hcju/proofwikips huggingface dataset:
Gelfond-Schneider Theorem/Lemma 2
 == Lemma ==  Let $\map f z$ be an analytic function in the disk $D \subseteq \C: D = \set {z : \size z < R}$ for some real number $R > 0$. Let $f$ also be continuous on the closure of $D$, that is, on $D^- = \set {z : \size z \le R}$. Then: :$\forall z \in D^-: \size {\map f z} \le \size f_R$, where we set $\size f_R = \max_{\size z = R} \size {\map f z}$.  This lemma is essentially a restatement of the maximum modulus principle for analytic functions. == Proof == Since $D^-$ is compact, the continuous function $\size f$ has a maximum point $z_0 \in D^-$, meaning that $\size {\map f {z_0} } = \max_{D^-} \size f$. If $z_0$ belongs to the open disk $D$, then the maximum modulus principle implies that $f$ is constant, in which case the statement is trivial. If instead $z_0$ belongs to the boundary $D^- \setminus D$, then $\size {z_0} = R$ and so $\size {\map f {z_0} } \le \size f_R$. Since $\size {\map f z} \le \size {\map f {z_0} }$ for any $z \in D^-$, the statement follows. {{qed|lemma}}
Gelfond-Schneider Theorem/Lemma 4
== Lemma ==  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 $\ge T^{−L}$.  == Proof == Observe that $T^L \Delta$ is an algebraic integer so that one of its conjugates has absolute value $\ge 1$. The result follows.  {{Proofread|There are plenty of definitions on this page which we may not have completely accurate.}}
Frobenius's Theorem/Lemma 1
== Lemma ==  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 $v^2 = -1$.  == 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.
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 ==  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: : $e_{k+1}^2 = -1$ : $\forall i \le k: e_i e_{k+1} = -e_{k+1} e_i$  == 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}}
Singleton Equality
{{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}}  == Sources == * {{BookReference|Set Theory and Its Logic|1963|Willard Van Orman Quine}}: $\S 7.7$
Cardinals are Totally Ordered
== 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 $\le$.  {{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
Idempotent Operators
{{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  it is self-adjoint. * An idempotent operator is an orthogonal projector  its norm is 1 (proof). {{refactor|The following to be done:<br/>1: Idempotent needs to be given its own definition page in this context, as do Projector and Orthogonal Projector<br/>2: The "immediate consequence" above needs to be given its own proof page<br>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.<br/>It is also worth reviewing the meaning of $\Rng P$ in this context, as "range" is ambiguously understood.}}
Squeeze Theorem for Real Sequences/Corollary
== Corollary to Squeeze Theorem for Real Sequences ==  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 $n \to \infty$.  == 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 $n \to \infty$.  == 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
Taylor's Theorem/One Variable
== Theorem ==  {{:Taylor's Theorem/One Variable/Statement of Theorem}}  == 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}}
Derivative of Exponential Function/Corollary 1
== Corollary to Derivative of Exponential Function == Let $\exp x$ be the exponential function.  Let $c \in \R$. Then: :$\map {\dfrac \d {\d x} } {\map \exp {c x} } = c \map \exp {c x}$  == 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}}  == 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: $3$.
Derivative of General Exponential Function
== Corollary to Derivative of Exponential Function ==  Let $a \in \R: a > 0$. Let $a^x$ be $a$ to the power of $x$. Then: :$\map {\dfrac \d {\d x} } {a^x} = a^x \ln a$  == Proof 1 == {{:Derivative of General Exponential Function/Proof 1}} == Proof 2 == {{:Derivative of General Exponential Function/Proof 2}}
Quaternion Group/Complex Matrices
== Representation of Quaternion Group ==  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''':  === 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: $(6) \ \text{(ii)}$
Axiom:Leibniz's Law
== 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  $x$ 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}}
Axiom:Axiom of Dependent Choice
== 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
== Corollary to Cartesian Product of Intersections ==  :$A \times \paren {B \cap C} = \paren {A \times B} \cap \paren {A \times C}$  == 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}}  == 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 $3 \ (1)$
Image of Set Difference under Relation/Corollary 1
== Corollary to Image of Set Difference under Relation ==  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.  == 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: :$\RR \sqbrk B \setminus \RR \sqbrk A \subseteq \RR \sqbrk {B \setminus A}$
Image of Set Difference under Relation/Corollary 2
== Corollary to Image of Set Difference under Relation ==  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 $\RR$.  == 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: :$\RR \sqbrk S \setminus \RR \sqbrk A \subseteq \RR \sqbrk {S \setminus A}$
Zassenhaus Lemma
== 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: :$(1): \quad N_1 \paren {H_1 \cap N_2} \lhd N_1 \paren {H_1 \cap H_2}$ :$(2): \quad \paren {H_1 \cap N_2} \paren {N_1 \cap H_2} \lhd H_1 \cap H_2$  === 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.  == 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
Axiom:Pasch's Axiom
{{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.}}  * Pasch's Axiom in Euclidean Geometry * Pasch's Axiom in Tarski's Geometry
Taylor's Theorem/One Variable/Proof by Cauchy Mean Value Theorem
== Theorem == {{:Taylor's Theorem/One Variable/Statement of Theorem}} == Proof ==  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 $\map G t = t - a$.   {{Namedfor|Brook Taylor}}