nyuuzyou/wikis huggingface dataset:
Bisection of Angle/Construction

== Bisection of Angle: Construction ==
:
Let $\angle BAC$ be the given angle to be bisected.
Let $D$ be an arbitrary point on $AB$.
From , let $AE$ be cut off from $AC$ such that $AE = AD$.
From , let the line segment $DE$ be constructed.
From , let an equilateral triangle $\triangle DEF$ be constructed on $AB$.
From , let the line segment $AF$ be constructed.
Then the angle $\angle BAC$ has been bisected by the straight line segment $AF$.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Bisection of Angle
Bisection of Angle in Cartesian Plane

== Theorem ==
Let $\theta$ be the angular coordinate of a point $P$ in a polar coordinate plane.
Let $QOR$ be a straight line that bisects the angle $\theta$.
Then the angular coordinates of $Q$ and $R$ are $\dfrac \theta 2$ and $\pi + \dfrac \theta 2$.
=== Corollary ===
== Proof ==
:
Let $A$ be a point on the polar axis.
By definition of bisection, $\angle AOQ = \dfrac \theta 2$.
This is the angular coordinate of $Q$.
Consider the conjugate angle $\map \complement {\angle AOP}$ of $\angle AOP$.
By definition of conjugate angle:
:$\map \complement {\angle AOP} = -2 \pi - \theta$
where the negative sign arises from the fact that it is measured clockwise.
Then the angle $\angle AOR$ is half of $\map \complement {\angle AOP}$:
The angular coordinate of point $R$ is the conjugate angle $\map \complement {\angle AOR}$ of $\angle AOR$:
Category:Analytic Geometry
Bisection of Angle in Cartesian Plane/Corollary

== Corollary to Bisection of Angle in Cartesian Plane ==
Let $\theta$ be the angular coordinate of a point $P$ in a polar coordinate plane.
Let $QOR$ be a straight line that bisects the angle $\theta$.
If $\theta$ is in quadrant I or quadrant II, then the angular coordinates of $Q$ and $R$ are in quadrant I and quadrant III.
If $\theta$ is in quadrant III or quadrant IV, then the angular coordinates of $Q$ and $R$ are in quadrant II and quadrant IV.
== Proof ==
:
From Bisection of Angle in Cartesian Plane, the angular coordinates of $Q$ and $R$ are $\dfrac \theta 2$ and $\pi + \dfrac \theta 2$.
, let $\angle Q = \dfrac \theta 2$ and $\angle R = \pi + \dfrac \theta 2$.
Let $\theta$ be in quadrant I or quadrant II.
Then $0 < \theta < \pi$.
Dividing each term in the inequality by $2$:
:$0 < \dfrac \theta 2 < \dfrac \pi 2$
and so $Q$ lies in quadrant I.
Adding $\pi$ to each expression in the inequality:
:$\pi < \pi + \dfrac \theta 2 < \dfrac {3 \pi} 2$
and so $R$ lies in quadrant I.
Let $\theta$ be in quadrant III or quadrant IV.
Then $\pi < \theta < 2 \pi$.
Dividing each term in the inequality by $2$:
:$\dfrac \pi 2 < \dfrac \theta 2 < \pi$
and so $Q$ lies in quadrant II.
Adding $\pi$ to each expression in the inequality:
:$\dfrac {3 \pi} 2 < \pi + \dfrac \theta 2 < 2 \pi$ and so $R$ lies in quadrant IV.
Category:Analytic Geometry
Bisection of Arc

== Theorem ==
It is possible to bisect an arc of a circle.
== Construction ==
:
Let $AB$ be the given arc.
Join the line $AB$ and bisect it at $C$.
Construct the line $CD$ perpendicular to $AB$ at $C$.
Then the arc $AB$ has been bisected at $D$.
== Proof ==
Join $AD$ and $BD$.
We have that $AC = CB$ and $CD$ is common.
We also have that $\angle ACD = \angle BCD$ as they are both right angles.
So from Triangle Side-Angle-Side Congruence $\triangle ACD = \triangle BCD$ and so $AD = BD$.
But from Straight Lines Cut Off Equal Arcs in Equal Circles, the arc $AD$ equals the arc $BD$.
That is, the arc $AB$ has been bisected at $D$.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Circles
Bisection of Straight Line

== Theorem ==
It is possible to bisect a straight line segment.
== Construction ==
:
Let $AB$ be the given straight line segment.
We construct an equilateral triangle $\triangle ABC$ on $AB$.
We bisect the angle $\angle ACB$ by the straight line segment $CD$.
Then $AB$ has been bisected at the point $D$.
== Proof ==
As $\triangle ABC$ is an equilateral triangle, it follows that $AC = CB$.
The two triangles $\triangle ACD$ and $\triangle BCD$ have side $CD$ in common, and side $AC$ of $\triangle ACD$ equals side $BC$ of $\triangle BCD$.
The angle $\angle ACD$ subtended by lines $AC$ and $CD$ equals the angle $\angle BCD$ subtended by lines $BC$ and $CD$, as $\angle ACB$ was bisected.
So triangles $\triangle ACD$ and $\triangle BCD$ are equal.
Therefore $AD = DB$.
So $AB$ has been bisected at the point $D$.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Lines
Bisector of Apex of Isosceles Triangle also Bisects Base

== Theorem ==
Let $\triangle ABC$ be an isosceles triangle whose apex is $A$.
Let $AD$ be the bisector of $\angle BAC$ such that $AD$ intersects $BC$ at $D$.
Then $AD$ bisects $BC$.
== Proof ==
By definition of isosceles triangle, $AB = AC$.
By definition of bisector, $\angle BAD = \angle CAD$.
By construction, $AD$ is common.
Thus by Triangle Side-Angle-Side Congruence, $\triangle ABD = \triangle ACD$.
Thus $AD = DC$.
The result follows by definition of bisection.
== Sources ==
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.12$: Corollary $1 \ \text{(i)}$
Category:Isosceles Triangles
Bisector of Apex of Isosceles Triangle is Perpendicular to Base

== Theorem ==
Let $\triangle ABC$ be an isosceles triangle whose apex is $A$.
Let $AD$ be the bisector of $\angle BAC$ such that $AD$ intersects $BC$ at $D$.
Then $AD$ is perpendicular to $BC$.
== Proof ==
By definition of isosceles triangle, $AB = AC$.
By definition of bisector, $\angle BAD = \angle CAD$.
By construction, $AD$ is common.
Thus by Triangle Side-Angle-Side Congruence, $\triangle ABD = \triangle ACD$.
Thus $\angle ADB = \angle ADC$.
By Two Angles on Straight Line make Two Right Angles, $\angle ADB + \angle ADC$ equals $2$ right angles.
Thus each of $\angle ADB$ and $\angle ADC$ are right angles.
The result follows by definition of perpendicular.
== Sources ==
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.12$: Corollary $1 \ \text{(ii)}$
Category:Isosceles Triangles
Bisectors of Adjacent Angles between Straight Lines Meeting at Point are Perpendicular

== Corollary to Two Angles on Straight Line make Two Right Angles ==
If a straight line meets another straight line, the bisectors of the two adjacent angles between them are perpendicular.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.1$: Corollary $2$
Category:Angle Bisectors
Category:Bisectors of Adjacent Angles between Straight Lines Meeting at Point are Perpendicular
Bisectors of Adjacent Angles between Straight Lines Meeting at Point are Perpendicular/Proof 1

== Corollary to Two Angles on Straight Line make Two Right Angles ==
== Proof ==
Let $AB$ and $CD$ be two straight lines that cross at $E$.
Let $\angle AEC$ be bisected by $EF$.
Let $\angle CEB$ be bisected by $EG$.
Thus:
:$2 \angle FEC = \angle AEC$
and:
:$2 \angle CEG = \angle CEB$
But from Two Angles on Straight Line make Two Right Angles, $\angle AEC + \angle CEB$ equal $2$ right angles.
Thus $2 \angle FEC + 2 \angle CEG$ equal $2$ right angles.
Hence $\angle FEG = \angle FEC + \angle CEG$ equals $1$ right angle.
That is $EF$ and $EG$ are perpendicular.
Category:Bisectors of Adjacent Angles between Straight Lines Meeting at Point are Perpendicular
Bisectors of Adjacent Angles between Straight Lines Meeting at Point are Perpendicular/Proof 2

== Theorem ==
== Proof ==
Let $\LL_1$ and $\LL_2$ be straight lines embedded in a cartesian plane $\CC$, expressed in normal form as:
From Bisectors of Angles between Two Straight Lines, the angle bisectors of the angles formed at the point of intersection of $\LL_1$ and $\LL_2$ are given by:
These are in the form $l x + m y + n = 0$.
We use Condition for Straight Lines in Plane to be Perpendicular to prove that $l_1 l_2 + m_1 m_2 = 0$, where:
Hence
Hence the result.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $12$. Bisectors of the angles between two straight lines
Category:Bisectors of Adjacent Angles between Straight Lines Meeting at Point are Perpendicular
Bisectors of Angles between Two Straight Lines/General Form

== Theorem ==
Let $\LL_1$ and $\LL_2$ be straight lines embedded in a cartesian plane $\CC$, expressed in general form as:
The angle bisectors of the angles formed at the point of intersection of $\LL_1$ and $\LL_2$ are given by:
:$\dfrac {l_1 x + m_1 y + n_1} {\sqrt { {l_1}^2 + {m_1}^2} } = \pm \dfrac {l_2 x + m_2 y + n_2} {\sqrt { {l_2}^2 + {m_2}^2} }$
== Proof ==
First we convert $\LL_1$ and $\LL_2$ into normal form:
Then from Bisectors of Angles between Two Straight Lines: Normal Form, the angle bisectors of the angles formed at the point of intersection of $\LL_1$ and $\LL_2$ are given by:
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $12$. Bisectors of the angles between two straight lines
Category:Angle Bisectors
Bisectors of Angles between Two Straight Lines/Homogeneous Quadratic Equation Form

== Theorem ==
Consider the homogeneous quadratic equation:
:$(1): \quad a x^2 + 2 h x y + b y^2 = 0$
representing two straight lines through the origin.
Then the homogeneous quadratic equation which represents the angle bisectors of the angles formed at their point of intersection is given by:
:$h x^2 - \paren {a - b} x y - h y^2 = 0$
== Proof ==
From Angle Bisectors are Harmonic Conjugates, the two angle bisectors are harmonic conjugates of the straight lines represented by $(1)$.
From Condition for Homogeneous Quadratic Equation to describe Perpendicular Straight Lines, these angle bisectors can be described by the homogeneous quadratic equation:
:$x^2 + 2 \lambda x y - y^2 = 0$
From Condition for Pairs of Lines through Origin to be Harmonic Conjugates: Homogeneous Quadratic Equation Form:
:$-a + b - 2 \lambda h = 0$
Hence:
:$\lambda = -\dfrac {a + b} {2 h}$
The result follows.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $22$. Equation of the bisectors of the angles between the two straight lines $a x^2 + 2 h x y + b y^2 = 0$
Category:Harmonic Ranges
Category:Bisectors of Angles between Two Straight Lines
Bisectors of Angles between Two Straight Lines/Normal Form

== Theorem ==
Let $\LL_1$ and $\LL_2$ be straight lines embedded in a cartesian plane $\CC$, expressed in normal form as:
The angle bisectors of the angles formed at the point of intersection of $\LL_1$ and $\LL_2$ are given by:
== Proof ==
Let $A'SA$ and $B'SB$ be the straight lines $\LL_1$ and $\LL_2$ respectively, intersecting at the point $S$.
Let $P = \tuple {x, y}$ be an arbitrary point on either of the angle bisectors of $\angle ASB$ or $\angle BSA'$.
:
Drop perpendiculars $PM$ from $P$ to $SA$ and $PN$ from $P$ to $SB$.
Because:
:$\angle PSM = \angle PSN$
:$\angle PMS = \angle PNS$
:$PS$ is common
we have that:
:$\triangle PSM = \triangle PSN$
and so:
:$PM = PN$
Having regard only for the magnitude of the perpendiculars:
which is the equation of a straight line in normal form.
This represents either the angle bisector of $\angle ASB$ or the angle bisector of $\angle BSA'$.
Considering now the signs of the perpendiculars, it is seen that while $PM$ and $PM'$ have the same sign, $PN$ and $PN'$ have opposite signs.
Hence:
:for one bisector we have $PM = PN$
:for the other bisector we have $PM = -PN$.
So the equation for the other bisector is $x \paren {\cos \alpha + \cos \beta} + y \paren {\sin \alpha + \sin \beta} = p + q$.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $12$. Bisectors of the angles between two straight lines
Category:Angle Bisectors
Blaschke-Lebesgue Theorem

== Theorem ==
Let $C$ be a closed contour in the complex plane.
Let the minimum width $w$ of $C$ be such that $w \ge 1$.
Then $C$ can contain a circle whose radius $\dfrac 1 3$.
== Proof ==
== Also known as ==
The Blaschke-Lebesgue Theorem is also seen as Blaschke's Theorem.
== Sources ==
* : $0,33333 33333 33 \ldots$
* : Blashke's Theorem
Category:Plane Geometry
Bloch's Theorem

== Theorem ==
Let $f: \C \to \C$ be a holomorphic function in the unit disk $\cmod z \le 1$.
Let $\cmod {\map {f'} 0} = 1$.
Then there exists:
:a disk $D$ of radius $B$
:an analytic function $\phi$ in $D$ such that $\map f {\map \phi z} = z$ for all $z$ in $D$
where $B > \dfrac 1 {72}$ is an absolute constant.
The lower bound $\dfrac 1 {72}$ is not the best possible.
== Proof ==
== Also see ==
* Definition:Bloch's Constant
* Lower Bound of Bloch's Constant
* Upper Bound of Bloch's Constant
* Landau's Theorem
== Sources ==
* : $0,433012701 \ldots < B < 0,4719 \ldots$
Category:Complex Analysis
Category:Bloch's Theorem
Block Copy Program

== Theorem ==
Let $k, m, n \in \N$ be natural numbers such that:
* $k \ge 1$;
* $\size {m - n} \ge k$.
The URM program defined as:
Line 		Command
$1$ 		$\map C {m, n}$
$2$ 		$\map C {m + 1, n + 1}$
$\vdots$		$\vdots$
$k$ 		$\map C {m + k - 1, n + k - 1}$
is called a block copy program.
It is abbreviated $\map C {m, n, k}$.
It has the effect of copying the contents of registers $R_m, R_{m + 1}, \ldots, R_{m + k - 1}$ into the registers $R_n, R_{n + 1}, \ldots, R_{n + k - 1}$ respectively.
It has length defined as $\map \lambda {\map C {m, n, k} } = k$.
== Proof ==
Immediate.
Category:URM Programs
Block Design/Examples/Arbitrary Example 1

== Example of Block Design ==
Consider a block design in which:
Then:
:$S = \set {1, 2, 3, 4}$
:each of the $B_r$ are the subsets of $S$ with exactly $3$ elements
:every subset of $S$ with exactly $2$ elements belong to exactly $\lambda = 2$ of the subsets $B_r$.
== Sources ==
* 
Category:Examples of Block Designs
Block Matrix/Examples/Arbitrary 1

== Example of Block Matrix ==
Let $\mathbf A = \sqbrk a_{m n}, \mathbf B = \sqbrk b_{m p}, \mathbf C = \sqbrk c_{r n}, \mathbf D = \sqbrk d_{r p}$.
The $\paren {m + r} \times \paren {n + p}$ block matrix $\mathbf M$: can then be created as:
:$\mathbf M = \begin {bmatrix}
\mathbf A & \mathbf B \\
\mathbf C & \mathbf D
\end {bmatrix}$
Category:Block Matrices
Block Matrix/Examples/Arbitrary 2

== Example of Block Matrix ==
Consider the matrices:
:$\mathbf A := \sqbrk \alpha_{m n}$: an $m \times n$ matrix
:$\mathbf B := \sqbrk \beta_{p m}$: a $p \times m$ matrix
over an arbitrary ring $R$.
Then:
:$\mathbf M = \begin {bmatrix}
\mathbf I_m & \mathbf A \\
\mathbf B & \mathbf 0
\end {bmatrix}$
is a matrix whose order is $\paren {m + p} \times \paren {m + n}$ where:
:$\mathbf I_m$ is the unit matrix of order $m$
:$\mathbf 0$ is the zero matrix of order $p \times n$.
$\mathbf M := \sqbrk \gamma_{\paren {m + p}, \paren {m + n} }$ can be represented as:
:$\gamma_{i j} = \begin {cases}
\delta_{i j} & : \tuple {i, j} \in \closedint 1 m \times \closedint 1 m \\
\alpha_{i, j - m} & : \tuple {i, j} \in \closedint 1 m \times \closedint {m + 1} {m + n} \\
\beta_{i - m, j} & : \tuple {i, j} \in \closedint {m + 1} {m + p} \times \closedint 1 m \\
0_R & : \tuple {i, j} \in \closedint {m + 1} {m + p} \times \closedint {m + 1} {m + n} \end {cases}$
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices
Category:Block Matrices
Block Matrix/Examples/Arbitrary 3

== Example of Block Matrix ==
The matrix of dimensions $4 \times 3$:
:$\paren {\begin {array} {rr|r} 1 & 2 & 3 \\ 1 & 0 & 1 \\ \hline -2 & 3 & -1 \\ 1 & 2 & 0 \end {array} }$
can be written as:
:$\mathbf M = \begin {bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end {bmatrix}$
where:
:$A_{11}$ and $A_{21}$ are $2 \times 2$ matrices
:$A_{12}$ and $A_{22}$ are $2 \times 1$ matrices.
== Sources ==
* 
Category:Block Matrices
Block Triangular Matrix/Examples/Arbitrary Example 1

== Examples of Block Triangular Matrices ==
This $4 \times 4$ square matrix is an example of a $2 \times 2$ block upper triangular matrix wth $2 \times 2$ submatrices:
:$\paren {\begin {array} {rr|rr} 1 & 2 & 5 & 4 \\ 1 & 1 & 1 & 2 \\ \hline 0 & 0 & -1 & 1 \\ 0 & 0 & 0 & 1 \end {array} }$
which is not an upper triangular matrix.
== Sources ==
* 
Category:Examples of Block Triangular Matrices
Boadicea/Linguistic Note

== Linguistic Note on ==
Yes, I know this is not about mathematics. It's far more important than that.
The name Boadicea is also presented as Boudicca or Boudica. Other spellings abound.
Its pronunciation is correctly Bo-dee-ca.
The name derives from the old English word for victory.
Hence the modern form of Boadicea or Boudicca can be argued as being Victoria.
Beware that in general usage there exists the incorrect pronunciation: Bow-er-di-see-a.
The explanation is that Boadicea is the Roman form, and Boudicca the English form.
In fact, the spelling Boadicea is the result of the medieval miscopying of progressive incorrect transliterations.
Hence it is nonsense to pronounce it as it is spelt from the standpoint of modern English.
Some prefer the pronunciation Boo-dee-cha, but while gloriously euphonious, it is unfortunately (probably) incorrect.
Category:Linguistic Notes
Body behaves as Particle under Gravitation

== Theorem ==
A body of finite volume under a gravitational force behaves identically to a point mass under the same gravitational force.
== Proof ==
== Historical Note ==
== Sources ==
* : Chapter $\text{VI}$: On the Seashore
* : Chapter $\text {B}.25$: Kepler's Laws and Newton's Law of Gravitation
Category:Gravitation
Body behaves as Particle under Gravitation/Historical Note

== Historical Note on Body behaves as Particle under Gravitation ==
The fact that a Body behaves as Particle under Gravitation was proved by after considerable work.
Some sources suggest that it took some $20$ years to resolve this, and this is what explains the delay in publishing his work.
Its truth justifies the mathematical derivation of Newton's Law of Universal Gravitation.
== Sources ==
* : Chapter $\text{VI}$: On the Seashore
* : Chapter $\text {B}.25$: Kepler's Laws and Newton's Law of Gravitation
Category:Historical Notes
Body in Free Fall moves in Parabolic Path

== Theorem ==
A body in free fall above the surface of the Earth follows a path approximating to a parabola.
== Proof ==
== Historical Note ==
== Sources ==
* : Chapter $\text{II}$: Modern Minds in Ancient Bodies
* : $\text {I}$: Motion, Sound and Heat: Chapter $4$: Gravitation: Combination of Forces
* : Chapter $8$: The System of the World: Galileo
Category:Mechanics
Body in Free Fall moves in Parabolic Path/Historical Note

== Historical Note on Body in Free Fall moves in Parabolic Path ==
According to tradition, Body in Free Fall moves in Parabolic Path was noted by by observation and reasoning.
However, it appears that in fact it was his pupil who first published it.
Because it upset to have had his pupil take that last step, agreed that the result should be accredited to himself, and so offered to withdraw his own publication.
It was confirmed mathematically by by application of his laws of motion.
== Sources ==
* : Chapter $\text {A}.14$: Cavalieri ()
* : $\text {I}$: Motion, Sound and Heat: Chapter $4$: Gravitation: Combination of Forces
* : Chapter $8$: The System of the World: Galileo
Category:Historical Notes
Body under Constant Acceleration

== Theorem ==
Let $B$ be a body under constant acceleration $\mathbf a$.
The following equations apply:
=== $(1):$ Velocity after Time ===
=== $(2):$ Distance after Time ===
=== $(3):$ Velocity after Distance ===
where:
:$\mathbf u$ is the velocity at time $t = 0$
:$\mathbf v$ is the velocity at time $t$
:$\mathbf s$ is the displacement of $B$ from its initial position at time $t$
:$\cdot$ denotes the dot product.
== Also known as ==
== Sources ==
* 
Category:Body under Constant Acceleration
Category:Kinematics
Category:Mechanics
Body under Constant Acceleration/Distance after Time

== Theorem ==
Let $B$ be a body under constant acceleration $\mathbf a$.
Then:
:$\mathbf s = \mathbf u t + \dfrac {\mathbf a t^2} 2$
where:
:$\mathbf s$ is the displacement of $B$ from its initial position at time $t$
:$\mathbf u$ is the velocity at time $t = 0$.
== Proof ==
From Body under Constant Acceleration: Velocity after Time:
:$\mathbf v = \mathbf u + \mathbf a t$
By definition of velocity, this can be expressed as:
:$\dfrac {\d \mathbf s} {\d t} = \mathbf u + \mathbf a t$
where both $\mathbf u$ and $\mathbf a$ are constant.
By Solution to Linear First Order Ordinary Differential Equation:
:$\mathbf s = \mathbf c + \mathbf u t + \dfrac {\mathbf a t^2} 2$
where $\mathbf c$ is a constant vector.
We are (implicitly) given the initial condition:
:$\bigvalueat {\mathbf s} {t \mathop = 0} = \mathbf 0$
from which it follows immediately that:
:$\mathbf s = \mathbf u t + \dfrac {\mathbf a t^2} 2$
== Also presented as ==
== Sources ==
* : Chapter $\text I$: Functions and their Graphs: $1$. Constants and Variables
* 
Category:Body under Constant Acceleration
Body under Constant Acceleration/Distance after Time/Also presented as

== Body under Constant Acceleration: Distance after Time: Also presented as == 
Some sources present Body under Constant Acceleration: Distance after Time in the form:
:$\mathbf s = \mathbf s_0 + \mathbf v_0 t + \dfrac {\mathbf a t^2} 2$
where:
:$\mathbf s_0$ is the displacement of $B$ at time $t = 0$
:$\mathbf v_0$ is the velocity of $B$ at time $t = 0$.
== Sources ==
* 
Category:Body under Constant Acceleration
Body under Constant Acceleration/Velocity after Distance

== Theorem ==
Let $B$ be a body under constant acceleration $\mathbf a$.
Then:
:$\mathbf v \cdot \mathbf v = \mathbf u \cdot \mathbf u + 2 \mathbf a \cdot \mathbf s$
where:
:$\mathbf v$ is the velocity at time $t$
:$\mathbf u$ is the velocity at time $t = 0$
:$\mathbf s$ is the displacement of $B$ from its initial position at time $t$
:$\cdot$ denotes the dot product.
== Proof ==
From Body under Constant Acceleration: Velocity after Time
:$\mathbf v = \mathbf u + \mathbf a t$
Then:
From Body under Constant Acceleration: Distance after Time:
:$\mathbf s = \mathbf u t + \dfrac {\mathbf a t^2} 2$
Substituting for $\mathbf s$ in $(1)$ gives:
:$\mathbf v \cdot \mathbf v = \mathbf u \cdot \mathbf u + 2 \mathbf a \cdot \mathbf s$
and the proof is complete.
Category:Body under Constant Acceleration
Body under Constant Acceleration/Velocity after Time

== Theorem ==
Let $B$ be a body under constant acceleration $\mathbf a$.
Then:
:$\mathbf v = \mathbf u + \mathbf a t$
where:
:$\mathbf u$ is the velocity at time $t = 0$
:$\mathbf v$ is the velocity at time $t$.
== Proof ==
By definition of acceleration:
:$\dfrac {\d \mathbf v} {\d t} = \mathbf a$
By Solution to Linear First Order Ordinary Differential Equation:
:$\mathbf v = \mathbf c + \mathbf a t$
where $\mathbf c$ is a constant vector.
We are given the initial condition:
:$\bigvalueat {\mathbf v} {t \mathop = 0} = \mathbf u$
from which it follows immediately that:
:$\mathbf v = \mathbf u + \mathbf a t$
== Sources ==
* 
Category:Body under Constant Acceleration
Bolzano-Weierstrass Theorem

== Theorem ==
Every bounded sequence of real numbers has a convergent subsequence.
=== General Form ===
== Proof 1 ==
== Proof 2 ==
== Also known as ==
== Also see ==
* Heine-Borel Theorem
== Historical Note ==
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 5$: Limits: Exercise $5$
Category:Limits of Sequences
Category:Real Analysis
Category:Bolzano-Weierstrass Theorem
Bolzano-Weierstrass Theorem/Also known as

== Bolzano-Weierstrass Theorem: Also known as ==
Some sources refer to the Bolzano-Weierstrass Theorem as the Weierstrass-Bolzano Theorem.
It is also known as Weierstrass's Theorem, but that name is also applied to a completely different result.
Category:Bolzano-Weierstrass Theorem
Bolzano-Weierstrass Theorem/Historical Note

== Historical Note on Bolzano-Weierstrass Theorem ==
The Bolzano-Weierstrass Theorem is a crucial property of the real numbers discovered independently by both and during their work on putting real analysis on a rigorous logical footing.
It was originally referred to as Weierstrass's Theorem until 's thesis on the subject was rediscovered.
== Sources ==
* : Chapter $\text {A}.33$: Weierstrass ()
Category:Historical Notes