avewright/proofwiki-math huggingface dataset:
Theorem: Expectation of Exponential Distribution

Let $X$ be a continuous random variable of the exponential distribution with parameter $\beta$ for some $\beta \in \R_{> 0}$
Then the expectation of $X$ is given by:
:$\expect X = \beta$

Proof:
From Moment Generating Function of Exponential Distribution, the moment generating function $M_X$ of $X$, is given by:
:$\map {M_X} t = \dfrac 1 {1 - \beta t}$
By Moment in terms of Moment Generating Function:
:$\expect X = \map {M_X'} 0$
We have: 
{{begin-eqn}}
{{eqn | l = \map {M_X'} t 
      | r = \map {\frac \d {\d t} } {\frac 1 {1 - \beta t} }
}}
{{eqn | r = \frac {-\beta} {-1} \frac 1 {\paren {1 - \beta t}^2}
      | c = Chain Rule for Derivatives, Derivative of Power
}}
{{eqn | r = \frac \beta {\paren {1 - \beta t}^2}
}}
{{end-eqn}} 
Setting $t = 0$ gives:
{{begin-eqn}}
{{eqn | l = \expect X
      | r = \frac \beta {\paren {1 - 0 \beta}^2}
}}
{{eqn | r = \beta
}}
{{end-eqn}}
{{qed}}
Theorem: Finite Group is p-Group iff Order is Power of p

Let $p$ be a prime number.
Let $G$ be a finite group.
Then $G$ is a $p$-group {{iff}} the order of $G$ is a power of $p$.

Proof:
=== Necessary Condition ===
Let $G$ be a finite group whose order is $p^n$ for some $n \in \Z_{>0}$.
Let $g \in G$.
From Order of Element Divides Order of Finite Group, the order of $g$ is a divisor of $p^n$.
That is, $x$ is a $p$-element by definition.
As $x$ is arbitrary, it follows that all elements of $G$ are $p$-elements.
Thus $G$ is a $p$-group.
{{qed|lemma}}
Theorem: Equivalence of Definitions of Semantic Equivalence for Boolean Interpretations

{{TFAE|def = Semantic Equivalence for Boolean Interpretations}}
Let $\mathbf A, \mathbf B$ be WFFs of propositional logic.

Proof:
=== Definition 1 implies Definition 2 ===
Let $\mathbf A, \mathbf B$ be equivalent according to definition 1.
Let $v$ be a boolean interpretation.
By definition. either $\map v {\mathbf A} = \T$ or $\map v {\mathbf A} = \F$.
In the first case, it follows {{hypothesis}} that $\map v {\mathbf B} = \T$.
In particular, then:
:$\map v {\mathbf A} = \map v {\mathbf B}$
In the second case, it must be that $\map v {\mathbf B} \ne \T$.
That is, $\map v {\mathbf B} = \F$, so that:
:$\map v {\mathbf A} = \map v {\mathbf B}$
Hence $\mathbf A$ and $\mathbf B$ are also equivalent in the sense of definition 2.
{{qed|lemma}}
Theorem: Von Mangoldt Equivalence

For $n \in \N_{>0}$, let $\map \Lambda n$ be the von Mangoldt function.
Then:
:$\ds \lim_{N \mathop \to \infty} \frac 1 N \sum_{n \mathop = 1}^N \map \Lambda n = 1$
is logically equivalent to the Prime Number Theorem.

Proof:
Observe:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = 1}^N \map \Lambda n
      | r = \map \Lambda 1 + \map \Lambda 2 + \cdots + \map \Lambda n
      | c=
}}
{{eqn | r = 0 + \map \ln 2 + \map \ln 3 + \map \ln 2 + \map \ln 5 + 0 + \map \ln 7 + \map \ln 2 + \map \ln 3 + 0 + \cdots
      | c =
}}
{{end-eqn}}
Notice this sum will have:
:as many $\map \ln 2$ terms as there are powers of $2$ less than or equal to $N$
:as many $\map \ln 3$ terms as there are powers of $3$ less than or equal to $N$
and in general, if $p$ is a prime less than $N$, $\map \ln p$ will occur in this sum $\floor {\map {\log_p} N}$ times.
But:
:$\map \ln p \floor {\map {\log_p} N} \sim \map \ln p \map {\log_p} N = \map \ln N$
so:
:$\ds \sum_{p \text{ prime} \mathop \le N} \map \ln p \floor {\map {\log_p} N} \sim \sum_{p \text{ prime} \mathop \le N} \map \ln N = \map \pi N \map \ln N$
Therefore:
:$\ds \sum_{n \mathop = 1}^N \map \Lambda n \sim \map \pi N \map \ln N$
and so if:
:$\ds \lim_{N \mathop \to \infty} \frac 1 N \sum_{n \mathop = 1}^N \map \Lambda n = 1$ 
then:
:$\ds \lim_{N \mathop \to \infty} \frac 1 N \map \pi N \map \ln N = 1$
and vice versa.
But this last equation is precisely the Prime Number Theorem.
Hence our statement regarding the von Mangoldt function is logically equivalent to the Prime Number Theorem.
{{qed}}
{{namedfor|Hans Carl Friedrich von Mangoldt|cat=von Mangoldt}}
Category:Prime Numbers
6bgqz0ehahbx5mqbfhasr88q309yzxt
Theorem: Zeta Equivalence to Prime Number Theorem

Let $\map \zeta z$ be the Riemann $\zeta$ function.
The Prime Number Theorem is logically equivalent to the statement that the average of the first $N$ coefficients of $\dfrac {\zeta'} {\zeta}$ tend to $-1$ as $N$ goes to infinity.
{{explain|What does $z$ range over, and what does it mean by "first $N$ coefficients" of $\dfrac {\zeta'} {\zeta}$?}}

Proof:
The Von Mangoldt Equivalence is equivalent (clearly) to the statement that the average of the coefficients of the function of $z$ defined as:
:$(1): \quad \ds \sum_{n \mathop = 1}^\infty \frac {\map \Lambda n} {n^z}$
tend to $1$.
{{handwaving|Needs to be explained in more detail.}}
Let $ \set {p_1, p_2, p_3, \dots}$ be an enumeration of the prime numbers:
:$\set { 2, 3, 5, 7, 11, \dots}$
In the proof of the Von Mangoldt Equivalence, in the sum of von Mangoldt function, the $\map \ln p$ term will appear once for each power of $p$.
So, we expand out $(1)$ as:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = 1}^\infty \frac{\map \Lambda n} {n^z}
      | r = \map \ln {p_1} \paren {\frac 1 {p_1^z} + \frac 1 {p_1^{2 z} } + \frac 1 {p_1^{3 z} } + \cdots} + \map \ln {p_2} \paren {\frac 1 {p_2^z} + \frac 1 {p_2^{2 z} } + \cdots} + \cdots
      | c = 
}}
{{eqn | r = \map \ln {p_1} \sum_{n \mathop = 1}^\infty \paren {\paren {p_1^{-z} }^n} + \map \ln {p_2} \sum_{n \mathop = 1}^\infty \paren {\paren {p_2^{-z} }^n} + \cdots
      | c = 
}}
{{eqn | r = \map \ln {p_1} \frac {p_1^{-z} } {1 - p_1^{-z} } + \map \ln {p_2} \frac {p_2^{-z} } {1 - p_2^{-z} } + \cdots
      | c = Sum of Infinite Geometric Sequence
}}
{{eqn | r = \sum_{p \text{ prime} } \map \ln p \frac {p^{-z} } {1 - p^{-z} }
      | c = 
}}
{{end-eqn}}
This function of $z$ can be recognized as:
{{begin-eqn}}
{{eqn | l = \sum_{p \text{ prime} } \map \ln p \frac {p^{-z} } {1 - p^{-z} }
      | r = \sum_{p \text{ prime} } \paren {1 - p^{-z} } \frac {-\paren {0 - \map \ln p p^{-z} } } {\paren {1 - p^{-z} }^2}
      | c = 
}}
{{eqn | r = \sum_{p \text{ prime} } \frac \d {\d z} \map \ln {\frac {-1} {1 - p^{-z} } }
      | c = 
}}
{{eqn | r = \map {\frac \d {\d z} } {\sum_{p \text{ prime} } \map \ln {\frac {-1} {1 - p^{-z} } } }
      | c = 
}}
{{eqn | r = \map {\frac \d {\d z} } {\ln \prod_{p \text{ prime} } \frac {-1} {1 - p^{-z} } }
      | c = 
}}
{{eqn | r = -\frac \d {\d z} \map \ln {\map \zeta z}
      | c = $\ds \prod_{p \text{ prime} } \frac 1 {1 - p^{-z} }$ is the Riemann zeta function
}}
{{eqn | r = -\frac {\map {\zeta'} z} {\map \zeta z}
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Category:Prime Numbers
kjgeedoe43e66qrsmszvmv3gqlycrf7
Theorem: Divisor of Product may not be Divisor of Factors

Let $a, b, c \in \Z_{>0}$ be (strictly) positive integers.
Let:
:$c \divides a b$
where $\divides$ expresses the relation of divisibility.
Then it is not necessarily the case that either $c \divides a$ or $c \divides b$.

Proof:
Proof by Counterexample:
Let $c = 6, a = 3, b = 4$.
Then $6 \times 2 = 12$ so $c \divides a b$.
But neither $6 \divides 4$ nor $6 \divides 3$.
{{qed}}
Theorem: Existence of Prime between Prime and Factorial

Let $p$ be a prime number.
Then there exists a prime number $q$ such that:
:$p < q \le p! + 1$
where $p!$ denotes the factorial of $p$.

Proof:
Let $N = p! + 1$.
If $N$ is prime, then the proof is complete.
Otherwise, from Positive Integer Greater than 1 has Prime Divisor, $N$ has a prime divisor $q$.
From Absolute Value of Integer is not less than Divisors:
:$q < N$
{{AimForCont}} $q \le p$.
Then by the definition of factorial:
:$q \divides p!$
But $q$ was chosen so that :$q \divides N$.
Let $p! = q m_1, N = q m_2$.
Then:
:$1 = N - p! = q \paren {m_2 - m_1}$
and so:
:$q \divides 1$
From Absolute Value of Integer is not less than Divisors:
:$q \le 1$
and so $q$ cannot be a prime number.
It follows by Proof by Contradiction that $q > p$.
{{qed}}
Theorem: Substitution Rule for Matrices

Let $\mathbf A$ be a square matrix of order $n$.
Then:
:$(1): \quad \ds \sum_{j \mathop = 1}^n \delta_{i j} a_{j k} = a_{i k}$
:$(2): \quad \ds \sum_{j \mathop = 1}^n \delta_{i j} a_{k j} = a_{k i}$
where:
:$\delta_{i j}$ is the Kronecker delta
:$a_{j k}$ is element $\tuple {j, k}$ of $\mathbf A$.

Proof:
By definition of Kronecker delta:
:$\delta_{i j} = \begin {cases} 1 & : i = j \\ 0 & : i \ne j \end {cases}$
Thus:
:$\delta_{i j} a_{j k} = \begin {cases} a_{i k} & : i = j \\ 0 & : i \ne j \end {cases}$
and:
:$\delta_{i j} a_{k j} = \begin {cases} a_{k i} & : i = j \\ 0 & : i \ne j \end {cases}$
from which the result follows.
{{qed}}
Theorem: Trace of Unit Matrix

Let $\mathbf I_n$ be the unit matrix of order $n$.
Then:
:$\map \tr {\mathbf I_n} = n$
where $\map \tr {\mathbf I_n}$ denotes the trace of $\mathbf I_n$.

Proof:
By definition:
:$\mathbf I_n := \sqbrk a_n: a_{i j} = \delta_{i j}$
That is: each of the elements on the main diagonal is equal to $1$.
There are $n$ such elements.
Hence the result.
{{qed}}
Theorem: Absorption Laws (Set Theory)/Corollary

:$S \cup \paren {S \cap T} = S \cap \paren {S \cup T}$

Proof:
{{begin-eqn}}
{{eqn | o = 
      | r = S \cup \paren {S \cap T}
}}
{{eqn | o = \leadstoandfrom
      | r = \paren {S \cup S} \cap \paren {S \cup T}
      | c = Union Distributes over Intersection
}}
{{eqn | o = \leadstoandfrom
      | r = S \cap \paren {S \cup T}
      | c = Set Union is Idempotent
}}
{{end-eqn}}
{{qed}}
Category:Set Intersection
Category:Set Union
Category:Absorption Laws
e52mu3o7kw28626wez0fbridd1yoac6
Theorem: Matrix is Nonsingular iff Determinant has Multiplicative Inverse/Necessary Condition

Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\mathbf A \in R^{n \times n}$ be an nonsingular square matrix of order $n$.
Let $\mathbf B = \mathbf A^{-1}$ be the inverse of $\mathbf A$.
Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$.
Then:
:$\map \det {\mathbf B} = \dfrac 1 {\map \det {\mathbf A} }$

Proof:
Let $\mathbf A$ be nonsingular with $\mathbf B = \mathbf A^{-1}$.
Let $1_R$ denote the unity of $R$.
Let $\mathbf I_n$ denote the unit matrix of order $n$.
Then:
{{begin-eqn}}
{{eqn | l = 1_R
      | r = \map \det {\mathbf I_n}
      | c = Determinant of Unit Matrix
}}
{{eqn | r = \map \det {\mathbf A \mathbf B}
      | c = {{Defof|Inverse Matrix}}
}}
{{eqn | r = \map \det {\mathbf A} \, \map \det {\mathbf B}
      | c = Determinant of Matrix Product
}}
{{end-eqn}} 
This shows that:
:$\map \det {\mathbf B} = \dfrac 1 {\map \det {\mathbf A} }$
{{qed|lemma}}
Theorem: Product of Orthogonal Matrix with Transpose is Identity

Let $\mathbf Q$ be an orthogonal matrix.
Then:
:$\mathbf Q \mathbf Q^\intercal = \mathbf I = \mathbf Q^\intercal \mathbf Q$
where:
:$\mathbf Q^\intercal$ is the transpose of $\mathbf Q$
:$\mathbf I$ is a unit (identity) matrix

Proof:
By definition, an orthogonal matrix is one such that:
:$\mathbf Q^\intercal = \mathbf Q^{-1}$
and so the result follows by definition of inverse.
{{qed}}
Theorem: Determinant of Orthogonal Matrix is Plus or Minus One

Let $\mathbf Q$ be an orthogonal matrix.
Then:
:$\det \mathbf Q = \pm 1$
where $\det \mathbf Q$ is the determinant of $\mathbf Q$.

Proof:
By Determinant of Transpose:
:$\det \mathbf Q^\intercal = \det \mathbf Q$
Then:
{{begin-eqn}}
{{eqn | l = \mathbf Q \mathbf Q^\intercal
      | r = \mathbf I
      | c = Product of Orthogonal Matrix with Transpose is Identity
}}
{{eqn | ll= \leadsto
      | l = \map \det {\mathbf Q \mathbf Q^\intercal}
      | r = \det \mathbf I
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \det {\mathbf Q \mathbf Q^\intercal}
      | r = 1
      | c = Determinant of Unit Matrix
}}
{{eqn | ll= \leadsto
      | l = \det \mathbf Q \det \mathbf Q^\intercal
      | r = 1
      | c = Determinant of Matrix Product
}}
{{eqn | ll= \leadsto
      | l = \paren {\det \mathbf Q}^2
      | r = 1
      | c = Determinant of Transpose
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Product of Proper Orthogonal Matrices is Proper Orthogonal Matrix

Let $\mathbf P$ and $\mathbf Q$ be proper orthogonal matrices.
Let $\mathbf {P Q}$ be the (conventional) matrix product of $\mathbf P$ and $\mathbf Q$.
Then $\mathbf {P Q}$ is a proper orthogonal matrix.

Proof:
By definition, $\mathbf {P Q}$ is a proper orthogonal matrix {{iff}} it is an orthogonal matrix with a determinant of $1$.
From Product of Orthogonal Matrices is Orthogonal Matrix, $\mathbf {P Q}$ is an orthogonal matrix.
By definition, $\mathbf P$ and $\mathbf Q$ both have a determinant of $1$.
From Determinant of Matrix Product:
:$\map \det {\mathbf {P Q} } = \map \det {\mathbf P} \map \det {\mathbf Q}$
Thus:
:$\map \det {\mathbf {P Q} } = 1$
Hence the result.
{{qed}}
Category:Proper Orthogonal Matrices
38yatr6xdm7kdy7ceb35oys7j7fs7fk
Theorem: Product of Orthogonal Matrices is Orthogonal Matrix

Let $\mathbf P$ and $\mathbf Q$ be orthogonal matrices.
Let $\mathbf P \mathbf Q$ be the (conventional) matrix product of $\mathbf P$ and $\mathbf Q$.
Then $\mathbf P \mathbf Q$ is an orthogonal matrix.

Proof:
From Determinant of Orthogonal Matrix is Plus or Minus One and Matrix is Nonsingular iff Determinant has Multiplicative Inverse it follows that both $\mathbf P$ and $\mathbf Q$ are nonsingular.
Thus:
{{begin-eqn}}
{{eqn | l = \paren {\mathbf P \mathbf Q}^{-1}
      | r = \mathbf Q^{-1} \mathbf P^{-1}
      | c = Inverse of Matrix Product
}}
{{eqn | r = \mathbf Q^\intercal \mathbf P^\intercal
      | c = {{Defof|Orthogonal Matrix}}
}}
{{eqn | r = \paren {\mathbf P \mathbf Q}^\intercal
      | c = Transpose of Matrix Product
}}
{{end-eqn}}
Hence the result, by definition of orthogonal matrix.
{{qed}}
Theorem: Internal Group Direct Product is Injective/General Result

Let $G$ be a group whose identity is $e$.
Let $\sequence {H_n}$ be a sequence of subgroups of $G$.
Let $\ds \phi_n: \prod_{j \mathop = 1}^n H_j \to G$ be a mapping defined by:
:$\ds \map {\phi_n} {h_1, h_2, \ldots, h_n} = \prod_{j \mathop = 1}^n h_j$
Then $\phi_n$ is injective {{iff}}:
:$\forall i, j \in \set {1, 2, \ldots, n}: i \ne j \implies H_i \cap H_j = \set e$
That is, {{iff}} $\sequence {H_n}$ is a sequence of independent subgroups.

Proof:
=== Necessary Condition ===
Let $\ds \phi_n: \prod_{j \mathop = 1}^n H_j \to G$ be a mapping defined by:
:$\ds \map {\phi_n} {h_1, h_2, \ldots, h_n} = \prod_{j \mathop = 1}^n h_j$
Let $\phi_n$ be an injection.
Let $\map {\phi_n} {h_1, h_2, \ldots, h_n} = \map {\phi_n} {g_1, g_2, \ldots, g_n}$.
As $\phi_n$ is injective:
:$\tuple {h_1, h_2, \ldots, h_n} = \tuple {g_1, g_2, \ldots, g_n}$
and thus
:$\forall i \in \set {1, 2, \ldots, n}: h_i = g_i$
From the definition of $\phi_n$:
:$\ds \prod_{j \mathop = 1}^n h_j = \prod_{j \mathop = 1}^n g_j$
Thus, each element of $G$ that can be expressed as a product of the form $\ds \prod_{j \mathop = 1}^n h_j$ can be thus expressed ''uniquely''.
Let $i, j \in \set {1, 2, \ldots, n}: i \ne j$.
Suppose $h \in H_i \cap H_j$.
{{begin-eqn}}
{{eqn | l = h = h e
      | o = :
      | r = h \in H_i, e \in H_j
      | c = 
}}
{{eqn | l = h = e h
      | o = :
      | r = e \in H_i, h \in H_j
      | c = 
}}
{{end-eqn}}
Thus we see that:
{{begin-eqn}}
{{eqn | l = \map {\phi_n} {e, e, \ldots, h, \ldots, e, \ldots, e, e}
      | r = \map {\phi_n} {e, e, \ldots, e, \ldots, h, \ldots, e, e}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \tuple {e, e, \ldots, h, \ldots, e, \ldots, e, e}
      | r = \tuple {e, e, \ldots, e, \ldots, h, \ldots, e, e}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = h
      | r = e
      | c = 
}}
{{end-eqn}}
Thus $H_i \cap H_j = \set e$.
This holds for ''all'' pairs of integers $i, j \in \set {1, 2, \ldots, n}$ where $i \ne j$.
Thus:
:$\forall i, j \in \set {1, 2, \ldots, n}: i \ne j \implies H_i \cap H_j = \set e$
{{qed|lemma}}
Theorem: Trace of Matrix Product

Let $\mathbf A$ and $\mathbf B$ be square matrices of order $n$.
Let $\mathbf A \mathbf B$ be the (conventional) matrix product of $\mathbf A$ and $\mathbf B$.
Then:
:$\ds \map \tr {\mathbf A \mathbf B} = \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n a_{i j} b_{j i}$
where $\map \tr {\mathbf A \mathbf B}$ denotes the trace of $\mathbf A \mathbf B$.
Using the Einstein summation convention, this can be expressed as:
:$\map \tr {\mathbf A \mathbf B} = a_{i j} b_{j i}$

Proof:
Let $\mathbf C := \mathbf A \mathbf B$.
By definition of matrix product:
:$\ds c_{i k} = \sum_{j \mathop = 1}^n a_{i j} b_{j k}$
Thus for the diagonal elements:
:$\ds c_{i i} = \sum_{j \mathop = 1}^n a_{i j} b_{j i}$
By definition of trace:
:$\ds \map \tr {\mathbf C} = \sum_{i \mathop = 1}^n c_{i i}$
Hence the result.
{{qed}}
Theorem: Trace of Matrix Product/General Result

Let $\mathbf A_1, \mathbf A_2, \ldots, \mathbf A_m$ be square matrices of order $n$.
Let $\mathbf A_1 \mathbf A_2 \cdots \mathbf A_m$ be the (conventional) matrix product of $\mathbf A_1, \mathbf A_2, \ldots, \mathbf A_m$.
Then:
:$(1): \quad \ds \map \tr {\mathbf A_1 \mathbf A_2 \cdots \mathbf A_m} = \map {a_1} {i_1, i_2} \map {a_2} {i_2, i_3} \cdots \map {a_{m - 1} } {i_{m - 1}, i_m} \map {a_m} {i_m, i_1}$
where:
:$\map {a_1} {i_1, i_2}$ (for example) denotes the element of $\mathbf A_1$ whose indices are $i_1$ and $i_2$
:$\map \tr {\mathbf A_1 \mathbf A_2 \cdots \mathbf A_m}$ denotes the trace of $\mathbf A_1 \mathbf A_2 \cdots \mathbf A_m$.
In $(1)$, the Einstein summation convention is used, with the implicit understanding that a summation is performed over each of the indices $i_1$ to $i_m$.

Proof:
Let $\mathbf C = \mathbf A_1 \mathbf A_2 \cdots \mathbf A_m$
From Product of Finite Sequence of Matrices, the general element of $\mathbf C$ is given in  the Einstein summation convention by:
:$\map c {i_1, j} = \map {a_1} {i_1, i_2} \map {a_2} {i_2, i_3} \cdots \map {a_{m - 1} } {i_{m - 1}, i_m} \map {a_m} {i_m, j}$
Thus for the diagonal elements:
:$\ds \map c {i_1, i_1} = \map {a_1} {i_1, i_2} \map {a_2} {i_2, i_3} \cdots \map {a_{m - 1} } {i_{m - 1}, i_m} \map {a_m} {i_m, i_1}$
which is the Einstein summation convention for the trace of $\mathbf C$.
{{Qed}}
Theorem: Product of Finite Sequence of Matrices

Let $\mathbf A_1, \mathbf A_2, \ldots, \mathbf A_n$ be matrices.
Let the order of $\mathbf A_j$ be $d_j \times d_{j + 1}$.
Let $\ds \mathbf C := \prod_{j \mathop = 1}^n \mathbf A_j = \mathbf A_1 \mathbf A_2 \cdots \mathbf A_n$ be the (conventional) matrix product of $\mathbf A_1, \mathbf A_2, \ldots, \mathbf A_n$.
Then:
:$(1): \quad \ds \map c {i_1, i_{n + 1} } = \sum_{i_n \mathop = 1}^{d_n} \dotsm \sum_{i_3 \mathop = 1}^{d_3} \sum_{i_2 \mathop = 1}^{d_2} \map {a_1} {i_1, i_2} \map {a_2} {i_2, i_3} \dotsm \map {a_{n - 1} } {i_{n - 1}, i_n} \map {a_n} {i_n, i_{n + 1} }$
where:
:$\map {a_1} {i_1, i_2}$ (for example) denotes the element of $\mathbf A_1$ whose indices are $i_1$ and $i_2$
:the order of $\mathbf C$ is $d_1 \times d_{n+1}$.

Proof:
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \map c {i_1, i_{n + 1} } = \sum_{i_n \mathop = 1}^{d_n} \dotsm \sum_{i_3 \mathop = 1}^{d_3} \sum_{i_2 \mathop = 1}^{d_2} \map {a_1} {i_1, i_2} \map {a_2} {i_2, i_3} \dotsm \map {a_{n - 1} } {i_{n - 1}, i_n} \map {a_n} {i_n, i_{n + 1} }$
$\map P 1$ is trivially true, as this just says:
:$\map c {i_1, i_2} = \map {a_1} {i_1, i_2}$
Theorem: Characteristic of Galois Field is Prime

Let $\GF$ be a Galois field. 
Then the characteristic of $\GF$ is a prime number.

Proof:
By Characteristic of Field is Zero or Prime, it follows that $\Char \GF$ is $0$ or a prime number. 
By Finite Field has Non-Zero Characteristic:
:$\Char \GF \ne 0$
Thus $\Char \GF$ is a prime number.
{{qed}}
Category:Galois Fields
j1df1dpfgmn7ywm6w4um79xpevopykt
Theorem: Finite Ring with No Proper Zero Divisors is Field

Let $\struct {R, +, \circ}$ be a finite non-null ring with no proper zero divisors.
Then $R$ is a field.

Proof:
As $R$ is non-null, there is at least one nonzero element in $R$. 
Consider the two maps from $R$ to itself, for each nonzero $a \in R$:
:$\varphi_R: x \mapsto a \circ x$
:$\varphi_L: x \mapsto x \circ a$
By Ring Element is Zero Divisor iff not Cancellable, all nonzero elements in $R$ are cancellable. Thus:
:$a \circ x = a \circ y \implies x=y$
:$x \circ a = y \circ a \implies x=y$
Therefore, both maps are by definition injective.
By Equivalence of Mappings between Finite Sets of Same Cardinality, the maps are then also surjective.
First, we show that $R$ has a unity. 
Since $\varphi_R$ is surjective, any element $b \in R$ can be written in the form:
:$b = a \circ x_b$ 
for some $x_b \in R$.
As $\varphi_L$ is surjective, there is some $x_L \in R$ such that:
:$a = x_L \circ a$
We now have:
{{begin-eqn}}
{{eqn | l = b
      | r = a \circ x_b
}}
{{eqn | r = x_L \circ a \circ x_b
      | c = as $a = x_L \circ a$
}}
{{eqn | r = x_L \circ b
      | c = as $b = a \circ x_b$
}}
{{end-eqn}}
Thus, $x_L$ is a multiplicative left identity. 
By similar arguments, $R$ also has a multiplicative right identity, which can be denoted $x_R$.
By Left and Right Identity are the Same, it follows that $x_L = x_R$, so they are the unity of $R$. 
Now we show that $R$ is a division ring, i.e. that each nonzero element of $R$ has an inverse.
Let $1_R$ denote the unity of $R$.
Since $\varphi_R$ is surjective, it follows that:
:$a \circ y_R = 1_R$
for some $y_R \in R$. 
Since $\varphi_L$ is surjective, it follows that:
:$y_L \circ a = 1_R$ 
for some $y_L \in R$.
Recalling that the maps were defined for each nonzero $a \in R$, this means that every nonzero element of $R$ has both a left inverse and a right inverse. 
By Left Inverse and Right Inverse is Inverse, each nonzero element of $R$ has an inverse, so $R$ is by definition a division ring. 
It now follows by Wedderburn's Theorem that $R$ is a field. 
{{qed}}
Category:Ring Theory
Category:Field Theory
Category:Galois Fields
9dxvo7b4qprmdep4g7oahfg35r5m4nj
Theorem: Equivalence of Definitions of Dot Product

{{TFAE|def = Dot Product|context = Real Euclidean Space}}
Let $\mathbf a$ and $\mathbf b$ be vectors in the real Euclidean space $\R^n$.

Proof:
=== General Context implies Definition by Cosine ===
Let $\mathbf a \cdot \mathbf b$ be a dot product in its general context.
From Cosine Formula for Dot Product:
:$\mathbf a \cdot \mathbf b = \norm {\mathbf a} \norm {\mathbf b} \cos \theta$
where $\theta$ is the angle between $\mathbf v$ and $\mathbf w$.
Thus $\cdot$ is a dot product by cosine definition.
{{qed|lemma}}
Theorem: Row Equivalent Matrix for Homogeneous System has same Solutions/Corollary

:$\set {\mathbf x: \mathbf A \mathbf x = \mathbf 0} = \set {\mathbf x: \map {\mathrm {ref} } {\mathbf A} \mathbf x = \mathbf 0}$
where $\map {\mathrm {ref} } {\mathbf A}$ is the reduced echelon form of $\mathbf A$.

Proof:
Follows from Row Equivalent Matrix for Homogeneous System has same Solutions and from Matrix is Row Equivalent to Reduced Echelon Matrix.
{{qed}}
Category:Matrix Theory
Category:Linear Algebra
dla471omeqku92w6a8j8ujatufa5qe7
Theorem: Continuous Mapping is Sequentially Continuous/Corollary

Let $f$ be continuous (everywhere) on $X$.
Then $f$ is sequentially continuous on $X$.

Proof:
This follows immediately from the definitions:
:$(1): \quad$ A mapping is sequentially continuous everywhere in $X$ {{iff}} if it is sequentially continuous at each point
:$(2): \quad$ A mapping is continuous everywhere on $X$ {{iff}} it is continuous at each point.
{{Qed}}
Category:Continuous Mappings
Category:Sequential Continuity
7rxtndvbxzgxxynpw07skzeofu0l4ny
Theorem: Overflow Theorem/Corollary

The class of finite models is not $\Delta$-elementary.
That is:
:there is no set of formulas $F$ such that $F$ is satisfied by a model $\MM$
{{iff}}:
:$\MM$ is finite.

Proof:
Follows directly from the Overflow Theorem.
{{qed}}
Category:Mathematical Logic
Category:Named Theorems
m2k1rzobtxxsdh4s6etfbvlsnaq57s4