nyuuzyou/wikis huggingface dataset:
Product of Rational Polynomials

== Theorem ==
Let $\Q \sqbrk X$ be the ring of polynomial forms over the field of rational numbers in the indeterminate $X$.
Let $\map f X, \map g X \in \Q \sqbrk X$.
Using Rational Polynomial is Content Times Primitive Polynomial, let these be expressed as:
:$\map f X = \cont f \cdot \map {f^*} X$
:$\map g X = \cont g \cdot \map {g^*} X$
where:
:$\cont f, \cont g$ are the content of $f$ and $g$ respectively
:$f^*, g^*$ are primitive.
Let $\map h X = \map f X \map g X$ be the product of $f$ and $g$.
Then:
:$\map {h^*} X = \map {f^*} X \map {g^*} X$
== Proof ==
From Rational Polynomial is Content Times Primitive Polynomial:
:$\cont h \cdot \map {h^*} X = \cont f \cont g \cdot \map {f^*} X \map {g^*} X$
and this expression is unique.
By Gauss's Lemma on Primitive Rational Polynomials we have that $\map {f^*} X \map {g^*} X$ is primitive.
From Content of Rational Polynomial is Multiplicative:
:$\cont f \cont g = \cont f \cont g > 0$.
The result follows.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 31$. Polynomials with Integer Coefficients: Theorem $62$
Category:Polynomial Theory
Product of Rationally Expressible Numbers is Rational

== Theorem ==
Let $a, b \in \set {x \in \R_{>0} : x^2 \in \Q}$ be the lengths of rational line segments.
Furthermore, let $\dfrac a b \in \Q$.
Then, $a b \in \Q$.
=== Lemma ===
== Proof 1 ==
:
Let the rectangle $AC$ be contained by the rational straight lines $AB$ and $BC$.
Let $AB$ and $BC$ be commensurable in length.
Let the square $AD$ be described on $AB$.
From , $AD$ is rational.
Since:
:$AB$ is commensurable in length with $BC$
and:
:$AB = BD$
it follows that
:$BD$ is commensurable in length with $BC$.
From Areas of Triangles and Parallelograms Proportional to Base:
:$BD : BC = DA : AC$
Therefore $DA$ is commensurable in length with $AC$.
But $DA$ is rational.
It follows from that $AC$ is also rational.
== Proof 2 ==
We have that $b^2 \in \Q$ by definition.
We also have that $\dfrac a b \in \Q$ by hypothesis.
Therefore, by Rational Multiplication is Closed, their product, $a b$, is also rational.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Product of Real Number with Quotient

== Theorem ==
:$\forall a, x \in \R, y \in \R_{\ne 0}: \dfrac {a \times x} y = a \times \dfrac x y$
== Proof ==
== Sources ==
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $1 \ \text{(s)}$
Category:Real Multiplication
Category:Real Division
Product of Real Numbers is Positive iff Numbers have Same Sign

== Theorem ==
The product of two real numbers is greater than $0$ either both are greater than $0$ or both are less than $0$.
:$\forall x, y \in \R: x \times y > 0 \iff \paren {x, y \in \R_{>0} } \lor \paren {x, y \in \R_{<0} }$
== Proof ==
=== Sufficient Condition ===
Let $x \times y > 0$.
 either $x = 0$ or $y = 0$.
Then from Real Zero is Zero Element:
:$x \times y = 0$
Therefore by Proof by Contradiction:
:$y \ne 0$ and $x \ne 0$
Let $x > 0$.
 $y < 0$.
But , $x \times y > 0$.
Therefore by Proof by Contradiction:
:$y > 0$
Let $x < 0$.
 $y > 0$.
But , $x \times y > 0$.
Therefore by Proof by Contradiction:
:$y > 0$
Thus:
:$x \times y > 0 \implies \paren {x > 0 \land y > 0} \lor \paren {x < 0 \land y < 0}$
=== Necessary Condition ===
Let $x > 0$ and $y > 0$.
Then from Strictly Positive Real Numbers are Closed under Multiplication:
:$x \times y > 0$
Let $x < 0$ and $y < 0$.
Thus if either $x, y > 0$ or $x, y < 0$:
:$x \times y > 0$
== Sources ==
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $2 \ \text{(h)}$
Category:Real Multiplication
Product of Reciprocals of Real Numbers

== Theorem ==
:$\forall x, y \in \R_{\ne 0}: \dfrac 1 x \times \dfrac 1 y = \dfrac 1 {x \times y}$
== Proof ==
== Sources ==
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $1 \ \text{(m)}$
Category:Real Multiplication
Category:Reciprocals
Product of Ring Negatives

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Then:
:$\forall x, y \in \struct {R, +, \circ}: \paren {-x} \circ \paren {-y} = x \circ y$
where $\paren {-x}$ denotes the negative of $x$.
== Proof ==
We have:
== Also see ==
* Product with Ring Negative
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 2$. Elementary Properties
* : Chapter $\text {IV}$: Rings and Fields: $20$. The Integers: Theorem $20.9$
* : Chapter $1$: Integral Domains: $\S 4$. Elementary Properties: Theorem $2 \ \text{(v)}$
* : Chapter $1$: Rings - Definitions and Examples: $3$: Some special classes of rings: Lemma $1.2 \ \text{(iii)}$
* : $\S 54.2$ The definition of a ring and its elementary consequences: $\text{(ii)}$
* : Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers: Exercise $4$
Category:Ring Theory
Product of Rook Matrices is Rook Matrix

== Theorem ==
Let $\mathbf A$ and $\mathbf B$ be rook matrices.
Their product $\mathbf {A B}$ is also a rook matrix.
== Proof ==
An element $a b_{ij}$ of $\mathbf {A B}$ is formed by multiplying each element of row $i$ of $\mathbf A$ by its corresponding element of column $j$ of $\mathbf B$.
No more than $1$ element of row $i$ equals $1$, and the rest equal $0$.
No more than $1$ column $k$ of $\mathbf B$ contains $1$ in its $i$th element, and the rest contain $0$.
So of all the elements of row $i$ of $\mathbf {A B}$, only $a b_{ik}$ is $1$, and the rest are $0$.
By the same argument, each row of $\mathbf {A B}$ contains no more than one $1$, and all the rest of the elements are $0$.
In $\mathbf B$, each column $j$ has no more than one element equal to $1$, and all are in a different row $k$.
Thus each row contains its $1$, if it has one, in a different column from all the other rows.
So there is no more than one $1$ in each column of $\mathbf {A B}$.
Hence the result by definition of rook matrix.
== Sources ==
* : Chapter $2$: 'And you do addition?': $\S 2.2$: Functions on vectors: $\S 2.2.4$: Multiplication and inverses: Ponderable $2.2.2 \ \text{(b)}$
Category:Rook Matrices
Product of Roots of Polynomial

== Theorem ==
Let $P$ be the polynomial equation:
:$a_n z^n + a_{n - 1} z^{n - 1} + \cdots + a_1 z + a_0 = 0$
such that $a_n \ne 0$.
The product of the roots of $P$ is $\dfrac {\paren {-1}^n a_0} {a_n}$.
== Proof ==
Let the roots of $P$ be $z_1, z_2, \ldots, z_n$.
Then $P$ can be written in factored form as:
:$\ds a_n \prod_{k \mathop = 1}^n \paren {z - z_k} = a_0 \paren {z - z_1} \paren {z - z_2} \dotsm \paren {z - z_n}$
Multiplying this out, $P$ can be expressed as:
:$a_n \paren {z^n - \paren {z_1 + z_2 + \dotsb + z_n} z^{n - 1} + \dotsb + \paren {-1}^n z_1 z_2 \dotsm z_n} = 0$
where the coefficients of $z^{n - 2}, z^{n - 3}, \ldots$ are more complicated and irrelevant.
Equating powers of $z$, it follows that:
:$a_n \paren {-1}^n z_1 z_2 \dotsm z_n = a_0$
from which:
:$z_1 z_2 \dotsm z_n = \dfrac {\paren {-1}^n a_0} {a_n}$
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Polynomial Equations: $35$
Category:Polynomial Equations
Product of Roots of Quadratic Equation

== Theorem ==
Let $P$ be the quadratic equation $a x^2 + b x + c = 0$.
Let $\alpha$ and $\beta$ be the roots of $P$.
Then:
:$\alpha \beta = \dfrac c a$
== Proof ==
== Also see ==
* Sum of Roots of Quadratic Equation
* Viète's Formulas
== Sources ==
* : $3$: Elementary Analytic Methods: $3.8$ Algebraic Equations: Solution of Quadratic Equations: $3.8.1$
* : $\S 9$: Solutions of Algebraic Equations: Quadratic Equation: $a x^2 + b x + c = 0$: $9.2$
* : $\S 5$: Solutions of Algebraic Equations: Quadratic Equation: $a x^2 + b x + c = 0$: $5.2.$
* 
Category:Quadratic Equations
Product of Row Sum Unity Matrices

== Theorem ==
Let $\mathbf A = \sqbrk a_{m n}$ be an $m \times n$ matrix.
Let $\mathbf B = \sqbrk b_{n p}$ be an $n \times p$ matrix.
Let the row sum of $\mathbf A$ and $\mathbf B$ be equal to $1$.
Then the row sum of their (conventional) product is also $1$.
== Proof ==
We have that:
:$\ds \sum_{i \mathop = 1}^n a_{i j} = \sum_{i \mathop = 1}^n b_{i j} = 1$
Then:
Category:Matrix Algebra
Product of Semigroup Element with Left Inverse is Idempotent

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup with a left identity $e_L$.
Let $x \in S$ such that $\exists x_L: x_L \circ x = e_L$, that is $x$ has a left inverse with respect to the left identity.
Then:
:$\paren {x \circ x_L} \circ \paren {x \circ x_L} = x \circ x_L$
That is, $x \circ x_L$ is idempotent.
== Proof ==
== Also see ==
* Product of Semigroup Element with Right Inverse is Idempotent
== Sources ==
* : Chapter $5$: Semigroups: Exercise $5$
Category:Semigroups
Category:Idempotence
Product of Semigroup Element with Right Inverse is Idempotent

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup with a right identity $e_R$.
Let $x \in S$ such that $\exists x_R: x \circ x_R = e_R$, that is, $x$ has a right inverse with respect to the right identity.
Then:
:$\paren {x_R \circ x} \circ \paren {x_R \circ x} = x_R \circ x$
That is, $x_R \circ x$ is idempotent.
== Proof ==
== Examples ==
== Also see ==
* Product of Semigroup Element with Left Inverse is Idempotent
== Sources ==
* : Chapter $4$. Groups: Exercise $10$
Category:Semigroups
Category:Idempotence
Category:Product of Semigroup Element with Right Inverse is Idempotent
Product of Semigroup Element with Right Inverse is Idempotent/Examples/Double and Half Mappings on Integers

== Examples of Use of Product of Semigroup Element with Right Inverse is Idempotent ==
Let $\struct {\Z^\Z, \circ}$ be the semigroup defined such that:
:$\Z$ is the set of all mappings on the integers.
:$\circ$ denotes composition of mappings.
Let $\rho, \sigma \in \Z^\Z$ such that:
:$\forall x \in \Z: \map \rho x = \begin{cases} \dfrac x 2 & : x \text { even} \\ 0 & : x \text { odd} \end{cases}$
:$\forall x \in \Z: \map \sigma x = 2 x$
Then:
:$\rho$ is a right inverse for $\sigma$
but:
:$\rho$ is not a left inverse for $\sigma$
As a result:
:$\paren {\sigma \circ \rho}^2 = \sigma \circ \rho$
== Proof ==
Thus $\map {\paren {\sigma \circ \rho} } x \ne \map {\paren {\rho \circ \sigma} } x$ $x$ is odd.
So $\rho$ is a right inverse but not a left inverse for $\sigma$.
Then we have that:
:$\map {\paren {\sigma \circ \rho} } x = \begin{cases} x & : x \text { even} \\ 0 & : x \text { odd} \end{cases}$
and it follows that:
:$\map {\paren {\sigma \circ \rho}^2} x = \begin{cases} x & : x \text { even} \\ 0 & : x \text { odd} \end{cases}$
== Sources ==
* : Chapter $4$. Groups: Exercise $10$
Category:Product of Semigroup Element with Right Inverse is Idempotent
Product of Sequence Converges to Zero with Cauchy Sequence Converges to Zero

== Theorem == 
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring with zero $0$.
Let $\sequence {x_n}$ and $\sequence {y_n} $ be sequences in $R$.
Let $\sequence {x_n}$ converge to $0$.
Let $\sequence {y_n}$ be a Cauchy sequence.
Then:
:$\sequence {x_n y_n}$ and $\sequence {y_n x_n}$ converge to $0$.
== Proof ==
By Cauchy Sequence is Bounded in Normed Division Ring:
:$\exists M \in \R_{>0}: \forall n, \norm {x_n} \le M$
Given $\epsilon > 0$.
Since $\sequence {x_n}$ converges to $0$ then: 
:$\exists N \in \N: \forall n > N, \norm {x_n} < \dfrac \epsilon M$
Hence:
Similarly, $\norm {y_n x_n - 0} < \epsilon$
The result follows by definition of convergence in normed division rings.
== Sources ==
* : $\S 3.2$: Completions
Category:Cauchy Sequences
Category:Normed Division Rings
Product of Sequence of 1 minus Reciprocal of Squares

== Theorem ==
For all $n \in \Z_{\ge 1}$:
:$\ds \prod_{j \mathop = 2}^n \paren {1 - \dfrac 1 {j^2} } = \dfrac {n + 1} {2 n}$
== Proof 1 ==
 
== Proof 2 ==
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $28$
Category:Product of Sequence of 1 minus Reciprocal of Squares
Category:Continued Products
Product of Sequence of 1 minus Reciprocal of Squares/Proof 1

== Theorem ==
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:
:$\ds \prod_{j \mathop = 2}^n \paren {1 - \dfrac 1 {j^2} } = \dfrac {n + 1} {2 n}$
It is first noted that $n = 0$ is excluded because in that case $\dfrac {n + 1} {2 n}$ is undefined.
$\map P 1$ is the other edge case:
Thus $\map P 1$ is seen to hold.
=== Basis for the Induction ===
$\map P 2$ is the case:
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown 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 the induction hypothesis:
:$\ds \prod_{j \mathop = 2}^k \paren {1 - \dfrac 1 {j^2} } = \dfrac {k + 1} {2 k}$
from which it is to be shown that:
:$\ds \prod_{j \mathop = 2}^{k + 1} \paren {1 - \dfrac 1 {j^2} } = \dfrac {k + 2} {2 \paren {k + 1} }$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{\ge 1}: \prod_{j \mathop = 2}^n \paren {1 - \dfrac 1 {j^2} } = \dfrac {n + 1} {2 n}$
Category:Product of Sequence of 1 minus Reciprocal of Squares
Product of Sequence of Fermat Numbers plus 2

== Theorem ==
Let $F_n$ denote the $n$th Fermat number.
Then:
=== Corollary ===
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$F_n = \ds \prod_{j \mathop = 0}^{n - 1} F_j + 2$
=== Basis for the Induction ===
$\map P 1$ is the case:
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown 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 the induction hypothesis:
:$F_k = \ds \prod_{j \mathop = 0}^{k - 1} F_j + 2$
from which it is to be shown that:
:$F_{k + 1} = \ds \prod_{j \mathop = 0}^k F_j + 2$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{>0}: F_n = \ds \prod_{j \mathop = 0}^{n - 1} F_j + 2$
== Sources ==
* : $257$
Category:Fermat Numbers
Product of Sequence of Fermat Numbers plus 2/Corollary

== Corollary to Product of Sequence of Fermat Numbers plus 2 ==
Let $F_n$ denote the $n$th Fermat number.
Let $m \in \Z_{>0}$ be a (strictly) positive integer.
Then:
:$F_n \divides F_{n + m} - 2$
where $\divides$ denotes divisibility.
== Proof ==
From Product of Sequence of Fermat Numbers plus 2:
and so all Fermat numbers of index less than $n + m$ are divisors of $F_{n + m} - 2$.
This of course includes $F_n$.
Category:Fermat Numbers
Product of Sequence of Fermat Numbers plus 2/Mistake

== Source Work ==
:
:The Dictionary
::$257$
== Mistake ==
:$F_{n + 1} = F_0 F_1 F_2 \ldots F_{n - 1} + 2$, from which it follows ...
== Correction ==
That expression should be:
:$F_n = F_0 F_1 F_2 \ldots F_{n - 1} + 2$
This statement (and indeed, the entire paragraph it is in) has been removed from of $1997$.
== Sources ==
* : $257$
Category:Mistakes/Curious and Interesting Numbers
Product of Sequentially Computable Real-Valued Functions is Sequentially Computable

== Theorem ==
Let $D \subseteq \R^n$ be a subset of real cartesian $n$-space.
Let $f, g : D \to \R$ be sequentially computable.
Then, $h : D \to \R$ defined as:
:$\map h \bsx = \map f \bsx \map g \bsx$
is sequentially computable.
== Proof ==
Follows immediately from:
* Real Multiplication is Sequentially Computable
* Composition of Sequentially Computable Real-Valued Functions is Sequentially Computable
Category:Computability Theory
Product of Similar Plane Numbers is Square

== Theorem ==
== Proof ==
Let $a$ and $b$ be similar plane numbers.
From :
:$a : b = a^2 : a b$
From :
:there exists one mean proportional between $a$ and $b$.
From :
:there exists one mean proportional between $a^2$ and $a b$.
From it follows that:
:$a b$ is square.
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Euclidean Number Theory
Product of Sines of Fractions of Pi

== Theorem ==
Let $m \in \Z$ such that $m > 1$.
Then:
:$\ds \prod_{k \mathop = 1}^{m - 1} \sin \frac {k \pi} m = \frac m {2^{m - 1} }$
== Proof ==
Consider the equation:
:$z^m - 1 = 0$
whose solutions are the complex roots of unity:
:$1, e^{2 \pi i / m}, e^{4 \pi i / m}, e^{6 \pi i / m}, \ldots, e^{2 \paren {m - 1} \pi i / m}$
Then:
== Also presented as ==
Some sources prefer to report this result as:
:$\ds \prod_{k \mathop = 1}^{m - 1} \sin \frac {k \pi} m = \frac {2 m} {2^m}$
on the grounds that it may be easier to remember.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Miscellaneous Problems: $52$
Category:Sine Function
Product of Smooth Function and Dirac Delta Distribution

== Theorem ==
Let $a \in \R^d$ be a vector in real Euclidean space.
Let $\delta_a \in \map {\DD'} {\R^d}$ be the Dirac delta distribution.
Let $\alpha \in \map {C^\infty} {\R^d}$ be a smooth real function.
Then in the distributional sense we have that:
:$\alpha \delta_a = \map \alpha a \delta_a$
== Proof ==
Let $\phi \in \map \DD {\R^d}$ be a test function.
Then:
== Sources ==
* : Chapter $\S 6.4$: Multiplication by $C^\infty$ functions
Category:Functional Analysis
Category:Distributions
Product of Strictly Negative Element with Strictly Positive Element is Strictly Negative

== Theorem ==
Let $\struct {D, +, \times}$ be an ordered integral domain, whose (strict) positivity property is denoted $P$.
Let $N$ be the (strict) negativity property on $D$:
:$\forall a \in D: \map N a \iff \map P {-a}$
Then for all $a \in D$:
:$\map N a, \map P b \implies \map N {a \times b}$
== Proof ==
== Sources ==
* : Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Theorem $10 \ \text {(v)}$
Category:Ordered Integral Domains
Product of Subgroup with Inverse

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Then:
:$\forall H \le \struct {G, \circ}:$
::$H^{-1} \circ H = H$
::$H \circ H^{-1} = H$
where $H \le G$ denotes that $H$ is a subgroup of $G$.
== Proof ==
From Inverse of Subgroup:
:$H = H^{-1}$
From Product of Subgroup with Itself:
:$H \circ H = H$
The result follows.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets: Exercise $9.10 \ \text {(a)}$
Category:Subgroups
Category:Subset Products
Product of Subgroup with Itself

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Then:
:$\forall H \le G: H \circ H = H$
where $\le$ denotes the subgroup relation.
== Proof ==
From Magma Subset Product with Self, we have:
:$H \circ H \subseteq H$
Let $e$ be the identity of $G$.
By Identity of Subgroup, it is also the identity of $H$.
So:
Hence the result from the definition of set equality.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets: Exercise $9.10 \ \text {(a)}$
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $2$
* : $\S 50.1$ Quotient groups
Category:Subgroups
Category:Subset Products
Product of Subgroups of Prime Power Order

== Theorem ==
Let $p$ be a prime number.
Let $G$ be a group of order $p^a k$, where:
:$a \in \Z_{>0}$ is a (strictly) positive integer
:$p$ is not a divisor of $k$.
Let $P \le G$ be a subgroup of $G$ of order $p^a$.
Let $Q \le G$ be a subgroup of $G$ of order $p^b$, where $0 < b \le a$.
Let it be the case that $Q$ is not a subgroup of $P$.
Then $P Q$ is not a subgroup of $G$.
== Proof ==
From Intersection of Subgroups is Subgroup, $P \cap Q$ is a subgroup of $P$.
Thus:
:$\order {P \cap Q} = p^c$ for some $c \in \Z$ such that $0 \le c \le a$
where $\order {P \cap Q}$ denotes the order of $P \cap Q$.
We have:
 $P Q$ is a subgroup of $G$.
By Lagrange's Theorem (Group Theory):
:$\order {P Q} \divides \order G$
where $\divides$ denotes divisibility.
We have that $\order {P Q}$ is a power of $p$.
The highest power of $p$ which divides $\order G$ is $p^a$.
Thus $P Q$ could have order $p^a$ at most.
Thus:
:$a + b - c \le a$
That is:
:$b \le c$
But $P \cap Q$ is a subgroup of $P$.
Hence it must be the case that $b = c$.
Thus:
:$P \cap Q = Q$
and so $Q$ is a subgroup of $P$.
This contradicts the fact that $Q$ is not a subgroup of $P$.
It must follow that $P Q$ is not a subgroup of $G$.
== Sources ==
* : Chapter $5$: Cosets and Lagrange's Theorem: Exercise $6$
Category:Subgroups
Product of Subset with Intersection

== Theorem ==
Let $\struct {G, \circ}$ be an algebraic structure.
Let $X, Y, Z \subseteq G$.
Then:
:$X \circ \paren {Y \cap Z} \subseteq \paren {X \circ Y} \cap \paren {X \circ Z}$
:$\paren {Y \cap Z} \circ X \subseteq \paren {Y \circ X} \cap \paren {Z \circ X}$
where $X \circ Y$ denotes the subset product of $X$ and $Y$.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Equality does not Hold ==
== Also see ==
* Product of Subset with Union
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets: Exercise $9.1$
Category:Subset Products
Category:Set Intersection
Category:Product of Subset with Intersection
Product of Subset with Intersection/Corollary

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $X, Y, Z \subseteq G$ such that $X$ is a singleton.
Then:
:$X \circ \paren {Y \cap Z} = \paren {X \circ Y} \cap \paren {X \circ Z}$
:$\paren {Y \cap Z} \circ X = \paren {Y \circ X} \cap \paren {Z \circ X}$
where $X \circ Y$ denotes the subset product of $X$ and $Y$.
== Proof ==
Let $X = \set x$.
We have from Product of Subset with Intersection that:
: $X \circ \paren {Y \cap Z} \subseteq \paren {X \circ Y} \cap \paren {X \circ Z}$
Let $g \in \paren {X \circ Y} \cap \paren {X \circ Z}$.
Then by definition of subset product:
: $\exists y \in Y, z \in Z: g = x \circ y = x \circ z$
By the Cancellation Laws it follows that $y = z$.
But:
: $g = x \circ y \implies g \in X \circ Y$
: $g = x \circ z \implies g \in X \circ Z$
and so as $x \circ y = x \circ z$ it follows that $g \in X \circ \paren {Y \cap Z}$.
Thus:
: $X \circ \paren {Y \cap Z} \supseteq \paren {X \circ Y} \cap \paren {X \circ Z}$
It follows by definition of set equality that:
:$X \circ \paren {Y \cap Z} = \paren {X \circ Y} \cap \paren {X \circ Z}$
By similar reasoning:
:$\paren {Y \cap Z} \circ X = \paren {Y \circ X} \cap \paren {Z \circ X}$
Hence the result.
== Sources ==
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $4 \ \text{(iii)}$
Category:Product of Subset with Intersection
Product of Subset with Intersection/Equality does not Hold

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $X, Y, Z \subseteq G$.
Let $X \circ Y$ denotes the subset product of $X$ and $Y$.
While it is the case that:
:$X \circ \paren {Y \cap Z} \subseteq \paren {X \circ Y} \cap \paren {X \circ Z}$
it is not necessarily the case that:
:$X \circ \paren {Y \cap Z} = \paren {X \circ Y} \cap \paren {X \circ Z}$
== Proof ==
Proof by Counterexample:
Let $a \in G$ such that $a \ne a^{-1}$.
Let $X = \set {a, a^{-1} }, Y = \set a, Z = \set {a^{-1} }$.
Then:
:$X \circ \paren {Y \cap Z} = X \circ \O = \O$
:$\paren {X \circ Y} \cap \paren {X \circ Z} = \set {a^2, e} \cap \set {e, a^{-2} } \ne \O$
so:
:$X \circ \paren {Y \cap Z} \ne \paren {X \circ Y} \cap \paren {X \circ Z}$
== Also see ==
* Product of Subset with Intersection/Corollary: equality holds when $X$ is a singleton.
== Sources ==
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $4 \ \text{(ii)}$
Category:Product of Subset with Intersection
Product of Subset with Intersection/Proof 1

== Theorem ==
== Proof ==
Let $x \in X, t \in Y \cap Z$.
By the definition of intersection, $t \in Y$ and $t \in Z$.
Consider $X \circ \paren {Y \cap Z}$.
We have $x \circ t \in X \circ \paren {Y \cap Z}$ by definition of subset product.
As $t \in Y$ and $t \in Z$, we also have $x \circ t \in X \circ Y$ and $x \circ t \in X \circ Z$.
The result follows.
Similarly, consider $\paren {Y \cap Z} \circ X$.
Then we have $t \circ x \in \paren {Y \cap Z} \circ X$ by definition of subset product.
As $t \in Y$ and $t \in Z$, we also have $t \circ x \in Y \circ X$ and $t \circ x \in Z \circ X$.
Again, the result follows.
== Sources ==
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $4 \ \text{(i)}$
Category:Product of Subset with Intersection