nyuuzyou/wikis huggingface dataset:
Piecewise Continuous Function with One-Sided Limits is Bounded

== Theorem ==
Let $f$ be a real function defined on a closed interval $\closedint a b$.
Let $f$ be a piecewise continuous function with one-sided limits.
Then $f$ is a bounded piecewise continuous function.
== Proof ==
Let $f$ be a piecewise continuous function with one-sided limits.
By definition, there exists a finite subdivision $\set {x_0, x_1, \ldots, x_n}$ of $\closedint a b$, where $x_0 = a$ and $x_n = b$, such that $f$ is continuous on $\openint {x_{i − 1} } {x_i}$ for every $i \in \set {1, 2, \ldots, n}$.
For every $i \in \set {1, 2, \ldots, n}$, we define a function $f_i$ with domain $\closedint {x_{i − 1} } {x_i}$, as follows:
$\quad\map {f_i} x := \begin{cases}
\ds \lim_{x \mathop \to x_{i − 1}^+} \map f x & : x = x_{i - 1} \\
\map f x & : x \in \openint {x_{i - 1} } {x_i} \\
\ds \lim_{x \mathop \to x_i^-} \map f x & : x = x_i
\end{cases}$
The one-sided limits in this definition exist because $f$ is piecewise continuous with one-sided limits.
We have that $f$ is continuous on $\openint {x_{i - 1} } {x_i}$.
We also have that $f_i = f$ on $\openint {x_{i - 1} } {x_i}$.
Therefore $f_i$ is also continuous on $\openint {x_{i - 1} } {x_i}$.
By definition, $f_i$ is right-continuous at $x_{i − 1}$ and left-continuous at $x_i$.
Therefore, $f_i$ is continuous throughout its domain $\closedint {x_{i − 1} } {x_i}$.
By Continuous Function on Compact Subspace of Euclidean Space is Bounded and Closed Real Interval is Compact in Metric Space, $f_i$ is bounded.
Therefore, $f_i$ is bounded on $\closedint {x_{i − 1} } {x_i}$.
We have that $\openint {x_{i - 1} } {x_i}$ constitutes a subset of $\closedint {x_{i − 1} } {x_i}$.
Thus $f_i$ is also bounded on $\openint {x_{i - 1} } {x_i}$.
As $f_i = f$ on $\openint {x_{i - 1} } {x_i}$, $f$ is also bounded on $\openint {x_{i - 1} } {x_i}$.
We have that:
:$f$ is bounded on the intervals $\openint {x_{i - 1} } {x_i}$
and
:the set of these intervals is finite.
Hence the set of bounds of $f$ on these intervals is itself bounded.
The bound of this set of bounds serves as a bound for $f$ on each of the intervals $\openint {x_{i - 1} } {x_i}$.
Therefore, this bound is a bound for $f$ on the union of these intervals.
That is, $f$ is bounded on the union of $\openint {x_{i - 1} } {x_i}$ for all $i \in \set {1, 2, \ldots, n}$.
The only points left to consider are the points in the set $\set {x_0, x_1, \ldots, x_n}$. 
Since this set is finite, the maximum $\map \max {\size {\map f {x_0} }, \size {\map f {x_1} }, \ldots, \size {\map f {x_n} } }$ is finite and serves as a bound for $f$ on $\set {x_0, x_1, \ldots, x_n}$.
Thus $f$ is bounded on the whole of its domain $\closedint a b$.
Hence $f$ is a bounded piecewise continuous function.
== Sources ==
* : Chapter Two: $\S 1$. Piecewise-Continuous Functions
Category:Piecewise Continuous Functions
Piecewise Continuous Function with One-Sided Limits is Darboux Integrable

== Theorem ==
Let $f$ be a real function defined on a closed interval $\closedint a b$.
Let $f$ be piecewise continuous with one-sided limits on $\closedint a b$.
Then $f$ is Darboux integrable on $\closedint a b$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter Two: $\S 1$. Piecewise-Continuous Functions
Category:Piecewise Continuous Functions
Category:Integral Calculus
Category:Piecewise Continuous Function with One-Sided Limits is Darboux Integrable
Piecewise Continuous Function with One-Sided Limits is Darboux Integrable/Proof 1

== Theorem ==
== Proof ==
We are given that $f$ is piecewise continuous with one-sided limits on $\closedint a b$.
From Piecewise Continuous Function with One-Sided Limits is Bounded, $f$ is a bounded piecewise continuous function.
The result follows from Bounded Piecewise Continuous Function is Darboux Integrable.
Category:Piecewise Continuous Function with One-Sided Limits is Darboux Integrable
Piecewise Continuous Function with One-Sided Limits is Darboux Integrable/Proof 2

== Theorem ==
== Proof ==
We are given that $f$ is piecewise continuous with one-sided limits on $\closedint a b$.
Therefore, there exists a finite subdivision $\set {x_0, x_1, \ldots, x_n}$ of $\closedint a b$, where $x_0 = a$ and $x_n = b$, such that for all $i \in \set {1, 2, \ldots, n}$:
:$f$ is continuous on $\openint {x_{i - 1} } {x_i}$ 
:$\ds \lim_{x \mathop \to {x_{i - 1} }^+} \map f x$ and $\ds \lim_{x \mathop \to {x_i}^-} \map f x$ exist.
Note that $n$ is the number of intervals $\openint {x_{i - 1} } {x_i}$ defined from the (finite) subdivision $\set {x_0, x_1, \ldots, x_n}$.
We shall use proof by induction on the intervals.
For all $k \in \set {1, 2, \ldots, n}$, let $\map P k$ be the proposition:
:$f$ is Darboux integrable on $\closedint {x_0} {x_k}$.
=== Basis for the Induction ===
$\map P 1$ is the case:
:$f$ is Darboux integrable on $\closedint {x_{i - 1} } {x_i}$
for an arbitrary $i \in \set {1, 2, \ldots, k}$. 
Piecewise continuity with one-sided limits of $f$ for the case $n = 1$ means that:
:$f$ is continuous on $\openint {x_{i - 1} } {x_i}$
:$\ds \lim_{x \mathop \to {x_{i - 1} }^+} \map f x$ and $\ds \lim_{x \mathop \to {x_i}^-} \map f x$ exist.
By Integrability Theorem for Functions Continuous on Open Intervals, $f$ is Darboux integrable on $\closedint {x_{i - 1} } {x_i}$.
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$f$ is Darboux integrable on $\openint {x_0} {x_k}$.
from which it is to be shown that:
:$f$ is Darboux integrable on $\openint {x_0} {x_{k + 1} }$.
=== Induction Step ===
This is the induction step:
By definition of a piecewise continuous function with one-sided limits, for every $i \in \set {1, 2, \ldots, k, k + 1}$:
:$f$ is continuous on $\openint {x_{i - 1} } {x_i}$
:the one-sided limits $\ds \lim_{x \mathop \to x_{i - 1}^+} \map f x$ and $\ds \lim_{x \mathop \to x_i^-} \map f x$ exist.
By the induction hypothesis, $f$ is Darboux integrable on $\closedint {x_0} {x_k}$.
From the basis for the induction, $f$ is Darboux integrable on $\closedint {x_k} {x_{k + 1} }$.
We have that $f$ is Darboux integrable on $\closedint {x_0} {x_k}$ and $\closedint {x_k} {x_{k + 1} }$.
Therefore, by Existence of Integral on Union of Adjacent Intervals:
:$f$ is Darboux integrable on $\closedint {x_0} {x_k} \cup \closedint {x_k} {x_{k + 1} }$.
We have that:
:$\closedint {x_0} {x_{k + 1} } = \closedint {x_0} {x_k} \cup \closedint {x_k} {x_{k + 1} }$
Accordingly, $f$ is Darboux integrable on $\closedint {x_0} {x_{k + 1} }$.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$f$ is Darboux integrable on $\closedint a b$.
Category:Piecewise Continuous Function with One-Sided Limits is Darboux Integrable
Piecewise Continuous Function with One-Sided Limits is Uniformly Continuous on Each Piece

== Theorem ==
Let $f$ be a real function defined on a closed interval $\closedint a b$.
Let $f$ be piecewise continuous with one-sided limits:
Then:
:for all $i \in \set {1, 2, \ldots, n}$, $f$ is uniformly continuous on $\openint {x_{i − 1} } {x_i}$.
== Proof ==
We have that $f$ is continuous on $\openint {x_{i − 1} } {x_i}$ for every $i \in \set {1, 2, \ldots, n}$.
Since $f$ is piecewise continuous with one-sided limits, the one-sided limits:
:$\ds \lim_{x \mathop \to {x_{i−1} }^+} \map f x$
and:
:$\ds \lim_{x \mathop \to {x_i}^-} \map f x$
exist for every $i \in \set {1, 2, \ldots, n}$.
Thus, from Extendability Theorem for Function Continuous on Open Interval, a function $f_i$ exists that satisfies:
:$f_i$ is defined on $\closedint {x_{i − 1} } {x_i}$
:$f_i$ is continuous on $\closedint {x_{i − 1} } {x_i}$
:$f_i$ equals $f$ on $\openint {x_{i − 1} } {x_i}$
for an arbitrary $i$ in $\set {1, 2, \ldots, n}$.
From Continuous Function on Closed Real Interval is Uniformly Continuous, $f_i$ is uniformly continuous on $\closedint {x_{i − 1} } {x_i}$.
We have that $\openint {x_{i − 1} } {x_i}$ is a subset of $\closedint {x_{i − 1} } {x_i}$.
Hence $f_i$ is uniformly continuous on $\openint {x_{i − 1} } {x_i}$.
As $f_i$ equals $f$ on $\openint {x_{i − 1} } {x_i}$, $f$ also is uniformly continuous on $\openint {x_{i − 1} } {x_i}$.
Since $i$ is arbitrary, this results holds for all $i \in \set {1, 2, \ldots, n}$.
Category:Piecewise Continuous Functions
Piecewise Continuously Differentiable Function/Definition 2 is Continuous

== Theorem ==
Let $f$ be a real function defined on a closed interval $\closedint a b$.
Let $f$ satisfy the definition Piecewise Continuously Differentiable Function/Closed Intervals.
Then $f$ is continuous.
== Proof ==
$f$ satisfies the conditions of Piecewise Continuously Differentiable Function/Closed Intervals.
Therefore, there exists a finite subdivision $\set {x_0, \ldots, x_n}$ of $\closedint a b$, $x_0 = a$ and $x_n = b$, such that $f$ is continuously differentiable on $\closedint {x_{i - 1} } {x_i}$ for every $i \in \set {1, \ldots, n}$.
Here, the derivatives at $x_{i - 1}$ and $x_i$ are understood as one-sided derivatives.
$f$ is differentiable on $\closedint {x_{i - 1} } {x_i}$ for every $i \in \set {1, \ldots, n}$.
Therefore, by Differentiable Function is Continuous, $f$ is continuous at every point of $\closedint {x_{i - 1} } {x_i}$ for every $i \in \set {1, \ldots, n}$.
Here, the continuities at $x_{i - 1}$ and $x_i$ are one-sided.
We use this result to go through every point of every $\closedint {x_{i - 1} } {x_i}$ in order to establish the continuity of $f$ there:
$f$ is continuous on $\closedint {x_{i - 1} } {x_i}$ for every $i \in \set {1, \ldots, n}$.
$\openint {x_{i - 1} } {x_i}$ is a subset of $\closedint {x_{i - 1} } {x_i}$ for every $i \in \set {1, \ldots, n}$.
Therefore, $f$ is continuous at every point of $\openint {x_{i - 1} } {x_i}$ for every $i \in \set {1, \ldots, n}$.
$x_0$ ($= a$) is the leftmost point in the domain of $f$.
$f$ is right-continuous at $x_0$.
Therefore, $f$ is continuous at $x_0$.
$x_n$ ($= b$) is the rightmost point in the domain of $f$.
$f$ is left-continuous at $x_n$.
Therefore, $f$ is continuous at $x_n$.
$f$ is continuous at every point $x_i$, $i \in \set {1, \ldots, n - 1}$, because $f$ is both left- and right-continuous at these points.
All in all, we have found that $f$ is continuous at every point of $\closedint {x_{i - 1} } {x_i}$ for every $i \in \set {1, \ldots, n}$.
$f$ is therefore continuous at every point of the union of those intervals: $\ds \bigcup_{i \mathop = 1}^n \closedint {x_{i - 1}} {x_i}$.
$\ds \bigcup_{i \mathop = 1}^n \closedint {x_{i - 1}} {x_i}$ equals $\closedint a b$, the domain of $f$.
Therefore, $f$ is continuous as $f$ is continuous at every point of its domain.
Category:Real Analysis
Piecewise Smooth Vector Field along Admissible Curve is Variation Field

== Theorem ==
Let $M$ be a smooth manifold.
Let $\gamma$ be an admissible curve on $M$.
Let $V$ be a piecewise smooth vector field along $\gamma$.
Then $V$ is the variation field of some variation of $\gamma$.
== Proof ==
== Sources ==
* : $\S 6$: Geodesics and Distance. Geodesics and Minimizing Curves
Category:Smooth Manifolds
Category:Vector Fields
Pitch of Helix/Examples/Circular Helix

== Example of Pitch of Helix ==
Consider the circular helix $\HH$ whose parametric equation is given as:
:$\begin {cases} x & = a \cos t \\ y & = a \sin t \\ z & = b t \end {cases}$
The pitch of $\HH$ is $\dfrac {2 \pi} b$.
== Proof ==
== Sources ==
* 
* 
Category:Examples of Pitch of Helix
Pizza Party Puzzle

== Puzzle ==
It was Pizza Party Night, and as usual Ariel, Brandon, Charlie and Delia had gathered in Charlie's apartment to watch movies and order one of the biggest pizzas ever.
They always ordered the same one, with the same toppings, which the gang's resident mathematician Charlie then split expertly with his pizza-cutter so as to distribute it evenly between the four of them:
:
Until one fateful night when the pizza company changed the shape of the pizza.
No longer was it round, it was oval.
Charlie then had the problem: at what angle to each other should the cuts across the pizza be, so that they all had identically-sized pieces?
:
:Charlie: "Hey guys, got a bit of a problem."
:Delia: "What's up, your pizza cutter broken?"
:Charlie: "No, it's just that I can't work out how to cut the pizza."
:Brandon: "Surely as a mathematician you can work it out. It's a simple ellipse, isn't it."
:Charlie: "Can't be certain it is actually a precise ellipse. It's oval, and it's symmetrical and all that, but unless I analyse exactly what shape of oval, I can't be completely certain exactly where to make the cuts so as to make sure we all get the same."
:Ariel: "Honestly, before I met you, Charlie, I would never have believed anyone could be so uselessly bone-headedly idiotically clueless. Good grief. Give me the stupid cutter."
And having worked out the position of the dead centre of the pizza, with two neat, straight cuts she deftly sliced the pizza into four parts which the rest of the gang acknowledged were, indeed, as far as they could tell, pretty much equal.
=== Click here for the solution ===
== Also see ==
* Second Pizza Theorem
== Historical Note ==
Category:Puzzles
Place where Sun is Closer at Noon

== Problem ==
:There is a certain village in the ,
:situated in a very low valley,
:and yet the sun is nearer to the inhabitants thereof,
:every noon by $3000$ miles and upwards,
:than when he either rises or sets,
:to those of the said village.
== Solution ==
By the physics of its rotation Earth, every point of Earth is closer to the sun at noon by the radius of the line of latitude on which that point is situated.
In the , that line of latitude happens to have a radius of $3000$ miles.
The specifics of the location of the puzzle are presented purely in order to mislead the person who is being asked to solve the puzzle.
== Sources ==
* : Geographical Paradoxes
* : Rational Amusements for Winter Evenings: $161$
Category:Geography Puzzles
Category:Trick Puzzles
Place where Wind always Blows from North

== Problem ==
:There is a particular place on Earth, where the winds (although frequently veering around the compass) always blow from the north.
== Solution ==
That point is the South Pole.
Every direction from the South Pole is north.
== Also see ==
* Man looking South in All Directions, of which this is a variant.
== Sources ==
* : Geographical Paradoxes
* : Rational Amusements for Winter Evenings: $160$
Category:Geography Puzzles
Category:Man looking South in All Directions
Place whose Visibility Changes with Tide

== Problem ==
:There is a certain place in , where, when the tide is in,
:the sheep may be seen feeding on a certain neighbouring island;
:yet, when the tide is out, and the water at the lowest,
:not one can be seen, though they be feeding there at the same instant.
== Solution ==
:The place may be the wharf at , the over against it,
:and the appearance caused by refraction, when the water is high.
== Sources ==
* : Geographical Paradoxes
* : Rational Amusements for Winter Evenings
Category:Geography Puzzles
Category:Physics
Plancherel Theorem

== Theorem ==
Let $n \in \Z_{>0}$.
Let $\map \BB {\R^n}$ be the Borel $\sigma$-algebra on $\R^n$.
Let $\lambda^n$ be the Lebesgue measure on $\R^n$.
For each $p \in \closedint 1 \infty$, let:
:$\map {L^p} {\R^n} := \map {L^p} {\R^n, \map \BB {\R^n}, \lambda ^n}$
be the $L^p$ space.
For all $f \in \map {L^1} {\R^n} \cap \map {L^2} {\R^n}$, we have:
:$\norm {\map \FF f}_2 = \norm {\map {\FF ^{-1} } f}_2 = \norm f_2$
where:
:$\map \FF f$ is the Fourier transform of $f$
:$\map {\FF ^{-1} } f$ is the inverse Fourier transform of $f$
:$\norm \cdot_2$ denotes the $L^2$ norm
== Proof ==
== Sources ==
* : $4.3.$ Transform Methods
Plane Angle between Lines from Center to Vertices of Tetrahedron

== Theorem ==
Let $T$ be a regular tetrahedron.
Let $L_1$ and $L_2$ be straight lines constructed from the center of $T$ to two of the vertices of $T$.
Then the plane angle between $L_1$ and $L_2$ is apppproximately equal to $108 \degrees \, 28 \minutes$.
This angle is sometimes referred to as the tetrahedral angle.
== Proof ==
== Sources ==
* 
* 
Category:Tetrahedra
Plane Curve Classification Theorem

== Theorem ==
Let $I = \closedint a b \subseteq \R$ be a closed real interval.
Let $\gamma, \tilde \gamma: I \to \R^2$ be smooth unit-speed curves on a plane.
Let $N$ and $\tilde N$ be their normal vector fields.
Let $\map {\kappa_N} t$ and $\map {\kappa_{\tilde N}} t$ be curvatures at $\map \gamma t$ and $\map {\tilde \gamma} t$ respectively.
Then $\gamma$ and $\tilde \gamma$ are congruent by a direction-preserving congruence iff:
:$\forall t \in I : \map {\kappa_N} t = \map {\kappa_{\tilde N}} t$
== Proof ==
== Sources ==
* : $\S 1$: What Is Curvature? The Euclidean Plane
Category:Curves
Category:Curvature
Plane Reflection is Involution

== Theorem ==
Let $M$ be a straight line in the plane passing through the origin.
Let $s_M$ be the reflection of $\R^2$ in $M$.
Then $s_M$ is an involution in the sense that:
:$s_M \circ s_M = I_{\R^2}$
where $I_{\R^2}$ is the identity mapping on $\R_2$.
That is:
:$s_M = {s_M}^{-1}$
== Proof ==
Let the angle between $M$ and the $x$-axis be $\alpha$.
Let $P = \tuple {x, y}$ be an arbitrary point in the plane.
Then from Equations defining Plane Reflection:
:$\map {s_M} P = \tuple {x \cos 2 \alpha + y \sin 2 \alpha, x \sin 2 \alpha - y \cos 2 \alpha}$
Thus:
Hence the result.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Example $28.4$
Category:Geometric Reflections
Category:Involutions
Plane Reflection is Space Rotation

== Theorem ==
Let $M$ be a straight line in the plane passing through the origin.
Let $s_M$ be the reflection of $\R^2$ in $M$.
Then $s_M$ is the rotation of the plane in space through one half turn about $M$ as an axis.
== Proof ==
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Example $28.4$
Category:Geometric Reflections
Category:Geometric Rotations
Plane Section of Hyperboloid of Revolution Parallel to Axis of Rotation is Hyperbola

== Theorem ==
Let $\HH$ be a hyperboloid of revolution.
Let $\SS$ be a plane section of $\HH$ taken parallel to the axis of rotation of $\HH$.
Then $\SS$ is a hyperbola.
== Proof ==
== Also see ==
* Cross-Section of Hyperboloid of Revolution is Circle
== Sources ==
* 
* 
Category:Hyperboloids of Revolution
Plane Section of Spheroid Perpendicular to Axis of Revolution is Circle

== Theorem ==
Let $\SS$ be a spheroid.
Let $\PP$ be a plane section of $\SS$ such that $\PP$ is perpendicular to the axis of revolution of $\SS$.
Then $\PP$ is a circle.
== Proof ==
A spheroid is defined as the solid of revolution formed by rotation of an ellipse about one of its axes.
Hence the result by definition of solid of revolution.
== Sources ==
* 
* 
Category:Spheroids
Plane contains Infinite Number of Lines

== Theorem ==
A plane contains an infinite number of distinct lines.
== Proof ==
A plane contains an infinite number of points.
Not all these points are collinear.
Let $A$, $B$ and $C$ be points in a plane $P$.
From Propositions of Incidence: Line in Plane, any two of these points determine a line.
Consider the lines $AB$, $AC$ and $BC$, all of which are distinct.
Let $X$ be one of the infinite number of points on $BC$ which is not $B$ or $C$.
Then $AX$ is a line in $P$ which is distinct from both $AB$ and $AC$.
As there is are infinite number of points on $BC$, there are an infinite number of lines incident to $A$ and $BC$.
All these lines are in $P$.
Hence the result.
== Sources ==
* : Chapter $1$: Introduction: The Propositions of Incidence: $1.2$: The projective method: The propositions of incidence
Category:Projective Geometry
Plane through Straight Line Perpendicular to other Plane is Perpendicular to that Plane

== Theorem ==
== Proof ==
:
Let $AB$ be an arbitrary straight line which is perpendicular to the plane of reference.
It is to be demonstrated that every plane holding $AB$ is perpendicular to the plane of reference.
Let an arbitrary plane $DE$ be drawn through $AB$.
Let $CE$ be the common section of the plane $DE$ and the plane of reference.
Let $F$ be an arbitrary point on $CE$.
From :
:let $FG$ be drawn in $DE$ perpendicular to $CE$.
We have that $AB$ is perpendicular to the plane of reference.
Therefore from : 
:$AB$ is perpendicular to all the straight lines which meet it and are in the plane of reference.
Therefore $AB$ is perpendicular to $CE$.
Therefore $\angle ABF$ is a right angle.
But $\angle GFB$ is also a right angle.
Therefore by :
:$AB \parallel FG$
But $AB$ is perpendicular to the plane of reference.
Therefore from :
: $FG$ is perpendicular to the plane of reference.
Thus the conditions are fulfilled for to apply:
:$DE$ is perpendicular to the plane of reference.
As $DE$ is arbitrary, the result follows.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Perpendiculars
Category:Planes
Planes Perpendicular to same Straight Line are Parallel

== Theorem ==
== Proof ==
:
Let $AB$ be a straight line which is perpendicular to each of the planes $CD$ and $EF$.
It is to be demonstrated that $CD$ and $EF$ are parallel.
Suppose, to the contrary, that $CD$ and $EF$ are not parallel.
Then when produced they will meet.
From :
:let the common section be the straight line $GH$.
Let $K$ be an arbitrary point on $GH$.
Let $AK$ and $BK$ be joined.
We have that $AB$ is perpendicular to $EF$.
So from :
:$AB$ is perpendicular to the straight line $BK$ in the planes $EF$ produced.
Therefore $\angle ABK$ is a right angle.
For the same reason $\angle BAK$ is a right angle.
Thus, in $\triangle ABK$, there are two angles which are right angles.
From this is impossible.
Therefore $CD$ and $EF$ do not meet when produced.
So from :
:$CD$ and $EF$ are parallel.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Parallel Planes
Category:Perpendiculars
Planes are Subspaces of Space

== Theorem ==
The two-dimensional subspaces of $\R^3$ are precisely the homogeneous planes of solid analytic geometry.
== Proof ==
Follows directly from Equivalent Statements for Vector Subspace Dimension One Less.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations
Category:Linear Algebra
Category:Solid Analytic Geometry
Planes through Parallel Pairs of Meeting Lines are Parallel

== Theorem ==
== Proof ==
:
Let $AB$ and $BC$ be two straight lines which meet each other at $B$.
Let $DE$ and $EF$ be two straight lines which meet each other at $E$ such that:
:$AB$ is parallel to $DE$
:$BC$ is parallel to $EF$.
It is to be demonstrated that the plane holding $AB$ and $BC$ is parallel to the plane holding $DE$ and $EF$.
From :
:let $BG$ be drawn from $B$ perpendicular to the plane holding $DE$ and $EF$.
From :
:let $GH$ be drawn from $G$ parallel to $ED$
and
:let $GK$ be drawn from $G$ parallel to $EF$.
We have that $BG$ is perpendicular to the plane holding $DE$ and $EF$.
Therefore from : 
:$BG$ is perpendicular to all the straight lines which meet it and are in the plane holding $DE$ and $EF$.
But each of $GH$ and $GK$ meets $BG$ and is in the plane holding $DE$ and $EF$.
Therefore each of $\angle BGH$ and $\angle BGK$ is a right angle.
From :
:$BA$ is parallel to $GH$.
Therefore from :
:$\angle GBA$ and $\angle BGH$ equal two right angles.
But $\angle BGH$ is a right angle.
Therefore $\angle GBA$ is a right angle.
Therefore $GB$ is perpendicular to $BA$.
For the same reason, $GB$ is also perpendicular to $BC$.
We have that the straight line $GB$ is set up perpendicular to the two straight lines $BA$ and $BC$ which cut one another.
Therefore from :
:$GB$ is perpendicular to the plane holding $AB$ and $BC$.
But from :
:the plane holding $AB$ and $BC$ is parallel to the plane holding $DE$ and $EF$.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Parallel Planes
Plastic Constant is Smallest Pisot-Vijayaraghavan Number/Historical Note

== Historical Note on Plastic Constant is Smallest Pisot-Vijayaraghavan Number ==
The plastic constant was identified by in $1944$ as being the smallest known Pisot-Vijayaraghavan number.
Later that year showed that it was also the smallest possible.
== Sources ==
* 
* 
* : $1,32471 795 \ldots$
* 
Category:Pisot-Vijayaraghavan Numbers