hcju/proofwikips huggingface dataset:
Hyperbolic Sine of Complex Number
Hyperbolic Sine of Complex Number 0 70971 365755 2018-09-15T09:06:14Z Prime.mover 59 Created page with "== Theorem ==  Let $a$ and $b$ be real numbers.  Let $i$ be the imaginary unit.  Then: :$\sinh \paren {a..." wikitext text/x-wiki == Theorem ==  Let $a$ and $b$ be real numbers. Let $i$ be the imaginary unit. Then: :$\sinh \paren {a + b i} = \sinh a \cos b + i \cosh a \sin b$ where: :$\sin$ denotes the real sine function :$\cos$ denotes the real cosine function :$\sinh$ denotes the hyperbolic sine function :$\cosh$ denotes the hyperbolic cosine function.  == Proof 1 == {{:Hyperbolic Sine of Complex Number/Proof 1}} == Proof 2 == {{:Hyperbolic Sine of Complex Number/Proof 2}} == Also see == * Hyperbolic Cosine of Complex Number * Hyperbolic Tangent of Complex Number * Hyperbolic Cosecant of Complex Number * Hyperbolic Secant of Complex Number * Hyperbolic Cotangent of Complex Number
Hyperbolic Cotangent of Complex Number/Formulation 1
Hyperbolic Cotangent of Complex Number/Formulation 1 0 70986 365788 2018-09-15T10:49:12Z Prime.mover 59 Created page with "== Theorem ==  Let $a$ and $b$ be real numbers.  Let $i$ be the imaginary unit.   Then:  :$\coth \paren {..." wikitext text/x-wiki == Theorem == Let $a$ and $b$ be real numbers. Let $i$ be the imaginary unit. Then:  :$\coth \paren {a + b i} = \dfrac {\cosh a \cos b + i \sinh a \sin b} {\sinh a \cos b + i \cosh a \sin b}$  where: :$\coth$ denotes the hyperbolic cotangent function :$\sin$ denotes the real sine function :$\cos$ denotes the real cosine function :$\sinh$ denotes the hyperbolic sine function :$\cosh$ denotes the hyperbolic cosine function. == Proof == {{begin-eqn}} {{eqn | l = \coth \paren {a + b i}       | r = \frac {\cosh \paren {a + b i} } {\sinh \paren {a + b i} }       | c = {{Defof|Hyperbolic Cotangent}} }} {{eqn | r = \dfrac {\cosh a \cos b + i \sinh a \sin b} {\sinh a \cos b + i \cosh a \sin b}       | c = Hyperbolic Sine of Complex Number and Hyperbolic Cosine of Complex Number }} {{end-eqn}}  == Also see == * Hyperbolic Sine of Complex Number * Hyperbolic Cosine of Complex Number * Hyperbolic Tangent of Complex Number * Hyperbolic Cosecant of Complex Number * Hyperbolic Secant of Complex Number
Cardinality/Examples/0 less than x less than 6
== Example of Cardinality == Let $S_2 = \set {x \in \Z: 0 < x < 6}$.  The cardinality of $S_2$ is given by: :$\card {S_2} = 5$  == Proof == The elements of $S_2$, by definition, are those integers greater than $0$ and less than $6$. That is: :$S_2 = \set {1, 2, 3, 4, 5}$ Thus $S_2$ has $5$ elements: $1, 2, 3, 4, 5$. Hence the result by definition of cardinality.  == Sources == * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Cardinality/Examples/-1,0,1|next = Cardinality/Examples/x^2-x}}: Chapter $1$: Sets and Logic: Exercise $4$
Fermat's Little Theorem/Corollary 3
 == Corollary to Fermat's Little Theorem ==  Let $p^k$ be a prime power for some prime number $p$ and $k \in \Z_{\gt 0}$. Then: :$\forall n \in \Z_{\gt 0}: n^{p^k} \equiv n \pmod p$  == Proof == The proof proceeds by induction. For all $k \in \Z_{\ge 1}$, let $P \paren {k}$ be the proposition: :$\forall n \in \Z_{\gt 0}: n^{p^k} \equiv n \pmod p$ === Basis for the Induction === $P \paren {1}$ is the case: :$\forall n \in \Z_{\gt 0}: n^p \equiv n \pmod p$ which follows from the corollary 1 to Fermat's Little Theorem. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \paren{k-1}$ is true, where $k \ge 2$, then it logically follows that $P \paren {k}$ is true. So this is the induction hypothesis: :$\forall n \in \Z_{\gt 0}: n^{p^{k - 1} } \equiv n \pmod p$ from which it is to be shown that: :$\forall n \in \Z_{\gt 0}: n^{p^k} \equiv n \pmod p$ === Induction Step === This is the induction step: For any $n \in \Z_{\gt 0}$ then: {{begin-eqn}} {{eqn | l = n^{p^k}       | r = \paren {n^{p^{k - 1} } }^p }} {{eqn | r = n^{p^{k - 1} } \pmod p       | o = \equiv       | c = Corollary 1 to Fermat's Little Theorem }} {{eqn | r = n \pmod p       | o = \equiv       | c = Induction Hypothesis }} {{end-eqn}}  == Sources ==
Fermat's Little Theorem/Corollary 4
 == Corollary to Fermat's Little Theorem ==  Let $p^k$ be a prime power for some prime number $p$ and $k \in \Z_{\gt 0}$. Let $n \in \Z_{\gt 0}$ with $p \nmid n$. Then: :$n^{p^k - 1} \equiv 1 \pmod p$  == Proof == By corollary 3 of Fermat's Little Theorem: :$n^{p^k} \equiv n \pmod p$ That is: :$p \divides \paren {n^{p^k} - n} = n \paren {n^{p^k - 1} - 1}$ Since $p \nmid n$, by Corollary to Divisors of Product of Coprime Integers: :$p \divides \paren {n^{p^k - 1} - 1}$ That is: :$n^{p^k-1} \equiv 1 \pmod p$  == Sources ==
Total Number of Set Partitions/Examples
== Examples of Total Number of Set Partitions ==  === Example: $\card S = 2$ === {{:Total Number of Set Partitions/Examples/2}} === Example: $\card S = 3$ === {{:Total Number of Set Partitions/Examples/3}} === Example: $\card S = 4$ === {{:Total Number of Set Partitions/Examples/4}}
Total Number of Set Partitions/Examples/2/Illustration
Total Number of Set Partitions/Examples/2/Illustration 0 71136 366804 2018-09-22T08:51:38Z Prime.mover 59 Created page with "== Example of Total Number of Set Partitions ==  Let a set $S$ of cardinality $2$ be exemplified by $S = \set {a..." wikitext text/x-wiki == Example of Total Number of Set Partitions ==  Let a set $S$ of cardinality $2$ be exemplified by $S = \set {a, b}$. Then the partitions of $S$ are: :$\set {a, b}$ :$\set {a \mid b}$
Total Number of Set Partitions/Examples/3/Illustration
Total Number of Set Partitions/Examples/3/Illustration 0 71137 366808 2018-09-22T08:54:10Z Prime.mover 59 Created page with "== Example of Total Number of Set Partitions ==  Let a set $S$ of cardinality $3$ be exemplified by $S = \set {a..." wikitext text/x-wiki == Example of Total Number of Set Partitions ==  Let a set $S$ of cardinality $3$ be exemplified by $S = \set {a, b, c}$. Then the partitions of $S$ are: :$\set {a, b, c}$ :$\set {a, b \mid c}$ :$\set {a, c \mid b}$ :$\set {b, c \mid a}$ :$\set {a \mid b \mid c}$
Total Number of Set Partitions/Examples/4/Illustration
== Example of Total Number of Set Partitions ==  Let a set $S$ of cardinality $4$ be exemplified by $S = \set {a, b, c, d}$. Then the partitions of $S$ are: :$\set {a, b, c, d}$ :$\set {\set a, \set {b, c, d} }$ :$\set {\set b, \set {a, c, d} }$ :$\set {\set c, \set {a, b, d} }$ :$\set {\set d, \set {a, b, c} }$ :$\set {\set {a, b}, \set {c, d} }$ :$\set {\set {a, c}, \set {b, d} }$ :$\set {\set {a, d}, \set {b, c} }$ :$\set {\set a, \set b, \set {c, d} }$ :$\set {\set a, \set c, \set {b, d} }$ :$\set {\set a, \set d, \set {b, c} }$ :$\set {\set b, \set c, \set {a, d} }$ :$\set {\set b, \set d, \set {a, c} }$ :$\set {\set c, \set d, \set {a, b} }$ :$\set {\set a, \set b, \set c, \set d}$  == Sources == * {{BookReference|Set Theory and Abstract Algebra|1975|T.S. Blyth|prev = Definition:Set Partition/Definition 1|next = Set Partition/Examples/Partition into Singletons}}: $\S 6$. Indexed families; partitions; equivalence relations: Example $6.2$
Modulo Addition/Cayley Table/Modulo 6
== Cayley Table for Modulo Addition == The additive group of integers modulo $m$ can be described by showing its Cayley table. This one is for modulo $6$:  :$\begin{array}{r|rrrrrr} \struct {\Z_6, +_6} & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 \\ \hline \eqclass 0 6 & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 \\ \eqclass 1 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 & \eqclass 0 6 \\ \eqclass 2 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 & \eqclass 0 6 & \eqclass 1 6 \\ \eqclass 3 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 \\ \eqclass 4 6 & \eqclass 4 6 & \eqclass 5 6 & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 \\ \eqclass 5 6 & \eqclass 5 6 & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 \\ \end{array}$  which can also be presented: :$\begin{array}{r|rrrrrr} +_6 & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline 0 & 0 & 1 & 2 & 3 & 4 & 5  \\ 1 & 1 & 2 & 3 & 4 & 5 & 0 \\ 2 & 2 & 3 & 4 & 5 & 0 & 1 \\ 3 & 3 & 4 & 5 & 0 & 1 & 2 \\ 4 & 4 & 5 & 0 & 1 & 2 & 3 \\ 5 & 5 & 0 & 1 & 2 & 3 & 4 \\ \end{array}$ == Sources == * {{BookReference|Sets and Groups|1965|J.A. Green|prev = Symmetric Group on 3 Letters/Cayley Table|next = Alternating Group on 4 Letters/Cayley Table}}: Tables: $2$. Cyclic group of order $6$ * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Definition:Multiplication/Modulo Multiplication/Definition 3|next = Modulo Multiplication/Cayley Table/Modulo 6}}: $\S 2$: Example $2.3$
Modulo Multiplication/Cayley Table/Modulo 6
== Cayley Table for Modulo Multiplication == The multiplicative monoid of integers modulo $m$ can be described by showing its Cayley table. This one is for modulo $6$:  :$\begin{array} {r|rrrrrr} \struct {\Z_6, \times_6} & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 \\ \hline \eqclass 0 6 & \eqclass 0 6 & \eqclass 0 6 & \eqclass 0 6 & \eqclass 0 6 & \eqclass 0 6 & \eqclass 0 6  \\ \eqclass 1 6 & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 \\ \eqclass 2 6 & \eqclass 0 6 & \eqclass 2 6 & \eqclass 4 6 & \eqclass 0 6 & \eqclass 2 6 & \eqclass 4 6 \\ \eqclass 3 6 & \eqclass 0 6 & \eqclass 3 6 & \eqclass 0 6 & \eqclass 3 6 & \eqclass 0 6 & \eqclass 3 6 \\ \eqclass 4 6 & \eqclass 0 6 & \eqclass 4 6 & \eqclass 2 6 & \eqclass 0 6 & \eqclass 4 6 & \eqclass 2 6 \\ \eqclass 5 6 & \eqclass 0 6 & \eqclass 5 6 & \eqclass 4 6 & \eqclass 3 6 & \eqclass 2 6 & \eqclass 1 6 \\ \end{array}$ which can also be presented: :$\begin{array} {r|rrrrrr} \times_6 & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline 0 & 0 & 0 & 0 & 0 & 0 & 0  \\ 1 & 0 & 1 & 2 & 3 & 4 & 5 \\ 2 & 0 & 2 & 4 & 0 & 2 & 4 \\ 3 & 0 & 3 & 0 & 3 & 0 & 3 \\ 4 & 0 & 4 & 2 & 0 & 4 & 2 \\ 5 & 0 & 5 & 4 & 3 & 2 & 1 \\ \end{array}$  == Sources == * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Modulo Addition/Cayley Table/Modulo 6|next = Modulo Multiplication is Associative/Proof 2}}: $\S 2$: Example $2.3$ * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Equivalence Relation on Square Matrices induced by Positive Integer Powers|next = Modulo Arithmetic/Examples/Solutions to x^2 = x Modulo 6}}: Chapter $3$: Equivalence Relations and Equivalence Classes: Exercise $7$
Cauchy Sequences form Ring with Unity/Corollary
== Corollary to Cauchy Sequences form Ring with Unity ==  Let $\struct {F, +, \circ, \norm {\, \cdot \,} }$ be a valued field. Let $\struct {F^\N, +, \circ}$ be the commutative ring of sequences over $F$ with unity $\tuple {1, 1, 1, \dotsc}$. {{explain|Has it been proved that $\struct {F^\N, +, \circ}$ is actually a commutative ring?}} Let $\CC \subset F^\N$ be the set of Cauchy sequences on $F$. Then: :$\struct {\CC, +, \circ}$ is a commutative subring of $F^\N$ with unity $\tuple {1, 1, 1, \dotsc}$.  == Proof == By Cauchy Sequences form Ring with Unity, $\struct {\CC, +, \circ}$ is a subring of $F^\N$. We have that $\circ$ is commutative on $F^\N$. Hence by Restriction of Commutative Operation is Commutative the restriction of $\circ$ to $\CC$ is commutative. == Sources == * {{BookReference|p-adic Numbers: An Introduction|1997|Fernando Q. Gouvea}}: $\S 3.2$: Completions
Modulo Arithmetic/Examples/Multiplicative Inverse of 41 Modulo 97
== Example of Modulo Arithmetic ==  The inverse of $41$ under multiplication modulo $97$ is given by: :${\eqclass {41} {97} }^{-1} = 71$  === Solution to $41 x \equiv 2 \pmod {97}$ === {{:Modulo Arithmetic/Examples/Multiplicative Inverse of 41 Modulo 97/41x = 2 Modulo 97}} == Proof == From Ring of Integers Modulo Prime is Field, multiplication modulo $97$ has an inverse for all $x \in \Z_{97}$ where $x \ne 0$. Using Euclid's Algorithm: {{begin-eqn}} {{eqn | n = 1       | l = 97       | r = 2 \times 41 + 15 }} {{eqn | n = 2       | l = 41       | r = 2 \times 15 + 11 }} {{eqn | n = 3       | l = 15       | r = 11 + 4 }} {{eqn | n = 4       | l = 11       | r = 2 \times 4 + 3 }} {{eqn | n = 5       | l = 4       | r = 3 + 1 }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = 1       | r = 4 - 3       | c = from $(5)$ }} {{eqn | r = 4 - \paren {11 - 2 \times 4}       | c = from $(4)$ }} {{eqn | r = 3 \times 4 - 11       | c =  }} {{eqn | r = 3 \times \paren {15 - 11} - 11       | c = from $(3)$ }} {{eqn | r = 3 \times 15 - 4 \times 11       | c =  }} {{eqn | r = 3 \times 15 - 4 \times \paren {41 - 2 \times 15}       | c = from $(2)$ }} {{eqn | r = 11 \times 15 - 4 \times 41       | c =  }} {{eqn | r = 11 \times \paren {97 - 2 \times 14} - 4 \times 41       | c = from $(1)$ }} {{eqn | r = 11 \times 97 - 26 \times 41       | c =  }} {{end-eqn}} So: : $\paren {-26} \times 41 \equiv 1 \pmod {97}$ {{begin-eqn}} {{eqn | l = \paren {-26} \times 41       | o = \equiv       | r = 1       | rr= \pmod {97}       | c =  }} {{eqn | l = 71 \times 41       | o = \equiv       | r = 1       | rr= \pmod {97}       | c = as $-26 \equiv 71 \pmod {97}$ because $26 + 71 = 97$ }} {{eqn | ll= \leadsto       | l = {\eqclass {41} {97} }^{-1}       | r = \eqclass {71} {97}       | c =  }} {{end-eqn}} Hence the result.  == Sources == * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Modulo Arithmetic/Examples/Solutions to x^2 = x Modulo 6|next = Modulo Arithmetic/Examples/Multiplicative Inverse of 41 Modulo 97/41x = 2 Modulo 97}}: Chapter $3$: Equivalence Relations and Equivalence Classes: Exercise $8$
Real Square Function is not Injective
== Example of Mapping which is Not an Injection ==  Let $f: \R \to \R$ be the real square function: :$\forall x \in \R: \map f x = x^2$ Then $f$ is not an injection.  == Proof == For $f$ to be an injection, it would be necessary that: :$\forall x_1, x_2 \in \R: \map f {x_1} = \map f {x_2} \implies x_1 = x_2$ By definition of the squaring operation, we have: :$\map f x = \map f {-x}$ But unless $x = 0$ it is not the case that $x = -x$. Hence $f$ is not an injection.  == Sources == * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Definition:Injection/Definition 1 a|next = Injection/Examples/Cube Function}}: $\S 22$: Injections; bijections; inverse of a bijection
Image of Mapping/Examples/Image of x^2-4x+5
== Example of Image of Element under Mapping ==  Let $f: \R \to \R$ be the mapping defined as: :$\forall x \in \R: \map f x = x^2 - 4 x + 5$ The image of $f$ is the unbounded closed interval: :$\Img f = \hointr 1 \to$ and so $f$ is not a surjection.  === Graphical Representation of $\map f x = x^2 - 4 x + 5$ === {{:Image of Mapping/Examples/Image of x^2-4x+5/Graph}} == Proof == By differentiating $x^2 - 4 x + 5$ twice {{WRT|Differentiation}} $x$: :$f' = 2 x - 4$ :$f' = 2 x - 4$ {{begin-eqn}} {{eqn | l = f'       | r = 2 x - 4 }} {{eqn | l = f''       | r = 2 }} {{end-eqn}} Equating $f'$ to $0$, a stationary point is found at $x = 2$. Inspecting the sign of $f''$, it is noted that $f'$ is increasing everywhere. Hence the stationary point at $x = 2$ is a minimum of $\Img f$. This is the only stationary point, so it can be stated that '''the''' minimum of $f$ occurs at $x = 2$. We have that: :$f \paren 2 = 2^2 - 4 \times 2 + 5 = 4 - 8 + 5 = 1$ As $f$ is strictly increasing on $x > 2$ and strictly decreasing on $x < 2$, it is seen that $f$ is unbounded above. Thus: :$\Img f = \hointr 1 \to$  == Also see == * Bijective Restrictions of $\map f x = x^2 - 4 x + 5$ == Sources == * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Cantor's Theorem|next = Bijective Restriction/Examples/x^2-4x+5}}: Chapter $4$: Mappings: Exercise $12 \ \text{(ii)}$
Subsemigroup/Examples/2x2 Matrices with One Non-Zero Entry
== Example of Subsemigroup ==  Let $\struct {S, \times}$ be the semigroup formed by the set of order $2$ square matrices over the real numbers $R$ under (conventional) matrix multiplication. Let $T$ be the subset of $S$ consisting of the matrices of the form $\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}$ for $x \in \R$. Then $\struct {T, \times}$ is a subsemigroup of $\struct {S, \times}$.  == Proof == From the Subsemigroup Closure Test it is sufficient to demonstrate that $\struct {T, \times}$ is closed. Let $A = \begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} y & 0 \\ 0 & 0 \end{bmatrix}$. Then: {{begin-eqn}} {{eqn | l = A B       | r = \begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} y & 0 \\ 0 & 0 \end{bmatrix}       | c =  }} {{eqn | r = \begin{bmatrix} x y + 0 \times 0 & x \times 0 + 0 \times 0 \\ 0 \times y + 0 \times 0 & 0 \times 0 + 0 \times 0 \end{bmatrix}       | c = {{Defof|Matrix Product (Conventional)}} }} {{eqn | r = \begin{bmatrix} x y & 0 \\ 0 & 0 \end{bmatrix}       | c =  }} {{eqn | 0 = \in       | r = T       | c =  }} {{end-eqn}} Hence the result.  == Sources == * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Subsemigroup Closure Test|next = Identity of Submonoid is not necessarily Identity of Monoid}}: $\S 32$ Identity element and inverses
Semigroup/Examples/x+y-xy on Integers
== Example of Semigroup ==  Let $\circ: \Z \times \Z$ be the operation defined on the integers $\Z$ as: :$\forall x, y \in \Z: x \circ y := x + y - x y$ Then $\struct {\Z, \circ}$ is a semigroup.  == Proof == We have that: :$\forall x, y \in \Z: x \circ y \in \Z$ and so $\struct {\Z, \circ}$ is closed. Now let $x, y, z \in \Z$. We have: {{begin-eqn}} {{eqn | l = x \circ \paren {y \circ z}       | r = x + \paren {y \circ z} - x \paren {y \circ z}       | c = Definition of $\circ$ }} {{eqn | r = x + \paren {y + z - y z} - x \paren {y + z - y z}       | c = Definition of $\circ$ }} {{eqn | r = x + y + z - y z - x y - x z + x y z       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \paren {x \circ y} \circ z       | r = \paren {x \circ y} + z - \paren {x \circ y} z       | c = Definition of $\circ$ }} {{eqn | r = \paren {x + y - x y} + z - \paren {x + y - x y} z       | c = Definition of $\circ$ }} {{eqn | r = x + y - x y + z - x z - y z + x y z       | c =  }} {{end-eqn}} As can be seen by inspection: :$x \circ \paren {y \circ z} = \paren {x \circ y} \circ z$ and so $\circ$ is associative. The result follows by definition of semigroup.  == Also see == * Inclusion-Exclusion Principle (think about why) == Sources == * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Identity of Submonoid is not necessarily Identity of Monoid|next = Subsemigroup/Examples/x+y-xy on Integers}}: Chapter $5$: Semigroups: Exercise $1$
Order of Group Element/Examples/Element of Multiplicative Group of Real Numbers
== Examples of Order of Group Element ==  Consider the multiplicative group of real numbers $\struct {\R_{\ne 0}, \times}$. The order of $2$ in $\struct {\R_{\ne 0}, \times}$ is infinite.  == Proof == From Real Multiplication Identity is One, the identity of $\struct {\R_{\ne 0}, \times}$ is $1$. There exists no $n \in \Z_{\ge 0}$ such that $2^n = 1$. Hence the result by definition of infinite order element.  == Sources == * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Definition:Order of Group Element/Also known as|next = Order of Group Element/Examples/Imaginary Unit in Multiplicative Group of Complex Numbers}}: $\S 38$. Period of an element: Illustrations: $\text{(i)}$
Order of Group Element/Examples/Imaginary Unit in Multiplicative Group of Complex Numbers
Order of Group Element/Examples/Imaginary Unit in Multiplicative Group of Complex Numbers 0 71675 370340 370337 2018-10-11T06:38:37Z Prime.mover 59 Prime.mover moved page Order of Group Element/Examples/Gaussian Integer in Multiplicative Group of Complex Numbers to Order of Group Element/Examples/Imaginary Unit in Multiplicative Group of Complex Numbers wikitext text/x-wiki == Examples of Order of Group Element ==  Consider the multiplicative group of complex numbers $\struct {\C_{\ne 0}, \times}$. The order of $i$ in $\struct {\C_{\ne 0}, \times}$ is $4$.  == Proof == We have: {{begin-eqn}} {{eqn | l = i^1       | r = 1       | c =  }} {{eqn | l = i^2       | r = -1       | c =  }} {{eqn | l = i^3       | r = -i       | c =  }} {{eqn | l = i^4       | r = 1       | c =  }} {{end-eqn}} Hence the result by definition of order of group element.  == Sources == * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Order of Group Element/Examples/Element of Multiplicative Group of Real Numbers|next = Identity is Only Group Element of Order 1}}: $\S 38$. Period of an element: Illustrations: $\text{(ii)}$
Quaternion Group/Cayley Table
== Cayley Table for Quaternion Group== The Cayley table for the quaternion group given with the group presentation: :$Q = \Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$ can be presented as:  :$\begin{array}{r|rrrrrrrr}       & e     & a     & a^2   & a^3   & b     & a b   & a^2 b & a^3 b \\ \hline e     & e     & a     & a^2   & a^3   & b     & a b   & a^2 b & a^3 b \\ a     & a     & a^2   & a^3   & e     & a b   & a^2 b & a^3 b & b     \\ a^2   & a^2   & a^3   & e     & a     & a^2 b & a^3 b & b     & a b   \\ a^3   & a^3   & e     & a     & a^2   & a^3 b & b     & a b   & a^2 b \\ b     & b     & a^3 b & a^2 b & a b   & a^2   & a     & e     & a^3   \\ a b   & a b   & b     & a^3 b & a^2 b & a^3   & a^2   & a     & e     \\ a^2 b & a^2 b & a b   & b     & a^3 b & e     & a^3   & a^2   & a     \\ a^3 b & a^3 b & a^2 b & a b   & b     & a     & e     & a^3   & a^2  \end{array}$  === Coset Decomposition of $\set {e, a^2}$ === Presenting the above Cayley table with respect to the coset decomposition of the normal subgroup $\gen a^2$ gives: {{:Quaternion Group/Cayley Table/Coset Decomposition of (e, a^2)}} == Sources == * {{BookReference|Elements of Abstract Algebra|1971|Allan Clark|prev = Definition:Quaternion Group|next = Quaternion Group is Hamiltonian}}: Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \iota$
Quaternion Group/Complex Matrices/Cayley Table
== Cayley Table for Quaternion Group== The Cayley table for the quaternion group: :$Q = \Dic 2 = \set {\mathbf 1, -\mathbf 1, \mathbf i, -\mathbf i, \mathbf j, -\mathbf j, \mathbf k, -\mathbf k}$ under the operation of conventional matrix multiplication, where: :$\mathbf 1 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \qquad \mathbf i = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} \qquad \mathbf j = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \qquad \mathbf k = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}$ can be presented as:  :$\begin{array}{r|rrrrrrrr}       &  \mathbf 1 &  \mathbf i & -\mathbf 1 & -\mathbf i &  \mathbf j &  \mathbf k & -\mathbf j & -\mathbf k \\ \hline  \mathbf 1 &  \mathbf 1 &  \mathbf i & -\mathbf 1 & -\mathbf i &  \mathbf j &  \mathbf k & -\mathbf j & -\mathbf k \\  \mathbf i &  \mathbf i & -\mathbf 1 & -\mathbf i &  \mathbf 1 &  \mathbf k & -\mathbf j & -\mathbf k &  \mathbf j \\ -\mathbf 1 & -\mathbf 1 & -\mathbf i &  \mathbf 1 &  \mathbf i & -\mathbf j & -\mathbf k &  \mathbf j &  \mathbf k \\ -\mathbf i & -\mathbf i &  \mathbf 1 &  \mathbf i & -\mathbf 1 & -\mathbf k &  \mathbf j &  \mathbf k & -\mathbf j \\  \mathbf j &  \mathbf j & -\mathbf k & -\mathbf j &  \mathbf k & -\mathbf 1 &  \mathbf i &  \mathbf 1 & -\mathbf i \\  \mathbf k &  \mathbf k &  \mathbf j & -\mathbf k & -\mathbf j & -\mathbf i & -\mathbf 1 &  \mathbf i &  \mathbf 1 \\ -\mathbf j & -\mathbf j &  \mathbf k &  \mathbf j & -\mathbf k &  \mathbf 1 & -\mathbf i & -\mathbf 1 &  \mathbf i \\ -\mathbf k & -\mathbf k & -\mathbf j &  \mathbf k &  \mathbf j &  \mathbf i &  \mathbf 1 & -\mathbf i & -\mathbf 1 \end{array}$  == Sources == * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Order of Power of Group Element/Examples/Powers of Element of Order 20|next = Order of Elements in Quaternion Group}}: Chapter $6$: An Introduction to Groups: Exercise $14$