nyuuzyou/wikis huggingface dataset:
Werner Formulas/Hyperbolic Sine by Hyperbolic Cosine

== Theorem ==
:$\sinh x \cosh y = \dfrac {\sinh \paren {x + y} + \sinh \paren {x - y} } 2$
where $\sinh$ denotes hyperbolic sine and $\cosh$ denotes hyperbolic cosine.
== Proof ==
== Also presented as ==
This result can also be seen presented as:
:$2 \sinh x \cosh y = \sinh \paren {x + y} + \sinh \paren {x - y}$
== Sources ==
* : $\S 4$. Elementary Functions of a Complex Variable: Exercise $8 \ \text{(ii)}$
* : $\S 8$: Hyperbolic Functions: $8.48$: Sum, Difference and Product of Hyperbolic Functions
Category:Hyperbolic Sine Function
Category:Hyperbolic Cosine Function
Category:Werner Formulas
Werner Formulas/Hyperbolic Sine by Hyperbolic Sine

== Theorem ==
:$\sinh x \sinh y = \dfrac {\map \cosh {x + y} - \map \cosh {x - y} } 2$
where $\sinh$ denotes hyperbolic sine and $\cosh$ denotes hyperbolic cosine.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also presented as ==
This result can also be seen presented as:
:$2 \sinh x \sinh y = \map \cosh {x + y} - \map \cosh {x - y}$
== Sources ==
* : $\S 4$. Elementary Functions of a Complex Variable: Exercise $8 \ \text{(iii)}$
* : $\S 8$: Hyperbolic Functions: $8.46$: Sum, Difference and Product of Hyperbolic Functions
Category:Werner Formula for Hyperbolic Sine by Hyperbolic Sine
Category:Werner Formulas
Category:Hyperbolic Sine Function
Werner Formulas/Sine by Cosine

== Theorem ==
:$\sin \alpha \cos \beta = \dfrac {\map \sin {\alpha + \beta} + \map \sin {\alpha - \beta} } 2$
where $\sin$ denotes sine and $\cos$ denotes cosine.
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
== Examples ==
== Also see ==
* Werner Formula for Sine by Sine
* Werner Formula for Cosine by Cosine
* Werner Formula for Cosine by Sine
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.67$
* 
* 
* : Appendix $12$: Trigonometric formulae: Product formulae
* : Appendix $14$: Trigonometric formulae: Product formulae
Category:Werner Formula for Sine by Cosine
Category:Werner Formulas
Category:Sine Function
Category:Cosine Function
Werner Formulas/Sine by Sine

== Theorem ==
:$\sin \alpha \sin \beta = \dfrac {\map \cos {\alpha - \beta} - \map \cos {\alpha + \beta} } 2$
where $\sin$ denotes sine and $\cos$ denotes cosine.
=== Corollary ===
== Proof ==
== Also presented as ==
== Examples ==
== Also see ==
* Werner Formula for Cosine by Cosine
* Werner Formula for Sine by Cosine
* Werner Formula for Cosine by Sine
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.65$
* 
* 
* : Appendix $12$: Trigonometric formulae: Product formulae
* : Appendix $14$: Trigonometric formulae: Product formulae
Category:Werner Formula for Sine by Sine
Category:Werner Formulas
Category:Sine Function
Westwood's Puzzle

== Theorem ==
:
Take any rectangle $ABCD$ and draw the diagonal $AC$.
Inscribe a circle $GFJ$ in one of the resulting triangles $\triangle ABC$.
Drop perpendiculars $IEF$ and $HEJ$ from the center of this incircle $E$ to the sides of the rectangle.
Then the area of the rectangle $DHEI$ equals half the area of the rectangle $ABCD$.
== Proof 1 ==
== Proof 2 ==
Category:Westwood's Puzzle
Category:Incircles of Triangles
Category:Euclidean Geometry
Westwood's Puzzle/Proof 2

== Theorem ==
== Proof ==
The crucial geometric truth to note is that:
:$CJ = CG, AG = AF, BF = BJ$
This follows from the fact that:
:$\triangle CEJ \cong \triangle CEG$, $\triangle AEF \cong \triangle AEG$ and $\triangle BEF \cong \triangle BEJ$
This is a direct consequence of the point $E$ being the center of the incircle of $\triangle ABC$.
Then it is just a matter of algebra.
Let $AF = a, FB = b, CJ = c$.
Category:Westwood's Puzzle
Wheat and Chessboard Problem

== Classic Problem ==
The squares of a chessboard are numbered in some way, from $1$st to $64$th.
A grain of wheat is placed on the $1$st squares of the chessboard.
$2$ grains of wheat are placed on the $2$nd square.
$4$ grains of wheat are placed on the $3$rd square.
The sequence continues: on the $n$th square, twice as many grains of wheat are placed as there are on the $n - 1$th square.
How much wheat is there on the chessboard when the $64$th square has been filled?
== Solution ==
There will be of the order of $1$ billion ($10^{12})$ cubic metres of grain on the chessboard.
== Proof ==
Let $W$ be the total volume of grain on the chessboard.
Let $N$ be the number of grains on the chessboard
$N$ is governed by a geometric sequence of $64$ terms whose initial term is $1$ and whose common ratio is $2$.
From Sum of Geometric Sequence $N$ is given by:
:$N = \dfrac {2^{64} - 1} {2 - 1} = 2^{64} - 1$
Let it be assumed that there are $2^{10} \times 3^2$ grains of wheat to the (imperial) pint.
This approximation is suggested in of $1881$.
There are $64$ (imperial) pints to the bushel.
Hence we have:
Using a conversion factor of $27.496$ bushels per cubic metre, this gives:
:$W \approx 1 \, 137 \, 000 \, 000 \, 000 \ \mathrm m^3$ 
That is quite a lot of grain.
== Historical Note ==
== Sources ==
* : Arithmetical Puzzles: The Progression of Numbers
* : Sissa and the Chessboard: $46$
Category:Classic Problems
Category:Wheat and Chessboard Problem
Category:Chessboard Puzzles
Wheat and Chessboard Problem/19th Century Version

== Wheat and Chessboard Problem: 19th Century Version ==
It is calculated that $200$ pins go to the ounce.
Suppose that for the $52$ weeks in the year:
:one pin were dropped into the hold during the first week
:two in the second
:four in the third
and so on.
By the end of the year the weight of the whole would be no less than $628 \, 292\, 358$ tons of pins.
As the Great Eastern steamship was built to carry $22 \, 500$ tons only, it follows that to carry all the pins there would be required $27 \, 924$ ships of the size of the Great Eastern.
== Sources ==
* : Arithmetical Puzzles: The Progression of Numbers
* : Sissa and the Chessboard: $46$
Category:Wheat and Chessboard Problem
Wheat and Chessboard Problem/Historical Note

== Historical Note on Wheat and Chessboard Problem ==
=== Note on $18 \, 446 \, 744 \, 073 \, 709 \, 551 \, 615$ ===
=== $19$th Century Version ===
The question was modernized in , as The Pin in the Hold of the "Great Eastern" Steamship.
== Sources ==
* : Arithmetical Puzzles: The Progression of Numbers
* : Sissa and the Chessboard: $46$
Category:Historical Notes
Who Drinks Hotter Coffee?

== Problem ==
Pinky and Perky both order coffee, and receive cups of it at the same temperature at the same time.
Pinky adds a small quantity of cool milk immediately, but does not drink his coffee immediately.
Perky waits some time before adding exactly the same quantity of milk of the same temperature as that which Pinky added.
Both Pinky and Perky then drink their coffee at the same time.
Who drinks the hotter coffee?
== Solution ==
By Newton's Law of Cooling, Perky's coffee loses heat faster than Pinky's.
When Perky adds his milk, it will have lost more heat to its surroundings than Pinky's.
Hence, after Perky has added his milk, Pinky's coffee will be hotter than Perky's coffee.
== Sources ==
* : $1$: The Nature of Differential Equations: Miscellaneous Problems for Chapter $1$: Problem $15$
Category:Classic Problems
Whole Sample Space represents Certain Event

== Theorem ==
Let the probability space of an experiment $\EE$ be $\struct {\Omega, \Sigma, \Pr}$.
The sample space $\Sigma$ represents an event which is certain.
== Proof ==
By definition, an event is a subset of the sample space $\Omega$.
Hence an outcome of $\EE$ is necessarily an element of $\Omega$.
That is, the probability that $\omega \in \Omega$ is $1$.
The result follows by definition of certain event.
== Sources ==
* 
* 
Category:Certain Events
Category:Sample Spaces
Whole Space is Open in Neighborhood Space

== Theorem ==
Let $\struct {S, \NN}$ be a neighborhood space.
Then $S$ itself is an open set of $\struct {S, \NN}$.
== Proof ==
Let $x \in S$.
Then by neighborhood space axiom $\text N 1$ there exists a neighborhood $N$ of $x$.
As $N \subseteq S$ it follows from neighborhood space axiom $\text N 3$ that $S$ is a neighborhood of $x$.
As this holds for all $x \in S$ it follows that $S$ is an open set of $\struct {S, \NN}$.
== Sources ==
* : Chapter $3$: Topological Spaces: $\S 3$: Neighborhoods and Neighborhood Spaces: Lemma $3.6$
Category:Neighborhood Spaces
Wholly Real Number and Wholly Imaginary Number are Linearly Independent over the Rationals

== Theorem ==
Let $z_1$ be a non-zero wholly real number.
Let $z_2$ be a non-zero wholly imaginary number.
Then, $z_1$ and $z_2$ are linearly independent over the rational numbers $\Q$, where the group is the complex numbers $\C$.
== Proof ==
From Rational Numbers form Subfield of Complex Numbers, the unitary module $\struct {\C, +, \times}_\Q$ over $\Q$ satisfies the unitary module axioms:
* $\Q$-Action: $\C$ is closed under multiplication, so $\Q \times \C \subset \C$.
* Distributive: $\times$ distributes over $+$.
* Associativity: $\times$ is associative.
* Multiplicative Identity: $1$ is the multiplicative identity in $\C$.
Let $a, b \in \Q$ such that:
:$a z_1 + b z_2 = 0$
By the definition of wholly imaginary, there is a real number $c$ such that:
:$c i = z_2$
where $i$ is the imaginary unit.
Therefore:
:$a z_1 + b c i = 0$
Equating real parts and imaginary parts:
:$a z_1 = 0$
:$b c = 0$
Since $z_1$ and $c$ are both non-zero, $a$ and $b$ must be zero.
The result follows from the definition of linear independence.
Category:Complex Analysis
Wieferich's Criterion

== Theorem ==
Suppose Fermat's equation:
:$x^p + y^p = z^p$
has a solution in which $p$ is an odd prime that does not divide any of $x$, $y$ or $z$.
Then $2^{p - 1} - 1$ is divisible by $p^2$.
== Proof ==
== Also known as ==
Some sources give this as Wieferich's theorem, but this is also used for his result concerning the Hilbert-Waring Theorem for cubes.
== Also see ==
* Definition:Wieferich Prime
== Historical Note ==
== Sources ==
* : $1093$
* : $1093$
Category:Fermat's Last Theorem
Category:Wieferich Primes
Wieferich's Criterion/Historical Note

== Historical Note on Wieferich's Criterion ==
 discovered what is now known as Wieferich's Criterion in $1909$.
It had profound implications for Fermat's Last Theorem, in that it demonstrated that the only cases that needed to be considered were those for the Wieferich primes, of which only $2$ are known less than $10^{17}$.
== Sources ==
* : $1093$
* : $1093$
Category:Historical Notes
Wilson's Theorem

== Theorem ==
A (strictly) positive integer $p$ is a prime :
:$\paren {p - 1}! \equiv -1 \pmod p$
=== Corollary $1$ ===
=== Corollary $2$ ===
== Proof ==
If $p = 2$ the result is immediate.
Therefore we assume that $p$ is an odd prime.
=== Necessary Condition ===
=== Sufficient Condition ===
== Examples ==
== Also known as ==
== Historical Note ==
== Sources ==
* : $\text {4-1}$ Basic Properties of Congruences: Example $\text {4-2}$
* : $24$
* : $561$
* : $24$
* : $563$
* 
* 
Category:Wilson's Theorem
Category:Prime Numbers
Category:Factorials
Category:Modulo Arithmetic
Wilson's Theorem/Corollary 1

== Theorem ==
Let $p$ be a prime number.
Then $p$ is the smallest prime number which divides $\paren {p - 1}! + 1$.
== Proof ==
From Wilson's Theorem, $p$ divides $\paren {p - 1}! + 1$.
Let $q$ be a prime number less than $p$.
Then $q$ is a divisor of $\paren {p - 1}!$ and so does not divide $\paren {p - 1}! + 1$.
== Sources ==
* : $\text {3-3}$ Wilson's Theorem: Exercise $1$
Category:Wilson's Theorem
Wilson's Theorem/Corollary 2

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $p$ be a prime factor of $n!$ with multiplicity $\mu$.
Let $n$ be expressed in a base $p$ representation as:
Then:
:$\dfrac {n!} {p^\mu} \equiv \paren {-1}^\mu a_0! a_1! \dotsb a_m! \pmod p$
== Proof 1 ==
== Proof 2 ==
== Historical Note ==
Category:Factorials
Category:Prime Numbers
Category:Modulo Arithmetic
Category:Wilson's Theorem
Wilson's Theorem/Corollary 2/Historical Note

== Historical Note on Corollary 2 to Wilson's Theorem ==
This corollary to Wilson's theorem was demonstrated, according to , by in $1890$.
It is supposed that this appears in 's $1890$ article during the course of his proof of Stickelberger's Theorem.
However, not only is the latter behind a paywall, it is also in a language the author of this page is not fluent in, and he has been disinclined to study it.
== Sources ==
* 
* : $\S 1.2.5$: Permutations and Factorials: Exercise $14$
Category:Historical Notes
Wilson's Theorem/Corollary 2/Proof 1

== Theorem ==
== Proof ==
Proof by induction:
Let $\map P n$ be the proposition:
:$\dfrac {n!} {p^\mu} \equiv \paren {-1}^\mu a_0! a_1! \dotsm a_k! \pmod p$
where $p, a_0, \dots, a_k, \mu$ are as defined above.
=== Basis for the Induction ===
For $n = 1$:
:$a_0 = 1, \mu = 0$
$\map P 1$ reduces to:
:$\dfrac {1!} {p^0} \equiv \paren {-1}^0 1! \pmod p$
which holds trivially.
This is the basis for the induction.
=== Induction Hypothesis ===
This is our induction hypothesis:
:$\dfrac {r!} {p^\mu} \equiv \paren {-1}^\mu a_0! a_1! \dots a_k! \pmod p$
Now we need to show true for $n = r + 1$:
:$\dfrac {\paren {r + 1}!} {p^{\mu'}} \equiv \paren {-1}^{\mu'} b_0! b_1! \dots b_k! \pmod p$
where $\ds r + 1 = \sum_{j \mathop = 0}^k b_j p^j$ is the base $p$ presentation of $r + 1$, and:
:$p^{\mu'} \divides \paren {r + 1}!$
:$p^{\mu' + 1} \nmid \paren {r + 1}!$
=== Induction Step ===
This is our induction step:
Let $p^m$ be the largest power of $p$ which divides $r + 1$, that is:
:$p^m \divides r + 1$
:$p^{m + 1} \nmid r + 1$
Then we must have:
:$b_m \ne 0$
:$b_j = 0$ for $0 \le j < m$
:$\mu' = \mu + m$
and consequently:
:$a_j = b_j$ for $j > m$
:$a_m + 1 = b_m$
:$a_j = p - 1$ for $0 \le j < m$
Thus:
By the Principle of Mathematical Induction, $\map P n$ is true for all $n \in \Z_{>0}$.
Category:Wilson's Theorem
Wilson's Theorem/Corollary 2/Proof 2

== Theorem ==
== Proof ==
Consider the numbers of $\set {1, 2, \ldots, n}$ which are not multiples of $p$.
There are $\floor {\dfrac n p}$ complete sets of $p - 1$ such consecutive elements of $\set {1, 2, \ldots, n}$.
Each one of these has a product which is congruent to $-1 \pmod p$ by Wilson's Theorem.
There are also $a_0$ left over which are congruent to $a_0! \pmod p$.
Thus:
:the contributions of the divisors which are not multiples of $p$ is $\paren {-1}^{\floor {n / p} } a_0!$
:the contributions of the divisors which are multiples of $p$ is the same as the contribution in $\floor {\dfrac n p}!$
Thus the argument can be repeated on $\floor {\dfrac n p}!$ until the formula is complete.
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials: Exercise $14$
Category:Wilson's Theorem
Wilson's Theorem/Examples/10 does not divide (n-1)!+1

== Example of Use of Wilson's Theorem ==
For all $n \in \Z_{>0}$, $10$ is not a divisor of $\paren {n - 1}! + 1$.
== Proof ==
For the first few $n$ we see:
Now consider $n > 5$.
We have that:
Hence $10 \nmid \paren {n - 1}! + 1$.
== Sources ==
* : $\text {3-3}$ Wilson's Theorem: Exercise $2$
Category:Wilson's Theorem
Wilson's Theorem/Examples/5 divides (5-1)!+1

== Example of Use of Wilson's Theorem ==
$5$ is a divisor of $\paren {5 - 1}! + 1$.
== Proof ==
For the first few $n$ we see:
Hence $5 \divides \paren {5 - 1}! + 1$.
== Sources ==
* 
* 
Category:Wilson's Theorem
Wilson's Theorem/Historical Note

== Historical Note on Wilson's Theorem ==
The proof of Wilson's Theorem was attributed to by in his $1770$ edition of Meditationes Algebraicae.
It was first stated by .
It appears also to have been known to in $1682$ or $1683$ (accounts differ).
It was in fact proved by in $1771$.
== Sources ==
* : $\text {3-3}$ Wilson's Theorem: Theorem $\text {3-5}$
* : $24$
* : $\S 1.2.5$: Permutations and Factorials: Exercise $13$
* : $24$
* 
* 
Category:Historical Notes
Wilson's Theorem/Necessary Condition

== Theorem ==
Let $p$ be a prime number.
Then:
:$\paren {p - 1}! \equiv -1 \pmod p$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also known as ==
== Historical Note ==
== Sources ==
* : $0$: Some Conventions and some Basic Facts: Exercise $9$
Category:Wilson's Theorem
Wilson's Theorem/Necessary Condition/Proof 1

== Theorem ==
== Proof ==
If $p = 2$ the result is obvious.
Therefore we assume that $p$ is an odd prime.
Let $p$ be prime.
Consider $n \in \Z, 1 \le n < p$.
As $p$ is prime, $n \perp p$.
From Law of Inverses (Modulo Arithmetic), we have:
:$\exists n' \in \Z, 1 \le n' < p: n n' \equiv 1 \pmod p$
By Solution of Linear Congruence, for each $n$ there is exactly one such $n'$, and $\paren {n'}' = n$.
So, provided $n \ne n'$, we can pair any given $n$ from $1$ to $p$ with another $n'$ from $1$ to $p$.
We are then left with the numbers such that $n = n'$.
Then we have $n^2 \equiv 1 \pmod p$.
Consider $n^2 - 1 = \paren {n + 1} \paren {n - 1}$ from Difference of Two Squares.
So either $n + 1 \divides p$ or $n - 1 \divides p$.
Observe that these cases do not occur simultaneously, as their difference is $2$, and $p$ is an odd prime.
From Negative Number is Congruent to Modulus minus Number‎:
:$p - 1 \equiv -1 \pmod p$
Hence $n = 1$ or $n = p - 1$.
So, we have that $\paren {p - 1}!$ consists of numbers multiplied together as follows:
:in pairs whose product is congruent to $1 \pmod p$
:the numbers $1$ and $p - 1$.
The product of all these numbers is therefore congruent to $1 \times 1 \times \cdots \times 1 \times p - 1 \pmod p$ by modulo multiplication.
From Negative Number is Congruent to Modulus minus Number:
:$\paren {p - 1}! \equiv -1 \pmod p$
== Historical Note ==
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials: Exercise $13$
Category:Wilson's Theorem
Wilson's Theorem/Necessary Condition/Proof 2

== Theorem ==
== Proof ==
If $p = 2$ the result is obvious.
Therefore we assume that $p$ is an odd prime.
Consider $p$ points on the circumference of a circle $C$ dividing it into $p$ equal arcs.
By joining these points in a cycle, we can create polygons, some of them stellated.
From Number of Different n-gons that can be Inscribed in Circle, the number of different such polygons is $\dfrac {\paren {p - 1}!} 2$.
When you rotate these polygons through an angle of $\dfrac {2 \pi} p$, exactly $\dfrac {p - 1} 2$ are unaltered.
These are the regular $p$-gons and regular stellated $p$-gons.
That there are $\dfrac {p - 1} 2$ of them follows from Number of Regular Stellated Odd n-gons.
The remaining $\dfrac {\paren {p - 1}!} 2 - \dfrac {p - 1} 2$ polygons can be partitioned into sets of $p$ elements: those which can be obtained from each other by rotation through multiples of $\dfrac {2 \pi} p$.
The total number of such sets is then:
:$\dfrac {\paren {p - 1}! - \paren {p - 1} } {2 p}$
Thus we have that:
:$2 p \divides \paren {\paren {p - 1}! - \paren {p - 1} }$
That is:
:$p \divides \paren {\paren {p - 1}! + 1}$
== Historical Note ==
== Sources ==
* : $\text {3-3}$ Wilson's Theorem: Theorem $\text {3-5}$
Category:Wilson's Theorem
Wilson's Theorem/Necessary Condition/Proof 3

== Theorem ==
== Proof ==
Let $p$ be prime.
Consider $\struct {\Z_p, +, \times}$, the ring of integers modulo $m$.
From Ring of Integers Modulo Prime is Field, $\struct {\Z_p, +, \times}$ is a field.
Hence, apart from $\eqclass 0 p$, all elements of $\struct {\Z_p, +, \times}$ are units
As $\struct {\Z_p, +, \times}$ is a field, it is also by definition an integral domain, we can apply:
From Product of Units of Integral Domain with Finite Number of Units, the product of all elements of $\struct {\Z_p, +, \times}$ is $-1$.
But the product of all elements of $\struct {\Z_p, +, \times}$ is $\paren {p - 1}!$
The result follows.
== Historical Note ==
== Sources ==
* : Chapter $9$: Rings: Exercise $12$
Category:Wilson's Theorem
Wilson's Theorem/Sufficient Condition

== Theorem ==
Let $p$ be a (strictly) positive integer such that:
:$\paren {p - 1}! \equiv -1 \pmod p$
Then $p$ is a prime number.
== Proof ==
Assume $p$ is composite, and $q$ is a prime such that $q \divides p$.
Then both $p$ and $\paren {p - 1}!$ are divisible by $q$.
If the congruence $\paren {p - 1}! \equiv -1 \pmod p$ were satisfied, we would have $\paren {p - 1}! \equiv -1 \pmod q$.
However, this amounts to $0 \equiv -1 \pmod q$, a contradiction.
Hence for $p$ composite, the congruence $\paren {p - 1}! \equiv -1 \pmod p$ cannot hold.
== Also known as ==
== Historical Note ==
Category:Wilson's Theorem
Wolstenholme's Conjecture

== Conjecture ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Suppose that:
:$\dbinom {2 n - 1} {n - 1} \equiv 1 \pmod {n^3}$
where $\dbinom {2 n - 1} {n - 1}$ denotes a binomial coefficient.
Then $n$ is a prime number
== Also see ==
* Wolstenholme's Theorem, of which this is the converse
== Sources ==
* 
Category:Binomial Coefficients
Category:Prime Numbers
Category:Unproven Hypotheses
Wolstenholme's Theorem

== Theorem ==
Let $p$ be a prime number such that $p \ge 5$.
Then:
:$\dbinom {2 p - 1} {p - 1} \equiv 1 \pmod {p^3}$
where $\dbinom {2 p - 1} {p - 1}$ denotes a binomial coefficient.
== Proof ==
== Also see ==
* Wolstenholme's Conjecture
* Babbage's Congruence
== Sources ==
* 
Category:Wolstenholme's Theorem
Category:Binomial Coefficients
Category:Prime Numbers
Word (Abstract Algebra)/Examples/Set with 2 Elements

== Examples of Words in the Context of Abstract Algebra ==
Let $G$ be a group.
Let $X \subseteq G$ be a subset of $G$ such that $X = \set {a, b}$.
Then some of the elements of the set of words $\map W S$ of $G$ are:
:$a, b, a b, b a, a b a, b a b, a^{-1} b, b a^{-1}, a b^{-1}, b^{-1} a, a b^{-1}, a^{-1} b^{-1}, a^{-1} b^{-1} a, \ldots$
== Sources ==
* : $\S 5.3$. Subgroup generated by a subset: Example $95$
Category:Examples of Words
Word (Abstract Algebra)/Examples/Symmetric Group on 3 Letters

== Examples of Words in the Context of Abstract Algebra ==
Let $S_3$ denote the Symmetric Group on 3 Letters, whose Cayley table is given as:
Consider the subset $\set {\tuple {1 2}, \tuple {132} }$ of $S_3$.
Then some of the elements of the set of words of $\set {\tuple {1 2}, \tuple {132} }$ are:
:$\tuple {123}^2, \tuple {123}^{-1} \tuple {1 2}^2, \tuple {123} \tuple {1 2} \tuple {123}^{-1} \tuple {1 2}^{-1}, \ldots$
== Sources ==
* : $\S 5.3$. Subgroup generated by a subset: Example $96$
Category:Examples of Words
Category:Symmetric Group on 3 Letters
Word Metric is Metric

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $S$ be a generating set for $G$ which is closed under inverses (that is, $x^{-1} \in S \iff x \in S$).
Let $d_S$ be the associated word metric.
Then $d_S$ is a metric on $G$.
== Proof ==
Let $g, h \in G$.
It is given that $S$ is a generating set for $G$.
It follows that there exist $s_1, \ldots, s_n \in S$ such that $g^{-1} \circ h = s_1 \circ \cdots \circ s_n$.
Therefore $\map {d_S} {g, h} \le n$, establishing that $\R$ is a valid codomain for the mapping $d_S$ with domain $G \times G$.
This is the form a mapping must have to be able to be a metric.
Now checking the other defining properties for a metric in turn:
=== ===
Clearly the empty sequence can be formed with elements from $S$.
It also has length zero.
Therefore, we have for any $g \in G$ that $\map {d_S} {g, g} = 0$.
Thus is seen to be fulfilled.
=== ===
Let $g, h, k \in G$.
Let $\map {d_S} {g, h} = n, \map {d_S} {h, k} = m$.
Let $s_1, \ldots, s_n, r_1, \ldots, r_m \in S$ be such that:
:$g^{-1} \circ h = s_1 \circ \cdots \circ s_n$
:$h^{-1} \circ k = r_1 \circ \cdots \circ r_m$
From these equations, obtain:
Therefore, $\map {d_S} {g, k} \le m + n = \map {d_S} {g, h} + \map {d_S} {h, k}$.
Thus is seen to be fulfilled.
=== ===
Let $g, h \in G$.
Let $\map {d_S} {g, h} = n$.
Furthermore, let $s_1, \ldots, s_n \in S$ be such that:
:$g^{-1} \circ h = s_1 \circ \cdots \circ s_n$
From Inverse of Group Product, obtain:
By assumption $S$ is closed under taking inverses, and hence the latter expression yields a valid sequence for the metric $d_S$.
It follows that $\map {d_S} {h, g} \le \map {d_S} {g, h}$.
Switching the roles of $g$ and $h$ in the above, we obtain the converse inequality, and hence equality.
Thus is seen to be fulfilled.
=== ===
There is only one word of length zero, namely the empty word.
However, the empty word sends an element $g$ to itself.
Hence $g, h \in G, g \ne h \implies \map {d_S} {g, h} > 0$.
Thus is seen to be fulfilled.
Thus $d_S$ is a metric.
Category:Group Theory
Category:Word Metric