nyuuzyou/wikis huggingface dataset:
Parallelogram Law for Vector Subtraction

== Theorem ==
Let $\mathbf u$ and $\mathbf v$ be vectors.
Consider a parallelogram, two of whose adjacent sides represent $\mathbf y$ and $\mathbf v$ (in magnitude and direction).
:
Then the diagonal of the parallelogram connecting the terminal points of $\mathbf u$ and $\mathbf v$ represents the magnitude and direction of $\mathbf u - \mathbf v$, the difference of $\mathbf u$ and $\mathbf v$.
== Proof ==
We can construct a parallelogram as follows:
:
and the construction is apparent.
== Sources ==
* : Chapter $1$ Vector Analysis $1.1$ Definitions, Elementary Approach
Category:Parallelogram Law
Category:Vector Subtraction
Parallelogram Similar and in Same Angle has Same Diameter

== Theorem ==
== Proof ==
From the parallelogram $\Box ABCD$ let there be taken away the parallelogram $\Box AF$ similar and similarly situated to the whole and having a common angle with it.
We need to show that $\Box ABCD$ is about the same diameter with $\Box AF$.
:
Suppose it is not, and that $AHC$ is the diameter of $\Box ABCD$.
Let $GF$ be produced to $H$.
From Construction of Parallel Line let $HK$ be drawn parallel to $AD$ and $BC$.
From Parallelograms About Diameter are Similar:
: $DA : AB = GA : AK$
But also because of the similarity of $\Box ABCD$ and $\Box EG$:
: $DA : AB = GA : AE$
So from Equality of Ratios is Transitive:
: $GA : AK = GA : AE$
Therefore from Magnitudes with Same Ratios are Equal:
: $AK = AE$
Hence the result.
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Parallelograms
Parallelogram on Same Base as Triangle has Twice its Area

== Theorem ==
A parallelogram on the same base as a triangle, and in the same parallels, has twice the area of the triangle.
== Proof ==
:
Let $ABCD$ be a parallelogram on the same base $BC$ as a triangle $EBC$, between the same parallels $BC$ and $AE$.
Join $AC$.
Then $\triangle ABC = \triangle EBC$ from Triangles with Same Base and Same Height have Equal Area.
But from Opposite Sides and Angles of Parallelogram are Equal, $AC$ bisects $ABCD$.
So the area of parallelogram $ABCD$ is twice the area of triangle $EBC$.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Area of Parallelogram
Category:Areas of Triangles
Parallelogram with One Right Angle is Rectangle

== Theorem ==
Let $\Box ABCD$ be a parallelogram.
Let one angle of $\Box ABCD$ be a right angle.
Then $\Box ABCD$ is a rectangle.
== Proof == 
, let $\angle ABC$ be a right angle.
Let $\angle CDA$ be opposite $\angle ABC$.
By Opposite Sides and Angles of Parallelogram are Equal:
:$\angle CDA = \angle ABC$
By Sum of Internal Angles of Polygon:
: the sum of the angles of $\Box ABCD$ is four right angles.
Hence:
$\angle DAB + \angle BCD$ together equal two right angles.
By Opposite Sides and Angles of Parallelogram are Equal:
:$\angle DAB = \angle BCD$
Hence, all four angles of $\Box ABCD$ are right angles.
By Opposite Sides and Angles of Parallelogram are Equal:
:$AB = CD$
:$AD = BC$
$\leadsto$
:$\Box ABCD$ is a rectangle.
The result follows.
Category:Parallelograms
Category:Rectangles
Parallelograms About Diameter are Similar

== Theorem ==
== Proof ==
Let $\Box ABCD$ be a parallelogram and $AC$ one of its diameters.
Let $EG, HK$ be parallelograms about $AC$.
:
We have that $EF \parallel CD$.
So from Parallel Transversal Theorem:
:$BE : EA = CF : FA$
Again, we have that:
:$FG \parallel CD$
So from Parallel Transversal Theorem:
:$CF : FA = DG : GA$
From Equality of Ratios is Transitive:
:$BE : EA = DG : GA$
So from Magnitudes Proportional Separated are Proportional Compounded:
:$BA : AE = DA : AG$
From Proportional Magnitudes are Proportional Alternately:
:$BA : AD = EA : AG$
Therefore in the parallelograms $\Box ABCD$ and $\Box EG$, the sides about the common angle $\angle BAD$ are proportional.
We have that $GF \parallel DC$, $\angle AFG = \angle DCA$ and $\angle DAC$ is common to $\triangle ADC$ and $\triangle AGF$.
Therefore $\triangle ADC$ is equiangular with $\triangle AGF$.
For the same reason, $\triangle ACB$ is equiangular with $\triangle AFE$.
Thus the whole parallelogram $\Box ABCD$ is equiangular with the parallelogram $\Box EG$.
Therefore:
:$AD : DC = AG : GF$
:$DC : CA = GF : FA$
:$AC : CB = AF : FE$
:$CB : BA = FE : FA$
Since we also have:
:$DC : CA = GF : FA$
:$AC : CB = AF : FE$
it follows from Equality of Ratios Ex Aequali that:
:$DC : CB = GF: FE$
Therefore in the parallelograms $\Box ABCD$ and $\Box EG$, sides about the equal angles are proportional.
Therefore from , $\Box ABCD$ is similar to $\Box EG$.
By the same argument, $\Box ABCD$ is similar to $\Box KH$.
Therefore by Similarity of Polygons is Equivalence Relation‎, $\Box EG$ is similar to $\Box HK$.
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Parallelograms
Parallelograms are Congruent if Two Adjacent Sides and Included Angle are respectively Equal

== Theorem ==
Let $ABCD$ and $EFGH$ be parallelograms.
Then $ABCD$ and $EFGH$ are congruent if:
: $2$ adjacent sides of $ABCD$ are equal to $2$ corresponding adjacent sides of $EFGH$
: the angle between those $2$ adjacent sides on both $ABCD$ and $EFGH$ are equal.
== Proof ==
 let the $2$ adjacent sides of $ABCD$ be $AB$ and $BC$.
Let the $2$ corresponding adjacent sides of $EFGH$ be $EF$ and $FG$ such that $AB = EF$ and $BC = FG$.
Hence let $\angle ABC = \angle EFG$.
From Triangle Side-Angle-Side Congruence:
:$\triangle ABC = \triangle EFG$
From Quadrilateral is Parallelogram iff Both Pairs of Opposite Angles are Equal:
:$\angle ABC = \angle ADC$
:$\angle EFG = \angle EHG$
From Quadrilateral is Parallelogram iff Both Pairs of Opposite Sides are Equal or Parallel:
:$AB = CD$
and:
:$EF = GF$
Thus again by Triangle Side-Angle-Side Congruence:
:$\triangle ADC = \triangle EHG$
So:
:$\triangle ABC + \triangle ADC = \triangle EFG + \triangle EHG$
and the result follows.
== Sources ==
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.24$
Category:Parallelograms
Parallelograms with Equal Base and Same Height have Equal Area

== Theorem ==
== Proof ==
:
Let $ABCD$ and $EFGH$ be parallelograms with equal bases $BC$ and $FG$, and in the same parallels $AH$ and $BG$.
Join $BE$ and $CH$.
We have $BC = FG$ and $FG = EH$
So by Common Notion 1 we have $BC = EH$.
But $EB$ and $HC$ join them.
So by Lines Joining Equal and Parallel Straight Lines are Parallel, $EB$ and $HC$ are equal and parallel.
So $EBCH$ is a parallelogram.
So, by Parallelograms with Same Base and Same Height have Equal Area, the area of $EBCH$ equals the area of $ABCD$.
Also by Parallelograms with Same Base and Same Height have Equal Area, the area of $EBCH$ equals the area of $EFGH$.
So by Common Notion 1, the area of $ABCD$ equals the area of $EFGH$.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Area of Parallelogram
Parallelograms with Same Base and Same Height have Equal Area

== Theorem ==
== Proof ==
:
Let $ABCD$ and $EBCF$ be parallelograms on the same base and in the same parallels $AF$ and $BC$.
As $ABCD$ is a parallelogram then $AD = BC$ from Opposite Sides and Angles of Parallelogram are Equal.
For the same reason, $EF = BC$.
So by Common Notion 1 we have that $AD = EF$.
Also, $DE$ is common, so the whole of $AE$ equals the whole of $DF$, by Common Notion 2.
But from Opposite Sides and Angles of Parallelogram are Equal, we have $AB = DC$.
From Parallelism implies Equal Corresponding Angles:
:$\angle FDC = \angle EAB$
So by Triangle Side-Angle-Side Congruence:
:$EB = FC$
and so $\triangle EAB = \triangle FDC$.
Subtract $\triangle DGE$ from each.
Then, by Common Notion 3, the trapezium $ABGD$ equals the trapezium $EGCF$.
Now we add $\triangle GBC$ to both.
Then, by Common Notion 2, $ABCD = EBCF$.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Area of Parallelogram
Parameter Independence of Riemannian Length of Admissible Curve

== Theorem ==
Let $\struct {M, g}$ be a Riemannian manifold.
Let $\closedint a b$ be a closed real interval.
Let $\gamma : \closedint a b \to M$ with $t \stackrel \gamma \mapsto \map \gamma t$ be an admissible curve.
Let $\tilde \gamma$ be a reparametrization of $\gamma$.
Let $\map {L_g} \gamma$ be the Riemannian length of $\gamma$ from $t = a$ to $t = b$.
Then:
:$\map {L_g} {\tilde \gamma} = \map {L_g} {\gamma}$
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Lengths and Distances
Category:Curves
Category:Riemannian Geometry
Parameter of Probability Distribution/Examples/Normal Distribution

== Examples of Parameters of Probability Distribution ==
Let $X$ be a continuous random variable with a normal distribution:
:$X \sim \Gaussian \mu {\sigma^2}$ 
The expectation $\mu$ and the variance $\sigma^2$ are the parameters of $\Gaussian \mu {\sigma^2}$.
== Sources ==
* 
* 
Category:Examples of Parameters of Probability Distributions
Category:Normal Distribution
Parameter of Stochastic Model/Examples/Linear Regression

== Examples of Parameters of Stochastic Model ==
Let $X$ and $Y$ be random variables.
Let $\expect {Y \mid x}$ be a linear regression of $Y$ on $X$:
:$\expect {Y \mid x_1, x_2} = \alpha + \beta_1 x_1 + \beta_1 x_1$
The parameters of $\expect {Y \mid x_1, x_2}$ are $\alpha$, $\beta_1$ and $\beta_2$.
== Sources ==
* 
* 
Category:Examples of Parameters of Stochastic Models
Category:Linear Regression
Parameterization of Unit Circle is Simple Loop

== Theorem ==
Let $\mathbb S^1$ denote the unit circle whose center is at the origin of the Euclidean space $\R^2$.
Let $p: \closedint 0 1 \to \R^2$ be defined by:
:$\forall t \in \closedint 0 1 : \map p t = \tuple {\map \cos {2 \pi t}, \map \sin {2 \pi t} }$
Then $p$ is a simple loop with image equal to $\mathbb S^1$.
== Proof ==
Parametric Equation of Circle shows that for all $r \in \R$, the point $\tuple {\map \cos r, \map \sin r}$ lies on the unit circle $\mathbb S^1$.
Parametric Equation of Circle also shows that for all points $\tuple {x, y}$ on the unit circle $\mathbb S^1$, the point can be expressed as $\tuple {x, y} = \tuple {\map \cos r, \map \sin r}$ for some $r \in \R$.
Sine and Cosine are Periodic on Reals shows that sine and cosine functions are periodic functions with period equal to $2 \pi$.
It follows that the image of $p$ is equal to $\mathbb S^1$.
Let $t_1, t_2 \in \closedint 0 1$ such that $\tuple {\map \cos {2 \pi t_1}, \map \sin {2 \pi t_1} } = \tuple {\map \cos {2 \pi t_2}, \map \sin {2 \pi t_2} }$ with $t_1 \ne t_2$.
Shape of Cosine Function shows that the cosine function is strictly decreasing on the interval $\closedint 0 \pi$, and strictly increasing on the interval $\closedint \pi {2 \pi}$.
Strictly Monotone Real Function is Bijective shows that with $t_1 \ne t_2$, we cannot have:
:both $t_1, t_2 \in \closedint 0 {\dfrac 1 2}$, as this would imply $\map \cos {2 \pi t_1} \ne \map \cos {2 \pi t_2}$
:or both $t_1, t_2 \in \closedint {\dfrac 1 2} 1$, as this would imply $\map \cos {2 \pi t_1} \ne \map \cos {2 \pi t_2}$
, we have $t_1 \in \closedint 0 {\dfrac 1 2}$, and $t_2 \in \closedint {\dfrac 1 2} 1$.
Sine and Cosine are Periodic on Reals/Corollary shows that the sine function is strictly positive on the interval $\openint 0 \pi$, and strictly negative on the interval $\openint \pi {2 \pi}$.
It follows that we have $\map \sin {2 \pi t_1} \ne \map \sin {2 \pi t_2}$, unless we have $t_1 = 0, t_2 = 1$, in which case:
:$\tuple {\map \cos 0, \map \sin 0} = \tuple {\map \cos {2 \pi}, \map \sin {2 \pi} } = \tuple {1, 0}$
by Cosine of Zero is One and Sine of Zero is Zero.
We have that Real Sine Function is Continuous and Cosine Function is Continuous.
Multiple Rule for Continuous Real Functions and Composite of Continuous Mappings between Metric Spaces is Continuous shows that the two coordinate projections:
:$\map \cos {2 \pi t}, \map \sin {2 \pi t}$
are continuous.
Hence from Continuity of Composite with Inclusion: Inclusion on Mapping and Continuous Mapping to Product Space, it follows that $p$ is continuous.
By definition of simple loop, it follows that $p$ is a simple loop.
Category:Circles
Category:Loops (Topology)
Parametric Equation/Examples/Circle

== Examples of Parametric Equations ==
Let $\CC$ be the circle embedded in a Cartesian plane with the equation:
:$x^2 + y^2 = 16$
This can be expressed in parametric equations as:
where $\theta$ is the parameter representing the angle between the $x$-axis and the point $\paren {x, y}$ on $\CC$.
Each point on $\CC$ corresponds exactly to a value of $\theta$ such that $\theta \in \hointr 0 {2 \pi}$.
== Sources ==
* 
* 
Category:Equation of Circle
Category:Examples of Parametric Equations
Parametric Equation/Examples/Ellipse

== Examples of Parametric Equations ==
Let $\EE$ be the ellipse embedded in a Cartesian plane with the equation:
:$\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$
This can be expressed in parametric equations as:
where $\phi$ is the parameter representing the eccentric angle of the point $\paren {x, y}$ on $\EE$.
Each point on $\CC$ corresponds exactly to a value of $\phi$ such that $\phi \in \hointr 0 {2 \pi}$.
== Sources ==
* 
* 
Category:Equation of Ellipse in Reduced Form
Category:Examples of Parametric Equations
Parametric Equation of Involute of Circle

== Theorem ==
Let $C$ be a circle of radius $a$ whose center is at the origin of a cartesian plane.
The involute $V$ of $C$ can be described by the parametric equation:
:$\begin {cases} x = a \paren {\cos \theta + \theta \sin \theta} \\ y = a \paren {\sin \theta - \theta \cos \theta} \end {cases}$
== Proof ==
By definition the involute of $C$ is described by the endpoint of a string unwinding from $C$.
Let that endpoint start at $\tuple {a, 0}$ on the circumference of $C$.
:
Let $P = \tuple {x, y}$ be an arbitrary point on $V$.
Let $Q$ be the point at which the cord is tangent to $C$.
Then $PQ$ equals the arc of $C$ from which the cord has unwound.
Thus:
:$PQ = a \theta$
where $\theta$ is the angle of $OQ$ to the $x$-axis.
Thus:
Hence the result.
== Sources ==
* : $\S 11$: Special Plane Curves: Involute of a Circle: $11.28$
* 
* 
* : $\S 9$: Special Plane Curves: Involute of a Circle: $9.28.$
Category:Involutes
Category:Spirals
Category:Circles
Parametric Equations for Evolute/Formulation 1

== Theorem ==
Let $C$ be a curve expressed as the locus of an equation $\map f {x, y} = 0$.
The parametric equations for the evolute of $C$ can be expressed as:
:$\begin{cases}
X = x - \dfrac {y' \paren {1 + y'^2} } {y' '} \\
Y = y + \dfrac {1 + y'^2} {y}
\end{cases}$
where:
:$\tuple {x, y}$ denotes the Cartesian coordinates of a general point on $C$
:$\tuple {X, Y}$ denotes the Cartesian coordinates of a general point on the evolute of $C$
:$y'$ and $y' '$ denote the derivative and second derivative respectively of $y$ $x$.
== Proof ==
:
Let $P = \tuple {x, y}$ be a general point on $C$.
Let $Q = \tuple {X, Y}$ be the center of curvature of $C$ at $P$.
From the above diagram:
:$x - X = \pm \rho \sin \psi$
:$Y - y = \pm \rho \cos \psi$
where:
:$\rho$ is the radius of curvature of $C$ at $P$
:$\psi$ is the angle between the tangent to $C$ at $P$ and the $x$-axis.
Whether the sign is plus or minus depends on whether the curve is convex or concave.
By definition of radius of curvature:
:$(1): \quad \begin {cases} x - X = \dfrac 1 k \sin \psi \\ Y - y = \dfrac 1 k \cos \psi \end {cases}$
where $k$ is the curvature of $C$ at $P$, given by:
:$k = \dfrac {y' '} {\paren {1 + y'^2}^{3/2} }$
We have that:
:$\sin \psi = \dfrac {\d y} {\d s} = \dfrac {y'} {\sqrt {1 + y'^2} }$
:$\cos \psi = \dfrac {\d x} {\d s} = \dfrac 1 {\sqrt {1 + y'^2} }$
Substituting for $k$ and $\psi$ in $(1)$ gives:
and:
== Sources ==
* : Chapter $\text {B}.23$: Evolutes and Involutes. The Evolute of a Cycloid
Category:Parametric Equations for Evolute
Parametric Equations for Evolute/Formulation 2

== Theorem ==
Let $C$ be a curve expressed as the locus of an equation $\map f {x, y} = 0$.
The parametric equations for the evolute of $C$ can be expressed as:
:$\begin {cases}
X = x - \dfrac {y' \paren {x'^2 + y'^2} } {x' y - y' x} \\
Y = y + \dfrac {x' \paren {x'^2 + y'^2} } {x' y - y' x}
\end {cases}$
where:
:$\tuple {x, y}$ denotes the Cartesian coordinates of a general point on $C$
:$\tuple {X, Y}$ denotes the Cartesian coordinates of a general point on the evolute of $C$
:$x'$ and $x' '$ denote the derivative and second derivative respectively of $x$ $t$
:$y'$ and $y' '$ denote the derivative and second derivative respectively of $y$ $t$.
== Proof ==
:
Let $P = \tuple {x, y}$ be a general point on $C$.
Let $Q = \tuple {X, Y}$ be the center of curvature of $C$ at $P$.
From the above diagram:
where:
:$\rho$ is the radius of curvature of $C$ at $P$
:$\psi$ is the angle between the tangent to $C$ at $P$ and the $x$-axis.
Whether the sign is plus or minus depends on whether the curve is convex or concave.
By definition of radius of curvature:
:$(1): \quad \begin {cases} x - X = \dfrac 1 k \sin \psi \\ Y - y = \dfrac 1 k \cos \psi \end {cases}$
where $k$ is the curvature of $C$ at $P$, given by:
:$k = \dfrac {x' y' ' - y' x' '} {\paren {x'^2 + y'^2}^{3/2} }$
We have that:
Substituting for $k$ and $\psi$ in $(1)$ gives:
and:
== Sources ==
* : Chapter $\text {B}.23$: Evolutes and Involutes. The Evolute of a Cycloid
Category:Parametric Equations for Evolute
Parenthesis/Examples/Arbitrary Example 1

== Examples of Parenthesis ==
Consider the following this formula of propositional logic:
:$p \land q \lor r$
This can mean either:
:the conjunction of $p$ with $q \lor r$
or:
:the disjunction of $p \land q$ with $r$.
Using parenthesis, the ambiguity is removed by presenting what is required either as:
:$p \land \paren {q \lor r}$
or:
:$\paren {p \land q} \lor r$
== Sources ==
* : $2$ Arguments Containing Compound Statements: $2.1$: Simple and Compound Statements
Category:Examples of Parenthesis
Parenthesis/Examples/Arbitrary Example 2

== Examples of Parenthesis ==
Consider the following this formula of propositional logic:
:$p \lor q \implies \neg \, r \implies p \land q$
This can be interpreted in several different ways:
:If either $p$ or $q$ is true, then it is not the case that the truth of $r$ implies the truth of both $p$ and $q$.
:Either $p$ is true, or if $q$ is true, then it is not the case that the truth of $r$ implies the truth of both $p$ and $q$.
:and so on.
So we need a way, for such a formula, to determine which of these interpretations is the one intended.
In the example above, the two different interpretations will be written in the style we have chosen as:
:$\paren {p \lor q} \implies \paren {\neg \paren {r \implies \paren {p \land q} } }$
:$p \lor \paren {q \implies \paren {\neg \paren {r \implies \paren {p \land q} } } }$
In these expressions, $\paren {p \lor q}$ and $\paren {\neg \paren {r \implies \paren {p \land q} } }$ are examples of formulas in parenthesis.
Note that while the latter expressions may in fact be WFFs of propositional logic, the ambiguous expression they were derived from is not.
Category:Examples of Parenthesis
Parenthesis/Natural Language/Examples/Use of Both

== Examples of Use of Parenthesis in Natural Language ==
Consider the sentence:
:Alice and Betty will not be elected.
This is of the form $\lnot p \land q$, where:
:$p$ is the simple statement Alice will be elected
:$q$ is the simple statement Betty will be elected.
The word both together with the word not can be used to clarify.
Hence:
:Alice and Betty will both not be elected
corresponds with the form $\paren {\lnot p} \land \paren {\lnot q}$ 
while:
:Alice and Betty will not both be elected
corresponds with the form $\lnot \paren {p \land q}$.
== Sources ==
* : $2$ Arguments Containing Compound Statements: $2.1$: Simple and Compound Statements
Category:Examples of Parenthesis
Parenthesis/Natural Language/Examples/Use of Either

== Examples of Use of Parenthesis in Natural Language ==
Consider the sentence:
:More stringent anti-pollution measures will be enacted and the laws will be strictly enforced or the quality of life will be degraded still further.
This is of the form $p \land q \lor r$, where:
:$p$ is the simple statement More stringent anti-pollution measures will be enacted
:$q$ is the simple statement The laws will be strictly enforced
:$r$ is the simple statement The quality of life will be degraded still further.
As it stands, it is not clear how the grouping is to be done.
The word either is often used together with the instance of or to tell the reader what the scope of the disjunction actually is.
Hence:
:Either more stringent anti-pollution measures will be enacted and the laws will be strictly enforced, or the quality of life will be degraded still further
corresponds with the form $\paren {p \land q} \lor r$ 
while:
:More stringent anti-pollution measures will be enacted, and either the laws will be strictly enforced or the quality of life will be degraded still further
corresponds with the form $p \land \paren {q \lor r}$.
== Sources ==
* : $2$ Arguments Containing Compound Statements: $2.1$: Simple and Compound Statements
Category:Examples of Parenthesis
Parenthesization/Examples/2

== Example of Parenthesization ==
A word of $2$ elements can be parenthesized in only $1$ distinct way:
:$\quad \paren {a_1 a_2}$
== Proof ==
From Number of Distinct Parenthesizations on Word, the number of distinct parenthesizations of a word $w$ of $n$ elements is the Catalan number $C_{n - 1}$:
:$C_{n - 1} = \dfrac 1 n \dbinom {2 \paren {n - 1} } {n - 1}$
For $n = 2$ we have:
== Sources ==
* : $42$
* : $42$
Category:Parenthesization
Category:Catalan Numbers
Parenthesization/Examples/3

== Example of Parenthesization ==
A word of $3$ elements can be parenthesized in $2$ distinct ways:
:$\quad a_1 \left({a_2 a_3}\right)$
:$\quad \left({a_1 a_2}\right) a_3$
== Proof ==
From Number of Distinct Parenthesizations on Word, the number of distinct parenthesizations of a word $w$ of $n$ elements is the Catalan number $C_{n - 1}$:
:$C_{n - 1} = \dfrac 1 n \dbinom {2 \paren {n - 1} } {n - 1}$
For $n = 3$ we have:
== Sources ==
* : $42$
* : $42$
Category:Parenthesization
Category:Catalan Numbers
Category:2
Parenthesization/Examples/4

== Example of Parenthesization ==
A word of $4$ elements can be parenthesized in $5$ distinct ways:
:$\quad a_1 \paren {a_2 \paren {a_3 a_4} }$
:$\quad a_1 \paren {\paren {a_2 a_3} a_4}$
:$\quad \paren {a_1 a_2} \paren {a_3 a_4}$
:$\quad \paren {a_1 \paren {a_2 a_3} } a_4$
:$\quad \paren {\paren {a_1 a_2} a_3} a_4$
== Proof ==
From Number of Distinct Parenthesizations on Word, the number of distinct parenthesizations of a word $w$ of $n$ elements is the Catalan number $C_{n - 1}$:
:$C_{n - 1} = \dfrac 1 n \dbinom {2 \paren {n - 1} } {n - 1}$
For $n = 4$ we have:
== Sources ==
* : $\text {3-4}$ Generating Functions
* : $42$
* : $42$
Category:Parenthesization
Category:Catalan Numbers
Category:5
Parity Addition is Associative

== Theorem ==
Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the parity ring.
The operation $+$ is associative:
:$\forall a, b, c \in R: \paren {a + b} + c = a + \paren {b + c}$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Example $2.2$
Category:Parity Ring
Category:Parity Addition is Associative
Parity Addition is Associative/Proof 1

== Theorem ==
== Proof ==
From Isomorphism between Ring of Integers Modulo 2 and Parity Ring:
:$\struct {\set {\text {even}, \text {odd} }, +, \times}$ is isomorphic with $\struct {\Z_2, +_2, \times_2}$
the ring of integers modulo $2$.
The result follows from:
:Modulo Addition is Associative
and:
:Isomorphism Preserves Associativity.
Category:Parity Addition is Associative