nyuuzyou/wikis huggingface dataset:
Product of Matrices is Nonsingular iff Matrices are Nonsingular

== Theorem ==
Let $\mathbf A$ and $\mathbf B$ be square matrices of order $n$.
Let $\mathbf A \mathbf B$ denote the matrix product of $\mathbf A$ and $\mathbf B$.
Let $\mathbf I$ be the $n \times n$ unit matrix.
Then:
:$\mathbf A \mathbf B$ is nonsingular
:both $\mathbf A$ and $\mathbf B$ are nonsingular.
== Proof ==
=== Necessary Condition ===
Let both $\mathbf A$ and $\mathbf B$ be nonsingular.
By Matrix is Nonsingular iff Determinant has Multiplicative Inverse:
:$\map \det {\mathbf A} \ne 0$ and $\map \det {\mathbf B} \ne 0$
where $\map \det {\mathbf A}$ denotes the determinant of $\mathbf A$.
By Determinant of Matrix Product:
:$\map \det {\mathbf A} \map \det {\mathbf B} = \map \det {\mathbf A \mathbf B}$
Thus as both $\map \det {\mathbf A} \ne 0$ and $\map \det {\mathbf B} \ne 0$, it follows that:
:$\map \det {\mathbf A \mathbf B} \ne 0$
Hence by Matrix is Nonsingular iff Determinant has Multiplicative Inverse, $\mathbf A \mathbf B$ is nonsingular.
=== Sufficient Condition ===
Let $\mathbf A \mathbf B$ be nonsingular.
 it is not the case that both $\mathbf A$ and $\mathbf B$ are nonsingular.
Then by Matrix is Nonsingular iff Determinant has Multiplicative Inverse, either:
:$\map \det {\mathbf A} = 0$
or:
:$\map \det {\mathbf B} = 0$
By Determinant of Matrix Product:
:$\map \det {\mathbf A} \map \det {\mathbf B} = \map \det {\mathbf A \mathbf B}$
and so:
:$\map \det {\mathbf A \mathbf B} = 0$
Hence by Matrix is Nonsingular iff Determinant has Multiplicative Inverse, $\mathbf A \mathbf B$ is not nonsingular.
This contradicts the assumption that $\mathbf A \mathbf B$ is nonsingular.
Hence by Proof by Contradiction it follows that both $\mathbf A$ and $\mathbf B$ are nonsingular.
== Also see ==
* Inverse of Matrix Product
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.3$ The inverse of a matrix: Proposition $1.2$
Category:Inverse Matrices
Category:Conventional Matrix Multiplication
Product of Metacompact Spaces is not always Metacompact

== Theorem ==
Let $I$ be an indexing set.
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.
Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be a metacompact space.
Then it is not necessarily the case that $\struct {S, \tau}$ is also metacompact space.
== Proof ==
Let $T$ be the Sorgenfrey line.
Let $T' = T \times T$ be Sorgenfrey's half-open square topology.
From Sorgenfrey Line is Metacompact Space, $T$ is a metacompact space.
From Sorgenfrey's Half-Open Square Topology is Not Metacompact Space, $T'$ is not a metacompact space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties
Category:Metacompact Spaces
Category:Product Spaces
Product of Multiplicative Functions is Multiplicative

== Theorem ==
Let $f: \N \to \C$ and $g: \N \to \C$ be multiplicative functions.
Then their pointwise product:
:$f \times g: \Z \to \Z: \forall s \in S: \map {\paren {f \times g} } s := \map f s \times \map g s$
is also multiplicative.
== Proof ==
Let $f$ and $g$ be multiplicative.
Let $m \perp n$.
Then:
Hence the result by definition of multiplicative function.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $29 \ \text{(d)}$
Category:Multiplicative Functions
Product of Negative Real Numbers is Positive

== Theorem ==
Let $a, b \in \R_{\le 0}$ be negative real numbers.
Then:
:$a \times b \in \R_{\ge 0}$
That is, their product $a \times b$ is a positive real number.
== Proof ==
From Real Numbers form Ring, the set $\R$ of real numbers forms a ring.
The result then follows from Product of Ring Negatives.
== Sources ==
* : $\S 1.1$. Number Systems
Category:Real Multiplication
Product of Negative with Product Inverse

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity.
Let $z \in U_R$, where $U_R$ is the set of units.
Then:
:$(1): \quad \forall x \in R: -\paren {x \circ z^{-1} } = \paren {-x} \circ z^{-1} = x \circ \paren {\paren {-z}^{-1} }$
:$(2): \quad \forall x \in R: -\paren {z^{-1} \circ x} = z^{-1} \circ \paren {-x} = \paren {\paren {-z}^{-1} } \circ x$
== Proof ==
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers: Theorem $23.7 \ (1)$
Category:Rings with Unity
Product of Number of Edges, Edges per Face and Faces of Cube

== Theorem ==
The product of the number of edges, edges per face and faces of a cube is $288$.
== Proof ==
A cube has:
:$12$ edges
and:
:$6$ faces.
Each face is a square, and so has $4$ edges.
Hence:
:$12 \times 6 \times 4 = 288$
== Sources ==
* : $72$
Category:Cubes
Category:288
Product of Number of Edges, Edges per Face and Faces of Regular Dodecahedron

== Theorem ==
The product of the number of edges, edges per face and faces of a regular dodecahedron is $1800$.
== Proof ==
A regular dodecahedron has:
:$30$ edges
and:
:$12$ faces.
Each face is a pentagon, and so has $5$ edges.
Hence:
:$30 \times 12 \times 5 = 1800$
== Sources ==
* : $72$
Category:Regular Dodecahedra
Category:1800
Product of Number of Edges, Edges per Face and Faces of Regular Icosahedron

== Theorem ==
The product of the number of edges, edges per face and faces of a regular icosahedron is $1800$.
== Proof ==
A regular icosahedron has:
:$30$ edges
and:
:$20$ faces.
Each face is a triangle, and so has $3$ edges.
Hence:
:$30 \times 20 \times 3 = 1800$
== Sources ==
* : $72$
Category:Regular Icosahedra
Category:1800
Product of Number of Edges, Edges per Face and Faces of Regular Octahedron

== Theorem ==
The product of the number of edges, edges per face and faces of a regular octahedron is $288$.
== Proof ==
A regular octahedron has:
:$12$ edges
and:
:$8$ faces.
Each face is a triangle, and so has $3$ edges.
Hence:
:$12 \times 8 \times 3 = 288$
== Sources ==
* : $72$
Category:Regular Octahedra
Category:288
Product of Number of Edges, Edges per Face and Faces of Tetrahedron

== Theorem ==
The product of the number of edges, edges per face and faces of a tetrahedron is $72$.
== Proof ==
A tetrahedron has:
:$6$ edges
and:
:$4$ faces.
Each face is a triangle, and so has $3$ edges.
Hence:
:$6 \times 4 \times 3 = 72$
== Sources ==
* : $72$
Category:Tetrahedra
Category:72
Product of One Plus Cotangent with One Plus Tangent

== Theorem ==
:$\paren {1 + \cot x} \paren {1 + \tan x} = 2 + \csc x \sec x$
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Exercise $\text {XXXI}$: $2.$
Category:Trigonometric Identities
Category:Tangent Function
Category:Cotangent Function
Product of Orthogonal Matrices is Orthogonal Matrix

== Theorem ==
Let $\mathbf P$ and $\mathbf Q$ be orthogonal matrices.
Let $\mathbf P \mathbf Q$ be the (conventional) matrix product of $\mathbf P$ and $\mathbf Q$.
Then $\mathbf P \mathbf Q$ is an orthogonal matrix.
== Proof ==
From Determinant of Orthogonal Matrix is Plus or Minus One and Matrix is Nonsingular iff Determinant has Multiplicative Inverse it follows that both $\mathbf P$ and $\mathbf Q$ are nonsingular.
Thus:
Hence the result, by definition of orthogonal matrix.
== Sources ==
* : $2.1$: Matrices
* 
Category:Orthogonal Matrices
Product of Orthogonal Matrix with Transpose is Identity

== Theorem ==
Let $\mathbf Q$ be an orthogonal matrix.
Then:
:$\mathbf Q \mathbf Q^\intercal = \mathbf I = \mathbf Q^\intercal \mathbf Q$
where:
:$\mathbf Q^\intercal$ is the transpose of $\mathbf Q$
:$\mathbf I$ is a unit (identity) matrix
== Proof ==
By definition, an orthogonal matrix is one such that:
:$\mathbf Q^\intercal = \mathbf Q^{-1}$
and so the result follows by definition of inverse.
== Sources ==
* : $2.1$: Matrices: $(2.12)$
Category:Orthogonal Matrices
Category:Transposes of Matrices
Category:Unit Matrices
Product of Outer Product with Multiplicand

== Theorem ==
Let $\mathbf u = \tuple {u_1, u_2, \ldots, u_m}$ and $\mathbf v = \tuple {v_1, v_2, \ldots, v_n}$ be vectors.
Let $\mathbf u \otimes \mathbf v$ denote the outer product of $\mathbf u$ and $\mathbf v$.
Then:
:$\paren {\mathbf u \otimes \mathbf v} \mathbf v = \mathbf u \norm {\mathbf v}^2$
where $\norm {\mathbf v}$ denotes the length of $\mathbf v$.
== Proof ==
Category:Outer Products
Product of Paracompact Spaces is not always Paracompact

== Theorem ==
Let $I$ be an indexing set.
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.
Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be a paracompact space.
Then it is not necessarily the case that $\struct {S, \tau}$ is also paracompact space.
== Proof ==
Let $T$ be the Sorgenfrey line.
Let $T' = T \times T$ be Sorgenfrey's half-open square topology.
From Sorgenfrey Line is Paracompact Space, $T$ is a paracompact space.
From Sorgenfrey's Half-Open Square Topology is Not Paracompact Space, $T'$ is not a paracompact space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties
Category:Paracompact Spaces
Category:Product Spaces
Product of Positive Cuts is Positive Cut

== Theorem ==
Let $0^*$ denote the rational cut associated with the (rational) number $0$.
Let $\alpha$ and $\beta$ be cuts such that $\alpha \ge 0^*$ and $\beta \ge 0^*$, where $\ge$ denotes the ordering on cuts.
Let $\gamma$ be the set of all rational numbers $r$ such that either:
:$r < 0$
or:
:$\exists p \in \alpha, q \in \beta: r = p q$
where $p \ge 0$ and $q \ge 0$.
Then $\gamma$ is also a cut.
Thus the operation of multiplication on the set of positive cuts is closed.
== Proof ==
By definition of $\gamma$, we have that $r < 0 \implies r \in \gamma$.
Hence $\gamma$ is not empty.
First suppose that either $\alpha = 0^*$ or $\beta = 0^*$.
Then by definition of cut:
:$p \in \alpha \implies p < 0$
:$q \in \beta \implies q < 0$
and so there exist no $r \in \gamma$ such that $r = p q$.
Thus $\gamma$ consists entirely of rational numbers $r$ such that $r < 0$.
That is:
:$\gamma = \set {r \in \Q: r < 0}$
and so:
:$\gamma = 0^*$
Hence $\gamma$ is a cut, as we needed to demonstrate.
Suppose that neither $\alpha$ nor $\beta$ are equal to $0^*$
That is:
:$\alpha > 0$
and:
:$\beta > 0$
Hence by definition of cut:
:$0 \in \alpha$
and:
:$0 \in \beta$
As a cut has no greatest element:
:$\exists p \in \alpha: p > 0$
:$\exists q \in \alpha: q > 0$
Let $s, t \in \Q$ such that $s \notin \alpha$ and $t \notin \beta$, where $\Q$ denotes the set of rational numbers.
Such $s$ and $t$ are bound to exist because by definition of cut, neither $\alpha$ nor $\beta$ equal $\Q$.
Note that as $\alpha \ge 0^*$ and $\beta \ge 0^*$, it follows by definition of cut that, both $s \ge 0$ and $t \ge 0$.
We have:
Thus it is demonstrated that $\gamma$ does not contain every rational number.
Thus condition $(1)$ of the definition of a cut is fulfilled.
Let $r \in \gamma$.
Let $s \in \Q$ such that $s < r$.
If $r < 0$, then $s < 0$ and so $s \in \gamma$ by definition of $\gamma$.
If $r > 0$ and $s < 0$, then $s \in \gamma$ by definition of $\gamma$.
Otherwise, we have the case where $r > 0$ and $s > 0$.
By definition of $\gamma$:
:$r = p q$ for some $p \in \alpha, q \in \beta$ such that $p > 0$ and $q > 0$.
Let $t \in \Q$ such that $s = t q$.
Then $0 < t < p$.
Hence $t \in \alpha$.
Hence by definition of $\gamma$, $t q = s \in \gamma$.
Thus we have that $r \in \gamma$ and $s < r$ implies that $s \in \gamma$.
Thus condition $(2)$ of the definition of a cut is fulfilled.
 $r \in \gamma$ is the greatest element of $\gamma$.
Then $r = p q$ for some $p \in \alpha, q \in \beta$ such that $p > 0$ and $q > 0$.
By definition of a cut, $\alpha$ has no greatest element:
Hence:
:$\exists s \in \Q: s > p: s \in \alpha$
But then $s q \in \gamma$ while $s q > r$.
This contradicts the supposition that $r$ is the greatest element of $\gamma$.
Hence $\gamma$ itself can have no greatest element.
Thus condition $(3)$ of the definition of a cut is fulfilled.
Thus it is seen that all the conditions are fulfilled for $\gamma$ to be a cut.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.22$. Theorem
Category:Cuts
Category:Multiplication
Product of Positive Element and Element Greater than One

== Theorem ==
Let $\struct {R, +, \circ, \le}$ be an ordered ring with unity $1_R$ and zero $0_R$.
Let $x, y \in R$.
Suppose that $x > 0_R$ and $y > 1_R$.
Then $x \circ y > x$ and $y \circ x > x$.
== Proof ==
A similar argument shows that $y \circ x > x$.
Category:Ordered Rings
Product of Positive Strictly Increasing Mappings is Strictly Increasing

== Theorem ==
Let $A$ be an ordered set.
Let $B$ be an ordered field.
Let $f, g: A \to B$ be strictly increasing mappings with positive values.
Let $h: A \to B$ be defined by $\map h x = \map f x \map g x$.
Then $h$ is strictly increasing.
== Proof ==
Let $x, y \in A$ such that $x < y$.
If $\map h x = 0$.
By the definition of strictly increasing:
:$\map f y > \map f x \ge 0$
and:
:$\map g y > \map g x \ge 0$
So:
:$\map h y > 0 = \map h x$
If $\map h x \ne 0$, then $\map h x > 0$
so:
:$\map f x > 0$
and:
:$\map g x > 0$
Also:
:$\map f x < \map f y$
and:
:$\map g x < \map g y$
by the definition of strictly increasing.
Because $\map g x > 0$:
:$\map f x \map g x < \map f y \map g x$
Because $\map f y > 0$:
:$\map f y \map g x < \map f y \map g y$
By transitivity:
:$\map h x = \map f x \map g x < \map f y \map g y = \map h y$
Product of Positive and Negative Parts is Zero

== Theorem ==
Let $X$ be a set.
Let $f : X \to \R$ be a real-valued function.
Let $f^+$ and $f^-$ be the positive part and negative part of $f$ respectively.
Then $f^+ f^- = 0$. 
== Proof ==
Let $x \in X$.
If $\map f x = 0$, then we have $\map {f^+} x = \map {f^-} x = 0$. 
Hence $\map {f^+} x = \map {f^-} x = 0$ in this case.
If $\map f x > 0$, then $\map {f^+} x = \map f x$ and $\map {f^-} x = 0$.
So $\map {f^+} x \map {f^-} x = 0$ in this case also.
Finally, if $\map f x < 0$, then $\map {f^+} x = 0$ and $\map {f^-} x = -\map f x$.
So $\map {f^+} x \map {f^-} x = 0$ in this case.
We have covered all possibilities, so we have $\map {f^+} x \map {f^-} x = 0$ for each $x \in X$.
Category:Positive Parts
Category:Negative Parts
Product of Powers in B-Algebra

== Theorem ==
Let $\struct {X, \circ}$ be a $B$-algebra.
Let $x \in X$ and $m, n \in \N$.
Then:
:$x^m \circ x^n = \begin {cases}
x^{m - n} & : m \ge n \\
0 \circ x^{n - m} & : n > m
\end {cases}$
== Proof ==
For $m \ge n$ the result follows from $B$-Algebra Power Law.
For $n > m$ the result follows from $B$-Algebra Power Law with Zero.
Category:B-Algebras
Product of Powers of Positive Element of Unital C*-Algebra

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $a \in A$ be positive elements.
Let $\alpha, \beta > 0$ be real numbers.
Let $a^\alpha$, $a^\beta$ and $a^{\alpha + \beta}$ be given by the continuous functional calculus.
Then:
:$a^{\alpha + \beta} = a^\alpha a^\beta$
== Proof ==
Let $\map {\sigma_A} a \subseteq \hointr 0 \infty$ be the spectrum of $a$.
Define $e_\alpha : \map {\sigma_A} a \to \hointr 0 \infty$, $e_\beta : \map {\sigma_A} a \to \hointr 0 \infty$ and $e_{\alpha + \beta} : \map {\sigma_A} a \to \hointr 0 \infty$ by:
:$\map {e_\alpha} t = t^\alpha$
:$\map {e_\beta} t = t^\beta$
:$\map {e_{\alpha + \beta} } t = t^{\alpha + \beta}$
for each $t \in \map {\sigma_A} a$. 
For $t \in \map {\sigma_A} a$ we have:
:$\map {e_\alpha} t \map {e_\beta} t = t^\alpha t^\beta = t^{\alpha + \beta} = \map {e_{\alpha + \beta} } t$
by Product of Powers.
Hence since the continuous functional calculus is an algebra homomorphism, we obtain:
:$\map {e_\alpha} a \map {e_\beta} a = \map {e_{\alpha + \beta} } a$
That is:
:$a^\alpha a^\beta = a^{\alpha + \beta}$
Category:Positive Elements of C*-Algebras
Product of Products

== Theorem ==
Let $R: \Z \to \set {\T, \F}$ be a propositional function on the set of integers.
Let $\ds \prod_{R \paren i} x_i$ denote a continued product over $R$.
Let the fiber of truth of $R$ be finite.
Then:
:$\ds \prod_{R \paren i} \paren {b_i c_i} = \paren {\prod_{R \paren i} b_i} \paren {\prod_{R \paren i} c_i}$
== Proof ==
Let $b_i =: a_{i 1}$ and $c_i =: a_{i 2}$.
Then:
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $22$
Category:Continued Products
Product of Products over Overlapping Domains

== Theorem ==
Let $R: \Z \to \set {\T, \F}$ and $S: \Z \to \set {\T, \F}$ be propositional functions on the set of integers.
Let $\ds \prod_{\map R i} x_i$ denote a continued product over $R$.
Then:
:$\ds \prod_{\map R j} a_j \prod_{\map S j} a_j = \paren {\prod_{\map R j \mathop \lor \map S j} a_j} \paren {\prod_{\map R j \mathop \land \map S j} a_j}$
where $\lor$ and $\land$ signify logical disjunction and logical conjunction respectively.
== Proof ==
Let:
:$A := \set {j \in \Z: \map R j}$
:$B := \set {j \in \Z: \map S j}$
The result then follows from Cardinality of Set Union.
== Sources ==
* : $\S 1.2.3$: Sums and Products
Category:Continued Products
Product of Projections

== Theorem ==
Let $H$ be a Hilbert space.
Let $P, Q$ be projections.
Then the following are equivalent:
:$(1): \quad P Q$ is a projection
:$(2): \quad P Q = Q P$
:$(3): \quad P + Q - P Q$ is a projection
== Proof ==
The proof proceeds by first showing that $(1)$ is equivalent to $(2)$.
Then, these are combined and shown equivalent to $(3)$.
=== $(1)$ implies $(2)$ ===
Let $P Q$ be a projection. 
Then by Characterization of Projections, statement $(4)$, one has:
:$P Q = \paren {P Q}^* = Q^* P^* = Q P$
where the penultimate equality follows from Adjoint of Composition of Linear Transformations is Composition of Adjoints.
=== $(2)$ implies $(1)$ ===
Let $P Q = Q P$.
Then:
:$\paren {P Q}^2 = P Q P Q = P^2 Q^2 = P Q$
as $P, Q$ are projections.
Hence $P Q$ is an idempotent.
Also from Adjoint of Composition of Linear Transformations is Composition of Adjoints:
:$\paren {P Q}^* = Q^* P^* = Q P = P Q$.
Hence, by Characterization of Projections, statement $(4)$, $P Q$ is a projection.
=== $(1), (2)$ imply $(3)$ ===
The above establishes that assuming either of $(1)$ and $(2)$ yields both to hold.
So assuming $(1)$, $P, Q$ and $P Q$ are all projections, and $P Q = Q P$.
Now compute:
It follows that $P + Q - P Q$ is an idempotent.
From:
:Adjoining is Linear
:Adjoint of Composition of Linear Transformations is Composition of Adjoints:
:$\paren {P + Q - P Q}^* = P^* + Q^* - Q^* P^* = P + Q - QP = P + Q - P Q$
Now applying Characterization of Projections, statement $(4)$, conclude that $P + Q - P Q$ is a projection.
=== $(3)$ implies $(2)$ ===
Let $P + Q - P Q$ be a projection.
Then by Characterization of Projections, statement $(4)$, compute:
Hence necessarily $P Q = Q P$.
== Also see ==
* Sum of Projections
* Difference of Projections
== Sources ==
* $\text {II}.3$ Exercises $4, 7$
Category:Linear Transformations on Hilbert Spaces
Product of Proper Divisors

== Theorem ==
Let $n$ be an integer such that $n \ge 1$.
Let $\map P n$ denote the product of the proper divisors of $n$.
Then:
:$\map P n = n^{\map {\sigma_0} n / 2 - 1}$
where $\map {\sigma_0} n$ denotes the divisor count Function of $n$.
== Proof ==
Let $\map D n$ denote the product of all the divisors of $n$.
From Product of Divisors:
:$\map D n = n^{\map {\sigma_0} n / 2}$
The proper divisors of $n$ are defined as being the divisors of $n$ excluding $n$ itself.
Thus:
:$\map P n = \dfrac {\map D n} n = \dfrac {n^{\map {\sigma_0} n / 2} } n = n^{\map {\sigma_0} n / 2 - 1}$
== Examples ==
Category:Product of Divisors
Category:Divisor Count Function
Category:Proper Divisors
Product of Proper Divisors/Examples/12

== Example of Product of Proper Divisors ==
The product of the proper divisors of $12$ is $144 = 12^2$.
== Proof ==
From Product of Proper Divisors:
:$\map P n = n^{\map {\sigma_0} n / 2 - 1}$
where $\map P n$ denotes the product of the proper divisors of $n$.
From :
:$\map {\sigma_0} {12} = 6$
and so:
:$\map P {12} = 12^{6 / 2 - 1} = 12^2 = 144$
== Sources ==
* : $12$
* : $12$
Category:Product of Divisors
Category:12
Product of Proper Divisors/Examples/20

== Example of Product of Proper Divisors ==
The product of the proper divisors of $20$ is $400 = 20^2$.
== Proof ==
From Product of Proper Divisors:
:$\map P n = n^{\map {\sigma_0} n / 2 - 1}$
where $\map P n$ denotes the product of the proper divisors of $n$.
From :
:$\map {\sigma_0} {20} = 6$
and so:
:$\map P {20} = 20^{6 / 2 - 1} = 20^2 = 144$
== Sources ==
* : $400$
* : $400$
Category:Product of Divisors
Category:20
Category:400
Product of Proper Orthogonal Matrices is Proper Orthogonal Matrix

== Theorem ==
Let $\mathbf P$ and $\mathbf Q$ be proper orthogonal matrices.
Let $\mathbf {P Q}$ be the (conventional) matrix product of $\mathbf P$ and $\mathbf Q$.
Then $\mathbf {P Q}$ is a proper orthogonal matrix.
== Proof ==
By definition, $\mathbf {P Q}$ is a proper orthogonal matrix it is an orthogonal matrix with a determinant of $1$.
From Product of Orthogonal Matrices is Orthogonal Matrix, $\mathbf {P Q}$ is an orthogonal matrix.
By definition, $\mathbf P$ and $\mathbf Q$ both have a determinant of $1$.
From Determinant of Matrix Product:
:$\map \det {\mathbf {P Q} } = \map \det {\mathbf P} \map \det {\mathbf Q}$
Thus:
:$\map \det {\mathbf {P Q} } = 1$
Hence the result.
Category:Orthogonal Matrices
Product of Quaternion Conjugates

== Theorem ==
Let $\mathbf x, \mathbf y \in \mathbb H$ be quaternions.
Let $\overline{\mathbf x}$ be the conjugate of $\mathbf x$.
Then:
:$\overline{\mathbf x \times \mathbf y} = \overline{\mathbf y} \times \overline{\mathbf x}$
but in general:
:$\overline{\mathbf x \times \mathbf y} \ne \overline{\mathbf x} \times \overline{\mathbf y}$
== Proof ==
Consider the matrix form of $\mathbf x$ and $\mathbf y$:
:$\mathbf x = \begin{bmatrix} a & b \\ -\overline b & \overline a \end{bmatrix}$
:$\mathbf y = \begin{bmatrix} c & d \\ -\overline d & \overline c \end{bmatrix}$
where $a, b, c, d \in \C$.
but:
== Sources ==
* : Chapter $\text {B}.26$: Extensions of the Complex Number System. Algebras, Quaternions, and Lagrange's Four Squares Theorem
Category:Quaternions
Product of Quaternion with Conjugate

== Theorem ==
Let $\mathbf x = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$ be a quaternion.
Let $\overline {\mathbf x}$ be the conjugate of $\mathbf x$.
Then their product is given by:
:$\mathbf x \overline {\mathbf x} = \paren {a^2 + b^2 + c^2 + d^2} \mathbf 1 = \overline {\mathbf x} \mathbf x$
== Proof ==
From the definition of quaternion multiplication:
We have that $a, b, c, d \in \R$.
So from Real Multiplication is Commutative, their products commute.
So the terms in $\mathbf i, \mathbf j, \mathbf k$ vanish.
The proof that $\overline {\mathbf x} \mathbf x = \paren {a^2 + b^2 + c^2 + d^2} \mathbf 1$ is similar.
Hence the result.
== Also see ==
* Definition:Field Norm of Quaternion
== Sources ==
* : Chapter $1$: Rings - Definitions and Examples: $2$: Some examples of rings: Ring Example $9$
* : Chapter $1$: Introductory Concepts: $1$. Basic Building Blocks: $3$. FIELD
* : Chapter $\text {B}.26$: Extensions of the Complex Number System. Algebras, Quaternions, and Lagrange's Four Squares Theorem
Category:Quaternions
Product of Quotients of Real Numbers

== Theorem ==
:$\forall x, w \in \R, y, z \in \R_{\ne 0}: \dfrac x y \times \dfrac w z = \dfrac {x \times w} {y \times z}$
== Proof ==
== Sources ==
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $1 \ \text{(n)}$
Category:Real Multiplication
Category:Real Division
Product of Rational Cuts is Rational Cut

== Theorem ==
Let $p \in\ Q$ and $q \in \Q$ be rational numbers.
Let $p^*$ and $q^*$ denote the rational cuts associated with $p$ and $q$.
Then:
:$p^* q^* = \paren {p q}^*$
Thus the operation of multiplication on the set of rational cuts is closed.
== Proof ==
From Product of Cuts is Cut, $p^* q^*$ is a cut.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.28$. Theorem: $\text {(b)}$
Category:Cuts
Category:Multiplication