nyuuzyou/wikis huggingface dataset:
Gelfond-Schneider Theorem/Lemma 1

== Lemma ==
Let $\map {a_1} t, \ldots, \map {a_n} t$ be non-zero polynomials in $\R \sqbrk t$ of degrees $d_1, \ldots, d_n$ respectively.
Let $w_1, \ldots, w_n$ be pairwise distinct real numbers.
Then:
:$\ds \map F t = \sum_{j \mathop = 1}^n \map {a_j} t e^{w_j t}$
has at most $d_1 + \cdots + d_n + n − 1$ real roots (counting multiplicities).
== Proof ==
By multiplying through by $e^{−w_n t}$ if necessary, we may suppose that $w_n = 0$ and that otherwise $w_j \ne 0$.
Let $k = d_1 + \cdots + d_n + n$.
We use strong induction on $k$.
If $k = 1$, then $n = 1$ and $d_1 = 0$, and the lemma easily follows.
Let $l \ge 2$ be such that the lemma holds whenever $k < l$.
Suppose $k = l$.
Let $N$ be the number of real roots of $\map F t$.
By Rolle's Theorem, the number of real roots of $\map {F'} t$ is at least $N − 1$.
On the other hand:
:$\ds \map {F'} t = \sum_{j \mathop = 1}^n \map {b_j} t e^{w_j t}$
where:
:$\map {b_j} t = \map { {a_j}'} t + w_j \map {a_j} t$
Note that for $1 \le j \le n − 1$, we have that $\map {b_j} t$ has degree exactly $d_j$.
Also, since $w_n = 0$, either:
:there are only $n − 1$ non-zero polynomials $\map {b_j} t$ in the expression for $\map {F'} t$ above
or:
:there are $n$ such polynomials and the degree of $\map {b_n} t$ is one less than the degree of $\map {a_n} t$.
It follows from the induction hypothesis that $\map {F'} t$ has at most $d_1 + \cdots + d_n + n − 2$ real roots.
Hence:
:$N − 1 \le d_1 + \cdots + d_n + n − 2$
and the result follows.
Category:Gelfond-Schneider Theorem
Gelfond-Schneider Theorem/Lemma 2

== Lemma ==
Let $\map f z$ be an analytic function in the open disk $D \subseteq \C: D = \set {z : \size z < R}$ for some real number $R > 0$.
Let $f$ also be continuous on the closure of $D$, that is, on $D^- = \set {z : \size z \le R}$.
Then:
:$\forall z \in D^-: \size {\map f z} \le \size f_R$
where we set:
:$\ds \size f_R = \max_{\size z \mathop = R} \size {\map f z}$
This lemma is essentially a restatement of the Maximum Modulus Principle for analytic functions.
== Proof ==
Since $D^-$ is compact, the continuous function $\size f$ has a maximum point $z_0 \in D^-$, meaning that $\size {\map f {z_0} } = \max_{D^-} \size f$.
If $z_0$ belongs to the open disk $D$, then the Maximum Modulus Principle implies that $f$ is constant, in which case the statement is trivial.
If instead $z_0$ belongs to the boundary $D^- \setminus D$, then $\size {z_0} = R$ and so $\size {\map f {z_0} } \le \size f_R$.
Since $\size {\map f z} \le \size {\map f {z_0} }$ for any $z \in D^-$, the statement follows.
Category:Gelfond-Schneider Theorem
Gelfond-Schneider Theorem/Lemma 3

== Lemma ==
Let $r$ and $R$ be two real numbers such that $0 < r \le R$.
Let $\map {f_1} z, \map {f_2} z, \ldots, \map {f_L} z$ be:
:analytic in $D \subseteq \C: D = \set {z : \size z < R}$
:continuous on the closure $D$, that is, $D^- = \set {z: \size z \le R}$.
Let $\zeta_1, \ldots, \zeta_L$ be complex numbers such that:
:$\forall j \in \set {1, 2, \ldots, L}: \size {\zeta_j} \le r$
Then the determinant:
:$\Delta = \det \begin {bmatrix}
\map {f_1} {\zeta_1} & \cdots & \map {f_L} {\zeta_1} \\
\vdots & \ddots & \vdots \\
\map {f_1} {\zeta_L} & \cdots & \map {f_L} {\zeta_L}
\end{bmatrix}$
satisﬁes:
:$\ds \size \Delta \le \paren {\frac R r}^{−L \paren {L − 1} / 2} L! \prod_{\lambda \mathop = 1}^L \size {f_λ}_R$
== Proof ==
Let $\map h z$ be the determinant of the $L \times L$ matrix $\sqbrk {\map {f_j} {\zeta_i z} }$.
Then $\map h z$ is:
:analytic in $D' = \set {z: \size z < R / r}$
:continuous on $D'^- = \set {z: \size z \le R / r}$.
Let $M = L \paren {L − 1} / 2$, and write:
:$\ds \map {f_j} {\zeta_i z} = \sum_{k \mathop = 0}^{M − 1} \map {b_k} j \zeta_i^k z^k + z^M \map{g_{i, j} } z$
where:
:$\map {b_k} j \in \C$ for each $k$
:$\map {g_{i, j} } z$ is analytic in $D'$
:$\map {g_{i, j} } z$ is continuous on $D'^-$
By evaluating along the columns it is seen that the determinant is linear in its columns.
So we can view $\map h z$ as $z^M$ times an analytic function plus terms involving the factor:
:$z^{n_1 + \cdots + n_L} \det \sqbrk {\zeta_i^{n_j} }$
where the $n_j$ denote non-negative integers.
Observe that the determinant in this last expression is zero if the $n_j$ are not distinct.
Therefore, the non-zero terms of this form satisfy:
:$n_1 + n_2 + \cdots + n_L \ge 0 + 1 + \cdots + \paren {L − 1} = \dfrac {L \paren {L − 1} } 2$
Hence, we deduce that $\map h z$ is divisible by $z^M$.
More precisely, $\dfrac {\map h z} {z^M}$ is analytic in $D'$ and continuous on $D'^-$.
Therefore, by , for any $w \in D'$:
:$\ds \size {\frac {\map h w} {w^M} } \le \size {\frac {\map h z} {z^M} }_{R / r} = \paren {\frac r R}^M \size {\map h z}_{R / r}$
For $\size z = R / r$, we get that $\size {\zeta_i z} \le R$.
We create a bound for $\size {\map h z}_{R / r}$ by multiplying the number of terms in $\det \sqbrk {\map {f_j} {\zeta_i z} }$ by an obvious upper bound on the maximum such term.
Thus:
:$\ds \size {\map h z}_{R / r} \le L! \prod_{\lambda \mathop = 1}^L \size {f_λ}_R$
Observe that $\size \Delta = \size {\map h 1}$ and $1 \le R / r$.
We deduce that
:$\ds \size \Delta \le \paren {\frac r R}^M \size {\map h z}_{R / r} \le \paren {\frac r R}^M L! \prod_{\lambda \mathop = 1}^L \size {f_λ}_R$
giving the desired conclusion.
Category:Gelfond-Schneider Theorem
Gelfond-Schneider Theorem/Lemma 4

== Lemma ==
Let:
:$\Delta = \det \sqbrk {\alpha_{i, j} }_{L \times L}$
where the $\alpha_{i, j}$ are algebraic numbers.
Suppose that $T$ is a positive (rational) integer for which $T \alpha_{i, j}$ is an algebraic integer for every $i, j \in \set {1, 2, \ldots, L}$.
Also, suppose that $\Delta \ne 0$.
Then there is a conjugate of $\Delta$ with absolute value $\ge T^{−L}$.
== Proof ==
Observe that $T^L \Delta$ is an algebraic integer so that one of its conjugates has absolute value $\ge 1$.
The result follows.
Category:Gelfond-Schneider Theorem
General Antiperiodicity Property

== Theorem ==
Let $f: X \to X$ be an antiperiodic function, where $X$ is either $\R$ or $\C$.
Let $L$ be an antiperiodic element of $f$.
Let $n \in \Z$ be an integer.
:If $n$ is even, then $n L$ is a periodic element of $f$.
:If $n$ is odd, then $n L$ is an antiperiodic element of $f$.
== Proof ==
Suppose that $X = \C$.
=== Case 1 ===
If $n$ is even, then:
=== Case 2 ===
If $n$ is odd, then:
The proof for when $X = \R$ is nearly identical.
Category:Antiperiodic Functions
General Associative Law for Ordinal Sum

== Theorem ==
Let $x$ be a finite ordinal.
Let $\sequence {a_i}$ be a sequence of ordinals.
Then:
:$\ds \sum_{i \mathop = 1}^{x + 1} a_i = a_1 + \sum_{i \mathop = 1}^x a_{i + 1}$
== Proof 1 ==
== Proof 2 ==
Category:Ordinal Arithmetic
Category:General Associative Law for Ordinal Sum
General Associative Law for Ordinal Sum/Proof 1

== Theorem ==
== Proof ==
The proof shall proceed by induction on $x$.
=== Basis for the Induction ===
If $x = 0$, then:
This proves the basis for the induction.
=== Induction Step ===
Suppose:
:$\ds \sum_{i \mathop = 1}^{x + 1} a_i = a_1 + \sum_{i \mathop = 1}^x a_{i + 1}$
Then:
This proves the induction step.
Category:General Associative Law for Ordinal Sum
General Associative Law for Ordinal Sum/Proof 2

== Theorem ==
== Proof ==
From Ordinal Addition is Associative we have that:
:$\forall a, b, c \in \On: a + \paren {b + c} = \paren {a + b} + c$
The result follows directly from the General Associativity Theorem.
Category:General Associative Law for Ordinal Sum
General Associativity Theorem

== Theorem ==
If an operation is associative on $3$ entities, then it is associative on any number of them.
=== Formulation 1 ===
=== Formulation 2 ===
=== Formulation 3 ===
== Also known as ==
Also known as the general (or generalized) associative law.
== Also see ==
* Associativity on Four Elements
== Motivation ==
== Sources ==
* : Chapter $\text{II}$: Groups: Exercise $\text{C}$
* : Part $\text I$: Prologue: Chapter $1$: Basic Properties of Numbers
:::in the context of addition
* : Chapter $1$: Rings - Definitions and Examples: $3$: Some special classes of rings: $1.3$
* : Chapter $2$: The Definition of Group Structure: $\S 27 \beta, \gamma$
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups: Theorem $1.6$
* : $2$: Examples of Groups and Homomorphisms: $2.1$ Definition
* : $\S 28$. Associativity and commutativity
Category:General Associativity Theorem
Category:Associativity
Category:Named Theorems
General Associativity Theorem/Formulation 2

== Theorem ==
Let $n \in \N_{>0}$ and let $a_1, \ldots, a_n$ be elements of a set $S$.
Let $\circ$ be an associative operation on $S$.
Let the set $\map {P_n} {a_1, a_2, \ldots, a_n}$ be defined inductively by:
:$\map {P_1} {a_1} = \set {a_1}$
:$\map {P_2} {a_1, a_2} = \set {a_1 \circ a_2}$
:$\map {P_n} {a_1, a_2, \ldots, a_n} = \set {x \circ y: x \in \map {P_r} {a_1, a_2, \ldots, a_r} \land y \in \map {P_s} {a_{r + 1}, a_{r + 2}, \ldots, a_{r + s} }, n = r + s}$
Then $\map {P_n} {a_1, a_2, \ldots, a_n}$ consists of a unique entity which we can denote $a_1 \circ a_2 \circ \ldots \circ a_n$.
== Proof 1 ==
== Proof 2 ==
Category:General Associativity Theorem
General Associativity Theorem/Formulation 2/Proof 1

== Theorem ==
== Proof ==
The cases where $n = 1$ and $n = 2$ are clear.
Let $a = x \circ y \in P_n: x \in P_r, y \in P_s$.
If $r > 1$ then we write $x = a_1 \circ z$ where $z = a_2 \circ a_3 \circ \ldots \circ a_r$ by induction.
Then $x \circ y = \paren {a_1 \circ z} \circ y = a_1 \circ \paren {z \circ y} = a_1 \circ \paren {a_2 \circ a_3 \circ \ldots \circ a_n}$ (again by induction).
If $r = 1$, then by induction $x \circ y = a_1 \circ y = a_1 \circ \paren {a_2 \circ a_3 \circ \ldots \circ a_n}$.
Thus in either case, $x \circ y = a_1 \circ \paren {a_2 \circ a_3 \circ \ldots \circ a_n}$ which is a single element of $P_n$.
Hence we see that $\map {P_n} {a_1, a_2, \ldots, a_n}$ consists of a single element.
== Sources ==
* : Chapter $3$: Elementary consequences of the definitions: Proposition $3.6$
Category:General Associativity Theorem
General Associativity Theorem/Formulation 2/Proof 2

== Theorem ==
== Proof ==
Proof by strong induction:
For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:
:The General Associativity Theorem holds for all composites $a_1 \circ a_2 \circ \cdots \circ a_r$ such that $r \le n$.
$P \left({1}\right)$ is trivially true, as this just says $a_1 = a_1$.
$P \left({2}\right)$ is the case:
:$a_1 \circ a_2 = a_1 \circ a_2$
for which there is also nothing to prove.
=== Basis for the Induction ===
$P \left({3}\right)$ is the case:
:$\left({a_1 \circ a_2}\right) \circ a_3 = a_1 \circ \left({a_2 \circ a_3}\right)$
which is the definition of associativity.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 3$, then it logically follows that $P \left({k+1}\right)$ is true.
So this is our induction hypothesis:
:The General Associativity Theorem holds for all composites $a_1 \circ a_2 \circ \cdots \circ a_r$ such that $r \le k$.
Then we need to show:
:The General Associativity Theorem holds for all composites $a_1 \circ a_2 \circ \cdots \circ a_r$ such that $r \le k+1$.
=== Induction Step ===
This is our induction step:
Consider the expressions:
:$(1): \quad \left({a_1 \circ a_2 \circ \cdots \circ a_i}\right) \circ \left({a_{i+1} \circ a_{i+2} \circ \cdots \circ a_{k+1}}\right)$
:$(2): \quad \left({a_1 \circ a_2 \circ \cdots \circ a_j}\right) \circ \left({a_{j+1} \circ a_{i+2} \circ \cdots \circ a_{k+1}}\right)$
for some $i, j \in \Z: 1 \le i, j \le k$.
We require to show that whatever the values of $i$ and $j$, these expressions are equal.
, suppose that $i < j$.
Then the above expressions can be written:
:$\left({a_1 \circ a_2 \circ \cdots \circ a_i}\right) \circ \left({a_{i+1} \circ a_{i+2} \circ \cdots \circ a_j}\right) \circ \left({a_{j+1} \circ a_{i+2} \circ \cdots \circ a_{k+1}}\right)$
By the basis for the induction, the General Associativity Theorem holds for each of the parts in parenthesis.
Let:
: $a = \left({a_1 \circ a_2 \circ \cdots \circ a_i}\right)$
: $b = \left({a_{i+1} \circ a_{i+2} \circ \cdots \circ a_j}\right)$
: $c = \left({a_{j+1} \circ a_{i+2} \circ \cdots \circ a_{k+1}}\right)$
By definition of associative operation:
:$\left({a \circ b}\right) \circ c = a \circ \left({b \circ c}\right)$
This demonstrates the equality of the expressions $(1)$ and $(2)$.
So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Second Principle of Mathematical Induction.
Therefore:
:The General Associativity Theorem holds for all composites $a_1 \circ a_2 \circ \cdots \circ a_n$ for $n \in \N$.
== Sources ==
* : $\S 3.1$: Monoids
Category:General Associativity Theorem
General Associativity Theorem/Formulation 3

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Let $a_i$ denote elements of $S$.
Let $\circ$ be associative.
Let $n \in \Z$ be a positive integer such that $n \ge 3$.
Then all possible parenthesizations of the expression:
:$a_1 \circ a_2 \circ \cdots \circ a_n$
are equivalent.
== Proof ==
Let $\circ$ be associative.
It will be shown that any parenthesization of $a_1 \circ a_2 \circ \dots \circ a_n$ is equal to the left-associated expression:
:$\paren {\paren {\paren {\cdots \paren {a_1 \circ a_2} \circ a_3} \circ \cdots} \circ a_n}$
The proof proceeds by induction on $n$.
=== Basis for the Induction ===
The base case, $n = 3$:
The product $a_1 \circ a_2 \circ a_3$ has two parenthesizations:
:$P_1: a_1 \circ \paren {a_2 \circ a_3}$
:$P_2: \paren {a_1 \circ a_2} \circ a_3$
$P_2$ is the left-associative parenthesization for $n = 3$.
From the associativity condition, $P_1$ and $P_2$ are identical.
This is the basis for the induction.
=== Induction Hypothesis ===
If for all $m < n$, all parenthesizations of the $m$-product are identical to its left-associated parenthesization:
:$\paren {\paren {\paren {\cdots \paren {a_1 \circ a_2} \circ a_3} \circ \cdots} \circ a_m}$
then all parenthesizations of the $n$-product are identical to its left-associated parenthesization:
:$\paren {\paren {\paren {\cdots \paren {a_1 \circ a_2} \circ a_3} \circ \cdots} \circ a_n}$
This is the induction hypothesis.
=== Induction Step ===
Let $P_i$ denote any parenthesization of $a_1 \circ a_2 \circ \dots \circ a_n$.
Then $P_i$ is always the product of two smaller products:
By the Principle of Mathematical Induction, the proof is complete.
It follows that all parenthesizations of $a_1 \circ a_2 \circ \dots \circ a_n$ yield identical results.
So the theorem holds.
== Sources ==
* : $\S 4.2$. Commutative and associative operations
Category:General Associativity Theorem
General Associativity Theorem/Motivation

== Motivation for General Associativity Theorem ==
The General Associativity Theorem answers the following question:
It has been proved that, for example, union and intersection are associative in Union is Associative and Intersection is Associative.
That is:
:$R \cup \paren {S \cup T} = \paren {R \cup S} \cup T$
and the same with intersection.
However, are we sure that there is only one possible answer to $\ds \bigcup_{i \mathop = 1}^n S_i$ and $\ds \bigcap_{i \mathop = 1}^n S_i$?
That is, is it completely immaterial where we put the brackets in an expression containing an arbitrary number of multiple instances of one of these operations?
The question is a larger one than that: given any associative operation, is it completely associative?
This result shows that it is. Always.
Category:General Associativity Theorem
General Binomial Theorem/Convergence

== Theorem ==
Recall the General Binomial Theorem:
The above binomial series:
:converges when $\size x < 1$
:diverges when $\size x > 1$
For the special case where $x = 1$, the binomial series converges if $n > -1$.
For the special case where $x = -1$, the binomial series converges if $n > 0$.
== Proof ==
== Sources ==
* 
* 
Category:Binomial Theorem
General Binomial Theorem/Examples

== Examples of Use of General Binomial Theorem ==
=== Example: $\paren {1 - x}^{-3}$ ===
=== Example: $\paren {1 + 2 x}^{-\frac 3 2}$ ===
=== Example: $\paren {1 + 5 x}^{\frac 1 5}$ ===
=== Example: $\paren {1 - 4 x}^{\frac 1 2}$ ===
Category:Examples of Use of General Binomial Theorem
General Binomial Theorem/Examples/(1+2x)^(-3/2)

== Example of Use of General Binomial Theorem ==
:$\paren {1 + 2 x}^{-\frac 3 2} = 1 - 3 x + \dfrac {15} 2 x^2 - \dfrac {35} 2 x^3 + \cdots$
== Proof ==
== Sources ==
* : $\text I$. Algebra: The Binomial Theorem: Exercises $\text {III}$: $1 \ \text {(b)}$
Category:Examples of Use of General Binomial Theorem
General Binomial Theorem/Examples/(1+5x)^(1/5)

== Example of Use of General Binomial Theorem ==
:$\paren {1 + 5 x}^{\frac 1 5} = 1 + x - 2 x^2 + 6 x^3 + \cdots$
== Proof ==
== Sources ==
* : $\text I$. Algebra: The Binomial Theorem: Exercises $\text {III}$: $1 \ \text {(c)}$
Category:Examples of Use of General Binomial Theorem
General Binomial Theorem/Examples/(1-4x)^(1/2)

== Example of Use of General Binomial Theorem ==
:$\paren {1 - 4 x}^{\frac 1 2} = 1 - 2 x - 2 x^2 + 4 x^3 + \cdots$
== Proof ==
== Sources ==
* : $\text I$. Algebra: The Binomial Theorem: Exercises $\text {III}$: $1 \ \text {(d)}$
Category:Examples of Use of General Binomial Theorem
General Binomial Theorem/Examples/(1-x)^(-3)

== Example of Use of General Binomial Theorem ==
:$\paren {1 - x}^{-3} = 1 + 3 x + 6 x^2 + 10 x^3 + \cdots$
== Proof ==
== Sources ==
* : $\text I$. Algebra: The Binomial Theorem: Exercises $\text {III}$: $1 \ \text {(a)}$
Category:Examples of Use of General Binomial Theorem
General Distributivity Theorem

== Theorem ==
Let $\struct {R, \circ, *}$ be a ringoid.
Then for every pair of sequences $\sequence {a_i}_{1 \mathop \le i \mathop \le m}, \sequence {b_j}_{1 \mathop \le j \mathop \le n}$ of elements of $R$:
:$\ds \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^n b_j} = \sum_{\substack {1 \mathop \le i \mathop \le m \\ 1 \mathop \le j \mathop \le n} } \paren {a_i * b_j}$
where:
:$\ds \sum_{i \mathop = 1}^m a_i$ is the summation $a_1 \circ a_2 \circ \cdots \circ a_m$
:$m$ and $n$ are strictly positive integers: $m, n \in \Z_{> 0}$
== Lemmata ==
The proof requires the following lemmata:
=== Lemma 1 ===
=== Lemma 2 ===
== Proof ==
Proof by induction:
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$\ds \forall m \in \Z_{> 0}: \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^n b_j} = \sum_{\substack {1 \mathop \le i \mathop \le m \\ 1 \mathop \le j \mathop \le n} } \paren {a_i * b_j}$
We have that $\struct {R, \circ, *}$ is a ringoid, and so:
:$\forall a, b, c \in S: \paren {a \circ b} * c = \paren {a * c} \circ \paren {b * c}$
:$\forall a, b, c \in R: a * \paren {b \circ c} = \paren {a * b} \circ \paren {a * c}$
=== Basis for the Induction ===
$\map P 1$ is the case:
:$\ds \forall m \in \Z_{>0}: \paren {\sum_{i \mathop = 1}^m a_i} * b_1 = \sum_{1 \mathop \le i \mathop \le m} \paren {a_i * b_1}$
This is demonstrated in Lemma 1.
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:
:$\ds \forall m \in \Z_{> 0}: \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^k b_j} = \sum_{\substack {1 \mathop \le i \mathop \le m \\ 1 \mathop \le j \mathop \le k} } \paren {a_i * b_j}$
Then we need to show:
:$\ds \forall m \in \Z_{> 0}: \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^{k + 1} b_j} = \sum_{\substack {1 \mathop \le i \mathop \le m \\ 1 \mathop \le j \mathop \le {k + 1} } } \paren {a_i * b_j}$
=== 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:
:$\ds \forall m, n \in \Z_{> 0}: \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^n b_j} = \sum_{\substack {1 \mathop \le i \mathop \le m \\ 1 \mathop \le j \mathop \le n} } \paren {a_i * b_j}$
The same result can be obtained by fixing $n$ and using induction on $m$, which requires Lemma 2 to be used for its base case.
== Examples ==
== Warning ==
== Also see ==
* Multiple of Ring Product
* Product Rule for Sequences for the infinite case in the context of the standard number fields
== Sources ==
* : Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers
* : Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers: Exercise $2$
* : $\S 1.2.3$: Sums and Products: $(4)$
Category:Abstract Algebra
Category:Summations
Category:Named Theorems
Category:Distributive Operations
Category:General Distributivity Theorem
General Distributivity Theorem/Examples/Sum of j from m to n by Sum of k from r to s

== Example of General Distributivity Theorem ==
:$\ds \sum_{j \mathop = m}^n \sum_{k \mathop = r}^s j k = \frac 1 4 \paren {n \paren {n + 1} - \paren {m - 1} m} \paren {s \paren {s + 1} - \paren {r - 1} r}$
for $m \le n, r \le s$.
== Proof ==
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $14$
Category:General Distributivity Theorem
General Distributivity Theorem/Lemma 1

== Lemma ==
Let $\struct {R, \circ, *}$ be a ringoid.
Then for every sequence $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ of elements of $R$, and for every $b \in R$:
:$\ds \paren {\sum_{j \mathop = 1}^n a_j} * b = \sum_{j \mathop = 1}^n \paren {a_j * b}$
where:
:$\ds \sum_{j \mathop = 1}^n a_j$ is the composite $a_1 \circ a_2 \circ \cdots \circ a_n$
:$n$ is a strictly positive integer: $n \in \Z_{> 0}$.
== Proof ==
The proof proceeds by the Principle of Mathematical Induction.
Recall that as $\struct {R, \circ, *}$ is a ringoid, $*$ is distributive over $\circ$:
:$\forall a, b, c \in S: \paren {a \circ b} * c = \paren {a * c} \circ \paren {b * c}$
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$\ds \paren {\sum_{j \mathop = 1}^n a_j} * b = \sum_{j \mathop = 1}^n \paren {a_j * b}$
=== Basis for the Induction ===
$P(1)$ is true, as this just says:
:$a_1 * b = a_1 * b$
$P(2)$ is the case:
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:
:$\ds \paren {\sum_{j \mathop = 1}^k a_j} * b = \sum_{j \mathop = 1}^k \paren {a_j * b}$
Then we need to show:
:$\ds \paren {\sum_{j \mathop = 1}^{k + 1} a_j} * b = \sum_{j \mathop = 1}^{k + 1} \paren {a_j * b}$
=== 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:
:$\ds \forall n \in \Z_{> 0}: \paren {\sum_{j \mathop = 1}^k a_j} * b = \sum_{j \mathop = 1}^k \paren {a_j * b}$
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Theorem $18.8$
Category:General Distributivity Theorem