nyuuzyou/wikis huggingface dataset:
Product is Right Identity Therefore Right Cancellable

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Let $e_R \in S$ be a right identity of $S$.
Let $a \in S$ such that:
:$\exists b \in S: a \circ b = e_R$
Then $a$ is right cancellable in $\struct {S, \circ}$.
== Proof ==
Let $x, y \in S$ be arbitrary.
Then:
The result follows by definition of right cancellable.
== Also see ==
* Product is Left Identity Therefore Left Cancellable
Category:Semigroups
Category:Cancellability
Category:Identity Elements
Product of 3 Consecutive Integers is Divisible by 6

== Theorem ==
Let $a, b, c \in \Z$ be consecutive integers.
Then their product $a b c$ is divisible by $6$.
== Proof ==
This is an application of Divisibility of Product of Consecutive Integers with $n = 3$.
By the theorem, the product of $3$ consecutive integers is divisible by $3! = 6$.
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $14$
Category:Examples of Divisors of Integers
Product of 4 Consecutive Integers is Divisible by 24

== Theorem ==
Let $a, b, c, d \in \Z$ be consecutive integers.
Then their product $a b c d$ is divisible by $24$.
== Proof ==
This is an application of Divisibility of Product of Consecutive Integers with $n = 4$.
By the theorem, the product of $4$ consecutive integers is divisible by $4! = 24$.
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $14$
Category:Examples of Divisors of Integers
Product of 4 Consecutive Integers is One Less than Square

== Theorem ==
Let $a$, $b$, $c$ and $d$ be consecutive integers.
Then:
:$\exists n \in \Z: a b c d = n^2 - 1$
That is, the product of $a$, $b$, $c$ and $d$ is one less than a square.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $8 \ \text {(b)}$
Category:Product of 4 Consecutive Integers is One Less than Square
Category:Square Numbers
Product of 4 Consecutive Integers is One Less than Square/Lemma

== Lemma for Product of 4 Consecutive Integers is One Less than Square ==
Let $a$, $b$, $c$ and $d$ be consecutive integers.
Let us wish to prove that the product of $a$, $b$, $c$ and $d$ is one less than a square.
Then it is sufficient to consider $a$, $b$, $c$ and $d$ all strictly positive.
== Proof ==
If $0 \in \set {a, b, c, d}$ then $a b c d = 0$ which is $1$ less than $1^2$.
If $a$, $b$, $c$ and $d$ are all negative, then:
:$a b c d = \paren {-a} \paren {-b} \paren {-c} \paren {-d}$
Hence it is sufficient to consider $a$, $b$, $c$ and $d$ all strictly positive.
Category:Product of 4 Consecutive Integers is One Less than Square
Product of 4 Consecutive Integers is One Less than Square/Proof 1

== Theorem ==
== Proof ==
=== Lemma ===
As $a$, $b$, $c$ and $d$ are all consecutive, we can express them as:
:$a$, $a + 1$, $a + 2$ and $a + 3$
where $a \ge 1$.
Hence:
Hence the result.
Category:Product of 4 Consecutive Integers is One Less than Square
Product of 4 Consecutive Integers is One Less than Square/Proof 2

== Theorem ==
== Proof ==
=== Lemma ===
As $a$, $b$, $c$ and $d$ are all consecutive, we can express them as:
:$a$, $a + 1$, $a + 2$ and $a + 3$
where $a \ge 1$.
Then:
Category:Product of 4 Consecutive Integers is One Less than Square
Product of 5 Consecutive Integers is Divisible by 120

== Theorem ==
Let $a, b, c, d, e \in Z$ be consecutive integers
Then their product $a b c d e$ is divisible by $120$.
== Proof ==
This is an application of Divisibility of Product of Consecutive Integers with $n = 5$.
By the theorem, the product of $5$ consecutive integers is divisible by $5! = 120$.
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $14$
Category:Examples of Divisors of Integers
Product of Absolute Values of Integers

== Theorem ==
Let $a, b \in \Z$ be integers.
Let $\size a$ denote the absolute value of $a$:
:$\size a = \begin {cases} a & : a \ge 0 \\ -a & : a < 0 \end {cases}$
Then:
:$\size a \times \size b = \size {a \times b}$
== Proof ==
From Integers form Ordered Integral Domain, $\Z$ is an ordered integral domain.
The result follows from Product of Absolute Values on Ordered Integral Domain.
== Sources ==
* : Introduction $\S 5$: The system of integers: Exercise $2$
Category:Absolute Value Function
Category:Integers
Product of Absolute Values on Ordered Integral Domain

== Theorem ==
Let $\struct {D, +, \times, \le}$ be an ordered integral domain whose zero is denoted by $0_D$.
For all $a \in D$, let $\size a$ denote the absolute value of $a$.
Then:
:$\size a \times \size b = \size {a \times b}$
== Proof ==
Let $P$ be the (strict) positivity property on $D$.
Let $<$ be the (strict) total ordering defined on $D$ as:
:$a < b \iff a \le b \land a \ne b$
Let $N$ be the strict negativity property on $D$.
We consider all possibilities in turn.
$(1): \quad a = 0_D$ or $b = 0_D$
In this case, both the $\size a \times \size b$ and the are equal to zero.
So:
:$\size a \times \size b = \size {a \times b}$
$(2): \quad \map P a, \map P b$
First:
Then:
So:
:$\size a \times \size b = \size {a \times b}$
 
$(3): \quad \map P a, \map N b$
First:
Then:
So:
:$\size a \times \size b = \size {a \times b}$
Similarly $\map N a, \map P b$.
$(4): \quad \map N a, \map N b$
First:
Then:
So:
:$\size a \times \size b = \size {a \times b}$
In all cases the result holds.
== Sources ==
* : Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Theorem $11 \ \text{(i)}$
Category:Absolute Value Function
Category:Ordered Integral Domains
Product of Absolutely Continuous Functions is Absolutely Continuous

== Theorem ==
Let $a, b$ be real numbers with $a < b$.
Let $f, g : \closedint a b \to \R$ be absolutely continuous real functions.
Then $f \times g$ is absolutely continuous.
== Proof ==
From Absolutely Continuous Real Function is Continuous:
:$f$ and $g$ are continuous.
From Closed Real Interval is Compact in Metric Space:
:$\closedint a b$ is compact.
Therefore, by Continuous Function on Compact Subspace of Euclidean Space is Bounded:
:$f$ and $g$ are bounded.
That is, there exists $M_f, M_g \in \R_{> 0}$ such that: 
:$\size {\map f x} \le M_f$
:$\size {\map g x} \le M_g$
for all $x \in \closedint a b$.
Let $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq \closedint a b$ be a set of disjoint closed real intervals.
Then:
Let $\epsilon$ be a positive real number.
Since $f$ is absolutely continuous, there exists $\delta_1 > 0$ such that for all sets of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq \closedint a b$ with: 
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta_1$
we have: 
:$\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \frac \epsilon {2 M_g}$
Similarly, since $g$ is absolutely continuous, there exists $\delta_2 > 0$ such that whenever:
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta_2$
we have:
:$\ds \sum_{i \mathop = 1}^n \size {\map g {b_i} - \map g {a_i} } < \frac \epsilon {2 M_f}$
Let:
:$\delta = \map \min {\delta_1, \delta_2}$
Then, whenever:
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$
We have both:
and hence: 
Since $\epsilon$ was arbitrary, we have:
:$f \times g$ is absolutely continuous.
Category:Absolutely Continuous Real Functions
Product of Absolutely Convergent Products is Absolutely Convergent

== Theorem ==
Let $\struct {\mathbb K, \norm {\,\cdot\,} }$ be a valued field.
Let $\ds \prod_{n \mathop = 1}^\infty a_n$ converge absolutely.
Let $\ds \prod_{n \mathop = 1}^\infty b_n$ converge absolutely.
Then $\ds \prod_{n \mathop = 1}^\infty a_nb_n$ converges absolutely.
== Proof ==
We have:
:$a_n b_n - 1 = \paren {a_n - 1} \paren {b_n - 1} + \paren {a_n - 1} + \paren {b_n - 1}$
By the Triangle Inequality:
:$\norm {a_n b_n - 1} \le \norm {a_n - 1} \norm {b_n - 1} + \norm {a_n - 1} + \norm {b_n -1}$
By the absolute convergence, $\ds \sum_{n \mathop = 1}^\infty \norm {a_n - 1}$ and $\ds \sum_{n \mathop = 1}^\infty \norm {b_n - 1}$ converge.
By Inner Product of Absolutely Convergent Series, $\ds \sum_{n \mathop = 1}^\infty \norm {a_n - 1} \norm{b_n -1}$ converges.
By the Comparison Test, $\ds \sum_{n \mathop = 1}^\infty \norm {a_n b_n - 1}$ converges.
== Also see ==
* Product of Convergent and Divergent Product is Divergent
* Product of Convergent Products is Convergent
Category:Infinite Products
Product of Absolutely Convergent Series

== Theorem ==
Let $\map f z = \ds \sum_{n \mathop = 1}^\infty a_n$ and $\map g z = \ds \sum_{n \mathop = 1}^\infty b_n$ be two real or complex series that are absolutely convergent.
Then $\map f z \map g z$ is an absolutely convergent series, and:
:$\map f z \map g z = \ds \sum_{n \mathop = 1}^\infty c_n$
where:
:$c_n = \ds \sum_{k \mathop = 1}^n a_k b_{n - k}$
== Proof ==
== Sources ==
* : $\S 4.4$. Power Series
Category:Absolute Convergence
Product of Affine Spaces is Affine Space

== Theorem ==
Let $\EE, \FF$ be affine spaces.
Let $\GG = \EE \times \FF$ be the product of $\EE$ and $\FF$.
Then $\GG$ is an affine space.
== Proof ==
Let $G = \vec \GG$ be the difference space of $\GG$.
We are required to show that the following axioms are satisfied:
Proof of $(1)$:
Let $p = \tuple {p', p' '}, q = \tuple {q', q' '} \in \GG$.
We have:
Proof of $(2)$:
Let $p = \tuple {p', p' '} \in \GG$.
Let $u = \tuple {u', u' '}, v = \tuple {v', v' '} \in G$.
We have:
Proof of $(3)$:
Let $p = \tuple {p', p' '}, q = \tuple {q', q' '} \in \GG$.
Let $u = \tuple {u', u' '} \in G$.
We have:
Category:Affine Geometry
Product of Big-O Estimates/Real Analysis

== Theorem ==
Let $c$ be a real number.
Let $f, g : \hointr c \infty \to \R$ be real functions.
Let $\OO$ denote big-$\OO$ notation.
Let $R_1 : \hointr c \infty \to \R$ be a real function such that $f = \map \OO {R_1}$.
Let $R_2 : \hointr c \infty \to \R$ be a real function such that $g = \map \OO {R_2}$.
Then: 
:$f g = \map \OO {R_1 R_2}$
== Proof ==
Since: 
:$f = \map \OO {R_1}$
there exists $x_1 \in \hointr c \infty$ and a real number $C_1$ such that: 
:$\size {\map f x} \le C_1 \size {\map {R_1} x}$
for $x \ge x_1$.
Similarly, since:
:$g = \map \OO {R_2}$
there exists $x_2 \in \hointr c \infty$ and a real number $C_2$ such that: 
:$\size {\map g x} \le C_2 \size {\map {R_2} x}$
Set:
:$x_0 = \max \set {x_1, x_2}$
Then for $x \ge x_0$ we have: 
That is, by the definition of big-$\OO$ notation, we have: 
:$f g = \map \OO {R_1 R_2}$
Category:Big-O Notation
Product of Big-O Estimates/Sequences

== Theorem ==
Let $\sequence {a_n}, \sequence {b_n}, \sequence {c_n}, \sequence {d_n}$ be sequences of real or complex numbers.
Let:
:$a_n = \map \OO {b_n}$
:$c_n = \map \OO {d_n}$
where $\OO$ denotes big-$\OO$ notation.
Then:
:$a_n c_n = \map \OO {b_n d_n}$
== Proof ==
Since:
:$a_n = \map \OO {b_n}$
there exists a positive real number $C_1$ and natural number $N_1$ such that: 
:$\size {a_n} \le C_1 \size {b_n}$
for all $n \ge N_1$. 
Similarly, since: 
:$c_n = \map \OO {d_n}$
there exists a positive real number $C_2$ and natural number $N_2$ such that: 
:$\size {c_n} \le C_2 \size {d_n}$
for all $n \ge N_2$. 
Let:
:$N = \max \set {N_1, N_2}$
Then, for $n \ge N$, we have: 
giving:
:$a_n c_n = \map \OO {b_n d_n}$
Category:Big-O Notation
Product of Bounded Mappings on Normed Algebra is Bounded

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,}_R}$ be a normed division ring.
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed algebra over $R$. 
Let $S$ be a set.
Let $f, g : S \to X$ be bounded mappings.
Then $f g$ is bounded.
== Proof ==
Since $f$ is bounded, there exists a real number $M_1 > 0$ such that:
:$\norm {\map f x} \le M_1$ for each $x \in X$. 
Since $g$ is bounded, there exists a real number $M_2 > 0$ such that:
:$\norm {\map g x} \le M_2$ for each $x \in X$. 
Since $\norm {\, \cdot \,}$ is a algebra norm, we have:
:$\norm {\map f x \map g x} \le \norm {\map f x} \norm {\map g x}$ for each $x \in X$.
So we have:
:$\norm {\map f x \map g x} \le M_1 M_2$
for each $x \in X$. 
Since $M_1 M_2 > 0$, we have that $f g$ is bounded.
Category:Bounded Mappings on Normed Vector Spaces
Product of Cardinals is Associative

== Theorem ==
Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be cardinals.
Then:
: $\mathbf a \paren {\mathbf {b c} } = \paren {\mathbf {a b} } \mathbf c$
where $\mathbf {a b}$ denotes the product of $\mathbf a$ and $\mathbf b$.
== Proof ==
Let $\mathbf a = \card A$, $\mathbf b = \card B$ and $\mathbf c = \card C$ for some sets $A$, $B$ and $C$.
By definition of product of cardinals:
:$\mathbf a \paren {\mathbf {b c} }$ is the cardinal associated with $A \times \paren {B \times C}$.
Consider the mapping $f: A \times \paren {B \times C} \to \paren {A \times B} \times C$ defined as:
:$\forall a \in A, b \in B, c \in C: \map f {a, \tuple {b, c} } = \tuple {\tuple {a, b}, c}$
Let $a_1, a_2 \in A, b_1, b_2 \in B, c_1, c_2 \in C$ such that:
:$\map f {a_1, \tuple {b_1, c_1} } = \map f {a_2, \tuple {b_2, c_2} }$
Then:
Thus $f$ is an injection.
Thus $f$ is a surjection.
So $f$ is both an injection and a surjection, and so by definition a bijection.
Thus a bijection has been established between $A \times \tuple {B \times C}$ and $\tuple {A \times B} \times C$.
It follows by definition that $A \times \tuple {B \times C}$ and $\tuple {A \times B} \times C$ are equivalent.
The result follows by definition of cardinal.
== Sources ==
* : $\S 8$: Theorem $8.4: \ (2)$
Category:Cardinals
Product of Cardinals is Commutative

== Theorem ==
Let $\mathbf a$ and $\mathbf b$ be cardinals.
Then:
: $\mathbf a \mathbf b = \mathbf b \mathbf a$
where $\mathbf a \mathbf b$ denotes the product of $\mathbf a$ and $\mathbf b$.
== Proof ==
Let $\mathbf a = \card A$ and $\mathbf b = \card B$ for some sets $A$ and $B$.
Then:
== Sources ==
* : $\S 8$: Theorem $8.4: \ (1)$
Category:Cardinals
Product of Change of Basis Matrices

== Theorem ==
Let $R$ be a ring with unity.
Let $M$ be a free $R$-module of finite dimension $n>0$.
Let $\AA$, $\BB$ and $\CC$ be ordered bases of $M$.
Let $\mathbf M_{\AA, \BB}$, $\mathbf M_{\BB, \CC}$ and $\mathbf M_{\AA, \CC}$ be the change of basis matrices from $\AA$ to $\BB$, $\BB$ to $\CC$ and $\AA$ to $\CC$ respectively.
Then:
:$\mathbf M_{\AA, \CC} = \mathbf M_{\AA, \BB} \cdot \mathbf M_{\BB, \CC}$
== Proof ==
Let $m \in M$.
Let $\sqbrk m_\AA$ be its coordinate vector relative to $\AA$, and similarly for $\BB$ and $\CC$.
On the one hand:
On the other hand:
Thus:
:$\forall m \in M: \paren {\mathbf M_{\AA, \CC} - \mathbf M_{\AA, \BB} \cdot \mathbf M_{\BB, \CC} } \cdot \sqbrk m_\CC = 0$
Because $m$ is arbitrary, the result follows.
== Also see ==
* Relative Matrix of Composition of Linear Transformations, an analogous result for linear transformations, of which this is a special case
Category:Change of Basis
Product of Closed Sets is Closed

== Theorem ==
Let $\family {\struct {S_i, \tau_i} }_{i \mathop \in I}$ be a family of topological spaces where $I$ is an arbitrary indexing set.
Let $\ds S = \prod_{i \mathop \in I} S_i$.
Let $T = \struct {S, \TT}$ be the product space of $\family {\struct {S_i, \tau_i} }_{i \mathop \in I}$ with the product topology $\TT$.
Suppose we have an indexed family of sets $\family {C_i}_{i \mathop \in I}$, where each $C_i$ is closed in $\struct {S_i, \tau_i}$.
Then $\ds \prod_{i \mathop \in I} C_i$ is closed in $\struct {S, \TT}$.
== Proof ==
First note that:
Thus by Intersection of Closed Sets is Closed in Topological Space, our result is proven if we can show that $\forall i \in I: \set {x \in S: x_i \in C_i}$ is closed in $T$.
Let $y \in I$.
We see that:
 
Since $C_y$ is closed, $\paren {S_y \setminus C_y}$ is open in $\struct {S_y, \tau_y}$ by definition.
Thus:
:$\pr_y^{-1} \sqbrk {S_y \setminus C_y} \in \BB$
where $\BB$ is the natural sub-basis of $T$:
:$\BB = \set {\pr_i^{-1} \sqbrk U: i \in I, U \in \tau_i}$
Therefore $\pr_y^{-1} \sqbrk {S_y \setminus C_y} = S \setminus \set {x \in S: x_y \in C_y}$ is open in $T$.
This implies that $\set {x \in S: x_y \in C_y}$ must be closed in $T$ by definition.
== Sources ==
* : $\text{I}: \ \S 11$: Product Spaces
Category:Product Spaces
Category:Closed Sets
Product of Closed and Half-Open Unit Intervals is Homeomorphic to Product of Half-Open Unit Intervals

== Theorem ==
Let $\closedint 0 1$ denote the closed unit interval $\set {x \in \R: 0 \le x \le 1}$.
Let $\hointr 0 1$ denote the half-open unit interval $\set {x \in \R: 0 \le x < 1}$.
Let both $\closedint 0 1$ and $\hointr 0 1$ have the Euclidean topology.
Then the product space:
:$\closedint 0 1 \times \hointr 0 1$
is homeomorphic to:
:$\hointr 0 1 \times \hointr 0 1$
== Proof ==
First we take the square $\Box ABCD$ embedded in the Cartesian plane such that $AD$ corresponds to $\closedint 0 1$ and $AB$ corresponds to $\hointr 0 1$:
:
This corresponds to the set $\closedint 0 1 \times \hointr 0 1$.
It is noted that the line segment $BC$ which corresponds to $\closedint 0 1 \times \set 1$ is not in the set $\closedint 0 1 \times \hointr 0 1$.
Then we apply a homeomorphism which maps the perimeter of $\closedint 0 1 \times \hointr 0 1$ to the circle whose center is $\tuple {\dfrac 1 2, \dfrac 1 2}$ and whose radius is $\dfrac {\sqrt 2} 2$.
The points $A$, $B$, $C$ and $D$ are fixed by this homeomorphism.
:
Note how the point $E$ is mapped to the point $E'$.
Then we apply a homeomorphism to the circle $\bigcirc ABE'CD$ which maps:
:the arc $BADC$ to the arc $BAD$
:the arc $BE'C$ to the arc $BEC$
which in the process maps:
:$E'$ to $E$ which is the same as $C$.
:$C$ to $C$ which is the same point as $D$.
:
Then we apply a homeomorphism to the circle $\bigcirc ABEC$ back to the square $\Box ABEC$.
It is seen that the line segment $EC$ is now identified with the $\set 1 \times \hointr 0 1$.
Hence $\closedint 0 1 \times \hointr 0 1$ has been transformed via $3$ homeomorphisms to $\hointr 0 1 \times \hointr 0 1$.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 21$
Category:Product Spaces
Category:Homeomorphisms (Topological Spaces)
Product of Commuting Elements in Monoid is Unit iff Each Element is Unit

== Theorem ==
Let $A$ be a monoid.
Let $\map G A$ be the group of units of $A$.
Let $n \ge 2$ be an integer. 
Let $x_1, \ldots, x_n$ be commuting elements in $A$.
Let:
:$\ds x = \prod_{i \mathop = 1}^n x_i$
Then:
:$x \in \map G A$ $x_i \in \map G A$ for each $1 \le i \le n$.
== Proof 1 ==
== Proof 2 ==
Category:Monoids
Product of Commuting Elements in Monoid is Unit iff Each Element is Unit/Proof 1

== Theorem ==
== Proof ==
=== Necessary Condition ===
We proceed by induction on $n$. 
For all $n \ge 2$, let $\map P n$ be the proposition:
:for every set of commuting elements $x_1, \ldots, x_n$ in $A$
::if $\ds \prod_{i \mathop = 1}^n x_i \in \map G A$, then $x_i \in \map G A$ for each $1 \le i \le n$.
==== Basis for the Induction ====
Let $x, y \in A$ be commuting elements in $A$ such that $x y \in \map G A$.
Let $z \in A$ be such that $\paren {x y} z = z \paren {x y} = e$. 
We then have:
So we have: 
:$\paren {y z} x = x \paren {y z} = e$
So we have $x \in \map G A$ with $x^{-1} = y z$. 
Then we have $y = x^{-1} z^{-1}$.
From Inverse of Product: Monoid: General Result, it follows that $y \in \map G A$ as well.
This is our basis for the induction.
==== Induction Hypothesis ====
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:for every set of commuting elements $x_1, \ldots, x_k$ in $A$
::if $\ds \prod_{i \mathop = 1}^k x_i \in \map G A$, then $x_i \in \map G A$ for each $1 \le i \le k$.
Then we need to show:
:for every set of commuting elements $x_1, \ldots, x_{k + 1}$ in $A$
::if $\ds \prod_{i \mathop = 1}^{k + 1} x_i \in \map G A$, then $x_i \in \map G A$ for each $1 \le i \le k + 1$.
==== Induction Step ====
This is our induction step.
Let $x_1, \ldots, x_{k + 1}$ be commuting elements of $A$ such that:
:$\ds \prod_{i \mathop = 1}^{k + 1} x_i \in \map G A$
We can write:
:$\ds \prod_{i \mathop = 1}^{k + 1} x_i = \paren {\prod_{i \mathop = 1}^k x_i} x_{k + 1}$
Using the basis for the induction, we have that:
:$\ds \prod_{i \mathop = 1}^k x_i \in \map G A$ and $x_{k + 1} \in \map G A$.
Using the induction hypothesis, we then have:
:$x_1, \ldots, x_k \in \map G A$ and $x_{k + 1} \in \map G A$
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
=== Sufficient Condition ===
This is precisely Inverse of Product: Monoid: General Result.
Category:Monoids
Product of Commuting Elements in Monoid is Unit iff Each Element is Unit/Proof 2

== Theorem ==
== Proof ==
If $x_1, \ldots, x_n \in \map G A$, then:
:$\ds \prod_{i \mathop = 1}^k x_i \in \map G A$
by Inverse of Product: Monoid: General Result.
Conversely, suppose:
:$\ds \prod_{i \mathop = 1}^k x_i \in \map G A$
That is, there is a $z \in A$ such that:
:$(1):\quad \ds z \paren {\prod_{i \mathop = 1}^k x_i} = \paren {\prod_{i \mathop = 1}^k x_i} z = e$
We shall show:
:$\forall i \in \set {1, \ldots, n} : x_i \in \map G A$
It suffices to show this for $i=1$, since $x_1, \ldots, x_n$ are commuting.
Define:
:$\ds z_1 := \paren {\prod_{i \mathop = 2}^k x_i} z$
Then:
:$x_1 z_1 = z_1 x_1 = e$
so that $x_1 \in \map G A$.
Indeed:
Moreover:
Category:Monoids
Product of Commuting Elements with Inverses

== Theorem ==
Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$.
Let $x, y \in S$ such that $x$ and $y$ are both invertible.
Then:
:$x \circ y \circ x^{-1} \circ y^{-1} = e_S = x^{-1} \circ y^{-1} \circ x \circ y$
 $x$ and $y$ commute.
== Proof ==
As $\struct {S, \circ}$ is a monoid, it is by definition a semigroup.
Therefore $\circ$ is associative, so we can dispense with parentheses.
From Invertible Element of Monoid is Cancellable, we also have that $x, y, x^{-1}, y^{-1}$ are cancellable.
So let :
Similarly:
Category:Monoids
Category:Commutativity
Product of Commuting Idempotent Elements is Idempotent

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Let $a, b \in S$ be idempotent elements of $S$.
Let $a$ and $b$ commute:
:$a \circ b = b \circ a$
Then $a \circ b$ is idempotent.
== Proof ==
From we take it for granted that $\struct {S, \circ}$ is closed under $\circ$.
Hence:
Thus $a \circ b$ is idempotent.
Category:Semigroups
Category:Idempotence
Product of Complex-Valued Functions Vanishing at Infinity Vanishes at Infinity

== Theorem ==
Let $X$ be a locally compact Hausdorff space. 
Let $f, g : X \to \C$ be complex-valued functions vanishing at infinity.
Then $f g$ vanishes at infinity.
== Proof ==
Let $\epsilon > 0$. 
Since $f$ vanishes at infinity, there exists a compact set $F_1 \subseteq X$ such that:
:$\cmod {\map f x} < \sqrt \epsilon$ for each $x \in X \setminus F_1$.
Since $g$ vanishes at infinity, there exists a compact set $F_2 \subseteq X$ such that:
:$\cmod {\map g x} < \sqrt \epsilon$ for each $x \in X \setminus F_2$. 
From Union of Two Compact Sets is Compact, $F_1 \cup F_2$ is compact and:
:$\cmod {\map f x} < \sqrt \epsilon$ and $\cmod {\map g x} < \sqrt \epsilon$ for each $x \in X \setminus \paren {F_1 \cup F_2}$.
Then, we have for $x \in X \setminus \paren {F_1 \cup F_2}$:
:$\cmod {\map f x \map g x} \le \cmod {\map f x} \cmod {\map g x} < \paren {\sqrt \epsilon}^2 = \epsilon$
Since again $F_1 \cup F_2$ is compact, we have that $f g$ vanishes at infinity.
Category:Complex-Valued Functions Vanishing at Infinity
Product of Complex Conjugates

== Theorem ==
Let $z_1, z_2 \in \C$ be complex numbers.
Let $\overline z$ denote the complex conjugate of the complex number $z$.
Then:
:$\overline {z_1 z_2} = \overline {z_1} \cdot \overline {z_2}$
=== General Result ===
== Proof ==
Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$, where $x_1, y_1, x_2, y_2 \in \R$.
Then:
== Examples ==
== Sources ==
* : Chapter $\text I$: Functions of a Complex Variable: $\S 2$. Conjugate Complex Numbers
* : $\S 1.2$. The Algebraic Theory: $(1.7)$
* : $1$: Complex Numbers: Supplementary Problems: Fundamental Operations with Complex Numbers: $55 \ \text {(a)}$
* : $1$ The complex plane: Complex numbers $\S 1.3$ Complex conjugation: $(4)$
* 
Category:Complex Conjugates
Category:Complex Multiplication
Category:Product of Complex Conjugates