nyuuzyou/wikis huggingface dataset:
Quadratic Equation for Parallel Straight Lines

== Theorem ==
Let $\LL_1$ and $\LL_2$ represent $2$ straight lines in the Cartesian plane which are represented by a quadratic equation $E$ in two variables:
:$a x^2 + b y^2 + 2 h x y + 2 g x + 2 f y + c = 0$
Let $\LL_1$ and $\LL_2$ be parallel.
Then:
:$h^2 - a b = 0$
== Proof ==
From Homogeneous Quadratic Equation representing Coincident Straight Lines, $\LL_1$ and $\LL_2$ are parallel respectively to the $2$ coincident straight lines through the origin $\LL'_1$ and $\LL'_2$ represented by the homogeneous quadratic equation:
:$a x^2 + 2 h x y + b y^2$
where:
:$h^2 - a b = 0$
Hence $\LL_1$ and $\LL_2$ are represented by the homogeneous quadratic equation:
:$a x^2 + b y^2 + 2 h x y + 2 g x + 2 f y + c = 0$
where:
:$h^2 - a b = 0$
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $17$.
Category:Straight Lines
Category:Parallel Lines
Quadratic Form (Polynomial Theory)/Examples/Arbitrary Example 3

== Example of Quadratic Form ==
An example of a quadratic form is:
:$a x^2 + 2 h x y + b y^2$
which can be written in matrix notation as:
:$\begin {pmatrix} x & y \end {pmatrix} \begin {pmatrix} a & h \\ h & b \end {pmatrix} \begin {pmatrix} x \\ y \end {pmatrix}$
== Sources ==
* 
* 
Category:Examples of Quadratic Forms (Polynomial Theory)
Quadratic Form in Two Variables represents Conic Section

== Theorem ==
A quadratic form in $2$ variables, when put equal to a constant
:$a x^2 + b x y + c y^2 = r$
represents a conic section.
== Proof ==
It can be seen directly that:
:$a x^2 + b x y + c y^2 = r$
is an instance of an equation for a conic section in Cartesian form:
:$a x^2 + b x y + c y^2 + d x + e y + f = 0$
by setting $f \gets -r$ and equating $d$ and $e$ to zero.
== Sources ==
* 
* 
Category:Quadratic Forms (Polynomial Theory)
Category:Conic Sections
Quadratic Integers over 2 are Not a Field

== Theorem ==
Let $\Z \sqbrk {\sqrt 2}$ denote the set of quadratic integers over $2$:
:$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$
that is, all (real) numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are integers.
Then the algebraic structure:
:$\struct {\Z \sqbrk {\sqrt 2}, +, \times}$
where $+$ and $\times$ are addition and multiplication on real numbers, is not a field.
== Proof ==
We start from the result Quadratic Integers over 2 form Subdomain of Reals.
We have from that result that $1 + 0 \sqrt 2$ is the unity.
Now consider a representative element $a + b \sqrt 2 \in \Z \sqbrk {\sqrt 2}$.
From Difference of Two Squares we have:
:$\paren {a + b \sqrt 2} \paren {a - b \sqrt 2} = a^2 - 2 b^2$
which leads to:
:$\paren {a + b \sqrt 2} \paren {\dfrac {a - b \sqrt 2} {a^2 - 2 b^2} } = 1$
so demonstrating that the product inverse of $\paren {a + b \sqrt 2}$ is $\dfrac a {a^2 - 2b^2} + \dfrac {b \sqrt 2} {a^2 - 2 b^2}$.
But for any two given integers $a$ and $b$, the values $\dfrac a {a^2 - 2 b^2}$ and $\dfrac b {a^2 - 2 b^2}$ are not in general integers.
For instance, when $a = 3$ and $b = 4$ we get:
:$\dfrac a {a^2 - 2 b^2} = \dfrac 3 {23}$ and $\dfrac b {a^2 - 2 b^2} = \dfrac 4 {23}$
(unless I've got my sums wrong).
So not every element of $\struct {\Z \sqbrk {\sqrt 2}, +, \times}$ has a product inverse, demonstrating that $\struct {\Z \sqbrk {\sqrt 2}, +, \times}$ is not a field.
== Sources ==
* : Chapter $4$: Fields: $\S 15$. Examples of Fields: Example $18$
Category:Field Theory
Category:Quadratic Integers
Quadratic Integers over 2 form Ordered Integral Domain

== Theorem ==
Let $\Z \sqbrk {\sqrt 2}$ denote the set of quadratic integers over $2$:
:$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$
that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are integers.
Then the algebraic structure:
:$\struct {\Z \sqbrk {\sqrt 2}, +, \times}$
where $+$ and $\times$ are conventional addition and multiplication on real numbers, is an ordered integral domain.
== Proof ==
We have that Quadratic Integers over 2 form Subdomain of Reals.
We also have that such numbers are real.
The result follows from Real Numbers form Ordered Integral Domain.
== Sources ==
* : Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Example $10$
Category:Examples of Ordered Integral Domains
Category:Quadratic Integers
Quadratic Integers over 2 form Subdomain of Reals

== Theorem ==
Let $\Z \sqbrk {\sqrt 2}$ denote the set of quadratic integers over $2$:
:$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$
That is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are integers.
Then the algebraic structure:
:$\struct {\Z \sqbrk {\sqrt 2}, +, \times}$
where $+$ and $\times$ are conventional addition and multiplication on real numbers, form an integral subdomain of the real numbers $\R$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* 
Category:Examples of Integral Domains
Category:Quadratic Integers
Category:Quadratic Integers over 2 form Subdomain of Reals
Quadratic Integers over 2 form Subdomain of Reals/Proof 1

== Theorem ==
== Proof ==
$\Z \sqbrk {\sqrt 2} \subseteq \R$ where $\R$ is the set of real numbers, so we immediately have that addition and multiplication are well-defined.
=== Closure ===
Let $a_1 + b_1 \sqrt 2, a_2 + b_2 \sqrt 2 \in \Z \sqbrk {\sqrt 2}$.
Then:
So both the operations $+$ and $\times$ are closed on $\Z \sqbrk {\sqrt 2}$.
=== Associativity ===
We have that addition and multiplication are associative on $\R$.
Therefore it follows from Restriction of Associative Operation is Associative that they are also associative on $\Z \sqbrk {\sqrt 2}$.
=== Commutativity ===
We have that addition and multiplication are commutative on $\R$.
Therefore it follows from Restriction of Commutative Operation is Commutative that they are also commutative on $\Z \sqbrk {\sqrt 2}$.
=== Identities ===
We have:
and similarly for $\paren {0 + 0 \sqrt 2} + \paren {a + b \sqrt 2}$.
So $\paren {0 + 0 \sqrt 2}$ is the identity for $+$ on $\Z \sqbrk {\sqrt 2}$.
Then:
and similarly for $\paren {a + b \sqrt 2} \times \paren {1 + 0 \sqrt 2}$.
So $\paren {1 + 0 \sqrt 2}$ is the identity for $\times$ on $\Z \sqbrk {\sqrt 2}$.
=== Inverses ===
We have:
and similarly for $\paren {-a + \paren {-b} \sqrt 2} + \paren {a + b \sqrt 2}$.
So $\paren {-a + \paren {-b} \sqrt 2}$ is the inverse of $\paren {a + b \sqrt 2}$ for $+$ on $\Z \sqbrk {\sqrt 2}$.
We have no need to investigate inverses for $\times$ (which is convenient as $\times$ happens not to be closed for inverses on $\Z \sqbrk {\sqrt 2}$).
=== Distributivity ===
We have that Real Multiplication Distributes over Addition, so by Restriction of Operation Distributivity, $\times$ is distributive over $+$ on $\Z \sqbrk {\sqrt 2}$.
=== Divisors of Zero ===
By Real Numbers form Field, $\R$ is a field and so by Field is Integral Domain is itself an integral domain.
Hence it has no proper zero divisors.
The result follows by putting all the pieces together.
== Sources ==
* : Chapter $1$: Integral Domains: $\S 5$. Further Examples of Integral Domains: Example $4$
Category:Quadratic Integers over 2 form Subdomain of Reals
Quadratic Integers over 2 form Subdomain of Reals/Proof 2

== Theorem ==
== Proof ==
From Integers form Subdomain of Reals, $\struct {\Z, +, \times}$ is an integral subdomain of the real numbers $\R$.
We have that $\sqrt 2 \in \R$.
Every expression of the form:
:$a_0 + a_1 \sqrt 2 + a_2 \paren {\sqrt 2}^2 + \cdots + a_n \paren {\sqrt 2}^n$
can be simplified to a number of the form $a + b \sqrt 2$, where $a, b \in \Z$.
The result follows from Set of Polynomials over Integral Domain is Subring.
== Sources ==
* : $\S 64.1$ Polynomial rings over an integral domain: Illustration
Category:Quadratic Integers over 2 form Subdomain of Reals
Quadratic Integers over 3 form Integral Domain

== Theorem ==
Let $\R$ denote the set of real numbers.
Let $\Z \sqbrk {\sqrt 3} \subseteq \R$ denote the set of quadratic integers over $3$:
:$\Z \sqbrk {\sqrt 3} = \set {a + b \sqrt 3: a, b \in \Z}$
Then $\struct {\Z \sqbrk {\sqrt 3}, +, \times}$ is an integral domain.
== Proof ==
From Real Numbers form Integral Domain we have that $\struct {\R, +, \times}$ is an integral domain.
Hence to demonstrate that $\struct {\Z \sqbrk {\sqrt 3}, +, \times}$ is an integral domain, we can use the Subdomain Test.
We have that the unity of $\struct {\R, +, \times}$ is $1$.
Then we note:
:$1 = 1 + 0 \times \sqrt 3$
and so $1 \in S$.
Thus property $(2)$ of the Subdomain Test is fulfilled.
It remains to demonstrate that $\struct {\Z \sqbrk {\sqrt 3}, +, \times}$ is a subring of $\struct {\R, +, \times}$, so fulfilling property $(2)$ of the Subdomain Test.
Hence we use the Subring Test.
We note that $\Z \sqbrk {\sqrt 3} \ne \O$ as $1 \in \Z \sqbrk {\sqrt 3}$.
This fulfils property $(1)$ of the Subring Test.
Let $x, y \in \Z \sqbrk {\sqrt 3}$ such that:
:$x = a + b \sqrt 3$
:$y = c + d \sqrt 3$
Then:
This fulfils property $(2)$ of the Subring Test.
Then: 
This fulfils property $(3)$ of the Subring Test.
Hence the result.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.7 \ \text {(a)}$
* : Exercises: Chapter $1$: Exercise $1 \ \text{(iv)}$
Category:Real Numbers
Category:Examples of Integral Domains
Category:Quadratic Integers
Quadratic Irrational is Root of Quadratic Equation

== Theorem ==
Let $x$ be a quadratic irrational.
Then $x$ is a solution to a quadratic equation with rational coefficients.
== Proof ==
Let $x = r + s \sqrt n$.
From Solution to Quadratic Equation, the solutions of $a x^2 + b x + c$ are:
:$x = \dfrac {-b \pm \sqrt {b^2 - 4 a c}} {2 a}$
given the appropriate condition on the discriminant.
So if $x = r + s \sqrt n$ is a solution, then so is $x = r - s \sqrt n$.
Hence we have:
As $r$ and $s$ are rational and $n$ is an integer, it follows that $-2 r$ and $r^2 - s^2 n$ are also rational from Rational Numbers form Field.
Hence the result.
Category:Quadratic Equations
Category:Quadratic Irrationals
Quadratic Representation of Pair of Straight Lines

== Theorem ==
Consider the general quadratic equation in $2$ variables:
:$(1): \quad a x^2 + b x y + c y^2 + d x + e y + f = 0$
Then $(1)$ is the locus of $2$ straight lines in the Cartesian plane it can be expressed in the form:
:$\paren {l_1 x + m_1 y + n_1} \paren {l_2 x + m_2 y + n_2} = 0$
where $l_1$, $m_1$, $n_1$, $l_2$, $m_2$ and $n_2$ are real numbers.
== Proof ==
=== Sufficient Condition ===
Let $\LL_1$ and $\LL_2$ be straight lines embedded in a cartesian plane $\CC$, expressed in general form as:
Let $\tuple {x, y}$ be a point on either $\LL_1$ or $\LL_2$.
Then because either $l_1 x + m_1 y + n_1 = 0$ or $l_2 x + m_2 y + n_2 = 0$, it is certainly the case that:
:$\paren {l_1 x + m_1 y + n_1} \paren {l_2 x + m_2 y + n_2} = 0$
If $\tuple {x, y}$ is not on either $\LL_1$ or $\LL_2$, then neither $l_1 x + m_1 y + n_1 = 0$ nor $l_2 x + m_2 y + n_2 = 0$ hold.
Hence $l_1 x + m_1 y + n_1 \ne 0$ and $l_2 x + m_2 y + n_2 \ne 0$, and so:
:$\paren {l_1 x + m_1 y + n_1} \paren {l_2 x + m_2 y + n_2} \ne 0$
Hence two straight lines embedded in $\CC$ can be expressed by an equation in the form:
:$\paren {l_1 x + m_1 y + n_1} \paren {l_2 x + m_2 y + n_2} = 0$
as required.
=== Necessary Condition ===
Let it be possible to express $(1)$ in the form:
:$(2): \quad \paren {l_1 x + m_1 y + n_1} \paren {l_2 x + m_2 y + n_2} = 0$
Let $\tuple {x, y}$ satisfy $(2)$.
Then either $l_1 x + m_1 y + n_1 = 0$ or $l_2 x + m_2 y + n_2 = 0$ or both.
Thus $\tuple {x, y}$ is either:
:on the straight line defined by the equation $l_1 x + m_1 y + n_1 = 0$
or:
:on the straight line defined by the equation $l_2 x + m_2 y + n_2 = 0$
It follows that $(2)$, and hence $(1)$, is an equation for two straight lines in $\CC$.
== Examples ==
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $13$. Quadratic equation representing a pair of straight lines
Category:Straight Lines
Category:Quadratic Equations
Category:Quadratic Representation of Pair of Straight Lines
Quadratic Representation of Pair of Straight Lines/Examples

== Examples of Quadratic Representation of Pair of Straight Lines ==
=== Example: $x^2 - x y - 2 y^2 + 2 x + 5 y = 3$ ===
=== Example: $x^2 + y^2 - 1 = 0$ ===
Category:Quadratic Representation of Pair of Straight Lines
Quadratic Representation of Pair of Straight Lines/Examples/x^2 + y^2 - 1 = 0

== Examples of Quadratic Representation of Pair of Straight Lines ==
The equation:
:$x^2 + y^2 - 1 = 0$
does not represent two straight lines embedded in the Cartesian plane.
== Proof ==
$x^2 + y^2 - 1$ cannot be expressed in the form:
:$\paren {l_1 x + m_1 y + n_1} \paren {l_2 x + m_2 y + n_2} = 0$
The result follows from Quadratic Representation of Pair of Straight Lines.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $13$. Quadratic equation representing a pair of straight lines
Category:Quadratic Representation of Pair of Straight Lines
Quadratic Representation of Pair of Straight Lines/Examples/x^2 - x y - 2 y^2 + 2 x + 5 y = 3

== Examples of Quadratic Representation of Pair of Straight Lines ==
The equation:
:$x^2 - x y - 2 y^2 + 2 x + 5 y = 3$
represents the two straight lines embedded in the Cartesian plane:
== Proof ==
The result follows from Quadratic Representation of Pair of Straight Lines.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $13$. Quadratic equation representing a pair of straight lines
Category:Quadratic Representation of Pair of Straight Lines
Quadratic Residue/Examples

== Examples of Quadratic Residues ==
=== Quadratic Residues Modulo $3$ ===
=== Quadratic Residues Modulo $5$ ===
=== Quadratic Residues Modulo $7$ ===
=== Quadratic Residues Modulo $11$ ===
=== Quadratic Residues Modulo $17$ ===
=== Quadratic Residues Modulo $29$ ===
=== Quadratic Residues Modulo $61$ ===
Category:Examples of Quadratic Residues
Quadratic Residue/Examples/11

== Example of Quadratic Residues ==
The set of quadratic residues modulo $11$ is:
:$\set {1, 3, 4, 5, 9}$
== Proof ==
To list the quadratic residues of $11$ it is enough to work out the squares $1^2, 2^2, \ldots, 10^2$ modulo $11$.
So the set of quadratic residues modulo $11$ is:
:$\set {1, 3, 4, 5, 9}$
The set of quadratic non-residues of $11$ therefore consists of all the other non-zero least positive residues:
:$\set {2, 6, 7, 8, 10}$
== Sources ==
* : $\text {3-5}$ The Use of Computers in Number Theory: Exercise $6$
* : Chapter $7$: Patterns in Numbers: Gauss
Category:Examples of Quadratic Residues
Category:11
Quadratic Residue/Examples/17

== Example of Quadratic Residues ==
The set of quadratic residues modulo $17$ is:
:$\set {1, 2, 4, 8, 9, 13, 15, 16}$
== Proof ==
To list the quadratic residues of $17$ it is enough to work out the squares $1^2, 2^2, \dotsc, 16^2$ modulo $17$.
So the set of quadratic residues modulo $17$ is:
:$\set {1, 2, 4, 8, 9, 13, 15, 16}$
The set of quadratic non-residues of $17$ therefore consists of all the other non-zero least positive residues:
:$\set {3, 5, 6, 7, 10, 11, 12, 14}$
== Sources ==
* : $\text {3-5}$ The Use of Computers in Number Theory: Exercise $6$
Category:Examples of Quadratic Residues
Category:17
Quadratic Residue/Examples/29

== Example of Quadratic Residues ==
The set of quadratic residues modulo $29$ is:
:$\set {1, 4, 5, 6, 7, 9, 13, 16, 20, 22, 23, 24, 25, 28}$
== Proof ==
From Square Modulo n Congruent to Square of Inverse Modulo n, to list the quadratic residues of $29$ it is sufficient to work out the squares $1^2, 2^2, \dotsc, \paren {\dfrac {28} 2}^2$ modulo $29$.
So:
So the set of quadratic residues modulo $29$ is:
:$\set {1, 4, 5, 6, 7, 9, 13, 16, 20, 22, 23, 24, 25, 28}$
The set of quadratic non-residues of $29$ therefore consists of all the other non-zero least positive residues:
:$\set {2, 3, 8, 10, 11, 12, 14, 15, 17, 18, 19, 21, 26, 27}$
== Sources ==
* : $\text {3-5}$ The Use of Computers in Number Theory: Exercise $6$
Category:Examples of Quadratic Residues
Category:29
Quadratic Residue/Examples/3

== Example of Quadratic Residues ==
There exists exactly $1$ quadratic residue modulo $3$, and that is $1$.
== Proof ==
To list the quadratic residues of $3$ it is enough to work out the squares $1^2$ and $2^2$ modulo $3$.
So the set of quadratic residues modulo $3$ is:
:$\set 1$
The set of quadratic non-residues of $3$ therefore consists of all the other non-zero least positive residues:
:$\set 2$
== Sources ==
* : $\text {3-5}$ The Use of Computers in Number Theory: Exercise $6$
Category:Examples of Quadratic Residues
Category:3
Quadratic Residue/Examples/5

== Example of Quadratic Residues ==
The set of quadratic residues modulo $5$ is:
:$\set {1, 4}$
== Proof ==
To list the quadratic residues of $5$ it is enough to work out the squares $1^2, 2^2, 3^2, 4^2$ modulo $5$.
So the set of quadratic residues modulo $5$ is:
:$\set {1, 4}$
The set of quadratic non-residues of $5$ therefore consists of all the other non-zero least positive residues:
:$\set {2, 3}$
== Sources ==
* : $\text {3-5}$ The Use of Computers in Number Theory: Exercise $6$
Category:Examples of Quadratic Residues
Category:5
Quadratic Residue/Examples/61

== Example of Quadratic Residues ==
The set of quadratic residues modulo $61$ is:
:$\set {1, 3, 4, 5, 9, 12, 13, 14, 15, 16, 19, 20, 22, 25, 27, 34, 36, 39, 41, 42, 45, 46, 47, 48, 49, 53, 56, 57, 58, 60}$
== Proof ==
From Square Modulo n Congruent to Square of Inverse Modulo n, to list the quadratic residues of $61$ it is sufficient to work out the squares $1^2, 2^2, \dotsc, \paren {\dfrac {60} 2}^2$ modulo $61$.
So:
So the set of quadratic residues modulo $61$ is:
:$\set {1, 3, 4, 5, 9, 12, 13, 14, 15, 16, 19, 20, 22, 25, 27, 34, 36, 39, 41, 42, 45, 46, 47, 48, 49, 53, 56, 57, 58, 60}$
The set of quadratic non-residues of $61$ therefore consists of all the other non-zero least positive residues:
:$\set {2, 6, 7, 8, 10, 11, 17, 18, 21, 23, 24, 26, 28, 29, 30, 31, 32, 33, 35, 37, 38, 40, 43, 44, 50, 51, 52, 54, 55, 59}$
== Sources ==
* : $\text {3-5}$ The Use of Computers in Number Theory: Exercise $6$
Category:Examples of Quadratic Residues
Category:61
Quadratic Residue/Examples/7

== Example of Quadratic Residues ==
The set of quadratic residues modulo $7$ is:
:$\set {1, 2, 4}$
== Proof ==
To list the quadratic residues of $7$ it is enough to work out the squares $1^2, 2^2, \dotsc, 6^2$ modulo $7$.
So the set of quadratic residues modulo $7$ is:
:$\set {1, 2, 4}$
The set of quadratic non-residues of $7$ therefore consists of all the other non-zero least positive residues:
:$\set {3, 5, 6}$
== Sources ==
* : $\text {3-5}$ The Use of Computers in Number Theory: Exercise $6$
* 
* 
Category:Examples of Quadratic Residues
Category:7
Quadrature of Parabola

== Theorem ==
Let $T$ be a parabola.
Consider the parabolic segment bounded by an arbitrary chord $AB$.
Let $C$ be the point on $T$ where the tangent to $T$ is parallel to $AB$.
Then the area $S$ of the parabolic segment $ABC$ of $T$ is given by:
:$S = \dfrac 4 3 \triangle ABC$
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
== Historical Note ==
== Sources ==
* : Chapter $\text {B}.3$: Archimedes' Quadrature of the Parabola
* 
Category:Quadrature of Parabola
Category:Parabolas
Category:Area Formulas
Quadrature of Parabola/Historical Note

== Historical Note on Quadrature of Parabola ==
The quadrature of a parabola was first given by in his book . It is also discussed in .
Two proofs are given. His purely geometric proof was the same as the one documented here, except that he used a different technique to prove:
:$\triangle ADC + \triangle CEB = \dfrac {\triangle ABC} 4$
== Sources ==
* : Chapter $\text{II}$: Modern Minds in Ancient Bodies
* : Chapter $\text {A}.5$: Archimedes (ca. B.C.)
* : Chapter $\text {B}.3$: Archimedes' Quadrature of the Parabola
Category:Historical Notes
Quadrature of Parabola/Proof 1

== Theorem ==
== Proof ==
, consider the parabola $y = a x^2$.
Let $A, B, C$ be the points:
:
The slope of the tangent at $C$ is given by using:
:$\dfrac {\d y} {\d x} 2 a x_1$
which is parallel to $AB$.
Thus:
:$2 a x_1 = \dfrac {a {x_0}^2 - a {x_2}^2} {x_0 - x_2}$
which leads to:
:$x_1 = \dfrac {x_0 + x_2} 2$
So the vertical line through $C$ is a bisector of $AB$, at point $P$.
Complete the parallelogram $CPBQ$.
Also, find $E$ which is the point where the tangent to $T$ is parallel to $BC$.
By the same reasoning, the vertical line through $E$ is a bisector of $BC$, and so it also bisects $BP$ at $H$.
Next:
 
At the same time:
 
So:
:$QB = 4 FE = FH$
and because $CB$ is the diagonal of a parallelogram:
:$2 FE = 2 EG = FG$
This implies that:
:$2 \triangle BEG = \triangle BGH$
and:
:$2 \triangle CEG = \triangle BGH$
So:
:$\triangle BCE = \triangle BGH$
and so as $\triangle BCP = 4 \triangle BGH$ we have that:
:$BCE = \dfrac {\triangle BCP} 4$
A similar relation holds for $\triangle APC$:
:
so it can be seen that:
:$\triangle ABC = 4 \paren {\triangle ADC + \triangle CEB}$
Similarly, we can create four more triangles underneath $\triangle ADC$ and $\triangle CEB$ which are $\dfrac 1 4$ the area of those combined, or $\dfrac 1 {4^2} \triangle ABC$.
This process can continue indefinitely.
So the area $S$ is given as:
:$S = \triangle ABC \paren {1 + \dfrac 1 4 + \dfrac 1 {4^2} + \cdots}$
But from Sum of Geometric Sequence it follows that:
:$S = \triangle ABC \paren {\dfrac 1 {1 - \dfrac 1 4} } = \dfrac 4 3 \triangle ABC$
Category:Quadrature of Parabola
Quadrature of Parabola/Proof 2

== Theorem ==
== Proof ==
Let $T$ be the parabola which is the locus of points $\tuple {x, y}$ satisfying $y = x^2$.
By Area of Triangle Inscribed in Parabola:
:the point $C$ where the tangent to $T$ at $C$ is parallel to $AB$ has $x$-coordinate $\dfrac 1 2 \paren {x_0 + x_2}$.
Let $d$ be the horizontal distance between $A$ and $B$.
By Area of Triangle Inscribed in Parabola:
:$\map \AA {\triangle ABC} = \dfrac 1 8 d^3$
Now construct two triangles. 
One is $BC$ bisected with point $E$ on the parabola, forming:
:$\triangle EBC$
The other is $AC$ bisected with point $D$ on the parabola, forming:
:$\triangle DAC$
For each triangle, the length along the $x$-coordinate $\Delta x$ is equal to $\dfrac d 2$.
The area of each triangle is then:
:$\AA = \dfrac 1 8 \paren {\dfrac d 2}^3 = \dfrac 1 8 \cdot \dfrac 1 8 d^3$
From above:
:$\map \AA {\triangle ABC} = \dfrac 1 8 d^3$
Thus
:$\map \AA {\triangle EBC} = \map \AA {\triangle DAC} = \dfrac 1 8 \map \AA {ABC}$
For two triangles, the area together is: 
:$\map \AA {\triangle DAC} + \map \AA {\triangle EBC} = \dfrac 1 4 \map \AA {ABC}$
By a variant of the method of exhaustion, the total area over the parabola and under $AB$ is:
:$\map \AA {ABC} \cdot \paren {1 + \dfrac 1 4 + \dfrac 1 {4^2} + \dots}$
By Sum of Infinite Geometric Sequence:
:$1 + \dfrac 1 4 + \dfrac 1 {4^2} + \dots = \dfrac 4 3$
Hence the total area is:
:$S = \map \AA {ABC} \cdot \dfrac 4 3$
Category:Quadrature of Parabola
Quadrilateral Tessellates the Plane/Mistake

== Source Work ==
:
:tessellation
== Mistake ==
:... any ... plane quadrilateral and its mirror image will tessellate the plane.
== Correction ==
While it is true to state that any Quadrilateral Tessellates the Plane, it is not necessarily the case that a tessellation is still possible with the quadrilateral together with its mirror image.
== Sources ==
* 
* 
Category:Mistakes/The Penguin Dictionary of Mathematics
Quadrilateral in Complex Plane is Cyclic iff Cross Ratio of Vertices is Real

== Theorem ==
Let $z_1, z_2, z_3, z_4$ be distinct complex numbers.
Then:
:$z_1, z_2, z_3, z_4$ define the vertices of a cyclic quadrilateral
 their cross-ratio:
:$\paren {z_1, z_3; z_2, z_4} = \dfrac {\paren {z_1 - z_2} \paren {z_3 - z_4} } {\paren {z_1 - z_4} \paren {z_3 - z_2} }$
is wholly real.
== Proof ==
Let $z_1 z_2 z_3 z_4$ be a cyclic quadrilateral.
By Geometrical Interpretation of Complex Subtraction, the four sides of $z_1 z_2 z_3 z_4$ can be defined as $z_1 - z_2$, $z_3 - z_2$, $z_3 - z_4$ and $z_1 - z_4$.
:
From Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles, the opposite angles of $z_1 z_2 z_3 z_4$ sum to $\pi$ radians.
By Complex Multiplication as Geometrical Transformation, it follows that:
Thus:
Hence the result.
== Sources ==
* : $\S 2$. Geometrical Representations: Exercise $6$.
Category:Geometry of Complex Plane
Category:Cyclic Quadrilaterals