nyuuzyou/wikis huggingface dataset:
Problem of Apollonius/Circles

== Classic Problem ==
Let there be three circles in the plane.
It is required to draw another circle tangent to each of the three.
== Solution ==
It is generically possible to construct such a circle in $8$ different ways.
Each tangent circle encloses a subset of the three original circles.
From Cardinality of Power Set of Finite Set, there are thus $2^3 = 8$ distinct such tangent circles
== Proof ==
== Historical Note ==
== Sources ==
* : Chapter $\text{III}$: Gentleman, Soldier and Mathematician
* : Chapter $\text {A}.6$: Apollonius (ca. B.C.)
* 
* 
Category:Problem of Apollonius
Category:Circles
Problem of Apollonius/Historical Note

== Historical Note on the Problem of Apollonius ==
The Problem of Apollonius was originally posed, and apparently solved, by in his work .
The cases of $3$ points and $3$ straight lines were given by in his Elements: Book $\text {IV}$.
The case of the $3$ circles was interesting enough to attract the attention of a number of mathematicians of the $17$th century, including and .
It was supposedly solved by during the course of a series of mathematical discussions with sometime around or after $1641$.
While this in itself is a remarkable feat, the fact that she achieved it using ' own methods makes it more impressive, as these are not the easiest tools to use to attack the problem.
However, appears to have been unfairly and cruelly dismissive of her achievement
== Sources ==
* : Chapter $\text{III}$: Gentleman, Soldier and Mathematician
* 
* 
Category:Historical Notes
Problem of Apollonius/Points

== Classic Problem ==
Let there be three points in the plane which are not collinear.
It is required to draw a circle passing through each of the three points.
== Solution ==
Let the points be $A$, $B$ and $C$.
As $A$, $B$ and $C$ are not collinear, the triangle $ABC$ can be constructed by forming the lines $AB$, $BC$ and $CA$.
== Proof ==
Follows from Circumscribing Circle about Triangle.
== Historical Note ==
== Sources ==
* : $3$
* : Chapter $\text {A}.6$: Apollonius (ca. B.C.)
* : $3$
* 
* 
Category:Problem of Apollonius
Category:Points
Category:Circles
Problem of Apollonius/Straight Lines

== Classic Problem ==
Let there be three straight lines in the plane which are not parallel.
It is required to draw a circle tangent to each of the three straight lines.
== Solution ==
As the straight lines are not parallel, they form a triangle .
== Proof ==
Follows from Inscribing Circle in Triangle.
== Historical Note ==
== Sources ==
* : Chapter $\text {A}.6$: Apollonius (ca. B.C.)
* 
* 
Category:Problem of Apollonius
Category:Straight Lines
Category:Circles
Problem of Pappus/Historical Note

== Historical Note on the Problem of Pappus ==
The Problem of Pappus was originally posed by in Book $\text {VII}$ of his .
He generalized a problem of known as the Four-Line Locus:
:Find the locus of a point $P$ such that the product of the distances from $P$ to any two of them is proportional to the product of the distances from $P$ to the other two.
This locus is in fact a conic section.
It was by considering this problem in $1631$ that was led to invent analytic geometry.
== Sources ==
* 
* 
Category:Historical Notes
Producer of Dedekind Cut/Examples/2

== Example of Producer of Dedekind Cut ==
Let $L = \set {l \in \Q: l \le 2}$ and $R = \set {r \in \Q: 2 < r}$.
Then $\tuple {L, R}$ is a dedekind cut of $\Q$, and $2$ is a producer of $\tuple {L, R}$
Category:Dedekind Cuts
Producer of Dedekind Cut/Examples/sqrt2

== Example of Producer of Dedekind Cut ==
Let $L = \set {l \in \Q: l < \sqrt 2}$ and $R = \set {r \in \Q: \sqrt 2 < r}$.
Then $\tuple {L, R}$ is a Dedekind cut of $\Q$, and $\sqrt 2 \notin \Q$ is a producer of $\tuple {L, R}$.
Category:Dedekind Cuts
Product (Category Theory) is Unique

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $A$ and $B$ be objects of $\mathbf C$.
Let $A \times B$ and $A \times' B$ both be products of $A$ and $B$.
Then there is a unique isomorphism $u: A \times B \to A \times' B$.
That is, products are unique up to unique isomorphism.
== Proof ==
Denote the implicit projections of the two products by:
$\begin{xy}\xymatrix@C=2em@L+4mu{
&
 A \times B
 \ar@/_/[ddl]_*{\textrm{pr}_1}
 \ar@/^/[ddr]^*{\textrm{pr}_2}
\\ &
 A \times' B
 \ar[dl]^*{\textrm{pr}'_1}
 \ar[dr]_*{\textrm{pr}'_2}
\\
 A
& &
 B
}\end{xy}$
By the UMP of $A \times' B$, there is a unique $u: A \times B \to A \times' B$ fitting the dashed arrow in:
$\begin{xy}\xymatrix@C=2em@L+4mu{
&
 A \times B
 \ar@/_/[ddl]_*{\textrm{pr}_1}
 \ar@/^/[ddr]^*{\textrm{pr}_2}
 \ar@{-->}[d]
\\ &
 A \times' B
 \ar[dl]^*{\textrm{pr}'_1}
 \ar[dr]_*{\textrm{pr}'_2}
\\
 A
& &
 B
}\end{xy}$
By the UMP of $A \times B$, there is a further unique $v: A \times' B \to A \times B$ fitting the dashed arrow in:
$\begin{xy}\xymatrix@C=2em@L+4mu{
&
 A \times B
 \ar@/_/[ddl]_*{\textrm{pr}_1}
 \ar@/^/[ddr]^*{\textrm{pr}_2}
 \ar[d]^*{u}
\\ &
 A \times' B
 \ar[dl]^*{\textrm{pr}'_1}
 \ar[dr]_*{\textrm{pr}'_2}
 \ar@{-->}[d]
\\
 A
&
 A \times B
 \ar[l]^*{\textrm{pr}_1}
 \ar[r]_*{\textrm{pr}_2}
&
 B
}\end{xy}$
We can now form the following diagram:
$\begin{xy}\xymatrix@C=2em@R=3em@L+4mu{
 &
A \times B
 \ar[dl]_*{\textrm{pr}_1}
 \ar[dr]^*{\textrm{pr}_2}
 \ar@{-->}[d]
\\
 A
&
 A \times B
 \ar[l]^*{\textrm{pr}_1}
 \ar[r]_*{\textrm{pr}_2}
&
 B
}\end{xy}$
where we see both the identity morphism $\operatorname{id}_{A \times B}$ and $v \circ u$ fit the dashed arrow.
The UMP of $A \times B$ however states that there is a unique fitting morphism.
Hence $v \circ u = \operatorname{id}_{A \times B}$.
That $u \circ v = \operatorname{id}_{A \times' B}$ follows, mutatis mutandis, by interchanging $A \times B$ and $A \times' B$.
Hence there is a unique isomorphism $u: A \times B \to A \times' B$, as desired.
== Sources ==
* : $\S 2.4$: Proposition $2.17$
Category:Category Theory
Product Category is Category

== Theorem ==
Let $\mathbf C$ and $\mathbf D$ be metacategories.
Then the product category $\mathbf C \times \mathbf D$ is a metacategory.
== Proof ==
Let $\left({X,Y}\right), \left({X',Y'}\right) \in \mathbf C \times \mathbf D$.
Let $\left({f,g}\right) : \left({X,Y}\right) \to \left({X',Y'}\right)$ and $\left({h,k}\right) : \left({X',Y'}\right) \to \left({X,Y}\right)$ be morphisms.
Let $\operatorname{id}_X$, $\operatorname{id}_Y$ be the identity morphisms for the objects $X$ and $Y$ respectively.
Then:
Similarly:
Therefore, $\left({\operatorname{id}_X, \operatorname{id}_Y}\right)$ satisfies the property of an identity morphism.
Let $\left({f, g}\right)$, $\left({h, k}\right)$ and $\left({\ell, m}\right)$ be composable morphisms of $\mathbf C \times \mathbf D$. We have:
Therefore, composition of morphisms in $\mathbf C \times \mathbf D$ is also associative.
Category:Category Theory
Product Category is Product in Category of Categories

== Theorem ==
Let $\mathbf{Cat}$ be the category of categories.
Let $\mathbf C$ and $\mathbf D$ be small categories, and let $\mathbf C \times \mathbf D$ be their product category.
Then $\mathbf C \times \mathbf D$ is a binary product of $\mathbf C$ and $\mathbf D$ in $\mathbf{Cat}$.
== Proof ==
Let $F: \mathbf A \to \mathbf C$ and $G: \mathbf A \to \mathbf D$ be morphisms in $\mathbf{Cat}$, i.e. functors.
Suppose $X: \mathbf A \to \mathbf C \times \mathbf D$ is a functor such that $\pr_1 \circ X = F$ and $\pr_2 \circ X = G$:
::$\begin{xy}\xymatrix@C=3em@R=3em@L+3mu{
&
 \mathbf A
 \ar[dl]_*{F}
 \ar[d]^*{X}
 \ar[dr]^*{G}
\\
 \mathbf C
&
 \mathbf C \times \mathbf D
 \ar[l]^*{\pr_1}
 \ar[r]_*{\pr_2}
&
 \mathbf D
}\end{xy}$
Here, $\pr_1$ and $\pr_2$ are projection functors.
Thus, for all objects $A$ of $\mathbf A$, it is required that:
:$\map {\pr_1} {X A} = F A$
:$\map {\pr_2} {X A} = G A$
meaning that $X A = \tuple {F A, G A}$ by definition of the objects of the product category $\mathbf C \times \mathbf D$.
Similarly, for all morphisms $f$ of $\mathbf A$, it is required that:
:$\map {\pr_1} {X f} = F f$
:$\map {\pr_2} {X f} = G f$
meaning that $X f = \tuple {F f, G f}$ by definition of the morphisms of the product category $\mathbf C \times \mathbf D$.
This uniquely determines the functor $X$, if it exists.
That this definition actually yields a functor is shown on Functor to Product Category.
Hence $\mathbf C \times \mathbf D$ is a binary product of $\mathbf C$ and $\mathbf D$ in $\mathbf{Cat}$.
== Sources ==
* : $\S 2.5.3$
Category:Category of Categories
Product Distributes over Modulo Operation

== Theorem ==
Let $x, y, z \in \R$ be real numbers.
Let $x \bmod y$ denote the modulo operation.
Then:
:$z \paren {x \bmod y} = \paren {z x} \bmod \paren {z y}$
== Proof ==
By definition of modulo operation:
:$x \bmod y := \begin {cases}
x - y \floor {\dfrac x y} & : y \ne 0 \\
x & : y = 0
\end {cases}$
If $y = 0$ we have that:
:$z \paren {x \bmod 0} = z x = \paren {z x} \bmod \paren {z 0}$
If $y \ne 0$ we have that:
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $15$
Category:Modulo Arithmetic
Product Equation for Riemann Zeta Function

== Theorem ==
There exists a constant $B$ such that:
:$\ds \frac {\map {\zeta'} s} {\map \zeta s} = B - \frac 1 {s - 1} + \frac 1 2 \ln \pi - \frac 1 2 \frac {\map {\Gamma'} {s / 2 + 1} } {\map \Gamma {s / 2 + 1} } + \sum_\rho \paren {\frac 1 {s - \rho} + \frac 1 \rho}$
where:
:$\zeta$ is the Riemann zeta function
:$\rho$ runs over the nontrivial zeros of $\zeta$
:$\Gamma$ is the gamma function.
== Proof ==
Let $\xi$ be the completed Riemann zeta function:
We have that the Completed Riemann Zeta Function has Order One.
So, by the Hadamard Factorisation Theorem, there exist constants $A$, $B$ such that:
:$\ds \map \xi s = \map \exp {A + B s} : \prod_\rho \paren {1 - \frac s \rho} \map \exp {\frac s \rho}$
where $\rho$ runs over the zeros of $\xi$, that is, the nontrivial zeros of $\zeta$.
Therefore:
Combining $(1)$ and $(2)$ we have:
== Warning ==
Category:Riemann Zeta Function
Category:Product Equation for Riemann Zeta Function
Product Equation for Riemann Zeta Function/Warning

== Warning concerning Product Equation for Riemann Zeta Function ==
== Warning ==
The sum $\ds \sum_\rho \frac 1 {\size \rho}$ diverges, so we must be careful of the order in which we take the terms.
By the Functional Equation for Riemann Zeta Function, the zeroes occur in complex conjugate pairs, and:
:$\ds \frac 1 \rho + \frac 1 {\overline \rho} = \frac {2 \map \Re \rho} {\size \rho^2} \le \frac 2 {\size \rho^2}$
and we see by the corollary to Zeroes of Functions of Finite Order that a sum of such terms does converge.
Category:Product Equation for Riemann Zeta Function
Product Form of Sum on Completely Multiplicative Function

== Theorem ==
Let $f$ be a completely multiplicative arithmetic function.
Let the series $\ds \sum_{n \mathop = 1}^\infty \map f n$ be absolutely convergent.
Then:
:$\ds \sum_{n \mathop = 1}^\infty \map f n = \prod_p \frac 1 {1 - \map f p}$
where the infinite product ranges over the primes.
== Proof ==
Define $P$ by:
Change the summing variable using:
The Fundamental Theorem of Arithmetic guarantees a unique factorization for each positive natural number.
Therefore this function is one to one:
:$\ds \map h v = \prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } p^{v_p}$
Then:
where $\map Q {A, K}$ is defined as:
:$\ds \map Q {A, K} := \set {\prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } p^{-v_p} : v \in \prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } \set {0 \,.\,.\, K} }$
Consider:
The construction defines it as the set of all possible products of positive powers of primes.
From the definition of a prime number, every positive natural number may be expressed as a prime or a product of powers of primes:
:$k \in \N^+ \implies k \in W$
and also every element of W is a positive natural number:
:$k \in W \implies k \in \N^+$
So $W = \N^+$.
Then taking limits on $\map P {A, K}$:
== Note ==
When the function $f$ is multiplicative but not completely multiplicative, the above derivation is still valid, except that we do not have the equality:
:$\dfrac 1 {1 - \map f p} = \paren {1 + \map f p + \map f {p^2} + \cdots}$
Therefore, in this case we may write:
:$\ds \sum_{n \mathop = 1}^\infty \map f n = \prod_p \paren {1 + \map f p + \map f {p^2} + \cdots}$
Category:Completely Multiplicative Functions
Product Formula for Norms on Non-zero Rationals

== Theorem ==
Let $\Q_{\ne 0}$ be the set of non-zero rational numbers.
Let $\Bbb P$ denote the set of prime numbers.
Let $a \in \Q_{\ne 0}$.
Then the following infinite product converges:
:$\size a \times \ds \prod_{p \mathop \in \Bbb P}^{} \norm a_p = 1$
where:
:$\size {\,\cdot\,}$ is the absolute value on $\Q$
:$\norm {\,\cdot\,}_p$ is the $p$-adic norm on $\Q$ for prime number $p$
== Proof ==
=== Lemma ===
Let $a = \dfrac b c$, where $b, c \in \Z_{\ne 0}$.
From the Lemma, the following infinite products converge:
:$\size b \ds \times \prod_{p \mathop \in \Bbb P} \norm b_p = 1$
:$\size c \ds \times \prod_{p \mathop \in \Bbb P} \norm c_p = 1$
From Quotient Rule for Real Sequences, the following infinite product converges:
:$\ds \dfrac {\size b} {\size c} \times \prod_{p \mathop \in \Bbb P } \dfrac {\norm b_p} {\norm c_p} = \dfrac 1 1 = 1 $
We have:
== Sources ==
* : $\S 3.1$ Absolute Values on $\Q$: Proposition $3.1.4$ 
* : $\S 1.9$ Metrics and norms on the rational numbers. Ostrowski’s Theorem: Proposition $1.51$
Category:P-adic Number Theory
Product Formula for Norms on Non-zero Rationals/Lemma

== Theorem ==
Let $\Q$ be the set of rational numbers.
Let $\Bbb P$ denote the set of prime numbers.
Let $z \in \Z_{\ne 0}$.
Then the following infinite product converges:
:$\size z \times \ds \prod_{p \mathop \in \Bbb P}^{} \norm z_p = 1$
where:
:$\size {\,\cdot\,}$ is the absolute value on $\Q$
:$\norm {\,\cdot\,}_p$ is the $p$-adic norm on $\Q$ for prime number $p$
== Proof ==
=== Case 1 : $z \in \Z_{>0}$ ===
Let $z \in \Z_{>0}$.
From Fundamental Theorem of Arithmetic, we can factor $z$ as a product of one or more primes:
:$z = p_1^{b_1} p_2^{b_2} \dots p_k^{b_k}$
Then for every prime number $q$:
:$\norm z_q = \begin{cases}
p_i^{-b_i} & : \exists i \in \closedint 1 k :q = p_i \\
1 & : \forall i \in \closedint 1 k : q \ne p_i \\
\end {cases}$
By definition of absolute value on $\Q$:
:$\size z = p_1^{b_1} p_2^{b_2} \dots p_k^{b_k} $
For $n \ge \max \set{p_1, p_2 \dots p_k}$:
Hence:
=== Case 2 : $z \in \Z_{<0}$ ===
Let $z \in \Z_{<0}$.
Hence
: $-z \in \Z_{>0}$.
We have:
In either case:
:$\size z \times \ds \prod_{p \mathop \in \Bbb P} \norm z_p = 1$
Category:P-adic Number Theory
Product Formula for Sine

== Theorem ==
:$\ds \map \sin {n z} = 2^{n - 1} \prod_{k \mathop = 0}^{n - 1} \map \sin {z + \frac {k \pi} n}$
=== Corollary ===
== Proof ==
We have:
Consider the equation:
:$x^n - 1 = 0$
whose solutions are the complex roots of unity:
:$1, e^{-2 \pi i / n}, e^{-4 \pi i / n}, e^{-6 \pi i / n}, \ldots, e^{-2 \paren {n - 1} \pi i / n}$
Then:
Let $x = e^{2 i z}$.
Then:
Plugging $\paren {2}$ into $\paren {1}$, we have:
== Also see ==
* Digamma Additive Formula
* Gauss Multiplication Formula
* Multiple Angle Formula for Sine
Category:Sine Function
Category:Complex Roots of Unity
Category:Named Theorems
Product Functor is Functor

== Definition ==
Let $\mathbf C$ be a metacategory with binary products.
Let $\mathbf C \times \mathbf C$ be the product category of $\mathbf C$ with itself.
Let $\times: \mathbf C \times \mathbf C \to \mathbf C$ be the product functor.
Then $\times$ is a functor.
== Proof ==
For any pair of objects $\tuple {C, D}$:
Thus $\times$ preserves identity morphisms.
For composable morphisms $\tuple {f, f'}$ and $\tuple {g, g'}$ of $\mathbf C \times \mathbf C$:
Hence $\times$ is a functor.
Category:Product Categories
Category:Functors
Product Inverse Operation Properties

== Theorem ==
Let $\struct {G, \oplus}$ be a closed algebraic structure on which the following properties hold:
Let $\circ$ be the operation on $G$ defined as:
:$\forall x, y \in G: x \circ y = x \oplus \paren {e \oplus y}$
The following lemmata hold:
=== Lemma 1 ===
=== Lemma 2 ===
=== Lemma 3 ===
=== Lemma 4 ===
=== Lemma 5 ===
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.7 \ \text {(b)}$
Category:Product Inverse Operation
Product Inverse Operation Properties/Lemma 1

== Theorem ==
Let $\struct {G, \oplus}$ be a closed algebraic structure on which the following properties hold:
Let $\circ$ be the operation on $G$ defined as:
:$\forall x, y \in G: x \circ y = x \oplus \paren {e \oplus y}$
Then:
:$\forall x, y, z \in G: \paren {x \circ z} \oplus \paren {y \circ z} = x \oplus y$
 
== Proof ==
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.7 \ \text {(b)}$
Category:Product Inverse Operation
Product Inverse Operation Properties/Lemma 2

== Theorem ==
Let $\struct {G, \oplus}$ be a closed algebraic structure on which the following properties hold:
Let $\circ$ be the operation on $G$ defined as:
:$\forall x, y \in G: x \circ y = x \oplus \paren {e \oplus y}$
Then:
:$\forall x, y, z \in G: \paren {x \oplus z} \circ \paren {z \oplus y} = x \oplus y$
 
== Proof ==
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.7 \ \text {(b)}$
Category:Product Inverse Operation
Product Inverse Operation Properties/Lemma 3

== Theorem ==
Let $\struct {G, \oplus}$ be a closed algebraic structure on which the following properties hold:
Let $\circ$ be the operation on $G$ defined as:
:$\forall x, y \in G: x \circ y = x \oplus \paren {e \oplus y}$
Then:
:$\forall x, y \in G: \quad x \oplus y = e \implies x = y$
 
== Proof ==
Let $x \oplus y = e$.
Then we have:
Then:
That is:
:$x = y$
Hence the result.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.7 \ \text {(b)}$
Category:Product Inverse Operation
Product Inverse Operation Properties/Lemma 4

== Theorem ==
Let $\struct {G, \oplus}$ be a closed algebraic structure on which the following properties hold:
Let $\circ$ be the operation on $G$ defined as:
:$\forall x, y \in G: x \circ y = x \oplus \paren {e \oplus y}$
Then:
:$\forall x, y, z \in G: x \oplus z = y \oplus z \implies x = y$
 
== Proof ==
Let $x \oplus z = y \oplus z$.
Then we have:
Then:
Thus we have:
:$x \oplus y = e$
as both are equal to $\paren {x \oplus z} \oplus \paren {y \oplus z}$.
Then from Lemma $3$:
:$x = y$
Hence the result.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.7 \ \text {(b)}$
Category:Product Inverse Operation
Product Inverse Operation Properties/Lemma 5

== Theorem ==
Let $\struct {G, \oplus}$ be a closed algebraic structure on which the following properties hold:
Let $\circ$ be the operation on $G$ defined as:
:$\forall x, y \in G: x \circ y = x \oplus \paren {e \oplus y}$
Then:
:$\forall x, y \in G: \paren {x \circ y} \oplus y = x$
 
== Proof ==
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.7 \ \text {(b)}$
Category:Product Inverse Operation
Product Inverse Operation Properties induce Group

== Theorem ==
Let $\struct {G, \oplus}$ be a closed algebraic structure on which the following properties hold:
Let $\circ$ be the operation on $G$ defined as:
:$\forall x, y \in G: x \circ y = x \oplus \paren {e \oplus y}$
Then $\struct {G, \circ}$ is a group.
== Proof ==
Taking the group axioms in turn:
=== ===
$\struct {G, \oplus}$ is closed by definition.
Hence $\struct {G, \circ}$ is likewise closed.
=== ===
and:
Thus $e$ is the identity element of $\struct {G, \circ}$.
=== ===
We have that $e$ is the identity element of $\struct {G, \circ}$.
Hence:
and:
Thus every element $x$ of $\struct {G, \circ}$ has inverse $e \oplus x$.
=== ===
We have:
Thus $\circ$ is associative.
All the group axioms are thus seen to be fulfilled, and so $\struct {G, \circ}$ is a group.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.7 \ \text {(b)}$
Category:Product Inverse Operation
Category:Examples of Groups
Product Inverse Operation is Self-Inverse

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\oplus: G \times G \to G$ be the product inverse of $\circ$ on $G$.
Then $\oplus$ is self inverse in the sense that:
:$\forall x \in G: x \oplus x = e$
== Proof ==
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.7 \ \text {(a)}: 1^\circ$
Category:Product Inverse Operation
Product Inverse in Ring is Unique

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity.
Let $x \in R$ be a unit of $R$.
Then the product inverse $x^{-1}$ of $x$ is unique.
== Proof ==
By definition of ring with unity, the algebraic structure $\struct {R, \circ}$ is a monoid.
The result follows from Inverse in Monoid is Unique.
== Sources ==
* : Chapter $4$: Fields: $\S 14$. Definition of a Field
Category:Rings with Unity
Product Manifold of Pseudo-Riemannian Manifolds is Pseudo-Riemannian Manifold

== Theorem ==
Let $\struct {M_1, g_1}$ and $\struct {M_2, g_2}$ be pseudo-Riemannian manifolds of signatures $\tuple {r_1, s_1}$ and $\tuple {r_2, s_2}$ respectively.
Then $\struct {M_1 \times M_2, g_1 \oplus g_2}$ is a pseudo-Riemannian manifold of signature $\tuple {r_1 + r_2, s_1 + s_2}$, where $\times$ denotes the cartesian product and $\oplus$ stands for the direct sum.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Pseudo-Riemannian Metrics
Category:Pseudo-Riemannian Geometry
Product Rule

* Analysis
** Product Rule for Sequences
*** Product Rule for Real Sequences
*** Product Rule for Complex Sequences
** Product Rule for Continuous Functions
*** Product Rule for Continuous Real Functions
* Calculus:
** Product Rule for Derivatives
** Product Rule for Complex Derivatives
** Product Rule for Differentials
* Vector Calculus:
** Product Rule for Divergence
** Product Rule for Curl
* Combinatorics:
** Product Rule for Counting
* Probability Theory:
** Product Rule for Probabilities, also known as Bayes' Theorem
Product Rule for Counting

== Theorem ==
Let it be possible to choose an element $\alpha$ from a given set $S$ in $m$ different ways.
Let it be possible to choose an element $\beta$ from a given set $T$ in $n$ different ways.
Then the ordered pair $\tuple {\alpha, \beta}$ can be chosen from the cartesian product $S \times T$ in $m n$ different ways.
=== General Theorem ===
== Proof ==
The validity of this rule follows directly from the definition of multiplication of integers.
The product $a b$ (for $a, b \in \N_{>0}$) is the number of sequences $\sequence {A, B}$, where $A$ can be any one of $a$ items and $B$ can be any one of $b$ items.
== Also known as ==
Some sources give this as the General Combinatorial Principle.
Some sources call it the $r$, $s$ principle: if one operation can be performed in $r$ different ways, and if another operation can be performed in $s$ different ways, the two operations can be performed in succession in $r \times s$ different ways.
== Examples ==
== Also see ==
* Fundamental Principle of Counting
== Sources ==
* : Chapter $\text I$ Permutations and Combinations: The $r$, $s$ Principle: Example $1$.
* : $\text {3-1}$ Permutations and Combinations
Category:Combinatorics
Category:Counting Arguments
Category:Product Rule for Counting
Product Rule for Counting/Examples

== Examples of Use of Product Rule for Counting ==
=== Choices from $2$ and $3$ ===
=== Choices from $4$, $3$ and $2$ ===
=== Choices from $5$ and $3$ ===
=== $6$ Football Matches ===
=== $7$ Choices from $4$ ===
Category:Product Rule for Counting
Product Rule for Counting/Examples/6 Football Matches

== Example of Use of Product Rule for Counting ==
Let it be understood that a football match between two teams $\text A$ and $\text B$ can end as:
:A win for team $\text A$
:A draw
:A loss for team $\text A$.
That being understood, then there are $729$ ways to predict the results of $6$ football matches.
== Proof ==
Each match as the above described $3$ possible outcomes.
Hence by the Product Rule for Counting, there are $3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6 = 729$ possible ways of forecasting the result of $6$ such matches.
== Sources ==
* : $\text I$. Algebra: Permutations and Combinations: Exercises $\text I$: $10$
Category:Product Rule for Counting