nyuuzyou/wikis huggingface dataset:
Transplant (Abstract Algebra)/Examples/Multiplication on Reals under 1-x

== Example of Transplant ==
Let $\struct {\R, \times}$ be the set of real numbers under multiplication.
Let $f: \R \to \R$ be the permutation defined as:
:$\forall x \in \R: \map f x = 1 - x$
The transplant $\otimes$ of $\times$ under $f$ is given by:
:$x \otimes y = x + y - x y$
== Proof ==
The fact that $f$ is a bijection is taken as read.
The inverse of $f$ is given as:
:$\forall x \in \R: \map {f^{-1} } x = 1 - x$
Hence from the Transplanting Theorem:
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.8 \ \text {(a)}$
Category:Transplants
Transplant (Abstract Algebra)/Examples/Multiplication on Reals under Tenth Power

== Example of Transplant ==
Let $\struct {\R, \times}$ be the set of real numbers under multiplication.
Let $\R_{>0}$ be the set of strictly positive real numbers.
Let $f: \R \to \R_{>0}$ be the mapping from $\R$ to $\R_{>0}$ defined as:
:$\forall x \in \R: \map f x = 10^x$
The transplant $\otimes$ of $\times$ on $\R$ under $f$ is given by:
:$\forall x, y \in \R_{>0}: x \otimes y = x^{\log_{10} y}$
== Proof ==
From Group Isomorphism Examples: Real Power Function, $f$ is an isomorphism.
Hence $f$ is a bijection.
The inverse of $f$ is given as:
:$\forall y \in \R_{>0}: \map {f^{-1} } y = \log_{10} y$
Hence from the Transplanting Theorem:
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures
Category:Transplants
Transplanting Theorem

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $f: S \to T$ be a bijection.
Then there exists one and only one operation $\oplus$ such that $f: \struct {S, \circ} \to \struct {T, \oplus}$ is an isomorphism.
The operation $\oplus$ is defined by:
:$\forall x, y \in T: x \oplus y = \map f {\map {f^{-1} } x \circ \map {f^{-1} } y}$
The operation $\oplus$ is called the transplant of $\circ$ under $f$.
=== Corollary ===
== Proof ==
=== Existence ===
To show that $\oplus$ as defined above exists:
Let $u, v \in S$, and let $x = \map f u, y = \map f v$.
Then as $f$ is a bijection, $u = \map {f^{-1} } x, v = \map {f^{-1} } y$.
Thus:
It is seen that $f$ is an isomorphism as required.
=== Uniqueness ===
Let $f$ be the isomorphism whose existence has been proven above.
Thus:
So, if $\oplus$ is an operation on $T$ such that $f$ is an isomorphism from $\struct {S, \circ} \to \struct {T, \oplus}$, then $\oplus$ must be defined as by this theorem, and there can be no other such operations.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Theorem $6.3$
Category:Transplanting Theorem
Category:Isomorphisms (Abstract Algebra)
Category:Transplants
Category:Named Theorems
Transplanting Theorem/Corollary

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $f: S \to S$ be an automorphism on $\struct {S, \circ}$.
Then the transplant of $\circ$ under $f$ is $\circ$ itself.
== Proof ==
From the Transplanting Theorem there exists one and only one operation $\circ$ such that $f: \struct {S, \circ} \to \struct {S, \circ}$ is an automorphism.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures
Category:Transplanting Theorem
Transpose of Column Matrix is Row Matrix

== Theorem ==
Let $\mathbf x = \sqbrk x_{1 n} = \begin {bmatrix} x_1 & x_2 & \cdots & x_n \end {bmatrix}$ be a column matrix.
Then $\mathbf x^\intercal$, the transpose of $\mathbf x$, is a row matrix:
:$\begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}^\intercal = \begin {bmatrix} x_1 & x_2 & \cdots & x_n \end {bmatrix}$
== Proof ==
Self-evident.
== Also see ==
* Transpose of Row Matrix is Column Matrix
== Sources ==
* 
* 
Category:Transposes of Matrices
Category:column Matrices
Category:Row Matrices
Transpose of Linear Transformation is a Linear Transformation

== Theorem ==
Let $R$ be a commutative ring.
Let $G$ and $H$ be $R$-modules.
Let $G^*$ and $H^*$ be the algebraic duals of $G$ and $H$ respectively.
Let $\map {\LL_R} {G, H}$ be the set of all linear transformations from $G$ to $H$.
Let $u \in \map {\LL_R} {G, H}$.
Let $u^\intercal: H^* \to G^*$ be the transpose of $u$.
Then $u^\intercal: H^* \to G^*$ is itself a linear transformation.
== Proof ==
By definition of evaluation linear transformation:
:$\forall x \in G: y \in H^*: \innerprod x {\map {u^\intercal} y} = \innerprod {\map u x} y$
Since we have:
and:
it follows that $u^\intercal: H^* \to G^*$ is a linear transformation.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations
Category:Linear Transformations
Transpose of Matrix Product

== Theorem ==
Let $\mathbf A$ and $\mathbf B$ be matrices over a commutative ring such that $\mathbf A \mathbf B$ is defined.
Then $\mathbf B^\intercal \mathbf A^\intercal$ is defined, and:
:$\paren {\mathbf A \mathbf B}^\intercal = \mathbf B^\intercal \mathbf A^\intercal$
where $\mathbf X^\intercal$ is the transpose of $\mathbf X$.
== Proof ==
Let $\mathbf A = \sqbrk a_{m n}$, $\mathbf B = \sqbrk b_{n p}$
Let $\mathbf A \mathbf B = \sqbrk c_{m p}$.
Then from the definition of matrix product:
:$\ds \forall i \in \closedint 1 m, j \in \closedint 1 p: c_{i j} = \sum_{k \mathop = 1}^n a_{i k} \circ b_{k j}$
So, let $\paren {\mathbf A \mathbf B}^\intercal = \sqbrk r_{p m}$.
The dimensions are correct, because $\mathbf A \mathbf B$ is an $m \times p$ matrix, thus making $\paren {\mathbf A \mathbf B}^\intercal$ a $p \times m$ matrix.
Thus:
:$\ds \forall j \in \closedint 1 p, i \in \closedint 1 m: r_{j i} = \sum_{k \mathop = 1}^n a_{i k} \circ b_{k j}$
Now, let $\mathbf B^\intercal \mathbf A^\intercal = \sqbrk s_{p m}$
Again, the dimensions are correct because $\mathbf B^\intercal$ is a $p \times n$ matrix and $\mathbf A^\intercal$ is an $n \times m$ matrix.
Thus:
:$\ds \forall j \in \closedint 1 p, i \in \closedint 1 m: s_{j i} = \sum_{k \mathop = 1}^n b_{k j} \circ a_{i k}$
As the underlying structure of $\mathbf A$ and $\mathbf B$ is a commutative ring, then $a_{i k} \circ b_{k j} = b_{k j} \circ a_{i k}$.
Note the order of the indices in the term in the summation sign on the of the above.
They are reverse what they would normally be because we are multiplying the transposes together.
Thus it can be seen that $r_{j i} = s_{j i}$ and the result follows.
== Also see ==
* Inverse of Matrix Product
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.4$ The transpose of a matrix: Proposition $1.3 \ \text {(a)}$
* : $\text {(iv)}$
* 
Category:Transposes of Matrices
Category:Conventional Matrix Multiplication
Transpose of Outer Product

== Theorem ==
Let $\mathbf u = \tuple {u_1, u_2, \ldots, u_m}$ and $\mathbf v = \tuple {v_1, v_2, \ldots, v_n}$ be vectors.
Let $\mathbf u \otimes \mathbf v$ denote the outer product of $\mathbf u$ and $\mathbf v$.
Then:
:$\mathbf u \otimes \mathbf v = \paren {\mathbf v \otimes \mathbf u}^\intercal$
where ${}^\intercal$ denotes the transpose operation.
== Proof ==
We have:
Hence the result.
Category:Outer Products
Category:Transposes of Matrices
Transpose of Row Matrix is Column Matrix

== Theorem ==
Let $\mathbf x = \sqbrk x_{1 n} = \begin {bmatrix} x_1 & x_2 & \cdots & x_n \end {bmatrix}$ be a row matrix.
Then $\mathbf x^\intercal$, the transpose of $\mathbf x$, is a column matrix:
:$\begin {bmatrix} x_1 & x_2 & \cdots & x_n \end{bmatrix}^\intercal = \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix}$
== Proof ==
Self-evident.
== Also see ==
* Transpose of Column Matrix is Row Matrix
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 30$. Linear Equations
* : $2.1$: Matrices
* 
* 
Category:Transposes of Matrices
Category:Row Matrices
Category:column Matrices
Transpose of Transpose of Matrix

== Theorem ==
Let $\mathbf A$ be a matrix.
Let $\mathbf A^\intercal$ be the transpose of $\mathbf A$.
Then:
:$\paren {\mathbf A^\intercal}^\intercal = \mathbf A$
== Proof ==
Follows directly from the definition of the transpose of a matrix.
== Sources ==
* : $\text {(i)}$
Category:Transposes of Matrices
Transpose of Upper Triangular Matrix is Lower Triangular

== Theorem ==
The transpose of an upper triangular matrix is a lower triangular matrix.
== Proof ==
Let $\mathbf U = \sqbrk a_{m n}$ be an upper triangular matrix.
By definition:
:$\forall a_{i j} \in \mathbf U: i > j \implies a_{i j} = 0$
Let $\mathbf U^\intercal = \sqbrk b_{n m}$ be the transpose of $\mathbf U$.
That is:
:$\mathbf U^\intercal = \sqbrk b_{n m}: \forall i \in \closedint 1 n, j \in \closedint 1 n: b_{i j} = a_{j i}$
Thus:
:$\forall b_{j i} \in \mathbf U^\intercal: i > j \implies b_{j i} = 0$
By exchanging $i$ and $j$ in the notation of the above:
:$\forall b_{i j} \in \mathbf U^\intercal: i < j \implies b_{i j} = 0$
Thus by definition it is seen that $\mathbf U^\intercal$ is a lower triangular matrix.
Category:Triangular Matrices
Category:Transposes of Matrices
Transposition Cryptography/Examples/Arbitrary Example 1

== Example of Transposition Cryptography ==
Let $\pi$ be the permutation described in two-row notation as:
:$\begin {pmatrix} 1 & 2 & 3 & 4 & 5 \\ 3 & 5 & 2 & 1 & 4 \end {pmatrix}$
The plaintext would then be split into blocks of $5$ characters upon which $\pi$ would be applied.
Hence:
:the corresponding ciphertext to the plaintext rugby would be bgryu
:the corresponding ciphertext to the plaintext wales would be elwsa.
== Sources ==
* 
Category:Examples of Transposition Cryptography
Transposition is Self-Inverse

== Theorem ==
All transpositions are self-inverse.
== Proof ==
Let $\pi = \begin{bmatrix} k_1 & k_2 \end{bmatrix}$ be a transposition.
Writing $\pi \pi$ in cycle notation gives:
:$\begin{bmatrix} k_1 & k_2 \end{bmatrix} \begin{bmatrix} k_1 & k_2 \end{bmatrix}$
from which we see that $k_1 \to k_2 \to k_1$ and $k_2 \to k_1 \to k_2$.
The result follows from the definition of self-inverse.
Category:Symmetric Groups
Transposition is of Odd Parity

== Theorem ==
Let $S_n$ denote the set of permutations on $n$ letters.
Let $\pi \in S_n$ be a transposition.
Then $\pi$ is of odd parity.
== Proof ==
Let $\pi = \begin{pmatrix} 1 & 2 \end{pmatrix}$ be a transposition.
Let $\Delta_n$ be defined as Product of Differences.
Then $\forall n \in \N_{>0}: \pi \cdot \Delta_n$ produces only one sign change in $\Delta_n$, that is, the one occurring in the factor $\paren {x_1 - x_2}$.
Thus:
:$\begin{pmatrix} 1 & 2 \end{pmatrix} \cdot \Delta_n = - \Delta_n$
and thus $\begin{pmatrix} 1 & 2 \end{pmatrix}$ is odd.
From Conjugates of Transpositions:
:$\begin{pmatrix} 1 & k \end{pmatrix} = \begin{pmatrix} 2 & k \end{pmatrix} \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & k \end{pmatrix}$
Thus as $\begin{pmatrix} 2 & k \end{pmatrix}$ is self-inverse:
:$\begin{pmatrix} 1 & k \end{pmatrix} = \begin{pmatrix} 2 & k \end{pmatrix} \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & k \end{pmatrix}^{-1}$
But from Parity of Conjugate of Permutation:
:$\map \sgn {\begin{pmatrix} 2 & k \end{pmatrix} \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & k \end{pmatrix}^{-1} } = \sgn {\begin{pmatrix} 1 & 2 \end{pmatrix} }$
Thus:
:$\sgn {\begin{pmatrix} 1 & k \end{pmatrix} } = \sgn {\begin{pmatrix} 1 & 2 \end{pmatrix} }$
and thus $\begin{pmatrix} 1 & k \end{pmatrix}$ is odd.
Finally:
But from Parity of Conjugate of Permutation:
:$\map \sgn {\begin{pmatrix} 1 & h \end{pmatrix} \begin{pmatrix} 1 & k \end{pmatrix} \begin{pmatrix} 1 & h \end{pmatrix}^{-1} } = \sgn {\begin{pmatrix} 1 & k \end{pmatrix} }$
Thus:
:$\sgn {\begin{pmatrix} h & k \end{pmatrix} } = \sgn {\begin{pmatrix} 1 & k \end{pmatrix} }$
and thus $\begin{pmatrix} h & k \end{pmatrix}$ is odd.
== Sources ==
* : Chapter $9$: Permutations: Corollary $9.18$
Category:Symmetric Groups
Transpositions of Adjacent Elements generate Symmetric Group

== Theorem ==
Let $n \in \Z: n > 1$.
Let $S_n$ denote the symmetric group on $n$ letters.
Then the transpositions $a_k = \begin{pmatrix} k & k + 1 \end{pmatrix}$ for $1 \le k < n$ are a set of generators for $S_n$.
They satisfy the relations:
== Proof ==
First, we show that each $\begin{pmatrix} i & j \end{pmatrix}$ where $i < j$ is in the subgroup $\gen {a_1, a_2, \ldots, a_{n - 1} }$.
From Cycle Decomposition of Conjugate, we can conjugate $a_i$ by $a_{i + 1}$ to give:
:$\begin{pmatrix} i & i + 2 \end{pmatrix}$
Conjugating $a_i$ by the product $a_{j - 1} a_{j - 2} \ldots a_{i + 1}$ will give:
:$\begin{pmatrix} i & j \end{pmatrix}$
Next, we note that:
:from K-Cycle can be Factored into Transpositions, every cycle is a product of transpositions.
and:
:from Existence and Uniqueness of Cycle Decomposition, every permutation is a product of cycles.
Thus, every permutation can be obtained from some product of the given transpositions.
Thus $\gen {a_1, a_2, \ldots, a_{n - 1} }$ is a generator of $S_n$.
From Transposition is Self-Inverse, we have:
:$a_k^2 = e$
From K-Cycle can be Factored into Transpositions, $a_i a_{i + 1}$ is the $3$-cycle $\begin{pmatrix} i & i + 1 & i + 2 \end{pmatrix}$.
Thus:
:$\paren {a_k a_{k + 1} }^3 = e$
If $\size {i - j} > 1$, then from Disjoint Permutations Commute, $a_i$ and $a_j$ are disjoint.
Thus:
:$\paren {a_i a_j}^2 = a_i^2 a_j^2 = e$
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 80 \gamma$
* : Chapter $9$: Permutations: Proposition $9.21$
Category:Symmetric Groups
Category:Examples of Generators of Groups
Transversal (Group Theory)/Examples

== Examples of Transversals in the Context of Group Theory ==
=== Reflection Subgroup in Equilateral Triangle ===
Consider the symmetry group of the equilateral triangle $D_3$.
:
Let $H \subseteq D_3$ be defined as:
:$H = \set {e, r}$
where:
:$e$ denotes the identity mapping
:$r$ denotes reflection in the line $r$.
=== Integer Multiples in Integers ===
Category:Examples of Transversals
Transversal (Group Theory)/Examples/Integer Multiples in Integers

== Example of Transversal in the Context of Group Theory ==
Let $\struct {\Z, +}$ denote the additive group of integers.
Let $\struct {n \Z, +}$ denote the additive group of integer multiples.
Then a transversal for $\struct {n \Z, +}$ in $\struct {\Z, +}$ is:
:$\set {0, 1, \ldots, n - 1}$
== Sources ==
* : $\S 6.3$. Index. Transversals: Example $116$
Category:Examples of Transversals
Category:Additive Groups of Integer Multiples
Trapezium Rule for Definite Integrals

== Theorem ==
Let $f$ be a real function which is integrable on the closed interval $\closedint a b$.
Let $P = \set {x_0, x_1, x_2, \ldots, x_{n - 1}, x_n}$ form a normal subdivision of $\closedint a b$:
:$\forall i \in \set {1, 2, \ldots, n}: x_i - x_{i - 1} = \dfrac {b - a} n$
Then the definite integral of $f$ $x$ from $a$ to $b$ can be approximated as:
:$\ds \int_a^b \map f x \rd x \approx \dfrac h 2 \paren {\map f {x_0} + \map f {x_n} + \sum_{i \mathop = 1}^{n - 1} 2 \map f {x_i} }$
where $h = \dfrac {b - a} n$.
=== Error Term ===
== Proof ==
The geometric interpretation of a definite integral states that the area between the $4$ lines $x = a$, $x = b$, $y = 0$ and $y = \map f x$ is equal to $\ds \int_a^b \map f x \rd x$.
We approximate this area by dividing it into trapezia:
:
Consider the trapezium $T_i$ whose vertices are $\tuple {x_i, \map f {x_i}, x_{i + 1}, \map f {x_{i + 1} } }$ for some $i \in \set {0, 1, \ldots, n}$.
The area of $T_i$ is given by Area of Trapezium as:
:$\map \Area {T_i} = \dfrac {\map f {x_i} + \map f {x_{i + 1} } } 2 h$
Now let us consider the summation of all such areas:
Hence the result.
== Also known as ==
== Also see ==
* Newton's Three-Eighths Rule
* Simpson's Rule
== Sources ==
* : $\S 15$: Approximate Formulas for Definite Integrals: $15.16$
* 
* 
* 
* : $\S 18$: Definite Integrals: Approximate Formulas for Definite Integrals: $18.16$
* 
Category:Trapezium Rule for Definite Integrals
Category:Newton-Cotes Rules
Category:Definite Integrals
Category:Trapezia
Trapezium Rule for Definite Integrals/Also known as

== Trapezium Rule for Definite Integrals: Also known as == 
US sources refer to this rule as:
:the trapezoidal formula
:the trapezoidal rule
:the trapezoid rule
as a result of the fact that, in the US, the terms trapezoid and trapezium have the opposite definitions.
== Sources ==
* : $\S 15$: Approximate Formulas for Definite Integrals: $15.16$
* 
* 
* 
* : $\S 18$: Definite Integrals: Approximate Formulas for Definite Integrals: $18.16$
* 
Category:Trapezium Rule for Definite Integrals
Trapezium Rule for Definite Integrals/Error Term

== Theorem ==
Let $f$ be a real function which is integrable on the closed interval $\closedint a b$.
Let $P = \set {x_0, x_1, x_2, \ldots, x_{n - 1}, x_n}$ form a normal subdivision of $\closedint a b$:
:$\forall r \in \set {1, 2, \ldots, n}: x_r - x_{r - 1} = \dfrac {b - a} n$
Let $f$ be approximated by the trapezium rule:
:$\ds \int_a^b \map f x \rd x \approx \dfrac h 2 \paren {\map f {x_0} + \map f {x_n} + \sum_{r \mathop = 1}^{n - 1} 2 \map f {x_r} }$
where $h = \dfrac {b - a} n$.
The error can be quantified as:
:$\dfrac {\paren {b - a}^3 \map {f' '} \xi} {12 n^2}$
where $\xi \in \closedint a b$.
== Proof ==
first, through integration by parts: 
:$\int_{x_{i+1}}^{x_i} f(x)\ dx \\ =\int_{0}^{h} f(t+x_i)\ dt    \\ = \left [ (t-\frac{h}{2})f(t+x_i) \right ]^h_0 -\int_{0}^{h} (t-h/2) f'(t+x_i)\ dt \\ = \left [ (t-\frac{h}{2})f(t+x_i) \right ]^h_0 -\left [ \left(\frac{(t-h/2)^2}{2}-\frac{h^2}{8}\right)f'(t+x_i) \right ]^h_0 + \int_{0}^{h} \left(\frac{(t-h/2)^2}{2}-\frac{h^2}{8}\right) f''(t+x_i)\ dt\\ =\frac{h}{2}(f(x_{i+1})+f(x_i))-0+\int_{0}^{h} \left(\frac{(t-h/2)^2}{2}-\frac{h^2}{8}\right) f''(t+x_i)\ dt\\ $
Note the left component when added for all $i=0...n-1$ is the approximation according to the trapezoidal rule. Now, to bound total error we consider the series of the right component in the above equation:
:$E_T\\ =\left | \sum^{n-1}_{i=0}\int_{0}^{h}\left(\frac{(t-h/2)^2}{2}-\frac{h^2}{8}\right) f''(t+x_i)\ dt \right |\\ =\left | \int_{0}^{h}\left(\frac{(t-h/2)^2}{2}-\frac{h^2}{8}\right) \sum^{n-1}_{i=0} f''(t+x_i)\ dt \right |\\ \leq \int_{0}^{h}\left |\frac{(t-h/2)^2}{2}-\frac{h^2}{8}\right | \sum^{n-1}_{i=0}\left |f''(t+x_i) \right |\ dt\\ \leq nK\int_{0}^{h}\left |\frac{(t-h/2)^2}{2}-\frac{h^2}{8}\right |\ dt\\ =nK\frac{h^3}{12}=\frac{(b-a)^3}{12n^2} K$
== Sources ==
* 
Category:Trapezium Rule for Definite Integrals
Traveller whose Head goes Further than Feet

== Problem ==
:A traveller sets out on a journey,
:and eventually returns to the place where he started.
:During his journey, his head has travelled $12$ yards further than his feet,
:and yet his head remains attached to his body.
:How is this possible?
== Solution ==
The traveller went around the world, at the equator.
== Proof ==
Let the radius of Earth at the equator be $R$.
Thus, by Perimeter of Circle, the equator measures $2 \pi R$ all the way round.
Let the height of the traveller be $h$ yards.
Consider the equation:
:$12 = 2 \pi \paren {R + h} - 2 \pi R$
Thus we have:
:$h = \dfrac {12} {2 \pi} \approx 1.91$
So, if the traveller:
:were $1.91$ yards, that is, approximately $5$ feet $9$ inches tall
:and walked the entire way around the world, along the equator,
his head would have travelled $12$ yards further than his feet.
== Historical Note ==
== Sources ==
* : Geographical Paradoxes
* : Rational Amusements for Winter Evenings: $164$
Category:Geography Puzzles
Tree Diagram/Examples/Arbitrary Example 1

== Example of Tree Diagram ==
Consider an urn in which there are $5$ balls and $3$ balls.
Suppose $2$ balls are to be removed without replacement.
Suppose we want to know the probability that there are then $4$ balls and $2$ balls left in the urn.
All possible outcomes are represented on the following tree diagram:
:
The root node $A$ corresponds to the urn in its original state, with $5$ balls and $3$ balls in it.
The first draw leads either to node $B$ or node $C$.
The edges $AB$ and $AC$ are labeled with the color of the ball drawn and the probability of the corresponding outcome.
The node itself contains a label specifying the current contents of the urn.
The second draw is then made from either $B$ or $C$.
Hence the possible outcomes are given at nodes $D$, $E$, $F$ or $H$.
Again:
:the edges $BD$, $BE$, $CF$ and $CH$ indicate the color of the ball drawn and the probability of the corresponding outcome
:the nodes indicate the current contents of the urn.
Here the experiment ends, as we have drawn $2$ balls from the urn.
Each of the draws are independent events.
Hence the probability of reaching each of the nodes $D$, $E$, $F$ and $H$ is the product of the probabilities along the edges on the path from $A$.
The condition that there are $4$ balls and $2$ balls left in the urn is satisfied by the paths $AE$ and $AF$.
The total probability of reaching either $E$ or $F$ is then equal to the sum of the probabilities of arriving at each.
Thus the probability that there are then $4$ balls and $2$ balls left in the urn is:
:$\dfrac {15} {56} + \dfrac {15} {56} = \dfrac {15} {28}$
== Sources ==
* 
* 
Category:Examples of Tree Diagrams
Tree has Center or Bicenter

== Theorem ==
Every tree has either:
:$(1): \quad$ Exactly one center
or:
:$(2): \quad$ Exactly one bicenter,
but never both.
That is, every tree is either central or bicentral.
== Proof ==
A tree whose order is $1$ or $2$ is already trivially central or bicentral.
Let $T$ be a tree of order at least $3$.
First we establish that the construction of a center or bicenter actually works.
From Finite Tree has Leaf Nodes, there are always at least two nodes of degree $1$ to be removed.
By the Handshake Lemma, it is clear that $T$ must also have at least one node whose degree is greater than $1$: 
:$\ds \sum_{i \mathop = 1}^n \map {\deg_G} {v_i} = 2 q$
where $q$ is the number of edges in $T$.
But from Finite Connected Simple Graph is Tree iff Size is One Less than Order:
:$q = n - 1$
So if each node has degree $1$, then $n = 2 \paren {n - 1}$ and so $n = 2$.
Therefore if $n > 2$ there must be at least one node in $T$ of degree is greater than $1$.
Next, from Connected Subgraph of Tree is Tree, after having removed those nodes, what is left is still a tree.
Therefore the construction is valid. 
We need to show the following:
:$(1): \quad T$ has only one center or bicenter
:$(2): \quad T$ has either a center or a bicenter.
Suppose $T$ has more than one center or bicenter.
It would follow that at least one of the iterations constructing the center or bicenter disconnects $T$ into more than one component.
That could only happen if we were to remove an edge between two nodes of degree greater than $1$.
Hence $T$ has at most one center or bicenter.
Now to show that $T$ has at least one center or bicenter.
The proof works by the Principle of Complete Induction.
We know that a tree whose order is $1$ or $2$ is already trivially central or bicentral. This is our base case.
Suppose that all tree whose order is $n$ have at most one center or bicenter. This is our induction hypothesis.
Take a tree $T$ whose order is $n+1$ where $n > 2$.
Let $T$ have $k$ nodes of degree $1$.
We remove all these $k$ nodes.
This leaves us with a tree with $n + 1 - k$ nodes.
As we have seen that $T$ has at least one node whose degree is greater than $1$, $n + 1 - k \ge 1$.
As there are always at least two nodes of degree $1$, $n + 1 - k \le n - 1$.
So after the first iteration, we are left with a tree whose order is between $1$ and $n - 1$ inclusive.
By the induction hypothesis, this tree has either a center or bicenter.
The result follows by the Principle of Complete Induction.
Category:Tree Theory
Trefoil Knot in Paper forms Pentagon

== Theorem ==
Take a strip of paper with parallel edges.
Tie a simple trefoil knot in it.
Flatten the knot and pull the paper tight.
The knot will be in the shape of a regular pentagon.
The edges of the paper strip trace out a pentagram.
:
== Proof ==
== Historical Note ==
== Sources ==
* : Solutions: $120$. -- The Ribbon Pentagon
* : $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
* : Henry van Etten: $126$
* : $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
Category:Pentagons
Category:Pentagrams