Gelfond-Schneider Theorem/Lemma 4

Gelfond-Schneider Theorem/Lemma 4 0 12002 490684 349511 2020-09-25T15:47:45Z Prime.mover 59 wikitext text/x-wiki == Lemma == <onlyinclude> Let: :$\Delta = \det \sqbrk {\alpha_{i, j} }_{L \times L}$ where the $\alpha_{i, j}$ are algebraic numbers. Suppose that $T$ is a positive (rational) integer for which $T \alpha_{i, j}$ is an algebraic integer for every $i, j \in \set {1, 2, \ldots, L}$. Also, suppose that $\Delta \ne 0$. Then there is a conjugate of $\Delta$ with absolute value $\ge T^{−L}$. </onlyinclude> == Proof == Observe that $T^L \Delta$ is an algebraic integer so that one of its conjugates has absolute value $\ge 1$. The result follows. {{qed}} {{Proofread|There are plenty of definitions on this page which we may not have completely accurate.}} Category:Transcendental Number Theory mjaqjter8d19jybn9fgakn1m2tff5bt


Frobenius's Theorem/Lemma 1

Frobenius's Theorem/Lemma 1 0 12012 489385 146286 2020-09-20T10:35:06Z Prime.mover 59 wikitext text/x-wiki == Lemma == <onlyinclude> Let $\struct {A, \oplus}$ be a quadratic real algebra. Then: :$(1): \quad U = \set {u \in A \setminus \R: u^2 \in \R} \cup \set 0$ is a linear subspace of $A$ :$(2): \quad \forall u, v \in U: u v + v u \in \R$ :$(3): \quad A = \R \oplus U$ :$(4): \quad$ If $A$ is also a division algebra, then every nonzero $u \in U$ can be written as $u = \alpha v$ with $\alpha \in \R$ and $v^2 = -1$. </onlyinclude> == Proof == === Proof of First Assertion === $U$ is closed under scalar multiplication. We have to show that $u, v \in U$ implies $u + v \in U$. Let $\set {u, v, 1}$ be a linearly dependent set. Then there exist constants $a, b \in \R$ such that $u = av + b$. Then $u^2 = a^2 v^2 + 2 a b v + b^2$. Since $u^2$, $a^2 v^2$, and $b^2$ are all real, it follows that $2 a b v \in \R$, that is, $v \in \R$. Since $0$ is the only real element of $U$, it follows that $v = 0$. Reversing $u$ and $v$ in the preceding argument shows that also $u = 0$, so that $u + v = 0 \in U$. Now let $\set {u, v, 1}$ be a linearly independent set. We have: :$\paren {u + v}^2 + \paren {u - v}^2 = 2 u^2 + 2 v^2 \in \R$ On the other hand, as $A$ is quadratic there exist $\lambda, \mu \in \R$ such that: :$\paren {u + v}^2 - \lambda \paren {u + v} \in \R$ :$\paren {u - v}^2 - \mu \paren {u - v} \in R$ Hence: :$\lambda \paren {u + v} + \mu \paren {u - v} \in \R$ However, we have that $\set {u, v, 1}$ is linearly independent. Therefore $\lambda + \mu = \lambda - \mu = 0$, and so $\lambda = \mu = 0$. This proves that $u \pm v \in U$. Thus $U$ is indeed a subspace of $A$. {{qed|lemma}} === Proof of Second Assertion === Accordingly, from the result of $(1)$: :$\forall u, v \in U: u v + v u = \paren {u + v}^2 - u^2 - v^2 \in \R$. {{qed|lemma}} === Proof of Third Assertion === Let $a \in A \setminus \R$. Then: :$\exists \nu \in \R: a^2 - \nu a \in \R$ Therefore, if we set :$u = a - \dfrac \nu 2 \in U$ then $u^2 = a^2 - \nu a + \nu^2/4 \in \R$, so :$a = \dfrac \nu 2 + u \in \R \oplus U$ which proves the assertion. {{qed|lemma}} === Proof of Fourth Assertion === This follows directly from the definition of division algebra. Also, we have that $u^2 \in \R$. Suppose $u^2 = 0$. Since $u$ is nonzero and since $A$ is a division algebra, there exists an inverse $u^{-1}$ such that $u u^{-1} = 1$. But then $u = u 1 = u u u^{-1} = 0 \cdot u^{-1} = 0$, which cannot be since $u$ was assumed to be nonzero. Now suppose $u^2 > 0$. Then there exists an $\alpha \in \R$ such that $\alpha^2 = u^2$. But then $\paren {u - \alpha} \paren {u + \alpha} = u^2 - \alpha^2$ would be $0$. Since $A$ is assumed to be a division algebra, this would imply that either $u = \alpha$ or $u = -\alpha$, which is impossible since $u \in U$. So $u^2  < 0$ and so $u^2 = -\alpha^2$ with $0 \ne \alpha \in \R$. Thus, $v = \alpha^{-1} u$ is a desired element. {{qed}} {{MissingLinks}} Category:Abstract Algebra f6o1q5guzy610v4t64l15321xx6akoh


Frobenius's Theorem/Lemma 2

Frobenius's Theorem/Lemma 2 0 12013 146288 79502 2013-05-06T11:49:08Z Nearyan 2167 /* Proof */ wikitext text/x-wiki == Lemma == <onlyinclude> Let $A$ be a quadratic real division algebra. Let: :$U = \left\{{u \in A \setminus \R: u^2 \in \R}\right\} \cup \left\{{0}\right\}$ where $\setminus$ denotes set difference. Suppose $e_1, \ldots, e_k \in U$ are such that: : $\forall i \le k: e_i^2 = -1$ : $\forall i, j \le k, i \ne j: e_i e_j = -e_j e_i$ If $U$ is not equal to the linear span of $e_1, \ldots, e_k$, then there exists $e_{k+1} \in U$ such that: : $e_{k+1}^2 = -1$ : $\forall i \le k: e_i e_{k+1} = -e_{k+1} e_i$ </onlyinclude> == Proof == From Lemma 1, $U$ is a linear subspace of $A$ such that: :$\forall u, v \in U: u v + v u \in \R$ :$A = \R \oplus U$ Pick $u \in U$ that is not contained in the linear span of $e_1, \ldots, e_k$. By Lemma 1, we can set $\alpha_i = \dfrac 1 2 \left({u e_i + e_i u}\right) \in \R$. Note that $v = u + \alpha_1 e_1 + \cdots + \alpha_k e_k$ satisfies $e_i v = v e_i$ for all $i \le k$. Let $e_{k+1}$ be a scalar multiple of $v$ such that $e_{k+1}^2 = -1$ (whose existence follows from Lemma 1: Assertion 4). Then $e_{k+1}$ has all desired properties. {{qed|lemma}} Category:Abstract Algebra 94euoen9b0d870ivfbxtg45pqb9x1y4


Singleton Equality

Singleton Equality 0 12160 444460 444459 2020-01-22T06:14:56Z Prime.mover 59 wikitext text/x-wiki {{refactor|Two results here.}} == Theorems == Let $x$ and $y$ be sets. Then: :$\set x \subseteq \set y \iff x = y$ :$\set x = \set y \iff x = y$ == Proof == {{begin-eqn}} {{eqn | o =        | rr= \set x \subseteq \set y       | c =  }} {{eqn | o = \leadstoandfrom       | r = \forall z:       | rr= \paren {z \in \set x \implies z \in \set y}       | c = {{Defof|Subset}} }} {{eqn | o = \leadstoandfrom       | r = \forall z:       | rr= \paren {z = x \implies z = y}       | c = {{Defof|Singleton}} }} {{eqn | o = \leadstoandfrom       | rr= x = y       | c = Equality implies Substitution }} {{end-eqn}} {{qed|lemma}} Then: {{begin-eqn}} {{eqn | l = x       | r = y       | c =  }} {{eqn | ll= \leadsto       | l = \set x       | r = \set y       | c = Substitutivity of Equality }} {{eqn | ll= \leadsto       | l = \set x       | o = \subseteq       | r = \set y       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = y       | c = by the first part }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Set Theory and Its Logic|1963|Willard Van Orman Quine}}: $\S 7.7$ Category:Set Theory Category:Subsets Category:Singletons 2wb5g0fjhg6xwbwjpofkvpe1yurrsse


Cardinals are Totally Ordered

Cardinals are Totally Ordered 0 12226 178891 123437 2014-03-07T07:11:21Z Prime.mover 59 wikitext text/x-wiki == Corollary to Zermelo's Theorem == Every set of cardinals is totally ordered under $\le$. == Proof == Let $S$ be a set of cardinals. From Zermelo's theorem, $S$ is well-ordered. Let $a, b \in S$. Consider the subset $X = \left\{{a, b}\right\}$ of $S$. Since $S$ is well-ordered, $\inf \left({X}\right)$ exists and belongs to $X$. So either $\inf \left({X}\right) = a$ or $\inf \left({X}\right) = b$. By definition of infimum, we have that $\inf \left({X}\right) \le a$ and $\inf \left({X}\right) \le b$. It follows that either $a \le b$ or $b \le a$. So, by definition, $S$ is totally ordered under $\le$. {{qed}} {{AoC|Zermelo's Theorem (Set Theory)}} == Sources == * {{BookReference|Set Theory and Abstract Algebra|1975|T.S. Blyth|prev=Zermelo's Theorem (Set Theory)|next=Definition:Zero (Cardinal)}}: $\S 8$: Theorem $8.3$: Corollary Category:Cardinals Category:Total Orderings tld7afth5rd0ezdwoarguq1q910izr8


Idempotent Operators

Idempotent Operators 0 12233 375015 374551 2018-11-06T09:31:44Z Leigh.Samphier 3031 wikitext text/x-wiki {{rename|something more descriptive}} Let $H$ be a Hilbert space $H$. A linear operator $P: H \to H$ is called '''idempotent''' if :$P^2 = P$ or equivalently :$P x = x$ for $x \in \Rng P$. An idempotent operator is called a '''projector''' or '''orthogonal projector''' if :$\forall x \in H: P x - x \perp \Rng P$. Especially in the context of linear algebra, many texts refer to ''all'' idempotent operators as "projectors" and use the same definition as above only for "orthogonal projectors." In such texts, idempotent operators that are ''not'' orthogonal projectors may be called '''oblique projectors'''. Orthogonal projectors are extremely important in applied linear algebra and spectral theory. They can be characterized in several additional ways: * An idempotent operator is an orthogonal projector {{iff}} it is self-adjoint. * An idempotent operator is an orthogonal projector {{iff}} its norm is 1 (proof). {{refactor|The following to be done:
1: Idempotent needs to be given its own definition page in this context, as do Projector and Orthogonal Projector
2: The "immediate consequence" above needs to be given its own proof page
3: A definition needs to be raised for "linear idempotent operator" unless its meaning is as straightforward as being a "linear operator" which is also an "idempotent operator" - and even then it might be worth giving it a page if the concept occurs frequently.
It is also worth reviewing the meaning of $\Rng P$ in this context, as "range" is ambiguously understood.}} {{MissingLinks}} Category:Hilbert Spaces lk84c9j9tuw0l9gubhntjfzjm8dh55d


Squeeze Theorem for Real Sequences/Corollary 0 12410 408180 408168 2019-06-15T13:22:04Z Prime.mover 59 wikitext text/x-wiki == Corollary to Squeeze Theorem for Real Sequences == <onlyinclude> Let $\sequence {y_n}$ be a sequence in $\R$ which is null, that is: :$y_n \to 0$ as $n \to \infty$ Let: :$\forall n \in \N: \size {x_n - l} \le y_n$ Then $x_n \to l$ as $n \to \infty$. </onlyinclude> == Proof == From Negative of Absolute Value: Corollary 2: :$\size {x_n - l} \le y_n \iff l - y_n \le x_n \le l + y_n$ From the Difference Rule for Real Sequences: :$l - y_n \to l$ as $n \to \infty$ and from the Sum Rule for Real Sequences: :$l + y_n \to l$ as $n \to \infty$ So by the Squeeze Theorem for Real Sequences, $x_n \to l$ as $n \to \infty$. {{qed}} == Sources == * {{BookReference|Mathematical Analysis: A Straightforward Approach|1977|K.G. Binmore|prev = Squeeze Theorem for Real Sequences|next = Sequence of Powers of Number less than One/Necessary Condition/Proof 1}}: $\S 4$: Convergent Sequences: $\S 4.11$: Corollary Category:Squeeze Theorem nsvfni1ehlrolqwsh5fywnkicxecjfk


Taylor's Theorem/One Variable

Taylor's Theorem/One Variable 0 12521 484044 446910 2020-08-30T05:16:42Z Prime.mover 59 wikitext text/x-wiki == Theorem == <onlyinclude> {{:Taylor's Theorem/One Variable/Statement of Theorem}} </onlyinclude> == Proof == === Integral Version === {{:Taylor's Theorem/One Variable/Integral Version}} === Proof using Cauchy Mean Value Theorem === An alternative proof, which holds under milder technical assumptions on the function $f$, can be supplied using the Cauchy Mean Value Theorem. {{:Taylor's Theorem/One Variable/Proof by Cauchy Mean Value Theorem}} === Proof using Rolle's Theorem directly === Yet another proof  for Lagrange Form of the Remainder can be constructed applying Rolle's theorem directly $n$ times; this proof might be easier to visualize geometrically. {{:Taylor's Theorem/One Variable/Proof by Rolle's Theorem}} == Also see == * Mean Value Theorem: Taylor's Theorem when $n = 0$ {{Namedfor|Brook Taylor}} == Sources == * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = Mathematician:Brook Taylor|next = Definition:Taylor Series|entry = Taylor's theorem}} * {{BookReference|The Penguin Dictionary of Mathematics|2008|David Nelson|ed = 4th|edpage = Fourth Edition|prev = Mathematician:Brook Taylor|next = Definition:Taylor Series|entry = Taylor's theorem}} * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Definition:Taylor Series|next = Mathematician:Pafnuty Lvovich Chebyshev|entry = Taylor's Theorem}} Category:Taylor's Theorem oys9195a6vm91w58n0uh1qgkx8y1of4


Derivative of Exponential Function/Corollary 1 0 12582 490881 485755 2020-09-26T09:47:50Z Prime.mover 59 wikitext text/x-wiki == Corollary to Derivative of Exponential Function == Let $\exp x$ be the exponential function. <onlyinclude> Let $c \in \R$. Then: :$\map {\dfrac \d {\d x} } {\map \exp {c x} } = c \map \exp {c x}$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = \map {\dfrac \d {\d x} } {\map \exp {c x} }       | r = c \map {\dfrac \d {\map \d {c x} } } {\map \exp {c x} }       | c = Derivative of Function of Constant Multiple }} {{eqn | r = c \map \exp {c x}       | c = Derivative of Exponential Function }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Integration|1944|R.P. Gillespie|ed = 2nd|edpage = Second Edition|prev = Primitive of Reciprocal|next = Primitive of Exponential of a x}}: Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $3$. Category:Derivatives involving Exponential Function 84gqrvoa9b6562l37zobp4zp7jf6n4m


Derivative of General Exponential Function 0 12583 485952 485756 2020-09-06T21:07:05Z Prime.mover 59 wikitext text/x-wiki == Corollary to Derivative of Exponential Function == <onlyinclude> Let $a \in \R: a > 0$. Let $a^x$ be $a$ to the power of $x$. Then: :$\map {\dfrac \d {\d x} } {a^x} = a^x \ln a$ </onlyinclude> == Proof 1 == {{:Derivative of General Exponential Function/Proof 1}} == Proof 2 == {{:Derivative of General Exponential Function/Proof 2}} Category:Derivatives involving Exponential Function pox8ott2n50d2p1ayac7snucrugq5xc


Quaternion Group/Complex Matrices

Quaternion Group/Complex Matrices 0 12762 379946 379139 2018-12-03T18:26:28Z Prime.mover 59 wikitext text/x-wiki == Representation of Quaternion Group == <onlyinclude> Let $\mathbf 1, \mathbf i, \mathbf j, \mathbf k$ denote the following four elements of the matrix space $\map {\mathcal M_\C} 2$: :$\mathbf 1 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \qquad \mathbf i = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} \qquad \mathbf j = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \qquad \mathbf k = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}$ where $\C$ is the set of complex numbers. The set: :$\Dic 2 = \set {\mathbf 1, -\mathbf 1, \mathbf i, -\mathbf i, \mathbf j, -\mathbf j, \mathbf k, -\mathbf k}$ under the operation of conventional matrix multiplication, forms the '''quaternion group''': </onlyinclude> === Cayley Table === Its Cayley table is given by: {{:Quaternion Group/Complex Matrices/Cayley Table}} == Also see == * Quaternions Defined by Matrices where it is shown that these have the appropriate properties. In Matrix Form of Quaternion it is shown that a general element $\mathbf x$ of $\mathbb H$ has the form: :$\mathbf x = \begin{bmatrix} a + b i & c + d i \\ -c + d i & a - bi \end{bmatrix}$ == Sources == * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Group of Gaussian Integer Units/Cayley Table|next = Quaternions Defined by Matrices}}: $\S 34$. Examples of groups: $(6) \ \text{(ii)}$ Category:Quaternion Group raxbuehbo9ts4rsu3mbper98sgofdv0


Axiom:Leibniz's Law

Axiom:Leibniz's Law 100 12959 425237 425236 2019-09-14T13:23:03Z Prime.mover 59 wikitext text/x-wiki == Axiom == Let $=$ represent the relation of equality and let $P$ be an arbitrary property. Then: :$x = y \dashv \vdash \map P x \iff \map P y$ for all $P$ in the universe of discourse. That is, two objects $x$ and $y$ are equal {{iff}} $x$ has every property $y$ has, and $y$ has every property $x$ has. === Application to Equality of Sets === {{:Leibniz's Law for Sets}} == Also see == * Axiom:Axioms of Equality * Definition:Equals {{NamedforAxiom|Gottfried Wilhelm von Leibniz|cat = Leibniz}} == Historical Note == {{:Axiom:Leibniz's Law/Historical Note}} == Sources == * {{BookReference|Introduction to Logic and to the Methodology of Deductive Sciences|1946|Alfred Tarski|ed = 2nd|edpage = Second Edition}}: $\S 3.17$ * {{BookReference|A Handbook of Terms used in Algebra and Analysis|1972|A.G. Howson|prev = Equality is Transitive|next = Definition:Proof System}}: $\S 1$: Some mathematical language: Equality: $\text{(d)}$ * {{BookReference|Dictionary of Mathematics|1989|Ephraim J. Borowski|author2 = Jonathan M. Borwein|prev = Leibniz's Alternating Series Test|next = Leibniz's Law for Sets|entry = Leibniz's law|index = 1}} Category:Axioms/Logic tlr9rhzko3h8021an1sh8r376yx3qpz


Axiom:Axiom of Dependent Choice

Axiom:Axiom of Dependent Choice 100 13124 453399 444071 2020-03-08T13:00:27Z Prime.mover 59 wikitext text/x-wiki == Axiom == === Left-Total Form === {{:Axiom:Axiom of Dependent Choice/Left-Total}} === Right-Total Form === {{:Axiom:Axiom of Dependent Choice/Right-Total}} == Also known as == Some sources call this the '''Axiom of Dependent Choices''', reflecting the infinitely many choices made. This axiom can be abbreviated '''ADC''' or simply '''DC'''. == Also see == * This axiom is a weaker form of the axiom of choice, as shown in Axiom of Choice Implies Axiom of Dependent Choice. * This axiom is also a stronger form of the axiom of countable choice, as shown in Axiom of Dependent Choice Implies Axiom of Countable Choice. * Dependent Choice (Fixed First Element) shows that it is possible to choose any element of the set to be the first element of the sequence. Dependent Choice Dependent Choice 007arhkjbrui5rbabctehqgx8t5u1m0


Cartesian Product of Intersections/Corollary 1 0 13257 417118 375824 2019-08-07T20:31:54Z Prime.mover 59 wikitext text/x-wiki == Corollary to Cartesian Product of Intersections == <onlyinclude> :$A \times \paren {B \cap C} = \paren {A \times B} \cap \paren {A \times C}$ </onlyinclude> == Proof == Take the result Cartesian Product of Intersections: :$\paren {S_1 \cap S_2} \times \paren {T_1 \cap T_2} = \paren {S_1 \times T_1} \cap \paren {S_2 \times T_2}$ Put $S_1 = S_2 = A, T_1 = B, T_2 = C$: {{begin-eqn}} {{eqn | l = A \times \paren {B \cap C}       | r = \paren {A \cap A} \times \paren {B \cap C}       | c = Intersection is Idempotent }} {{eqn | r = \paren {A \times B} \cap \paren {A \times C}       | c = Cartesian Product of Intersections }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Sets and Groups|1965|J.A. Green|prev = Intersection Distributes over Symmetric Difference|next = Cartesian Product of Unions/Corollary}}: Chapter $1$. Sets: Exercise $8 \ \text{(i)}$ * {{BookReference|Undergraduate Topology|1971|Robert H. Kasriel|prev = Cartesian Product is Anticommutative|next = Cartesian Product of Subsets/Corollary 1}}: $\S 1.9$: Cartesian Product: Theorem $9.2$ * {{BookReference|Set Theory and Abstract Algebra|1975|T.S. Blyth|prev = Definition:Unbounded Open Real Interval|next = Cartesian Product Distributes over Union}}: $\S 3$. Ordered pairs; cartesian product sets: Exercise $3 \ (1)$ Category:Cartesian Product of Intersections ky822jywuu7e08ht1awdjabx97s3xsd


Image of Set Difference under Relation/Corollary 1 0 13584 495690 226802 2020-10-21T07:52:33Z Prime.mover 59 wikitext text/x-wiki == Corollary to Image of Set Difference under Relation == <onlyinclude> Let $\RR \subseteq S \times T$ be a relation. Let $A \subseteq B \subseteq S$. Then: :$\relcomp {\RR \sqbrk B} {\RR \sqbrk A} \subseteq \RR \sqbrk {\relcomp B A}$ where: :$\RR \sqbrk B$ denotes the image of $B$ under $\RR$ :$\complement$ (in this context) denotes relative complement. </onlyinclude> == Proof == We have that $A \subseteq B$. Then by definition of relative complement: :$\relcomp B A = B \setminus A$ :$\relcomp {\RR \sqbrk B} {\RR \sqbrk A} = \RR \sqbrk B \setminus \RR \sqbrk A$ Hence, when $A \subseteq B$: :$\relcomp {\RR \sqbrk B} {\RR \sqbrk A} \subseteq \RR \sqbrk {\relcomp B A}$ means exactly the same thing as: :$\RR \sqbrk B \setminus \RR \sqbrk A \subseteq \RR \sqbrk {B \setminus A}$ {{qed}} Category:Image of Set Difference under Relation 5058rrwlmkf3j5x4um5uteu6svor7p0


Image of Set Difference under Relation/Corollary 2 0 13585 495692 495691 2020-10-21T07:54:37Z Prime.mover 59 wikitext text/x-wiki == Corollary to Image of Set Difference under Relation == <onlyinclude> Let $\RR \subseteq S \times T$ be a relation. Let $A$ be a subset of $S$. Then: :$\relcomp {\Img \RR} {\RR \sqbrk A} \subseteq \RR \sqbrk {\relcomp S A}$ where: :$\Img \RR$ denotes the image of $\RR$ :$\RR \sqbrk A$ denotes the image of $A$ under $\RR$. </onlyinclude> == Proof == By definition of the image of $\RR$: :$\Img \RR = \RR \sqbrk S$ So, when $B = S$ in Image of Set Difference under Relation: Corollary 1: :$\relcomp {\Img \RR} {\RR \sqbrk A} = \relcomp {\RR \sqbrk S} {\RR \sqbrk A}$ Hence: :$\relcomp {\Img \RR} {\RR \sqbrk A} \subseteq \RR \sqbrk {\relcomp S A}$ means exactly the same thing as: :$\relcomp {\RR \sqbrk S} {\RR \sqbrk A} \subseteq \RR \sqbrk {\relcomp S A}$ that is: :$\RR \sqbrk S \setminus \RR \sqbrk A \subseteq \RR \sqbrk {S \setminus A}$ {{Qed}} Category:Image of Set Difference under Relation t36jcol89y1tjockt9oh052ij7wfcrh


Zassenhaus Lemma

Zassenhaus Lemma 0 13788 404925 360091 2019-05-15T21:41:09Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $G$ be a group. Let $H_1$ and $H_2$ be subgroups of $G$. Let: :$N_1 \lhd H_1$ :$N_2 \lhd H_2$ where $\lhd$ denotes the relation of being a normal subgroup. Then: :$\dfrac {N_1 \paren {H_1 \cap H_2} } {N_1 \paren {H_1 \cap N_2} } \cong \dfrac {H_1 \cap H_2} {\paren {H_1 \cap N_2} \paren {N_1 \cap H_2} } \cong \dfrac {N_2 \paren {H_1 \cap H_2} } {N_2 \paren {N_1 \cap H_2} }$ where: :$N_1 \paren {H_1 \cap H_2}$, and so on, denotes subset product :$\cong$ denotes group isomorphism. == Proof == Because of symmetry, only the first of the isomorphisms needs to be proved. === Proof of Normality === In order for the expressions to make sense, the expressions on the bottom of the fractions need to be normal subgroups. That is, we need to show that: :$(1): \quad N_1 \paren {H_1 \cap N_2} \lhd N_1 \paren {H_1 \cap H_2}$ :$(2): \quad \paren {H_1 \cap N_2} \paren {N_1 \cap H_2} \lhd H_1 \cap H_2$ {{finish}} === Proof of Isomorphism === Let: :$H = H_1 \cap H_2$ :$N = N_1 \paren {H_1 \cap N_2}$ Then: {{begin-eqn}} {{eqn | l = \frac {N_1 \paren {H_1 \cap N_2} \paren {H_1 \cap H_2} } {N_1 \paren {H_1 \cap N_2} }       | r = \frac {N H} N       | c =  }} {{eqn | o = \cong       | r = \frac H {H \cap N}       | c = Second Isomorphism Theorem }} {{eqn | r = \frac {H_1 \cap H_2} {H_1 \cap H_2 \cap N_1 \paren {H_1 \cap N_2} }       | c =  }} {{end-eqn}} From Intersection with Subgroup Product of Superset: :If $X, Y, Z$ are subgroups of a group $\struct {G, \circ}$, and $Y \subseteq X$, then: ::$X \cap \paren {Y \circ Z} = Y \circ \paren {X \cap Z}$ Thus: {{begin-eqn}} {{eqn | l = N_1 \paren {H_1 \cap N_2} \paren {H_1 \cap H_2}       | r = N_1 \paren {H_1 \cap H_2}       | c = by taking $X = H_1, Y = H_1 \cap N_2, Z = H_2$ }} {{eqn | l = H_1 \cap H_2 \cap N_1 \paren {H_1 \cap N_2}       | r = \paren {H_1 \cap N_2} \paren {N_1 \cap H_2}       | c = by taking $X = H_1 \cap H_2, Y = H_1 \cap N_2, Z = N_1$ }} {{end-eqn}} Hence the result. {{qed}} == Also known as == This lemma is also known as: :the '''butterfly lemma''', so named because the Hasse diagram of the various subgroups involved can be drawn to resemble a butterfly :either the '''third isomorphism theorem''' or the '''fourth isomorphism theorem''', which names are not recommended because of the lack of any consistent naming convention for the Isomorphism Theorems in the literature. {{Namedfor|Hans Julius Zassenhaus|cat = Zassenhaus}} == Sources == * {{BookReference|Elements of Abstract Algebra|1971|Allan Clark|prev = Second Isomorphism Theorem for Groups|next = Definition:Normal Series}}: Chapter $2$: Group Homomorphism and Isomorphism: $\S 70$. The Third Isomorphism Theorem Category:Group Isomorphisms Category:Normal Subgroups c393a6nodonyl3uuq3tba352zktj110


Axiom:Pasch's Axiom

Axiom:Pasch's Axiom 100 14315 230283 158844 2015-10-29T23:20:44Z Prime.mover 59 wikitext text/x-wiki {{refactor|These axioms state the same truth, but in different language. A disambiguation page is for documenting multiple uses of a term which are completely different. This paqe needs to be converted into a parent page with transclusions, and then a separate page which proves the logical equivalence of these two axioms.}} {{disambig}} * Pasch's Axiom in Euclidean Geometry * Pasch's Axiom in Tarski's Geometry Category:Axioms/Geometry 4tpavwovpquem1a758myy9mzy4c53ry


Taylor's Theorem/One Variable/Integral Version 0 14611 483449 362648 2020-08-28T10:12:49Z Prime.mover 59 wikitext text/x-wiki == Theorem == {{:Taylor's Theorem/One Variable/Statement of Theorem}} == Proof == <onlyinclude> This proof requires $f^{\paren n}$ to be absolutely continuous on $\closedint a x$, so that the Fundamental Theorem of Calculus holds. Except at the end when the Mean Value Theorem is invoked, differentiability of $f^{\paren n}$ need not be assumed, since absolute continuity implies: :differentiability almost everywhere :the validity of the Fundamental Theorem of Calculus provided the integrals involved are understood as Lebesgue integrals. Consequently, the integral form of the remainder holds with this particular weakening of the assumptions on $f$. We first prove Taylor's Theorem with the integral remainder term. The Fundamental Theorem of Calculus states that: :$\displaystyle \int_a^x \map {f'} t \rd t = \map f x - \map f a$ which can be rearranged to: :$\displaystyle \map f x = \map f a + \int_a^x \map {f'} t \rd t$ Now we can see that an application of Integration by Parts yields: {{begin-eqn}} {{eqn | l = \map f x       | r = \map f a + x \map {f'} x - a \map {f'} a - \int_a^x t \map {f''} t \rd t       | c = $u = \map {f'} t$ and $\d v = \d t$ }} {{eqn | r = \map f a + \int_a^x x \map {f''} t \rd t + x \map {f'} a - a \map {f'} a - \int_a^x t \map {f''} t \rd t       | c = $\displaystyle \int_a^x x \map {f''} t \rd t = x \map {f'} x - x \map {f'} a$ }} {{eqn | r = \map f a + \paren {x - a} \map {f'} a + \int_a^x \paren {x - t} \map {f''} t \rd t       | c = factoring out some common terms }} {{end-eqn}} Another application yields: :$\displaystyle \map f x = \map f a + \paren {x - a} \map {f'} a + \frac 1 2 \paren {x - a}^2 \map {f''} a + \frac 1 2 \int_a^x \paren {x - t}^2 \map {f'''} t \rd t$ By repeating this process, we may derive Taylor's theorem for higher values of $n$. This can be formalized by applying the technique of Principle of Mathematical Induction. So, suppose that Taylor's theorem holds for a $n$, that is, suppose that: {{begin-eqn}} {{eqn | l = \map f x       | r = \map f a       | c =  }} {{eqn | o = +       | r = \frac {\map {f'} a} {1!} \paren {x - a}       | c =  }} {{eqn | o = +       | r = \cdots       | c =  }} {{eqn | o = +       | r = \frac {\map {f^{\paren n} } a} {n!} \paren {x - a}^n       | c =  }} {{eqn | o = +       | r = \int_a^x \frac {\map {f^{\paren {n + 1} } } t} {n!} \paren {x - t}^n \rd t       | c = $*$ }} {{end-eqn}} We can rewrite the integral using Integration by Parts. A primitive of $\paren {x - t}^n$ as a function of $t$ is given by $\dfrac {-\paren {x - t}^{n + 1} } {n + 1}$. So: {{begin-eqn}} {{eqn | o =        | r = \int_a^x \map {\frac {f^{\paren {n + 1} } } t} {n!} \paren {x - t}^n \rd t       | c =  }} {{eqn | r = -\intlimits {\map {\frac {f^{\paren {n + 1} } } t} {\paren {n + 1} n!} \paren {x - t}^{n + 1} } a x + \int_a^x \frac {\map {f^{\paren {n + 2} } } t} {\paren {n + 1} n!} \paren {x - t}^{n + 1} \rd t       | c =  }} {{eqn | r = \frac {\map {f^{\paren {n + 1} } } a} {\paren {n + 1}!} \paren {x - a}^{n + 1} + \int_a^x \frac {\map {f^{\paren {n + 2} } } t} {\paren {n + 1}!} \paren {x - t}^{n + 1} \rd t       | c =  }} {{end-eqn}} The last integral can be solved immediately, which leads to :$R_n = \dfrac {\map {f^{\paren {n + 1} } } \xi} {\paren {n + 1}!} \paren {x - a}^{n + 1}$ {{qed}} </onlyinclude> {{Namedfor|Brook Taylor}} Category:Taylor's Theorem 6vf2wc06etjrisnr8v6d25ktn1upas4


Taylor's Theorem/One Variable/Proof by Cauchy Mean Value Theorem 0 14612 484062 483450 2020-08-30T06:10:15Z Prime.mover 59 wikitext text/x-wiki == Theorem == {{:Taylor's Theorem/One Variable/Statement of Theorem}} == Proof == <onlyinclude> Let $G$ be a real-valued function continuous on $\closedint a x$ and differentiable with non-vanishing derivative on $\openint a x$. Let: :$\map F t = \map f t + \dfrac {\map {f'} t} {1!} \paren {x - t} + \dotsb + \dfrac {\map {f^{\paren n} } t} {n!} \paren {x - t}^n$ By the Cauchy Mean Value Theorem: :$(1): \quad \dfrac {\map {F'} x} {\map {G'} \xi} = \dfrac {\map F x - \map F a} {\map G x - \map G a}$ for some $\xi \in \openint a x$. Note that the numerator: :$\map F x - \map F a = R_n$ is the remainder of the Taylor polynomial for $\map f x$. On the other hand, computing $\map {F'} \xi$: :$\map {F'} \xi = \map {f'} \xi - \map {f'} \xi + \dfrac {\map {f''} \xi} {1!} \paren {x - \xi} - \dfrac {\map {f''} \xi} {1!} \paren {x - \xi} + \dotsb + \dfrac {\map {f^{\paren {n + 1} } } t} {n!} \paren {x - \xi}^n = \dfrac {\map {f^{\paren {n + 1} } } \xi} {n!} \paren {x - \xi}^n$ Putting these two facts together and rearranging the terms of $(1)$ yields: :$R_n = \dfrac {\map {f^{\paren {n + 1} } } \xi} {n!} \paren {x - \xi}^n \dfrac {\map G x - \map G a} {\map {G'} \xi}$ which was to be shown. {{refactor|Find a way of neatly handling the two different remainder forms}} Note that the Lagrange Form of the Remainder comes from taking $\map G t = \paren {x - t}^{n + 1}$ and the given Cauchy Form of the Remainder comes from taking $\map G t = t - a$. {{qed}} </onlyinclude> {{Namedfor|Brook Taylor}} Category:Taylor's Theorem bp8i1weofd8nv7m8rgfrqubreirojaz


Taylor's Theorem/One Variable/Proof by Rolle's Theorem 0 14613 484136 483451 2020-08-30T11:45:11Z Prime.mover 59 wikitext text/x-wiki == Theorem == {{:Taylor's Theorem/One Variable/Statement of Theorem}} == Proof == <onlyinclude> Let the function $g$ be defined as: :$\map g t = \map {R_n} t - \dfrac {\paren {t - a}^{n + 1} } {\paren {x - a}^{n + 1} } \map {R_n} x$ Then: :$\map {g^{\paren k} } a = 0$ for $k = 0, \dotsc, n$, and $\map g x = 0$. Apply Rolle's Theorem successively to $g, g', \dotsc, g^{\paren n}$. Then there exist: :$\xi_1, \ldots, \xi_{n + 1}$ between $a$ and $x$ such that: :$\map {g'} {\xi_1} = 0, \map {g''} {\xi_2} = 0, \ldots, \map {g^{\paren {n + 1} } } {\xi_{n + 1} } = 0$ Let $\xi = \xi_{n + 1}$. Then: :$0 = \map {g^{\paren {n + 1} } } \xi = \map {f^{\paren {n + 1} } } \xi - \dfrac {\paren {n + 1}!} {\paren {x - a}^{n + 1} } \map {R_n} x$ and the formula for $\map {R_n} x$ follows. {{qed}} </onlyinclude> {{Namedfor|Brook Taylor}} == Sources == * {{BookReference|Mathematical Analysis: A Straightforward Approach|1977|K.G. Binmore|prev = Taylor's Theorem/One Variable/Statement of Theorem/Also presented as|next = Definition:Increasing Real Function}}: $\S 11.10$ Category:Taylor's Theorem 5ca6mx3oeb2lvib8nmi92h6dw3zvvwg


Matrix Inverse Algorithm

Matrix Inverse Algorithm 0 14899 476345 476210 2020-06-29T13:42:09Z Prime.mover 59 wikitext text/x-wiki == Algorithm == <onlyinclude> The '''matrix inverse algorithm''' is an algorithm which either: :$(1): \quad$ converts a matrix into its inverse, if it exists or: :$(2): \quad$ determines that such an inverse does not exist. Let $\mathbf A$ be the $n \times n$ square matrix in question. Let $\mathbf I$ be the unit matrix of order $n$. :'''Step $0$''': Create the augmented matrix $\begin {bmatrix} \mathbf A & \mathbf I \end {bmatrix}$. :'''Step $1$''': Perform elementary row operations until $\begin {bmatrix} \mathbf A & \mathbf I \end {bmatrix}$ is in reduced echelon form. By Matrix is Row Equivalent to Reduced Echelon Matrix, this is possible. By Reduced Echelon Matrix is Unique, this process is well-defined. Call this new augmented matrix $\begin {bmatrix} \mathbf H & \mathbf C \end {bmatrix}$. :'''Step $2$''':  ::If $\mathbf H = \mathbf I$, then take $\mathbf C = \mathbf A^{-1}$. ::If $\mathbf H \ne \mathbf I$, $\mathbf A$ is not invertible. </onlyinclude> == Proof == Suppose $\begin {bmatrix} \mathbf A & \mathbf I \end {bmatrix}$ can be transformed into an upper triangular matrix with no zeroes on its main diagonal. Then from Identity Matrix from Upper Triangular Matrix, it can further be transformed into $\begin {bmatrix} \mathbf H & \mathbf C \end {bmatrix}$ where $\mathbf H = \mathbf I$. From Row Operation is Equivalent to Pre-Multiplication by Product of Elementary Matrices, the row operation to convert $\begin {bmatrix} \mathbf A & \mathbf I \end {bmatrix}$ to $\begin {bmatrix} \mathbf H & \mathbf C \end {bmatrix}$ is equivalent to the matrix product of the elementary row matrices corresponding to the sequence of elementary row operations that compose that row operation. Let $\mathbf R$ be that matrix corresponding to that row operation. Because $\mathbf H = \mathbf I$, it follows that: :$\mathbf R \mathbf A = \mathbf I$ and so $\mathbf R$ is the inverse of $\mathbf A$. That is: :$\mathbf R = \mathbf A^{-1}$ We also have, from the action of $\mathbf R$ on the {{RHS}} of the augmented matrix $\begin {bmatrix} \mathbf A & \mathbf I \end {bmatrix}$ that: :$\mathbf R \mathbf I = \mathbf C$ and so: :$\mathbf C = \mathbf R = \mathbf A^{-1}$ {{qed|lemma}} Now suppose that $\mathbf H \ne \mathbf I$. {{finish|It remains to be shown that in such a case $\mathbf A$ is not invertible.}} == Examples == {{:Matrix Inverse Algorithm/Examples}} == Sources == * {{BookReference|Linear Algebra|1998|Richard Kaye|author2 = Robert Wilson|prev = Definition:Augmented Matrix|next = Matrix Inverse Algorithm/Examples/Arbitrary Matrix 1}}: Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations Category:Inverse Matrices Category:Matrix Inverse Algorithm 258v7l48cdgzlyr3lvydeos4hc8ha87


Convergence of Generalized Sum of Complex Numbers/Corollary 0 14938 178094 178091 2014-03-02T17:20:16Z Prime.mover 59 wikitext text/x-wiki == Corollary to Convergence of Generalized Sum of Complex Numbers == Let $\left({z_i}\right)_{i \in I}$ be an $I$-indexed family of complex numbers. That is, let $z_i \in \C$ for all $i \in I$. <onlyinclude> Suppose that $\displaystyle \sum \left\{{ z_i: i \in I }\right\}$ converges to $z \in \C$. Then $\displaystyle \sum \left\{{ \overline{z_i}: i \in I }\right\}$ converges to $\overline z$, where $\overline z$ denotes the complex conjugate of $z$. </onlyinclude> Here, the $\sum$ denote generalized sums. == Proof == Using Convergence of Generalized Sum of Complex Numbers, one has: :$\displaystyle \sum \left\{{\operatorname{Re} z_i : i \in I}\right\} = \operatorname{Re} z$ :$\displaystyle \sum \left\{{\operatorname{Im} z_i : i \in I}\right\} = \operatorname{Im} z$ Now, observe that, from the definition of complex conjugate: {{begin-eqn}} {{eqn|l = \overline z      |r = \operatorname{Re} z - i \operatorname{Im} z }} {{eqn|r = \sum \left\{ {\operatorname{Re} z_i : i \in I}\right\} - i \sum \left\{ {\operatorname{Im} z_i : i \in I}\right\} }} {{eqn|r = \sum \left\{ {\operatorname{Re} z_j - i \operatorname{Im} z_j : j \in I}\right\}      |c = Generalized Sum is Linear }} {{eqn|r = \sum \left\{ {\overline{z_j}: j \in I}\right\} }} {{end-eqn}} {{qed}} Category:Generalized Sums Category:Complex Numbers mtmfc9a1rt4ibp1ms8sld6zo6yverm5


Bijection iff Left and Right Inverse/Corollary 0 14953 447362 368399 2020-02-08T12:26:59Z Prime.mover 59 wikitext text/x-wiki == Corollary to Bijection iff Left and Right Inverse == <onlyinclude> Let $f: S \to T$ and $g: T \to S$ be mappings such that: : $g \circ f = I_S$ : $f \circ g = I_T$ Then both $f$ and $g$ are bijections. </onlyinclude> == Proof  == Suppose we have such mappings $f$ and $g$ with the given properties. From Bijection iff Left and Right Inverse, we have that $f$ is a bijection, by considering $g = g_1$ and $g = g_2$. It directly follows that by setting $g = f, f = g_1, f = g_2$, the result Bijection iff Left and Right Inverse can be used the other way about. {{qed}} == Sources == * {{BookReference|Lectures in Abstract Algebra|1951|Nathan Jacobson|volume = I|subtitle = Basic Concepts|prev = Composite of Bijection with Inverse is Identity Mapping|next = Definition:Binary Operation}}: Introduction $\S 2$: Product sets, mappings * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Composite of Bijection with Inverse is Identity Mapping|next = Composite of Bijections is Bijection/Proof 1}}: $\S 25.1$: Some further results and examples on mappings Category:Bijections Category:Inverse Mappings 09zl59d76hre8zvavs3lkuhcvyw266n


Exponential of Sum/Real Numbers/Lemma

Exponential of Sum/Real Numbers/Lemma 0 14995 396492 396466 2019-03-21T07:02:29Z Prime.mover 59 wikitext text/x-wiki == Lemma == <onlyinclude> Let $x, y \in \R$. Let $n \in \N_{> 0}$ such that $n > -\paren {x + y}$. Then: :$1 + \dfrac {x + y} n + \dfrac {x y} {n^2} = \paren {1 + \dfrac {x + y} n} \paren {1 + \dfrac {\paren {\frac {x y} {n + x + y} } } n}$ {{qed|lemma}} </onlyinclude> == Proof == As $n \in \N_{> 0}$ we have that $n \ne 0$ and so the fractions in the expressions are defined. {{begin-eqn}} {{eqn | l = 1 + \frac {x + y} n + \frac {x y} {n^2}       | r = \frac {\paren {1 + \frac {x + y} n} \paren {1 + \frac {x + y} n + \frac {x y} {n^2} } } {1 + \frac {x + y} n}        | c = multiplying and dividing by $1 + \dfrac {x + y} n$ }} {{eqn | r = \paren {1 + \frac {x + y} n} \frac {\paren {1 + \frac {x + y} n + \frac {x y} {n^2} } } {\paren {1 + \frac {x + y} n} }        | c = extracting a factor }} {{eqn | r = \paren {1 + \frac {x + y} n} \paren {\frac {n^2 + n x + n y + x y} {n^2 + n x + n y} }       | c = multiplying top and bottom by $n^2$ }} {{eqn | r = \paren {1 + \frac {x + y} n} \paren {1 + \frac {x y} {n^2 + n x + n y} }       | c = Polynomial Long Division }} {{eqn | r = \paren {1 + \frac {x + y} n} \paren {1 + \frac {\paren {\frac {x y} {n + x + y} } } n}       | c = dividing top and bottom by $n + x + y$ }} {{end-eqn}} That final step is justified, as we have that $n > -\paren {x + y}$ and so $n + x + y \ne 0$. {{qed}} == Also see == This lemma is used in: :Proof 2 of (Real) Exponential of Sum :Proof 3 of (Real) Exponential of Sum :$\map \exp {x + y} = \paren {\exp x} \paren {\exp y}$ for real $x$ and $y$. Category:Exponential of Sum mkhqezzsgxcvr5s3a3ntuqluxryficm


Natural Logarithm of 2 is Greater than One Half 0 15131 356186 267867 2018-05-19T09:36:22Z Prime.mover 59 wikitext text/x-wiki == Lemma == <onlyinclude> :$\ln 2 \ge \dfrac 1 2$ where $\ln$ denotes the natural logarithm function. </onlyinclude> == Proof 1 == {{:Natural Logarithm of 2 is Greater than One Half/Proof 1}} == Proof 2 == {{:Natural Logarithm of 2 is Greater than One Half/Proof 2}} == Also see == * Logarithm Tends to Infinity == Sources == * {{BookReference|Calculus|2005|Roland E. Larson|author2 = Robert P. Hostetler|author3 = Bruce H. Edwards|ed = 8th|edpage = Eighth Edition}}: Appendix $A$: ''Properties of the Natural Logarithmic Function'' Category:Natural Logarithms Category:Inequalities Category:Natural Logarithm of 2 is Greater than One Half nmpmqpkbmuc8l8xfz3lo3jsxua2ki78


Schur's Lemma (Representation Theory)/Corollary 0 15357 497538 316079 2020-11-02T06:59:29Z Prime.mover 59 wikitext text/x-wiki {{MissingLinks}} == Corollary to Schur's Lemma == <onlyinclude> Let $\struct {G, \cdot}$ be a finite group. Let $\struct {V, \phi}$ be a $G$-module. Let the underlying field $k$ of $V$ be algebraically closed. Let: :$\map {\mathrm {End}_G} V := \leftset {f: V \to V: f}$ is a homomorphism of $G$-modules$\rightset {}$ Then: :$\map {\mathrm {End}_G} V$ is a field, with the same structure as $k$. </onlyinclude> == Proof == Denote the identity mapping on $V$ as $I_V: V \to V$. If $f = 0$, since $0\in k$ it can be written $f = 0 I_V$. {{explain|Not clear what the above line means.}} Let $f$ be an automorphism. We have that $k$ is algebraically closed. Therefore the characteristic polynomial of $f$ is complete reducible in $k \sqbrk x$. Hence $f$ has all eigenvalue in $k$. Let $\lambda \in k$ be an eigenvalue of $f$. Consider the endomorphism: :$f - \lambda I_V: V \to V$ Because $\lambda$ is an eigenvalue: :$\map \ker {f - \lambda I_V} \ne \set 0$ From Schur's Lemma: :$f = \lambda I_V$ :$\paren {\lambda I_V} \circ \paren {\mu I_V} = \paren {\lambda \mu} I_V$ :$\lambda I_V + \paren {-\mu I_V} = \paren {\lambda - \mu} I_V$ From Subring Test: :$\map {\mathrm {End}_G} V$ is a subring of the ring endomorphisms of $V$ as an abelian group. Let $\phi: \map {\mathrm {End}_G} V \to k$ be defined as: :$\map \phi {\lambda I_V} = \lambda$ Then: :$\map \phi {\lambda I_V + \mu I_V} = \lambda + \mu = \map \phi {\lambda I_V} + \map \phi {\mu I_V}$ :$\map \phi {\paren {\lambda I_V} \circ \paren {\mu I_V} } = \lambda \mu = \map \phi {\lambda I_V} \map \phi {\mu I_V}$ Hence $\phi$ is a ring isomorphism. But since $k$ is a field it is a field isomorphism. {{qed}}   {{proofread}} Category:Representation Theory 0iluo82z6mhpuey50216wc2typ8r5hj


De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection/Corollary 0 15601 352420 225769 2018-05-01T20:45:53Z Prime.mover 59 wikitext text/x-wiki == Corollary to De Morgan's Laws: Difference with Intersection == Let $S, T_1, T_2$ be sets. <onlyinclude> Suppose that $T_1 \subseteq S$. Then: :$S \setminus \left({T_1 \cap T_2}\right) = \left({S \setminus T_1}\right) \cup \left({T_1 \setminus T_2}\right)$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = S \setminus \left({T_1 \cap T_2}\right)       | r = \left({S \setminus T_1}\right) \cup \left({S \setminus T_2}\right)       | c = De Morgan's Laws: Difference with Intersection }} {{eqn | r = \left({S \setminus T_1}\right) \cup \left({\left({\left({S \setminus T_1}\right) \cup \left({S \cap T_1}\right)}\right) \setminus T_2}\right)       | c = Set Difference Union Intersection }} {{eqn | r = \left({S \setminus T_1}\right) \cup \left({\left({S \setminus T_1}\right) \setminus T_2}\right) \cup \left({\left({S \cap T_1}\right) \setminus T_2}\right)       | c = Set Difference is Right Distributive over Union }} {{eqn | r = \left({S \setminus T_1}\right) \cup \left({\left({S \setminus T_1}\right) \setminus T_2}\right) \cup \left({T_1 \setminus T_2}\right)       | c = Intersection with Subset is Subset: $T_1 \subseteq S$ }} {{eqn | r = \left({S \setminus T_1}\right) \cup \left({T_1 \setminus T_2}\right)       | c = Set Difference Union First Set is First Set }} {{end-eqn}} {{qed}} Category:Set Difference Category:De Morgan's Laws scwakh2xporyr40vaai5w15c3j18g9n


Image of Preimage under Mapping/Corollary 0 15653 417770 414514 2019-08-11T11:14:31Z Prime.mover 59 wikitext text/x-wiki == Corollary to Image of Preimage under Mapping == Let $f: S \to T$ be a mapping. Then: <onlyinclude> :$B \subseteq \Img S \implies \paren {f \circ f^{-1} } \sqbrk B = B$ </onlyinclude> == Proof == From Image of Subset under Relation is Subset of Image/Corollary 3 we have: :$B \subseteq \Img S \implies f^{-1} \sqbrk B \subseteq f^{-1} \sqbrk {\Img S}$ and from Intersection with Subset is Subset we have: :$f^{-1} \sqbrk B \subseteq f^{-1} \sqbrk {\Img S} \implies f^{-1} \sqbrk B \cap f^{-1} \sqbrk {\Img S} = f^{-1} \sqbrk B$ Hence the result. {{qed}} == Sources == * {{BookReference|Rings, Modules and Linear Algebra|1970|B. Hartley|author2 = T.O. Hawkes|prev = Definition:Direct Image Mapping of Mapping|next = Image of Subset under Relation is Subset of Image/Corollary 2}}: $\S 2.2$: Homomorphisms: $\text{(i)}$ Category:Preimages under Mappings evr46sk5tjwomo75w6b2tyk4gkdm20i


Axiom:Axiom of Countable Choice

Axiom:Axiom of Countable Choice 100 15701 453398 444063 2020-03-08T13:00:08Z Prime.mover 59 wikitext text/x-wiki == Axiom == <onlyinclude> === Form 1 === {{:Axiom:Axiom of Countable Choice/Form 1}} === Form 2 === {{:Axiom:Axiom of Countable Choice/Form 2}} </onlyinclude> == Also known as == This axiom can be abbreviated $\mathrm{ACC}$, $\mathrm{CC}$, $\mathrm{AC}_\omega$, or $\mathrm{AC}_\N$. == Also see == * Equivalence of Forms of Axiom of Countable Choice * This axiom is a weaker form of the axiom of dependent choice, as shown in Axiom of Dependent Choice Implies Axiom of Countable Choice. Countable Choice Countable Choice 4g8kezgx135n752bu4a8j7r5ij9wckk


Kernel of Linear Transformation contains Zero Vector 0 16446 417034 349654 2019-08-07T12:48:10Z Prime.mover 59 wikitext text/x-wiki == Corollary to Linear Transformation Maps Zero Vector to Zero Vector == Let $\mathbf V$ be a vector space, with zero $\mathbf 0$. Likewise let $\mathbf V'$ be another vector space, with zero $\mathbf 0'$. Let $T: \mathbf V \to \mathbf V'$ be a linear transformation. Then: <onlyinclude> :$\mathbf 0 \in \map \ker T$ where $\map \ker T$ is the kernel of $T$. </onlyinclude> == Proof == Follows from Linear Transformation Maps Zero Vector to Zero Vector and the definition of kernel. {{qed}} Category:Linear Transformations Category:Linear Transformation Maps Zero Vector to Zero Vector orb8mqtbdb8a78fe6cg9tz5otd2n0qi


Fermat's Little Theorem/Corollary 1

Fermat's Little Theorem/Corollary 1 0 16621 486213 445854 2020-09-07T21:56:28Z Prime.mover 59 wikitext text/x-wiki == Corollary to Fermat's Little Theorem == <onlyinclude> If $p$ is a prime number, then $n^p \equiv n \pmod p$. </onlyinclude> == Proof 1 == {{:Fermat's Little Theorem/Corollary 1/Proof 1}} == Proof 2 == {{:Fermat's Little Theorem/Corollary 1/Proof 2}} == Also known as == Some sources call this '''Fermat's Little Theorem''', and from it derive that theorem as a corollary. == Also reported as == This result can also be reported as: :If $p$ is a prime number, then $n^p - n$ is divisible by $p$. {{Namedfor|Pierre de Fermat}} == Historical Note == {{:Fermat's Little Theorem/Historical Note}} == Sources == * {{BookReference|Men of Mathematics|1937|Eric Temple Bell|prev = Construction of Regular Prime p-Gon Exists iff p is Fermat Prime/Historical Note|next = Fermat's Little Theorem/Historical Note}}: Chapter $\text{IV}$: The Prince of Amateurs * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Fermat's Little Theorem|next = Prime Group has no Proper Subgroups}}: Exercise $25.6$ * {{BookReference|Number Theory|1971|George E. Andrews|prev = Binomial Theorem/Integral Index|next = Alternating Summation of Binomial Coefficient of Summation of Binomial Coefficient of Sequence}}: $\text {3-1}$ Permutations and Combinations: Exercise $11$ * {{BookReference|Number Theory|1971|George E. Andrews|prev = Congruence Modulo Integer/Examples/11 equiv -1 mod 12|next = Wilson's Theorem}}: $\text {4-1}$ Basic Properties of Congruences: Example $\text {4-2}$ * {{BookReference|Differential Equations|1972|George F. Simmons|prev = Fermat's Two Squares Theorem|next = Fermat's Last Theorem}}: $1$: The Nature of Differential Equations: $\S 6$: The Brachistochrone. Fermat and the Bernoullis * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Chinese Remainder Theorem|next = Fermat's Little Theorem/Proof 1}}: $\S 1.2.4$: Integer Functions and Elementary Number Theory: Theorem $\text{F}$ * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = Fermat's Little Theorem|next = Fermat's Little Theorem/Historical Note|entry = Fermat's theorem}} * {{BookReference|The Penguin Dictionary of Mathematics|2008|David Nelson|ed = 4th|edpage = Fourth Edition|prev = Fermat's Little Theorem|next = Fermat's Little Theorem/Historical Note|entry = Fermat's theorem}} * {{BookReference|Taming the Infinite|2008|Ian Stewart|prev = Lagrange's Four Square Theorem|next = Fermat's Last Theorem}}: Chapter $7$: Patterns in Numbers: Fermat * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Fermat's Little Theorem|next = Feuerbach's Theorem|entry = Fermat's Little Theorem}} Category:Fermat's Little Theorem 9s54cx1hptsxhfr3264m8jfyymogv7s


Generated Sigma-Algebra by Generated Monotone Class/Corollary 0 16680 179757 169487 2014-03-11T16:49:28Z Prime.mover 59 wikitext text/x-wiki == Corollary to Generated Sigma-Algebra by Generated Monotone Class == <onlyinclude> Let $X$ be a set, and let $\mathcal G \subseteq \mathcal P \left({X}\right)$ be a nonempty collection of subsets of $X$. Define $\complement_X \left({\mathcal G}\right)$ by: :$\complement_X \left({\mathcal G}\right) := \left\{{\complement_X \left({A}\right): A \in \mathcal G}\right\}$ Then: :$\sigma \left({\mathcal G}\right) = \mathfrak m \left({\mathcal G \cup \complement_X \left({\mathcal G}\right)}\right)$ </onlyinclude> Here, $\mathfrak m$ denotes generated monotone class, and $\sigma$ denotes generated $\sigma$-algebra. == Proof == From Set is Subset of Union: : $\mathcal G \subseteq \mathcal G \cup \complement_X \left({\mathcal G}\right)$ Further, as $\sigma \left({\mathcal G}\right)$ is a $\sigma$-algebra: :$S \in \sigma \left({\mathcal G}\right) \implies \complement_X \left({X}\right) = X \setminus S \in \sigma \left({\mathcal G}\right)$ from Set Difference as Intersection with Relative Complement. Since $\mathcal G \subseteq \sigma \left({\mathcal G}\right)$: :$\mathcal G \cup \complement_X \left({\mathcal G}\right) \subseteq \sigma \left({\mathcal G}\right)$ By Condition on Equality of Generated Sigma-Algebras: :$\sigma \left({\mathcal G}\right) = \sigma \left({\mathcal G \cup \complement_X \left({\mathcal G}\right)}\right)$ Applying Generated Sigma-Algebra by Generated Monotone Class: :$\sigma \left({\mathcal G}\right) = \mathfrak m \left({\mathcal G \cup \complement_X \left({\mathcal G}\right)}\right)$ {{qed}} Category:Sigma-Algebras Category:Monotone Classes 034okkbyk5kjaao4x9mh0k72o8wyjjb


Nth Derivative of Nth Power

Nth Derivative of Nth Power 0 16748 484121 466592 2020-08-30T10:52:22Z Prime.mover 59 wikitext text/x-wiki == Corollary to Nth Derivative of Mth Power == Let $n \in \Z$ be an integer such that $n \ge 0$. The $n$th derivative of $x^n$ {{WRT|Differentiation}} $x$ is: <onlyinclude> :$\dfrac {\d^n} {\d x^n} x^n = n!$ where $n!$ denotes $n$ factorial. </onlyinclude> == Proof == From Nth Derivative of Mth Power, we have: :$\dfrac {\d^n} {\d x^n} x^m = \begin {cases} m^\underline n \, x^{m - n} & : n \le m \\ 0 & : n > m \end {cases}$ where $m^\underline n$ denotes the falling factorial. Putting $m = n$: :$\dfrac {\d^n} {\d x^n} x^n = n^\underline n$ where from the definition of the falling factorial: :$n^\underline n = n!$ Hence the result. {{Qed}} Category:Derivatives n5qs4fn8tymcfoj9djk165f81mdu141


Unique Linear Transformation Between Vector Spaces 0 16781 147340 93014 2013-05-21T00:14:28Z Dfeuer 1672 Add to vector spaces category wikitext text/x-wiki == Corollary to Unique Linear Transformation Between Modules == <onlyinclude> Let $G$ be a finite-dimensional $K$-vector space. Let $H$ be a $K$-vector space (not necessarily finite-dimensional). Let $\left \langle {a_n} \right \rangle$ be a linearly independent sequence of vectors of $G$. Let $\left \langle {b_n} \right \rangle$ be a sequence of vectors of $H$. Then there is a unique linear transformation $\phi: G \to H$ satisfying $\forall k \in \left[{1 \,.\,.\, n}\right]: \phi \left({a_k}\right) = b_k$ </onlyinclude> == Proof == From Generator of Vector Space Contains Basis, $\left\{{a_1, \ldots, a_m}\right\}$ is contained in a basis for $G$. The result then follows from Unique Linear Transformation Between Modules. {{Qed}} == Sources == * {{BookReference|Modern Algebra|1965|Seth Warner|prev=Unique Linear Transformation Between Modules|next=Definition:Rank of Linear Transformation}}: $\S 28$: Theorem $28.4$: Corollary Category:Linear Transformations Category:Vector Spaces 0ke7kn6ysxeidzaioq8u2u93x6fr11x


Linear Transformations Isomorphic to Matrix Space/Corollary 0 16835 399279 182109 2019-04-06T22:05:54Z Prime.mover 59 wikitext text/x-wiki === Corollary to Linear Transformations Isomorphic to Matrix Space === <onlyinclude> Let $R$ be a commutative ring with unity. Let $M: \struct {\map {\mathcal L_R} G, +, \circ} \to \struct {\map {\mathcal M_R} n, +, \times}$ be defined as: :$\forall u \in \map {\mathcal L_R} G: \map M u = \sqbrk {u; \sequence {a_n} }$ Then $M$ is an isomorphism. </onlyinclude> == Proof == Follows directly from Linear Transformations Isomorphic to Matrix Space. {{qed}} {{Proofread}} == Sources == * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Definition:Order of Square Matrix|next = Ring of Square Matrices over Commutative Ring with Unity}}: $\S 29$: Theorem $29.2$ Category:Linear Algebra Category:Matrix Algebra 5fse8w10os2p8180ahpxdbqva6jv3ib


Fundamental Theorem of Calculus/First Part/Corollary 0 16873 486124 362632 2020-09-07T20:28:01Z Prime.mover 59 wikitext text/x-wiki == Corollary to Fundamental Theorem of Calculus (First Part) == Let $f$ be a real function which is continuous on the closed interval $\closedint a b$. Let $F$ be a real function which is defined on $\closedint a b$ by: :$\displaystyle \map F x = \int_a^x \map f t \rd t$ Then: <onlyinclude> :$\displaystyle \frac \d {\d x} \int_a^x \map f t \rd t = \map f x$ </onlyinclude> == Proof == Follows from the Fundamental Theorem of Calculus (First Part) and the definition of primitive. {{qed}} == Sources == * {{BookReference|Calculus|2005|Roland E. Larson|author2 = Robert P. Hostetler|author3 = Bruce H. Edwards|ed = 8th|edpage = Eighth Edition}}: $\S 4.4$ Category:Fundamental Theorem of Calculus e7l3fggbx4owixdxnxj43t6fskbolvm


Ring of Integers Modulo Prime is Integral Domain 0 16893 399547 399546 2019-04-09T06:50:31Z Prime.mover 59 wikitext text/x-wiki == Corollary to Ring of Integers Modulo Prime is Field == <onlyinclude> Let $m \in \Z: m \ge 2$. Let $\struct {\Z_m, +, \times}$ be the ring of integers modulo $m$. Then: :$m$ is prime {{iff}}: :$\struct {\Z_m, +, \times}$ is an integral domain. </onlyinclude> == Proof 1 == {{:Ring of Integers Modulo Prime is Integral Domain/Proof 1}} == Proof 2 == {{:Ring of Integers Modulo Prime is Integral Domain/Proof 2}} Category:Ring of Integers Modulo m Category:Integral Domains Category:Ring of Integers Modulo Prime is Field Category:Ring of Integers Modulo Prime is Integral Domain 7uqzb2473za599um7rmo9jc9r5orpp7


Reduced Residue System under Multiplication forms Abelian Group/Corollary 0 16903 374438 374436 2018-11-01T07:47:11Z Prime.mover 59 wikitext text/x-wiki == Corollary of Reduced Residue System under Multiplication forms Abelian Group == <onlyinclude> Let $p$ be a prime number. Let $\Z_p$ be the set of integers modulo $p$. Let $\struct {\Z'_p, \times}$ denote the multiplicative group of reduced residues modulo $p$. Then $\struct {\Z'_p, \times}$ is an abelian group. </onlyinclude> == Proof == Suppose $p \in \Z$ be a prime number. From the definition of reduced residue system modulo $p$, as $p$ is prime, $\Z'_p$ becomes: :$\set {\eqclass 1 p, \eqclass 2 p, \ldots, \eqclass {p - 1} p}$ This is precisely $\Z_p \setminus \set {\eqclass 0 p}$ which is what we wanted to show. The result follows from Reduced Residue System under Multiplication forms Abelian Group. {{qed}} == Sources == * {{BookReference|The Theory of Groups|1968|Ian D. Macdonald|prev = Modulo Multiplication is Well-Defined/Proof 1|next = Modulo Multiplication is Associative}}: $\S 1$: Some examples of groups: Example $1.11$ * {{BookReference|Elements of Abstract Algebra|1971|Allan Clark|prev = Multiplicative Inverse in Monoid of Integers Modulo m|next = Definition:Multiplicative Group of Reduced Residues}}: Chapter $2$: Examples of Group Structure: $\S 34$ * {{BookReference|Algebra|1974|Thomas W. Hungerford|prev = Reduced Residue System under Multiplication forms Abelian Group/Proof 2|next = Definition:Group of Rationals Modulo One}}: $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups * {{BookReference|Algebra|1974|Thomas W. Hungerford|prev = Klein Four-Group/Cayley Table|next = Modulo One is Congruence Relation on Rational Numbers}}: $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups: Exercise $7$ * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Integers Modulo m under Addition form Abelian Group|next = Definition:Symmetric Group}}: $\S 34$. Examples of groups: $(3)$ Category:Reduced Residue System under Multiplication forms Abelian Group 1o7zkplb21p4hmrgrkf9o439mnrtkva


Product with Ring Negative/Corollary

Product with Ring Negative/Corollary 0 16907 399490 346892 2019-04-08T21:27:04Z Prime.mover 59 wikitext text/x-wiki == Corollary to Product with Ring Negative == <onlyinclude> Let $\struct {R, +, \circ}$ be a ring with unity $1_R$. Then: :$\forall x \in R: \paren {-1_R} \circ x = -x$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = \paren {-1_R} \circ x       | r = -\paren {1_R \circ x}       | c = Product with Ring Negative }} {{eqn | r = -x       | c = {{Defof|Unity of Ring}} }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Product of Ring Negatives|next = Cancellation Law of Ring Product of Integral Domain}}: Chapter $1$: Integral Domains: $\S 4$. Elementary Properties: Theorem $2 \ \text{(vi)}$ Category:Ring Theory Category:Rings with Unity mxg1scb688xhderm1zwzc5b3x88sxs9


Ring Homomorphism Preserves Subrings/Corollary 0 16919 398070 397449 2019-03-29T08:17:53Z Prime.mover 59 wikitext text/x-wiki == Corollary to Ring Homomorphism Preserves Subrings == Let $\struct {R_1, +_1, \circ_1}$ and $\struct {R_2, +_2, \circ_2}$ be rings. <onlyinclude> The image of a ring homomorphism $\phi: R_1 \to R_2$ is a subring of $R_2$. </onlyinclude> == Proof == From Null Ring and Ring Itself Subrings, $R_1$ is a subring of itself. The result then follows from Ring Homomorphism Preserves Subrings. {{qed}} == Sources == * {{BookReference|Rings, Modules and Linear Algebra|1970|B. Hartley|author2 = T.O. Hawkes|prev = Kernel is Trivial iff Monomorphism/Ring|next = Definition:Coset}}: $\S 2.2$: Homomorphisms: Lemma $2.6 \ \text{(ii)}$ * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Ring Homomorphism Preserves Subrings/Proof 3|next = Kernel of Ring Homomorphism is Subring}}: $\S 57.3$ Ring homomorphisms: $\text{(i)}$ Category:Ring Homomorphisms Category:Subrings Category:Ring Homomorphism Preserves Subrings i0g4hpd4lq79nx8j6qjddrzqtpndsfh


Handshake Lemma/Corollary

Handshake Lemma/Corollary 0 16955 482233 482230 2020-08-17T21:01:36Z Prime.mover 59 wikitext text/x-wiki == Corollary to the Handshake Lemma == Let $G$ be a $\tuple {p, q}$-graph, which may be a multigraph or a loop-graph, or both. <onlyinclude> The number of odd vertices in $G$ is even. </onlyinclude> == Proof == Let $G$ be a $\tuple {p, q}$-graph. Consider the sum of the degrees of its vertices: :$\displaystyle K = \sum_{v \mathop \in V} \map {\deg_G} v$ From the Handshake Lemma: :$K = 2 \card E$ which is an even integer. Subtracting from $K$ the degrees of all even vertices, we are left with the sum of all degrees of odd vertices in $V$. That is: :$\displaystyle \paren {\sum_{v \mathop \in V} \map {\deg_G} v} - \paren {\sum_{v \mathop \in V : \map {\deg_G} v \mathop = 2 k} \map {\deg_G} v} = \paren {\sum_{v \mathop \in V : \map {\deg_G} v \mathop = 2 k + 1} \map {\deg_G} v}$ This must still be an even number, as it is equal to the difference of two even numbers. Because this is a sum of exclusively odd terms, there must be an even number of such terms for the sum on the {{RHS}} to be even. Hence the number of odd vertices in $G$ must be even. {{qed}} == Sources == * {{BookReference|Introductory Graph Theory|1977|Gary Chartrand|prev = Definition:Odd Vertex (Graph Theory)|next = Definition:Regular Graph}}: Chapter $2$: Elementary Concepts of Graph Theory: $\S 2.1$: The Degree of a Vertex: Theorem $2.2$ * {{BookReference|Calculus Gems|1992|George F. Simmons|prev = Definition:Eulerian Circuit|next = Characteristics of Eulerian Graph}}: Chapter $\text {A}.21$: Euler ({{DateRange|1707|1783}}) * {{BookReference|Curious and Interesting Puzzles|1992|David Wells|prev = Definition:Odd Vertex (Graph Theory)|next = Orthogonal Latin Squares of Order 6 do not Exist/Historical Note}}: The Bridges of Königsberg: $134$ Category:Handshake Lemma ih6kbddv63eqeggf5ra0uqq495xnrue


Image of Composite Mapping/Corollary

Image of Composite Mapping/Corollary 0 16980 416117 226853 2019-08-03T07:46:11Z Prime.mover 59 wikitext text/x-wiki == Corollary to Image of Composite Mapping == <onlyinclude> Let $f: S \to T$ and $g: R \to S$ be mappings. Then: :$\Img {f \circ g} \subseteq \Img f$ </onlyinclude> where: :$f \circ g$ denotes composition of $g$ and $f$ :$\Img f$ denotes image of $f$. == Proof == From Image of Composite Mapping, it holds that: :$\Img {f \circ g} = f \sqbrk {\Img g}$ where $f \sqbrk {\, \cdot \,}$ denotes image of subset. By definition of composite mapping: :$\Img g \subseteq \Dom f$ where $\Dom f$ denotes the domain of $f$. Now Image of Subset under Relation is Subset of Image: Corollary 2 yields: :$\Img {f \circ g} \subseteq \Img f$ {{qed}} == Sources == * {{BookReference|Undergraduate Topology|1971|Robert H. Kasriel|prev = Domain of Composite Mapping|next = Composition of Mappings is Associative}}: $\S 1.14$: Composition of Functions Category:Composite Mappings g66e40owxmfalc7v3i6ikwsrwrp9x7k


Relation between Two Ordinals/Corollary

Relation between Two Ordinals/Corollary 0 17233 144451 144447 2013-04-18T21:36:33Z Prime.mover 59 wikitext text/x-wiki == Corollary to Relation between Two Ordinals == <onlyinclude> Let $S$ and $T$ be ordinals. If $S \ne T$, then either $S$ is an initial segment of $T$, or vice versa. </onlyinclude> == Proof 1 == {{:Relation between Two Ordinals/Corollary/Proof 1}} == Proof 2 == {{:Relation between Two Ordinals/Corollary/Proof 2}} == Proof 3 == {{:Relation between Two Ordinals/Corollary/Proof 3}} Category:Ordinals 7fyjylzo2mfuc548txqu06h1pl3bg7e


Complement Union with Superset is Universe/Corollary 0 17270 416790 416716 2019-08-05T20:59:45Z Prime.mover 59 wikitext text/x-wiki == Corollary to Complement Union with Superset is Universe == <onlyinclude> :$S \cup T = \mathbb U \iff \map \complement S \subseteq T$ </onlyinclude> where: :$S \subseteq T$ denotes that $S$ is a subset of $T$ :$S \cup T$ denotes the union of $S$ and $T$ :$\complement$ denotes set complement :$\mathbb U$ denotes the universal set. == Proof == Let $X = \map \complement S$. Then: {{begin-eqn}} {{eqn | l = X \subseteq T       | o = \iff       | r = \map \complement X \cup T = \mathbb U       | c = Complement Union with Superset is Universe }} {{eqn | ll= \leadsto       | l = \map \complement S \subseteq T       | o = \iff       | r = \map \complement {\map \complement S} \cup T = \mathbb U       | c = substituting $X = \map \complement S$ }} {{eqn | ll= \leadsto       | l = \map \complement S \subseteq T       | o = \iff       | r = S \cup T = \mathbb U       | c = Complement of Complement }} {{end-eqn}} {{qed}} == Also see == * Empty Intersection iff Subset of Complement == Sources == * {{BookReference|Set Theory and Abstract Algebra|1975|T.S. Blyth|prev = Cardinality of Set Union/Examples/Student Subjects/Mathematics and Chemistry|next = Empty Intersection iff Subset of Complement}}: $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $11 \ \text{(a)}$ * {{BookReference|Introduction to Topology|1975|Bert Mendelson|ed = 3rd|edpage = Third Edition|prev = Empty Intersection iff Subset of Complement|next = Set Complement inverts Subsets}}: Chapter $1$: Theory of Sets: $\S 3$: Set Operations: Union, Intersection and Complement: Exercise $1 \ \text{(d)}$ Category:Subsets Category:Set Union Category:Set Complement Category:Empty Set qe36vxibbova8xblwt4zz2947z63op2


Image of Subset under Relation is Subset of Image/Corollary 1 0 17281 484938 226706 2020-09-03T12:31:06Z Prime.mover 59 wikitext text/x-wiki == Corollary to Image of Subset under Relation is Subset of Image == Let $S$ and $T$ be sets. Let $\RR \subseteq S \times T$ be a relation from $S$ to $T$. <onlyinclude> Let $C, D \subseteq T$. Then: :$C \subseteq D \implies \RR^{-1} \sqbrk C \subseteq \RR^{-1} \sqbrk D$ where $\RR^{-1} \sqbrk C$ is the preimage of $C$ under $\RR$. </onlyinclude> == Proof == We have that $\RR^{-1}$ is itself a relation, by definition of inverse relation. The result follows directly from Image of Subset under Relation is Subset of Image. {{qed}} Category:Subsets Category:Relation Theory Category:Mapping Theory i5ia4t6iycfmfoqbm1uamvevrgw1yjp


Image of Subset under Mapping is Subset of Image 0 17282 417838 417397 2019-08-11T14:24:15Z Prime.mover 59 wikitext text/x-wiki == Corollary to Image of Subset under Relation is Subset of Image == Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping from $S$ to $T$. <onlyinclude> Let $A, B \subseteq S$ such that $A \subseteq B$. Then the image of $A$ is a subset of the image of $B$: :$A \subseteq B \implies f \sqbrk A \subseteq f \sqbrk B$ This can be expressed in the language and notation of direct image mappings as: :$\forall A, B \in \powerset S: A \subseteq B \implies \map {f^\to} A \subseteq \map {f^\to} B$ </onlyinclude> == Proof == As $f: S \to T$ is a mapping, it is also a relation, and thus: :$f \subseteq S \times T$ The result follows directly from Image of Subset under Relation is Subset of Image. {{qed}} == Sources == * {{BookReference|Rings, Modules and Linear Algebra|1970|B. Hartley|author2 = T.O. Hawkes|prev = Image of Preimage under Mapping/Corollary|next = Image of Subset under Relation is Subset of Image/Corollary 3}}: $\S 2.2$: Homomorphisms: $\text{(ii)}$ * {{BookReference|Undergraduate Topology|1971|Robert H. Kasriel|prev = Mapping Images are Disjoint only if Domains are Disjoint|next = Image of Union under Mapping}}: $\S 1.10$: Functions: Exercise $5 \ \text{(c)}$ * {{BookReference|Set Theory and Abstract Algebra|1975|T.S. Blyth|prev = Direct Image Mapping of Mapping is Empty iff Argument is Empty|next = Image of Subset under Relation is Subset of Image/Corollary 3}}: $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.1: \ \text{(i)}$ * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Definition:Surjection/Also known as|next = Image of Intersection under Mapping/Proof 1}}: $\S 21.3$: The image of a subset of the domain; surjections * {{BookReference|Topology|2000|James R. Munkres|ed = 2nd|edpage = Second Edition|prev = Preimage of Set Difference under Mapping|next = Image of Union under Mapping}}: $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.2 \ \text{(e)}$ Category:Subsets Category:Mapping Theory ovvbvaj2tmbfp9jxrpv24wvoupdglyq


Image of Subset under Relation is Subset of Image/Corollary 3 0 17283 417837 417445 2019-08-11T14:24:08Z Prime.mover 59 wikitext text/x-wiki == Corollary of Image of Subset under Relation is Subset of Image == Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping from $S$ to $T$. <onlyinclude> Let $C, D \subseteq T$. Then: :$C \subseteq D \implies f^{-1} \sqbrk C \subseteq f^{-1} \sqbrk D$ This can be expressed in the language and notation of inverse image mappings as: :$\forall C, D \in \powerset T: C \subseteq D \implies \map {f^\gets} C \subseteq \map {f^\gets} D$ </onlyinclude> == Proof == As $f: S \to T$ is a mapping, it is also a relation, and thus so is its inverse: :$f^{-1} \subseteq T \times S$ The result follows directly from Image of Subset under Relation is Subset of Image. {{qed}} == Sources == * {{BookReference|Rings, Modules and Linear Algebra|1970|B. Hartley|author2 = T.O. Hawkes|prev = Image of Subset under Relation is Subset of Image/Corollary 2|next = Fourth Isomorphism Theorem}}: $\S 2.2$: Homomorphisms: $\text{(ii)}$ * {{BookReference|Set Theory and Abstract Algebra|1975|T.S. Blyth|prev = Image of Subset under Relation is Subset of Image/Corollary 2|next = Image of Union under Mapping/Proof 1}}: $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.1: \ \text{(j)}$ * {{BookReference|Topology|2000|James R. Munkres|ed = 2nd|edpage = Second Edition|prev = Image of Preimage of Subset under Surjection equals Subset|next = Preimage of Union under Mapping}}: $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.2 \ \text{(a)}$ Category:Subsets Category:Preimages under Mappings 5xj29ze8j94smaws01ws6c6xapyr8bu


Subset of Preimage under Relation is Preimage of Subset/Corollary 0 17289 417845 417454 2019-08-11T14:38:44Z Prime.mover 59 wikitext text/x-wiki == Corollary to Subset of Preimage under Relation is Preimage of Subset == Let $f: S \to T$ be a mapping. Let $X \subseteq S, Y \subseteq T$. Then: <onlyinclude> :$X \subseteq f^{-1} \sqbrk Y \iff f \sqbrk X \subseteq Y$ This can be expressed in the language and notation of direct image mappings and inverse image mappings as: :$\forall X \in \powerset S, Y \in \powerset T: X \subseteq \map {f^\gets} Y \iff \map {f^\to} X \subseteq Y$ </onlyinclude> == Proof == Let $f: S \to T$ be a mapping. As a mapping is also a relation, it follows that $f$ is a relation and so: :$X \subseteq f^{-1} \sqbrk Y \iff f \sqbrk X \subseteq Y$ holds on the strength of Subset of Preimage under Relation is Preimage of Subset. {{qed}} == Sources == * {{BookReference|Set Theory and Abstract Algebra|1975|T.S. Blyth|prev = Complement of Preimage equals Preimage of Complement|next = Definition:Composition of Mappings/Definition 1}}: $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.1 \ \text{(v)}$ Category:Subsets Category:Preimages under Mappings j3spfa22m9v2t52ky5c0280xviuz2f1


Monotone Convergence Theorem

Monotone Convergence Theorem 0 17378 497022 97090 2020-10-30T13:59:00Z Prime.mover 59 wikitext text/x-wiki {{Disambiguation}} * Monotone Convergence Theorem (Real Analysis) * Monotone Convergence Theorem (Measure Theory) mbc419yt8xmiqu5qcn58fczbv7yeyti


Cardinality of Cartesian Product/Corollary 0 17491 376852 342139 2018-11-15T22:21:46Z Prime.mover 59 wikitext text/x-wiki == Corollary to Cardinality of Cartesian Product == Let $S \times T$ be the cartesian product of two sets $S$ and $T$ which are both finite. Then: <onlyinclude> :$\card {S \times T} = \card {T \times S}$ </onlyinclude> where $\card {S \times T}$ denotes the cardinality of $S \times T$. == Proof 1 == {{:Cardinality of Cartesian Product/Corollary/Proof 1}} == Proof 2 == {{:Cardinality of Cartesian Product/Corollary/Proof 2}} == Sources == * {{BookReference|Sets and Groups|1965|J.A. Green|prev = Surjection from Finite Set to Itself is Permutation|next = Definition:Binary Operation}}: Chapter $3$. Mappings: Exercise $9 \ \text {(i)}$ * {{BookReference|Topology|2000|James R. Munkres|ed = 2nd|edpage = Second Edition|prev = Definition:Countable-Dimensional Real Cartesian Space|next = Definition:Finite Set}}: $1$: Set Theory and Logic: $\S 5$: Cartesian Products: Exercise $1$ Category:Cardinality of Cartesian Product jget1nbvchrxhkc9v1ziu3iypml1nxj


Modulo Subtraction is Well-Defined

Modulo Subtraction is Well-Defined 0 17673 451161 392592 2020-02-28T16:57:46Z Prime.mover 59 wikitext text/x-wiki == Corollary to Modulo Addition is Well-Defined == Let $m \in \Z$ be an integer. Let $\Z_m$ be the set of integers modulo $m$. The modulo subtraction operation on $\Z_m$, defined by the rule: <onlyinclude> :$\eqclass a m -_m \eqclass b m = \eqclass {a - b} m$ is a well-defined operation. </onlyinclude> That is: :If $a \equiv b \pmod m$ and $x \equiv y \pmod m$, then $a - x \equiv b - y \pmod m$. == Proof == We have: {{begin-eqn}} {{eqn | l = \eqclass a m -_m \eqclass b m       | r = \eqclass {a - b} m       | c =  }} {{eqn | r = \eqclass {a + \paren {-b} } m       | c =  }} {{eqn | r = \eqclass a m +_m \eqclass {-b} m       | c =  }} {{end-eqn}} The result follows from the fact that Modulo Addition is Well-Defined for all integers. {{qed}} == Examples == {{:Modulo Subtraction is Well-Defined/Examples}} == Sources == * {{BookReference|Sets and Groups|1965|J.A. Green|prev = Modulo Addition is Well-Defined/Proof 2|next = Modulo Multiplication is Well-Defined}}: $\S 2.6$. Algebra of congruences: $\text{(ii)}$ * {{BookReference|Number Theory|1971|George E. Andrews|prev = Modulo Addition is Well-Defined/Proof 1|next = Modulo Multiplication is Well-Defined/Proof 1}}: $\text {4-1}$ Basic Properties of Congruences: Theorem $\text {4-2}$ * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Modulo Addition is Well-Defined|next = Modulo Multiplication is Well-Defined}}: $\S 14.3 \ \text {(ii)}$: Congruence modulo $m$ ($m \in \N$) * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Modulo Addition is Well-Defined|next = Modulo Multiplication is Well-Defined}}: $\S 1.2.4$: Integer Functions and Elementary Number Theory: Law $\text{A}$ * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Modulo Addition is Well-Defined|next = Modulo Multiplication is Well-Defined|entry = congruence}} Category:Modulo Subtraction Category:Modulo Addition Category:Modulo Addition is Well-Defined k8f18w2wfijpyrj6un3079cm8tje8hj


Order Isomorphism between Ordinals and Proper Class/Lemma 0 17766 335895 297355 2018-01-11T20:45:11Z Barto 3079 notation for image clarified wikitext text/x-wiki {{MissingLinks|one or two}} == Lemma for Order Isomorphism between Ordinals and Proper Class == <onlyinclude> Suppose the following conditions are met: Let $A$ be a class. We allow $A$ to be a proper class or a set. Let $\left({A, \prec}\right)$ be a strict well-ordering. Let every $\prec$-initial segment be a set, not a proper class. Let $\operatorname{Im} \left({x}\right)$ denote the image of a subclass $x$. Let $G$ equal the class of all ordered pairs $\left({x, y}\right)$ satisfying: ::$y \in \left({A \setminus \operatorname{Im} \left({x}\right)}\right)$ ::The initial segment $A_y$ of $\left({A, \prec}\right)$ is a subset of $\operatorname{Im} \left({x}\right)$ Let $F$ be a mapping with a domain of $\operatorname{On}$. Let $F$ also satisfy: :$F \left({x}\right) = G \left({F \restriction x}\right)$ Then: : $G$ is a mapping : $G \left({x}\right) \in \left({A \setminus \operatorname{Im} \left({x}\right)}\right) \iff \left({A \setminus \operatorname{Im} \left({x}\right)}\right) \ne \varnothing$ </onlyinclude> Note that only the first four conditions need hold: we may construct classes $F$ and $G$ satisfying the other conditions using transfinite recursion. {{explain|Far too woolly. If it is important to the statement of the theorem, then it needs to be expounded properly. If it is not, then it needs to be put somewhere else, either in a separate section, or as a separate proof, or be deleted.}} == Proof == {{begin-eqn}} {{eqn | o =        | r = \left({\left({x, y}\right) \in G \land \left({x, z}\right) \in G}\right)       | c =  }} {{eqn | o = \implies        | r = \left({y \in \left({A \setminus \operatorname{Im} \left({x}\right)}\right) \land z \in \left({A \setminus \operatorname{Im} \left({x}\right)}\right)}\right)       | c = Definition of $G$ }} {{eqn | o = \implies       | r = \left({y \notin A_z \land y \notin A_y}\right)       | c = $A_y$ is disjoint with $\left({A \setminus \operatorname{Im} \left({x}\right)}\right)$.  Same with $A_z$. }} {{eqn | o = \implies       | r = \left({y \not \prec z \land z \not \prec y}\right)       | c = Definition of initial segment }} {{eqn | o = \implies       | r = y = z       | c = $\prec$ is a strict well-ordering }} {{end-eqn}} Therefore, we may conclude, that $G$ is a single-valued relation and therefore a mapping. For the second part: {{begin-eqn}} {{eqn | o =        | r = \left({A \setminus \operatorname{Im} \left({x}\right)}\right) \ne \varnothing       | c =  }} {{eqn | n = 1       | o = \implies       | r = \exists y \in \left({A \setminus \operatorname{Im} \left({x}\right)}\right): \left({A \cap A_y}\right) \setminus \operatorname{Im} \left({x}\right) = \varnothing       | c = Proper Well-Ordering Determines Smallest Elements }} {{eqn | o = \implies       | r = G \left({x}\right) = y       | c = Conditions are satisfied for $\left({x, y}\right) \in G$.  Follows from first part. }} {{eqn | o = \implies       | r = G \left({x}\right) \in \left({A \setminus \operatorname{Im} \left({x}\right)}\right)       | c = equation $(1)$, $y \in \left({A \setminus \operatorname{Im} \left({x}\right)}\right)$ }} {{end-eqn}} Furthermore: :$G \left({x}\right) \in \left({A \setminus \operatorname{Im} \left({x}\right)}\right) \implies \left({A \setminus \operatorname{Im} \left({x}\right)}\right) \ne \varnothing$ by the definition of nonempty. {{qed}} == Also see == * Transfinite Recursion * Condition for Injective Mapping on Ordinals * Maximal Injective Mapping from Ordinals to a Set == Sources == * {{BookReference|Introduction to Axiomatic Set Theory|1971|Gaisi Takeuti|author2 = Wilson M. Zaring}}: $\S 7.48$ Category:Ordinals Category:Order Theory jcc4dvmpmjxol5bxzw6uuc74ozq9vsq


Single Instruction URM Programs/Identity Function 0 17866 99431 2012-07-29T21:49:10Z Prime.mover 59 Created page with "== Theorem == <onlyinclude> The identity function $I_\N: \N \to \N$ defined as: :$\forall n \in \N: I_\N \l..." wikitext text/x-wiki == Theorem == <onlyinclude> The identity function $I_\N: \N \to \N$ defined as: :$\forall n \in \N: I_\N \left({n}\right) = n$ </onlyinclude> is URM computable by a single-instruction URM program. == Proof == Any of the following URM programs compute the identity function: {| |- ! align="right" | Line !! ! align="left" | Command |- | align="right" | $1$ || | align="left" | $Z \left({m}\right)$ |} ... where $m \ne 1$. This sets the value $0$ into $R_m$ and then stops. {| |- ! align="right" | Line !! ! align="left" | Command |- | align="right" | $1$ || | align="left" | $S \left({m}\right)$ |} ... where $m \ne 1$. The input $n$ is in $R_1$ when the program starts. The program adds $1$ to $r_m$ and then stops. {| |- ! align="right" | Line !! ! align="left" | Command |- | align="right" | $1$ || | align="left" | $C \left({j, m}\right)$ |} ... where $m \ne 1$. The input $n$ is in $R_1$ when the program starts. The program copies $r_j$ to $r_m$ and then stops. {| |- ! align="right" | Line !! ! align="left" | Command |- | align="right" | $1$ || | align="left" | $C \left({1, 1}\right)$ |} The input $n$ is in $R_1$ when the program starts. The program copies $r_1$ to $r_1$, i.e. to itself, and then stops. In none of these programs is $R_1$ affected, and so no change is effected to the input, which is returned as the output unchanged. Hence they all compute the identity function. {{qed}} Note that the latter program, consisting entirely of $C \left({1, 1}\right)$, is a direct implementation of the projection function program $\operatorname{pr}^1_1: \N \to \N$. This latter function is the usual way of implementing the identity function as it is well-defined and obvious, and guaranteed to have no other side-effects when embedded in a larger program. Category:URM Programs Category:Primitive Recursive Functions qtu82uqpzhzzsxzp19zei4qyxl8tw20


Cancellability of Congruences/Corollary 1 0 17943 392725 392713 2019-02-18T22:04:24Z Prime.mover 59 wikitext text/x-wiki == Corollary to Cancellability of Congruences == <onlyinclude> Let $c$ and $n$ be coprime integers, that is: :$c \perp n$ Then: :$c a \equiv c b \pmod n \implies a \equiv b \pmod n$ </onlyinclude> where $\equiv$ denotes congruence. == Proof 1 == {{:Cancellability of Congruences/Corollary 1/Proof 1}} == Proof 2 == {{:Cancellability of Congruences/Corollary 1/Proof 2}} == Proof 3 == {{:Cancellability of Congruences/Corollary 1/Proof 3}} Category:Cancellability of Congruences 6ulnth22u6ya1pzd0h89ftd3c7ohsry


Euler Phi Function of Integer/Corollary

Euler Phi Function of Integer/Corollary 0 18047 491187 342514 2020-09-27T13:00:53Z Prime.mover 59 wikitext text/x-wiki == Corollary to Euler Phi Function of Integer == Let $n \in \Z_{>0}$ be a (strictly) positive integer. Let $p$ be a prime number. Let $\map \phi n$ denote the Euler $\phi$ function of $n$. <onlyinclude> Let $p$ be a divisor of $n$. Then $p - 1$ is a divisor of $\phi \left({n}\right)$. </onlyinclude> == Proof == From Euler Phi Function of Integer: :$\displaystyle \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$ Let $n$ be expressed as its prime decomposition: :$n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}, p_1 < p_2 < \ldots < p_r$ Then: {{begin-eqn}} {{eqn | l = \phi \left({n}\right)       | r = \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i} \paren {1 - \dfrac 1 {p_i} }       | c =  }} {{eqn | r = \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i} \paren {\dfrac {p_i - 1} {p_i} }       | c =  }} {{eqn | r = \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i - 1} \paren {p_i - 1}       | c =  }} {{eqn | r = \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i - 1} \paren {p_i - 1}       | c =  }} {{end-eqn}} Hence the result. {{qed}} Category:Euler Phi Function 1dua6m2t9769x60o0hprendhv8zk4de


Euler Phi Function of Prime Power/Corollary 0 18049 371220 102037 2018-10-14T12:28:50Z Prime.mover 59 wikitext text/x-wiki == Corollary to Euler Phi Function of Prime Power == Let $\phi: \Z_{>0} \to \Z_{>0}$ be the Euler $\phi$ function. Then: <onlyinclude> :$\map \phi {2^k} = 2^{k-1}$ </onlyinclude> == Proof == We have that: :$\displaystyle 1 - \frac 1 2 = \frac {2 - 1} 2 = \frac 1 2$ It follows from Euler Phi Function of Prime Power: :$\map \phi {2^k} = \paren {\dfrac 1 2} 2^k = 2^{k - 1}$ {{qed}} Category:Euler Phi Function Category:Prime Numbers kcxcsg07owdm84w44f3o5tfdjisapfd


Abelian Group Factored by Prime/Corollary 0 18066 389112 389034 2019-01-22T21:08:48Z Prime.mover 59 wikitext text/x-wiki == Corollary to Abelian Group Factored by Prime == <onlyinclude> Any finite abelian group $G$ can be factored as follows: Let $\order G = \displaystyle \prod_{i \mathop = 1}^k p_i^{n_i}$ be the prime factorisation of the order of $G$. Then we have $G = \displaystyle \prod_{i \mathop = 1}^k H_i$, where $H_i = \set {x \in G : x^{p_i^{n_i} } = e}$. This factorisation is unique up to ordering of the factors. </onlyinclude> == Proof == Let $\displaystyle \prod_{i \mathop = 1}^k p_i^{n_i}$ be the prime factorisation of $\order G$. We proceed by induction on $k$. === Basis for the induction === For $n = 1$, the statement is trivial. === Induction Hypothesis === Now we assume the theorem is true for abelian groups whose order has $k - 1$ distinct prime factors. === Induction Step === Apply Abelian Group Factored by Prime to $G$ and $p_1$.  By definition, $H = H_1$. Also, the resulting $K$ has $\order K = \displaystyle \prod_{i \mathop = 2}^k p_i^{n_i}$. Therefore, it satisfies the induction hypothesis. It follows that $G = H_1 \times \displaystyle \prod_{i \mathop = 2}^k H_i$. From: :Subgroup of Abelian Group is Normal :the definition of Sylow $p$-subgroup all the $H_i$ are normal Sylow $p$-subgroups. From Sylow $p$-Subgroup is Unique iff Normal, the factorisation is unique up to ordering of the factors. {{qed}} Category:Abelian Groups shkkcod24tx5x3qpry41iu7ssb6xpeg


Measure is Subadditive/Corollary

Measure is Subadditive/Corollary 0 18086 100774 2012-08-06T07:40:12Z Lord Farin 560 Created page with "== Theorem ==  Let $\left({X, \Sigma, \mu}\right)$ be a measure space. <onlyinclude> Let $E_1, \ldots, E_n \in \Sigma$.   Then:  :$\displaystyle \..." wikitext text/x-wiki == Theorem == Let $\left({X, \Sigma, \mu}\right)$ be a measure space. <onlyinclude> Let $E_1, \ldots, E_n \in \Sigma$. Then: :$\displaystyle \mu \left({\bigcup_{k \mathop = 1}^n E_k}\right) \le \sum_{k \mathop = 1}^n \mu \left({E_k}\right)$. </onlyinclude> == Proof == We have Measure is Subadditive. The result follows by an application of Finite Union of Sets in Subadditive Function. {{qed}} Category:Measure Theory 4yu4a6sg5yzpmyk0x79qp2m3gzyxi8v


Composite Functor is Functor

Composite Functor is Functor 0 18201 439677 102954 2019-12-16T08:44:17Z Prime.mover 59 wikitext text/x-wiki == Definition == Let $\mathbf C, \mathbf D$ and $\mathbf E$ be metacategories. Let $F: \mathbf C \to \mathbf D$ and $G: \mathbf D \to \mathbf E$ be covariant functors. Let $GF: \mathbf C \to \mathbf E$ be the composition of $G$ with $F$. Then $G F$ is also a covariant functor. == Proof == Let $f, g$ be morphisms of $\mathbf C$ such that $g \circ f$ is defined. Then: {{begin-eqn}} {{eqn | l = \map {G F} {g \circ f}       | r = \map G {\map F {g \circ f} }       | c = {{Defof|Composition of Functors}} }} {{eqn | r = \map G {F g \circ F f}       | c = $F$ is a Covariant Functor }} {{eqn | r = \map G {F g} \circ \map G {F f}       | c = $G$ is a functor }} {{eqn | r = G F g \circ G F f       | c = {{Defof|Composition of Functors}} }} {{end-eqn}} Also, for any object $C$ of $\mathbf C$: {{begin-eqn}} {{eqn | l = G F I_C       | r = \map G {F I_C}       | c = {{Defof|Composition of Functors}} }} {{eqn | r = G I_{F C}       | c = $F$ is a Covariant Functor }} {{eqn | r = I_{\map G {F C} }       | c = $G$ is a Covariant Functor }} {{eqn | r = I_{G F C}       | c = {{Defof|Composition of Functors}} }} {{end-eqn}} Hence $G F$ is shown to be a covariant functor. {{qed}} Category:Category Theory d5uraaxxfmvvnznnmua2jo3e8jfd94v


Transitive Set is Proper Subset of Ordinal iff Element of Ordinal/Corollary 0 18556 175087 107940 2014-01-26T01:40:13Z Prime.mover 59 wikitext text/x-wiki == Corollary to Transitive Set is Proper Subset of Ordinal iff Element of Ordinal == <onlyinclude> Let $A$ and $B$ be ordinals. Then: :$A \subsetneq B \iff A \in B$ </onlyinclude> == Proof == We have that an ordinal is transitive. The result follows directly from Transitive Set is Proper Subset of Ordinal iff Element of Ordinal. {{qed}} == Sources == * {{BookReference|Introduction to Axiomatic Set Theory|1971|Gaisi Takeuti|author2=Wilson M. Zaring}}: $\S 7.8$ Category:Ordinals qcqrzek318h7uxv2h5ih4a7edxf8ns9


Absolute Value of Integer is not less than Divisors/Corollary 0 18700 497629 401283 2020-11-02T22:58:51Z Prime.mover 59 wikitext text/x-wiki == Corollary to Absolute Value of Integer is not less than Divisors == <onlyinclude> Let $a, b \in \Z_{>0}$ be (strictly) positive integers. Let $a \divides b$. Then: :$a \le b$ </onlyinclude> == Proof == Follows directly from Absolute Value of Integer is not less than Divisors. {{qed}} == Sources == * {{BookReference|Elements of Abstract Algebra|1966|Richard A. Dean|prev = Definition:Integer|next = Definition:Divisor of Integer}}: $\S 0.1$. Arithmetic: Theorem $1$ * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Divisor Relation is Antisymmetric/Corollary/Proof 2|next = Division Theorem/Positive Divisor/Existence/Proof 3}}: Chapter $3$: The Integers: $\S 10$. Divisibility: Theorem $17 \ \text{(iii)}$ Category:Number Theory Category:Divisors 5u0zsn3k7vf61zx6tz78aj2617jkl42


Euclid's Lemma for Prime Divisors/General Result 0 18713 462840 462839 2020-04-18T17:06:14Z Prime.mover 59 wikitext text/x-wiki == Lemma == <onlyinclude> Let $p$ be a prime number. Let $\displaystyle n = \prod_{i \mathop = 1}^r a_i$. Then if $p$ divides $n$, it follows that $p$ divides $a_i$ for some $i$ such that $1 \le i \le r$. That is: :$p \divides a_1 a_2 \ldots a_n \implies p \divides a_1 \lor p \divides a_2 \lor \cdots \lor p \divides a_n$ </onlyinclude> == Proof 1 == {{:Euclid's Lemma for Prime Divisors/General Result/Proof 1}} == Proof 2 == {{:Euclid's Lemma for Prime Divisors/General Result/Proof 2}} == Proof 3 == {{:Euclid's Lemma for Prime Divisors/General Result/Proof 3}} {{Namedfor|Euclid}} == Sources == * {{BookReference |An Introduction to the Theory of Numbers|1979|G.H. Hardy|author2 = E.M. Wright|ed = 5th|edpage = Fifth Edition|prev = Euclid's Lemma for Prime Divisors|next = Expression for Integer as Product of Primes is Unique/Proof 1}}: $\text I$: The Series of Primes: $1.3$ Statement of the fundamental theorem of arithmetic Category:Euclid's Lemma for Prime Divisors qcekxkp6nvapbtae89pe99sdjj27qw2


Euclid's Lemma for Prime Divisors/General Result/Proof 2 0 18721 490838 401317 2020-09-25T22:13:55Z Prime.mover 59 wikitext text/x-wiki == Lemma == {{:Euclid's Lemma for Prime Divisors/General Result}} == Proof == <onlyinclude> Proof by induction: For all $r \in \N_{>0}$, let $\map P r$ be the proposition: :$\displaystyle p \divides \prod_{i \mathop = 1}^r a_i \implies \exists i \in \closedint 1 r: p \divides a_i$ $\map P 1$ is true, as this just says $p \divides a_1 \implies p \divides a_1$. === Basis for the Induction === $\map P 2$ is the case: :$p \divides a_1 a_2 \implies p \divides a_2$ or $p \divides a_2$ which is proved in Euclid's Lemma for Prime Divisors. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\displaystyle p \divides \prod_{i \mathop = 1}^k a_i \implies \exists i \in \closedint 1 k: p \divides a_i$ Then we need to show: :$\displaystyle p \divides \prod_{i \mathop = 1}^{k + 1} a_i \implies \exists i \in \closedint 1 {k + 1}: p \divides a_i$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = p       | o = \divides       | r = a_1 a_2 \ldots a_{k + 1}       | c =  }} {{eqn | ll= \leadsto       | l = p       | o = \divides       | r = \paren {a_1 a_2 \ldots a_k} \paren {a_{k + 1} }       | c = }} {{eqn | ll= \leadsto       | l = p       | o = \divides       | r = a_1 a_2 \ldots a_k \lor p \divides a_{k + 1}       | c = Basis for the Induction }} {{eqn | ll= \leadsto       | l = p       | o = \divides       | r = a_1 \lor p \divides a_2 \lor \ldots \lor p \divides a_k \lor p \divides a_{k + 1}       | c = Induction Hypothesis }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall r \in \N: p \divides \prod_{i \mathop = 1}^r a_i \implies \exists i \in \closedint 1 r: p \divides a_i$ {{qed}} </onlyinclude> {{Namedfor|Euclid}} == Sources == * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Euclid's Lemma for Prime Divisors/Proof 2|next = Fundamental Theorem of Arithmetic}}: Chapter $3$: The Integers: $\S 12$. Primes: Theorem $21 \ \text{(ii)}$ * {{BookReference|Number Theory|1971|George E. Andrews|prev = Euclid's Lemma for Prime Divisors/Proof 2|next = Euclidean Algorithm/Examples/527 and 765}}: $\text {2-2}$ Divisibility: Corollary $\text {2-4}$ * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Euclid's Lemma for Prime Divisors/Proof 2|next = Fundamental Theorem of Arithmetic}}: $\S 12.6$: Highest common factors and Euclid's algorithm * {{BookReference|Computability and Unsolvability|1982|Martin Davis|ed = 2nd|edpage = Second Edition|prev = Euclid's Lemma for Prime Divisors/Proof 2|next = Fundamental Theorem of Arithmetic}}: Appendix $1$: Some Results from the Elementary Theory of Numbers: Corollary $9$ Category:Euclid's Lemma for Prime Divisors Category:Proofs by Induction j804jinwhibh8z6fiajssv98bjlprim


Euclid's Lemma for Irreducible Elements/General Result 0 18722 402087 402086 2019-04-22T08:02:44Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $\struct {D, +, \times}$ be a Euclidean domain whose unity is $1$. <onlyinclude> Let $p$ be an irreducible element of $D$. Let $n \in D$ such that: :$\displaystyle n = \prod_{i \mathop = 1}^r a_i$ where $a_i \in D$ for all $i: 1 \le i \le r$. If $p$ divides $n$, then $p$ divides $a_i$ for some $i$. That is: :$p \divides a_1 a_2 \ldots a_n \implies p \divides a_1 \lor p \divides a_2 \lor \cdots \lor p \divides a_n$ </onlyinclude> == Proof == Proof by induction: For all $r \in \N_{>0}$, let $\map P r$ be the proposition: :$\displaystyle p \divides \prod_{i \mathop = 1}^r a_i \implies \exists i \in \closedint 1 r: p \divides a_i$ $\map P 1$ is true, as this just says $p \divides a_1 \implies p \divides a_1$. === Basis for the Induction === $\map P 2$ is the case: :$p \divides a_1 a_2 \implies p \divides a_2$ or $p \divides a_2$ which is proved in Euclid's Lemma for Irreducible Elements. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\displaystyle p \divides \prod_{i \mathop = 1}^k a_i \implies \exists i \in \closedint 1 k: p \divides a_i$ Then we need to show: :$\displaystyle p \divides \prod_{i \mathop = 1}^{k + 1} a_i \implies \exists i \in \closedint 1 {k + 1}: p \divides a_i$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = p       | o = \divides       | r = a_1 a_2 \ldots a_{k + 1}       | c =  }} {{eqn | ll= \leadsto       | l = p       | o = \divides       | r = \paren {a_1 a_2 \ldots a_k} \paren {a_{k + 1} }       | c =  }} {{eqn | ll= \leadsto       | l = p       | o = \divides       | r = a_1 a_2 \ldots a_k \lor p \divides a_{k + 1}       | c = Basis for the Induction }} {{eqn | ll= \leadsto       | l = p       | o = \divides       | r = a_1 \lor p \divides a_2 \lor \ldots \lor p \divides a_k \lor p \divides a_{k + 1}       | c = Induction Hypothesis }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall r \in \N: p \divides \prod_{i \mathop = 1}^r a_i \implies \exists i \in \closedint 1 r: p \divides a_i$ {{qed}} {{Namedfor|Euclid}} == Also see == * Euclid's Lemma for Prime Divisors, for the usual statement of this result, which is this lemma as applied specifically to the integers. == Sources == * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Euclid's Lemma for Irreducible Elements|next = Polynomial Forms over Field form Principal Ideal Domain/Corollary 2}}: Chapter $6$: Polynomials and Euclidean Rings: $\S 29$. Irreducible elements: Theorem $56 \ \text{(ii)}$ Category:Euclidean Domains Category:Euclid's Lemma 2sd682kw559rrgpxie3yybdzq0nfccp


Common Divisor Divides Integer Combination 0 18747 493818 462684 2020-10-10T21:57:07Z Prime.mover 59 wikitext text/x-wiki == Corollary to Common Divisor in Integral Domain Divides Linear Combination == <onlyinclude> Let $c$ be a common divisor of two integers $a$ and $b$. That is: :$a, b, c \in \Z: c \divides a \land c \divides b$ Then $c$ divides any integer combination of $a$ and $b$: :$\forall p, q \in \Z: c \divides \paren {p a + q b}$ </onlyinclude> === Corollary === {{:Common Divisor Divides Integer Combination/Corollary}} === General Result === {{:Common Divisor Divides Integer Combination/General Result}} == Proof 1 == {{:Common Divisor Divides Integer Combination/Proof 1}} == Proof 2 == {{:Common Divisor Divides Integer Combination/Proof 2}} == Sources == * {{BookReference |An Introduction to the Theory of Numbers|1979|G.H. Hardy|author2 = E.M. Wright|ed = 5th|edpage = Fifth Edition|prev = Multiple of Divisor Divides Multiple|next = Definition:Prime Number/Definition 4}}: $\text I$: The Series of Primes: $1.1$ Divisibility of integers * {{BookReference|Elementary Number Theory|1980|David M. Burton|ed = revised|edpage = Revised Printing|prev = Absolute Value of Integer is not less than Divisors|next = Common Divisor Divides Integer Combination/General Result}}: Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Theorem $2 \text{-} 2 \ (7)$ * {{BookReference|A Course in Number Theory|1994|H.E. Rose|ed = 2nd|edpage = Second Edition|prev = Divisor Relation is Transitive|next = Definition:Greatest Common Divisor/Integers/Definition 2}}: $1$ Divisibility: $1.1$ The Euclidean algorithm and unique factorization: $\text {(v)}$ Category:Divisors Category:Integer Combinations Category:Common Divisor Divides Integer Combination b2qsir1pvol6l0ql0sfpxscdcek3kax


Divisor Relation is Antisymmetric/Corollary/Proof 2 0 18762 447447 411262 2020-02-08T13:30:51Z Prime.mover 59 wikitext text/x-wiki == Corollary to Divisor Relation is Antisymmetric == {{:Divisor Relation is Antisymmetric/Corollary}} == Proof == <onlyinclude> Let $a \divides b$ and $b \divides a$. Then by definition of divisor: :$\exists c, d \in \Z: a c = b, b d = a$ Thus: {{begin-eqn}} {{eqn | l = a c d       | r = a       | c =  }} {{eqn | ll= \leadsto       | l = c d       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = d       | r = \pm 1       | c = Divisors of One }} {{eqn | ll= \leadsto       | l = a       | r = \pm b       | c = as $a = b d$ }} {{end-eqn}} {{qed}} </onlyinclude> == Sources == * {{BookReference|Lectures in Abstract Algebra|1951|Nathan Jacobson|volume = I|subtitle = Basic Concepts|prev = Divisor Relation is Transitive|next = Definition:Greatest Common Divisor/Integers/Definition 2}}: Introduction $\S 6$: The division process in $I$ * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Divisors of One|next = Absolute Value of Integer is not less than Divisors/Corollary}}: Chapter $3$: The Integers: $\S 10$. Divisibility: Theorem $17 \ \text{(ii)}$ * {{BookReference|Algebra Volume 1|1982|P.M. Cohn|edpage = Second Edition|ed = 2nd|prev = Divisor Divides Multiple|next = Definition:Associate of Integer}}: Chapter $2$: Integers and natural numbers: $\S 2.2$: Divisibility and factorization in $\mathbf Z$ Category:Divisor Relation is Antisymmetric d5drfre7at4ibdme38oh5qjbo896q6d


Polynomial Forms over Field form Principal Ideal Domain/Corollary 3 0 18766 489212 488790 2020-09-19T13:30:43Z Prime.mover 59 wikitext text/x-wiki == Corollary to Polynomial Forms over Field form Principal Ideal Domain == Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$. Let $X$ be transcendental over $F$. Let $F \sqbrk X$ be the ring of polynomials in $X$ over $F$. Then <onlyinclude>$F \sqbrk X$ is a unique factorization domain. </onlyinclude> == Proof == We have the result Principal Ideal Domain is Unique Factorization Domain. The result then follows from Polynomial Forms over Field form Principal Ideal Domain. {{qed}} == Sources == * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Polynomial Forms over Field form Principal Ideal Domain/Proof 1|next = Polynomial Forms over Field form Principal Ideal Domain/Corollary 1}}: $\S 65.2$ Some properties of $F \sqbrk X$, where $F$ is a field Category:Polynomial Forms over Field form Principal Ideal Domain Category:Unique Factorization Domains eh1gxv09ue7w3wdhy4cflj5ey11daxz