nyuuzyou/wikis huggingface dataset:
Diagonal Relation is Smallest Equivalence Relation

== Theorem ==
The diagonal relation $\Delta_S$ on $S$ is the smallest equivalence in $S$, in the sense that:
:$\forall \EE \subseteq S \times S: \Delta_S \subseteq \EE$
where $\EE$ denotes a general equivalence relation.
== Proof ==
It is confirmed that, from Diagonal Relation is Equivalence, $\Delta_S$ is an equivalence relation.
Let $\EE$ be an arbitrary equivalence relation.
By definition, $\EE$ is reflexive.
From Relation Contains Diagonal Relation iff Reflexive it follows that as $\Delta_S \subseteq \EE$.
== Sources ==
* : $\S 7$: Relations
Category:Equivalence Relations
Diagonal Relation is Universally Congruent

== Theorem ==
The diagonal relation $\Delta_S$ on a set $S$ is universally congruent on $S$.
== Proof ==
We have that the diagonal relation is an equivalence relation.
Let $\struct {S, \circ}$ be any algebraic structure.
$\Delta_S$ can therefore be described as universally congruent.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Example $11.3$
Category:Examples of Congruence Relations
Diagonal Relation on Ring is Ordering Compatible with Ring Structure

== Theorem ==
Let $\struct {R, +, \circ, \preceq}$ be a ring whose zero is $0_R$.
Then the diagonal relation $\Delta_R$ on $R$ is an ordering compatible with the ring structure of $R$.
== Proof ==
From Diagonal Relation is Ordering and Equivalence, we have that $\Delta_R$ is actually an ordering on $R$.
From the definition of the diagonal relation:
:$\tuple {x, y} \in \Delta_R \iff x = y$
Thus:
Similarly:
So $\Delta_R$ is compatible with $+$.
Then note that:
Hence the result, from the definition of an ordering compatible with the ring structure of $R$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers
Category:Ring Theory
Category:Ordered Rings
Diagonal of Square is Axis of Rotational Symmetry of Order 2

== Theorem ==
Let $\SS$ be a square.
Let $\AA$ be a diagonal of $\SS$.
Then $\AA$ is an axis of rotational symmetry of $\SS$ of order $2$.
== Proof ==
== Sources ==
* 
* 
Category:Axes of Rotational Symmetry
Category:Order of Rotational Symmetry
Category:Squares
Diagonalizable Operator Bounded iff Value Set Bounded

== Theorem ==
Let $H$ be a Hilbert space.
Let $A: H \to H$ be a diagonalizable operator.
Let $\family {\alpha_e}_{e \mathop \in E}$ be the value set of $A$, a suitable basis $E$ for $H$.
Then $A$ is bounded $\family {\alpha_e}_{e \mathop \in E}$ is bounded.
That is, :
:$\exists M \in \R: \forall e \in E: \cmod {\alpha_e} \le M$
== Proof ==
== Sources ==
* $\text {II}.4.6$
Category:Linear Transformations on Hilbert Spaces
Diagonalizable Operator Compact iff Value Set Converges to Zero

== Theorem ==
Let $H$ be a Hilbert space of countable dimension.
Let $A: H \to H$ be a diagonalizable operator.
Let $\sequence {\alpha_n}_{n \mathop \in \N}$ be the value set of $A$, with respect to a suitable basis $E = \sequence {e_n}_{n \mathop \in \N}$ for $H$.
Then $A$ is compact :
:$\ds \lim_{n \mathop \to \infty} \alpha_n = 0$
== Proof ==
== Sources ==
* $II.4.6$
Category:Linear Transformations on Hilbert Spaces
Diagonally Dominant Matrix/Examples/Arbitrary Example 1

== Example of Diagonally Dominant Matrix ==
Consider the square matrix $\mathbf A$:
:$\mathbf A = \begin {pmatrix} 2 & 1 & 0 \\ 1 & -3 & 1 \\ -2 & 1 & 4 \end {pmatrix}$
$\mathbf A$ is diagonally dominant by rows, but not diagonally dominant by columns.
== Sources ==
* 
Category:Examples of Diagonally Dominant Matrices
Diagonals of Kite are Perpendicular

== Theorem ==
Let $ABCD$ be a kite such that $AC$ and $BD$ are its diagonals.
Then $AC$ and $BD$ are perpendicular.
== Proof ==
:
Let $AC$ and $BD$ meet at $E$.
Consider the triangles $\triangle ABD$ and $\triangle CBD$.
We have that:
:$AB = CB$
:$AD = CD$
:$BD$ is common.
Hence by Triangle Side-Side-Side Congruence, $\triangle ABD$ and $\triangle CBD$ are congruent.
Consider the triangles $\triangle ABE$ and $\triangle CBE$.
We have from the congruence of $\triangle ABD$ and $\triangle CBD$ that:
:$\angle ABE = \angle CBE$
:$AB = CB$
and $BE$ is common.
Hence by Triangle Side-Angle-Side Congruence, $\triangle ABE$ and $\triangle CBE$ are congruent.
We have that $AC$ is a straight line.
We have from the congruence of $\triangle ABE$ and $\triangle CBE$ that:
:$\angle BEC = \angle BEA$
From Two Angles on Straight Line make Two Right Angles, $\angle BEC + \angle BEA$ make two right angles.
Thus:
:$2 \angle BEC = 2 \angle BEA = 2$ right angles
and so:
:$\angle BEC = \angle BEA$ are both right angles.
That is, $AC$ and $BD$ are perpendicular.
== Sources ==
* 
Category:Kites
Diagonals of Rectangle are Equal

== Theorem ==
The diagonals of a rectangle are equal.
== Proof ==
:
Let $ABCD$ be a rectangle.
The diagonals of $ABCD$ are $AC$ and $BD$.
Then $\angle ADC = \angle DAB$ as both are right angles by definition of rectangle.
By Rectangle is Parallelogram, $ABCD$ is also a type of parallelogram.
Thus by Opposite Sides and Angles of Parallelogram are Equal $AB = DC$.
Thus we have:
:$AB = DC$
:$\angle ADC = \angle DAB$
:$AD$ is common to both $\triangle ADC$ and $\triangle DAB$
and so by Triangle Side-Angle-Side Congruence:
:$\triangle ADC = \triangle DAB$
Thus:
:$AC = BD$
Category:Rectangles
Diagonals of Rhombus Bisect Angles

== Theorem ==
Let $OABC$ be a rhombus.
Then:
:$(1): \quad OB$ bisects $\angle AOC$ and $\angle ABC$
:$(2): \quad AC$ bisects $\angle OAB$ and $\angle OCB$
== Proof 1 ==
== Proof 2 ==
Category:Euclidean Geometry
Category:Vector Algebra
Category:Parallelograms
Category:Diagonals of Rhombus Bisect Angles
Diagonals of Rhombus Bisect Angles/Proof 1

== Theorem ==
== Proof ==
, we will only prove $OB$ bisects $\angle AOC$.
We have:
Comparing corresponding angles gives:
:$\angle AOB = \angle COB$
hence $OB$ bisects $\angle AOC$.
Category:Euclidean Geometry
Category:Parallelograms
Category:Diagonals of Rhombus Bisect Angles
Diagonals of Rhombus Bisect Angles/Proof 2

== Theorem ==
== Proof ==
, we will only prove $OB$ bisects $\angle AOC$.
Let the position vectors of $A$, $B$ and $C$ with respect to $O$ be $\mathbf a$, $\mathbf b$ and $\mathbf c$ respectively.
By definition of rhombus, we have:
From the above we have:
By definition of dot product, the angle between the vectors is between $0$ and $\pi$.
From Shape of Cosine Function, cosine is injective on this interval.
Hence:
:$\angle \mathbf a, \mathbf b = \angle \mathbf c, \mathbf b$
The result follows.
Category:Euclidean Geometry
Category:Vector Algebra
Category:Parallelograms
Category:Diagonals of Rhombus Bisect Angles
Diagonals of Rhombus Bisect Each Other at Right Angles

== Theorem ==
Let $ABCD$ be a rhombus.
The diagonals $AC$ and $BD$ of $ABCD$ bisect each other at right angles.
== Proof ==
By the definition of a rhombus, $AB = AD = BC = DC$.
, consider the diagonal $BD$.
Thus:
: $\triangle ABD$ is an isosceles triangle whose apex is $A$ and whose base is $BD$.
By Diagonals of Rhombus Bisect Angles, $AC$ bisects $\angle BAD$.
From Bisector of Apex of Isosceles Triangle also Bisects Base, $AC$ bisects $BD$.
From Bisector of Apex of Isosceles Triangle is Perpendicular to Base, $AC$ bisects $BD$ at right angles.
Hence the result.
== Sources ==
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.22$: Corollary $2$
Category:Parallelograms
Diagonals of Rhombus Intersect at Right Angles

== Theorem ==
Let $ABCD$ be a rhombus.
The diagonals $AC$ and $BD$ of $ABCD$ intersect each other at right angles.
== Proof ==
:
, let $ABCD$ be embedded in the complex plane so that vertex $A$ coincides with the origin $0 + 0 i$.
Let $AB$ and $AD$ be represented by the complex numbers $a$ and $b$ respectively, expressed as vectors $\mathbf a$ and $\mathbf b$ respectively.
By Geometrical Interpretation of Complex Addition, the diagonal $AC$ is represented by the complex number $a + b$, expressed as a vector $\mathbf a + \mathbf b$.
By Geometrical Interpretation of Complex Subtraction, the diagonal $BD$ is represented by the complex number $a - b$, expressed as a vector $\mathbf a - \mathbf b$.
Let $a - b = w \paren {a + b}$ for some complex number $w \in \C$.
For $A \perp BD$, it is necessary and sufficient from Complex Multiplication as Geometrical Transformation for the argument of $w$ to be $\dfrac \pi 2$.
That is, for $w$ to be wholly imaginary.
From Complex Number equals Negative of Conjugate iff Wholly Imaginary, that is for $w + \overline w = 0$.
It remains to demonstrate this.
So:
The result follows.
== Sources ==
* : $\S 2$. Geometrical Representations: Example $1$.
Category:Parallelograms
Diameter of Closure of Subset is Diameter of Subset

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $S \subseteq A$ be bounded in $M$.
Then:
:$\map \diam S = \map \diam {S^-}$
where $\map \diam S$ denotes the diameter of $S$, and $S^-$ denotes the closure of $S$ in $M$.
== Proof ==
 that $\map \diam S \ne \map \diam {S^-}$.
$S \subseteq S^-$ by Subset of Metric Space is Subset of its Closure, so it then follows that:
:$\map \diam S < \map \diam {S^-}$
Then there exists $x, y \in S^-$ such that $\map d {x, y} > \map \diam S$.
By Point in Closure of Subset of Metric Space iff Limit of Sequence there exists sequences $\sequence {x_n}$ and $\sequence {y_n}$ that converge to $x$ and $y$ respectively.
Let $d_\infty: \paren {A \times A} \times \paren {A \times A} \to \R$ be the metric on $A \times A$ defined as:
:$\map {d_\infty} {\tuple {x, y}, \tuple {x', y'} } = \max \set {\map d {x, x'}, \map d {y, y'} }$
From this it can be said that $\sequence {\tuple {x_n, y_n} }$ is a sequence of points in $A \times A$ that converge to $\tuple {x, y}$.
Let $\epsilon = \map d {x, y} - \map \diam S$.
From Distance Function of Metric Space is Continuous, there exists a $N \in \N_{> 0}$ such that:
:$n \ge N \implies \size {\map d {x_n, y_n} - \map d {x, y} } < \epsilon$
If $\map d {x_n, y_n} - \map d {x, y}$ is non-negative, then:
:$\map d {x_n, y_n} \ge \map d {x, y} > \map \diam S$
which is a contradiction.
If $\map d {x_n, y_n} - \map d {x, y}$ is a negative number, then:
which is again a contradiction.
From Proof by Contradiction it follows that $\map \diam S = \map \diam {S^-}$.
Hence the result.
== Sources ==
* $\S 6.5$
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 32$
Category:Metric Spaces
Category:Set Closures
Diameter of N-Cube

== Theorem ==
Let $Q_n = \closedint {c - R} {c + R}^n$ be an $n$-cube in Euclidean $n$-Space equipped with the usual metric.
Then the diameter of $Q_n$ is given by:
:$\map \diam {Q_n} = 2 R \sqrt n$
=== Corollary ===
== Proof ==
Write: 
:$Q_n = \ds \prod_{i \mathop = 1}^n \closedint {c - R} {c + R}_i$
Let $x, y \in Q_n$
By the definition of the usual metric, the distance between any two points $x$ and $y$ is given by:
:$\ds \map d {y - x} = \paren {\sum_{i \mathop = 1}^n \paren {y_i - x_i}^2}^{1 / 2}$
By Positive Power Function on Non-negative Reals is Strictly Increasing, this quantity is maximal when each summand is maximal.
Consider $x_i, y_i$ in the $i$th interval:
:$\closedint {c - R} {c + R}_i$
To maximize $\size {y_i - x_i}$, take $x_i = \min \closedint {c - R} {c + R}_i$ and $y_i = \max \closedint {c - R} {c + R}_i$.
Then:
:$\size {y_i - x_i} = \size {c + R - \paren {c - R} } = 2$
By the definition of an $n$-cube, each interval is of the same length.
Thus:
Category:Real Analysis
Diameter of N-Cube/Corollary

== Corollary to Diameter of N-Cube ==
Let $Q_n = \closedint {c - R} {c + R}^n$ be an $n$-cube in Euclidean $n$-Space equipped with the usual metric.
The diameter of $Q_n$ is the length of some diagonal of $Q_n$.
== Proof ==
To minimize the sum in question, we chose each coordinate $y_i$, $x_i$ of $x$ and $y$ to be endpoints.
Thus any $x, y$ so chosen is a vertex, by the definition of vertex.
Certainly $x \ne y$ because were they equal, the distance between them would be zero, and the sum would not be maximal.
The result follows from the definition of a diagonal.
Category:Real Analysis
Diameters of Parallelogram Bisect each other

== Theorem ==
Let $\Box ABCD$ be a parallelogram with diameters $AC$ and $BD$.
Let $AC$ and $BD$ intersect at $E$.
Then $E$ is the midpoint of both $AC$ and $BD$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Quadrilateral with Bisecting Diagonals is Parallelogram
Category:Parallelograms
Category:Diameters of Parallelogram Bisect each other
Diameters of Parallelogram Bisect each other/Proof 1

== Theorem ==
== Proof ==
:
By definition of parallelogram:
By Opposite Sides and Angles of Parallelogram are Equal:
:$AB = CD$
:$AD = BC$
:$\angle ABC = \angle ADC$
:$\angle BAD = \angle BCD$
Therefore by Triangle Side-Angle-Side Congruence:
:$\triangle ABC = \triangle ADC$
:$\triangle BAD = \triangle BCD$
Thus:
:$\angle ADE = \angle CBE$
:$\angle DAE = \angle BCE$
We have $AD = BC$.
So from Triangle Angle-Side-Angle Congruence:
:$\triangle ADE = \triangle CBE$
and so:
:$DE = BE$
:$AE = CE$
Hence the result.
== Sources ==
* 
Category:Diameters of Parallelogram Bisect each other
Diameters of Parallelogram Bisect each other/Proof 2

== Theorem ==
== Proof ==
:
Let $\Box ABCD$ be embedded in the complex plane such that $B$ is identified with the origin $0 + 0 i$.
Let $A$ be identified with the complex number $z_1$.
Let $C$ be identified with the complex number $z_2$.
By Geometrical Interpretation of Complex Subtraction:
:$z_1 - z_2 = AC$
Then:
:$AE = m \paren {z_1 - z_2}$
for some $m$ where $0 \le m \le 1$.
Similarly, by Geometrical Interpretation of Complex Addition:
:$z_1 + z_2 = BD$
Then:
:$BE = n \paren {z_1 + z_2}$
for some $n$ where $0 \le n \le 1$.
Then:
and so $E$ is at the midpoint of the diameters of $\Box ABCD$.
== Sources ==
* : $1$: Solved Problems: Graphical Representations of Complex Numbers. Vectors: $10$
Category:Diameters of Parallelogram Bisect each other
Dichotomy Paradox

== Paradox ==
Motion is impossible.
Suppose $B$ is a body which is to move from $P$ to $Q$.
Before $B$ can reach $Q$, it must first move halfway to $Q$.
But before it can reach the midpoint between $P$ and $Q$, it must first move halfway to that midpoint.
And before it can reach the point half way to the midpoint, it has to move halfway to that point too.
So on, indefinitely.
It is therefore impossible for motion to happen, because you cannot even get started. Every time you try to reach a point, you have to reach the point half way before it first.
== Resolution ==
:$\ds \sum_{n \mathop \ge 1} 2^{-n} = 1$
== Historical Note ==
== Sources ==
* : Chapter $\text{II}$: Modern Minds in Ancient Bodies
* 
* 
* 
* 
Category:Zeno's Paradoxes
Category:Antinomies
Dicyclic Group Dic3/Matrix Representation

== Matrix Representation of Dicyclic Group $\Dic 3$ ==
Let $\omega$ denote the complex number $\map \exp {\dfrac {2 \pi i} 6}$, so that $\omega^6 = 1$.
Let $\mathbf X$ be the matrix defined as:
:$\mathbf X = \begin{bmatrix} \omega & 0 \\ 0 & \omega^{-1} \end{bmatrix}$
Let $\mathbf Y$ be the matrix defined such that:
:$\mathbf X \mathbf Y = \mathbf Y \mathbf X^{-1}$
and:
:$\mathbf Y^2 = \mathbf X^3$
Then the set:
:$G = \set {\mathbf X^i, \mathbf Y \mathbf X^j: 1 \le i, j \le 6}$
defines the dicyclic group $\Dic 3$.
== Proof ==
By calculation, we have:
and it is seen that:
:$\mathbf X^{-1} = \begin{bmatrix} \omega^{-1} & 0 \\ 0 & \omega \end{bmatrix}$
Now consider $\mathbf Y = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}$
We have:
and:
Thus $\mathbf X$ and $\mathbf Y$ fulfil the criteria given.
== Sources ==
* : Chapter $3$: Elementary consequences of the definitions: Exercise $8$
Category:Dicyclic Group Dic3
Dicyclic Group is Non-Abelian Group

== Theorem ==
The dicyclic group $\Dic n$ is a non-abelian group on two generators.
=== Corollary ===
== Proof ==
The dicyclic group $\Dic n$ is defined as follows:
First it is to be demonstrated that $\Dic n$ is a group.
First we deduce the following:
$(1): \quad b^4 = e$:
$(2): \quad b^2 a^k = a^{k + n} = a^k b^2$:
$(3): \quad j = \pm 1 \implies b^j a^k = a^{-k} b^j$
Thus, every element of $\Dic n$ can be uniquely written as $a^k b^j$, where $0 \le k < 2 n$ and $j \in \set {0, 1}$. 
The definition of the group product gives:
:$a^k a^m = a^{k + m}$
:$a^k a^m b = a^{k + m} b$
:$a^k b a^m = a^{k - m} b$
:$a^k b a^m x = a^{k - m + n}$
Taking the group axioms in turn:
=== ===
Let $x, y \in \Dic n$.
Thus $\Dic n$ is closed.
=== ===
Thus $\Dic n$ is associative.
=== ===
Thus $e$ is the identity element of $\Dic n$.
=== ===
We have that $e$ is the identity element of $\Dic n$.
Thus every element $...$ of $\Dic n$ has an inverse $...$.
All the group axioms are thus seen to be fulfilled, and so $...$ is a group.
=== Generators ===
The generator of $\Dic n$ is seen to be $\set {a, b}$.
Category:Dicyclic Groups
Dicyclic Group is Non-Abelian Group/Corollary

== Theorem ==
The quaternion group $Q_4$ is a non-abelian group.
== Proof ==
The quaternion group $Q_4$ is an example of the dicyclic group with $4$ elements.
The result follows from Dicyclic Group is Non-Abelian Group.
== Sources ==
* : Chapter $1$: Introductory Concepts: $1$. Basic Building Blocks: $3$. FIELD
Category:Dicyclic Groups
Dido's Problem

== Classic Problem ==
The new is to be founded.
Its area is to be determined by enclosing as much land as possible within a line whose length is fixed.
What shape should it be to make sure the area is a maximum?
== Solution ==
The classical answer is that the enclosure was in the shape of a semicircle whose diameter is the coastline.
However, it is clear that this depends upon the shape of the coast, and a better solution may be to cut off a .
== Proof ==
Let the shoreline be assumed to be a straight line.
Imagine the enclosure takes some geometric figure $S$.
Let $S$ be reflected in the shoreline.
Then the entire geometric figure formed by $S$ along with its reflection $S'$ encloses the largest area for double the length of the boundary line.
This largest area is a circle.
== Historical Note ==
== Sources ==
* : Chapter $\text{VIII}$: Nature or Nurture?
* 
* : The Area Enclosed Against The Seashore: $30$
* 
* 
Category:Classic Problems
Category:Calculus of Variations
Category:Dido's Problem
Dido's Problem/Historical Note

== Historical Note on Dido's Problem ==
There are several more or less romantic legends of the origin of Dido's Problem.
One of the more prosaic is that the founders were granted as much land as a man could plough a furrow round it in a day, given that a ploughman works at a constant rate.
A more fanciful one concerns , who was given a 's , and told she could take as much land for her as she could enclose. According to the legend, she cut it (or arranged to have it cut -- she was a queen, after all) into one long strip of .
In the words of :
:So they reached the place where you will now behold mighty walls and the rising towers of the new town of Carthage;
:and they bought a plot of ground named Byrsa ...
:for they were to have as much as they could enclose within a bull's hide.
::::-- , , Book $\text I$, ll. 360 -- 70
In both cases, the resulting length was used to measure out a semicircle whose diameter formed the coastline.
The legend of the 's is repeated throughout history in the contexts of the founding of several cities.
== Sources ==
* : Chapter $\text{VIII}$: Nature or Nurture? (Footnote $1$)
* 
* : The Area Enclosed Against The Seashore
* 
* 
Category:Historical Notes
Dido's Problem/Variant 1

== Problem ==
Consider a frame consisting of $4$ rods freely hinged at their ends:
:
When will the area enclosed by the frame be a maximum?
== Solution ==
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : The Area Enclosed Against The Seashore: $31$
Category:Cyclic Quadrilaterals
Category:Dido's Problem
Dido's Problem/Variant 1/Proof 1

== Problem ==
== Solution ==
== Proof ==
Let the frame be adjusted so that the vertices of the quadrilateral it forms lie on the circumference of a circle.
Suppose the $4$ corresponding arcs that form the circle are also hinged, like the frame, at the vertices.
Suppose the shape of the frame be changed by bending it at the hinges.
Then the arcs no longer form a circle.
Thus the arcs no longer enclose the maximum area.
But the areas of the segments formed by the arcs and the sides of the frame are the same.
So the area enclosed by the frame is not as great as it was when the frame formed a cyclic quadrilateral.
Hence the result.
== Sources ==
* : The Area Enclosed Against The Seashore: $31$
Category:Cyclic Quadrilaterals
Category:Dido's Problem
Dido's Problem/Variant 2

== Problem ==
Consider the corner of a room which is separated from the rest of the room by a screen of two identical halves, $a$ and $b$ where the length of $a$ equals the length of $b$.
The screen meets the walls of the room at $A$ and $B$.
:
How can the screen be arranged so that the area enclosed by the screen is a maximum?
== Solution ==
When the screen forms two sides of a regular octagon:
:
Thus the sides of the screen must make an angle of $67 \frac 1 2 \degrees$ with the walls.
== Proof ==
When the area enclosed by the screen is a maximum, so will be the area of the octagon so formed by the construction above.
The angles the screen makes with the walls is half the internal angle of the regular octagon.
From Internal Angles of Regular Polygon, that is:
:$A = \dfrac 1 2 \paren {\dfrac {\paren {n - 2} 180 \degrees} n}$
where:
:$A$ is the angle the screen makes with the wall
:$n$ is the number of sides of the regular polygon, in this case $8$.
Hence the result.
== Sources ==
* : The Area Enclosed Against The Seashore: $32$
Category:Dido's Problem
Difference Between Adjacent Polygonal Numbers is Triangular Number

== Theorem ==
Let $\map P {k, n}$ be the $n$th $k$-gonal number.
Then:
:$\map P {k + 1, n} - \map P {k, n} = T_{n - 1}$
where $T_n$ is the $n$th triangular number.
== Proof ==
From Closed Form for Polygonal Numbers:
:$\map P {k, n} = \dfrac n 2 \paren {\paren {k - 2} n - k + 4}$
Thus:
== Sources ==
* : $45$
* : $45$
Category:Polygonal Numbers
Category:Triangular Numbers
Difference Between Adjacent Square Roots Converges

== Theorem ==
Let $\sequence {x_n}$ be the sequence in $\R$ defined as $x_n = \sqrt {n + 1} - \sqrt n$.
Then $\sequence {x_n}$ converges to a zero limit.
== Proof ==
We have:
But from Sequence of Powers of Reciprocals is Null Sequence, $\dfrac 1 {\sqrt n} \to 0$ as $n \to \infty$.
The result follows by the Squeeze Theorem.
== Sources ==
* : $\S 4$: Convergent Sequences: Exercise $\S 4.20 \ (3)$
Category:Limits of Sequences
Difference Equation/Examples/Arbitrary Example 1

== Example of Difference Equations ==
Let us wish to find a sequence $y_0, y_1, y_2, \ldots$ such that the first forward difference $\Delta y_n = n$ has the solution:
:$(1): \quad y_n = \dfrac {n \paren {n - 1} } 2 + k$
where $k$ is an arbitrary constant.
There is an initial condition:
:$k = 1$ when $y_0 = 1$
$(1)$ is an example of a difference equation.
As $\Delta y_n = y_{n + 1} - y_n$, $\Delta y_n = n$ may be written as:
:$y_{n + 1} - y_n = n$
== Sources ==
* 
* 
Category:Examples of Difference Equations
Difference between 2 Consecutive Cubes is Odd

== Theorem ==
Let $a$ and $b$ be consecutive integers.
Then $b^3 - a^3$ is odd.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $9$
Category:Difference between 2 Consecutive Cubes is Odd
Category:Cube Numbers
Difference between 2 Consecutive Cubes is Odd/Proof 1

== Theorem ==
== Proof ==
Let $a, b \in \Z$ such that $b = a + 1$.
Then:
From Product of Consecutive Integers is Even, $a \paren {a + 1}$ is even.
Hence $3 a \paren {a + 1} + 1$ is odd.
Hence the result.
Category:Difference between 2 Consecutive Cubes is Odd
Difference between 2 Consecutive Cubes is Odd/Proof 2

== Theorem ==
== Proof ==
Let $a, b \in \Z$ such that $b = a + 1$.
Either:
:$a$ is even and $b$ is odd
or:
:$b$ is even and $a$ is odd.
Hence from Parity of Integer equals Parity of Positive Power either:
:$a^3$ is even and $b^3$ is odd
or:
:$b^3$ is even and $a^3$ is odd.
The result follows.
Category:Difference between 2 Consecutive Cubes is Odd