nyuuzyou/wikis huggingface dataset:
Distribution Function of Finite Borel Measure is Right-Continuous

== Theorem ==
Let $\mu$ be a finite Borel measure on $\R$.
Let $F_\mu$ be the distribution function of $\mu$.
Then $F_\mu$ is right-continuous.
== Proof ==
From Monotonic Sequential Right-Continuity is Equivalent to Right-Continuity in the Reals, it suffices to show that: 
:for all monotone sequences $\sequence {x_n}_{n \mathop \in \N}$, with $x_n > x$ for each $n$, that converge to $x$ we have:
::$\ds \map {F_\mu} x = \lim_{n \mathop \to \infty} \map {F_\mu} {x_n}$
Now let $\sequence {x_n}_{n \mathop \in \N}$ be a monotone sequence with $x_n > x$ for each $n \in \N$ and $x_n \to x$.
Then $\sequence {\hointl {-\infty} {x_n} }_{n \mathop \in \N}$ is a decreasing sequence of sets.
Then from Limit of Decreasing Sequence of Unbounded Below Closed Intervals, we have: 
:$\ds \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} = \hointl {-\infty} x$
Since $\mu$ is finite, we have: 
:$\map \mu {\hointl {-\infty} {x_1} } < \infty$
so Measure of Limit of Decreasing Sequence of Measurable Sets is admissable and gives: 
:$\ds \map \mu {\bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} } = \lim_{n \mathop \to \infty} \map \mu {\hointl {-\infty} {x_n} }$
That is: 
:$\ds \map \mu {\bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} } = \lim_{n \mathop \to \infty} \map {F_\mu} {x_n}$
so:
:$\ds \map \mu {\hointl {-\infty} x} = \lim_{n \mathop \to \infty} \map {F_\mu} {x_n}$
Hence: 
:$\ds \map {F_\mu} x = \lim_{n \mathop \to \infty} \map {F_\mu} {x_n}$
Since the sequence $\sequence {x_n}_{n \mathop \in \N}$ was arbitrary, we have: 
:$F_\mu$ is right-continuous.
from Monotonic Sequential Right-Continuity is Equivalent to Right-Continuity in the Reals.
== Sources ==
* : $1.3$: Outer Measures
Category:Right-Continuous Functions
Category:Distribution Function of Finite Borel Measure
Distribution Function of Finite Signed Borel Measure is of Bounded Variation

== Theorem ==
Let $\mu$ be a finite signed Borel measure on $\R$.
Let $F_\mu$ be the distribution function of $\mu$. 
Then $F_\mu$ is of bounded variation.
== Proof ==
Let $\SS$ be a non-empty finite subset of $\R$. 
Write: 
:$\SS = \set {x_0, x_1, \ldots, x_n}$
with:
:$x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n$
Then, we have: 
Note that: 
:$\set {\hointl {-\infty} {x_0}, \hointl {x_0} {x_1}, \hointl {x_1} {x_2}, \ldots, \hointl {x_{n - 1} } {x_n}, \hointr {x_n} \infty}$ is a partition of $\R$ into Borel sets.
So from Definition 2 of the variation of a signed measure, we have: 
:$\ds \size {\map \mu {\hointl {-\infty} {x_0} } } + \sum_{i \mathop = 1}^n \size {\map \mu {\hointl {x_{i - 1} } {x_i} } } + \size {\map \mu {\hointr {x_n} \infty} } \le \map {\size \mu} \R$ 
where $\size \mu$ is the variation of $\mu$.
From Signed Measure Finite iff Finite Total Variation, we have: 
:$\map {\size \mu} \R < \infty$
So, using the notation from the definition of bounded variation, for all non-empty finite subset of $\R$, $\SS$, we have:
:$\map {V_f^\ast} {\SS; I} \le M = \map {\size \mu} \R$
So $f$ is of bounded variation.
== Sources ==
* : $4.4$: Functions of Finite Variation
Category:Distribution Function of Finite Signed Borel Measure
Category:Bounded Variation
Distribution Space over Smooth Functions is Unitary Module

== Theorem ==
The distribution space over smooth functions is a unitary module.
== Proof ==
Let $\phi \in \map \DD {\R^d}$ be a test function.
=== ===
Hence:
:$\alpha \paren {T_1 + T_2} = \alpha T_1 + \alpha T_2$
=== ===
Hence:
:$\paren {\alpha_1 + \alpha_2} T = \alpha_1 T + \alpha_2 T$
=== ===
Hence:
:$\paren {\alpha \beta} T = \alpha \paren {\beta T}$
=== ===
Let $\mathbf 1 \in \map {C^\infty} {\R^d}$ be a constant mapping such that:
:$\R^d \stackrel {\mathbf 1} {\mapsto} 1$
Then:
Hence:
:$1 \cdot T = T$
== Sources ==
* : Chapter $\S 6.4$: A glimpse of distribution theory. Multiplication by $C^\infty$ functions
Category:Distributions
Category:Examples of Unitary Modules
Distribution Space with Pointwise Addition and Pointwise Scalar Multiplication forms Vector Space

== Theorem ==
Let $\map {\DD'} {\R^d}$ be the distribution space.
Let $\struct {\C, +_\C, \times_\C}$ be the field of complex numbers.
Let $\paren +$ be the pointwise addition of distributions.
Let $\paren {\, \cdot \,}$ be the pointwise scalar multiplication of distributions over $\C$.
Then $\struct {\map {\DD'} {\R^d}, +, \, \cdot \,}_\C$ is a vector space.
== Proof ==
== Sources ==
* : Chapter $\S 6.1$: A glimpse of distribution theory. Test functions, distributions, and examples
Category:Examples of Vector Spaces
Category:Functional Analysis
Distribution acting on Sequence of Test Functions without common Support is not Continuous

== Theorem ==
Let $T \in \map {\DD'} \R$ be a distribution.
Let $\sequence {\phi_n}_{n \mathop \in \N} \in \map \DD \R$ be a sequence of test functions.
Suppose $\sequence {\phi_n}_{n \mathop \in \N}$ does not have the common compact support.
Then $T$ acting on $\sequence {\phi_n}_{n \mathop \in \N}$ is not continuous.
== Proof ==
Let $\operatorname {III} \in \map {\DD'} \R$ be the Dirac comb.
Let $\mathbf 0 : \R \to 0$ be the zero mapping.
Let $\phi \in \map \DD \R$ be a test function with compact support $K = \closedint 0 1$ such that:
:$\forall x \in K : \map \phi x > 0$
Let $\phi_n \in \map \DD \R$ be a test function sequence such that:
:$\ds \forall x \in \R : \forall n \in \N : \map {\phi_n} x = \frac 1 n \map \phi {\frac x n}$
We have that:
:$\ds \forall n \in \N : \map {\phi_{n + 1} } n = \frac 1 {n + 1} \map \phi {\frac n {n + 1} } > 0$
Thus, with each $n$ we generate a new test function with a larger support.
By taking the $k$th derivative of $\map {\phi_n} x$ and applying the chain rule we get that:
:$\ds \forall k \in \N : \map {\phi_n^{\paren k} } x = \frac 1 {n^{k + 1}} \map {\phi^{\paren k} } {\frac x n}$
Hence:
Fix $k \in \N$.
Let $\epsilon \in \R_{\mathop > 0}$.
Let $N \in \R$ be such that $\ds \frac {M_k}{N^k} < \epsilon$.
Then:
:$\ds \forall n \in \N : n \ge N \implies \frac {M_k}{n^k} \le \frac {M_k}{N^k} < \epsilon$
Altogether:
:$\ds \forall \epsilon \in \R_{\mathop > 0} : \exists N \in \R : \forall n \in \N : n \ge N \implies \sup_{x \mathop \in \R} \size {\map {\phi_n^{\paren k} } x - 0} \le \frac {M_k}{n^k} \le \frac {M_k}{N^k} < \epsilon$
Hence, as $n \rightarrow \infty$ we get that $\phi_n^{\paren k}$ converges to $\mathbf 0$ uniformly. 
However:
We have that Test Function is Riemann Integrable.
Let the subdivision of the interval $\closedint 0 1$ be normal:
:$\ds \forall i \in \N_{\mathop \le n} : x_{i + 1} - x_i = \frac 1 n$
Then $\map T {\phi_n}$ is a Riemann sum with $\ds \Delta x_k = \frac 1 n$, $\ds c_k = \frac k n$, $x_0 = 0$ and $x_n = 1$.
Taking $n \to \infty$ yields:
:$\ds \map T {\phi_n} \stackrel {n \mathop \to \infty} \longrightarrow \int_0^1 \map \phi x \rd x$
However, we assumed that:
:$\forall x \in K : \map \phi x > 0$
Thus:
:$\ds \int_0^1 \map \phi x \rd x > 0$
So we have that, as $n \to \infty$, $\sequence {\phi_n}$ converges to $\mathbf 0$ uniformly, but $\map T {\phi_n}$ does not converge to $0$.
== Sources ==
* : Chapter $\S 6.1$: A glimpse of distribution theory. Test functions, distributions, and examples
Category:Convergence
Category:Continuity
Distribution of Macroscopic Electric Charge within Conductor

== Theorem ==
Let $B$ be a body made out of an electrically conducting substance.
Then it is not possible for different volume elements of $B$ to have different (macroscopic) electric fields caused by intrinsic imbalance of the electric charges distributed throughout $B$.
== Proof ==
In an electrical conductor, electric charges can flow, because of the action of conduction electrons within the substance.
Hence an electric charge in one point within $B$ will flow to another point within $B$ under a potential difference caused by differences in (macroscopic) electric fields within $B$.
Hence any imbalances caused by internal (macroscopic) electric fields will not be possible.
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.3$ Electric Fields in Matter: $1.3.3$ The macroscopic electric field
Category:Electrical Conductors
Category:Macroscopic Electric Fields
Distribution of Macroscopic Electric Charge within Insulator

== Theorem ==
Let $B$ be a body made out of an electrically insulating substance.
Then it is possible for different volume elements of $B$ to have different (macroscopic) electric fields caused by intrinsic imbalance of the electric charges distributed throughout $B$.
== Proof ==
In an electrical insulator, electric charges cannot flow, because of the nature of the substance.
Hence an electric charge in one point within $B$ cannot flow to another point within $B$ under a potential difference caused by the presence of those (macroscopic) electric fields.
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.3$ Electric Fields in Matter: $1.3.3$ The macroscopic electric field
Category:Electrical Insulators
Category:Macroscopic Electric Fields
Distribution of Numbers with More than 2 Prime Factors

== Theorem ==
For sufficiently large $x$, there always exists an integer with more than $2$ prime factors between $\paren {x - x^\alpha}$ and $x$, where:
:$\alpha \ge 0 \cdotp 477 \ldots$
== Proof ==
== Sources ==
* 
* : $0,477 \ldots$
Category:Number Theory
Distribution of Random Sample

== Theorem ==
Let $S = \set {x_1, x_2, \ldots, x_n}$ be a random sample from a probability distribution either discrete or continuous.
Then the distribution of $S$ is discrete, and its frequency function given by:
:$\map \Pr {X = x_i} = p_i = \dfrac 1 n$
for $i = 1, 2, \ldots, n$. 
== Proof ==
== Sources ==
* 
* 
Category:Random Samples (Probability Theory)
Distributional Derivative of Absolute Value Function

== Theorem ==
Let $H: \R \to \closedint 0 1$ be the Heaviside step function.
Let $\size x$ be the absolute value of $x$.
Let $T_{\size x}$ be the distribution associated with $\size x$.
Then the distributional derivative of $T_{\size x}$ is $T_{2 H - 1}$
== Proof ==
Furthermore:
:$\ds \lim_{x \mathop \to 0^+} \size x = \lim_{x \mathop \to 0^-} \size x = 0$
By the Jump Rule:
:$T_{\size x}' = T_{2H - 1}$
== Also see ==
*Derivative of Absolute Value Function
== Sources ==
* : Chapter $\S 6.2$: A glimpse of distribution theory. Derivatives in the distributional sense
Category:Absolute Value Function
Category:Distributional Derivatives
Distributional Derivative of Floor Function

== Theorem ==
Let $\floor x$ be the floor function.
Let $\map {\operatorname {III} } x$ be the Dirac comb.
Then the distributional derivative of $\floor x$ is $\map {\operatorname {III} } 0$.
== Proof ==
By definition:
:$\floor x := \sup \set {m \in \Z: m \le x}$
Hence, $\forall m \in \Z : \forall x \in \openint m {m + 1}$ the floor function is constant.
Therefore:
:$\forall m \in \Z : \forall x \in \openint m {m + 1} : \dfrac {\d \floor x} {\d x} = 0$
Every $x \in \Z$ is a discontinuity of $\floor x$.
Hence, the jump rule has to be applied to each such $x$.
Suppose $k \in \Z$.
Then:
:$\ds \lim_{x \mathop \to k^+} \floor x = k$
:$\ds \lim_{x \mathop \to k^-} \floor x = k - 1$
By the Jump Rule:
== Sources ==
* : Chapter $\S 6.2$: A glimpse of distribution theory. Derivatives in the distributional sense
Category:Floor Function
Category:Distributional Derivatives
Distributional Derivative of Heaviside Step Function

== Theorem ==
Let $H : \R \to \closedint 0 1$ be the Heaviside step function.
Let $T \in \map {\DD'} \R$ be a distribution corresponding to $H$.
Let $\delta \in \map {\DD'} \R$ be the Dirac delta distribution.
Then the distributional derivative of $T$ is $\delta$.
== Proof ==
Let $\phi \in \map \DD \R$ be a test function.
Then:
== Sources ==
* : Chapter $\S 6.2$: A glimpse of distribution theory. Derivatives in the distributional sense
Category:Examples of Distributional Derivatives
Category:Examples of Distributions
Distributional Derivative of Heaviside Step Function times Cosine

== Theorem ==
Let $H$ be the Heaviside step function.
Let $\delta$ be the Dirac delta distribution.
Then in the distributional sense:
:$T'_{\map H x \cos x} = T_{- \map H x \sin x} + \delta$
== Proof ==
$x \stackrel f {\longrightarrow} \map H x \cos x$ is a continuously differentiable real function on $\R \setminus \set 0$ and has a discontinuity at $x = 0$.
By Differentiable Function as Distribution we have that $T'_f = T_{f'}$.
Moreover:
:$x < 0 \implies \paren {{\map H x} \map \cos x}' = 0$.
:$x > 0 \implies \paren {{\map H x} \map \cos x}' = -\sin x$.
Altogether:
:$\forall x \in \R \setminus \set 0 : \paren {{\map H x} \map \cos x}' = - \map H x \sin x$
Furthermore:
:$\ds \map f {0^+} - \map f {0^-} = \map H {0^+} \map \cos {0^+} - \map H {0^-} \map \cos {0^-} = 1$
The result follows from the Jump Rule.
== Sources ==
* : Chapter $\S 6.2$: A glimpse of distribution theory. Derivatives in the distributional sense
Category:Examples of Distributional Derivatives
Category:Piecewise Continuously Differentiable Functions
Distributional Derivative of Heaviside Step Function times Sine

== Theorem ==
Let $H$ be the Heaviside step function.
Let $\delta$ be the Dirac delta distribution.
Then in the distributional sense:
:$T'_{\map H x \sin x} = T_{\map H x \cos x} $
== Proof ==
$x \stackrel f {\longrightarrow} \map H x \sin x$ is a continuously differentiable real function on $\R \setminus \set 0$ and possibly has a discontinuity at $x = 0$.
By Differentiable Function as Distribution we have that $T'_f = T_{f'}$.
Moreover:
:$x < 0 \implies \paren {{\map H x} \map \sin x}' = 0$.
:$x > 0 \implies \paren {{\map H x} \map \sin x}' = \cos x$.
Altogether:
:$\forall x \in \R \setminus \set 0 : \paren {{\map H x} \map \sin x}' = \map H x \cos x$
Furthermore:
:$\ds \map f {0^+} - \map f {0^-} = \map H {0^+} \map \sin {0^+} - \map H {0^-} \map \sin {0^-} = 0$
The result follows from the Jump Rule.
== Sources ==
* : Chapter $\S 6.2$: A glimpse of distribution theory. Derivatives in the distributional sense
Category:Examples of Distributional Derivatives
Category:Piecewise Continuously Differentiable Functions
Distributional Derivative on Distributions is Continuous Operator

== Theorem ==
The distributional derivative on distributions is a continuous operator.
== Proof ==
Let $\mathbf 0$ be the zero mapping.
Let $\sequence {\phi_n}_{n \mathop \in \N} \in \map \DD \R$ be a sequence of test functions.
Let $\sequence {\phi_n}_{n \mathop \in \N}$ converge to $\mathbf 0$ in the test function space:
:$\phi_n \stackrel \DD {\longrightarrow} \mathbf 0$
By definition, a test function is a smooth real function with a compact support $K$.
By differentiating a test function we get another smooth real function with a compact support not larger than $K$.
Thus, $\phi_n'$ is also a test function.
Since $\phi_n \stackrel \DD {\longrightarrow} \mathbf 0$, we have that for every multiindex $k$ the sequence $\sequence {D^k \phi_n}_{n \mathop \in \N}$ converges uniformly to $\mathbf 0$.
Obviously, all the sequences of $\sequence {D^k \phi_n}_{n \mathop \in \N}$ with $k \ge 1$ converge uniformly to $\mathbf 0$.
By definition, $\sequence {\phi_n'}_{n \mathop \in \N}$ converges to $\mathbf 0$ in the test function space:
:$\phi_n' \stackrel \DD {\longrightarrow} \mathbf 0$
Let $T \in \map {\DD'} \R$ be a distribution.
Then:
:$\paren {\phi_n' \stackrel \DD {\longrightarrow} \mathbf 0} \implies \paren {\map T {\phi_n'} \to \map T {\mathbf 0} = 0}$
By definition of the distributional derivative:
:$\map T {\phi_n'} = - \map {T'} {\phi_n}$
Therefore:
:$\paren {\phi_n' \stackrel \DD {\longrightarrow} \mathbf 0} \implies \paren {\map {T'} {\phi_n} \to 0}$
By definition, the distributional derivative on distributions is a continuous operator.
== Sources ==
* : Chapter $\S 6.2$: A glimpse of distribution theory. Derivatives in the distributional sense
Category:Distributional Derivatives
Category:Continuous Operators
Distributional Derivative on Distributions is Linear Operator

== Theorem ==
The Distributional derivative on distributions is a linear operator.
== Proof ==
Let $\phi, \psi \in \map \DD \R$ be test functions.
Let $\alpha \in \C$ be a complex number.
Let $T \in \map {\DD'} \R$ be a distribution.
Then:
Thus the distributional derivative is a linear operator.
== Sources ==
* : Chapter $\S 6.2$: A glimpse of distribution theory. Derivatives in the distributional sense
Category:Distributional Derivatives
Category:Linear Operators
Distributional Derivatives of Dirac Delta Distribution do not Vanish

== Theorem ==
Let $\delta \in \map {\DD'} \R$ be the Dirac delta distribution.
Then for any $n \in \N$ the distributional derivative $\delta^{\paren n}$ does not vanish.
== Proof ==
Let $\phi \in \map \DD \R$ be a test function such that $\map \phi 0 \ne 0$.
Then:
:$\forall n \in \N : \forall x \in \R : x^n \phi \in \map \DD \R$
By the definition of the distributional derivative:
== Sources ==
* : Chapter $\S 6.2$: A glimpse of distribution theory. Derivatives in the distributional sense
Category:Dirac Delta Distribution
Category:Distributional Derivatives
Distributional Partial Derivatives Commute

== Theorem ==
Let $T \in \map {\DD'} {\R^d}$ be a distribution.
Then in the distributional sense:
:$\dfrac {\partial^2 T} {\partial x_i \partial x_j} = \dfrac {\partial^2 T} {\partial x_j \partial x_i}$
where:
:$i, j \in \N : 1 \le i, j \le d$
== Proof ==
Let $\phi \in \map \DD {\R^d}$ be a test function.
== Sources ==
* : Chapter $\S 6.2$: A glimpse of distribution theory. Derivatives in the distributional sense
Category:Examples of Commutative Operations
Category:Distributional Derivatives
Distributional Solution to x T = 0

== Theorem ==
Let $T \in \map {\DD'} \R$ be a distribution.
Let $\delta$ be the Dirac delta distribution.
Let $\mathbf 0$ be the zero distribution.
Suppose $T$ satisfies the following equation in the distributional sense:
:$x T = \mathbf 0$
Then $T = \alpha \delta$ where $c \in \C$.
== Proof ==
Let $\phi \in \map \DD \R$ be a test function.
Let $c \in \C$.
Suppose:
:$T = c \delta$
Then:
=== Direct inclusion ===
Suppose:
:$T \in \map {\DD'} \R : x T = 0$
Then:
So:
:$\set {x \phi : \phi \in \map \DD \R} \subseteq \ker T$
where $\ker$ denotes the kernel of $T$.
Let $\psi = x \phi$.
Suppose $\phi \in \map \DD \R$.
Then $\psi \in \map \DD \R$.
Furthermore:
:$\paren {\map \phi 0 = 0} \implies \paren {\map \psi 0 = 0}$
Hence:
:$\set {x \phi : \phi \in \map \DD \R} \subseteq \set {\psi \in \map \DD \R : \map \psi 0 = 0}$
=== Reverse inclusion ===
Let $\psi \in \map \DD \R$ such that $\map \psi 0 = 0$.
Then:
Hence:
:$\ker \delta = \set {\psi \in \map \DD \R : \map \psi 0 = 0}$
By the fundamental theorem of calculus:
Let $\ds \map \phi x := \int_0^1 \map {\psi'} {tx} \rd t$.
Then $\map \psi x = x \map \phi x$
By assumption, $\psi$ is a test function.
Hence, $\psi$ is smooth:
:$\psi \in \map {C^\infty} \R$
By differentiating under integral sign:
:$\phi \in \map {C^\infty} \R$
By definition, $\psi$ has a compact support.
Let $a \in \R_{\mathop > 0}$.
Suppose:
:$\forall x \notin \closedint {-a} a : \map \psi x = 0$
Then:
:$\ds \forall x \notin \closedint {-a} a : \map \phi x = \frac {\map \psi x}x = 0$
We have that $\phi$ is smooth and compactly supported.
By definition, $\phi \in \map \DD \R$.
Altogether:
:$\set {\psi \in \map \DD \R : \map \psi 0 = 0} \subset \set {x \phi : \phi \in \map \DD \R}$
Thus:
:$\ker \delta \subseteq \set {x \phi : \phi \in \map \DD \R} \subseteq \ker T$
By Kernel of Linear Transformation contained in Kernel of different Linear Transformation implies Transformations are Proportional:
:$\exists c \in \C : T = c \delta$
== Sources ==
* : Chapter $\S 6.4$: A glimpse of distribution theory. Multiplication by $C^\infty$ functions
Category:Examples of Distributional Solutions
Distributional Solution to y' - k y = 0

== Theorem ==
Let $f \in \map {C^1} \R$ be a continuously differentiable function.
Let $T \in \map {\DD'} \R$ be a distribution.
Let $T_f$ be a distribution associated with $f$.
Let $\mathbf 0 \in \map {\DD'} \R$ be the zero distribution.
Let $T$ be a distributional solution to the following distributional differential equation:
:$\paren {\dfrac \d {\d x} - k} T = \mathbf 0$
Then $T = T_f$ where $f$ is a classical solution to the ordinary differential equation:
:$\paren {\dfrac \d {\d x} - k} f = 0$
== Proof ==
In the distributional sense we have:
By Vanishing Distributional Derivative of Distribution implies Distribution is Constant:
:$\exists c \in \C : \map \exp {-k x} T = T_c$
In other words:
If $f = c \map \exp {k x}$, then:
:$\paren {\dfrac \d {\d x} - k} f = 0$
== Sources ==
* : Chapter $\S 6.4$: A glimpse of distribution theory. Multiplication by $C^\infty$ functions
Category:Examples of Distributional Solutions
Category:Distributional Derivatives
Distributive Laws/Arithmetic

== Theorem ==
On all the number systems:
:natural numbers $\N$
:integers $\Z$
:rational numbers $\Q$
:real numbers $\R$
:complex numbers $\C$
the operation of multiplication is distributive over addition:
:$m \paren {n + p} = m n + m p$
:$\paren {m + n} p = m p + n p$
== Proof ==
This is demonstrated in these pages:
:Natural Number Multiplication Distributes over Addition
:Integer Multiplication Distributes over Addition
:Rational Multiplication Distributes over Addition
:Real Multiplication Distributes over Addition
:Complex Multiplication Distributes over Addition
== Examples ==
== Also known as ==
== Also see ==
* Modulo Multiplication Distributes over Modulo Addition
* Matrix Multiplication Distributes over Matrix Addition
* 's proofs:
** Multiplication of Numbers is Left Distributive over Addition
** Multiplication of Numbers is Right Distributive over Addition
Beware:
* Addition of Numbers is not Distributive over Multiplication
== Sources ==
* : Chapter $\text I$: Definitions. Elements of Vector Algebra: $1$. Scalar and Vector Quantities
* : $\S 1.1$. Number Systems: $\text{VII}.$
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers
* 
* 
* 
* : Chapter $1$: Elementary, my dear Watson: $\S 1.1$: You have a logical mind if...
* 
* 
* 
Category:Distributive Laws of Arithmetic
Category:Distributive Laws
Category:Multiplication
Category:Addition
Category:Examples of Distributive Operations
Category:Arithmetic
Category:Algebra
Distributive Laws/Set Theory

== Theorem ==
=== Intersection Distributes over Union ===
=== Union Distributes over Intersection ===
== Examples ==
== Also known as ==
== Sources ==
* : Chapter $0$: Subsets and Complements; Union and Intersection: Theorem $2 \ \text{(d)}$
* : $\S 4$: Unions and Intersections
* : A Note on Set-Theoretic Concepts: $(3)$
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
* : Chapter $1$: Set Theory: $1.3$: Set operations
* : $1$ Set Theory: $1$. Sets and Functions: $1.2$: Operations on sets: $(1)$
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 7 \ \text{(a)}$
* : $\S 7.4 \ \text{(i)}$: Unions and Intersections
* 
* 
* : $\text {(vi)}$
Category:Distributive Laws of Set Theory
Category:Set Union
Category:Set Intersection
Distributive Laws/Set Theory/Examples/A cap B cap (C cup D) subset of (A cap D) cup (B cap C)

== Example of Use of Distributive Laws of Set Theory ==
Let:
:$P = A \cap B \cap \paren {C \cup D}$
:$Q = \paren {A \cap D} \cup \paren {B \cap C}$
Then:
:$P \subseteq Q$
=== Corollary ===
== Proof ==
Let $P$ be expressed as:
:$P = X \cup Y$
where:
:$X = A \cap B \cap D$
:$Y = A \cap B \cap C$
Then:
and:
and so:
== Sources ==
* : Chapter $1$: Sets and Logic: Exercise $11$
Category:Examples of Use of Distributive Laws of Set Theory
Distributive Laws/Set Theory/Examples/A cap B cap (C cup D) subset of (A cap D) cup (B cap C)/Corollary

== Corollary to $A \cap B \cap \paren {C \cup D} \subseteq \paren {A \cap D} \cup \paren {B \cap C}$ ==
Let:
:$P = A \cap B \cap \paren {C \cup D}$
:$Q = \paren {A \cap D} \cup \paren {B \cap C}$
Then:
:$P = Q$
:
:both $B \cap C \subseteq A$ and $A \cap D \subseteq B$
== Proof ==
=== Sufficient Condition ===
Let $P = Q$.
and:
Thus we have both:
:$B \cap C \subseteq A$
and
:$A \cap D \subseteq B$
=== Necessary Condition ===
Let $B \cap C \subseteq A$ and $A \cap D \subseteq B$.
We have:
and:
and so:
and so $P = Q$ as required.
== Sources ==
* : Chapter $1$: Sets and Logic: Exercise $11$
Category:Examples of Use of Distributive Laws of Set Theory