nyuuzyou/wikis huggingface dataset:
Product of Complex Conjugates/Examples/3 Arguments

== Theorem ==
Let $z_1, z_2, z_3 \in \C$ be complex numbers.
Let $\overline z$ denote the complex conjugate of the complex number $z$.
Then:
:$\overline {z_1 z_2 z_3} = \overline {z_1} \cdot \overline {z_2} \cdot \overline {z_3}$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Fundamental Operations with Complex Numbers: $55 \ \text {(b)}$
Category:Product of Complex Conjugates
Product of Complex Conjugates/Examples/3 Arguments/Proof 2

== Theorem ==
== Proof ==
From Product of Complex Conjugates: General Result:
:$\ds \overline {\prod_{j \mathop = 1}^n z_j} = \prod_{j \mathop = 1}^n \overline {z_j}$
The result follows by setting $n = 3$.
Category:Product of Complex Conjugates
Product of Complex Conjugates/General Result

== Theorem ==
Let $z_1, z_2, \ldots, z_n \in \C$ be complex numbers.
Let $\overline z$ be the complex conjugate of the complex number $z$.
Then:
:$\ds \overline {\prod_{j \mathop = 1}^n z_j} = \prod_{j \mathop = 1}^n \overline {z_j}$
That is: the conjugate of the product equals the product of the conjugates.
== Proof ==
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \overline {\prod_{j \mathop = 1}^n z_j} = \prod_{j \mathop = 1}^n \overline {z_j}$
$\map P 1$ is trivially true, as this just says $\overline {z_1} = \overline {z_1}$.
=== Basis for the Induction ===
$\map P 2$ is the case:
:$\overline {z_1 z_2} = \overline {z_1} \cdot \overline {z_2}$
which has been proved in Product of Complex Conjugates.
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:
:$\ds \overline {\prod_{j \mathop = 1}^k z_j} = \prod_{j \mathop = 1}^k \overline {z_j}$
Then we need to show:
:$\ds \overline {\prod_{j \mathop = 1}^{k + 1} z_j} = \prod_{j \mathop = 1}^{k + 1} \overline {z_j}$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \N: \overline {\prod_{j \mathop = 1}^n z_j} = \prod_{j \mathop = 1}^n \overline {z_j}$
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Fundamental Operations with Complex Numbers: $55$
Category:Product of Complex Conjugates
Product of Complex Number with Conjugate

== Theorem ==
Let $z = a + i b \in \C$ be a complex number.
Let $\overline z$ denote the complex conjugate of $z$.
Then:
:$z \overline z = a^2 + b^2 = \cmod z^2$
and thus is wholly real.
== Proof ==
By the definition of a complex number, let $z = a + i b$ where $a$ and $b$ are real numbers.
Then:
As $a^2 + b^2$ is wholly real, the result follows.
== Sources ==
* : Chapter $\text I$: Functions of a Complex Variable: $\S 2$. Conjugate Complex Numbers
* : $\S 1.2$. The Algebraic Theory
* : Chapter $\text{II}$: Groups: Subgroups
* : $1$ The complex plane: Complex numbers $\S 1.3$ Complex conjugation: $(3)$
* 
Category:Complex Conjugates
Category:Complex Modulus
Category:Complex Multiplication
Product of Complex Number with Conjugate by Dot and Cross Product

== Theorem ==
Let $z_1$ and $z_2$ be complex numbers.
Then:
:$\overline {z_1} z_2 = \paren {z_1 \circ z_2} + i \paren {z_1 \times z_2}$
where:
:$\overline {z_1}$ denotes the complex conjugate of $z_1$
:$z_1 \circ z_2$ denotes the complex dot product of $z_1$ with $z_2$
:$z_1 \times z_2$ denotes the complex cross product of $z_1$ with $z_2$.
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Dot and Cross Product
Category:Complex Conjugates
Category:Complex Dot Product
Category:Complex Cross Product
Product of Complex Number with Conjugate in Exponential Form

== Theorem ==
Let $z_1$ and $z_2$ be complex numbers.
Then:
:$\overline {z_1} z_2 = \cmod {z_1} \, \cmod {z_2} e^{i \theta}$
where:
:$\overline {z_1}$ denotes the complex conjugate of $z_1$
:$\cmod {z_1}$ denotes the complex modulus of $z_1$
:$\theta$ denotes the angle from $z_1$ to $z_2$, measured in the positive direction.
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Dot and Cross Product
Category:Complex Conjugates
Product of Complex Numbers in Exponential Form

== Theorem ==
Let $z_1 := r_1 e^{i \theta_1}$ and $z_2 := r_2 e^{i \theta_2}$ be complex numbers expressed in exponential form.
Then:
:$z_1 z_2 = r_1 r_2 e^{i \paren {\theta_1 + \theta_2} }$
== Proof ==
== Sources ==
* : $\S 7$: Operations with Complex Numbers in Polar Form: $7.25$
* : $1$: Complex Numbers: Solved Problems: De Moivre's Theorem: $19$
* : $1$ The complex plane: Complex numbers $\S 1.2$ The algebraic structure of the complex numbers
Category:Complex Multiplication
Category:Exponential Form of Complex Number
Product of Complex Numbers in Polar Form

== Theorem ==
Let $z_1 := \polar {r_1, \theta_1}$ and $z_2 := \polar {r_2, \theta_2}$ be complex numbers expressed in polar form.
Then:
:$z_1 z_2 = r_1 r_2 \paren {\map \cos {\theta_1 + \theta_2} + i \map \sin {\theta_1 + \theta_2} }$
=== General Result ===
== Proof ==
== Examples ==
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 1$: The Language of Set Theory
* : $\S 6$: Complex Numbers: Multiplication and Division of Complex Numbers in Polar Form: $6.7$
* : $1$: Complex Numbers: De Moivre's Theorem
* : $1$: Complex Numbers: Solved Problems: De Moivre's Theorem: $19 \ \text{(a)}$
* 
* 
* : $\S 4$: Complex Numbers: Multiplication and Division of Complex Numbers in Polar Form: $4.8.$
Category:Product of Complex Numbers in Polar Form
Category:Complex Multiplication
Category:Polar Form of Complex Number
Product of Complex Numbers in Polar Form/Examples

== Examples of Use of Product of Complex Numbers in Polar Form ==
=== Example: $3 \cis 40 \degrees \times 4 \cis 80 \degrees$ ===
=== Example: $5 \cis 20 \degrees \times 3 \cis 40 \degrees$ ===
Category:Examples of Use of Product of Complex Numbers in Polar Form
Product of Complex Numbers in Polar Form/Examples/3 cis 40 x 4 cis 80

== Example of Use of Product of Complex Numbers in Polar Form ==
:$3 \cis 40 \degrees \times 4 \cis 80 \degrees = -6 + 6 \sqrt 3 i$
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Solved Problems: De Moivre's Theorem: $26 \ \text {(a)}$
Category:Examples of Use of Product of Complex Numbers in Polar Form
Product of Complex Numbers in Polar Form/Examples/5 cis 20 x 3 cis 40

== Example of Use of Product of Complex Numbers in Polar Form ==
:$5 \cis 20 \degrees \times 3 \cis 40 \degrees = \dfrac {15} 2 + \dfrac {15 \sqrt 3} 2 i$
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: De Moivre's Theorem: $89 \ \text{(a)}$
Category:Examples of Use of Product of Complex Numbers in Polar Form
Product of Complex Numbers in Polar Form/General Result

== Theorem ==
Let $z_1, z_2, \ldots, z_n \in \C$ be complex numbers.
Let $z_j = \polar {r_j, \theta_j}$ be $z_j$ expressed in polar form for each $j \in \set {1, 2, \ldots, n}$.
Then:
:$z_1 z_2 \cdots z_n = r_1 r_2 \cdots r_n \paren {\map \cos {\theta_1 + \theta_2 + \cdots + \theta_n} + i \map \sin {\theta_1 + \theta_2 + \cdots + \theta_n} }$
== Proof ==
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$z_1 z_2 \cdots z_n = r_1 r_2 \cdots r_n \paren {\map \cos {\theta_1 + \theta_2 + \cdots + \theta_n} + i \map \sin {\theta_1 + \theta_2 + \cdots + \theta_n} }$
Let this be expressed as:
:$\ds \prod_{j \mathop = 1}^n z_j = \prod_{j \mathop = 1}^n r_j \sum_{j \mathop = 1}^n \paren {\cos \theta_j + i \sin \theta_j}$
$\map P 1$ is the case:
:$r_1 \paren {\cos x + i \sin x} = r_1 \paren {\cos x + i \sin x }$
which is trivially true.
=== Basis for the Induction ===
$\map P 2$ is the case:
:$z_1 z_2 = r_1 r_2 \paren {\map \cos {\theta_1 + \theta_2} + i \map \sin {\theta_1 + \theta_2} }$
which is proved in Product of Complex Numbers in Polar Form.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\ds \prod_{j \mathop = 1}^k z_j = \prod_{j \mathop = 1}^k r_j \sum_{j \mathop = 1}^k \paren {\cos \theta_j + i \sin \theta_j}$
Then we need to show:
:$\ds \prod_{j \mathop = 1}^{k + 1} z_j = \prod_{j \mathop = 1}^{k + 1} r_j \sum_{j \mathop = 1}^{k + 1} \paren {\cos \theta_j + i \sin \theta_j}$
=== Induction Step ===
This is our induction step:
Hence, by induction, for all $n \in \N_{>0}$:
:$z_1 z_2 \cdots z_n = r_1 r_2 \cdots r_n \paren {\map \cos {\theta_1 + \theta_2 + \cdots + \theta_n} + i \map \sin {\theta_1 + \theta_2 + \cdots + \theta_n} }$
== Sources ==
* : Chapter $\text I$: Functions of a Complex Variable: $\S 3$. Geometric Representation of Complex Numbers
* : $\S 2$. Geometrical Representations: $(2.15)$
* : $1$: Complex Numbers: De Moivre's Theorem
* : $1$: Complex Numbers: Supplementary Problems: Axiomatic Foundations of Complex Numbers: $79$
Category:Complex Multiplication
Product of Composite Morphisms

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $f \times f': A \times A' \to B \times B'$ and $g \times g': B \times B' \to C \times C'$ be two composable products of morphisms in $\mathbf C$.
Then:
:$\paren {g \circ f} \times \paren {g' \circ f'} = \paren {g \times g'} \circ \paren {f \times f'}$
where $\times$ signifies product of morphisms.
== Proof ==
The situation is efficiently captured in the following commutative diagram:
$\quad\quad \begin{xy}
<-5em,0em>*+{A} = "A",
<0em,0em>*+{A \times A'} = "P",
<5em,0em>*+{A'} = "A2",
<-5em,-5em>*+{B} = "B",
<0em,-5em>*+{B \times B'} = "Q",
<5em,-5em>*+{B'} = "B2",
<-5em,-10em>*+{C} = "C",
<0em,-10em>*+{C \times C'} = "R",
<5em,-10em>*+{C'} = "C2",
"P";"A" **@{-} ?>*@{>} ?*!/^.8em/{\pr_1},
"P";"A2" **@{-} ?>*@{>} ?*!/_.8em/{\pr_2},
"Q";"B" **@{-} ?>*@{>} ?*!/^.8em/{\pr_1},
"Q";"B2" **@{-} ?>*@{>} ?*!/_.8em/{\pr_2},
"R";"C" **@{-} ?>*@{>} ?*!/_.8em/{\pr_1},
"R";"C2" **@{-} ?>*@{>} ?*!/^.8em/{\pr_2},
"A";"B" **@{-} ?>*@{>} ?*!/^.8em/{f},
"A2";"B2" **@{-} ?>*@{>} ?*!/_.8em/{f'},
"P";"Q" **@{--} ?>*@{>} ?*!/_1.4em/{f \times f'},
"B";"C" **@{-} ?>*@{>} ?*!/^.8em/{g},
"B2";"C2" **@{-} ?>*@{>} ?*!/_.8em/{g'},
"Q";"R" **@{--} ?>*@{>} ?*!/_1.4em/{g \times g'},
\end{xy}$
The result is immediate from deleting the $B$s:
$\quad\quad \begin{xy}
<-5em,0em>*+{A} = "A",
<0em,0em>*+{A \times A'} = "P",
<5em,0em>*+{A'} = "A2",
<-5em,-10em>*+{C} = "C",
<0em,-10em>*+{C \times C'} = "R",
<5em,-10em>*+{C'} = "C2",
"P";"A" **@{-} ?>*@{>} ?*!/^.8em/{\pr_1},
"P";"A2" **@{-} ?>*@{>} ?*!/_.8em/{\pr_2},
"R";"C" **@{-} ?>*@{>} ?*!/_.8em/{\pr_1},
"R";"C2" **@{-} ?>*@{>} ?*!/^.8em/{\pr_2},
"A";"C" **@{-} ?>*@{>} ?*!/^1.1em/{g \circ f},
"A2";"C2" **@{-} ?>*@{>} ?*!/_1.3em/{g' \circ f'},
"P";"R" **@{--} ?>*@{>} ?*!/_2em/{\paren {g \circ f} \times \\ \paren {g' \circ f'} },
\end{xy}$
The uniqueness of $\paren {g \circ f} \times \paren {g' \circ f'}$ implies it equals $\paren {g \times g'} \circ \paren {f \times f'}$.
Category:Product Categories
Category:Category Theory
Product of Composite Number with Number is Solid Number

== Theorem ==
Let $a, b \in \Z$ be positive integers.
Let $a$ be a composite number.
Then $a b$ is a solid number.
== Proof ==
By definition of composite number:
:$\exists p, q \in \Z_{>1}: a = p q$
Then:
:$a b = p q b$
Hence the result by definition of solid number.
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Euclidean Number Theory
Product of Computable Rational Sequences is Computable

== Theorem ==
Let $\sequence {x_n}$ and $\sequence {y_n}$ be computable rational sequences.
Then, $\sequence {x_n y_n}$ is a computable rational sequence.
== Proof ==
By definition of computable rational sequence, there exist total recursive $f_x, g_x, f_y, g_y : \N \to \N$ such that:
:$x_n = \dfrac {k_n} {\map {g_x} n + 1}$
:$y_n = \dfrac {\ell_n} {\map {g_y} n + 1}$
where:
:$\map {f_x} n$ codes the integer $k_n$
:$\map {f_y} n$ codes the integer $\ell_n$
We define:
:$\map f n = \map {f_x} n \times_Z \map {f_y} n$
:$\map g n = \map {g_x} n \times \map {g_y} n + \map {g_x} n + \map {g_y} n$
which are total recursive by:
* Multiplication of Integers is Primitive Recursive
* Addition is Primitive Recursive
* Multiplication is Primitive Recursive
Thus, we have:
Category:Computability Theory
Product of Computable Real Sequences is Computable

== Theorem ==
Let $\sequence {x_m}$ and $\sequence {y_m}$ be computable real sequences.
Then $\sequence {x_m y_m}$ is a computable real sequence.
== Proof ==
By definition of computable real sequence, there exist total recursive $f, g : \N^2 \to \N$ such that, for all $m, n \in \N$:
:$\dfrac {k_{m,n} - 1} {n + 1} < x_m < \dfrac {k_{m,n} + 1} {n + 1}$
:$\dfrac {\ell_{m,n} - 1} {n + 1} < y_m < \dfrac {\ell_{m,n} + 1} {n + 1}$
By Computable Real Sequence iff Limits of Computable Rational Sequences, there exist:
* Computable rational sequences $\sequence {a_k}, \sequence {b_k}$
* Total recursive functions $\phi_x, \psi_x, \phi_y, \psi_y : \N^2 \to \N$
such that:
:$\forall m, p \in \N: \forall n \ge \map {\psi_x} {m, p}: \size {a_{\map {\phi_x} {m, n}} - x_m} < \dfrac 1 {p + 1}$
:$\forall m, p \in \N: \forall n \ge \map {\psi_y} {m, p}: \size {a_{\map {\phi_y} {m, n}} - y_m} < \dfrac 1 {p + 1}$
By Computable Subsequence of Computable Rational Sequence is Computable/Corollary, there exist:
:Computable rational sequences $\sequence {a'_k}, \sequence {b'_k}$
such that, for all $m, n \in \N$:
:$a_{\map {\phi_x} {m, n}} = a'_{\map \pi {m, n}}$
:$b_{\map {\phi_y} {m, n}} = b'_{\map \pi {m, n}}$
By Product of Computable Rational Sequences is Computable:
:$\sequence {a'_k b'_k}$
is a computable rational sequence.
Define $\psi : \N^2 \to \N$ as:
:$\map \psi {m, p} = \map \max {\map {\psi_x} {\paren {2 p + 2} \size {\ell_{m,0}} + 2 p + 1}, \map {\psi_y} {\paren {2 p + 2} \size {k_{m,0}} + 4 p + 3}}$
which is total recursive by:
* Maximum Function is Primitive Recursive
* Absolute Value of Integer is Primitive Recursive
* Addition is Primitive Recursive
* Multiplication is Primitive Recursive
* Constant Function is Primitive Recursive
* Primitive Recursive Function is Total Recursive Function
Let $m, n, p \in \N$ be arbitrary, and suppose $n \ge \map \psi {m, p}$.
The following inequalities hold:
:$n \ge \map {\psi_x} {\paren {2 p + 2} \size {\ell_{m,0}} + 2 p + 1}$
:$n \ge \map {\psi_y} {\paren {2 p + 2} \size {k_{m,0}} + 4 p + 3}$
And thus:
:$\size {a_{\map {\psi_x} {m, n}} - x_m} < \dfrac 1 {\paren {2 p + 2} \size {\ell_{m,0}} + 2 p + 2} = \dfrac 1 {\paren {2 p + 2} \paren {\size {\ell_{m,0}} + 1}}$
:$\size {b_{\map {\psi_y} {m, n}} - y_m} < \dfrac 1 {\paren {2 p + 2} \size {k_{m,0}} + 4 p + 4} = \dfrac 1 {\paren {2 p + 2} \paren {\size {k_{m,0}} + 2}}$
We also have:
:$\size {x_m - k_{m,0}} < 1$
:$\size {y_m - \ell_{m,0}} < 1$
so it follows that:
:$\size {x_m} < \size {k_{m,0}} + 1$
:$\size {y_m} < \size {\ell_{m,0}} + 1$
Finally, we also have:
:$\size {a_{\map {\phi_x} {m, n}} - x_m} < \dfrac 1 {q + 1} \le 1$
for some $q \in \N$, so:
:$\size {a_{\map {\phi_x} {m, n}}} < \size {x_m} + 1 < \size {k_{m,0}} + 2$
Then:
Thus, the conditions are satisfied to apply Computable Real Sequence iff Limits of Computable Rational Sequences.
Category:Computability Theory
Product of Conjugates equals Conjugate of Products

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Then:
:$\forall a, x, y \in G: \paren {a \circ x \circ a^{-1} } \circ \paren {a \circ y \circ a^{-1} } = a \circ \paren {x \circ y} \circ a^{-1}$
That is, the product of conjugates is equal to the conjugate of the product.
== Proof ==
Follows directly from the group axioms.
== Sources ==
* : $\S 6.6$. Normal subgroups: Example $124 \ \text{(i)}$
Category:Conjugacy
Product of Consecutive Integers is Even

== Theorem ==
Let $a$ and $b$ be consecutive integers.
Then $a b$ is even.
== Proof ==
 let $a < b$.
Then $b$ can be expressed as $a + 1$.
Hence:
From Parity of Integer equals Parity of its Square, $a$ and $a^2$ are either both even or both odd.
The result follows from:
:Sum of Even Integers is Even
and:
:Sum of Even Number of Odd Numbers is Even
Category:Even Integers
Product of Continuous Functions on Topological Ring is Continuous

== Theorem ==
Let $X$ be a topological space. 
Let $R$ be a topological ring. 
Let $f, g : X \to R$ be continuous functions.
Then $f \cdot g : X \to R$ is continuous.
== Proof ==
Equip the Cartesian product $R \times R$ with its product topology.
Define $h : X \to R \times R$ by: 
:$\map h x = \tuple {\map f x, \map g x}$ 
for each $x \in X$ and $p : R \times R \to R$ by: 
:$\map p {x, y} = x y$
for each $\tuple {x, y} \in R \times R$.
From Continuous Mapping to Product Space, $h$ is continuous.
From the definition of a topological ring, $p$ is continuous.
Then from Composite of Continuous Mappings is Continuous, $p \circ h : X \to R$ is continuous, while: 
for $x \in X$.
Category:Topological Rings
Product of Convergent Products is Convergent

== Theorem ==
Let $\struct {\mathbb K, \norm {\,\cdot\,} }$ be a valued field.
Let $\ds \prod_{n \mathop = 1}^\infty a_n$ converge to $a$.
Let $\ds \prod_{n \mathop = 1}^\infty b_n$ converge to $b$.
Then $\ds \prod_{n \mathop = 1}^\infty a_n b_n$ converges to $ab$.
== Proof ==
Let $n_0 \in \N$ such that $a_n \ne 0$ for $n> n_0$.
Let $n_1 \in \N$ such that $b_n \ne 0$ for $n> n_1$.
Then $a_n b_n \ne 0$ for $n > n_2 = \max \set {n_0, n_1}$.
Let $p_n$ be the $n$th partial product of $\ds \prod_{n \mathop = n_2 + 1}^\infty a_n$.
Let $q_n$ be the $n$th partial product of $\ds \prod_{n \mathop = n_2 + 1}^\infty b_n$.
Then $p_n q_n$ is the $n$th partial product of $\ds \prod_{n \mathop = n_2 + 1}^\infty a_n b_n$.
Because $p_n$ and $q_n$ converge to a nonzero limit, so does $p_n q_n$.
Thus $\ds \prod_{n \mathop = 1}^\infty a_n b_n$ converges.
Let $P_n$ be the $n$th partial product of $\ds \prod_{n \mathop = 1}^\infty a_n$.
Let $Q_n$ be the $n$th partial product of $\ds \prod_{n \mathop = 1}^\infty b_n$.
Then $P_n Q_n$ is the $n$th partial product of $\ds \prod_{n \mathop = 1}^\infty a_n b_n$.
By Limit of Product of Sequences, $P_n Q_n\to ab$.
Thus $\ds \prod_{n \mathop = 1}^\infty a_n b_n$ converges to $a b$.
== Also see ==
* Product of Convergent and Divergent Product is Divergent
* Product of Absolutely Convergent Products is Absolutely Convergent
Category:Infinite Products
Product of Convergent and Divergent Product is Divergent

== Theorem ==
Let $\struct {\mathbb K, \norm {\, \cdot \,} }$ be a valued field.
Let $\ds \prod_{n \mathop = 1}^\infty a_n$ be convergent.
Let $\ds \prod_{n \mathop = 1}^\infty b_n$ be divergent.
Then $\ds \prod_{n \mathop = 1}^\infty a_n b_n$ is divergent.
== Proof ==
== Also see ==
* Product of Convergent Products is Convergent
* Product of Absolutely Convergent Products is Absolutely Convergent
Category:Infinite Products
Product of Coprime Factors

== Theorem ==
Let $a, b, c \in \Z$ such that $a$ and $b$ are coprime.
Let both $a$ and $b$ be divisors of $c$.
Then $a b$ is also a divisor of $c$.
That is:
:$a \perp b \land a \divides c \land b \divides c \implies a b \divides c$
== Proof ==
By definition of divisor:
:$a \divides c \implies \exists r \in \Z: c = a r$
:$b \divides c \implies \exists s \in \Z: c = b s$
So:
== Sources ==
* : Chapter $2$: Some Properties of $\Z$: Exercise $2.6$
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Theorem $2 \text{-} 4$: Corollary $2$
Category:Coprime Integers
Product of Coprime Ideals equals Intersection

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $\mathfrak a, \mathfrak b \subseteq A$ be coprime ideals.
Then their product equals their intersection:
:$\mathfrak a \mathfrak b = \mathfrak a \cap \mathfrak b$
== Proof ==
By Intersection of Ideals of Ring contains Product:
:$\mathfrak a \mathfrak b \subseteq \mathfrak a \cap \mathfrak b$
It remains to show that $\mathfrak a \mathfrak b \supseteq \mathfrak a \cap \mathfrak b$.
Let $c \in \mathfrak a \cap \mathfrak b$.
Because $\mathfrak a$ and $\mathfrak b$ are coprime, there exist $x \in \mathfrak a$, $y \in \mathfrak b$ with $x + y = 1$.
Then:
:$c = c x + c y$
Because:
:$c \in \mathfrak b$ and $x \in \mathfrak a$
:$c \in \mathfrak a$ and $y \in \mathfrak b$
we have:
:$c \in \mathfrak a \mathfrak b$
== Converse theorem ==
In a Dedekind domain, the conditions
:$\mathfrak a + \mathfrak b = \ideal 1$
and
:$\mathfrak a \cap \mathfrak b = \mathfrak a \mathfrak b$
are equivalent, both expressing the fact that $\mathfrak a$ and $\mathfrak b$ have no common ideal factors except $\ideal 1$.
In a general commutative ring with unity, these conditions are not equivalent.
For example, in the polynomial ring $\Z\sqbrk T$ we have
:$\ideal 2 + \ideal T = \ideal {2, T} \ne \ideal 1$
but
:$\ideal 2 \cap \ideal T = \ideal {2T} = \ideal 2 \ideal T$
== Sources ==
* : Chapter $\text {III}$, $\S 3$ Exercises: Problem $12 \ \text{(b)}$
Category:Coprime Ideals
Product of Coprime Numbers whose Divisor Sum is Square has Square Divisor Sum

== Theorem ==
Let $m, n \in \Z_{>0}$ be a positive integer.
Let the divisor sum of $m$ and $n$ both be square.
Let $m$ and $n$ be coprime.
Then the divisor sum of $m n$ is square.
== Proof ==
Let $\map {\sigma_1} m = k^2$.
Let $\map {\sigma_1} n = l^2$.
Thus:
Category:Numbers whose Divisor Sum is Square
Category:Coprime Integers
Product of Coprime Pairs is Coprime

== Theorem ==
Let $a, b, c, d$ be integers.
Let: 
:$a \perp c, b \perp c, a \perp d, b \perp d$
where $a \perp c$ denotes that $a$ and $c$ are coprime.
Then:
:$a b \perp c d$
== Proof ==
Let $e = a b, f = c d$.
== Sources ==
* : Book $\text{VII}$. Propositions
Category:Coprime Integers
Product of Cosets of Ideals is Subset of Product in Quotient Ring

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $\powerset R$ be the power set of $R$.
Let $J$ be an ideal of $R$.
Let $X$ and $Y$ be cosets of $J$.
Let $X \circ_\PP Y$ be the product of $X$ and $Y$, where $\circ_\PP$ is the operation induced on $\powerset R$ by $\circ$.
The product $X \circ_\PP Y$ in $\powerset R$ is a subset of their product in $R / J$.
== Proof ==
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old
Category:Ring Operations on Coset Space of Ideal
Product of Cotangents of Fractions of Pi

== Theorem ==
Let $m \in \Z$ such that $m > 1$.
Then:
:$\ds \prod_{k \mathop = 1}^{m - 1} \cot \frac {k \pi} {2 m} = 1$
== Proof ==
We have:
That means:
:$(1): \quad \dfrac \pi 2 - \dfrac {k \pi} {2 m} = \dfrac {\paren {m - k} \pi} {2 m}$
Let $m$ be odd.
Then $m - 1$ is even, and so:
Now suppose $m$ is even.
Then $m - 1$ is odd, and so:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $148$
Category:Cotangent Function
Product of Countable Discrete Space with Sierpiński Space is Paracompact

== Theorem ==
Let $T_X = \struct {S, \tau}$ be a countable discrete space.
Let $T_Y = \struct {\set {0, 1}, \tau_0}$ be a Sierpiński space.
Let $T_X \times T_Y$ be the product space of $T_X$ and $T_Y$.
Then $T_X \times T_Y$ is paracompact.
== Proof ==
From Discrete Space is Paracompact, $T_X$ is paracompact.
We have that the Sierpiński space $T_Y$ is a finite topological space.
From Finite Topological Space is Compact, $T_Y$ is a compact space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $11$. Sierpinski Space: $19$
Category:Sierpiński Space
Category:Discrete Topology
Category:Paracompact Spaces