nyuuzyou/wikis huggingface dataset:
Product of Subset with Intersection/Proof 2

== Theorem ==
== Proof ==
Consider the relation $\RR \subseteq G \times G$ defined as:
:$\forall g, h \in G: \tuple {g, h} \in \RR \iff \exists g \in X$
Then:
:$\forall S \subseteq G: X \circ S = \RR \sqbrk S$
Then:
Next, consider the relation $\RR \subseteq G \times G$ defined as:
:$\forall g, h \in G: \tuple {g, h} \in \RR \iff \exists h \in X$
Then:
:$\forall S \subseteq G: S \circ X = \RR \sqbrk S$
Then:
Category:Product of Subset with Intersection
Product of Subset with Union

== Theorem ==
Let $\struct {G, \circ}$ be an algebraic structure.
Let $X, Y, Z \subseteq G$.
Then:
:$X \circ \paren {Y \cup Z} = \paren {X \circ Y} \cup \paren {X \circ Z}$
:$\paren {Y \cup Z} \circ X = \paren {Y \circ X} \cup \paren {Z \circ X}$
where $X \circ Y$ denotes the subset product of $X$ and $Y$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Product of Subset with Intersection
== 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 Union
Category:Product of Subset with Union
Product of Subset with Union/Proof 1

== Theorem ==
== Proof ==
Let $x \circ t \in X \circ \paren {Y \cup Z}$.
We have $x \in X, t \in Y \cup Z$ by definition of subset product.
By definition of set union, it follows that $t \in Y$ or $t \in Z$.
So we also have $x \circ t \in X \circ Y$ or $x \circ t \in X \circ Z$.
That is:
:$x \circ t \in \paren {X \circ Y} \cup \paren {X \circ Z}$
and so:
:$X \circ \paren {Y \cup Z} \subseteq \paren {X \circ Y} \cup \paren {X \circ Z}$
Now let $x \circ t \in \paren {X \circ Y} \cup \paren {X \circ Z}$.
By definition of set union, it follows that $x \circ t \in X \circ Y$ or $x \circ t \in X \circ Z$.
So $x \in X$, and $y \in Y$ or $y \in Z$.
That is, $x \in X$, and $y \in Y \cup Z$ by definition of set union.
Hence:
:$x \circ t \in X \circ \paren {Y \cup Z}$
and so:
:$\paren {X \circ Y} \cup \paren {X \circ Z} \subseteq X \circ \paren {Y \cup Z}$
That is:
:$X \circ \paren {Y \cup Z} = \paren {X \circ Y} \cup \paren {X \circ Z}$
The result:
:$\paren {Y \cup Z} \circ X = \paren {Y \circ X} \cup \paren {Z \circ X}$
follows similarly.
Category:Product of Subset with Union
Product of Subset with Union/Proof 2

== Theorem ==
== Proof ==
Consider the relation $\RR \subseteq G \times G$ defined as:
:$\forall g, h \in G: \tuple {g, h} \in \RR \iff \exists g \in X$
Then:
:$\forall S \subseteq G: X \circ S = \map \RR S$
Then:
Next, consider the relation $\RR \subseteq G \times G$ defined as:
:$\forall g, h \in G: \tuple {g, h} \in \RR \iff \exists h \in X$
Then:
:$\forall S \subseteq G: S \circ X = \map \RR S$
Then:
Category:Product of Subset with Union
Product of Summations

== Theorem ==
:$\ds \prod_{j \mathop = 1}^n \sum_{i \mathop = 1}^m a_{i j} = \sum_{1 \mathop \le i_1 \mathop , \mathop \ldots \mathop , i_n \mathop \le m} a_{i_1 1} \cdots a_{i_n n}$
== Proof ==
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $32$
Category:Summations
Category:Continued Products
Product of Sums

== Theorem ==
Let $\ds \sum_{n \mathop \in A} a_n$ and $\ds \sum_{n \mathop \in B} b_n$ be absolutely convergent sequences.
Then:
:$\ds \paren {\sum_{i \mathop \in A} a_i} \paren {\sum_{j \mathop \in B} b_j} = \sum_{\tuple {i, j} \mathop \in A \times B} a_i b_j$
=== Corollary ===
== Proof ==
We have that both series are absolutely convergent.
Thus by Manipulation of Absolutely Convergent Series, it is permitted to expand the product as:
:$\ds \paren {\sum_{i \mathop \in A} a_i} \paren {\sum_{j \mathop \in B} b_j} = \sum_{i \mathop \in A} \paren {a_i \sum_{j \mathop \in B} b_j}$
But since $a_i$ is a constant, it may be brought into the summation, to obtain:
:$\ds \sum_{i \mathop \in A} \sum_{j \mathop \in B} a_i b_j$
Hence the result.
Category:Number Theory
Category:Analysis
Product of Sums/Corollary

== Corollary to Product of Sums ==
Let $\ds \sum_{i \mathop \in X} a_{i j}$ be absolutely convergent sequences for all $j \in Y$.
Then:
:$\ds \prod_{j \mathop \in Y} \paren {\sum_{i \mathop \in X} a_{i j} } = \sum_{f \mathop: Y \mathop \to X} \paren {\prod_{j \mathop \in Y} a_{\map f j j} }$
where $f$ runs over all mappings from $Y$ to $X$.
== Proof ==
We will prove the case $X = Y = \N$ to avoid the notational inconvenience of enumerating the elements of $Y$ as $j_1, j_2, j_3 \dots$.
The general case where $X, Y$ are arbitrary sets has the same proof, but with more indices and notational distractions.
Consider that by the main theorem:
:$\ds \prod_{j \mathop = 1, 2} \paren {\sum_{i \mathop \in \N} a_{i j} } = \sum_{x, y \mathop \in \N} a_{x_1}a_{y_2}$
and continuing in this vein:
:$\ds \prod_{j \mathop = 1, 2, 3} \paren {\sum_{i \mathop \in \N} a_{i j} } = \paren {\sum_{x, y \mathop \in \N} a_{x_1} a_{y_2} } \paren {\sum_{z \mathop \in \N} a_{z_3} } = \sum_{x, y, z \mathop \in \N} a_{x_1} a_{y_2} a_{z_3}$
For an inductive proof of this concept for finite $n$, we assume that for some $n \in \N$:
:$\ds \prod_{j \mathop = 1}^n \paren {\sum_{i \mathop \in \N} a_{i j} } = \sum_{u, v, \ldots, x, y \mathop \in \N} a_{u_1} a_{v_2} \dotsb a_{x_{\paren {n - 1} } } a_{y_n}$
Then:
:$\ds \prod_{j \mathop = 1}^{n + 1} \paren {\sum_{i \mathop \in \N} a_{i j} } = \paren {\sum_{u, v, \dots, x, y \mathop \in \N} a_{u_1} a_{v_2} \dotsb a_{x_{\paren {n - 1} } } a_{y_n} } \paren {\sum_{z \mathop \in \N} a_{z_n} }$
which by Product of Sums is simply:
:$\ds \sum_{u, v, \ldots, x, y, z \mathop \in \N} a_{u_1} a_{v_2} \ldots a_{x_{\paren {n - 1} } a_{y_n} a_{z_{\paren {n + 1} } } }$
completing the induction for finite $n$.
Category:Number Theory
Category:Analysis
Product of Sums of Four Squares/Corollary

== Theorem ==
Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n, c_1, c_2, \ldots, c_n, d_1, d_2, \ldots, d_n$ be integers.
Then:
:$\ds \exists w, x, y, z \in \Z: \prod_{j \mathop = 1}^n \paren {a_j^2 + b_j^2 + c_j^2 + d_j^2} = w^2 + x^2 + y^2 + z^2$
That is, the product of any number of sums of four squares is also a sum of four squares.
== Proof ==
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \exists w, x, y, z \in \Z: \prod_{j \mathop = 1}^n \paren {a_j^2 + b_j^2 + c_j^2 + d_j^2} = w^2 + x^2 + y^2 + z^2$
$\map P 1$ is true, as this just says:
:$\exists w, x, y, z \in \Z: a^2 + b^2 + c^2 + d^2 = w^2 + x^2 + y^2 + z^2$
which is trivially true.
=== Basis for the Induction ===
$\map P 2$ is the case:
:$\exists w, x, y, z \in \Z: \paren {a_1^2 + b_1^2 + c_1^2 + d_1^2} \paren {a_2^2 + b_2^2 + c_2^2 + d_2^2} = w^2 + x^2 + y^2 + z^2$
which follows from Product of Sums of Four Squares.
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:
:$\ds \exists w, x, y, z \in \Z: \prod_{j \mathop = 1}^k \paren {a_j^2 + b_j^2 + c_j^2 + d_j^2} = w^2 + x^2 + y^2 + z^2$
Then we need to show that it directly implies:
:$\ds \exists w, x, y, z \in \Z: \prod_{j \mathop = 1}^{k + 1} \paren {a_j^2 + b_j^2 + c_j^2 + d_j^2} = w^2 + x^2 + y^2 + z^2$
=== 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:
:$\ds \forall n \in \Z_{>0}: \exists w, x, y, z \in \Z: \prod_{j \mathop = 1}^n \paren {a_j^2 + b_j^2 + c_j^2 + d_j^2} = w^2 + x^2 + y^2 + z^2$
Category:Product of Sums of Four Squares
Category:Proofs by Induction
Product of Sums of Four Squares/Proof 2

== Theorem ==
== Proof ==
Let:
:$\mathbf m = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$
:$\mathbf n = -w \mathbf 1 + x \mathbf i + y \mathbf j + z \mathbf k$
be two quaternions.
Then:
== Sources ==
* : Chapter $\text {B}.26$: Extensions of the Complex Number System. Algebras, Quaternions, and Lagrange's Four Squares Theorem
Category:Product of Sums of Four Squares
Product of Supremum and Infimum in Lattice-Ordered Group

== Theorem ==
Let $\struct {G, \odot}$ be a group.
Let $\preccurlyeq$ be a lattice ordering on $G$.
Let $x, y \in G$ such that $x$ commutes with $y$.
Then:
:$\sup \set {x, y} \odot \inf \set {x, y} = x \odot y$
== Proof ==
First we show that $x$ commutes with $\sup \set {x, y}$.
Indeed:
Similarly:
:$y$ commutes with $\sup \set {x, y}$
Then using Infima in Ordered Group in the same way as above:
:$x$ commutes with $\inf \set {x, y}$
:$y$ commutes with $\inf \set {x, y}$
Then we have:
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.11$
Category:Ordered Groups
Category:Lattice Theory
Category:Suprema
Category:Infima
Product of Three Consecutive Integers is never Perfect Power

== Theorem ==
Let $n \in \Z_{> 1}$ be a (strictly) positive integer.
Then:
:$\paren {n - 1} n \paren {n + 1}$
cannot be expressed in the form $a^k$ for $a, k \in \Z$ where $k \ge 2$.
That is, the product of $3$ consecutive (strictly) positive integers can never be a perfect power.
== Proof ==
 $\paren {n - 1} n \paren {n + 1} = a^k$ for $a, k \in \Z$ where $k \ge 2$.
We have that:
:$\gcd \set {n − 1, n} = 1 = \gcd \set {n, n + 1}$
Thus $n$ must itself be a perfect power of the form $z^k$ for some $z \in \Z$.
That means $\paren {n - 1} \paren {n + 1} = n^2 - 1$ must also be a perfect power of the same form.
Let:
:$n = r^k$ and $n^2 − 1 = s^k$
for $r, s \in \Z$.
Then:
:$\paren {r^2}^k = 1 + s^k$
But the only consecutive integers that are $k$th powers are (trivially) $0$ and $1$.
Hence by Proof by Contradiction there can be no such $n$.
== Sources ==
* 
* : $3$
Category:Number Theory
Product of Triangular Matrices

== Theorem ==
Let $\mathbf A = \sqbrk a_n, \mathbf B = \sqbrk b_n$ be upper triangular matrices of order $n$.
Let $\mathbf C = \mathbf A \mathbf B$.
Then
:$(1): \quad$ the diagonal elements of $\mathbf C$ are given by:
::::$\forall j \in \closedint 1 n: c_{j j} = a_{j j} b_{j j}$
:::That is, the diagonal elements of $\mathbf C$ are those of the factor matrices multiplied together.
:$(2): \quad$ The matrix $\mathbf C$ is itself upper triangular.
The same applies if both $\mathbf A$ and $\mathbf B$ are lower triangular matrices.
== Proof ==
From the definition of matrix product, we have:
:$\ds \forall i, j \in \closedint 1 n: c_{i j} = \sum_{k \mathop = 1}^n a_{i k} b_{k j}$
Now when $i = j$ (as on the main diagonal):
:$\ds c_{j j} = \sum_{k \mathop = 1}^n a_{j k} b_{k j}$
Now both $\mathbf A$ and $\mathbf B$ are upper triangular.
Thus:
:if $k > j$, then $b_{k j} = 0$ and thus $a_{j k} b_{k j} = 0$
:if $k < j$, then $a_{j k} = 0$ and thus $a_{j k} b_{k j} = 0$.
So $a_{j k} b_{k j} \ne 0$ only when $j = k$.
So:
:$\ds c_{j j} = \sum_{k \mathop = 1}^n a_{j k} b_{k j} = a_{j j} b_{j j}$
Now if $i > j$, it follows that either $a_{i k}$ or $b_{k j}$ is zero for all $k$, and thus $c_{i j} = 0$.
Thus $\mathbf C$ is upper triangular.
The same argument can be used for when $\mathbf A$ and $\mathbf B$ are both lower triangular matrices.
Category:Triangular Matrices
Category:Conventional Matrix Multiplication
Product of Two Distinct Primes has 4 Positive Divisors

== Theorem ==
Let $n \in \Z_{>0}$ be a positive integer which is the product of $2$ distinct primes.
Then $n$ has exactly $4$ positive divisors.
== Proof ==
Let $n = p \times q$ where $p$ and $q$ are primes.
We have by definition of divisor:
where $\divides$ denotes divisibility.
Suppose $a \divides n$ such that $1 \le a < n$.
Suppose $a \ne p$.
By definition of prime number:
:$a \perp p$
where $\perp$ denotes coprimality.
From Euclid's Lemma:
:$a \divides q$
and so by definition of prime number:
:$a = q$
or:
:$a = 1$
Similarly, suppose $a \ne q$.
By the same reasoning:
:$a = p$
or:
:$a = 1$
Thus the only positive divisors are as above.
== Sources ==
* : $33$
Category:Semiprimes
Product of Two Distinct Primes is Multiplicatively Perfect

== Theorem ==
Let $n \in \Z_{>0}$ be a positive integer which is the product of $2$ distinct primes.
Then $n$ is multiplicatively perfect.
== Proof ==
Let $n = p \times q$ where $p$ and $q$ are primes.
From Product of Two Distinct Primes has 4 Positive Divisors, the positive divisors of $n$ are:
:$1, p, q, pq$
Thus the product of all the divisors of $n$ is:
:$1 \times p \times q \times p q = p^2 q^2 = n^2$
Hence the result, by definition of multiplicatively perfect.
== Sources ==
* : $33$
Category:Semiprimes
Category:Multiplicatively Perfect Numbers
Product of Two Strictly Negative Elements is Strictly Positive

== 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 $a \in D$:
:$\map N a, \map N b \implies \map P {a \times b}$
== Proof ==
== Sources ==
* : Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Theorem $10 \ \text {(iv)}$
Category:Ordered Integral Domains
Product of Two Triangular Numbers to make Square

== Theorem ==
Let $T_n$ be a triangular number.
Then there is an infinite number of $m \in \Z_{>0}$ such that $T_n \times T_m$ is a square number.
== Proof ==
Since $n^2 < n \paren {n + 1} < \paren {n + 1}^2$, $n \paren {n + 1}$ cannot be a square number.
Thus there are infinitely many distinct integer solutions to Pell's equation:
:$x^2 - n \paren {n + 1} y^2 = 1$
and for each solution:
Hence the result.
== Examples ==
== Sources ==
* : No. $241$
* : $15$
* : $15$
Category:Product of Two Triangular Numbers to make Square
Category:Triangular Numbers
Category:Square Numbers
Product of Two Triangular Numbers to make Square/Examples/T2 by T24

== Example of Product of Two Triangular Numbers to make Square ==
:$T_2 \times T_{24} = 30^2$
where $T_2, T_{24}$ denote the $2$nd and $24$th triangular numbers.
== Proof ==
== Sources ==
* : $15$
* : $15$
::although the example chosen, $T_3 \times T_{24}$, is erroneous.
Category:Product of Two Triangular Numbers to make Square
Product of Two Triangular Numbers to make Square/Mistake

== Source Work ==
:
:The Dictionary
::$15$
:
:The Dictionary
::$15$
== Mistake ==
:For every triangular number, $T_n$, there are an infinite number of other triangular numbers, $T_m$, such that $T_n T_m$ is a square. For example, $T_3 \times T_{24} = 30^2$.
== Correction ==
This is incorrect.
It is supposed that this is a misprint for:
:$T_2 \times T_{24} = 30^2$
which is correct.
== Sources ==
* : $15$
* : $15$
Category:Mistakes/Curious and Interesting Numbers
Product of Uniformly Convergent Sequences of Bounded Functions is Uniformly Convergent

== Theorem ==
Let $X = \struct {A, d}$ and $Y = \struct {B, \rho}$ be metric spaces.
Let $\sequence {f_n}$ and $\sequence {g_n}$ be sequences of mappings from $X$ to $Y$.
Let $\sequence {f_n}$ and $\sequence {g_n}$ be uniformly convergent on some subspace $S$ of $X$. 
$\forall n \in \N$, let $f_n$ and $g_n$ be bounded.
Then the sequence $\sequence {f_n g_n}$ is uniformly convergent on $S$.
== Proof ==
Category:Metric Spaces
Product of Units of Integral Domain with Finite Number of Units

== Theorem ==
Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$ and whose unity is $1_D$.
Let $D$ have a finite number of units.
Let $U_D$ be the set of units of $\struct {D, +, \circ}$.
Then:
:$\ds \prod_{x \mathop \in U_D} x = -1_D$
== Proof ==
Consider the set $S$ defined as:
:$S = U_R \setminus \set {1_D, -1_D}$
If $S$ has even cardinality, it can be partitioned into doubletons of the form $\set {u, u^{-1} }$.
Each of these doubletons has a product of $1_D$.
The product of all these with $1_D$ and $-1_D$ is $-1_D$.
It remains to be shown that $S$ cannot be of odd cardinality.
 $S$ has an odd number of elements.
Then after pairing off each $u \in D$ with its product inverse $u^{-1}$, we are left with one element of $S$ which is self-inverse.
But from Self-Inverse Element of Integral Domain is Unity or its Negative, this self-inverse element is either $1_D$ or $-1_D$.
These have already been counted.
Hence there cannot be an odd number of elements of $S$.
== Sources ==
* : Chapter $9$: Rings: Exercise $12$
Category:Integral Domains
Product of Wholly Real Numbers is Wholly Real

== Theorem ==
Let $x = \tuple {a, 0}$ and $y = \tuple {b, 0}$ be wholly real complex numbers.
Then $x y$ is also wholly real.
== Proof ==
We have:
== Sources ==
* : Chapter $\text I$: Functions of a Complex Variable: $\S 1$. Complex Numbers: The abbreviated notation
Category:Complex Multiplication
Product of nth Lucas and Fibonacci Numbers

== Theorem ==
Let $L_k$ be the $k$th Lucas number.
Let $F_k$ be the $k$th Fibonacci number.
Then:
:$\forall n \in \N_{>0}: F_n L_n = F_{2 n}$
== Proof ==
By definition of Lucas numbers:
:$L_n = F_{n - 1} + F_{n + 1}$
Hence:
:$F_n L_n = F_n \paren {F_{n - 1} + F_{n + 1} }$
From Honsberger's Identity:
:$\forall m, n \in \Z_{>0}: F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$
The result follows by setting $m = n$.
== Sources ==
* : $\text {1-1}$ Principle of Mathematical Induction: Exercise $15$
Category:Lucas Numbers
Category:Fibonacci Numbers
Product of nth Roots of Unity

== Theorem ==
Let $n \in \Z$ be an integer such that $n > 0$.
Let $z \in \C$ be a complex number such that $z^n = 1$.
Then:
:$U_n = \set {e^{2 i k \pi / n}: k \in \N_n}$
where $U_n$ is the set of $n$th roots of unity.
That is:
:$z \in \set {1, e^{2 i \pi / n}, e^{4 i \pi / n}, \ldots, e^{2 \paren {n - 1} i \pi / n} }$
Then the product of all of the elements of $U_n$ is $(-1)^{n-1}$
== Proof ==
Category:Complex Roots of Unity
Product of r Choose k with r Minus Half Choose k/Formulation 1

== Theorem ==
Let $k \in \Z$, $r \in \R$.
:$\dbinom r k \dbinom {r - \frac 1 2} k = \dfrac {\dbinom {2 r} k \dbinom {2 r - k} k} {4^k}$
where $\dbinom r k$ denotes a binomial coefficient.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $47$
Category:Product of r Choose k with r Minus Half Choose k
Product of r Choose k with r Minus Half Choose k/Formulation 1/Proof 2

== Theorem ==
== Proof ==
From Binomial Coefficient expressed using Beta Function:
:$(1): \quad \dbinom r k \dbinom {r - \frac 1 2} k = \dfrac 1 {\paren {r + 1} \map \Beta {k + 1, r - k + 1} \paren {r + \frac 1 2} \map \Beta {k + 1, r - k + \frac 1 2} }$
Then:
Again from Binomial Coefficient expressed using Beta Function:
:$(3): \quad \dbinom {2 r} k \dbinom {2 r - k} k = \dfrac 1 {\paren {2 r + 1} \map \Beta {k + 1, 2 r - k + 1} \paren {2 r - k + 1} \map \Beta {k + 1, 2 r - 2 k + 1} }$
Then:
Category:Product of r Choose k with r Minus Half Choose k
Product of r Choose k with r Minus Half Choose k/Formulation 2

== Theorem ==
Let $k \in \Z$, $r \in \R$.
:$\dbinom r k \dbinom {r - \frac 1 2} k = \dfrac {\dbinom {2 r} {2 k} \dbinom {2 k} k} {4^k}$
where $\dbinom r k$ denotes a binomial coefficient.
== Proof ==
From Binomial Coefficient expressed using Beta Function:
:$(1): \quad \dbinom r k \dbinom {r - \frac 1 2} k = \dfrac 1 {\paren {r + 1} \map \Beta {k + 1, r - k + 1} \paren {r + \frac 1 2} \map \Beta {k + 1, r - k + \frac 1 2} }$
Then:
Then:
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $47$
Category:Product of r Choose k with r Minus Half Choose k
Product of r Choose m with m Choose k

== Theorem ==
Let $r \in \R, m \in \Z, k \in \Z$.
Then:
:$\dbinom r m \dbinom m k = \dbinom r k \dbinom {r - k} {m - k}$
where $\dbinom r m$ is a binomial coefficient.
=== Complex Numbers ===
== Proof 1 ==
== Proof 2 ==
Category:Binomial Coefficients
Category:Product of r Choose m with m Choose k
Product of r Choose m with m Choose k/Complex Numbers

== Theorem ==
For all $z, w \in \C$ such that it is not the case that $z$ is a negative integer and $t, w$ integers:
:$\dbinom z t \dbinom t w = \dbinom z w \dbinom {z - w} {t - w}$
where $\dbinom z w$ is a binomial coefficient.
== Proof ==
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $42$ (Solution)
Category:Product of r Choose m with m Choose k
Product of r Choose m with m Choose k/Proof 1

== Theorem ==
== Proof ==
=== Integral Index ===
Let $r \in \Z$.
Then:
=== Real Index ===
Both sides of the above equation are polynomials in $r$.
Since these polynomials agree for all $r \in \Z$, they must agree for all $r \in \R$.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $\text {H} \ (20)$
Category:Product of r Choose m with m Choose k
Product of r Choose m with m Choose k/Proof 2

== Theorem ==
== Proof ==
Consider the trinomial coefficient:
:$\dbinom r {k, m - k, r - m}$
We use Multinomial Coefficient expressed as Product of Binomial Coefficients:
:$\dbinom {k_1 + k_2 + k_3} {k_1, k_2, k_3} = \dbinom {k_1 + k_2} {k_1} \dbinom {k_1 + k_2 + k_3} {k_1 + k_2}$
and substitute as appropriate for $k_1, k_2, k_3$.
We have:
Similarly:
The result follows on equating the two expressions.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients
Category:Product of r Choose m with m Choose k
Product on Left with Idempotent Element under Left Self-Distributive Operation is Idempotent

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\circ$ be left self-distributive.
Let $a \in S$ be an idempotent element of $\struct {S, \circ}$.
Then for all $b \in S$, $b \circ a$ is an idempotent element of $\struct {S, \circ}$
== Proof ==
Let $a \in S$ be an idempotent element of $\struct {S, \circ}$.
We have:
The result follows by definition of idempotent element.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.23 \ \text{(d)}$
Category:Idempotence
Category:Self-Distributive Operations
Product on Right with Idempotent Element under Right Self-Distributive Operation is Idempotent

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\circ$ be right self-distributive.
Let $a \in S$ be an idempotent element of $\struct {S, \circ}$.
Then for all $b \in S$, $a \circ b$ is an idempotent element of $\struct {S, \circ}$
== Proof ==
Let $a \in S$ be an idempotent element of $\struct {S, \circ}$.
We have:
The result follows by definition of idempotent element.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.23 \ \text{(d)}$
Category:Idempotence
Category:Self-Distributive Operations
Product to n of Product to Index

== Theorem ==
:$\ds \prod_{i \mathop = 0}^n \prod_{j \mathop = 0}^i a_i a_j = \prod_{i \mathop = 0}^n {a_i}^{n + 2}$
== Proof ==
Let:
Then:
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $26$
Category:Continued Products
Product with Degenerate Linear Transformation is Degenerate

== Theorem ==
Let $U, V, W$ be vector spaces over a field $K$.
Let $G: U \to V$ be a degenerate linear transformation.
Let $N: W \to U$ be a linear transformation.
Then $G \circ N$ is degenerate.
== Proof ==
Recall that the dimension of subspace is not greater than its super space.
Thus the claim follows from:
:$\Img {G \circ N} \subseteq \Img G$
== Sources ==
* : $2.2$: Index of a Linear Map
Category:Linear Algebra
Product with Field Negative

== Theorem ==
Let $\struct {F, +, \times}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $a, b \in F$.
Then:
:$-\paren {a \times b} = a \times \paren {-b} = \paren {-a} \times b$
=== Corollary ===
== Proof ==
Similarly:
== Sources ==
* : Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Theorem $2 \ \text {(iv)}$
Category:Field Theory
Product with Field Negative/Corollary

== Theorem ==
Let $\struct {F, +, \times}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $a \in F$.
Then:
:$\paren {-1_F} \times a = \paren {-a}$
== Proof ==
== Sources ==
* : Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 87 \beta$
* : Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Theorem $2 \ \text {(vi)}$
Category:Field Theory