hcju/proofwikips huggingface dataset:
Multiplicative Group of Reduced Residues Modulo 5/Cayley Table
== Cayley Table for Multiplicative Group of Reduced Residues Modulo 5 == The multiplicative group of reduced residues modulo $5$: :$\Z'_5 = \set {\eqclass 1 5, \eqclass 2 5, \eqclass 3 5, \eqclass 4 5}$ can be described completely by showing its Cayley table:  :$\begin{array}{r|rrrr} \times_5 & \eqclass 1 5 & \eqclass 2 5 & \eqclass 3 5 & \eqclass 4 5 \\ \hline \eqclass 1 5 & \eqclass 1 5 & \eqclass 2 5 & \eqclass 3 5 & \eqclass 4 5 \\ \eqclass 2 5 & \eqclass 2 5 & \eqclass 4 5 & \eqclass 1 5 & \eqclass 3 5 \\ \eqclass 3 5 & \eqclass 3 5 & \eqclass 1 5 & \eqclass 4 5 & \eqclass 2 5 \\ \eqclass 4 5 & \eqclass 4 5 & \eqclass 3 5 & \eqclass 2 5 & \eqclass 1 5 \\ \end{array}$ By arranging the rows and columns into a different order, its cyclic nature becomes clear: :$\begin{array}{r|rrrr} \times_5 & \eqclass 1 5 & \eqclass 2 5 & \eqclass 4 5 & \eqclass 3 5 \\ \hline \eqclass 1 5 & \eqclass 1 5 & \eqclass 2 5 & \eqclass 4 5 & \eqclass 3 5 \\ \eqclass 2 5 & \eqclass 2 5 & \eqclass 4 5 & \eqclass 3 5 & \eqclass 1 5 \\ \eqclass 4 5 & \eqclass 4 5 & \eqclass 3 5 & \eqclass 1 5 & \eqclass 2 5 \\ \eqclass 3 5 & \eqclass 3 5 & \eqclass 1 5 & \eqclass 2 5 & \eqclass 4 5 \\ \end{array}$  == Sources == * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Modulo Addition/Cayley Table/Modulo 4|next = Modulo Multiplication is Well-Defined}}: Chapter $2$: Maps and relations on sets: Exercise $5$
Group Generated by Reciprocal of z and 1 minus z/Cayley Table
== Cayley Table for Group Generated by Reciprocal of $1 / z$ and $1 - z$ == We have: {{begin-eqn}} {{eqn | l = \map {f_1} z       | r = z }} {{eqn | l = \map {f_2} z       | r = \dfrac 1 {1 - z} }} {{eqn | l = \map {f_3} z       | r = \dfrac {z - 1} z }} {{eqn | l = \map {f_4} z       | r = \dfrac 1 z }} {{eqn | l = \map {f_5} z       | r = 1 - z }} {{eqn | l = \map {f_6} z       | r = \dfrac z {z - 1} }} {{end-eqn}} Hence from Group Generated by Reciprocal of z and 1 minus z:  :$\begin{array}{r|rrrrrr} \circ & f_1 & f_2 & f_3 & f_4 & f_5 & f_6 \\ \hline f_1 & f_1 & f_2 & f_3 & f_4 & f_5 & f_6 \\ f_2 & f_2 & f_3 & f_1 & f_6 & f_4 & f_5 \\ f_3 & f_3 & f_1 & f_2 & f_5 & f_6 & f_4 \\ f_4 & f_4 & f_5 & f_6 & f_1 & f_2 & f_3 \\ f_5 & f_5 & f_6 & f_4 & f_3 & f_1 & f_2 \\ f_6 & f_6 & f_4 & f_5 & f_2 & f_3 & f_1 \\ \end{array}$ Expressing the elements in full: :$\begin{array}{c|cccccc} \circ & z & \dfrac 1 {1 - z} & \dfrac {z - 1} z & \dfrac 1 z & 1 - z & \dfrac z {z - 1} \\ \hline z & z & \dfrac 1 {1 - z} & \dfrac {z - 1} z & \dfrac 1 z & 1 - z & \dfrac z {z - 1} \\ \dfrac 1 {1 - z} & \dfrac 1 {1 - z} & \dfrac {z - 1} z & z & \dfrac z {z - 1} & \dfrac 1 z & 1 - z \\ \dfrac {z - 1} z & \dfrac {z - 1} z & z & \dfrac 1 {1 - z} & 1 - z & \dfrac z {z - 1} & \dfrac 1 z \\ \dfrac 1 z & \dfrac 1 z & 1 - z & \dfrac z {z - 1} & z & \dfrac 1 {1 - z} & \dfrac {z - 1} z \\ 1 - z & 1 - z & \dfrac z {z - 1} & \dfrac 1 z & \dfrac {z - 1} z & z & \dfrac 1 {1 - z} \\ \dfrac z {z - 1} & \dfrac z {z - 1} & \dfrac 1 z & 1 - z & \dfrac 1 {1 - z} & \dfrac {z - 1} z & z \\ \end{array}$  == Sources == * {{BookReference|Introduction to the Theory of Finite Groups|1964|Walter Ledermann|ed = 5th|edpage = Fifth Edition|prev = Group Generated by Reciprocal of z and 1 minus z|next = Definition:Congruence (Number Theory)/Integers/Integer Multiple}}: Chapter $\text {I}$: The Group Concept: $\S 6$: Examples of Finite Groups: $\text{(ii)}$ * {{BookReference|Elements of Abstract Algebra|1971|Allan Clark|prev = Group Generated by Reciprocal of z and 1 minus z|next = Isomorphism between Group Generated by Reciprocal of z and 1 minus z and Symmetric Group on 3 Letters}}: Introduction
Group Generated by Reciprocal of z and Minus z/Cayley Table
== Cayley Table for Group Generated by Reciprocal of $1 / z$ and $-z$ == We have: :$\map {f_1} z = z$ :$\map {f_2} z = -z$ :$\map {f_3} z = \dfrac 1 z$ :$\map {f_4} z = -\dfrac 1 z$ Hence from Group Generated by Reciprocal of z and Minus z:  :$\begin{array}{r|rrrr} \circ & f_1 & f_2 & f_3 & f_4 \\ \hline f_1 & f_1 & f_2 & f_3 & f_4 \\ f_2 & f_2 & f_1 & f_4 & f_3 \\ f_3 & f_3 & f_4 & f_1 & f_2 \\ f_4 & f_4 & f_3 & f_2 & f_1 \\ \end{array}$ Expressing the elements in full: :$\begin{array}{c|cccc} \circ & z & -z & \dfrac 1 z & -\dfrac 1 z \\ \hline z & z & -z & \dfrac 1 z & -\dfrac 1 z \\ -z & -z & z & -\dfrac 1 z & \dfrac 1 z \\ \dfrac 1 z & \dfrac 1 z & -\dfrac 1 z & z & -z \\ -\dfrac 1 z & -\dfrac 1 z & \dfrac 1 z & -z & z \\ \end{array}$
Group of Reflection Matrices Order 4/Cayley Table
== Cayley Table for Group of Reflection Matrices Order $4$ == Consider the group of reflection matrices order $4$ :$R_4 = \set {\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} }$ $R_4$ can be described completely by showing its Cayley table. Let: :$r_0 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ :$r_1 = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$ :$r_2 = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$ :$r_3 = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$ Then we have:  :$\begin{array}{r|rrrr} \times & r_0 & r_1 & r_2 & r_3 \\ \hline r_0 & r_0 & r_1 & r_2 & r_3 \\ r_1 & r_1 & r_0 & r_3 & r_2 \\ r_2 & r_2 & r_3 & r_0 & r_1 \\ r_3 & r_3 & r_2 & r_1 & r_0 \\ \end{array}$
Multiplicative Group of Reduced Residues Modulo 8/Cayley Table
== Cayley Table for Multiplicative Group of Reduced Residues Modulo 8 == The multiplicative group of reduced residues modulo $8$: :$\Z'_8 = \set {\eqclass 1 8, \eqclass 3 8, \eqclass 5 8, \eqclass 7 8}$ can be described completely by showing its Cayley table:  :$\begin{array}{r|rrrr} \times_8 & \eqclass 1 8 & \eqclass 3 8 & \eqclass 5 8 & \eqclass 7 8 \\ \hline \eqclass 1 8 & \eqclass 1 8 & \eqclass 3 8 & \eqclass 5 8 & \eqclass 7 8 \\ \eqclass 3 8 & \eqclass 3 8 & \eqclass 1 8 & \eqclass 7 8 & \eqclass 5 8 \\ \eqclass 5 8 & \eqclass 5 8 & \eqclass 7 8 & \eqclass 1 8 & \eqclass 3 8 \\ \eqclass 7 8 & \eqclass 7 8 & \eqclass 5 8 & \eqclass 3 8 & \eqclass 1 8 \\ \end{array}$
Klein Four-Group/Cayley Table
== Cayley Table for Klein $4$-Group == The Klein $4$-group can be described completely by showing its Cayley table:  :$\begin{array}{c|cccc}   & e & a & b & c \\ \hline e & e & a & b & c \\ a & a & e & c & b \\ b & b & c & e & a \\ c & c & b & a & e \\ \end{array}$  {{NamedforDef|Felix Christian Klein}} == Sources == * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Definition:Klein Four-Group|next = Symmetric Group on 3 Letters/Group Presentation}}: $\S 25$ * {{BookReference|Elements of Abstract Algebra|1966|Richard A. Dean|prev = Exponential on Real Numbers is Group Isomorphism/Proof 1|next = Definition:Klein Four-Group}}: $\S 1.5$: Example $15$ * {{BookReference|Elements of Abstract Algebra|1971|Allan Clark|prev = Definition:Klein Four-Group|next = Klein Four-Group is Group}}: Chapter $2$: The Definition of Group Structure: $\S 26 \iota$ * {{BookReference|Algebra|1974|Thomas W. Hungerford|prev = Definition:Klein Four-Group|next = Reduced Residue System under Multiplication forms Abelian Group/Corollary}}: $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups: Exercise $6$ * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Definition:Klein Four-Group|next = Definition:Symmetry Group of Rectangle}}: $\S 44.5$ Some consequences of Lagrange's Theorem
Complex Roots of Unity/Examples/Cube Roots
== Example of Complex Roots of Unity ==  The '''complex cube roots of unity''' are the elements of the set: :$U_3 = \set {z \in \C: z^3 = 1}$ They are: {{begin-eqn}} {{eqn | m = e^{0 i \pi / 3}       | o =       | r = 1       | mo= =       | c =  }} {{eqn | l = \omega        | mo= =       | m = e^{2 i \pi / 3}       | r = -\frac 1 2 + \frac {i \sqrt 3} 2 }} {{eqn | l = \omega^2        | mo= =       | m = e^{4 i \pi / 3}       | r = -\frac 1 2 - \frac {i \sqrt 3} 2 }} {{end-eqn}} The notation $\omega$ for, specifically, the complex '''cube''' roots of unity is conventional.  === Conjugate Form === {{:Complex Roots of Unity/Examples/Cube Roots/Conjugate Form}} == Proof == {{:Complex Roots of Unity/Examples/Cube Roots/Proof}} == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Sum of Powers of Primitive Complex Roots of Unity|next = Difference of Two Cubes/Corollary}}: $\S 3$. Roots of Unity * {{BookReference|Elements of Abstract Algebra|1971|Allan Clark|prev = Roots of Complex Number/Examples/Cube Roots|next = Roots of Complex Number/Corollary/Examples/Cube Roots}}: Introduction * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Definition:Cube Root|next = Definition:Cubic (Geometry)|entry = cube root of unity}}
Primitive Complex Roots of Unity/Examples/Cube Roots
== Examples of Primitive Complex Roots of Unity ==  The '''primitive complex cube roots of unity''' are: {{begin-eqn}} {{eqn | l = \omega        | mo= =       | m = e^{2 \pi i / 3}       | r = -\frac 1 2 + \frac {i \sqrt 3} 2 }} {{eqn | l = \omega^2        | mo= =       | m = e^{4 \pi i / 3}       | r = -\frac 1 2 - \frac {i \sqrt 3} 2 }} {{end-eqn}}  == Proof == There are $3$ (complex) cube roots of unity: :$1, \omega, \omega^2$ We have: {{begin-eqn}} {{eqn | l = \omega        | r = e^{2 \pi i / 3} }} {{eqn | l = \omega^2        | r = e^{4 \pi i / 3} }} {{eqn | l = \omega^3        | r = 1       | c = Cube Roots of Unity }} {{end-eqn}} Also: {{begin-eqn}} {{eqn | l = \omega^2       | r = e^{4 \pi i / 3} }} {{eqn | l = \paren {\omega^2}^2       | r = \omega^3 \times \omega }} {{eqn | r = 1 \times \omega       | c = Cube Roots of Unity }} {{eqn | r = \omega       | c =  }} {{eqn | l = \paren {\omega^2}^3        | r = \paren {\omega^3}^2       | c = Cube Roots of Unity }} {{eqn | r = 1 \times 1       | c = Cube Roots of Unity }} {{eqn | r = 1       | c =  }} {{end-eqn}} Trivially, $1$ is not a primitive complex cube root of unity because you cannot make either $\omega$ or $\omega^2$ by multiplying $1$ by itself as many times as you like. Hence the result.
Symmetric Group on 3 Letters
== Group Example ==  Let $S_3$ denote the set of permutations on $3$ letters. The '''symmetric group on $3$ letters''' is the algebraic structure: :$\struct {S_3, \circ}$ where $\circ$ denotes composition of mappings.  It is usually denoted, when the context is clear, without the operator: $S_3$. === Cycle Notation === It can be expressed in the form of permutations given in cycle notation as follows: {{:Symmetric Group on 3 Letters/Cycle Notation}} === Cayley Table === {{:Symmetric Group on 3 Letters/Cayley Table}} === Group Presentation === Its group presentation is: {{:Symmetric Group on 3 Letters/Group Presentation}} == Order of Elements == {{:Symmetric Group on 3 Letters/Order of Elements}} == Subgroups == {{:Symmetric Group on 3 Letters/Subgroups}} == Normal Subgroups == {{:Symmetric Group on 3 Letters/Normal Subgroups}} == Generators == {{:Symmetric Group on 3 Letters/Generators}} == Centralizers == {{:Symmetric Group on 3 Letters/Centralizers}} == Normalizers of Subgroups == {{:Symmetric Group on 3 Letters/Normalizers}} == Center == {{:Symmetric Group on 3 Letters/Center}} == Conjugacy Classes == {{:Symmetric Group on 3 Letters/Conjugacy Classes}} == Also see == * Symmetric Group is Group, which demonstrates that this is a (finite) group.
Solution to Card Game with Bluffing
== Solution to Card Game with Bluffing == {{:Definition:Card Game with Bluffing}} == Proof ==  From the payoff table: {{:Definition:Card Game with Bluffing/Payoff Table}} The solution is: : $A$ takes strategy $A_1$ for $2/3$ of the time, and $A_2$ for $1/3$ of the time. : $B$ takes strategy $B_1$ for $2/3$ of the time, and $B_2$ for $1/3$ of the time. {{explain|It is not made clear in the source work why this is. You just get: "This is indeed a solution, because no player can do better if the other sticks to his strategy, but could gain or lose more than $1/3$ (which is what $A$ gains now) if he departed from the optimal solution and his opponent took advantage of it." This is handwaving. A source work with a higher level of precision is needed here.}}  == Sources == * {{BookReference|The Theory of Games and Linear Programming|1956|Steven Vajda|prev = Definition:Card Game with Bluffing/Payoff Table|next = Two-Person Zero-Sum Game with Multiple Solutions}}: Chapter $\text{I}$: An Outline of the Theory of Games: $3$
Induction of Finite Set
== Theorem Scheme == Let $A$ be finite set. Let $\map P -$ be a predicate. Let $\map P \O$. Let :$\forall B \subseteq A, x \in A: \paren {\map P B \implies \map P {B \cup \set x} }$ Then: :$\map P A$ == Proof == We will prove the result by induction on cardinality of argument. === Base Case === :$\forall X \subseteq A: \paren {\size X = 0 \implies \map P X}$ Let $X \subseteq A$ such that: :$\size X = 0$ By Cardinality of Empty Set: :$X = \O$ Thus by assumption: :$\map P X$ === Induction Hypothesis === :$\forall X \subseteq A: \paren {\size X = n \implies \map P X}$ === Induction Step === :$\forall X \subseteq A: \paren {\size X = n + 1 \implies \map P X}$ Let $X \subseteq A$ such that: :$\size X = n + 1$ By definition of cardinality: :$X = \set {x_1, \dots, x_n, x_{n + 1} }$ By Union of Unordered Tuples: :$X = \set {x_1, \dots, x_n} \cup \set {x_{n + 1} }$ By definition of cardinality: :$\size {\set {x_1, \dots, x_n} } = n$ By Set is Subset of Union: :$\set {x_1, \dots, x_n} \subseteq X \subseteq A$ Then by Induction Hypothesis: :$\map P {\set {x_1, \dots, x_n} }$ By definition of subset: :$x_{n + 1} \in A$ Thus by assumption: :$\map P X$ {{qed|lemma}} By the Principle of Mathematical Induction: :$\forall X \subseteq A: \paren {\size X = \size A \implies \map P X}$ Hence: :$\map P A$  == Sources == * {{Mizar|link = finset_1|sublink = S2|display = FINSET_1:sch 2}}
Infimum of Real Subset
== Lemma == Let $S$ be a set of extended real numbers. Let $S$ be bounded below (in $\R$). Let $T = S \cap \R$. Then: :$S$ admits an infimum (in $\R$) if and only if $T$ admits an infimum (in $\R$) and, if $\inf S$ and $\inf T$ exist as real numbers: :$\inf S = \inf T$ == Proof == We observe that $T$ constitutes the real numbers of $S$. Since there is a real number that is a lower bound for $S$, $-\infty$ is not an element of $S$. Accordingly, $\infty$ is the only possible element of $S \setminus T$. Therefore: :$S$ is a subset of $T \cup \set \infty$ First, we show that $S$ and $T$ have the same set of lower bounds. Let $b$ be a lower bound (in $\R$) for $S$. Then $b$ is a lower bound for $T$ as $T$ is a subset of $S$. Therefore: :the set of lower bounds for $S$ is a subset of the set of lower bounds for $T$ Assume that $c$ is a lower bound (in $\R$) for $T$. Then $c$ is a lower bound for $T \cup \left\{{\infty}\right\}$ as well since $c < \infty$. Accordingly, $c$ is a lower bound for $S$ since $S$ is a subset of $T \cup \set \infty$. Therefore: :the set of lower bounds for $T$ is a subset of the set of lower bounds for $S$ We have: :the set of lower bounds for $T$ is a subset of the set of lower bounds for $S$ :the set of lower bounds for $S$ is a subset of the set of lower bounds for $T$ Therefore: :the set of lower bounds for $T$ equals the set of lower bounds for $S$ by definition Next, we show that $S$ and $T$ have the same infima. We have that $S$ and $T$ have the same set of lower bounds. Therefore, $S$ and $T$ have the same greatest lower bound in $\overline \R$. Accordingly, as a corollary, if one of the sets $S$ and $T$ admits an infimum (in $\R$), so does the other. Furthermore, these infima are equal.
Infimum of Set of Oscillations on Set
== Lemma == Let $f: D \to \R$ be a real function where $D \subseteq \R$. Let $x$ be a point in $D$. Let $S_x$ be a set of real sets that contain (as an element) $x$. Let: :$\map {\omega_f} x = \inf \set {\map {\omega_f} I: I \in S_x}$ where $\map {\omega_f} I$ denotes the oscillation of $f$ on the set $I$: :$\map {\omega_f} I = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$ Then: :$\map {\omega_f} x \in \R$  $\set {\map {\omega_f} I: I \in S_x}$ contains a real number. == Proof == Let: :$S = \set {\map {\omega_f} I: I \in S_x}$ We observe that: :$\inf S = \map {\omega_f} x$ === Necessary Condition === Let $\inf S \in \R$. We need to prove that $S$ contains a real number. Note that $S$ is non-empty as the empty set does not admit an infimum (in $\R$). Therefore, $S$ has at least one element. Accordingly, there is an $I \in S_x$ such that $\map {\omega_f} I \in S$. Let $I \in S_x$. Therefore, $x \in I$. From this follows by Oscillation on Set is an Extended Real Number that $\map {\omega_f} I$ is an extended real number. Therefore $S$ is a set of extended real numbers as $S = \set {\map {\omega_f} I: I \in S_x}$. Accordingly, $S$ contains a real number by Infimum of Subset of Extended Real Numbers is Arbitrarily Close as $\inf S \in \R$. {{qed|lemma}} === Sufficient Condition === Let $S$ contain a real number. We need to prove that $\inf S \in \R$. We have: :$S \cap \R$ is non-empty as $S$ contains a real number. Let $I \in S_x$. Therefore, $x \in I$. From this follows by Oscillation on Set is an Extended Real Number that $\map {\omega_f} I \in \overline \R_{\ge 0}$. Therefore: :$S$ is a subset of $\overline \R_{\ge 0}$ as $S = \set {\map {\omega_f} I: I \in S_x}$ Accordingly: :$S$ is bounded below. From this follows that: :$S \cap \R$ is bounded below as $SR$ is a subset of $S$ We have: :$S \cap \R$ is bounded below :$S \cap \R$ is not empty Therefore: :$\inf S \cap \R \in \R$ Continuum Property We have: :$S$ is a set of extended real numbers as $S$ is a subset of $\overline \R_{\ge 0}$ :$S$ is bounded below Therefore: :$\inf S \in \R$ by Infimum of Real Subset as $\inf S \cap \R \in \R$
Infimum of Set of Oscillations on Set is Arbitrarily Close
Infimum of Set of Oscillations on Set is Arbitrarily Close 0 50507 272883 272273 2016-10-10T07:37:43Z Ivar Sand 2302 Generalized wikitext text/x-wiki == Lemma == Let $f: D \to \R$ be a real function where $D \subseteq \R$. Let $x$ be a point in $D$. Let $S_x$ be a set of real sets that contain (as an element) $x$. Let: :$\omega_f \left({x}\right) = \displaystyle \inf \left\{{\omega_f \left({I}\right): I \in S_x}\right\}$ where $\omega_f \left({I}\right)$ is the oscillation of $f$ on a real set $I$: :$\omega_f \left({I}\right) = \displaystyle \sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ Let $\epsilon \in \R_{>0}$. Let $\omega_f \left({x}\right) \in \R$. Then an $I \in S_x$ exists such that: :$\omega_f \left({I}\right) - \omega_f \left({x}\right) < \epsilon$ == Proof == Let $\epsilon \in \R_{>0}$. Let $\omega_f \left({x}\right) \in \R$. We need to prove that an $I \in S_x$ exists such that: :$\omega_f \left({I}\right) - \omega_f \left({x}\right) < \epsilon$ We have that $\omega_f \left({I}\right) \in \overline{\R}_{\ge 0}$ for every $I \in S_x$ by Oscillation on Set is an Extended Real Number. Therefore: :$\left\{{\omega_f \left({I}\right): I \in S_x}\right\}$ is a subset of $\overline{\R}$ We have also: :$\displaystyle \inf \left\{{\omega_f \left({I}\right): I \in S_x}\right\} \in \R$ as $\displaystyle \inf \left\{{\omega_f \left({I}\right): I \in S_x}\right\} = \omega_f \left({x}\right)$ Therefore, an $I \in S_x$ exists such that: {{begin-eqn}} {{eqn | l = \omega_f \left({I}\right) - \displaystyle \inf \left\{ {\omega_f \left({I'}\right): I' \in S_x}\right\}       | o = <        | r = \epsilon       | c = by Infimum of Subset of Extended Real Numbers is Arbitrarily Close }} {{eqn | ll = \iff       | l = \omega_f \left({I}\right) - \omega_f \left({x}\right)       | o = <        | r = \epsilon       | c = as $\omega_f \left({x}\right) = \displaystyle \inf \left\{ {\omega_f \left({I'}\right): I' \in S_x}\right\}$ }} {{end-eqn}} {{qed|lemma}}
Reciprocal/Examples/Euler's Number
== Example of Reciprocal == The reciprocal of Euler's Number $e$ is approximately:  :$\dfrac 1 e \approx 0 \cdotp 36787 \, 94411 \, 71442 \, 32159 \, 55237 \, 70161 \, 46086 \, 74458 \, 11131 \, 031 \ldots$  {{OEIS|A068985}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Reciprocal/Examples/Pi|next = Envelope Problem}}: $0 \cdotp 367 \, 879 \, 441 \, 171 \, 442 \, 321 \, 595 \, 523 \, 770 \, 161 \, 460 \, 867 \, 445 \, 811 \, 131 \, 031$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = One Third as Quotient of Sequences of Odd Numbers|next = Euler's Number as Sum of Egyptian Fractions}}: $0 \cdotp 36787 \, 94411 \, 71442 \, 32159 \, 55237 \, 70161 \, 46086 \, 74458 \, 11131 \, 031$
Continued Fraction Expansion of Irrational Square Root/Examples/2
Continued Fraction Expansion of Irrational Square Root/Examples/2 0 50722 470631 433258 2020-05-26T06:33:17Z Prime.mover 59 Prime.mover moved page Continued Fraction Expansion of Irrational Square Root/Example/2 to Continued Fraction Expansion of Irrational Square Root/Examples/2 wikitext text/x-wiki == Examples of Continued Fraction Expansion of Irrational Square Root ==  The continued fraction expansion of the square root of $2$ is given by: :$\sqrt 2 = \sqbrk {1, \sequence 2}$  {{OEIS|A040000}} == Proof == {{begin-eqn}} {{eqn | l = \sqrt 2       | r = 1 + \paren {\sqrt 2 − 1}       | c =  }} {{eqn | r = 1 + \frac {\paren {\sqrt 2 − 1} \paren {\sqrt 2 + 1} } {\sqrt 2 + 1}       | c = multiplying top and bottom by $\sqrt 2 + 1$ }} {{eqn | r = 1 + \frac {\paren {\sqrt 2}^2 − 1^2} {\sqrt 2 + 1}       | c = Difference of Two Squares }} {{eqn | r = 1 + \frac 1 {1 + \sqrt 2}       | c = as $\paren {\sqrt 2}^2 − 1^2 = 2 - 1 = 1$ }} {{end-eqn}} Thus it is possible to replace $\sqrt 2$ recursively: {{begin-eqn}} {{eqn | l = \sqrt 2       | r = 1 + \frac 1 {1 + \sqrt 2}       | c =  }} {{eqn | r = 1 + \frac 1 {1 + \paren {1 + \cfrac 1 {1 + \sqrt 2} } }       | c =  }} {{eqn | r = 1 + \frac 1 {2 + \cfrac 1 {1 + \sqrt 2} }       | c =  }} {{eqn | r = 1 + \frac 1 {2 + \cfrac 1 {1 + \paren {1 + \cfrac 1 {1 + \sqrt 2} } } }       | c =  }} {{eqn | r = 1 + \frac 1 {2 + \cfrac 1 {2 + \cfrac 1 {1 + \sqrt 2} } }       | c =  }} {{end-eqn}} The pattern repeats indefinitely, producing the continued fraction expansion: :$\sqrt 2 = \sqbrk {1, 2, 2, 2, \ldots} = \sqbrk {1, \sequence 2}$
Riemann Zeta Function at Even Integers/Examples/2
== Example of Riemann Zeta Function at Even Integers ==  The Riemann zeta function of $2$ is given by: {{begin-eqn}} {{eqn | l = \map \zeta 2       | r = \dfrac 1 {1^2} + \dfrac 1 {2^2} + \dfrac 1 {3^2} + \dfrac 1 {4^2} + \cdots       | c =  }} {{eqn | r = \dfrac {\pi^2} 6       | c =  }} {{eqn | o = \approx       | r = 1 \cdotp 64493 \, 4066 \ldots       | c =  }} {{end-eqn}}  {{OEIS|A013661}} == Proof == {{:Basel Problem/Proof 6}} The decimal expansion can be found by an application of arithmetic. == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Definition:Wythoff's Game|next = Square Root/Examples/3}}: $1 \cdotp 644 \, 934 \, 066 \ldots$ * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Riemann Zeta Function at Even Integers|next = Riemann Zeta Function of 4}}: $\S 1.2.7$: Harmonic Numbers: $(7)$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Definition:Wythoff's Game|next = Zeta of 2 as Product of Fractions with Prime Numerators}}: $1 \cdotp 64493 \, 4066 \ldots$
Axiom:Real Number as Complex Number
{{refactor|obviously needs refactoring}} {{MissingLinks|Needs a category}} == Axiom == Let $x \in \R$ be a real number. Then: : $x = \left({x, 0}\right)$ where $\left({a, b}\right)$ where $a, b \in R$ denotes a complex number. Also: : $x = x + 0 i$ where $a + b i$ where $a, b \in R$ denotes a complex number. 0ceu0zey8cofgzlvd94lyogii14qmr8