nyuuzyou/wikis huggingface dataset:
Scalar Multiplication Corresponds to Multiplication by 1x1 Matrix

== Theorem ==
Let $\map \MM 1$ denote the matrix space of square matrices of order $1$.
Let $\map \MM {1, n}$ denote the matrix space of order $1 \times n$.
Let $\mathbf A = \begin {pmatrix} a \end {pmatrix} \in \map \MM 1$ and $\mathbf B = \begin {pmatrix} b_1 & b_2 & \cdots & b_n \end{pmatrix} \in \map \MM {1, n}$.
Let $\mathbf C = \mathbf A \mathbf B$ denote the (conventional) matrix product of $\mathbf A$ with $\mathbf B$.
Let $\mathbf D = a \mathbf B$ denote the matrix scalar product of $a$ with $\mathbf B$.
Then $\mathbf C = \mathbf D$.
== Proof ==
By definition of (conventional) matrix product, $\mathbf C$ is of order $1 \times n$.
By definition of matrix scalar product, $\mathbf D$ is also of order $1 \times n$.
Consider arbitrary elements $c_i \in \mathbf C$ and $d_i \in \mathbf D$ for some index $i$ where $1 \le i \le n$.
We have:
and: 
== Sources ==
* : Chapter $\text I$: Definitions and Fundamental Operations of Matrices: $4$. Matrices, Row Vectors, Column Vectors, Scalars
Category:Matrix Scalar Product
Category:Conventional Matrix Multiplication
Scalar Multiplication by Minus One reverses Preordering in Vector Space

== Definition ==
Let $\GF \in \set {\R, \C}$.
Let $\tuple {X, \le}$ be a preordered vector space over $\GF$.
Let $x, y \in X$ be such that:
:$x \le y$
Then:
:$-y \le -x$ 
== Proof ==
From Characterization of Preordered Vector Spaces, there exists a convex cone $X^+$ such that for each $u, v \in X$:
:$u \le v$ $v - u \in X^+$.
Since $x \le y$ we have $y - x \in X^+$.
We can write $-x - \paren {-y} = y - x \in X^+$.
Hence we also obtain $-y \le -x$. 
Category:Preordered Vector Spaces
Scalar Multiplication by Zero gives Zero Vector

== Theorem ==
Let $\mathbf a$ be a vector quantity.
Let $0 \mathbf a$ denote the scalar product of $\mathbf a$ with $0$.
Then:
:$0 \mathbf a = \bszero$
where $\bszero$ denotes the zero vector.
== Proof ==
By definition of scalar product:
:$\size {0 \mathbf a} = 0 \size {\mathbf a}$
where $\size {\mathbf a}$ denotes the magnitude of $\mathbf a$.
Thus:
:$\size {0 \mathbf a} = 0$
That is: $0 \mathbf a$ is a vector quantity whose magnitude is zero.
Hence, by definition, $0 \mathbf a$ is the zero vector.
== Sources ==
* : $\S 22$: Fundamental Definitions: $2.$
Category:Zero Vectors
Category:Scalar Multiplication
Scalar Multiplication is Continuous in Weak Topology

== Theorem ==
Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$ with weak topology $w$. 
Define $m : K \times \struct {X, w} \to \struct {X, w}$ by: 
:$\map m {\lambda, x} = \lambda x$
for each $\lambda \in K$, $x \in X$. 
Then $m$ is continuous. 
That is, scalar multiplication remains continuous when restricting to the weak topology.
== Proof ==
Let $X^\ast$ be the topological dual space of $X$.
From Continuity in Initial Topology, it suffices to show that for each $f \in X^\ast$ we have: 
:$f \circ m : K \times \struct {X, w} \to K$ is continuous.
Define the projections $\pr_1 : K \times {X, w} \to \struct {X, w}$ and $\pr_2 : K \times {X, w} \to \struct {X, w}$ as the projection onto the first and second factors.
Then for each $\lambda \in K$ and $x \in X$ we have: 
That is: 
:$f \circ m = \pr_1 \cdot \paren {f \circ \pr_2}$
From the definition of the product topology:
:$\pr_1 : K \times \struct {X, w} \to K$
and:
:$\pr_2 : K \times \struct {X, w} \to \struct {X, w}$ 
are continuous.
From the definition of the weak topology:
:$f : \struct {X, w} \to K$ is continuous.
From Composite of Continuous Mappings is Continuous:
:$f \circ \pr_2 : K \times \struct {X, w} \to K$ is continuous.
From Product of Continuous Functions on Topological Ring is Continuous:
:$\pr_1 \cdot \paren {f \circ \pr_2} : K \times \struct {X, w} \to K$ is continuous.
So we have that:
:$f \circ m : K \times \struct {X, w} \to K$ is continuous for each $f \in X^\ast$.
So from Continuity in Initial Topology:
:$m : K \times \struct {X, w} \to \struct {X, w}$ is continuous in the weak topology. 
Category:Weak Topologies on Topological Vector Spaces
Scalar Multiplication of Vectors is Associative

== Theorem ==
Let $\mathbf a$ be a vector quantity.
Let $m, n$ be scalar quantities.
Then:
:$m \paren {n \mathbf a} = \paren {m n} \mathbf a = n \paren {m \mathbf a}$
== Proof ==
== Sources ==
* : Chapter $\text I$. Addition and Subtraction of Vectors. Centroids: Addition and Subtraction of Vectors: $5$. Multiplication by a number
* : Introduction: Vector Notation and Formulae
* : Chapter $\text I$: Vector Algebra: $\S 1$.
* : $\S 22$: Laws of Vector Algebra: $22.3$
* 
Category:Vectors
Category:Vector Algebra
Category:Scalar Multiplication
Scalar Multiplication of Vectors is Distributive over Scalar Addition

== Theorem ==
Let $\mathbf a$ be a vector quantity.
Let $m, n$ be scalar quantities.
Then:
:$\paren {m + n} \mathbf a = m \mathbf a + n \mathbf a$
== Proof ==
== Sources ==
* : Chapter $\text I$. Addition and Subtraction of Vectors. Centroids: Addition and Subtraction of Vectors: $5$. Multiplication by a number
* : Introduction: Vector Notation and Formulae
* : Chapter $\text I$: Vector Algebra: $\S 1$.
* : $\S 22$: Laws of Vector Algebra: $22.4$
* 
Category:Vectors
Category:Vector Algebra
Category:Scalar Multiplication
Category:Vector Addition
Scalar Multiplication of Vectors is Distributive over Vector Addition

== Theorem ==
Let $\mathbf a, \mathbf b$ be a vector quantities.
Let $m$ be a scalar quantity.
Then:
:$m \paren {\mathbf a + \mathbf b} = m \mathbf a + m \mathbf b$
== Proof ==
:
Let $\mathbf a = \vec {OP}$ and $\mathbf b = \vec {PQ}$.
Then:
:$\vec {OQ} = \mathbf a + \mathbf b$
Let $P'$ and $Q'$ be points on $OP$ and $OQ$ respectively so that:
:$OP' : OP = OQ' : OQ = m$
Then $P'Q'$ is parallel to $PQ$ and $m$ times it in length.
Thus:
:$\vec {P'Q'} = m \mathbf b$
which shows that:
== Sources ==
* : Chapter $\text I$. Addition and Subtraction of Vectors. Centroids: Addition and Subtraction of Vectors: $5$. Multiplication by a number
* : Introduction: Vector Notation and Formulae
* : Chapter $\text I$: Vector Algebra: $\S 1$.
* : $\S 22$: Laws of Vector Algebra: $22.5$
* 
Category:Vectors
Category:Vector Algebra
Category:Scalar Multiplication
Category:Vector Addition
Scalar Multiplication on Normed Vector Space is Continuous

== Theorem ==
Let $\struct { K, +, \circ_K }$ be a normed division ring with norm $\norm {\, \cdot \,}_K$.
Let $\struct {X, \norm {\, \cdot \,}_X }$ be a normed vector space over $K$.
Let $\struct {K \times X, \norm {\, \cdot \,}_P }$ be the direct product of $K$ and $X$ with the direct product norm $\norm {\, \cdot \,}_P$. 
Let $\circ : K \times X \to X$ be the scalar multiplication defined on $X$.
Then $\circ: K \times X \to X$ is a continuous mapping.
== Proof ==
Let $\tuple { \lambda_0 , x_0 } \in K \times X$.
Let $\epsilon' \in \R_{>0}$.
Set $\epsilon = \map \min { \epsilon', 1 }$
To show that $\circ$ is continuous, let $\tuple { \lambda , x } \in K \times X$ such that $\norm { \lambda_0 - \lambda }_K < \dfrac \epsilon { 1 + \norm { \lambda_0 }_K + \norm { x_0 }_X }$, and $\norm { x_0 - x }_X < \dfrac \epsilon { 1 + \norm { \lambda_0 }_K + \norm { x_0 }_X }$.
By definition of direct product norm, it follows that:
:$\norm { \tuple {\lambda_0, x_0} - \tuple {\lambda, x} }_P = \map \max {\norm {\lambda_0 - \lambda}_K, \norm {x_0 - x}_X } < \dfrac \epsilon { 1 + \norm { \lambda_0 }_K + \norm { x_0 }_X }$.
To show that $\circ$ is continuous at $\tuple {\lambda_0, x_0}$, we calculate:
It follows that $\circ$ is continuous from $K \times X$ to $X$.
== Sources ==
* : $\S 5$: Normed Spaces
Category:Continuous Mappings on Normed Vector Spaces
Category:Scalar Multiplication
Scalar Product of Magnitude by Unit Vector Quantity

== Theorem ==
Let $\mathbf a$ be a vector quantity.
Let $m$ be a scalar quantity.
Then:
:$m \mathbf a = m \paren {\size {\mathbf a} \hat {\mathbf a} } = \paren {m \size {\mathbf a} } \hat {\mathbf a}$
where:
:$\size {\mathbf a}$ denotes the magnitude of $\mathbf a$
:$\hat {\mathbf a}$ denotes the unit vector in the direction $\mathbf a$.
== Proof ==
Then:
== Sources ==
* : Chapter $\text I$. Addition and Subtraction of Vectors. Centroids: Addition and Subtraction of Vectors: $5$. Multiplication by a number
Category:Scalar Multiplication
Category:Unit Vectors
Scalar Product with Identity

== Theorem ==
Let $\struct {G, +_G}$ be an abelian group whose identity is $e$.
Let $\struct {R, +_R, \times_R}$ be a ring whose zero is $0_R$.
Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Let $x \in G, \lambda \in R$.
Then:
:$\lambda \circ e = 0_R \circ x = e$
== Proof ==
From , $y \to \lambda \circ y$ is an endomorphism of $\struct {G, +_G}$.
From , $\mu \to \mu \circ x$ is a homomorphism from $\struct {R, +_R}$ to $\struct {G, +_G}$.
The result follows from Homomorphism with Cancellable Codomain Preserves Identity.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 4$. Vector Spaces
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Theorem $26.2 \ (1)$
Category:Module Theory
Scalar Product with Inverse

== Theorem ==
Let $\struct {G, +_G}$ be an abelian group.
Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Let $x \in G, \lambda \in R$.
Then:
:$\lambda \circ \struct {-x} = \struct {-\lambda} \circ x = -\struct {\lambda \circ x}$
== Proof ==
From , $y \to \lambda \circ y$ is an endomorphism of $\struct {G, +_G}$.
From , $\mu \to \mu \circ x$ is a homomorphism from $\struct {R, +_R}$ to $\struct {G, +_G}$.
The result follows from Homomorphism with Identity Preserves Inverses.
== Also see ==
* Basic Results about Modules
* Basic Results about Unitary Modules
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 4$. Vector Spaces
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Theorem $26.2 \ (2)$
Category:Module Theory
Scalar Product with Inverse Unity

== Theorem ==
Let $\struct {G, +_G}$ be an abelian group whose identity is $e$.
Let $\struct {R, +_R, \times_R}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $\struct {G, +_G, \circ}_R$ be an unitary $R$-module.
Let $x \in G$.
Then:
:$\paren {-1_R} \circ x = - x$
== Proof ==
Follows directly from Scalar Product with Inverse.
== Also see ==
* Basic Results about Modules
* Basic Results about Unitary Modules
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Theorem $26.2 \ (6)$
Category:Unitary Modules
Scalar Product with Multiple of Unity

== Theorem ==
Let $\struct {G, +_G}$ be an abelian group whose identity is $e$.
Let $\struct {R, +_R, \times_R}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $\struct {G, +_G, \circ}_R$ be an unitary $R$-module.
Let $x \in G, n \in \Z$.
Then:
:$\paren {n \cdot 1_R} \circ x = n \cdot x$
that is:
:$\paren {\map {\paren {+_R}^n} {1_R} } \circ x = \map {\paren {+_G}^n} x$
== Proof ==
Follows directly from Scalar Product with Product.
== Also see ==
* Basic Results about Modules
* Basic Results about Unitary Modules
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Theorem $26.2 \ (7)$
Category:Unitary Modules
Scalar Product with Product

== Theorem ==
Let $\struct {G, +_G}$ be an abelian group.
Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Let $x \in G, \lambda \in R, n \in \Z$.
Then:
:$\lambda \circ \paren {n \cdot x} = n \cdot \paren {\lambda \circ x} = \paren {n \cdot \lambda} \circ x$
== Proof ==
First let $n = 0$.
The assertion follows directly from Scalar Product with Identity.
Next, let $n > 0$.
The assertion follows directly from Scalar Product with Sum and Product with Sum of Scalar, by letting $m = n$ and making all the $\lambda$'s and $x$'s the same.
Finally, let $n < 0$.
The assertion follows from Scalar Product with Product for positive $n$, Scalar Product with Inverse, and from Negative Index Law for Monoids.
== Also see ==
* Basic Results about Modules
* Basic Results about Unitary Modules
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Theorem $26.2 \ (5)$
Category:Module Theory
Scalar Product with Sum

== Theorem ==
Let $\struct {G, +_G}$ be an abelian group whose identity is $e$.
Let $\struct {R, +_R, \times_R}$ be a ring whose zero is $0_R$.
Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Let $x \in G, \lambda \in R$.
Let $\sequence {x_m}$ be a sequence of elements of $G$.
Then:
:$\ds \lambda \circ \paren {\sum_{k \mathop = 1}^m x_k} = \sum_{k \mathop = 1}^m \paren {\lambda \circ x_k}$
== Proof ==
This follows by induction from , as follows:
For all $m \in \N_{>0}$, let $\map P m$ be the proposition:
:$\ds \lambda \circ \paren {\sum_{k \mathop = 1}^m x_k} = \sum_{k \mathop = 1}^m \paren {\lambda \circ x_k}$
=== Basis for the Induction ===
$\map P 1$ is true, as this just says:
:$\lambda \circ x_1 = \lambda \circ x_1$
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P n$ is true, where $n \ge 1$, then it logically follows that $\map P {n + 1}$ is true.
So this is our induction hypothesis:
:$\ds \lambda \circ \paren {\sum_{k \mathop = 1}^n x_k} = \sum_{k \mathop = 1}^n \paren {\lambda \circ x_k}$
Then we need to show:
:$\ds \lambda \circ \paren {\sum_{k \mathop = 1}^{n + 1} x_k} = \sum_{k \mathop = 1}^{n + 1} \paren {\lambda \circ x_k}$
=== Induction Step ===
This is our induction step:
So $\map P n \implies \map P {n + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall m \in \N_{>0}: \lambda \circ \paren {\sum_{k \mathop = 1}^m x_k} = \sum_{k \mathop = 1}^m \paren {\lambda \circ x_k}$
== Also see ==
* Basic Results about Modules
* Basic Results about Unitary Modules
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Theorem $26.2 \ (3)$
Category:Module Theory
Scalar Triple Product as Product of Magnitudes

== Theorem ==
Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be vectors in a Cartesian $3$-space:
where $\tuple {\mathbf i, \mathbf j, \mathbf k}$ is the standard ordered basis of $\mathbf V$.
Let $\sqbrk {\mathbf a, \mathbf b, \mathbf c}$ be the scalar triple product of $\mathbf a$, $\mathbf b$ and $\mathbf c$.
Then:
:$\sqbrk {\mathbf a, \mathbf b, \mathbf c} = \size {\mathbf a} \size {\mathbf b} \size {\mathbf c} \sin \theta \cos \alpha$
where:
:$\theta$ is the angle between $\mathbf b$ and $\mathbf c$
:$\alpha$ is the angle between $\mathbf b \times \mathbf c$ and $\mathbf a$.
== Proof ==
== Sources ==
* 
* 
Category:Scalar Triple Product
Scale Parameter of Cauchy Distribution is Half-Width at Half-Height

== Definition ==
Let $X$ be a continuous random variable with a Cauchy distribution:
:$\map {f_X} x = \dfrac 1 {\pi \lambda \paren {1 + \paren {\frac {x - \gamma} \lambda }^2} }$
for:
:$\lambda \in \R_{>0}$
:$\gamma \in \R$
Then $\lambda$ can be referred to as the half-width at half-height:
:$\map \Pr {\gamma - \lambda < X \le \gamma + \lambda} = \dfrac 1 2$
== Proof == 
== Sources == 
* 
Category:Cauchy Distribution
Scaled Euclidean Metric is Metric

== Theorem ==
Let $\R_{>0}$ be the set of (strictly) positive integers.
Let $\delta: \R_{>0} \times \R_{>0} \to \R$ be the metric on $\R_{>0}$ defined as:
:$\forall x, y \in \R_{>0}: \map \delta {x, y} = \dfrac {\size {x - y} } {x y}$
Then $\delta$ is a metric.
== Proof ==
=== Proof of ===
So holds for $\delta$.
=== Proof of ===
So holds for $\delta$.
=== Proof of ===
So holds for $\delta$.
=== Proof of ===
So holds for $\delta$.
Category:Scaled Euclidean Metric
Scaled Real Function that Decreases Without Bound

== Theorem ==
Let $f: \R \to \R$ be a real function.
Let $\lambda \in \R_{\ne 0}$ be a nonzero constant.
Then:
For $\lambda > 0$:
:$\ds \lim_{x \mathop \to +\infty} \map f x = -\infty \implies \lim_{x \mathop \to +\infty} \lambda \map f x = -\infty$
:$\ds \lim_{x \mathop \to -\infty} \map f x = -\infty \implies \lim_{x \mathop \to -\infty} \lambda \map f x = -\infty$
For $\lambda < 0$:
:$\ds \lim_{x \mathop \to +\infty} \map f x = -\infty \implies \lim_{x \mathop \to +\infty} \lambda \map f x = +\infty$
:$\ds \lim_{x \mathop \to -\infty} \map f x = -\infty \implies \lim_{x \mathop \to -\infty} \lambda \map f x = +\infty$
== Proof ==
Let $\ds \lim_{x \mathop \to +\infty} \map f x = -\infty$.
From the definition of infinite limits at infinity, this means that:
:$\forall M < 0: \exists N > 0: x > N \implies \map f x < M$
Suppose $\lambda > 0$.
Then $M < 0 \iff \lambda^{-1}M < 0$.
Also, $\map f x < \lambda^{-1}M \iff \lambda \map f x < M$
So:
:$\forall M < 0: \exists N > 0: x > N \implies \lambda \map f x < M$
From the definition of infinite limits at infinity:
:$\ds \lim_{x \mathop \to +\infty} \lambda \map f x = -\infty$
The proof for $\ds \lim_{x \mathop \to -\infty} \map f x = -\infty$ is analogous.
Now, suppose $\lambda < 0$.
Then $-\lambda f < 0$, and so $-\lambda f \to -\infty$, from above.
Write $\lambda f = -\paren {-\lambda f}$ and the result follows from Negative of Real Function that Decreases Without Bound.
== Also see ==
* Scaled Real Function that Increases Without Bound
Category:Unbounded Mappings
Scaled Real Function that Increases Without Bound

== Theorem ==
Let $f: \R \to \R$ be a real function.
Let $\lambda \in \R_{\ne 0}$ be a nonzero constant.
Then:
For $\lambda > 0$:
:$\ds \lim_{x \mathop \to +\infty} \map f x = +\infty \implies \lim_{x \mathop \to +\infty} \lambda \map f x = +\infty$
:$\ds \lim_{x \mathop \to -\infty} \map f x = +\infty \implies \lim_{x \mathop \to -\infty} \lambda \map f x = + \infty$
For $\lambda < 0$:
:$\ds \lim_{x \mathop \to +\infty} \map f x = +\infty \implies \lim_{x \mathop \to +\infty} \lambda \map f x = -\infty$
:$\ds \lim_{x \mathop \to -\infty} \map f x = +\infty \implies \lim_{x \mathop \to -\infty} \lambda \map f x = -\infty$
== Proof ==
Let $\ds \lim_{x \mathop \to +\infty} \map f x = +\infty$.
From the definition of infinite limit at infinity, this means that:
:$\forall M > 0: \exists N > 0: x > N \implies \map f x > M$.
Suppose $\lambda > 0$.
Then $M > 0 \iff \lambda^{-1} M > 0$.
Also, $\map f x > \lambda^{-1} M \iff \lambda f\map f x > M$.
So:
:$\forall M > 0: \exists N > 0: x > N \implies \lambda \map f x > M$
From the definition of infinite limit at infinity:
:$\ds \lim_{x \mathop \to +\infty} \lambda \map f x = +\infty$.
The proof for $\ds \lim_{x \mathop \to -\infty} \map f x = +\infty$ is analogous.
Now, suppose $\lambda < 0$.
Then $-\lambda f > 0$, and so $-\lambda f \to + \infty$, from above.
Write $\lambda f = -\paren {-\lambda f}$ and the result follows from Negative of Real Function that Increases Without Bound.
== Also see ==
* Scaled Real Function that Decreases Without Bound
Category:Unbounded Mappings
Scaled Sine Functions of Integer Multiples form Orthonormal Set

== Theorem ==
For all $n \in \Z_{>0}$, let $\map {\phi_n} x$ be the real function defined on the interval $\openint 0 \lambda$ as:
:$\map {\phi_n} x = \sqrt {\dfrac 2 \lambda} \sin \dfrac {n \pi x} \lambda$
Let $S$ be the set:
:$S = \set {\phi_n: n \in \Z_{>0} }$
Then $S$ is an orthonormal set.
== Proof ==
Consider the definite integral:
:$I_{m n} = \ds \int_0^\lambda \map {\phi_m} x \map {\phi_n} x \rd x$
From Sine Function is Odd, each of $\map {\phi_n} x$ is an odd function.
From Odd Function Times Odd Function is Even, $\map {\phi_m} x \map {\phi_n} x$ is even.
That is:
:$\paren {\sqrt {\dfrac 2 \lambda} \sin \dfrac {m \pi x} \lambda} \paren {\sqrt {\dfrac 2 \lambda} \sin \dfrac {n \pi x} \lambda}$
is an even function.
Let $u = \dfrac {\pi x} \lambda$.
We have:
So:
Hence the result by definition of orthonormal set.
== Sources ==
* : Chapter One: $\S 8$. Orthonormal Sets of Functions
Category:Orthonormal Sets
Category:Sine Function
Scalene Triangle Tessellates the Plane

== Theorem ==
Let $T$ be a triangle, which can be any shape at all, even scalene.
Then $T$ can tessellate the plane.
== Proof ==
By placing $2$ copies of $T$ together so their corresponding sides coincide, it is possible to form a quadrilateral.
The result follows from Quadrilateral Tessellates the Plane.
== Sources ==
* 
* 
Category:Scalene Triangles
Category:Tessellations
Scaling Property of Dirac Delta Function

== Theorem ==
Let $\map \delta t$ be the Dirac delta function.
Let $a$ be a non zero constant real number.
Then:
:$\map \delta {a t} = \dfrac {\map \delta t} {\size a}$
== Proof ==
The equation can be rearranged as:
:$\size a \map \delta {a t} = \map \delta t$
We will check the definition of Dirac delta function in turn.
Definition of Dirac delta function:
:$\paren 1:\map \delta t = \begin{cases}
+\infty & : t = 0 \\
0 & : \text{otherwise}
\end{cases}$
:$\paren 2:\ds \int_{-\infty}^{+\infty} \map \delta t \rd t = 1$
$\paren 1:$
$\paren 2:$
The proof of this part will be split into two parts, one for positive $a$ and one for negative $a$.
For $a > 0$:
For $a < 0$:
Therefore, by definition, $\size a \map \delta {a t} = \map \delta t$.
The result follows after rearrangement.
Category:Dirac Delta Function
Scaling preserves Modulo Addition

== Theorem ==
Let $m \in \Z_{> 0}$.
Let $x, y, c \in \Z$.
Let $x \equiv y \pmod m$.
Then:
:$c x \equiv c y \pmod m$
== Proof ==
Let $x \equiv y \pmod m$.
Then by definition of congruence:
:$\exists k \in Z: x - y = k m$
Hence:
:$c x - c y = c k m$
and so by definition of congruence:
:$c x \equiv c y \pmod m$
Category:Modulo Addition
Scattered Space is T0

== Theorem ==
Let $T = \struct {S, \tau}$ be a scattered topological space.
Then $T$ is also a $T_0$ (Kolmogorov) space.
== Proof ==
Suppose $T$ is not a $T_0$ (Kolmogorov) space.
From Equivalence of Definitions of $T_0$ Space, there exist $x, y \in S$ such that $x$ and $y$ are both limit points of each other.
So by definition of isolated point, neither $x$ nor $y$ are isolated in $\set {x, y}$.
Thus we have found a subset $\set {x, y} \subseteq T$ such that $\set {x, y}$ is by definition dense-in-itself.
So $T$ is not scattered.
Hence the result by Rule of Transposition.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Disconnectedness
Category:T0 Spaces
Category:Scattered Spaces
Scattered Space is not necessarily T1

== Theorem ==
Let $T = \struct {S, \tau}$ be a scattered topological space.
Then $T$ is not necessarily a $T_1$ (Fréchet) space.
== Proof ==
Let $T = \struct {S, \tau}$ be a non-trivial particular point space.
From Particular Point Space is Scattered, $T$ is a scattered space.
From Non-Trivial Particular Point Topology is not $T_1$, $T$ is not a $T_1$ (Fréchet) space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Disconnectedness
Category:T1 Spaces
Category:Scattered Spaces
Scattered T1 Space is Totally Disconnected

== Theorem ==
Let $T = \struct {S, \tau}$ be a scattered topological space which is also a $T_1$ (Fréchet) space.
Then $T$ is totally disconnected.
== Proof ==
Let $T = \struct {S, \tau}$ be a scattered space which is also a $T_1$ (Fréchet) space.
We have that every Non-Trivial Connected Set in $T_1$ Space is Dense-in-itself.
As $T$ is scattered, every $H \subseteq S$ contains at least one point which is isolated in $H$.
So $H$ is not dense-in-itself and so if $H$ has more than one element it can not be connected.
As $H$ is arbitrary, it follows that $T$ is totally disconnected.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Disconnectedness
Category:T1 Spaces
Category:Scattered Spaces
Category:Totally Disconnected Spaces
Schanuel's Conjecture

== Conjecture ==
Let $z_1, \cdots, z_n$ be complex numbers that are linearly independent over the rational numbers $\Q$.
Then:
:the extension field $\map \Q {z_1, \cdots, z_n, e^{z_1}, \cdots, e^{z_n} }$ has transcendence degree at least $n$ over $\Q$
where $e^z$ is the complex exponential of $z$.
Category:Transcendental Number Theory
Category:Unproven Hypotheses
Category:Schanuel's Conjecture