nyuuzyou/wikis huggingface dataset:
General Distributivity Theorem/Lemma 2

== 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 $b \in R$:
:$\ds b * \paren {\sum_{j \mathop = 1}^n a_j} = \sum_{j \mathop = 1}^n \paren {b * a_j}$
where:
:$\ds \sum_{j \mathop = 1}^n a_j$ is the summation $a_1 \circ a_2 \circ \cdots \circ a_n$
:$n$ 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 $\circ$:
:$\forall a, b, c \in R: a * \paren {b \circ c} = \paren {a * b} \circ \paren {a * c}$
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$\ds b * \paren {\sum_{j \mathop = 1}^n a_j} = \sum_{j \mathop = 1}^n \paren {b * a_j}$
We have that $\struct {R, \circ, *}$ is a ringoid, and so:
:$\forall a, b, c \in R: a * \paren {b \circ c} = \paren {a * b} \circ \paren {a * c}$
=== Basis for the Induction ===
$\map P 1$ is true, as this just says:
:$b * a_1 = b * a_1$
$\map P 2$ 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:
:$\ds b * \paren {\sum_{j \mathop = 1}^k a_j} = \sum_{j \mathop = 1}^k \paren {b * a_j}$
Then we need to show:
:$\ds b * \paren {\sum_{j \mathop = 1}^{k + 1} a_j} = \sum_{j \mathop = 1}^{k + 1} \paren {b * a_j}$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{> 0}: b * \paren {\sum_{j \mathop = 1}^k a_j} = \sum_{j \mathop = 1}^k \paren {b * a_j}$
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Theorem $18.8$
* : Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers: Exercise $2$
Category:General Distributivity Theorem
General Distributivity Theorem/Warning

== Warning about use of of General Distributivity Theorem ==
When using the General Distributivity Theorem, be careful not to conflate the indices.
The following is not a valid argument:
== Proof ==
The second step uses the same index $i$ for two different purposes.
Thus elements of the form $\dfrac {a_i} {a_j}$ where $i \ne j$ are not represented in the summation.
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $25$
Category:General Distributivity Theorem
General Fibonacci Number in terms of Fibonacci Numbers

== Theorem ==
Let $r$ and $s$ be numbers, usually integers but not necessarily so limited.
Let $\sequence {a_n}$ be the general Fibonacci sequence:
:$a_n = \begin{cases}
r & : n = 0 \\
s & : n = 1 \\
a_{n - 2} + a_{n - 1} & : n > 1
\end{cases}$
Then $a_n$ can be expressed in Fibonacci numbers as:
:$a_n = F_{n - 1} r + F_n s$
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$a_n = F_{n - 1} r + F_n s$
=== Basis for the Induction ===
$\map P 0$ is the case:
Thus $\map P 0$ is seen to hold.
$\map P 1$ 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 $\map P {k - 1}$ and $\map P k$ are true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$a_{k - 1} = F_{k - 2} r + F_{k - 1} s$
:$a_k = F_{k - 1} r + F_k s$
from which it is to be shown that:
:$a_{k + 1} = F_k r + F_{k + 1} s$
=== Induction Step ===
This is the induction step:
So $\map P {k - 1} \land \map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: a_n = F_{n - 1} r + F_n s$
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $13 \ \text{(a)}$
Category:Fibonacci Numbers
General Fibonacci Sequence whose Terms are all Composite

== Theorem ==
The general Fibonacci sequence $\left\langle{a_n}\right\rangle$ defined as:
:$a_n = \begin{cases}
r & : n = 0 \\
s & : n = 1 \\
a_{n - 2} + a_{n - 1} & : n > 1
\end{cases}$
where:
:$r = 62 \, 638 \, 280 \, 004 \, 239 \, 857$
:$s = 49 \, 463 \, 435 \, 743 \, 205 \, 655$
is such that:
: $r$ and $s$ are coprime
: all its terms are composite.
== Proof ==
== Sources ==
* 
* : $62,638,280,004,239,857$
Category:Fibonacci Numbers
General Fibonacci Sequence whose Terms are all Composite/Mistake 1

== Source Work ==
:
:The Dictionary
::$62,638,280,004,239,857$
== Mistake ==
:The first term of a generalized Fibonacci sequence, in which each term is the sum of the previous two, in which every term is composite. The second term is $49,463,435,743,205,655$.
It is important also to note that $62,638,280,004,239,857$ and $49,463,435,743,205,655$ are coprime.
Otherwise, for example, such generalized Fibonacci sequences would be trivial to devise, for example: $4, 4, 8, 12, 20, \ldots$
== Sources ==
* : $62,638,280,004,239,857$
Category:Mistakes/Curious and Interesting Numbers
General Fibonacci Sequence whose Terms are all Composite/Mistake 2

== Source Work ==
:
:The Dictionary
::$1,786,772,701,928,802,632,268,715,130,455,793$
:
:The Dictionary
::$1,786,772,701,928,802,632,268,715,130,455,793$
== Mistake ==
:Together with $1,059,683,225,053,915,111,058,165,141,686,996$, the start of a generalized Fibonacci sequence (in which each term is the sum of the previous two) in which every member is composite although the first $2$ terms have no common factor.
== Correction ==
This pair was originally presented by in a paper of $1964$.
However, in $1990$ pointed out that:
:Incidentally the values ... are not the same as the $34$-digit values in 's original paper. A minor slip caused his original numbers to be respectively congruent to $F_{32}$ and $F_{33} \pmod {1087}$, not to $F_{33}$ and $F_{34}$, although all the other conditions were satisfied. Therefore the sequences defined by his published starting values may contain a prime number $A_{64 n + 31}$.
While the second edition of has added new pair from his $1990$ paper ($62,638,280,004,239,857$ and $49,463,435,743,205,655$), has neglected to remove the original paragraph that he published concerning $1,786,772,701,928,802,632,268,715,130,455,793$ and $1,059,683,225,053,915,111,058,165,141,686,996$.
Also note the tentative nature of note:
::may contain a prime number $A_{64 n + 31}$.
So even at this point it is not certain that the two numbers presented in 's original paper do in fact fail to form a generalized Fibonacci sequence with these properties.
And even then, the mistake may have originated with . As states in his $1964$ paper:
:I am grateful to Mr. for his assistance in obtaining an explicit solution to $(2)$. In particular ...
== Sources ==
* 
* : $1,786,772,701,928,802,632,268,715,130,455,793$
* 
* : $1,786,772,701,928,802,632,268,715,130,455,793$
Category:Mistakes/Curious and Interesting Numbers
General Fundamental Matrix

== Theorem ==
Let $\map \Phi t$ be a fundamental matrix of the system $x' = \map A t x$.
Then:
:$\map \Phi t C$
is a general fundamental matrix of $x' = \map A t x$, where $C$ is any nonsingular matrix.
== Proof ==
$\map \Phi t C$ is a fundamental matrix as follows:
:$\dfrac \d {\d t} \map \Phi t C = \map {\Phi'} t C = \map A t \map \Phi t C$
:$\map \det {\map \Phi t C} = \map \det {\map \Phi t} \map \det C \ne 0$
Let $\map \Psi t$ be an arbitrary fundamental matrix.
Then from General Vector Solution of Fundamental Matrix $\map \Phi t \map {\Phi^{-1} } {t_0} \map \Psi {t_0}$ solves the same matrix equation and has the same initial conditions.
Hence by uniqueness:
:$\map \Phi t \, \map {\Phi^{-1} } {t_0} \map \Psi {t_0}$ is equal to $\map \Psi t$
Letting $C = \map {\Phi^{-1} } {t_0} \map \Psi {t_0}$ finishes the proof.
Category:Differential Equations
General Harmonic Number Additive Formula

== Theorem == 
Let $n \in \N_{>0}$ where $\N_{>0}$ denotes the non-zero natural numbers.
Let $x \in \C \setminus \Z_{< 0}$
Then: 
:$\ds \harm 1 {n x} = \frac 1 n \sum_{k \mathop = 0}^{n - 1} \paren {\harm 1 {x - \dfrac k n } } + \ln n$
 
where:
:$\map {H^{\paren r} } x$ denotes the general harmonic number of order $1$ evaluated at $x$.
:$\ln$ is the complex natural logarithm.
=== Corollary ===
== Proof ==
== Also see ==
* Digamma Additive Formula
* Cotangent Additive Formula
Category:General Harmonic Number Additive Formula
Category:General Harmonic Numbers
General Harmonic Number Additive Formula/Corollary

== Corollary to General Harmonic Number Additive Formula == 
Let $n \in \N_{>0}$ be a non-zero natural number.
Then: 
:$\ds \sum_{k \mathop = 1}^{n - 1} \paren {\harm 1 {-\dfrac k n } } = -n \ln n$
where:
:$\harm 1 x$ denotes the general harmonic number of order $1$ evaluated at $x$.
:$\ln$ is the complex natural logarithm.
== Proof ==
Category:General Harmonic Number Additive Formula
General Harmonic Number Reflection Formula

== Theorem ==
:$\harm r {x - 1} + \paren {-1}^r \harm r {-x} = \paren {1 + \paren {-1}^r} \map \zeta r + \dfrac {\paren {-1}^r} {\paren {r - 1}!} \map {\dfrac {\d^{r - 1} } {\d x^{r - 1} } } {\pi \map \cot {\pi x} }$
where:
:$\harm r x$ denotes the general harmonic number of order $r$ evaluated at $x$
:$\map \zeta r$ is the Riemann zeta function
:$r \in \Z_{>0}$ and $x \in \C$ and $x \notin \Z$
== Proof ==
=== Lemma 1 ===
=== Lemma 2 ===
Finally, from the , we note that:
:$\harm r {x - 1} + \paren {-1}^r \harm r {-x}$
is defined on the domain $\C \setminus \Z$.
== Sources == 
* : Chapter $7$. Sums of Powers, Bernoulli Numbers and the Gamma Function
Category:General Harmonic Number Reflection Formula
Category:General Harmonic Numbers
Category:Reflection Formulas
General Harmonic Number Reflection Formula/Lemma 1

== General Harmonic Number Reflection Formula: Lemma 1 ==
Let $\harm r x$ denote the general harmonic number of order $r$ evaluated at $x$
The expression:
:$\harm r {x - 1} + \paren {-1}^r \harm r {-x}$
is defined on the domain $\C \setminus \Z$.
That is, on the set of complex numbers but specifically excluding the integers.
== Proof ==
From the definition of the general harmonic numbers:
:$\harm r {-x}$ is defined for $-x \in \C \setminus \Z_{<0}$
Therefore:
:$x \in \C \setminus \Z_{>0}$
and:
:$\harm r {x - 1}$ is defined for $\paren {x - 1} \in \C \setminus \Z_{<0}$.
Therefore:
:$x \in \C \setminus \Z_{\le 0}$
Therefore, $\harm r {x - 1} + \paren {-1}^r \harm r {-x}$ is defined for $z \in \C \setminus \Z$.
Category:General Harmonic Number Reflection Formula
General Harmonic Number Reflection Formula/Lemma 2

== General Harmonic Number Reflection Formula: Lemma 2 ==
:$\ds \harm 1 {x - 1} - \harm 1 {-x} = -\pi \map \cot {\pi x}$
where:
:$\harm 1 x$ denotes the general harmonic number of order $1$ evaluated at $x$
:$\map \zeta r$ is the Riemann zeta function
:$x \in \C$ and $x \notin \Z$
== Proof ==
Category:General Harmonic Number Reflection Formula
General Harmonic Numbers/Examples

== Examples of General Harmonic Numbers ==
=== General Harmonic Number of Order $1$ at $\dfrac 1 2$ ===
=== General Harmonic Number of Order $1$ at $-\dfrac 1 2$ ===
=== General Harmonic Number of Order $1$ at $\dfrac 1 3$ ===
=== General Harmonic Number of Order $1$ at $-\dfrac 1 3$ ===
=== General Harmonic Number of Order $1$ at $\dfrac 2 3$ ===
=== General Harmonic Number of Order $1$ at $-\dfrac 2 3$ ===
=== General Harmonic Number of Order $2$ at $\dfrac 1 2$ ===
=== General Harmonic Number of Order $2$ at $-\dfrac 1 2$ ===
Category:Examples of General Harmonic Numbers
General Harmonic Numbers/Examples/Order 1/Half

== Example of General Harmonic Number ==
:$\harm 1 {\dfrac 1 2} = 2 - 2 \ln 2$
where $\harm 1 {\dfrac 1 2}$ denotes the general harmonic number of order $1$ evaluated at $\dfrac 1 2$.
== Proof ==
Category:Examples of Use of Reciprocal times Derivative of Gamma Function
Category:Examples of General Harmonic Numbers
General Harmonic Numbers/Examples/Order 1/Minus One Half

== Example of General Harmonic Number ==
:$\harm 1 {-\dfrac 1 2} = -2 \ln 2$
where $\harm 1 {-\dfrac 1 2}$ denotes the general harmonic number of order $1$ evaluated at $-\dfrac 1 2$.
== Proof ==
Category:Examples of Use of Reciprocal times Derivative of Gamma Function
Category:Examples of General Harmonic Numbers
General Harmonic Numbers/Examples/Order 1/Minus One Third

== Example of General Harmonic Number ==
:$\harm 1 {-\dfrac 1 3} = -\dfrac 3 2 \ln 3 + \dfrac \pi {2 \sqrt 3}$
where $\harm 1 {-\dfrac 1 3}$ denotes the general harmonic number of order $1$ evaluated at $-\dfrac 1 3$.
== Proof ==
Category:Examples of Use of Reciprocal times Derivative of Gamma Function
Category:Examples of General Harmonic Numbers
General Harmonic Numbers/Examples/Order 1/Minus Two Thirds

== Example of General Harmonic Number ==
:$\harm 1 {-\dfrac 2 3} = -\dfrac 3 2 \ln 3 - \dfrac \pi {2 \sqrt 3}$
where $\harm 1 {-\dfrac 2 3}$ denotes the general harmonic number of order $1$ evaluated at $-\dfrac 2 3$.
== Proof ==
Category:Examples of Use of Reciprocal times Derivative of Gamma Function
Category:Examples of General Harmonic Numbers
General Harmonic Numbers/Examples/Order 1/One Third

== Example of General Harmonic Number ==
:$\harm 1 {\dfrac 1 3} = 3 - \dfrac 3 2 \ln 3 - \dfrac \pi {2 \sqrt 3}$
where $\harm 1 {\dfrac 1 3}$ denotes the general harmonic number of order $1$ evaluated at $\dfrac 1 3$.
== Proof ==
Category:Examples of Use of Reciprocal times Derivative of Gamma Function
Category:Examples of General Harmonic Numbers
General Harmonic Numbers/Examples/Order 1/Two Thirds

== Example of General Harmonic Number ==
:$\harm 1 {\dfrac 2 3} = \dfrac 3 2 - \dfrac 3 2 \ln 3 + \dfrac \pi {2 \sqrt 3}$
where $\harm 1 {\dfrac 2 3}$ denotes the general harmonic number of order $1$ evaluated at $\dfrac 2 3$.
== Proof ==
Category:Examples of Use of Reciprocal times Derivative of Gamma Function
Category:Examples of General Harmonic Numbers
General Harmonic Numbers/Examples/Order 2/Half

== Example of General Harmonic Number ==
:$\harm 2 {\dfrac 1 2} = 4 - 2 \map \zeta 2$
where $\harm 2 {\dfrac 1 2}$ denotes the general harmonic number of order $2$ evaluated at $\dfrac 1 2$.
== Proof ==
Category:Examples of General Harmonic Numbers
General Harmonic Numbers/Examples/Order 2/Minus One Half

== Example of General Harmonic Number ==
:$\harm 2 {-\dfrac 1 2} = -2 \map \zeta 2$
where:
:$\harm 2 {-\dfrac 1 2}$ denotes the general harmonic number of order $2$ evaluated at $-\dfrac 1 2$
:$\map \zeta 2$ denotes the Riemann zeta function evaluated at $2$.
== Proof ==
Category:Examples of General Harmonic Numbers
General Harmonic Numbers in terms of Riemann Zeta and Hurwitz Zeta Functions

== Theorem ==
:$\harm r x = \map \zeta r - \map \zeta {r, x + 1}$
where:
:$\harm r x$ denotes the general harmonic number of order $r$ evaluated at $x$
:$\map \zeta r$ is the Riemann zeta function
:$\map \zeta {r, x}$ is the Hurwitz zeta function
:$r$ and $x$ are complex numbers with $\map \Re r > 1$ and $x \notin \Z_{<0}$
== Proof ==
== Sources ==
* : Chapter $7$. Sums of Powers, Bernoulli Numbers, and the Gamma Function
Category:General Harmonic Numbers
Category:Riemann Zeta Function
Category:Hurwitz Zeta Function
General Implicit Function/Examples/u^2 + x^2 + y^2 = a^2

== Example of General Implicit Function ==
Consider the equation:
:$(1): \quad u^2 + x^2 + y^2 = a^2$
where $u, x, y \in \R$ are real variables.
Then $(1)$ defines $u$ as an implicit function of $x$ and $y$ on the region $x^2 + y^2 \le a^2$.
== Proof ==
Solving for $u$, we obtain:
:$y = \pm \sqrt {a^2 - x^2 - y^2}$
As it stands, $(1)$ does not define a real function, because:
:$\text{(a)}: \quad u$ is not defined for values of $x^2 + y^2$ outside the range $-a \le x \le a$
:$\text{(b)}: \quad$ For all $x^2 + y^2$ within the range $-a \le x \le a$, there are two possible values of $u$ that can be taken.
This can be corrected by:
:$\text{(a)}: \quad$ Specifying the domain to be the region $x^2 + y^2 \le a^2$ (or a subset thereof)
:$\text{(b)}: \quad$ Specifying which value of $x^2 + y^2$ that is to be chosen, for example:
:::$\forall x, y: x^2 + y^2 \le a^2: u = \sqrt {a^2 - x^2 - y^2}$ (where $\sqrt {}$ in this context specifically means the positive square root)
:::$\forall x, y: x^2 + y^2 \le a^2: u = -\sqrt {a^2 - x^2 - y^2}$
== Sources ==
* : $1$ Partial Differentiation: $\S 1$. Introduction
Category:Examples of Implicit Functions
General Intersection Property of Topological Space

== Theorem ==
Let $\struct {S, \tau}$ be a topological space.
Let $S_1, S_2, \ldots, S_n$ be open sets of $\struct {S, \tau}$.
Then:
:$\ds \bigcap_{i \mathop = 1}^n S_i$
is also an open set of $\struct {S, \tau}$.
That is, the intersection of any finite number of open sets of a topology is also in $\tau$.
Conversely, if the intersection of any finite number of open sets of a topology is also in $\tau$, then:
:$(1): \quad$ The intersection of any two elements of $\tau$ is an element of $\tau$
:$(2): \quad S$ is itself an element of $\tau$.
== Proof ==
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:For any sets $S_1, S_2, \ldots, S_n \in \tau$, it follows that $\ds \bigcap_{i \mathop = 1}^n S_i \in \tau$.
Let $\mathbb S$ be any finite subset of $\tau$.
From Intersection of Empty Set, we have that:
:$\mathbb S = \O \implies \bigcap \mathbb S = S$
From the definition of a topology, we have that $S \in \tau$.
Hence $\map P 0$ is true.
From Intersection of Singleton, we have that:
:$\mathbb S = S_1 \implies \bigcap \mathbb S = S_1$
Thus $\map P 1$ is trivially true.
=== Basis for the Induction ===
$\map P 2$ is the case $S_1 \cap S_2 \in \O$, which is our axiom:
:The intersection of any two elements of $\tau$ is an element of $\tau$.
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:
:For any sets $S_1, S_2, \ldots, S_k \in \tau$, it follows that $\ds \bigcap_{i \mathop = 1}^k S_i \in \tau$.
Then we need to show:
:For any sets $S_1, S_2, \ldots, S_{k + 1} \in \tau$, it follows that $\ds \bigcap_{i \mathop = 1}^{k + 1} S_i \in \tau$.
=== Induction Step ===
This is our induction step:
Consider the set $\ds \bigcap_{i \mathop = 1}^{k + 1} S_i$.
This is $\ds \paren {\bigcap_{i \mathop = 1}^k S_i} \cap S_{k + 1}$.
But from the induction hypothesis, we know that:
:$\ds \paren {\bigcap_{i \mathop = 1}^k S_i} \in \tau$
So from the base case, it follows that:
:$\ds \bigcap_{i \mathop = 1}^{k + 1} S_i \in \tau$
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore, for any sets $S_1, S_2, \ldots, S_n \in \tau$, it follows that $\ds \bigcap_{i \mathop = 1}^n S_i \in \tau$.
The converse follows directly from the above.
In particular, the fact that $S$ is itself an element of $\tau$ follows from the definition of Intersection of Empty Set.
== Sources ==
* : Chapter $\text {I}$: Topological Spaces: $1$. Open Sets and Closed Sets
* : $3$: Continuity generalized: topological spaces: $3.1$: Topological Spaces: Definition $3.1.1$
Category:Topology
General Linear Group is Group

== Theorem ==
Let $K$ be a field.
Let $\GL {n, K}$ be the general linear group of order $n$ over $K$.
Then $\GL {n, K}$ is a group.
== Proof ==
Taking the group axioms in turn:
=== ===
The matrix product of two $n \times n$ matrices is another $n \times n$ matrix.
The matrix product of two nonsingular matrices is another nonsingular matrix.
Thus $\GL {n, K}$ is closed.
=== ===
Matrix Multiplication is Associative.
=== ===
From Unit Matrix is Unity of Ring of Square Matrices, the unit matrix serves as the identity of $\GL {n, K}$.
=== ===
From the definition of nonsingular matrix, the inverse of any nonsingular matrix $\mathbf A$ is $\mathbf A^{-1}$.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 3$: Examples of Infinite Groups: $\text{(iv) (a)}$
* : Chapter $1$: Introductory Concepts: $1$. Basic Building Blocks: $2$. GROUP: Example $6$
* : $\S 34$. Examples of groups: $(2)$
* : Chapter $1$: Definitions and Examples: Example $1.7$
Category:General Linear Group
General Linear Group is not Abelian

== Theorem ==
Let $K$ be a field whose zero is $0_K$ and unity is $1_K$.
Let $\GL {n, K}$ be the general linear group of order $n$ over $K$.
Then $\GL {n, K}$ is not an abelian group.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: Introductory Concepts: $1$. Basic Building Blocks: $2$. GROUP
Category:General Linear Group is not Abelian
Category:General Linear Group
Category:Abelian Groups
General Linear Group is not Abelian/Proof 1

== Theorem ==
== Proof ==
From Special Linear Group is Subgroup of General Linear Group we have that the special linear group $\SL {n, K}$ is a subgroup of $\GL {n, K}$.
From Special Linear Group is not Abelian, $\SL {n, K}$ is not abelian.
From Subgroup of Abelian Group is Abelian it follows by the Rule of Transposition that $\GL {n, K}$ is not abelian.
Category:General Linear Group is not Abelian
General Linear Group to Determinant is Homomorphism

== Theorem ==
Let $\GL {n, \R}$ be the general linear group over the field of real numbers.
Let $\struct {\R_{\ne 0}, \times}$ denote the multiplicative group of real numbers.
Let $\det: \GL {n, \R} \to \struct {\R_{\ne 0}, \times}$ be the mapping:
:$\mathbf A \mapsto \map \det {\mathbf A}$
where $\map \det {\mathbf A}$ is the determinant of $\mathbf A$.
Then $\det$ is a group homomorphism.
=== Corollary ===
== Proof ==
From Determinant of Matrix Product:
: $\map \det {\mathbf A \mathbf B} = \map \det {\mathbf A} \, \map \det {\mathbf B}$
which is seen to be a group homomorphism by definition.
== Sources ==
* : $\S 47 \ \text {(a)}$ Homomorphisms and their elementary properties
* : Chapter $8$: The Homomorphism Theorem: Example $8.2$
Category:Examples of Group Homomorphisms
Category:General Linear Group
Category:Determinants
General Linear Group to Determinant is Homomorphism/Corollary

== Theorem ==
Let $\GL {n, \R}$ be the general linear group over the field of real numbers.
Let $\struct {\R_{\ne 0}, \times}$ denote the multiplicative group of real numbers.
Let $\det: \GL {n, \R} \to \struct {\R_{\ne 0}, \times}$ be the group homomorphism:
:$\mathbf A \mapsto \map \det {\mathbf A}$
where $\map \det {\mathbf A}$ is the determinant of $\mathbf A$.
The kernel of the $\det$ mapping is the special linear group $\SL {n, \R}$.
== Proof ==
From General Linear Group to Determinant is Homomorphism:
:$\det$ is a group homomorphism.
The special linear group $\SL {n, \R}$ is the subset of $\GL {n, \R}$ such that:
:$\forall \mathbf A \in \SL {n, \R}: \map \det {\mathbf A} = 1$
From Real Multiplication Identity is One:
: $1$ is the identity of the multiplicative group of real numbers.
It follows by definition that $\SL {n, \R}$ is the kernel of the $\det$ mapping.
== Sources ==
* : $\S 47.7$ Homomorphisms and their elementary properties
* : Chapter $8$: The Homomorphism Theorem: Theorem $8.13: \ (3)$
Category:Examples of Group Homomorphisms
Category:General Linear Group
Category:Special Linear Group
Category:Determinants
General Logarithm/Examples/Base Pi of Pi

== Example of General Logarithm ==
The logarithm base $\pi$ of $\pi$ is:
:$\log_\pi \pi = 1$
== Proof ==
An instance of Logarithm to Own Base equals 1:
:$\forall b \in \R_{>0}, b \ne 1: \log_b b = 1$
== Sources ==
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $17$
Category:Examples of General Logarithms
General Logarithm/Examples/Base b of -1

== Theorem ==
Let $b \in \R_{>0}$ be a strictly positive real number such that $b \ne 1$.
Let $\log_b$ denote the logarithm to base $b$.
Then:
:$\log_b \left({-1}\right)$ is undefined in the real number line.
== Proof ==
 $\log_b \left({-1}\right) = y \in \R$.
Then:
:$b^y = -1 < 0$
But from Power of Positive Real Number is Positive:
:$b^y > 0$
The result follows by Proof by Contradiction.
== Sources ==
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $17$
Category:Examples of General Logarithms