nyuuzyou/wikis huggingface dataset:
Klein Four-Group and Group of Cyclic Group of Order 4 are not Isomorphic/Proof 1

== Theorem ==
== Proof ==
Recall the Cayley table for $K_4$:
From Finite Cyclic Group is Isomorphic to Integers under Modulo Addition, $C_4$ can be exemplified using the additive group of integers modulo $4$.
Recall the Cayley table for $\struct {\Z_4, +_4}$:
Note that all elements of $K_4$ are self-inverse.
However, for example:
:$\eqclass 1 4 +_4 \eqclass 1 4 = \eqclass 2 4$
and so $\eqclass 1 4$ is not self-inverse.
Thus there can be no bijection between $K_4$ and $\struct {\Z_4, +_4}$ such that:
:$\forall a, b \in K_4: \map \phi a +_4 \map \phi b = \map \phi {a b}$
Hence the result.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 7$: Isomorphic Groups
Category:Klein Four-Group and Group of Cyclic Group of Order 4 are not Isomorphic
Klein Four-Group and Group of Cyclic Group of Order 4 are not Isomorphic/Proof 2

== Theorem ==
== Proof ==
Note that both $C_4$ and $K_4$ are of order $4$.
Also note that both $C_4$ and $K_4$ are abelian.
By definition, $C_4$ has elements of order $4$.
From Group Isomorphism Preserves Order of Group Element, the image of an element of $C_4$ under an isomorphism to $K_4$ would also be of order $4$.
But $K_4$ has no elements of order $4$.
Hence $C_4$ and $K_4$ are not isomorphic.
== Sources ==
* : $\S 7.3$. Isomorphism: Example $138$
Category:Klein Four-Group and Group of Cyclic Group of Order 4 are not Isomorphic
Klein Four-Group as Order 2 Matrices

== Theorem ==
Let $G$ be the set of order $2$ square matrices:
:$G = \set {I, A, B, C}$
where:
:$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \quad A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, \quad B = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}, \quad C = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$
Then the algebraic structure $\struct {G, \times}$, where $\times$ denotes (conventional) matrix multiplication, forms the Klein four-group.
== Proof ==
From Unit Matrix is Unity of Ring of Square Matrices, $I$ can be identified as the unit matrix of order $2$.
Then:
:$A^2 = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$
:$A B = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = C$
:$A C = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} = B$
:$B^2 = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$
:$B A = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = C$
:$B C = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = A$
:$C^2 = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$
:$C A = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} = B$
:$C B = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = A$
Putting this together into a Cayley table:
:$\begin{array}{c|cccccc}
 & I & A & B & C \\
\hline
I & I & A & B & C \\
A & A & I & C & B \\
B & B & C & I & A \\
C & C & B & A & I \\
\end{array}$
it is verified by inspection that this is an instance of the Klein four-group.
== Sources ==
* : Chapter $1$: Definitions and Examples: Example $1.8$
Category:Matrix Algebra
Category:Klein Four-Group
Klein Four-Group as Subgroup of S4

== Theorem ==
Let $G$ be the following subset of the symmetric group on $4$ letters $S_4$, expressed in two-row notation:
Then $G$ is an example of the Klein $4$-group.
== Proof ==
By inspection, the Cayley table is constructed:
:$\begin{array}{c|cccc}
 & e & a & b & c \\
\hline
e & e & a & b & c \\
a & a & e & c & b \\
b & b & c & e & a \\
c & c & b & a & e \\
\end{array}$
Again by inspection this can be seen to be the same as the Cayley table for the Klein $4$-group.
== Sources ==
* : $\S 44$. Some consequences of Lagrange's Theorem: Illustration $2$
Category:Klein Four-Group
Klein Four-Group is Group

== Theorem ==
The Klein $4$-group $K_4$ is a group.
== Proof ==
From Klein Four-Group as Subgroup of $S_4$ it is demonstrated that $K_4$ is a subgroup of the $4$th symmetric group.
Hence the result.
== Sources ==
* : Chapter $2$: The Definition of Group Structure: $\S 26 \iota$
Category:Klein Four-Group
Klein Four-Group is Normal in A4

== Theorem ==
Let $A_4$ denote the alternating group on $4$ letters, whose Cayley table is given as:
<onlyinclude>
The subsets of $A_4$ which form subgroups of $A_4$ are as follows:
Consider the order $4$ subgroup $V$ of $A_4$, presented by Cayley table:
:$\begin{array}{c|cccc}
\circ & e & t & u & v \\
\hline
e & e & t & u & v \\
t & t & e & v & u \\
u & u & v & e & t \\
v & v & u & t & e \\
\end{array}$
Then $V$ is normal in $A_4$.
Its index is:
:$\index {A_4} V = \dfrac {\order {A_4} } {\order V} = \dfrac {12} 4 = 3$
The (left) cosets of $V$ are:
:$V$
:$A := a V$
:$P := p V$
and the Cayley table of the quotient group $A_4 / V$ is given by:
:$\begin{array}{c|ccc}
\circ & V & A & P \\
\hline
V & V & A & P \\
A & A & P & V \\
P & P & V & A \\
\end{array}$
Note that while $A_4 / V$ is Abelian, $A_4$ is not.
== Proof ==
:$\index {A_4} V = 3$ follows from Lagrange's Theorem.
== Sources ==
* : $\S 6.7$. Quotient groups: Example $127$
Category:Alternating Group on 4 Letters
Category:Examples of Normal Subgroups
Kluyver's Formula for Ramanujan's Sum

== Theorem ==
Let $q \in \N_{>0}$.
Let $n \in \N$.
Let $\map {c_q} n$ be Ramanujan's sum.
Let $\mu$ denote the Möbius function.
Then:
:$\ds \map {c_q} n = \sum_{d \mathop \divides \gcd \set {q, n} } d \map \mu {\frac q d}$
where $\divides$ denotes divisibility.
== Proof ==
Let $\alpha \in \R$.
Let $e: \R \to \C$ be the mapping defined as:
:$\map e \alpha := \map \exp {2 \pi i \alpha}$
Let $\zeta_q$ be a primitive $q$th root of unity.
Let:
:$\ds \map {\eta_q} n := \sum_{1 \mathop \le a \mathop \le q} \map e {\frac {a n} q}$
By Complex Roots of Unity in Exponential Form this is the sum of $n$th powers of all $q$th roots of unity.
Therefore:
:$\ds \map {\eta_q} n = \sum_{d \mathop \divides q} \map {c_d} n$
By the Möbius Inversion Formula, this gives:
:$\ds \map {c_q} n = \sum_{d \mathop \divides q} \map {\eta_d} n \map \mu {\frac q d}$
Now by Sum of Powers of Primitive Complex Roots of Unity, we have:
:$\ds \map {\eta_d} n = \begin{cases} d & : d \divides n \\ 0 & : d \nmid n \end{cases}$
Therefore:
as required.
Category:Analytic Number Theory
Knaster-Tarski Lemma

== Theorem ==
Let $\struct {L, \preceq}$ be a complete lattice.
Let $f: L \to L$ be an increasing mapping.
Then $f$ has a smallest fixed point and a greatest fixed point.
=== Power Set ===
=== Corollary ===
== Proof ==
Let $P = \set {x \in L: x \preceq \map f x}$.
Let $p = \bigvee P$, the supremum of $P$.
Let $x \in P$.
Then by the definition of supremum:
:$x \preceq p$
Since $f$ is increasing:
:$\map f x \preceq \map f p$
By the definition of $P$:
:$x \preceq \map f x$
Thus because $\preceq$ is an ordering, and therefore transitive:
:$x \preceq \map f p$
As this holds for all $x \in P$, $\map f p$ is an upper bound of $P$.
By the definition of supremum:
:$p \preceq \map f p$
As $f$ is increasing:
:$\map f p \preceq \map f {\map f p}$
Thus by the definition of $P$:
:$\map f p \in P$
Since $p$ is the supremum of $P$:
:$\map f p \preceq p$
Since we already know that $p \preceq \map f p$:
:$\map f p = p$
because $\preceq$ is an ordering and therefore antisymmetric.
Thus $p$ is a fixed p
oint of $f$.
We have that $\preceq$ is an ordering, and therefore reflexive.
Thus every fixed point of $f$ is in $P$.
So $p$ is the greatest fixed point of $f$.
Now note that $f$ is also increasing in the dual ordering.
Thus $f$ also has a greatest fixed point in the dual ordering.
That is, it has a least fixed point in the original ordering.
== Sources ==
* 
* : Lemma $9.2.2$
Category:Complete Lattices
Category:Knaster-Tarski Lemma
Knaster-Tarski Lemma/Corollary

== Theorem ==
Let $\struct {L, \preceq}$ be a complete lattice.
Let $f: L \to L$ be an increasing mapping.
Then $f$ has a fixed point
=== Power Set ===
== Proof ==
By the Knaster-Tarski Lemma, $f$ has a least fixed point.
Thus it has a fixed point.
Category:Knaster-Tarski Lemma
Knaster-Tarski Lemma/Corollary/Power Set

== Theorem ==
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $f: \powerset S \to \powerset S$ be a $\subseteq$-increasing mapping.
That is, suppose that for all $T, U \in \powerset S$:
:$T \subseteq U \implies \map f T \subseteq \map f U$
Then $f$ has a fixed point.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Lemma $9.2.2$
Category:Knaster-Tarski Lemma
Knaster-Tarski Lemma/Corollary/Power Set/Proof 2

== Theorem ==
== Proof ==
Let $P = \set {x \in \powerset S: x \subseteq \map f x}$.
Let $\ds \bigcup P$ be the union of $P$.
Let $x \in P$.
Then by Set is Subset of Union: General Result:
:$\ds x \subseteq \bigcup P$
Since $f$ is increasing, $\map f x \subseteq \map f {\bigcup P}$.
By definition of $P$, $x \subseteq \map f x$.
Thus $\ds x \subseteq \map f {\bigcup P}$ by Subset Relation is Transitive:
As this holds for all $x \in P$, by Union is Smallest Superset: General Result:
:$\ds \bigcup P \subseteq \map f {\bigcup P}$
Now since $f$ is increasing:
:$\ds \map f {\bigcup P} \subseteq \map f {\map f {\bigcup P} }$
Thus $\ds \map f {\bigcup P} \in P$ by definition of $P$.
Therefore by Set is Subset of Union: General Result:
:$\ds \map f {\bigcup P} \subseteq \bigcup P$
We have that:
:$\ds \bigcup P \subseteq \map f {\bigcup P}$
Thus by definition of set equality:
:$\ds \map f {\bigcup P} = \bigcup P$
Thus $\ds \bigcup P$ is a fixed point of $f$.
== Sources ==
* : Lemma $9.2.2$
Category:Knaster-Tarski Lemma
Knaster-Tarski Lemma/Power Set

== Theorem ==
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $f: \powerset S \to \powerset S$ be a $\subseteq$-increasing mapping.
That is, suppose that for all $T, U \in \powerset S$:
:$T \subseteq U \implies \map f T \subseteq \map f U$
Then $f$ has a greatest fixed point and a least fixed point.
== Proof ==
By Power Set is Complete Lattice, $\struct {\powerset S, \cap, \cup, \subseteq}$ is a complete lattice.
Thus the theorem holds by the Knaster-Tarski Lemma.
== Sources ==
* : Footnote $2$
* : Lemma $9.2.2$
Category:Knaster-Tarski Lemma
Category:Power Set
Knaster-Tarski Theorem

== Theorem ==
Let $\struct {L, \preceq}$ be a complete lattice.
Let $f: L \to L$ be an increasing mapping.
Let $F$ be the set (or class) of fixed points of $f$.
Then $\struct {F, \preceq}$ is a complete lattice.
== Proof ==
Let $S \subseteq F$.
Let $s = \bigvee S$ be the supremum of $S$.
We wish to show that there is an element of $F$ that succeeds all elements of $S$ and is the smallest element of $F$ to do so.
By the definition of supremum, an element succeeds all elements of $S$ it succeeds $s$.
Let $U = s^\succeq$ be the upper closure of $s$.
Thus we seek the smallest fixed point of $f$ that lies in $U$.
Note that $U = \closedint s \top$, the closed interval between $s$ and the top element of $L$.
First we show that $U$ is closed under $f$.
We have that:
:$\forall a \in S: a \preceq s$
so:
:$a = \map f a \preceq \map f s$
Thus $\map f s$ is an upper bound of $S$.
So by the definition of supremum, $s \preceq \map f s$.
Let $x \in U$.
Then $s \preceq x$.
So:
:$\map f s \preceq \map f x$
Since $s \preceq \map f s$, it follows that:
:$s \preceq \map f x$
so:
:$\map f x \in U$
Thus the restriction of $f$ to $U$ is an increasing mapping from $U$ to $U$.
By Closed Interval in Complete Lattice is Complete Lattice, $\struct {U, \preceq}$ is a complete lattice.
Thus by Knaster-Tarski Lemma, $f$ has a smallest fixed point in $U$.
Thus $S$ has a supremum in $F$.
A similar argument shows that $S$ has an infimum in $F$.
Since this holds for all $S \subseteq F$, it follows that $\struct {F, \preceq}$ is a complete lattice.
== Also see ==
* Knaster-Tarski Lemma
* Knaster-Tarski Lemma: Power Set form
== Historical Note ==
== Sources ==
* : Theorem $1$
Category:Complete Lattices
Category:Increasing Mappings
Koebe Quarter Theorem

== Theorem ==
Let $f: \C \to \C$ be a schlicht function, that is, a univalent complex function such that $\map f 0 = 0$ and $\map {f'} 0 = 1$.
Then the image of the unit disk contains the closed disk of radius $\dfrac 1 4$.
Hence for any $w \in f \sqbrk {\Bbb D}⁢$ we have that $\cmod w \le \dfrac 1 4$.
The constant $\dfrac 1 4$ is sharp and so cannot be improved.
== Proof ==
== Sources ==
* : $0,25$
Category:Complex Analysis
Kolmogorov's Law

== Theorem ==
Let $P$ be a population.
Let $P$ have mean $\mu$ and finite variance. 
Let $\sequence {X_n}_{n \mathop \ge 1}$ be a sequence of random variables forming a random sample from $P$.
Let: 
:$\ds {\overline X}_n = \frac 1 n \sum_{i \mathop = 1}^n X_i$
Then:
:$\ds {\overline X}_n \xrightarrow {\text {a.s.} } \mu$
where $\xrightarrow {\text {a.s.} }$ denotes almost sure convergence.
== Proof ==
We may assume that $X_n \ge 0$ for all $n \ge 1$.
Indeed, otherwise consider:
:${X_n}^+ := \max \set {X_n, 0}$
and:
:${X_n}^- := \max \set {- X_n, 0}$
instead of $X_n$.
Let $\varepsilon \in \R_{>0}$.
For $k \ge 1$ let:
:$\ell_k := \floor {\paren {1 + \epsilon}^k }$
be the floor of $\paren {1 + \epsilon}^k$.
Then:
Thus:
:$\ds \sum_{k \mathop = 1}^\infty \map \Pr {\size { {\overline X}_{\ell_k} - \mu} \ge \paren {1 + \epsilon}^{-k/4} } < \infty$
By Borel-Cantelli Lemma:
:$\ds \map \Pr { \limsup_{k \mathop \to \infty} \set {\size { {\overline X}_{\ell_k} - \mu} \ge \paren {1 + \epsilon}^{-k/4} } } = 0$
That is:
:$\ds \map \Pr { \liminf_{k \mathop \to \infty} \set {\size { {\overline X}_{\ell_k} - \mu} < \paren {1 + \epsilon}^{-k/4} } } = 1$
In particular:
:$\ds \map \Pr { \lim_{k \mathop \to \infty} \size { {\overline X}_{\ell_k} - \mu} = 0} = 1$
As:
:$\ds \lim_{k \mathop \to \infty} \frac {\ell_{k + 1} } {\ell_k} = 1 + \epsilon$
there exists an $k_0 \ge 1$ such that:
:$\forall k \ge k_0 : \ell_{k + 1} \le \paren {1 + 2 \epsilon} \ell_k$
For all $n > \ell_{k_0}$ we have:
:$\ds \paren {1 - 2 \epsilon} {\overline X}_{\ell_k} \le \frac 1 {1 + 2 \epsilon} {\overline X}_{\ell_k} \le {\overline X}_n \le \paren {1 + 2 \epsilon} {\overline X}_{\ell_{k + 1} }$
and therefore:
:$\ds \paren { {\overline X}_{\ell_k} - \mu} - 2 \epsilon {\overline X}_{\ell_k} \le {\overline X}_n - \mu \le \paren { {\overline X}_{\ell_{k + 1} } - \mu} + 2 \epsilon {\overline X}_{\ell_{k + 1} }$
where:
:$k := \max \set {i \ge 1 : n > \ell_i}$
As $k \to \infty$, when $n \to \infty$, we have:
:$\ds \set {\lim_{k \mathop \to \infty} \size { {\overline X}_{\ell_k} - \mu} = 0} \subseteq \set {\limsup_{n \mathop \to \infty} \size { {\overline X}_n - \mu} \le 2 \epsilon \mu}$
Finally, as $\epsilon \in \R_{>0}$ was arbitrary, we have:
Hence:
:$\ds \map \Pr {\lim_{n \mathop \to \infty} {\overline X}_n = \mu} = 1$
== Also known as ==
== Also see ==
* Bernoulli's Theorem, also known as the Law of Large Numbers
* Khinchin's Law, also known as the Weak Law of Large Numbers
== Sources ==
* 
* 
Category:Kolmogorov's Law
Category:Laws of Large Numbers
Category:Probability Theory
Kolmogorov-Sinai Entropy/Examples/Identity Mapping

== Example of Kolmogorov-Sinai Entropy ==
Let $\struct {X, \BB, \mu}$ be a probability space.
Let $I_X: X \to X$ be the identity mapping.
Then $I_X$ is $\mu$-preserving and:
:$ \map h {I_X} = 0$
where $\map h {I_X} $ is the entropy with respect to $I_X$.
== Proof ==
Category:Probability Theory
Category:Ergodic Theory
Category:Kolmogorov-Sinai Entropy
Koopman Operator is Isometry

== Theorem ==
Let $\struct {X, \BB, \mu, T}$ be a measure-preserving dynamical system.
Let $\map {L^2_\C} \mu$ be the complex-valued $L^2$ space of $\mu$.
Let $U_T : \map {L^2_\C} \mu \to \map {L^2_\C} \mu$ of $T$ be the Koopman operator.
Let $\innerprod \cdot \cdot$ denote the inner product on $\map {L^2_\C} \mu$, i.e.
:$\ds \forall f, g \in \map {L^2_\C} \mu : \innerprod f g := \int \overline f g \rd \mu$
where $\overline f$ denotes the complex conjugate of $f$.
Then $U_T$ is isometry on the $L^2$ Hilbert space $\struct {\map {L^2_\C} \mu, \innerprod \cdot \cdot}$.
That is:
:$\ds \forall f, g \in \map {L^2_\C} \mu : \innerprod {U_T f} {U_T g} = \innerprod f g$
== Proof ==
Let $f, g \in \map {L^2_\C} \mu $.
Then:
Category:Operator Theory
Korselt's Theorem

== Theorem ==
Let $n \ge 2$ be an integer.
Then $n$ is a Carmichael number :
: $(1): \quad n$ is odd
and the following conditions hold for every prime factor $p$ of $n$:
: $(2): \quad p^2 \nmid n$
: $(3): \quad \paren {p - 1} \divides \paren {n - 1}$
where:
:$\divides$ denotes divisibility
:$\nmid$ denotes non-divisibility.
== Proof ==
=== Sufficient Condition ===
Let $n$ be a Carmichael number:
:$(4): \quad \forall a \in \Z: a \perp n: a^n \equiv a \pmod n$
where $\perp$ denotes coprimality.
Suppose $n$ is even.
Set $a = -1$ in $(4)$.
Then $\paren {-1}^n = 1$ and so:
:$1 \equiv -1 \pmod n$
resulting in $n = 2$.
But as $2$ is not a Carmichael number, it follows that for $n$ to be Carmichael it must be odd.
Thus:
:$(1): \quad n$ is odd
holds.
From the Fundamental Theorem of Arithmetic, let $n$ be expressed as:
:$n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$
where:
:the $p_i$'s are distinct odd primes
:for all $i$, $k_i \ge 1$.
By Conditions for Integer to have Primitive Root there exists a primitive root $a_i$ modulo $p_i^{k_i}$ for each $i$.
In particular:
:$a_i \perp p_i$
By the Chinese Remainder Theorem:
:$\exists a: \forall i: a \equiv a_i \pmod {p_i^{k_i} }, a \perp n$
Consider a particular $i$.
We have:
We have that $a_i$ is a primitive root modulo $p_i^{k_i}$.
Thus:
But because $p_i \divides n$:
:$p_i \perp n - 1$
Thus:
:$k_i = 1$
giving that $n$ is square-free, and:
:$p_i - 1 \divides n - 1$
As $i$ is arbitrary, both conditions $(1)$ and $(2)$ of the statement of the theorem are thus fulfilled:
That is:
: $(1): \quad n$ is odd
and for all primes $p$ such that $p \divides n$:
: $(2): \quad p^2 \nmid n$
: $(3): \quad \paren {p - 1} \divides \paren {n - 1}$
=== Necessary Condition ===
Suppose $n$ is such that:
: $(1): \quad n$ is odd
and for all primes $p$ such that $p \divides n$:
: $(2): \quad p^2 \nmid n$
: $(3): \quad \paren {p - 1} \divides \paren {n - 1}$
Thus:
: $n = p_1 p_2 \cdots p_r$
is the product of $r$ distinct odd primes such that:
:$\paren {p_i - 1} \divides \paren {n - 1}$
for all $i$.
Let $a \in \Z$.
Let $a \perp p_i$.
Then by Fermat's Little Theorem:
:$a^{p_i - 1} \equiv 1 \pmod {p_i}$
Thus:
:$a^{n - 1} \equiv 1 \pmod {p_i}$
and so:
:$a^n \equiv a \pmod {p_i}$
Let $\gcd \set {a, p_i} > 1$.
Then as $p_i$ is prime:
:$p_i \divides a$
and so:
:$a^n \equiv a \equiv 0 \pmod {p_i}$
Thus for all $i$ and for all $a \in \Z$:
:$a^n \equiv a \pmod {p_i}$
From the Chinese Remainder Theorem:
:$a^n \equiv a \pmod n$
for all $a$.
Thus, by definition, $n$ is a Carmichael number.
== Also known as ==
Korselt's theorem is also seen referred to as Korselt's criterion.
== Historical Note ==
== Sources ==
* : $561$
* : $561$
Category:Carmichael Numbers
Korselt's Theorem/Historical Note

== Historical Note on Korselt's Theorem==
 first demonstrated this result in $1899$, but did not actually identify any integers that fulfilled the criterion.
That did not happen until found that $561$ was such a number in $1910$.
It turns out that had already found the first $7$ Carmichael numbers in $1885$, but his work was not noticed at the time.
Category:Historical Notes
Krattenthaler's Identity

== Theorem ==
:$\begin{vmatrix}
\paren {x + q_2} \paren {x + q_3} & \paren {x + p_1} \paren {x + q_3} & \paren {x + p_1} \paren {x + p_2} \\
\paren {y + q_2} \paren {y + q_3} & \paren {y + p_1} \paren {y + q_3} & \paren {y + p_1} \paren {y + p_2} \\
\paren {z + q_2} \paren {z + q_3} & \paren {z + p_1} \paren {z + q_3} & \paren {z + p_1} \paren {z + p_2}
\end{vmatrix} = \paren {x - y} \paren {x - z} \paren {y - z} \paren {p_1 - q_2} \paren {p_1 - q_3} \paren {p_2 - q_3}$
where $\size {\, \cdot \,}$ denotes determinant.
== Proof ==
== Sources ==
* 
* : $\S 1.2.3$: Sums and Products: Exercise $47$
Category:Determinants
Krein-Milman Theorem

== Theorem ==
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\R$.
Let $K$ be a non-empty compact convex subset of $X$.
Let $\map E K$ be the set of extreme points in $K$. 
Then $K$ is the closed convex hull of $\map E K$.
== Proof ==
Let $K'$ be the closed convex hull of $\map E K$.
We show that $K' = K$.
From Convex Hull is Smallest Convex Set containing Set, we have: 
:$\map E K \subseteq \map {\operatorname {conv} } {\map E K}$
From Set is Subset of its Topological Closure, we therefore have: 
:$\map E K \subseteq K'$
We also have, from Convex Hull preserves Subsets:
:$\map {\operatorname {conv} } {\map E K} \subseteq \map {\operatorname {conv} } K$ 
From Convex Hull is Smallest Convex Set containing Set: Corollary, we have: 
:$\map {\operatorname {conv} } {\map E K} \subseteq K$ 
Then, from Set Closure Preserves Set Inclusion, we have: 
:$K' \subseteq K^-$
where $K^-$ is the closure of $K$.
From Compact Subspace of Hausdorff Space is Closed, we have that: 
:$K$ is closed.
So we have: 
:$K' \subseteq K$ 
 suppose that $K' \ne K$.
Then there exists $a \in K \setminus K'$. 
Note that $K'$ is closed from Topological Closure is Closed.
Since $K$ is compact and $K' \subseteq K$, we have: 
:$K'$ is compact
from Closed Subspace of Compact Space is Compact. 
From Closed Convex Hull in Normed Vector Space is Convex, we have that: 
:$K'$ is convex.
Finally:
:$\set a$ is closed
from Singleton in Normed Vector Space is Closed, and:
:$\set a$ is convex
from Singleton is Convex Set.
So, from Hahn-Banach Separation Theorem in the case of a compact convex set and a closed convex set, there exists $f \in X^\ast$ such that: 
:$\map f x < \map f a$
for each $x \in K'$. 
Now let: 
:$\ds K^f = \set {x \in K : \map f x = \max_{y \mathop \in K} \map f y}$ 
From Preimage of Maximum of Bounded Linear Functional on Extreme Set in Convex Compact Set is Extreme Set:
:$K^f$ is an extreme set in $K$. 
Since: 
:$\map f x < \map f a$ for each $x \in K'$
and $a \in K$, we have: 
:$\ds \map f x < \max_{y \mathop \in K} \map f y$ for each $x \in K'$
from the definition of maximum.
So, we have: 
:$K^f \cap K' = \O$ 
However, from Extreme Set in Compact Convex Set contains Extreme Point, we also have: 
:$K^f$ contains an extreme point of $K$, say $x \in \map E K$
while: 
:$\map E K \subseteq K$ 
So we have: 
:$x \in K^f \cap K'$
contradicting that $K^f \cap K$ is empty.
So, there exists no such $a \in K \setminus K'$, and we have: 
:$K = K'$
== Sources ==
* $21.6$: The Krein-Milman Theorem
Category:Functional Analysis
Category:Closed Convex Hulls
Kronecker's Lemma

== Theorem ==
Let $\sequence {x_n}$ be an infinite sequence of real numbers such that:
:$\ds \sum_{n \mathop = 1}^\infty x_n = s$ 
exists and is finite.
Then for $0 < b_1 \le b_2 \le b_3 \le \ldots$ and $b_n \to \infty$: 
:$\ds \lim_{n \mathop \to \infty} \frac 1 {b_n} \sum_{k \mathop = 1}^n b_k x_k = 0$
== Proof ==
Let $S_k$ denote the partial sums of the $x$s.
Using Summation by Parts:
:$\ds \frac 1 {b_n} \sum_{k \mathop = 1}^n b_k x_k = S_n - \frac 1 {b_n} \sum_{k \mathop = 1}^{n - 1} \paren {b_{k + 1} - b_k} S_k$
Now, pick any $\epsilon \in \R_{>0}$.
Choose $N$ such that $S_k$ is $\epsilon$-close to $s$ for $k > N$.
This can be done, as the sequence $S_k$ converges to $s$.
Then the is:
Now, let $n \to \infty$.
The first term goes to $s$, which cancels with the third term.
The second term goes to zero (as the sum is a fixed value).
Since the $b$ sequence is increasing, the last term is bounded by $\epsilon \dfrac {b_n - b_N} {b_n} \le \epsilon$.
== Sources ==
* 
Category:Real Analysis