nyuuzyou/wikis huggingface dataset:
Cardinality of Union not greater than Product

== Theorem ==
Let $\FF$ be a set of sets.
Let:
:$\size \FF \le \mathbf m$
where 
:$\size \FF$ denotes the cardinality of $\FF$
:$\mathbf m$ is cardinal number (possibly infinite).
Let:
:$\forall A \in \FF: \size A \le \mathbf n$
where
:$\mathbf n$ is cardinal number (possibly infinite).
Then:
:$\ds \size {\bigcup \FF} \le \size {\mathbf m \times \mathbf n} = \mathbf m \mathbf n$
== Proof ==
$\FF = \O$ or $\FF = \set \O$ or $\O \ne \FF \ne \set \O$.
In case when $\FF = \O$ or $\FF = \set \O$:
In case when $\O \ne \FF \ne \{\O\}$:
:by Surjection iff Cardinal Inequality there exists a surjection $f: \mathbf m \to \FF$ as $\size {\mathbf m} = \mathbf m$ by Cardinal of Cardinal Equal to Cardinal.
$\FF$ contains non empty set $A_0$.
:$\size {A_0} > \mathbf 0$.
By assumption:
:$\size {A_0} \le \mathbf n$.
Then:
:$\mathbf 0 < \mathbf n$.
Hence:
:$\size {\set 0} = \mathbf 1 \le \mathbf n$.
Define a family $\family {B_A}_{A \mathop \in \FF}$:
:$B_A = \begin{cases} A & A \ne \O \\ \set 0 & A = \O \end {cases}$
By Surjection iff Cardinal Inequality define a family
$\family {g_A}_{A \mathop \in \FF}$ of surjections:
:$\forall A \in \FF: g_A: \mathbf n \to B_A$ is a surjection.
Define a mapping $h:\mathbf m \times \mathbf n \to \ds \bigcup_{A \mathop \in \FF} B_A$:
:$\forall \alpha \in \mathbf m: \forall \beta \in \mathbf n: \map h {\alpha, \beta} = \map {g_{\map f \alpha} } \beta$
We will show by definition that $h$ is a surjection.
Let $x \in \ds \bigcup_{A \mathop \in \FF} B_A$.
Then by definition of union:
:$\exists A \in \FF: x \in B_A$.
By definition of surjection:
:$\exists \alpha \in \mathbf m: \map f \alpha = A$.
By definition of surjection:
:$\exists \beta \in \mathbf n: \map {g_A} \beta = x$
So:
:$\map h {\alpha, \beta} = \map {g_{\map f \alpha} } \beta = x$
This ends the proof of surjection.
Hence by Surjection iff Cardinal Inequality:
:$\ds \size {\bigcup_{A \mathop \in \FF} B_A} \le \size {\mathbf m \times \mathbf n}$
By definition of subset:
:$\forall A \in \FF: A \subseteq B_A$.
Then by Set Union Preserves Subsets:
:$\ds \bigcup \FF \subseteq \bigcup_{A \mathop \in \FF} B_A$.
Hence by Subset implies Cardinal Inequality:
:$\ds \size {\bigcup \FF} \le \ds \size {\bigcup_{A \mathop \in \FF} B_A}$
Thus the result:
:$\size {\bigcup \FF} \le \size {\mathbf m \times \mathbf n}$
Thus by definition of product of cardinals:
:$\size {\mathbf m \times \mathbf n} = \mathbf m \mathbf n$
== Sources ==
* 
Category:Cardinals
Category:Set Union
Category:Cardinality
Cardinals are Totally Ordered

== Corollary to Zermelo's Theorem ==
Every set of cardinals is totally ordered under $\le$.
== Proof ==
Let $S$ be a set of cardinals.
From Zermelo's theorem, $S$ is well-ordered.
Let $a, b \in S$.
Consider the subset $X = \set {a, b}$ of $S$.
Since $S$ is well-ordered, $\map \inf X$ exists and belongs to $X$.
So either $\map \inf X = a$ or $\map \inf X = b$.
By definition of infimum, we have that $\map \inf X \le a$ and $\map \inf X \le b$.
It follows that either $a \le b$ or $b \le a$.
So, by definition, $S$ is totally ordered under $\le$.
== Sources ==
* : $\S 8$: Theorem $8.3$: Corollary
Category:Cardinals
Category:Total Orderings
Cardinals form Equivalence Classes

== Theorem ==
Let $\map \Card S$ denote the cardinal of the set $S$.
Then $\map \Card S$ induces an equivalence class which contains all sets which have the same cardinality as $S$.
== Proof ==
Follows directly from:
:The definition of a cardinal as $S \sim T \iff \map \Card S = \map \Card T$
:Set Equivalence behaves like Equivalence Relation
:Relation Partitions Set iff Equivalence.
Category:Cardinals
Cardioid is Special Case of Pascal's Snail

== Theorem ==
Let $C$ be a circle of diameter $a$ with a distinguished point $O$ on the circumference.
Let $OQ$ be a chord of $C$.
Let $\LL$ be a limaçon of Pascal defined as the locus of points $P$ in the plane such that:
:$PQ = b$
where:
:$OPQ$ is a straight line
:$b$ is a real constant.
Let $a = b$.
Then $\LL$ is a cardioid
== Proof ==
From Equation of Limaçon of Pascal, the polar equation of $\LL$ can be expressed as:
:$r = b + a \cos \theta$
Setting $b = a$ results in the polar equation:
:$r = a \paren {1 + \cos \theta}$
which from Equation of Cardioid is the polar equation of the cardioid
== Sources ==
* 
* 
Category:Limaçons of Pascal
Category:Cardioids
Carmichael's Theorem

== Theorem ==
Let $n \in \Z$ such that $n > 12$.
Then the $n$th Fibonacci number $F_n$ has at least one prime factor which does not divide any smaller Fibonacci number.
The exceptions for $n \le 12$ are:
:$F_1 = 1, F_2 = 1$: neither have any prime factors
:$F_6 = 8$ whose only prime factor is $2$ which is $F_3$
:$F_{12} = 144$ whose only prime factors are $2$ (which is $F_3$) and $3$ (which is $F_4$).
== Proof ==
We have that:
:$1$ has no prime factors.
Hence, vacuously, $1$ has no primitive prime factors.
:$8 = 2^3$
and $2 \divides 2 = F_3$
:$144 = 2^4 3^2$
and:
:$2 \divides 8 = F_6$
:$3 \divides 21 = F_8$
for example.
== Also see ==
* Zsigmondy's Theorem
== Sources ==
* 
* : $144$
* : $144$
Category:Fibonacci Numbers
Category:Primitive Prime Factors
Carmichael's Theorem/Mistake

== Source Work ==
:
:The Dictionary
::$144$
:
:The Dictionary
::$144$
== Mistake ==
:A divisor of a Fibonacci number is called proper if it does not divide any smaller Fibonacci number. The only Fibonacci numbers that do not possess a proper divisor are $1$, $8$ and $144$.
The following points about the above are questionable.
:$(1): \quad$ The concept of a proper divisor of a Fibonacci number defined in this way has not been corroborated by means of an internet search. A proper divisor of an integer has a different definition.
:$(2): \quad$ Having defined a proper divisor of a Fibonacci number as above, it needs to be pointed out that $4$ is a proper divisor of $F_6 = 8$, as no Fibonacci number smaller than $8$ has $4$ as a divisor. Similarly, $F_{12} = 144$ has the proper divisor $72$, which, similarly, is not the divisor of any Fibonacci number smaller than $144$.
:$(3): \quad$ In its definition, a proper divisor of a Fibonacci number is not directly specified as having to be a proper divisor, in the conventional sense that a proper divisor of $n$ is a divisor of $n$ which is not $1$ and is not $n$.
:$(4): \quad$ The only divisor of $1$ is $1$. But $1$ vacuously is a divisor of no smaller Fibonacci numbers, and so, again by the above definition, is a proper divisor of $1$.
The conventional presentation of this result is Carmichael's Theorem, to discuss the primitive prime factors of a Fibonacci Number:
== Sources ==
* : $144$
* : $144$
Category:Mistakes/Curious and Interesting Numbers
Carmichael Number/Examples

== Examples of Carmichael Numbers==
=== $561$ is a Carmichael Number ===
=== $1105$ is a Carmichael Number ===
=== $1729$ is a Carmichael Number ===
=== $2465$ is a Carmichael Number ===
=== $41 \, 041$ is a Carmichael Number ===
=== $294 \, 409$ is a Carmichael Number ===
=== $509 \, 033 \, 161$ is a Carmichael Number ===
Category:Examples of Carmichael Numbers
Carmichael Number/Examples/1105

== Example of Carmichael Number==
$1105$ is a Carmichael number:
:$\forall a \in \Z: a \perp 1105: a^{1105} \equiv a \pmod {1105}$
while $1105$ is composite.
== Proof ==
We have that:
:$1105 = 5 \times 13 \times 17$
and so:
We also have that:
The result follows by Korselt's Theorem.
Category:Examples of Carmichael Numbers
Category:1105
Carmichael Number/Examples/1729

== Example of Carmichael Number==
$1729$ is a Carmichael number:
:$\forall a \in \Z: a \perp 1729: a^{1729} \equiv a \pmod {1729}$
while $1729$ is composite.
== Proof ==
We have that:
:$1729 = 7 \times 13 \times 19$
and so:
We also have that:
The result follows by Korselt's Theorem.
== Sources ==
* : $509,033,161$
Category:Examples of Carmichael Numbers
Category:1729
Carmichael Number/Examples/2465

== Example of Carmichael Number==
$2465$ is a Carmichael number:
:$\forall a \in \Z: a \perp 2465: a^{2465} \equiv a \pmod {2465}$
while $2465$ is composite.
== Proof ==
We have that:
:$2465 = 5 \times 17 \times 29$
and so:
We also have that:
The result follows by Korselt's Theorem.
== Sources ==
* : $2465$
Category:Examples of Carmichael Numbers
Category:2465
Carmichael Number/Examples/294,409

== Example of Carmichael Number ==
$294 \, 409$ is a Carmichael number:
:$\forall a \in \Z: a \perp 294 \, 409: a^{294 \, 409} \equiv a \pmod {294 \, 409}$
while $294 \, 409$ is composite.
== Proof ==
We have that:
:$294 \, 409 = 37 \times 73 \times 109$
First note that $294 \, 409$ is square-free.
Hence the square of none of its prime factors is a divisor of $294 \, 409$:
:$\forall p \divides 294 \, 409: p^2 \nmid 294 \, 409$
We also see that:
Thus $294 \, 409$ is a Carmichael number by Korselt's Theorem.
== Sources ==
* : $509,033,161$
Category:Examples of Carmichael Numbers
Category:294,409
Carmichael Number/Examples/41,041

== Example of Carmichael Number==
$41 \, 041$ is a Carmichael number:
:$\forall a \in \Z: a \perp 41 \, 041: a^{41 \, 041} \equiv a \pmod {41 \, 041}$
while $41 \, 041$ is composite.
== Proof ==
We have that:
:$41,041 = 7 \times 11 \times 13 \times 41$
and so:
We also have that:
The result follows by Korselt's Theorem.
Category:Examples of Carmichael Numbers
Category:41,041
Carmichael Number/Examples/509,033,161

== Example of Carmichael Number ==
$509 \, 033 \, 161$ is a Carmichael number:
:$\forall a \in \Z: a \perp 509 \, 033 \, 161: a^{509 \, 033 \, 161} \equiv a \pmod {509 \, 033 \, 161}$
while $509 \, 033 \, 161$ is composite.
Also:
:$509 \, 033 \, 161 = 1729 \times 294 \, 409$
while both $1729$ and $294 \, 409$ are themselves Carmichael numbers.
== Proof ==
We have that:
:$509 \, 033 \, 161 = 7 \times 13 \times 19 \times 37 \times 73 \times 109$
First note that $509 \, 033 \, 161$ is square-free.
Hence the square of none of its prime factors is a divisor of $509 \, 033 \, 161$:
:$\forall p \divides 509 \, 033 \, 161: p^2 \nmid 509 \, 033 \, 161$
We also see that:
Thus $509 \, 033 \, 161$ is a Carmichael number by Korselt's Theorem.
Then we have:
:$1729$ is a Carmichael number
and:
:$294 \, 409$ is a Carmichael number.
== Sources ==
* : $509,033,161$
Category:Examples of Carmichael Numbers
Category:509,033,161
Carmichael Number/Examples/561

== Example of Carmichael Number==
$561$ is a Carmichael number:
:$\forall a \in \Z: a \perp 561: a^{561} \equiv a \pmod {561}$
while $561$ is composite.
== Proof ==
We have that:
:$561 = 3 \times 11 \times 17$
and so:
We also have that:
The result follows by Korselt's Theorem.
Category:Examples of Carmichael Numbers
Category:561
Carmichael Number has 3 Odd Prime Factors

== Theorem ==
Let $n$ be a Carmichael number.
Then $n$ has at least $3$ distinct odd prime factors.
== Proof ==
By Korselt's Theorem, $n$ is odd.
Therefore $n$ has at least $1$ odd prime factor.
By Korselt's Theorem, for each prime factor of $n$:
:$p^2 \nmid n$
:$\paren {p - 1} \divides \paren {n - 1}$
Suppose $n = p^k$ for some odd prime $p$.
By Korselt's Theorem, $k = 1$.
However by definition of a Carmichael Number, $n$ cannot be prime.
Therefore $n$ has at least $2$ distinct odd prime factors.
Suppose $n = p^a q^b$ for distinct odd primes $p$ and $q$.
By Korselt's Theorem, the following holds:
:$a = b = 1$
:$n = p q$
:$\paren {p - 1} \divides \paren {n - 1}$
:$\paren {q - 1} \divides \paren {n - 1}$
Hence:
Swapping $p$ and $q$ yields $\paren {q - 1} \divides \paren {p - 1}$.
Hence $p - 1 = q - 1$ and $p = q$, which is a contradiction.
Therefore $n$ has at least $3$ distinct odd prime factors.
== Historical Note ==
== Sources ==
* : $561$
* : $561$
Category:Carmichael Numbers
Carmichael Number has 3 Odd Prime Factors/Historical Note

== Historical Note on Carmichael Number has 3 Odd Prime Factors ==
 proved that a Carmichael Number has at least 3 distinct odd prime factors in $1912$, at around the same time that he discovered that $561$ was the smallest one.
== Sources ==
* : $561$
* : $561$
Category:Historical Notes
Carmichael Number with 4 Prime Factors

== Theorem ==
$41 \, 041$ is the smallest Carmichael number with $4$ prime factors:
:$41 \, 041 = 7 \times 11 \times 13 \times 41$
== Proof ==
From Sequence of Carmichael Numbers:
The sequence continues:
:$\ldots, 29 \, 341, 41 \, 041$
We now investigate their prime factors:
== Sources ==
* 
* : $41,041$
Category:Carmichael Numbers
Category:41,041
Carmichael Number with 4 Prime Factors/Mistake

== Source Work ==
:
:The Dictionary
::$41,041$
== Mistake ==
:The smallest Carmichael number with $4$ factors. It equals $7 \times 11 \times 13 \times 41$.
It should say:
:The smallest Carmichael number with $4$ prime factors.
== Sources ==
* : $41,041$
Category:Mistakes/Curious and Interesting Numbers
Carroll Paradox

== Paradox ==
Modus Ponendo Ponens leads to infinite regress.
== Proof ==
To be proven: $q$.
:$(1).\quad$ Assume $p \implies q$.
:$(2).\quad$ Assume $p$.
:$(3).\quad$ $p \land \paren {p \implies q} \vdash q$ by Modus Ponendo Ponens.
:$(4).\quad$ From $(2)$ and $(1)$, $p \land \paren {p \implies q}$.
:$(5).\quad \paren {p \land \paren {p \implies q} \land \paren {p \land \paren {p \implies q} } \vdash q} \vdash q$.
:$(6).\quad$ From $(4)$ and $(3)$, $\paren {p \land \paren {p \implies q} } \land \paren {\paren {p \land \paren {p \implies q} } \vdash q}$.
$\ldots$
and so (or, as put it, ad nauseaum).
== Resolution ==
This is an antinomy.
It arises because of confusion between an logical axiom and a rule of inference.
In this context, Modus Ponendo Ponens is a rule of inference.
== Also known as ==
== Historical Note ==
== Sources ==
* : Two-Part Invention, quoting What the Tortoise Said to Achilles by 
* 
* 
Category:Carroll Paradox
Category:Antinomies
Cartan-Hadamard Theorem

== Theorem ==
Let $M$ be a complete connected $n$-dimensional Riemannian manifold.
Suppose all sectional curvatures of $M$ are less than or equal to zero.
Then the universal covering space of $M$ is diffeomorphic to $\R^n$.
== Proof ==
== Sources ==
* : $\S 1$: What Is Curvature? Curvature in Higher Dimensions
Category:Riemannian Manifolds
Cartesian Definition of Gradient defines Gradient Operator

== Theorem ==
Let $R$ be a region of Cartesian $3$-space $\R^3$.
Let $\map F {x, y, z}$ be a scalar field acting over $R$.
Let $\tuple {i, j, k}$ be the standard ordered basis on $\R^3$.
Let $\grad F$ be defined according to the Cartesian $3$-space definition of the gradient of $F$:
Then $\grad F$ is a gradient operator as defined by the geometrical representation:
:$\grad F = \dfrac {\partial F} {\partial n} \mathbf {\hat n}$
where:
:$\mathbf {\hat n}$ denotes the unit normal to the equal surface $S$ of $F$ at $A$
:$n$ is the magnitude of the normal vector to $S$ at $A$.
== Proof ==
The vector rates of increase of $F$ in the directions of the $3$ axes are:
:$\dfrac {\partial F} {\partial x} \mathbf i$, $\dfrac {\partial F} {\partial y} \mathbf j$, $\dfrac {\partial F} {\partial z} \mathbf k$
Their sum will be a vector with the magnitude and direction of the most rapid rate of increase of $F$.
It remains to show that this expression:
:$\dfrac {\partial F} {\partial x} \mathbf i + \dfrac {\partial F} {\partial y} \mathbf j + \dfrac {\partial F} {\partial z} \mathbf k$
is equivalent to:
:$\dfrac {\partial F} {\partial n} \mathbf {\hat n}$
Let us take the dot product of both sides of the gradient equation with the position vector $\d \mathbf r$ of an arbitrary point $A$ on an equal surface $S$ of $F$.
Thus:
In Cartesian coordinates:
Thus the operations $\grad$ and $\nabla$ applied to a point in a scalar field are identical.
== Sources ==
* : Chapter $\text {IV}$: The Operator $\nabla$ and its Uses: $2 a$. The Operation $\nabla S$: $(4.3)$
Category:Gradient Operator
Cartesian Equation of Conchoid of Nicomedes

== Theorem ==
Let $a \in \R$, $b \in \R_{>0}$ be real constants.
Let the focus point of a conchoid of Nicomedes $\KK$ be located at the origin of a Cartesian plane.
Let the directrix of $\KK$ be the straight line $x = a$.
Then $\KK$ can be expressed in Cartesian coordinates as:
:$\paren {x - a}^2 \paren {x^2 + y^2} = b^2 x^2$
== Proof ==
== Also presented as ==
== Also see ==
* Polar Equation of Conchoid of Nicomedes
== Sources ==
* 
* 
* 
* 
Category:Conchoid of Nicomedes
Cartesian Equation of Conchoid of Nicomedes/Also presented as

== Cartesian Equation of Conchoid of Nicomedes: Also presented as ==
The directrix of $\KK$ can also be seen expressed in Cartesian coordinates in the form:
:$\paren {a - b - x} \paren {a + b - x} x^2 + \paren {a - x}^2 y^2 = 0$ 
== Proof ==
from which it follows that the given form is equivalent to:
:$\paren {x - a}^2 \paren {x^2 + y^2} = b^2 x^2$
== Sources ==
* 
Category:Conchoid of Nicomedes
Cartesian Equation of Witch of Agnesi/Also presented as

== Theorem ==
The cartesian equation of the Witch of Agnesi can also be seen presented as:
:$x^2 y = 4 a^2 \paren {2 a - y}$
Some sources present the curve with the $y$-axis as the asymptote:
:$x y^2 = 4 a^2 \paren {2 a - x}$
== Sources ==
* 
* 
* 
Category:Witch of Agnesi
Cartesian Form of Continuous Moment of Inertia

== Theorem ==
Let $B$ be a rigid body which is rotating in space about some axis $\LL$.
Let $B$ be embedded in Cartesian $3$-space whose origin lies on $\LL$.
Let $I_{xx}$, $I_{yy}$, and $I_{zz}$ be the moments of inertia of $B$ about the coordinate axes $\mathbf O x$, $\mathbf O y$ and $\mathbf O z$ respectively.
Then:
where:
:$\rho$ denotes the density of $B$
:$\d V$ denotes a volume element of $B$.
== Proof ==
== Sources ==
* 
* 
Category:Moments of Inertia
Cartesian Form of Discrete Moment of Inertia

== Theorem ==
Let $B$ be a rigid body which is rotating in space about some axis $\LL$.
Let $B$ be embedded in Cartesian $3$-space whose origin lies on $\LL$.
Let $I_{xx}$, $I_{yy}$, and $I_{zz}$ be the moments of inertia of $B$ about the coordinate axes $\mathbf O x$, $\mathbf O y$ and $\mathbf O z$ respectively.
Let $B$ be composed of a countable number of particles $P_1, P_2, \ldots$ such that:
:each $P_i$ has mass $m_i$
:each $P_i$ has perpendicular distance $r_i$ from $\LL$.
:each $P_i$ has coordinates $\tuple {x_i, y_i, z_i}$.
Then:
== Proof ==
== Sources ==
* 
* 
Category:Moments of Inertia
Cartesian Form of Moment of Inertia about Axis of Rotation

== Theorem ==
Let $B$ be a rigid body which is rotating in space about some axis $\LL$.
Let $B$ be embedded in Cartesian $3$-space whose origin lies on $\LL$.
Let $I_{xx}$, $I_{yy}$, and $I_{zz}$ be the moments of inertia of $B$ about the coordinate axes $\mathbf O x$, $\mathbf O y$ and $\mathbf O z$ respectively.
Let $I_{xy}$, $I_{yz}$, and $I_{zx}$ be the products of inertia of $B$ about the coordinate axes $\mathbf O x$, $\mathbf O y$ and $\mathbf O z$ respectively.
Then the moment of inertia $I$ of $B$ about $\LL$ is given by:
:$I = I_{xx} l^2 + I_{yy} m^2 + I_{zz} n^2 - 2 I_{yz} m n - 2 I_{zx} n l - 2 I_{xy} l m$
where $l$, $m$ and $n$ are the direction cosines of $\LL$ $\mathbf O x$, $\mathbf O y$ and $\mathbf O z$.
== Proof ==
== Sources ==
* 
* 
Category:Moments of Inertia
Cartesian Metric is Rotation Invariant

== Theorem ==
The cartesian metric does not change under rotation.
== Proof ==
Let the cartesian metric be $\delta_{ij} = \innerprod {e_i} {e_j}$.
Also, let $\delta_{ij}'$ be the metric of the coordinate system of $\delta_{ij}$ rotated by a rotation matrix $A$.
Then, $\delta_{ij}' = \innerprod {A e_i} {A e_j}$.
$\innerprod {A x} y = \innerprod x {A^T y}$, see Factor Matrix in the Inner Product, so
$\delta_{ij}' = \innerprod {A e_i} {A e_j} = \innerprod {e_i} {A^T A e_j}$.
For rotation matrices, we have $A^T = A^{-1}$, so $\delta_{ij}'$ reduces to:
:$\delta_{ij}' = \innerprod {e_i} {A^T A e_j} = \innerprod {e_i} {A^{-1} A e_j} = \innerprod {e_i} {I e_j} = \innerprod {e_i} {e_j}$
where $I$ is the unit matrix.
Thus:
:$\delta_{ij}' = \delta_{ij}$
The result follows.
This proof holds for any finite-dimensional Euclidean space.
Category:Metric Spaces
Cartesian Plane Rotated with respect to Another

== Theorem ==
Let $\mathbf r$ be a position vector embedded in a Cartesian plane $\CC$ with origin $O$.
Let $\CC$ be rotated anticlockwise through an angle $\varphi$ about the axis of rotation $O$.
Let $\CC'$ denote the Cartesian plane in its new position.
Let $\mathbf r$ be kept fixed during this rotation.
Let $\tuple {x, y}$ denote the components of $\mathbf r$ with respect to $\CC$.
Let $\tuple {x', y'}$ denote the components of $\mathbf r$ with respect to $\CC'$.
Then:
== Proof ==
:
Let $\mathbf r$ be represented by a directed line segment whose initial point coincides with the origin $O$.
Let the terminal point of $\mathbf r$ be identified with the point $P$.
Let $\CC$ be rotated to $\CC'$ through an angle $\varphi$ as shown, keeping $P$ fixed.
We have that:
hence the result.
== Sources ==
* : Chapter $1$ Vector Analysis $1.2$ Rotation of Coordinates
* 
Category:Coordinate Systems
Category:Geometric Rotations
Category:Vectors
Cartesian Product/Examples/Arbitrary Example 1

== Example of Cartesian Product ==
Let:
Then:
:$A \times B = \set {\tuple {1, 2}, \tuple {1, 3}, \tuple {2, 2}, \tuple {2, 3} }$
== Sources ==
* : Chapter $1$: Preliminaries: $1.4$ Set Notation: Example $1.2$
Category:Examples of Cartesian Products
Cartesian Product Distributes over Set Difference

== Theorem ==
Cartesian product is distributive over set difference:
:$(1): \quad S \times \paren {T_1 \setminus T_2} = \paren {S \times T_1} \setminus \paren {S \times T_2}$
:$(2): \quad \paren {T_1 \setminus T_2} \times S = \paren {T_1 \times S} \setminus \paren {T_2 \times S}$
== Proof ==
== Sources ==
* : $\S 6$: Ordered Pairs: Exercise $\text{(iii)}$
* : $1$: Set Theory and Logic: $\S 1$: Fundamental Concepts: Exercise $1.2 \ \text{(o)}$
Category:Cartesian Product
Category:Set Difference
Category:Examples of Distributive Operations
Cartesian Product Distributes over Union

== Theorem ==
Cartesian product is distributive over union:
:$A \times \paren {B \cup C} = \paren {A \times B} \cup \paren {A \times C}$
:$\paren {B \cup C} \times A = \paren {B \times A} \cup \paren {C \times A}$
== Proof ==
Take the result Cartesian Product of Unions:
:$\paren {S_1 \cup S_2} \times \paren {T_1 \cup T_2} = \paren {S_1 \times T_1} \cup \paren {S_2 \times T_2} \cup \paren {S_1 \times T_2} \cup \paren {S_2 \times T_1}$
Put $S_1 = S_2 = A, T_1 = B, T_2 = C$:
Thus:
:$A \times \paren {B \cup C} = \paren {A \times B} \cup \paren {A \times C}$
The other result is proved similarly.
== Sources ==
* : $\S 6$: Ordered Pairs: Exercise $\text{(i)}$
* : Chapter $1$. Sets: Exercise $8 \ \text{(ii)}$
* : $\S 3$. Ordered pairs; cartesian product sets: Exercise $3 \ (2)$
* : Chapter $1$: Sets and Logic: Exercise $15$
Category:Cartesian Product of Unions
Category:Cartesian Product
Category:Set Union
Category:Examples of Distributive Operations
Cartesian Product Exists and is Unique

== Theorem ==
Let $A$ and $B$ be classes.
Let $A \times B$ be the cartesian product of $A$ and $B$.
Then $A \times B$ exists and is unique.
== Proof ==
Let $A \times B$ be the cartesian product of $A$ and $B$.
Let $\tuple {x, y} \in A \times B$ such that $\tuple {x, y}$ satisfies the Kuratowski ordered pair formulation.
By Binary Cartesian Product in Kuratowski Formalization contained in Power Set of Power Set of Union:
:$A \times B \subseteq \powerset {\powerset {A \cup B} }$
By Axiom of Specification, there exists a set:
:$C = \set {\tuple {x, y} \in \powerset {\powerset {A \cup B} }: x \in A \land y \in B}$
By definition, $C$ is the cartesian product of $A$ and $B$.
Thus $A \times B$ exists.
By Cartesian Product is Unique, $A \times B$ is unique.
== Sources ==
* : Chapter $1$: Power Set
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 7$ Cartesian products
Category:Cartesian Product