hcju/proofwikips huggingface dataset:
Symmetry Group of Square/Cayley Table
== Cayley Table of Symmetry Group of Square ==  The Cayley table of the symmetry group of the square can be written: :$\begin{array}{c|cccccc}        & e & r & r^2 & r^3 & t_x & t_y & t_{AC} & t_{BD} \\ \hline e      & e & r & r^2 & r^3 & t_x & t_y & t_{AC} & t_{BD} \\ r      & r & r^2 & r^3 & e & t_{AC} & t_{BD} & t_y & t_x \\ r^2    & r^2 & r^3 & e & r & t_y & t_x & t_{BD} & t_{AC} \\ r^3    & r^3 & e & r & r^2 & t_{BD} & t_{AC} & t_x & t_y \\ t_x    & t_x & t_{BD} & t_y & t_{AC} & e & r^2 & r^3 & r \\ t_y    & t_y & t_{AC} & t_x & t_{BD} & r^2 & e & r & r^3 \\ t_{AC} & t_{AC} & t_x & t_{BD} & t_y & r & r^3 & e & r^2 \\ t_{BD} & t_{BD} & t_y & t_{AC} & t_x & r^3 & r & r^2 & e\\ \end{array}$  where the various symmetry mappings of the square $\mathcal S = ABCD$ are: :the identity mapping $e$ :the rotations $r, r^2, r^3$ of $90^\circ, 180^\circ, 270^\circ$ counterclockwise respectively :the reflections $t_x$ and $t_y$ are reflections about the $x$ and $y$ axis respectively :the reflection $t_{AC}$ is a reflection about the diagonal through vertices $A$ and $C$ :the reflection $t_{BD}$ is a reflection about the diagonal through vertices $B$ and $D$. == Sources == * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Composition of Symmetries is Associative|next = Count of Binary Operations on Set}}: $\S 2$: Example $2.5$ * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Right Operation is Associative|next = Symmetry Group of Equilateral Triangle/Cayley Table}}: Exercise $2.10$ * {{BookReference|Elements of Abstract Algebra|1966|Richard A. Dean|prev = Symmetry Group of Equilateral Triangle/Cayley Table|next = All Elements Self-Inverse then Abelian}}: $\S 1.4$: Exercise $2.1$ * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Dihedral Group D4/Matrix Representation/Formulation 2/Cayley Table|next = Definition:Symmetry Group of Rectangle}}: Chapter $1$: Definitions and Examples: Exercise $3$
Symmetry Group of Equilateral Triangle/Cayley Table
== Cayley Table of Symmetry Group of Equilateral Triangle == The Cayley table of the symmetry group of the equilateral triangle can be written:  :$\begin{array}{c|ccc|ccc} \circ & e & p & q & r & s & t \\ \hline e & e & p & q & r & s & t \\ p & p & q & e & s & t & r \\ q & q & e & p & t & r & s \\ \hline r & r & t & s & e & q & p \\ s & s & r & t & p & e & q \\ t & t & s & r & q & p & e \\ \end{array}$  where: {{begin-eqn}} {{eqn | l = e       | o = :       | r = (A) (B) (C)       | c = Identity mapping }} {{eqn | l = p       | o = :       | r = (ABC)       | c = Rotation of $120 \degrees$ anticlockwise about center }} {{eqn | l = q       | o = :       | r = (ACB)       | c = Rotation of $120 \degrees$ clockwise about center }} {{eqn | l = r       | o = :       | r = (BC)       | c = Reflection in line $r$ }} {{eqn | l = s       | o = :       | r = (AC)       | c = Reflection in line $s$ }} {{eqn | l = t       | o = :       | r = (AB)       | c = Reflection in line $t$ }} {{end-eqn}} :400px == Sources == * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Symmetry Group of Square/Cayley Table|next = Symmetry Group of Rectangle/Cayley Table}}: Exercise $2.11$ * {{BookReference|Elements of Abstract Algebra|1966|Richard A. Dean|prev = Cayley Table for Commutative Operation is Symmetrical about Main Diagonal|next = Symmetry Group of Square/Cayley Table}}: $\S 1.4$: Exercise $2.1$ * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Definition:Symmetry Group of Equilateral Triangle|next = Definition:Finite Group}}: $\S 34$. Examples of groups: $(5)$ * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Definition:Symmetry Group of Equilateral Triangle|next = Symmetry Group of Equilateral Triangle is Group}}: Chapter $1$: Definitions and Examples: Example $1.9$
Symmetry Group of Rectangle/Cayley Table
== Cayley Table of Symmetry Group of Rectangle == === Definition === {{:Definition:Symmetry Group of Rectangle}} === Cayley Table ===  The Cayley table of the symmetry group of the (non-square) rectangle can be written: :$\begin{array}{c|cccc}   & e & r & h & v \\ \hline e & e & r & h & v \\ r & r & e & v & h \\ h & h & v & e & r \\ v & v & h & r & e \\ \end{array}$  == Sources == * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Symmetry Group of Equilateral Triangle/Cayley Table|next = Max is Associative}}: Exercise $2.11$
Parity Ring/Cayley Tables
== Cayley Tables for Parity Ring == The parity ring can be described completely by showing its Cayley tables:  :$\begin{array}{r|rr} + & \text{even} &  \text{odd} \\ \hline \text{even} & \text{even} & \text{odd} \\ \text{odd} & \text{odd} & \text{even} \\ \end{array} \qquad \begin{array}{r|rr} \times & \text{even} &  \text{odd} \\ \hline \text{even} & \text{even} & \text{even} \\ \text{odd} & \text{even} & \text{odd} \\ \end{array}$  == Sources == * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Definition:Parity Ring|next = Parity Addition is Associative}}: $\S 2$: Example $2.2$
Ring of Integers Modulo 2/Cayley Tables
{{refactor|Transclude the separate tables for Addition and Multiplication}} == Cayley Tables for Ring of Integers Modulo $2$ == The Ring of Integers Modulo $2$: :$\struct {\Z_2, +_2, \times_2}$ can be described completely by showing its Cayley tables:  :$\begin{array} {r|rr} \struct {\Z_2, +_2} & \eqclass 0 2 & \eqclass 1 2 \\ \hline \eqclass 0 2 & \eqclass 0 2 & \eqclass 1 2 \\ \eqclass 1 2 & \eqclass 1 2 & \eqclass 0 2 \\ \end{array} \qquad \begin{array}{r|rr} \struct {\Z_2, \times_2} & \eqclass 0 2 & \eqclass 1 2 \\ \hline \eqclass 0 2 & \eqclass 0 2 & \eqclass 0 2 \\ \eqclass 1 2 & \eqclass 0 2 & \eqclass 1 2 \\ \end{array}$ They can be presented more simply as: :$\begin{array}{r|rr} + & 0 & 1 \\ \hline 0 & 0 & 1 \\ 1 & 1 & 0 \\ \end{array} \qquad \begin{array}{r|rr} \times & 0 & 1 \\ \hline 0 & 0 & 0 \\ 1 & 0 & 1 \\ \end{array}$  == Sources == * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Isomorphism between Ring of Integers Modulo 2 and Parity Ring|next = Isomorphism between Roots of Unity under Multiplication and Integers under Modulo Addition}}: $\S 6$: Example $6.1$
Group of Gaussian Integer Units/Cayley Table
== Cayley Table for Group of Gaussian Integer Units == The group of Gaussian integer units: :$\struct {U_\C, \times}$ can be described completely by showing its Cayley table:  :$\begin{array}{r|rrrr} \times & 1 & i & -1 & -i \\ \hline 1 & 1 & i & -1 & -i \\ i & i & -1 & -i & 1 \\ -1 & -1 & -i & 1 & i \\ -i & -i & 1 & i & -1 \\ \end{array}$  == Sources == * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Modulo Addition/Cayley Table/Modulo 4|next = Set of Integers under Addition is Isomorphic to Set of Even Integers under Addition}}: $\S 6$: Example $6.2$ * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Units of Gaussian Integers form Group|next = Quaternion Group/Complex Matrices}}: $\S 34$. Examples of groups: $(6) \ \text{(i)}$
Modulo Addition/Cayley Table/Modulo 4
== Cayley Table for Addition Modulo $4$ == The additive group of integers modulo $4$ can be described by showing its Cayley table:  :$\begin{array}{r|rrrr} \struct {\Z_4, +_4} & \eqclass 0 4 & \eqclass 1 4 & \eqclass 2 4 & \eqclass 3 4 \\ \hline \eqclass 0 4 & \eqclass 0 4 & \eqclass 1 4 & \eqclass 2 4 & \eqclass 3 4 \\ \eqclass 1 4 & \eqclass 1 4 & \eqclass 2 4 & \eqclass 3 4 & \eqclass 0 4  \\ \eqclass 2 4 & \eqclass 2 4 & \eqclass 3 4 & \eqclass 0 4 & \eqclass 1 4 \\ \eqclass 3 4 & \eqclass 3 4 & \eqclass 0 4 & \eqclass 1 4 & \eqclass 2 4 \\ \end{array}$  It can also be presented: :$\begin{array}{r|rrrr} +_4 & 0 & 1 & 2 & 3 \\ \hline 0 & 0 & 1 & 2 & 3  \\ 1 & 1 & 2 & 3 & 0 \\ 2 & 2 & 3 & 0 & 1 \\ 3 & 3 & 0 & 1 & 2 \\ \end{array}$ == Sources == * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Isomorphism between Gaussian Integer Units and Integers Modulo 4 under Addition/Proof 2|next = Group of Gaussian Integer Units/Cayley Table}}: $\S 6$: Example $6.2$ * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Modulo Multiplication has Identity|next = Modulo Multiplication/Cayley Table/Modulo 4}}: Chapter $1$: Integral Domains: $\S 6$. The Residue Classes * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Equivalence Relation/Examples/Non-Equivalence/Sum of Integers is Divisible by 3|next = Multiplicative Group of Reduced Residues Modulo 5/Cayley Table}}: Chapter $2$: Maps and relations on sets: Exercise $5$
Symmetric Group on 3 Letters/Cayley Table
== Cayley Table of Symmetric Group on $3$ Letters == The Cayley table of the symmetric group on $3$ letters can be presented in cycle notation as:  :$\begin{array}{c|cccccc} \circ & e & (123) & (132) & (23) & (13) & (12) \\ \hline e & e & (123) & (132) & (23) & (13) & (12) \\ (123) & (123) & (132) & e & (13) & (12) & (23) \\ (132) & (132) & e & (123) & (12) & (23) & (13) \\ (23) & (23) & (12) & (13) & e & (132) & (123) \\ (13) & (13) & (23) & (12) & (123) & e & (132) \\ (12) & (12) & (13) & (23) & (132) & (123) & e \\ \end{array}$  It can also often be seen presented as: :$\begin{array}{c|cccccc}   & e & p & q & r & s & t \\ \hline e & e & p & q & r & s & t \\ p & p & q & e & s & t & r \\ q & q & e & p & t & r & s \\ r & r & t & s & e & q & p \\ s & s & r & t & p & e & q \\ t & t & s & r & q & p & e \\ \end{array}$ == Sources == * {{BookReference|Sets and Groups|1965|J.A. Green|prev = Center of Symmetric Group is Trivial|next = Modulo Addition/Cayley Table/Modulo 6}}: Tables: $1$. Symmetric Group $\map S 3$ * {{BookReference|Modern Algebra|1965|Seth Warner|prev = Left and Right Operation Closed for All Subsets|next = Centralizer of Group Element is Subgroup}}: Exercise $8.6$ * {{BookReference|Topology: An Introduction with Application to Topological Groups|1967|George McCarty|prev = Definition:Cayley Table|next = Cayley Table for Commutative Operation is Symmetrical about Main Diagonal}}: Chapter $\text{II}$: Groups: Exercise $\text{A}$ * {{BookReference|A Course in Group Theory|1996|John F. Humphreys|prev = Order of Symmetric Group/Examples/Degree 3|next = Symmetry Group of Equilateral Triangle is Symmetric Group}}: Chapter $2$: Maps and relations on sets: Example $2.19$
Lower Bound is Upper Bound for Inverse Ordering
== Definition == Let $\struct {S, \preceq}$ be an ordered set. Let $T \subseteq S$. Let $m$ be a lower bound for $\struct {T, \preceq}$. Let $\succeq$ be the dual ordering of $\preceq$. Then $m$ is an upper bound for $\struct {T, \succeq}$. == Proof == Let $m$ be a lower bound for $\struct {T, \preceq}$. That is: :$\forall a \in T: m \preceq a$ By definition of dual ordering, it follows that: :$\forall a \in T: a \succeq m$ That is, $M$ is an upper bound for $\struct {T, \succeq}$.  == Also see == * Upper Bound is Lower Bound for Inverse Ordering
Real Addition Identity is Zero/Corollary
== Corollary to Real Addition Identity is Zero ==  :$\forall x, y \in \R: x + y = x \implies y = 0$  == Proof == {{begin-eqn}} {{eqn | l = x + y       | r = x       | c =  }} {{eqn | ll= \leadsto       | l = \paren {-x} + \paren {x + y}       | r = \paren {-x} + x       | c =  }} {{eqn | ll= \leadsto       | l = \paren {\paren {-x} + x} + y       | r = \paren {-x} + x       | c = Real Number Axioms: $\R \text A 1$ }} {{eqn | ll= \leadsto       | l = 0 + y       | r = 0       | c = Real Number Axioms: $\R \text A 4$ }} {{eqn | ll= \leadsto       | l = y       | r = 0       | c = Real Addition Identity is Zero }} {{end-eqn}}  == Sources == * {{BookReference|Calculus|1967|Michael Spivak|prev = Definition:Additive Inverse of Number|next = Definition:Subtraction}}: Part $\text I$: Prologue: Chapter $1$: Basic Properties of Numbers * {{BookReference|Topology|2000|James R. Munkres|ed = 2nd|edpage = Second Edition|prev = Archimedean Principle|next = Real Zero is Zero Element}}: $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $1 \ \text{(a)}$
Multiplication by Negative Real Number/Corollary
== Corollary to Multiplication by Negative Real Number ==  : $\forall x \in \R: \paren {-1} \times x = -x$  == Proof == {{begin-eqn}} {{eqn | l = \paren {-1} \times x       | r = -\paren {1 \times x}       | c = Multiplication by Negative Real Number }} {{eqn | r = -x       | c = Real Number Axioms: $\R M 3$: Identity for Multiplication }} {{end-eqn}}  == Sources == * {{BookReference|Topology|2000|James R. Munkres|ed = 2nd|edpage = Second Edition|prev = Multiplication by Negative Real Number|next = Multiplication of Real Numbers is Left Distributive over Subtraction/Proof 2}}: $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $1 \ \text{(f)}$
Cartesian Product of Bijections is Bijection/General Result
Cartesian Product of Bijections is Bijection/General Result 0 39315 215582 2015-04-30T08:06:02Z Prime.mover 59 Created page with "== Theorem ==  Let $I$ be an indexing set.  Let $\left\langle{S_i}\right\rangle_{i \mathop \in I}$ and $\left\langle{T_i}\right\rangle..." wikitext text/x-wiki == Theorem ==  Let $I$ be an indexing set. Let $\left\langle{S_i}\right\rangle_{i \mathop \in I}$ and $\left\langle{T_i}\right\rangle_{i \mathop \in I}$ be families of sets indexed by $I$. Let $\displaystyle \mathcal S := \prod_{i \mathop \in I} S_i$ and $\displaystyle \mathcal T := \prod_{i \mathop \in I} T_i$ be the Cartesian products of $\left\langle{S_i}\right\rangle_{i \mathop \in I}$ and $\left\langle{T_i}\right\rangle_{i \mathop \in I}$ respectively. For all $i \in I$, let $f_i: S_i \to T_i$ be a bijection. Let $\displaystyle \mathcal F: \mathcal S \to \mathcal T := \prod_{i \mathop \in I} \left({f_i: S_i \to T_i}\right)$ be the Cartesian product of $\left\langle{f_i}\right\rangle_{i \mathop \in I}$ defined as: :$\displaystyle \forall \mathbf s \in \mathcal S: \mathcal F \left({\mathbf s}\right) := \prod_{i \mathop \in I} \left({f_i \left({s_i}\right)}\right)$ where $\mathbf s := \left\langle{s_i}\right\rangle_{i \mathop \in I}$ is an arbitrary element of $\left\langle{S_i}\right\rangle_{i \mathop \in I}$. Then $\mathcal F$ is a bijection.  == Proof == Because $f_i$ are bijections, it follows by definition that they are surjections. Let $\mathbf t := \left\langle{t_i}\right\rangle_{i \mathop \in I} \in \mathcal T$. Then as $f_i$ is a surjection: :$\forall i \in I: \exists s_i \in S_i: f_i \left({s_i}\right) = t_i$ Thus: :$\exists \mathbf s \in \mathcal S: \mathcal F \left({\mathbf s}\right) = \mathbf t$ So $\mathcal F$ is a surjection. Because $f_i$ are bijections, it follows by definition that they are injections. Let: :$\mathbf t_1 := \left\langle{t_{i 1} }\right\rangle_{i \mathop \in I} \in \mathcal T$ :$\mathbf t_2 := \left\langle{t_{i 2} }\right\rangle_{i \mathop \in I} \in \mathcal T$ Let: : $\mathcal F \left({\mathbf s_1}\right) = \left({\mathbf t_1}\right), \mathcal F \left({\mathbf s_2}\right) = \left({\mathbf t_2}\right)$ for some $\mathbf s_1, \mathbf s_2 \in \mathcal S$ where: :$\mathbf s_1 := \left\langle{s_{i 1} }\right\rangle_{i \mathop \in I}$ :$\mathbf s_2 := \left\langle{s_{i 2} }\right\rangle_{i \mathop \in I}$ Suppose $\mathbf t_1 = \mathbf t_2$. Then: : $\forall i \in I: t_{i 1} = t_{i 2}$ By definition of $\mathcal F$: :$\forall i \in I: f_i \left({s_{i 1} }\right) = t_{i 1}$ :$\forall i \in I: f_i \left({s_{i 2} }\right) = t_{i 2}$ As $f_i$ is an injection for all $i \in I$: :$\forall i \in I: t_{i 1} = t_{i 2 }\implies s_{i 1} = s_{i 2}$ Thus it follows that: :$\mathbf t_1 = \mathbf t_2 \implies \mathbf s_1 = \mathbf s_2$ and so $\mathcal F$ is an injection. So $\mathcal F$ is a surjection and also an injection. Hence by definition, $\mathcal F$ is a bijection.
Invertible Element of Monoid is Cancellable
== Corollary to Invertible Element of Associative Structure is Cancellable ==  Let $\left({S, \circ}\right)$ be a monoid whose identity is $e_S$. An element of $\left({S, \circ}\right)$ which is invertible is also cancellable.  == Proof == By definition of monoid, $\left({S, \circ}\right)$ is an algebraic structure in which $\circ$ is associative. The result follows from Invertible Element of Associative Structure is Cancellable.
Group/Examples/inv x = 1 - x/Lemma 1
== Lemma for Group Example: $x^{-1} = 1 - x$ == Define $f: \openint 0 1 \to \R$ by: :$\map f x := \map \ln {\dfrac {1 - x} x}$ and $g: \R \to \openint 0 1$: :$\map g z := \dfrac 1 {1 + \exp z}$ Then:  :$\map {f \circ g} x = x$  == Proof == {{begin-eqn}} {{eqn | l = \map {f \circ g} x       | r = \map f {\map g x}       | c = }} {{eqn | r = \map \ln {\frac {1 - \map g x} {\map g x} }       | c =  }} {{eqn | r = \map \ln {\frac {1 - \frac 1 {1 + \exp x} } {\frac 1 {1 + \exp x} } }       | c =  }} {{eqn | r = \map \ln {\frac {1 + \exp x - 1} 1}       | c = multiplying both numerator and denominator by $1 + \exp x$ }} {{eqn | r = \map \ln {\exp x}       | c = Simplifying }} {{eqn | r = x       | c = Exponential of Natural Logarithm }} {{end-eqn}} {{qed|lemma}}
Group/Examples/inv x = 1 - x/Lemma 2
== Lemma for Group Example: $x^{-1} = 1 - x$ == Define $f: \openint 0 1 \to \R$ by: :$\map f x := \map \ln {\dfrac {1 - x} x}$ and $g: \R \to \openint 0 1$: :$\map g z := \dfrac 1 {1 + \exp z}$ Then:  :$\map {g \circ f} x = x$  == Proof == {{begin-eqn}} {{eqn | l = \map {g \circ f} x       | r = \map g {\map f x}       | c = }} {{eqn | r = \frac 1 {1 + \exp \map f x}       | c = }} {{eqn | r = \frac 1 {1 + \map \exp {\ln \dfrac {1 - x} x} }       | c = }} {{eqn | r = \frac 1 {1 + \dfrac {1 - x} x}       | c = Exponential of Natural Logarithm }} {{eqn | r = \frac x {x + 1 - x}       | c = multiplying both numerator and denominator by $x$ }} {{eqn | r = x       | c = simplifying }} {{end-eqn}} {{qed|lemma}}
Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group/Lemma 2
== Lemma ==  :$\paren {\pm a}^2 = \paren {\pm b}^2 = \paren {\pm c}^2 = -1$  == Proof == , $a$ is checked. The proofs for other $5$ elements are similar. {{begin-eqn}} {{eqn | l = \paren {a^2}^2       | r = a^4       | c = Powers of Group Elements: Product of Indices }} {{eqn | r = 1       | c = Definition of Order of Group Element }} {{end-eqn}} So $a^2 = 1$ or $a^2 = -1$. As the order of $a = 4$: :$a^2 \ne 1$ Hence: :$a^2 = -1$ as required.
Euler Triangle Formula/Lemma
== Lemma to Euler Triangle Formula ==  :400px Let the bisector of angle $C$ of triangle $\triangle ABC$ be produced to the circumcircle at $P$. Let $I$ be the incenter of $\triangle ABC$. Then: :$AP = BP = IP$  == Proof == , it will be demonstrated that $BP = IP$. Let $CP$ be the bisector of $\angle ACB$. We have therefore that: :$\angle ACP = \angle ICB$ From Angles on Equal Arcs are Equal: :$\angle ACP = \angle ABP$ and so: :$\angle ABP = \angle ICB$ By the construction of the incircle, $IB$ is the bisector of $B$. Then: :$\angle IBA = \angle IBC$ and so: {{begin-eqn}} {{eqn | l = \angle IBP       | r = \angle IBA + \angle ABP       | c =  }} {{eqn | r = \angle IBC + \angle ICB       | c =  }} {{end-eqn}} By Sum of Angles of Triangle equals Two Right Angles: :$\angle CIB + \angle IBC + \angle ICB = \angle CIB + \angle PIB$ as $\angle CIB$ and $\angle PIB$ are supplementary. Thus: {{begin-eqn}} {{eqn | l = \angle PIB       | r = \angle IBC + \angle ICB       | c =  }} {{eqn | r = \angle IBP       | c =  }} {{end-eqn}} and from Triangle with Two Equal Angles is Isosceles: :$IP = BP$ The proof that $IP = AP$ follows the same lines.
Cassini's Identity/Lemma
== Lemma for Cassini's Identity ==  :$\forall n \in \Z_{>1}: \begin{bmatrix} F_{n + 1} & F_n \\ F_n       & F_{n - 1} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n$  == Proof == === Basis for the Induction === :$\begin{bmatrix} F_2 & F_1     \\ F_1 & F_0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^1$ === Induction Hypothesis === For $k \in \Z_{>1}$, it is assumed that: :$\begin{bmatrix} F_{k + 1} & F_k \\ F_k       & F_{k - 1} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^k$ It remains to be shown that: :$\begin{bmatrix} F_{k + 2} & F_{k + 1}  \\ F_{k + 1} & F_k \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{k + 1}$ === Induction Step === The induction step follows from conventional matrix multiplication: {{begin-eqn}} {{eqn | l = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{k+1}       | r = \begin{bmatrix} F_{k + 1} & F_k \\ F_k       & F_{k - 1} \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}       | c = by the induction hypothesis }} {{eqn | r = \begin{bmatrix} F_{k + 1} + F_k & F_{k + 1} \\ F_k + F_{k - 1} & F_k \end{bmatrix}       | c = Definition of Matrix Product }} {{eqn | r = \begin{bmatrix} F_{k + 2} & F_{k + 1} \\ F_{k + 1} & F_k \end{bmatrix} }} {{end-eqn}} So by induction: :$\begin{bmatrix} F_{n + 1} & F_n \\ F_n       & F_{n - 1} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n$  == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Fibonacci Numbers which equal the Square of their Index|next = Prime Fibonacci Number has Prime Index except for 3}}: $\S 1.2.8$: Fibonacci Numbers: Exercise $6$
Real Numbers are Uncountable/Proof 2 using Ternary Notation/Lemma
== Lemma to Real Numbers are Uncountable: Proof 2 using Ternary Notation ==  Let $\left\langle{d_n}\right\rangle$ and $\left\langle{e_n}\right\rangle$ be infinite sequences in $\left\{{0, 1}\right\}$ such that: :$\exists m \in \N: d_m \ne e_m$ That is, the sequences $\left\langle{d_n}\right\rangle$ and $\left\langle{e_n}\right\rangle$ are different in at least one term. Then the ternary representations $D = 0.d_1 d_2 \ldots$ and $E = 0.e_1 e_2 \dots$ represent distinct real numbers.  == Proof == Let $\left\langle{d_n}\right\rangle \ne \left\langle{e_n}\right\rangle$. By the Well-Ordering Principle, there is a smallest $n \in \N_{>0}$ such that $d_n \ne e_n$. , suppose that $d_n = 0$ and $e_n = 1$. Let $K = 0.d_1 d_2 \ldots d_{n-1} = \sum_{i \mathop = 1}^{n-1} d_i 3^{-i}$. Let: : $\displaystyle D := K + \sum_{i \mathop = n+1}^\infty d_i 3^{-i}$ : $\displaystyle E := K + 3^{-n} + \sum_{i \mathop = n+1}^\infty e_i 3^{-i} \ge K + 3^{-n}$ But then: {{begin-eqn}} {{eqn | l = D \le K + \sum_{i \mathop = n + 1}^\infty 3^{-i}       | r = K + 3^{-n-1} \sum_{i \mathop = 0}^\infty 3^{-i}       | c =  }} {{eqn | r = K + 3^{-n - 1} \dfrac 3 2       | c =  }} {{eqn | r = K + \frac {3^{-n} } 2       | c =  }} {{end-eqn}} Thus $D < E$, so $D \ne E$.