous at Point|left-continuous]] at $a$. By [[Right-Hand Differentiable Function is Right-Continuous]], $f$ is [[Definition:Right-Continuous at Point|right-continuous]] at $a$. By [[Continuous at Point iff Left-Continuous and R...
Left-Hand and Right-Hand Differentiable Function is Continuous
https://proofwiki.org/wiki/Left-Hand_and_Right-Hand_Differentiable_Function_is_Continuous
https://proofwiki.org/wiki/Left-Hand_and_Right-Hand_Differentiable_Function_is_Continuous
[ "Continuous Real Functions", "Differentiable Real Functions" ]
[ "Definition:Real Function", "Definition:Real Interval", "Definition:Left-Hand Derivative", "Definition:Right-Hand Derivative", "Definition:Continuous Real Function/Point" ]
[ "Left-Hand Differentiable Function is Left-Continuous", "Definition:Continuous Real Function/Left-Continuous", "Right-Hand Differentiable Function is Right-Continuous", "Definition:Continuous Real Function/Right-Continuous", "Continuous at Point iff Left-Continuous and Right-Continuous", "Definition:Conti...
proofwiki-8371
Universal Affirmative and Negative are both False iff Particular Affirmative and Negative are both True
Consider the categorical statements: {{begin-axiom}} {{axiom | q = \map {\mathbf A} {S, P} | lc= The universal affirmative: | ml= \forall x: \map S x | mo= \implies | mr= \map P x }} {{axiom | q = \map {\mathbf E} {S, P} | lc= The universal negative: | ml= \forall x: \map...
=== Necessary Condition === Let $\map {\mathbf A} {S, P}$ and $\map {\mathbf E} {S, P}$ both be false. {{begin-eqn}} {{eqn | n = 1 | l = \neg \map {\mathbf A} {S, P} | o = \land | r = \neg \map {\mathbf E} {S, P} }} {{eqn | n = 2 | ll= \leadsto | l = \neg \map {\mathbf A} {S, P} | o ...
Consider the [[Definition:Categorical Statement|categorical statements]]: {{begin-axiom}} {{axiom | q = \map {\mathbf A} {S, P} | lc= The [[Definition:Universal Affirmative|universal affirmative]]: | ml= \forall x: \map S x | mo= \implies | mr= \map P x }} {{axiom | q = \map {\mathbf E}...
=== Necessary Condition === Let $\map {\mathbf A} {S, P}$ and $\map {\mathbf E} {S, P}$ both be [[Definition:False|false]]. {{begin-eqn}} {{eqn | n = 1 | l = \neg \map {\mathbf A} {S, P} | o = \land | r = \neg \map {\mathbf E} {S, P} }} {{eqn | n = 2 | ll= \leadsto | l = \neg \map {\math...
Universal Affirmative and Negative are both False iff Particular Affirmative and Negative are both True
https://proofwiki.org/wiki/Universal_Affirmative_and_Negative_are_both_False_iff_Particular_Affirmative_and_Negative_are_both_True
https://proofwiki.org/wiki/Universal_Affirmative_and_Negative_are_both_False_iff_Particular_Affirmative_and_Negative_are_both_True
[ "Categorical Statements" ]
[ "Definition:Categorical Statement", "Definition:Universal Affirmative", "Definition:Universal Negative", "Definition:Particular Affirmative", "Definition:Particular Negative", "Definition:False", "Definition:True" ]
[ "Definition:False", "Rule of Simplification", "Universal Affirmative and Particular Negative are Contradictory", "Rule of Simplification", "Particular Affirmative and Universal Negative are Contradictory", "Rule of Conjunction/Proof Rule", "Definition:True", "Definition:True", "Rule of Simplificatio...
proofwiki-8372
Law of Simple Conversion of I
Consider the particular affirmative categorical statement ''Some $S$ is $P$'': :$\map {\mathbf I} {S, P}: \exists x: \map S x \land \map P x$ Then ''Some $P$ is $S$'': :$\map {\mathbf I} {P, S}$
{{begin-eqn}} {{eqn | o = | r = \map {\mathbf I} {S, P} | c = }} {{eqn | ll= \leadsto | o = | q = \exists x | r = \map S x \land \map P x | c = {{Defof|Particular Affirmative}} }} {{eqn | ll= \leadsto | o = | q = \exists x | r = \map P x \land \map S x | c...
Consider the [[Definition:Particular Affirmative|particular affirmative]] [[Definition:Categorical Statement|categorical statement]] ''Some $S$ is $P$'': :$\map {\mathbf I} {S, P}: \exists x: \map S x \land \map P x$ Then ''Some $P$ is $S$'': :$\map {\mathbf I} {P, S}$
{{begin-eqn}} {{eqn | o = | r = \map {\mathbf I} {S, P} | c = }} {{eqn | ll= \leadsto | o = | q = \exists x | r = \map S x \land \map P x | c = {{Defof|Particular Affirmative}} }} {{eqn | ll= \leadsto | o = | q = \exists x | r = \map P x \land \map S x | c...
Law of Simple Conversion of I
https://proofwiki.org/wiki/Law_of_Simple_Conversion_of_I
https://proofwiki.org/wiki/Law_of_Simple_Conversion_of_I
[ "Laws of Conversion", "Particular Affirmative", "Categorical Statements" ]
[ "Definition:Particular Affirmative", "Definition:Categorical Statement" ]
[ "Rule of Commutation/Conjunction" ]
proofwiki-8373
Law of Simple Conversion of E
Consider the universal negative categorical statement ''No $S$ is $P$'': :$\map {\mathbf E} {S, P}: \forall x: \map S x \implies \neg \map P x$ Then ''No $P$ is $S$'': :$\map {\mathbf E} {P, S}$
{{begin-eqn}} {{eqn | o = | r = \map {\mathbf E} {S, P} | c = }} {{eqn | ll= \leadsto | q = \forall x | o = | r = \map S x \implies \neg \map P x | c = {{Defof|Universal Negative}} }} {{eqn | ll= \leadsto | q = \forall x | o = | r = \neg \paren {\map S x \land \...
Consider the [[Definition:Universal Negative|universal negative]] [[Definition:Categorical Statement|categorical statement]] ''No $S$ is $P$'': :$\map {\mathbf E} {S, P}: \forall x: \map S x \implies \neg \map P x$ Then ''No $P$ is $S$'': :$\map {\mathbf E} {P, S}$
{{begin-eqn}} {{eqn | o = | r = \map {\mathbf E} {S, P} | c = }} {{eqn | ll= \leadsto | q = \forall x | o = | r = \map S x \implies \neg \map P x | c = {{Defof|Universal Negative}} }} {{eqn | ll= \leadsto | q = \forall x | o = | r = \neg \paren {\map S x \land \...
Law of Simple Conversion of E
https://proofwiki.org/wiki/Law_of_Simple_Conversion_of_E
https://proofwiki.org/wiki/Law_of_Simple_Conversion_of_E
[ "Laws of Conversion", "Universal Negative", "Categorical Statements" ]
[ "Definition:Universal Negative", "Definition:Categorical Statement" ]
[ "Modus Ponendo Tollens/Variant", "Rule of Commutation/Conjunction", "Modus Ponendo Tollens/Variant" ]
proofwiki-8374
Conversion per Accidens
Consider the categorical statements: {{begin-axiom}} {{axiom | ll = \map {\mathbf A} {S, P}: | lc= The universal affirmative: | q = \forall x | m = \map S x \implies \map P x }} {{axiom | ll= \map {\mathbf I} {P, S}: | lc= The particular affirmative: | q = \exists x | m =...
From Universal Affirmative implies Particular Affirmative iff First Predicate is not Vacuous: :$\exists x: \map S x \iff \paren {\paren {\forall x: \map S x \implies \map P x} \implies \paren {\exists x: \map S x \land \map P x} }$ From Law of Simple Conversion of I: :$\paren {\exists x: \map S x \land \map P x} \impli...
Consider the [[Definition:Categorical Statement|categorical statements]]: {{begin-axiom}} {{axiom | ll = \map {\mathbf A} {S, P}: | lc= The [[Definition:Universal Affirmative|universal affirmative]]: | q = \forall x | m = \map S x \implies \map P x }} {{axiom | ll= \map {\mathbf I} {P, S}: ...
From [[Universal Affirmative implies Particular Affirmative iff First Predicate is not Vacuous]]: :$\exists x: \map S x \iff \paren {\paren {\forall x: \map S x \implies \map P x} \implies \paren {\exists x: \map S x \land \map P x} }$ From [[Law of Simple Conversion of I]]: :$\paren {\exists x: \map S x \land \map P ...
Conversion per Accidens
https://proofwiki.org/wiki/Conversion_per_Accidens
https://proofwiki.org/wiki/Conversion_per_Accidens
[ "Categorical Statements" ]
[ "Definition:Categorical Statement", "Definition:Universal Affirmative", "Definition:Particular Affirmative", "Definition:Symbolic Logic", "Definition:Predicate Logic" ]
[ "Universal Affirmative implies Particular Affirmative iff First Predicate is not Vacuous", "Law of Simple Conversion of I" ]
proofwiki-8375
Number of Standard Instances of Categorical Syllogism
There are $256$ distinct standard instances of the categorical syllogism.
Recall the four figures of the categorical syllogism: :<nowiki>$\begin{array}{r|rl} \text I & & \\ \hline \\ \text{Major Premise}: & \mathbf \Phi_1 & \tuple {M, P} \\ \text{Minor Premise}: & \mathbf \Phi_2 & \tuple {S, M} \\ \hline \\ \text{Conclusion}: & \mathbf \Phi_3 & \tuple {S, P} \\ \end{array} \qquad \begin{arra...
There are $256$ [[Definition:Distinct Elements|distinct]] [[Definition:Standard Instance of Categorical Syllogism|standard instances]] of the [[Definition:Categorical Syllogism|categorical syllogism]].
Recall the four [[Definition:Figure of Categorical Syllogism|figures]] of the [[Definition:Categorical Syllogism|categorical syllogism]]: :<nowiki>$\begin{array}{r|rl} \text I & & \\ \hline \\ \text{Major Premise}: & \mathbf \Phi_1 & \tuple {M, P} \\ \text{Minor Premise}: & \mathbf \Phi_2 & \tuple {S, M} \\ \hline \\ ...
Number of Standard Instances of Categorical Syllogism
https://proofwiki.org/wiki/Number_of_Standard_Instances_of_Categorical_Syllogism
https://proofwiki.org/wiki/Number_of_Standard_Instances_of_Categorical_Syllogism
[ "Categorical Syllogisms", "256" ]
[ "Definition:Distinct/Plural", "Definition:Standard Instance of Categorical Syllogism", "Definition:Categorical Syllogism" ]
[ "Definition:Figure of Categorical Syllogism", "Definition:Categorical Syllogism", "Definition:Categorical Syllogism", "Definition:Categorical Statement", "Definition:Categorical Syllogism/Premises/Major Premise", "Definition:Categorical Syllogism/Premises/Minor Premise", "Definition:Categorical Syllogis...
proofwiki-8376
Valid Patterns of Categorical Syllogism
The following categorical syllogisms are valid: :<nowiki>$\begin{array}{rl} \text{I} & AAA \\ \text{I} & AII \\ \text{I} & EAE \\ \text{I} & EIO \\ * \text{I} & AAI \\ * \text{I} & EAO \\ \end{array} \qquad \begin{array}{rl} \text{II} & EAE \\ \text{II} & AEE \\ \text{II} & AOO \\ \text{II} & EIO \\ * \text{II} & EAO \...
From Elimination of all but 24 Categorical Syllogisms as Invalid, all but these $24$ patterns have been shown to be invalid. It remains to be shown that these remaining syllogisms are in fact valid. {{ProofWanted|Considerable work to be done yet.}}
The following [[Definition:Categorical Syllogism|categorical syllogisms]] are valid: :<nowiki>$\begin{array}{rl} \text{I} & AAA \\ \text{I} & AII \\ \text{I} & EAE \\ \text{I} & EIO \\ * \text{I} & AAI \\ * \text{I} & EAO \\ \end{array} \qquad \begin{array}{rl} \text{II} & EAE \\ \text{II} & AEE \\ \text{II} & AOO \\ ...
From [[Elimination of all but 24 Categorical Syllogisms as Invalid]], all but these $24$ patterns have been shown to be [[Definition:Invalid Argument|invalid]]. It remains to be shown that these remaining syllogisms are in fact [[Definition:Valid Argument|valid]]. {{ProofWanted|Considerable work to be done yet.}}
Valid Patterns of Categorical Syllogism
https://proofwiki.org/wiki/Valid_Patterns_of_Categorical_Syllogism
https://proofwiki.org/wiki/Valid_Patterns_of_Categorical_Syllogism
[ "Categorical Syllogisms" ]
[ "Definition:Categorical Syllogism", "Definition:Figure of Categorical Syllogism", "Definition:Categorical Syllogism", "Definition:Universal Affirmative", "Definition:Universal Negative", "Definition:Particular Affirmative", "Definition:Particular Negative", "Definition:Categorical Syllogism/Shorthand"...
[ "Elimination of all but 24 Categorical Syllogisms as Invalid", "Definition:Invalid Argument", "Definition:Valid Argument" ]
proofwiki-8377
No Valid Categorical Syllogism contains two Particular Premises
Let $Q$ be a valid categorical syllogism. Then at least one of the premises of $Q$ is universal.
Suppose both premises of $Q$ are particular. Then the pattern of $Q$ is one of $\text{II}x$, $\text{IO}x$, $\text{OI}x$ or $\text{OO}x$, where $x$ is the conclusion. $\text{I}$ is neither universal nor negative. Thus the $\text{II}x$ pattern does not distribute the middle term of $Q$. So $\text{II}x$ violates the rule ...
Let $Q$ be a [[Definition:Valid Argument|valid]] [[Definition:Categorical Syllogism|categorical syllogism]]. Then at least one of the [[Definition:Premise of Syllogism|premises]] of $Q$ is [[Definition:Universal Categorical Statement|universal]].
Suppose both [[Definition:Premise of Syllogism|premises]] of $Q$ are [[Definition:Particular Categorical Statement|particular]]. Then the pattern of $Q$ is one of $\text{II}x$, $\text{IO}x$, $\text{OI}x$ or $\text{OO}x$, where $x$ is the [[Definition:Conclusion of Syllogism|conclusion]]. $\text{I}$ is neither [[Defi...
No Valid Categorical Syllogism contains two Particular Premises
https://proofwiki.org/wiki/No_Valid_Categorical_Syllogism_contains_two_Particular_Premises
https://proofwiki.org/wiki/No_Valid_Categorical_Syllogism_contains_two_Particular_Premises
[ "Categorical Syllogisms" ]
[ "Definition:Valid Argument", "Definition:Categorical Syllogism", "Definition:Categorical Syllogism/Premises", "Definition:Universal Categorical Statement" ]
[ "Definition:Categorical Syllogism/Premises", "Definition:Particular Categorical Statement", "Definition:Categorical Syllogism/Conclusion", "Definition:Universal Categorical Statement", "Definition:Negative Categorical Statement", "Definition:Distributed Term of Categorical Syllogism", "Definition:Catego...
proofwiki-8378
No Valid Categorical Syllogism with Particular Premise has Universal Conclusion
Let $Q$ be a valid categorical syllogism. Let one of the premises of $Q$ be particular. Then the conclusion of $Q$ is also particular.
Let the major premise of $Q$ be denoted $\text{Maj}$. Let the minor premise of $Q$ be denoted $\text{Min}$. Let the conclusion of $Q$ be denoted $\text{C}$. From No Valid Categorical Syllogism contains two Particular Premises, either $\text{Maj}$ or $\text{Min}$ has to be universal. Let the other premise of $Q$ be part...
Let $Q$ be a [[Definition:Valid Argument|valid]] [[Definition:Categorical Syllogism|categorical syllogism]]. Let one of the [[Definition:Premise of Syllogism|premises]] of $Q$ be [[Definition:Particular Categorical Statement|particular]]. Then the [[Definition:Conclusion of Syllogism|conclusion]] of $Q$ is also [[De...
Let the [[Definition:Major Premise of Syllogism|major premise]] of $Q$ be denoted $\text{Maj}$. Let the [[Definition:Minor Premise of Syllogism|minor premise]] of $Q$ be denoted $\text{Min}$. Let the [[Definition:Conclusion of Syllogism|conclusion]] of $Q$ be denoted $\text{C}$. From [[No Valid Categorical Syllogis...
No Valid Categorical Syllogism with Particular Premise has Universal Conclusion
https://proofwiki.org/wiki/No_Valid_Categorical_Syllogism_with_Particular_Premise_has_Universal_Conclusion
https://proofwiki.org/wiki/No_Valid_Categorical_Syllogism_with_Particular_Premise_has_Universal_Conclusion
[ "Categorical Syllogisms" ]
[ "Definition:Valid Argument", "Definition:Categorical Syllogism", "Definition:Categorical Syllogism/Premises", "Definition:Particular Categorical Statement", "Definition:Categorical Syllogism/Conclusion", "Definition:Particular Categorical Statement" ]
[ "Definition:Categorical Syllogism/Premises/Major Premise", "Definition:Categorical Syllogism/Premises/Minor Premise", "Definition:Categorical Syllogism/Conclusion", "No Valid Categorical Syllogism contains two Particular Premises", "Definition:Universal Categorical Statement", "Definition:Categorical Syll...
proofwiki-8379
Elimination of all but 48 Categorical Syllogisms as Invalid
Of the $256$ different types of categorical syllogism, all but $48$ can immediately be identified as invalid by consideration of the Rules of Quantity and the Rules of Quality.
There are $64$ patterns of categorical syllogism per figure: :<nowiki>$\begin{array}{cccc} AAA & AAE & AAI & AAO \\ AEA & AEE & AEI & AEO \\ AIA & AIE & AII & AIO \\ AOA & AOE & AOI & AOO \\ \end{array} \qquad \begin{array}{cccc} EAA & EAE & EAI & EAO \\ EEA & EEE & EEI & EEO \\ EIA & EIE & EII & EIO \\ EOA & EOE & EOI...
Of the $256$ different types of [[Definition:Categorical Syllogism|categorical syllogism]], all but $48$ can immediately be identified as [[Definition:Invalid Argument|invalid]] by consideration of the [[Rules of Quantity]] and the [[Rules of Quality]].
There are $64$ patterns of [[Definition:Categorical Syllogism|categorical syllogism]] per [[Definition:Figure of Categorical Syllogism|figure]]: :<nowiki>$\begin{array}{cccc} AAA & AAE & AAI & AAO \\ AEA & AEE & AEI & AEO \\ AIA & AIE & AII & AIO \\ AOA & AOE & AOI & AOO \\ \end{array} \qquad \begin{array}{cccc} EAA &...
Elimination of all but 48 Categorical Syllogisms as Invalid
https://proofwiki.org/wiki/Elimination_of_all_but_48_Categorical_Syllogisms_as_Invalid
https://proofwiki.org/wiki/Elimination_of_all_but_48_Categorical_Syllogisms_as_Invalid
[ "Categorical Syllogisms" ]
[ "Definition:Categorical Syllogism", "Definition:Invalid Argument", "Rules of Quantity", "Rules of Quality" ]
[ "Definition:Categorical Syllogism", "Definition:Figure of Categorical Syllogism", "No Valid Categorical Syllogism contains two Negative Premises", "No Valid Categorical Syllogism contains two Particular Premises", "Conclusion of Valid Categorical Syllogism is Negative iff one Premise is Negative", "No Val...
proofwiki-8380
Conjunction implies Disjunction
:$\vdash \paren {p \land q} \implies \paren {p \lor q}$
{{BeginTableau|\paren {p \land q} \implies \paren {p \lor q} }} {{Assumption|1|p \land q}} {{Simplification|2|1|p|1|1}} {{Addition|3|1|p \lor q|2|1}} {{Implication|4||\paren {p \land q} \implies \paren {p \lor q}|1|3}} {{EndTableau}} {{qed}}
:$\vdash \paren {p \land q} \implies \paren {p \lor q}$
{{BeginTableau|\paren {p \land q} \implies \paren {p \lor q} }} {{Assumption|1|p \land q}} {{Simplification|2|1|p|1|1}} {{Addition|3|1|p \lor q|2|1}} {{Implication|4||\paren {p \land q} \implies \paren {p \lor q}|1|3}} {{EndTableau}} {{qed}}
Conjunction implies Disjunction/Proof 2
https://proofwiki.org/wiki/Conjunction_implies_Disjunction
https://proofwiki.org/wiki/Conjunction_implies_Disjunction/Proof_2
[ "Conjunction", "Disjunction" ]
[]
[]
proofwiki-8381
Conjunction implies Disjunction
:$\vdash \paren {p \land q} \implies \paren {p \lor q}$
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations. :<nowiki>$\begin{array}{|ccc|c|ccc|} \hline (p & \land & q) & \implies & (p & \lor & q) \\ \hline \F & \F & \F & \T & \F & \F & \F \\ \F & \F & \T & \T & \F & \T & \T \\...
:$\vdash \paren {p \land q} \implies \paren {p \lor q}$
We apply the [[Method of Truth Tables]]. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] is [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]]. :<nowiki>$\begin{array}{|...
Conjunction implies Disjunction/Proof by Truth Table
https://proofwiki.org/wiki/Conjunction_implies_Disjunction
https://proofwiki.org/wiki/Conjunction_implies_Disjunction/Proof_by_Truth_Table
[ "Conjunction", "Disjunction" ]
[]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:True", "Definition:Boolean Interpretation" ]
proofwiki-8382
Angle Bisector Vector
Let $\mathbf u$ and $\mathbf v$ be vectors of non-zero length. Let $\norm {\mathbf u}$ and $\norm {\mathbf v}$ be their respective lengths. Then $\norm {\mathbf u} \mathbf v + \norm {\mathbf v} \mathbf u$ is the angle bisector of $\mathbf u$ and $\mathbf v$.
Let $\mathbf a = \norm {\mathbf u} \mathbf v + \norm {\mathbf v} \mathbf u$. Then: {{begin-eqn}} {{eqn | l = \cos \angle \mathbf u, \mathbf a | r = \frac {\mathbf u \cdot \mathbf a} {\norm {\mathbf u} \norm {\mathbf a} } | c = Cosine Formula for Dot Product }} {{eqn | r = \frac {\mathbf u \cdot \paren {\nor...
Let $\mathbf u$ and $\mathbf v$ be [[Definition:Vector (Real Euclidean Space)|vectors]] of non-zero [[Definition:Vector Length|length]]. Let $\norm {\mathbf u}$ and $\norm {\mathbf v}$ be their respective [[Definition:Vector Length|lengths]]. Then $\norm {\mathbf u} \mathbf v + \norm {\mathbf v} \mathbf u$ is the [[...
Let $\mathbf a = \norm {\mathbf u} \mathbf v + \norm {\mathbf v} \mathbf u$. Then: {{begin-eqn}} {{eqn | l = \cos \angle \mathbf u, \mathbf a | r = \frac {\mathbf u \cdot \mathbf a} {\norm {\mathbf u} \norm {\mathbf a} } | c = [[Cosine Formula for Dot Product]] }} {{eqn | r = \frac {\mathbf u \cdot \paren...
Angle Bisector Vector/Algebraic Proof
https://proofwiki.org/wiki/Angle_Bisector_Vector
https://proofwiki.org/wiki/Angle_Bisector_Vector/Algebraic_Proof
[ "Angle Bisector Vector", "Vector Algebra", "Angle Bisectors", "Euclidean Geometry" ]
[ "Definition:Vector/Real Euclidean Space", "Definition:Vector Length", "Definition:Vector Length", "Definition:Angle Bisector" ]
[ "Cosine Formula for Dot Product", "Dot Product Associates with Scalar Multiplication", "Dot Product of Vector with Itself", "Cosine Formula for Dot Product", "Dot Product Associates with Scalar Multiplication", "Dot Product of Vector with Itself", "Definition:Cosine", "Definition:Injection", "Shape ...
proofwiki-8383
Angle Bisector Vector
Let $\mathbf u$ and $\mathbf v$ be vectors of non-zero length. Let $\norm {\mathbf u}$ and $\norm {\mathbf v}$ be their respective lengths. Then $\norm {\mathbf u} \mathbf v + \norm {\mathbf v} \mathbf u$ is the angle bisector of $\mathbf u$ and $\mathbf v$.
:400px As shown above: :Let $\gamma$ be the angle between $\mathbf u$ and $\mathbf v$. :Let $\alpha$ be the angle between $\norm {\mathbf u} \mathbf v$ and $\norm {\mathbf v} \mathbf u$. :Let $\beta$ be the angle between $\mathbf u$ and $\norm {\mathbf u} \mathbf v + \norm {\mathbf v} \mathbf u$. Note that $\norm {\mat...
Let $\mathbf u$ and $\mathbf v$ be [[Definition:Vector (Real Euclidean Space)|vectors]] of non-zero [[Definition:Vector Length|length]]. Let $\norm {\mathbf u}$ and $\norm {\mathbf v}$ be their respective [[Definition:Vector Length|lengths]]. Then $\norm {\mathbf u} \mathbf v + \norm {\mathbf v} \mathbf u$ is the [[...
:[[File:Angular Bisector Vector Diagram.png|400px]] As shown above: :Let $\gamma$ be the [[Definition:Angle|angle]] between $\mathbf u$ and $\mathbf v$. :Let $\alpha$ be the angle between $\norm {\mathbf u} \mathbf v$ and $\norm {\mathbf v} \mathbf u$. :Let $\beta$ be the angle between $\mathbf u$ and $\norm {\math...
Angle Bisector Vector/Geometric Proof 1
https://proofwiki.org/wiki/Angle_Bisector_Vector
https://proofwiki.org/wiki/Angle_Bisector_Vector/Geometric_Proof_1
[ "Angle Bisector Vector", "Vector Algebra", "Angle Bisectors", "Euclidean Geometry" ]
[ "Definition:Vector/Real Euclidean Space", "Definition:Vector Length", "Definition:Vector Length", "Definition:Angle Bisector" ]
[ "File:Angular Bisector Vector Diagram.png", "Definition:Angle", "Vector Times Magnitude Same Length As Magnitude Times Vector", "Definition:Triangle (Geometry)/Isosceles", "Definition:Parallel Vectors", "Parallelism implies Equal Corresponding Angles" ]
proofwiki-8384
Angle Bisector Vector
Let $\mathbf u$ and $\mathbf v$ be vectors of non-zero length. Let $\norm {\mathbf u}$ and $\norm {\mathbf v}$ be their respective lengths. Then $\norm {\mathbf u} \mathbf v + \norm {\mathbf v} \mathbf u$ is the angle bisector of $\mathbf u$ and $\mathbf v$.
The vectors $\norm {\mathbf u} \mathbf v$ and $\norm {\mathbf v} \mathbf u$ have equal length from Vector Times Magnitude Same Length As Magnitude Times Vector. Thus $\norm {\mathbf u} \mathbf v + \norm {\mathbf v} \mathbf u$ is the diagonal of a rhombus. The result follows from Diagonals of Rhombus Bisect Angles. {{qe...
Let $\mathbf u$ and $\mathbf v$ be [[Definition:Vector (Real Euclidean Space)|vectors]] of non-zero [[Definition:Vector Length|length]]. Let $\norm {\mathbf u}$ and $\norm {\mathbf v}$ be their respective [[Definition:Vector Length|lengths]]. Then $\norm {\mathbf u} \mathbf v + \norm {\mathbf v} \mathbf u$ is the [[...
The vectors $\norm {\mathbf u} \mathbf v$ and $\norm {\mathbf v} \mathbf u$ have equal length from [[Vector Times Magnitude Same Length As Magnitude Times Vector]]. Thus $\norm {\mathbf u} \mathbf v + \norm {\mathbf v} \mathbf u$ is the diagonal of a [[Definition:Rhombus|rhombus]]. The result follows from [[Diagona...
Angle Bisector Vector/Geometric Proof 2
https://proofwiki.org/wiki/Angle_Bisector_Vector
https://proofwiki.org/wiki/Angle_Bisector_Vector/Geometric_Proof_2
[ "Angle Bisector Vector", "Vector Algebra", "Angle Bisectors", "Euclidean Geometry" ]
[ "Definition:Vector/Real Euclidean Space", "Definition:Vector Length", "Definition:Vector Length", "Definition:Angle Bisector" ]
[ "Vector Times Magnitude Same Length As Magnitude Times Vector", "Definition:Quadrilateral/Rhombus", "Diagonals of Rhombus Bisect Angles" ]
proofwiki-8385
Subgroup of Order 1 is Trivial
Let $\struct {G, \circ}$ be a group. Then $\struct {G, \circ}$ has exactly $1$ subgroup of order $1$: the trivial subgroup.
From Trivial Subgroup is Subgroup, $\struct {\set e, \circ}$ is a subgroup of $\struct {G, \circ}$. Suppose $\struct {\set g, \circ}$ is a subgroup of $\struct {G, \circ}$. From Group is not Empty, $e \in \set g$. Thus it follows trivially that $\struct {\set g, \circ} = \struct {\set e, \circ}$. That is, $\struct {\se...
Let $\struct {G, \circ}$ be a [[Definition:Group|group]]. Then $\struct {G, \circ}$ has exactly $1$ [[Definition:Subgroup|subgroup]] of [[Definition:Order of Structure|order]] $1$: the [[Definition:Trivial Subgroup|trivial subgroup]].
From [[Trivial Subgroup is Subgroup]], $\struct {\set e, \circ}$ is a [[Definition:Subgroup|subgroup]] of $\struct {G, \circ}$. Suppose $\struct {\set g, \circ}$ is a [[Definition:Subgroup|subgroup]] of $\struct {G, \circ}$. From [[Group is not Empty]], $e \in \set g$. Thus it follows trivially that $\struct {\set ...
Subgroup of Order 1 is Trivial
https://proofwiki.org/wiki/Subgroup_of_Order_1_is_Trivial
https://proofwiki.org/wiki/Subgroup_of_Order_1_is_Trivial
[ "Subgroups" ]
[ "Definition:Group", "Definition:Subgroup", "Definition:Order of Structure", "Definition:Trivial Subgroup" ]
[ "Trivial Subgroup is Subgroup", "Definition:Subgroup", "Definition:Subgroup", "Group is not Empty", "Definition:Subgroup", "Definition:Order of Structure" ]
proofwiki-8386
Injection from Finite Set to Itself is Surjection/Corollary
Let $S$ be a finite set. Let $f: S \to S$ be an injection. Then $f$ is a permutation.
From Injection from Finite Set to Itself is Surjection, $f$ is a surjection. As $f$ is thus both an injection and a surjection, $f$ is a bijection by definition. Thus as $f$ is a bijection to itself, it is by definition a permutation. {{qed}}
Let $S$ be a [[Definition:Finite Set|finite set]]. Let $f: S \to S$ be an [[Definition:Injection|injection]]. Then $f$ is a [[Definition:Permutation|permutation]].
From [[Injection from Finite Set to Itself is Surjection]], $f$ is a [[Definition:Surjection|surjection]]. As $f$ is thus both an [[Definition:Injection|injection]] and a [[Definition:Surjection|surjection]], $f$ is a [[Definition:Bijection|bijection]] by definition. Thus as $f$ is a [[Definition:Bijection|bijection]...
Injection from Finite Set to Itself is Surjection/Corollary
https://proofwiki.org/wiki/Injection_from_Finite_Set_to_Itself_is_Surjection/Corollary
https://proofwiki.org/wiki/Injection_from_Finite_Set_to_Itself_is_Surjection/Corollary
[ "Injections", "Permutations" ]
[ "Definition:Finite Set", "Definition:Injection", "Definition:Permutation" ]
[ "Injection from Finite Set to Itself is Surjection", "Definition:Surjection", "Definition:Injection", "Definition:Surjection", "Definition:Bijection", "Definition:Bijection", "Definition:Permutation" ]
proofwiki-8387
Surjection from Finite Set to Itself is Permutation
Let $S$ be a finite set. Let $f: S \to S$ be an surjection. Then $f$ is a permutation.
From Surjection iff Right Inverse, $f$ has a right inverse $g: S \to S$. From Right Inverse Mapping is Injection, $g$ is an injection. From Injection from Finite Set to Itself is Permutation, $g$ is a permutation and so a bijection. From Inverse of Bijection is Bijection, $f$ is also a bijection. Thus as $f$ is a bijec...
Let $S$ be a [[Definition:Finite Set|finite set]]. Let $f: S \to S$ be an [[Definition:Surjection|surjection]]. Then $f$ is a [[Definition:Permutation|permutation]].
From [[Surjection iff Right Inverse]], $f$ has a [[Definition:Right Inverse Mapping|right inverse]] $g: S \to S$. From [[Right Inverse Mapping is Injection]], $g$ is an [[Definition:Injection|injection]]. From [[Injection from Finite Set to Itself is Permutation]], $g$ is a [[Definition:Permutation|permutation]] and ...
Surjection from Finite Set to Itself is Permutation
https://proofwiki.org/wiki/Surjection_from_Finite_Set_to_Itself_is_Permutation
https://proofwiki.org/wiki/Surjection_from_Finite_Set_to_Itself_is_Permutation
[ "Surjections", "Permutations" ]
[ "Definition:Finite Set", "Definition:Surjection", "Definition:Permutation" ]
[ "Surjection iff Right Inverse", "Definition:Right Inverse Mapping", "Right Inverse Mapping is Injection", "Definition:Injection", "Injection from Finite Set to Itself is Surjection/Corollary", "Definition:Permutation", "Definition:Bijection", "Inverse of Bijection is Bijection", "Definition:Bijectio...
proofwiki-8388
Order of Power of Group Element
Let $\struct {G, \circ}$ be a group whose identity is $e$. Let $g \in G$ be an element of $G$ such that: :$\order g = n$ where $\order g$ denotes the order of $g$. Then: :$\forall m \in \Z: \order {g^m} = \dfrac n {\gcd \set {m, n} }$ where $\gcd \set {m, n}$ denotes the greatest common divisor of $m$ and $n$.
Let $\gcd \set {m, n} = d$. From Integers Divided by GCD are Coprime: there exists $m', n' \in \Z$ such that $m = d m'$, $n = d n'$. Then: {{begin-eqn}} {{eqn | l = \paren {g^m}^{n'} | r = \paren {g^{d m'} }^{n'} | c = Definition of $m'$ }} {{eqn | r = \paren {g^{d n'} }^{m'} | c = }} {{eqn | r = \pa...
Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$. Let $g \in G$ be an [[Definition:Element|element]] of $G$ such that: :$\order g = n$ where $\order g$ denotes the [[Definition:Order of Group Element|order]] of $g$. Then: :$\forall m \in \Z: \order {g^m} =...
Let $\gcd \set {m, n} = d$. From [[Integers Divided by GCD are Coprime]]: there exists $m', n' \in \Z$ such that $m = d m'$, $n = d n'$. Then: {{begin-eqn}} {{eqn | l = \paren {g^m}^{n'} | r = \paren {g^{d m'} }^{n'} | c = Definition of $m'$ }} {{eqn | r = \paren {g^{d n'} }^{m'} | c = }} {{eqn | ...
Order of Power of Group Element
https://proofwiki.org/wiki/Order_of_Power_of_Group_Element
https://proofwiki.org/wiki/Order_of_Power_of_Group_Element
[ "Order of Group Elements" ]
[ "Definition:Group", "Definition:Identity (Abstract Algebra)/Two-Sided Identity", "Definition:Element", "Definition:Order of Group Element", "Definition:Greatest Common Divisor/Integers" ]
[ "Integers Divided by GCD are Coprime", "Definition:Order of Group Element", "Element to Power of Multiple of Order is Identity", "Bézout's Identity", "Definition:Order of Group Element", "Definition:Order of Group Element", "Definition:Contradiction", "Definition:Order of Group Element" ]
proofwiki-8389
Existence of Group of Finite Order
Let $n \in \Z_{>0}$. Then there exists at least one group whose order is $n$.
From Existence of Cyclic Group of Order n, there exists a cyclic group whose order is $n$. In particular, a concrete example of such a group is demonstrated in Roots of Unity under Multiplication form Cyclic Group. {{qed}}
Let $n \in \Z_{>0}$. Then there exists at least one [[Definition:Group|group]] whose [[Definition:Order of Structure|order]] is $n$.
From [[Existence of Cyclic Group of Order n]], there exists a [[Definition:Cyclic Group|cyclic group]] whose [[Definition:Order of Structure|order]] is $n$. In particular, a concrete example of such a [[Definition:Group|group]] is demonstrated in [[Roots of Unity under Multiplication form Cyclic Group]]. {{qed}}
Existence of Group of Finite Order
https://proofwiki.org/wiki/Existence_of_Group_of_Finite_Order
https://proofwiki.org/wiki/Existence_of_Group_of_Finite_Order
[ "Order of Groups" ]
[ "Definition:Group", "Definition:Order of Structure" ]
[ "Existence of Cyclic Group of Order n", "Definition:Cyclic Group", "Definition:Order of Structure", "Definition:Group", "Roots of Unity under Multiplication form Cyclic Group" ]
proofwiki-8390
Vitali Set Existence Theorem
There exists a set of real numbers which is not Lebesgue measurable.
=== Lemma === {{:Vitali Set Existence Theorem/Lemma}}{{qed|lemma}} Let $\map \mu X$ denote the Lebesgue measure of a set $X$ of real numbers. We have that: {{begin-itemize}} {{item|(1):|$\map \mu X$ is a countably additive function}} {{item|(2):|$\map \mu X$ is translation invariant}} {{item|(3):|From Measure of Interv...
There exists a [[Definition:Set|set]] of [[Definition:Real Number|real numbers]] which is not [[Definition:Lebesgue Measure|Lebesgue measurable]].
=== [[Vitali Set Existence Theorem/Lemma|Lemma]] === {{:Vitali Set Existence Theorem/Lemma}}{{qed|lemma}} Let $\map \mu X$ denote the [[Definition:Lebesgue Measure|Lebesgue measure]] of a [[Definition:Set|set]] $X$ of [[Definition:Real Number|real numbers]]. We have that: {{begin-itemize}} {{item|(1):|[[Lebesgue Mea...
Vitali Set Existence Theorem/Proof 1
https://proofwiki.org/wiki/Vitali_Set_Existence_Theorem
https://proofwiki.org/wiki/Vitali_Set_Existence_Theorem/Proof_1
[ "Vitali Set Existence Theorem", "Vitali Sets", "Non-Measurable Sets", "Measure Theory" ]
[ "Definition:Set", "Definition:Real Number", "Definition:Lebesgue Measure" ]
[ "Vitali Set Existence Theorem/Lemma", "Definition:Lebesgue Measure", "Definition:Set", "Definition:Real Number", "Lebesgue Measure is Countably Additive", "Lebesgue Measure is Invariant under Translations", "Measure of Interval is Length", "Definition:Real Interval/Closed", "Definition:Real Number",...
proofwiki-8391
Vitali Set Existence Theorem
There exists a set of real numbers which is not Lebesgue measurable.
We construct such a set. For $x, y \in \hointr 0 1$, define the sum modulo 1: :$x +_1 y = \begin {cases} x + y & : x + y < 1 \\ x + y - 1 & : x + y \ge 1 \end {cases}$ Let $E \subset \hointr 0 1$ be a measurable set. Let $E_1 = E \cap \hointr 0 {1 - x}$ and $E_2 = E \cap \hointr {1 - x} 1$. By Measure of Interval is Le...
There exists a [[Definition:Set|set]] of [[Definition:Real Number|real numbers]] which is not [[Definition:Lebesgue Measure|Lebesgue measurable]].
We construct such a [[Definition:Set|set]]. For $x, y \in \hointr 0 1$, define the [[Definition:Modulo Addition|sum modulo 1]]: :$x +_1 y = \begin {cases} x + y & : x + y < 1 \\ x + y - 1 & : x + y \ge 1 \end {cases}$ Let $E \subset \hointr 0 1$ be a [[Definition:Measurable Set|measurable set]]. Let $E_1 = E \cap \...
Vitali Set Existence Theorem/Proof 2
https://proofwiki.org/wiki/Vitali_Set_Existence_Theorem
https://proofwiki.org/wiki/Vitali_Set_Existence_Theorem/Proof_2
[ "Vitali Set Existence Theorem", "Vitali Sets", "Non-Measurable Sets", "Measure Theory" ]
[ "Definition:Set", "Definition:Real Number", "Definition:Lebesgue Measure" ]
[ "Definition:Set", "Definition:Modulo Addition", "Definition:Measurable Set", "Measure of Interval is Length", "Definition:Disjoint Sets", "Definition:Real Interval", "Definition:Measurable Set", "Measurable Sets form Algebra of Sets", "Definition:Set Intersection", "Lebesgue Measure is Invariant u...
proofwiki-8392
Finite Number of Groups of Given Finite Order
Let $n \in \Z_{>0}$ be a (strictly) positive integer. Then there exists a finite number of types of group of order $n$.
For any group $\struct {G, \circ}$ of order $n$ and any set of $n$ elements, $X$ can be the underlying set of a group which is isomorphic to $\struct {G, \circ}$, as follows: Choose a bijection $\phi: G \to X$. Define the group operation $*$ on $X$ by the rule: :$\map \phi {g_1} * \map \phi {g_2} = \map \phi {g_1 \circ...
Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]]. Then there exists a [[Definition:Finite|finite number]] of [[Definition:Group Type|types of group]] of [[Definition:Order of Structure|order $n$]].
For any [[Definition:Group|group]] $\struct {G, \circ}$ of [[Definition:Order of Structure|order $n$]] and any [[Definition:Set|set]] of $n$ [[Definition:Element|elements]], $X$ can be the [[Definition:Underlying Set of Structure|underlying set]] of a [[Definition:Group|group]] which is [[Definition:Group Isomorphism|i...
Finite Number of Groups of Given Finite Order
https://proofwiki.org/wiki/Finite_Number_of_Groups_of_Given_Finite_Order
https://proofwiki.org/wiki/Finite_Number_of_Groups_of_Given_Finite_Order
[ "Order of Groups" ]
[ "Definition:Strictly Positive/Integer", "Definition:Finite", "Definition:Group Type", "Definition:Order of Structure" ]
[ "Definition:Group", "Definition:Order of Structure", "Definition:Set", "Definition:Element", "Definition:Underlying Set/Abstract Algebra", "Definition:Group", "Definition:Isomorphism (Abstract Algebra)/Group Isomorphism", "Definition:Bijection", "Definition:Group Product/Group Law", "Definition:Is...
proofwiki-8393
Nu of Prime Number is 1
Let $p$ be a prime number. Then: :$\map \nu p = 1$ where $\nu$ denotes the $\nu$ function: the number of types of group of a given order.
Let $G_1$ and $G_2$ be groups of order $p$. From Prime Group is Cyclic, $G_1$ and $G_2$ are both cyclic groups. From Cyclic Groups of Same Order are Isomorphic, $G_1$ and $G_2$ are isomorphic. Thus by definition, $G_1$ and $G_2$ are of the same type. Hence the result. {{qed}}
Let $p$ be a [[Definition:Prime Number|prime number]]. Then: :$\map \nu p = 1$ where $\nu$ denotes the [[Definition:Nu Function|$\nu$ function]]: the number of [[Definition:Group Type|types of group]] of a given [[Definition:Order of Group|order]].
Let $G_1$ and $G_2$ be [[Definition:Group|groups]] of [[Definition:Order of Group|order]] $p$. From [[Prime Group is Cyclic]], $G_1$ and $G_2$ are both [[Definition:Cyclic Group|cyclic groups]]. From [[Cyclic Groups of Same Order are Isomorphic]], $G_1$ and $G_2$ are [[Definition:Group Isomorphism|isomorphic]]. Thus...
Nu of Prime Number is 1
https://proofwiki.org/wiki/Nu_of_Prime_Number_is_1
https://proofwiki.org/wiki/Nu_of_Prime_Number_is_1
[ "Prime Groups", "Nu Function" ]
[ "Definition:Prime Number", "Definition:Nu Function", "Definition:Group Type", "Definition:Order of Structure" ]
[ "Definition:Group", "Definition:Order of Structure", "Prime Group is Cyclic", "Definition:Cyclic Group", "Cyclic Groups of Same Order are Isomorphic", "Definition:Isomorphism (Abstract Algebra)/Group Isomorphism", "Definition:Group Type" ]
proofwiki-8394
Image of Subset under Relation equals Union of Images of Elements
Let $S$ and $T$ be sets. Let $\RR \subseteq S \times T$ be a relation on $S \times T$. Let $X \subseteq S$ be a subset of $S$. Then: :$\ds \RR \sqbrk X = \bigcup_{x \mathop \in X} \map \RR x$ where: :$\RR \sqbrk X$ is the image of the subset $X$ under $\RR$ :$\map \RR x$ is the image of the element $x$ under $\RR$.
By definition: :$\RR \sqbrk X = \set {y \in T: \exists x \in X: \tuple {x, y} \in \RR}$ :$\map \RR x = \set {y \in T:\tuple {x, y} \in \RR}$ First: {{begin-eqn}} {{eqn | l = y | o = \in | r = \RR \sqbrk X }} {{eqn | ll= \leadsto | q = \exists x \in X | l = \tuple {x, y} | o = \in | ...
Let $S$ and $T$ be [[Definition:Set|sets]]. Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]] on $S \times T$. Let $X \subseteq S$ be a [[Definition:Subset|subset]] of $S$. Then: :$\ds \RR \sqbrk X = \bigcup_{x \mathop \in X} \map \RR x$ where: :$\RR \sqbrk X$ is the [[Definition:Image of Subset ...
By definition: :$\RR \sqbrk X = \set {y \in T: \exists x \in X: \tuple {x, y} \in \RR}$ :$\map \RR x = \set {y \in T:\tuple {x, y} \in \RR}$ First: {{begin-eqn}} {{eqn | l = y | o = \in | r = \RR \sqbrk X }} {{eqn | ll= \leadsto | q = \exists x \in X | l = \tuple {x, y} | o = \in ...
Image of Subset under Relation equals Union of Images of Elements
https://proofwiki.org/wiki/Image_of_Subset_under_Relation_equals_Union_of_Images_of_Elements
https://proofwiki.org/wiki/Image_of_Subset_under_Relation_equals_Union_of_Images_of_Elements
[ "Relation Theory", "Subsets", "Set Union" ]
[ "Definition:Set", "Definition:Relation", "Definition:Subset", "Definition:Image (Set Theory)/Relation/Subset", "Definition:Image (Set Theory)/Relation/Element" ]
[ "Definition:Set Equality/Definition 2", "Category:Relation Theory", "Category:Subsets", "Category:Set Union" ]
proofwiki-8395
Preimage of Subset under Relation equals Union of Preimages of Elements
Let $S$ and $T$ be sets. Let $\RR \subseteq S \times T$ be a relation on $S \times T$. Let $\RR^{-1} \subseteq T \times S$ be the inverse relation to $\RR$ Let $Y \subseteq T$ be a subset of $T$. Then: :$\RR^{-1} \sqbrk Y = \ds \bigcup_{y \mathop \in Y} \map {\RR^{-1} } y$ where: :$\RR^{-1} \sqbrk Y$ is the preimage of...
By definition, $\RR^{-1} \subseteq T \times S$ is a relation on $T \times S$. Thus Image of Subset under Relation equals Union of Images of Elements can be applied directly. {{qed}} Category:Relation Theory h2kcezt8nswgbhvmdszqs4q0igjmyeo
Let $S$ and $T$ be [[Definition:Set|sets]]. Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]] on $S \times T$. Let $\RR^{-1} \subseteq T \times S$ be the [[Definition:Inverse Relation|inverse relation]] to $\RR$ Let $Y \subseteq T$ be a [[Definition:Subset|subset]] of $T$. Then: :$\RR^{-1} \sqb...
By definition, $\RR^{-1} \subseteq T \times S$ is a [[Definition:Relation|relation]] on $T \times S$. Thus [[Image of Subset under Relation equals Union of Images of Elements]] can be applied directly. {{qed}} [[Category:Relation Theory]] h2kcezt8nswgbhvmdszqs4q0igjmyeo
Preimage of Subset under Relation equals Union of Preimages of Elements
https://proofwiki.org/wiki/Preimage_of_Subset_under_Relation_equals_Union_of_Preimages_of_Elements
https://proofwiki.org/wiki/Preimage_of_Subset_under_Relation_equals_Union_of_Preimages_of_Elements
[ "Relation Theory" ]
[ "Definition:Set", "Definition:Relation", "Definition:Inverse Relation", "Definition:Subset", "Definition:Preimage/Relation/Subset", "Definition:Subset", "Definition:Preimage/Relation/Element", "Definition:Element" ]
[ "Definition:Relation", "Image of Subset under Relation equals Union of Images of Elements", "Category:Relation Theory" ]
proofwiki-8396
Preimage of Subset under Mapping equals Union of Preimages of Elements
Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping from $S$ to $T$. Let $f^{-1} \subseteq T \times S$ be the inverse of $f$, defined as: :$f^{-1} = \set {\tuple {t, s}: \map f s = t}$ Let $Y \subseteq T$ be a subset of $T$. Then: :$\ds f^{-1} \sqbrk Y = \bigcup_{y \mathop \in Y} \map {f^{-1} } y$ where: :$f^{-1} \s...
By definition, $f^{-1} \subseteq T \times S$ is a relation on $T \times S$. Thus Image of Subset under Relation equals Union of Images of Elements can be applied directly. {{qed}} Category:Preimages under Mappings Category:Subsets Category:Set Union 3dwfblsxgav2kxtnvq5h5v003e9y11i
Let $S$ and $T$ be [[Definition:Set|sets]]. Let $f: S \to T$ be a [[Definition:Mapping|mapping]] from $S$ to $T$. Let $f^{-1} \subseteq T \times S$ be the [[Definition:Inverse of Mapping|inverse]] of $f$, defined as: :$f^{-1} = \set {\tuple {t, s}: \map f s = t}$ Let $Y \subseteq T$ be a [[Definition:Subset|subset...
By definition, $f^{-1} \subseteq T \times S$ is a [[Definition:Relation|relation]] on $T \times S$. Thus [[Image of Subset under Relation equals Union of Images of Elements]] can be applied directly. {{qed}} [[Category:Preimages under Mappings]] [[Category:Subsets]] [[Category:Set Union]] 3dwfblsxgav2kxtnvq5h5v003e9y...
Preimage of Subset under Mapping equals Union of Preimages of Elements
https://proofwiki.org/wiki/Preimage_of_Subset_under_Mapping_equals_Union_of_Preimages_of_Elements
https://proofwiki.org/wiki/Preimage_of_Subset_under_Mapping_equals_Union_of_Preimages_of_Elements
[ "Preimages under Mappings", "Subsets", "Set Union" ]
[ "Definition:Set", "Definition:Mapping", "Definition:Inverse of Mapping", "Definition:Subset", "Definition:Preimage/Mapping/Subset", "Definition:Preimage/Mapping/Element" ]
[ "Definition:Relation", "Image of Subset under Relation equals Union of Images of Elements", "Category:Preimages under Mappings", "Category:Subsets", "Category:Set Union" ]
proofwiki-8397
Image of Domain of Mapping is Image Set
Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping. The image of $S$ is the image set of $f$: :$f \sqbrk S = \Img f$
By definition, a mapping is a relation. Thus Image of Domain of Relation is Image Set applies. {{qed}}
Let $S$ and $T$ be [[Definition:Set|sets]]. Let $f: S \to T$ be a [[Definition:Mapping|mapping]]. The [[Definition:Image of Subset under Mapping|image of $S$]] is the [[Definition:Image of Mapping|image set of $f$]]: :$f \sqbrk S = \Img f$
By definition, a [[Definition:Mapping|mapping]] is a [[Definition:Relation|relation]]. Thus [[Image of Domain of Relation is Image Set]] applies. {{qed}}
Image of Domain of Mapping is Image Set
https://proofwiki.org/wiki/Image_of_Domain_of_Mapping_is_Image_Set
https://proofwiki.org/wiki/Image_of_Domain_of_Mapping_is_Image_Set
[ "Images" ]
[ "Definition:Set", "Definition:Mapping", "Definition:Image (Set Theory)/Mapping/Subset", "Definition:Image (Set Theory)/Mapping/Mapping" ]
[ "Definition:Mapping", "Definition:Relation", "Image of Domain of Relation is Image Set" ]
proofwiki-8398
Image of Singleton under Mapping
Let $f: S \to T$ be a mapping. Then the image of an element of $S$ is equal to the image of a singleton containing that element, the singleton being a subset of $S$: :$\forall s \in S: \set {\map f s} = f \sqbrk {\set s}$
By definition, a mapping is a relation. Thus Image of Singleton under Relation applies. {{Qed}}
Let $f: S \to T$ be a [[Definition:Mapping|mapping]]. Then the [[Definition:Image of Element under Mapping|image]] of an [[Definition:Element|element]] of $S$ is equal to the [[Definition:Image of Subset under Mapping|image]] of a [[Definition:Singleton|singleton]] containing that element, the singleton being a [[Defi...
By definition, a [[Definition:Mapping|mapping]] is a [[Definition:Relation|relation]]. Thus [[Image of Singleton under Relation]] applies. {{Qed}}
Image of Singleton under Mapping
https://proofwiki.org/wiki/Image_of_Singleton_under_Mapping
https://proofwiki.org/wiki/Image_of_Singleton_under_Mapping
[ "Images", "Singletons" ]
[ "Definition:Mapping", "Definition:Image (Set Theory)/Mapping/Element", "Definition:Element", "Definition:Image (Set Theory)/Mapping/Subset", "Definition:Singleton", "Definition:Subset" ]
[ "Definition:Mapping", "Definition:Relation", "Image of Singleton under Relation" ]
proofwiki-8399
Image of Subset under Mapping equals Union of Images of Elements
Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping from $S$ to $T$. Let $X \subseteq S$ be a subset of $S$. Then: :$\ds f \sqbrk X = \bigcup_{x \mathop \in X} \map f x$ where: :$f \sqbrk X$ is the image of the subset $X$ under $f$ :$\map f x$ is the image of the element $x$ under $f$.
By definition, a mapping is a relation. Thus Image of Subset under Relation equals Union of Images of Elements applies. {{Qed}}
Let $S$ and $T$ be [[Definition:Set|sets]]. Let $f: S \to T$ be a [[Definition:Mapping|mapping]] from $S$ to $T$. Let $X \subseteq S$ be a [[Definition:Subset|subset]] of $S$. Then: :$\ds f \sqbrk X = \bigcup_{x \mathop \in X} \map f x$ where: :$f \sqbrk X$ is the [[Definition:Image of Subset under Mapping|image of...
By definition, a [[Definition:Mapping|mapping]] is a [[Definition:Relation|relation]]. Thus [[Image of Subset under Relation equals Union of Images of Elements]] applies. {{Qed}}
Image of Subset under Mapping equals Union of Images of Elements
https://proofwiki.org/wiki/Image_of_Subset_under_Mapping_equals_Union_of_Images_of_Elements
https://proofwiki.org/wiki/Image_of_Subset_under_Mapping_equals_Union_of_Images_of_Elements
[ "Images", "Set Union" ]
[ "Definition:Set", "Definition:Mapping", "Definition:Subset", "Definition:Image (Set Theory)/Mapping/Subset", "Definition:Image (Set Theory)/Mapping/Element" ]
[ "Definition:Mapping", "Definition:Relation", "Image of Subset under Relation equals Union of Images of Elements" ]