nyuuzyou/wikis huggingface dataset:
Component of Vector is Scalar Projection on Standard Ordered Basis Element

== Theorem ==
Let $\tuple {\mathbf e_1, \mathbf e_2, \mathbf e_3}$ be the standard ordered basis of Cartesian $3$-space $S$.
Let $\mathbf a = a_1 \mathbf e_1 + a_2 \mathbf e_2 + a_3 \mathbf e_3$ be a vector quantity in $S$.
Then:
:$\mathbf a \cdot \mathbf e_i = a_i$
== Proof ==
Using the Einstein summation convention
== Sources ==
* : Volume One: Chapter $1$ Vectors in Classical Physics: $1.3$ The Scalar Product
Category:Scalar Projections
Components are Open iff Union of Open Connected Sets

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
:$(1): \quad$ The components of $T$ are open.
:$(2): \quad S$ is a union of open connected sets of $T$.
== Proof ==
=== Condition (1) implies Condition (2) ===
=== Condition (2) implies Condition (1) ===
== Also see ==
* Path Components are Open iff Union of Open Path-Connected Sets, an analogous result for path components
Category:Connected Spaces
Category:Components
Components are Open iff Union of Open Connected Sets/Components are Open implies Space is Union of Open Connected Sets

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let the components of $T$ be open sets.
Then:
:$S$ is a union of open connected sets of $T$.
== Proof ==
Let the components of $T$ be open.
By definition, the components of $T$ are a partition of $S$.
Hence $S$ is the union of the open components of $T$.
Since a component is a maximal connected set by definition, then $S$ is a union of open connected sets of $T$.
Category:Connected Spaces
Category:Components
Components are Open iff Union of Open Connected Sets/Lemma 1

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $C$ be a component of $T$.
Let $U$ be a connected set of $T$.
Then:
::$U \cap C \ne \O$ $U \ne \O$ and $U \subseteq C$
== Proof ==
=== Necessary Condition ===
Let $U \cap C \ne \O$.
From Union of Connected Sets with Common Point is Connected, $U \cup C$ is a connected set of $T$.
From Set is Subset of Union, $C \subseteq U \cup C$.
By definition of a component, $C$ is a maximal connected set.
Hence $C = U \cup C$.
From Union with Superset is Superset, $U \subseteq C$.
Then:
The result follows.
=== Sufficient Condition ===
Let $U \ne \O$ and $U \subseteq C$.
Then:
The result follows.
Category:Connected Spaces
Category:Components
Components are Open iff Union of Open Connected Sets/Space is Union of Open Connected Sets implies Components are Open

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $S$ be a union of open connected sets of $T$.
Then:
:The components of $T$ are open sets.
== Proof ==
Let $S = \ds \bigcup \set {U \subseteq S : U \in \tau \text { and } U \text { is connected} }$.
Let $C$ be a component of $T$.
=== Lemma ===
:For any connected set $U$ then:
Then:
 
Hence $C$ is the union of open sets.
By definition of a topology then $C$ is an open set.
The result follows.
Category:Connected Spaces
Category:Components
Components of Integer Reciprocal Space with Zero are Single Points

== Theorem ==
Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
:$A := \set 0 \cup \set {\dfrac 1 n : n \in \Z_{>0} }$
Let $\struct {A, \tau_d}$ be the integer reciprocal space with zero under the usual (Euclidean) topology.
Then the components of $A$ are singletons.
== Proof ==
From Integer Reciprocal Space with Zero is Totally Separated, $A$ is totally separated.
From Totally Separated Space is Totally Disconnected, $A$ is totally disconnected. 
The result follows by definition of totally disconnected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $2 \ \text{(c)}$
Category:Integer Reciprocal Space
Components of Metric at Origin of Normal Neighborhood of Pseudo-Riemannian Neighborhood

== Theorem ==
Let $\struct {M, g}$ be an $n$-dimensional Riemannian manifold.
Let $U_p$ be the normal neighborhood for $p \in M$.
Let $\struct {U_p, \tuple {x^i}}$ be a normal coordinate chart.
Suppose $g_{ij}$ are the components of metric $g$ in coordinates $\tuple {x^i}$ at $p \in M$.
Then:
:$g_{ij} = \pm_{ij} \delta_{ij}$
where $\delta_{ij}$ denotes the Kronecker delta, and $\pm_{ij}$ is a sign that depends on $i$ and $j$.
== Proof ==
== Sources ==
* : $\S 5$: The Levi-Civita Connection. Normal Neighborhoods and Normal Coordinates
Category:Normal Neighborhoods
Category:Pseudo-Riemannian Metrics
Components of Metric at Origin of Normal Neighborhood of Riemannian Neighborhood

== Theorem ==
Let $\struct {M, g}$ be an $n$-dimensional Riemannian manifold.
Let $U_p$ be the normal neighborhood for $p \in M$.
Let $\struct {U_p, \tuple {x^i}}$ be a normal coordinate chart.
Suppose $g_{ij}$ are the components of metric $g$ in coordinates $\tuple {x^i}$ at $p \in M$.
Then:
:$g_{ij} = \delta_{ij}$
where $\delta_{ij}$ denotes the Kronecker delta.
== Proof ==
== Sources ==
* : $\S 5$: The Levi-Civita Connection. Normal Neighborhoods and Normal Coordinates
Category:Normal Neighborhoods
Category:Riemannian Metrics
Components of Separation are Clopen

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $A \mid B$ be a separation of $T$.
Then both $A$ and $B$ are clopen in $T$.
== Proof ==
From Set with Relative Complement forms Partition:
:$A = \complement_S \left({B}\right)$
and:
:$B = \complement_S \left({A}\right)$
where $\complement_S$ denotes the complement relative to $S$.
As $A$ and $B$ are both open, it follows by definition that $A$ and $B$ are also both closed.
That is, by definition, they are clopen.
== Sources ==
* : $6.2$: Connectedness: Definition $6.2.2$
Category:Disconnected Spaces
Components of Separation are Separated Sets

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A \mid B$ be a separation of $T$.
Then $A$ and $B$ are separated sets of $T$.
== Proof ==
By definition of closure, $A^-$ is the smallest closed set of $T$ that contains $A$.
Components of Separation are Clopen shows that $A$ and $B$ are closed.
It follows that $A^- = A$, and $B^- = B$.
Definition of separation shows that $A \cap B = \O$, so we have:
Hence the result by definition of separated sets.
Category:Separated Sets
Category:Separations
Components of Vector between two Points

== Theorem ==
Let $A, B$ be points in the Euclidean space $\R^n$.
Let their Cartesian coordinates be given by:
Let $\vec {AB}$ be the vector quantity that represents the directed line segment with startpoint $A$ and endpoint $B$.
Then the vector components of $\vec {AB}$ are:
:$\vec {AB} = \tuple {b_1 - a_1, b_2 - a_2, \ldots, b_n - a_n}$
== Proof ==
Let $O$ denote the origin of $\R^n$.
Let $\vec {OA}$ and $\vec {OB}$ be the positions vectors of $A$ and $B$.
By definition of positions vectors, their components are:
for some $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \in \R$.
Let the components of $\vec {AB}$ be denoted as:
:$\vec {AB} := \tuple {v_1 , v_2, \ldots, v_n}$
By the definition of Triangle Law of Vector Addition, it follows that $\vec {OB}$ can be represented as:
:$\vec {OB} = \vec{OA} + \vec{AB}$
From Equivalence of Definitions of Vector Sum, it follows that:
:$\tuple {b_1 , b_2, \ldots, b_n} = \tuple {a_1 , a_2, \ldots, a_n} + \tuple {v_1 , v_2, \ldots, v_n}$
By the definition of vector addition of components, it follows that:
:$\tuple {b_1 - a_1, b_2 - a_2, \ldots, b_n - a_n} = \tuple {v_1 , v_2, \ldots, v_n}$
Hence the result.
Category:Vectors
Components of Vector in Terms of Dot Product

== Theorem ==
Let $\mathbf a$ be a vector quantity embedded in Cartesian $3$-space $S$.
Let $\mathbf i$, $\mathbf j$ and $\mathbf k$ be the unit vectors in the positive directions of the $x$-axis, $y$-axis and $z$-axis respectively.
Let $\mathbf a$ be expressed in component form:
:$\mathbf a = x \mathbf i + y \mathbf j + z \mathbf k$
Then:
== Proof ==
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $2$. The Scalar Product: $(2.9)$
Category:Dot Product
Components of Vector in terms of Direction Cosines

== Theorem ==
Let $\mathbf r$ be a vector quantity embedded in a Cartesian $3$-space.
Let $\mathbf i$, $\mathbf j$ and $\mathbf k$ be the unit vectors in the positive directions of the $x$-axis, $y$-axis and $z$-axis respectively.
Let $\cos \alpha$, $\cos \beta$ and $\cos \gamma$ be the direction cosines of $\mathbf r$ with respect to the $x$-axis, $y$-axis and $z$-axis respectively.
Let $x$, $y$ and $z$ be the components of $\mathbf r$ in the $\mathbf i$, $\mathbf j$ and $\mathbf k$ directions respectively.
Let $r$ denote the magnitude of $\mathbf r$, that is:
:$r := \size {\mathbf r}$
Then:
== Proof ==
:
By definition, the direction cosines are the cosines of the angles that $\mathbf r$ makes with the coordinate axes.
By definition of the components of $\mathbf r$:
:$\mathbf r = x \mathbf i + y \mathbf j + z \mathbf k$
Thus:
:$\mathbf r = r \cos \alpha \mathbf i + r \cos \beta \mathbf j + r \cos \gamma \mathbf k$
and the result follows.
== Also presented as ==
== Sources ==
* : Chapter $\text I$. Addition and Subtraction of Vectors. Centroids: Components of a Vector: $7$. The unit vectors $\mathbf i$, $\mathbf j$, $\mathbf k$
* : Chapter $1$ Vector Analysis $1.1$ Definitions, Elementary Approach: $(1.4)$
Category:Components of Vector in terms of Direction Cosines
Category:Direction Cosines
Components of Zero Vector Quantity are Zero

== Theorem ==
Let $\mathbf r$ be a vector quantity embedded in a Cartesian $3$-space.
Let $\mathbf r$ be expressed in terms of its components:
:$\mathbf r = x \mathbf i + y \mathbf j + z \mathbf k$
Let $\mathbf r$ be the zero vector.
Then:
:$x = y = z = 0$
== Proof ==
By definition of the zero vector, the magnitude of $\mathbf r$ is equal to zero.
By Magnitude of Vector Quantity in terms of Components:
:$\size {\mathbf r} = \sqrt {x^2 + y^2 + z^2} = 0$
where $\size {\mathbf r}$ denotes the magnitude of $\mathbf r$.
As each of $x$, $y$ and $z$ are real numbers, each of $x^2$, $y^2$ and $z^2$ is non-negative.
so in order for $\sqrt {x^2 + y^2 + z^2} = 0$, it must follow that each of $x$, $y$ and $z$ is zero.
Hence the result.
== Sources ==
* : Chapter $1$ Vector Analysis $1.1$ Definitions, Elementary Approach
Category:Zero Vectors
Composite Fibonacci Numbers with Prime Index

== Theorem ==
The sequence of composite Fibonacci numbers with a prime index begins:
:$4181, 1 \, 346 \, 269, 24 \, 157 \, 817, 165 \, 580 \, 141, \ldots$
The corresponding sequence of prime indices begins:
:$19, 31, 37, 41, 53, 59, 61, 67, 71, 73, 79, \ldots$
== Proof ==
By observation:
== Sources ==
* : $4181$
* : $4181$
Category:Fibonacci Numbers
Category:Prime Numbers
Composite Frame Homomorphism is Frame Homomorphism

== Theorem ==
Let $L_1 = \struct{S_1, \preceq_1}$, $L_2 = \struct{S_2, \preceq_2}$ and $L_3 = \struct{S_3, \preceq_3}$ be frames.
Let $\phi_1: L_1 \to L_2$ and $\phi_2: L_2 \to L_3$ be frame homomorphisms.
Let $\phi_2 \circ \phi_1 : S_1 \to S_3$ be the composite mapping of $\phi_1$ and $\phi_2$ 
Then:
:$\phi_2 \circ \phi_1$ is a frame homomorphism of $L_1$ to $L_3$
== Proof ==
=== $\phi_2 \circ \phi_1$ is Finite Meet Preserving ===
Let $F \subseteq S_1$ be a finite subset.
We have:
Since $F$ was arbitrary, it follows that $\phi_2 \circ \phi_1$ is finite meet preserving by definition.
=== $\phi_2 \circ \phi_1$ is Arbitrary Join Preserving ===
Let $A \subseteq S_1$ be any subset of $S$.
We have:
Since $A$ was arbitrary, it follows that $\phi_2 \circ \phi_1$ is arbitrary join preserving by definition.
By definition, $\phi_2 \circ \phi_1$ is a frame homomorphism
Category:Frame Homomorphisms
Composite Functor is Functor

== Theorem ==
Let $\mathbf C, \mathbf D$ and $\mathbf E$ be metacategories.
Let $F: \mathbf C \to \mathbf D$ and $G: \mathbf D \to \mathbf E$ be covariant functors.
Let $GF: \mathbf C \to \mathbf E$ be the composition of $G$ with $F$.
Then $G F$ is also a covariant functor.
== Proof ==
Let $f, g$ be morphisms of $\mathbf C$ such that $g \circ f$ is defined.
Then:
Also, for any object $C$ of $\mathbf C$:
Hence $G F$ is shown to be a covariant functor.
Category:Category Theory
Category:Functors
Composite Hypothesis Test/Examples/Arbitrary Example 1

== Example of Composite Hypothesis Test ==
Let $\delta$ be a hypothesis test.
Let $H_0$ be the null hypothesis that the probability distribution $X$ of $\delta$ is:
:$X$ is normally distributed with an expectation of $2$, but whose variance is unknown.
Then $\delta$ is a composite hypothesis test.
== Sources ==
* 
* 
Category:Examples of Composite Hypothesis Tests
Composite Lattice Homomorphisms is Lattice Homomorphism

== Theorem ==
Let $L_1 = \struct{A_1, \vee_1, \wedge_1, \preceq_1}$, $L_2 = \struct{A_2, \vee_2, \wedge_2, \preceq_2}$ and $L_3 = \struct{A_3, \vee_3, \wedge_3, \preceq_3}$ be lattices.
Let $\phi_1: L_1 \to L_2$ and $\phi_2: L_2 \to L_3$ be lattice homomorphisms.
Let $\phi_2 \circ \phi_1 : A_1 \to A_3$ be the composite mapping of $\phi_1$ and $\phi_2$ 
Then:
:$\phi_2 \circ \phi_1$ is a lattice homomorphism of $L_1$ to $L_3$
== Proof ==
Follows immediately from Composite of Homomorphisms is Homomorphism
Category:Lattice Homomorphisms
Composite Localic Mapping is Localic Mapping

== Theorem ==
Let $L_1 = \struct{S_1, \preceq_1}, L_2 = \struct{S_2, \preceq_2}$ and $L_3 = \struct{S_3, \preceq_3}$ be locales.
Let $f_1 : L_1 \to L_2$ and $f_2 : L_2 \to L_3$ be localic mappings.
Then:
:the composite $f_2 \circ f_1 : L_1 \to L_3$ is a localic mapping.
== Proof ==
By definition of a localic mapping:
:$f_1, f_2$ are the upper adjoint of Galois connections $\tuple{f_1, d_1}, \tuple{f_2, d_2}$ respectively
where $d_1 : L_2 \to L_1$ and $d_2 : L_3 \to L_2$ are frame homomorphsims.
From Composite Frame Homomorphism is Frame Homomorphism:
:$d_1 \circ d_2 : L_3 \to L_2$ is a frame homomorphsims
From Composition of Galois Connections is Galois Connection:
:$\tuple {f_2 \circ f_1, d_1 \circ d_2}$ is a Galois connection
Hence:
:$f_2 \circ f_1$ is the upper adjoint of a Galois connection
By definition of localic mapping:
:$f_2 \circ f_1$ is a localic mapping
== Sources ==
* : Chapter II: Frames and Locales. Spectra, $\S 2.2$
Category:Localic Mappings
Composite Mapping is Mapping

== Theorem ==
Let $S_1, S_2, S_3$ be sets.
Let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be mappings.
Then the composite mapping $g \circ f$ is also a mapping.
== Proof ==
The composite of $f$ and $g$ is defined as:
:$g \circ f := \set {\tuple {x, z} \in S_1 \times S_3: \exists y \in S_2: \tuple {x, y} \in f \land \tuple {y, z} \in g}$
It is necessary to show that $g \circ f$ is both left-total and many-to-one.
=== Left-total ===
From the definition of a mapping:
That is, $g \circ f$ is left-total.
=== Many-to-one ===
Suppose $x_1, x_2 \in S_1: x_1 = x_2$.
Then as $f$ is a mapping and so by definition many-to-one:
:$\map f {x_1} = \map f {x_2}$
and so:
:$\map g {\map f {x_1} } = \map g {\map f {x_2} }$
demonstrating that $g \circ f$ is similarly many-to-one.
== Sources ==
* : Chapter $0$: Functions
* : Appendix $\text{A}.4$: Functions
Category:Composite Mappings
Composite Mersenne Number/Examples

== Examples of Composite Mersenne Numbers ==
=== Mersenne Number $M_{67}$ ===
=== Mersenne Number $M_{157}$ ===
=== Mersenne Number $M_{167}$ ===
=== Mersenne Number $M_{193}$ ===
=== Mersenne Number $M_{199}$ ===
=== Mersenne Number $M_{227}$ ===
=== Mersenne Number $M_{229}$ ===
=== Mersenne Number $M_{257}$ ===
Category:Mersenne Numbers
Composite Mersenne Number/Examples/M157/Historical Note

== Historical Note on Mersenne Number $M_{157}$ ==
Mersenne number $M_{167}$ was one of a set of $6$ demonstrated to be composite by using a manual desk calculator in the $1940$s, in what turned out to be a vain attempt to find the next Mersenne prime after $M_{127}$.
== Sources ==
* : $2^{229} - 1$
* : $2^{229} - 1$
Category:Historical Notes
Composite Mersenne Number/Examples/M167/Historical Note

== Historical Note on Mersenne Number $M_{167}$ ==
Mersenne number $M_{167}$ was one of a set of $6$ demonstrated to be composite by using a manual desk calculator in the $1940$s, in what turned out to be a vain attempt to find the next Mersenne prime after $M_{127}$.
== Sources ==
* : $2^{229} - 1$
* : $2^{229} - 1$
Category:Historical Notes
Composite Mersenne Number/Examples/M193/Historical Note

== Historical Note on Mersenne Number $M_{193}$ ==
Mersenne number $M_{193}$ was one of a set of $6$ demonstrated to be composite by using a manual desk calculator in the $1940$s, in what turned out to be a vain attempt to find the next Mersenne prime after $M_{127}$.
== Sources ==
* : $2^{229} - 1$
* : $2^{229} - 1$
Category:Historical Notes
Composite Mersenne Number/Examples/M199/Historical Note

== Historical Note on Mersenne Number $M_{199}$ ==
Mersenne number $M_{199}$ was one of a set of $6$ demonstrated to be composite by using a manual desk calculator in the $1940$s, in what turned out to be a vain attempt to find the next Mersenne prime after $M_{127}$.
== Sources ==
* : $2^{229} - 1$
* : $2^{229} - 1$
Category:Historical Notes
Composite Mersenne Number/Examples/M227/Historical Note

== Historical Note on Mersenne Number $M_{227}$ ==
Mersenne number $M_{227}$ was one of a set of $6$ demonstrated to be composite by using a manual desk calculator in the $1940$s.
== Sources ==
* : $2^{229} - 1$
* : $2^{229} - 1$
Category:Historical Notes
Composite Mersenne Number/Examples/M229/Historical Note

== Historical Note on Mersenne Number $M_{229}$ ==
Mersenne number $M_{229}$ was the largest of a set of $6$ demonstrated to be composite by using a manual desk calculator in the $1940$s.
The next Mersenne prime after $M_{127}$, proved prime in $1896$ by , does not occur until $M_{521}$.
As remarked, nobody had any idea that the next one would be so far away.
It is instructive to note that, in $2017$, it took a freely-available online factorization tool $4 \cdotp 8$ seconds to perform this exact calculation.
== Sources ==
* : $2^{229} - 1$
* : $2^{229} - 1$
Category:Historical Notes
Composite Mersenne Number/Examples/M257/Historical Note

== Historical Note on Mersenne Number $M_{257}$ ==
Mersenne number $M_{257}$ was the largest Mersenne number conjectured by to be prime, in his of $1644$.
In $1922$, demonstrated that it was composite, but did not find any actual factors.
Its full factorisation was eventually achieved by and .
It is instructive to note that, in $2017$, it took a freely-available online factorization tool $14 \cdotp 8$ seconds to perform this exact calculation.
== Sources ==
* : $2^{257} - 1$
* : $2^{257} - 1$
Category:Historical Notes
Composite Number has Prime Factor

== Theorem ==
Let $a$ be a composite number.
Then there exists a prime number $p$ such that:
:$p \divides a$
where $\divides$ means is a divisor of.
== Proof ==
By definition of composite number:
:$\exists b \in \Z: b \divides a$
If $b$ is a prime number then the proof is complete.
Otherwise $b$ is composite.
By definition of composite number:
:$\exists c \in \Z: c \divides b$
and so:
:$c \divides a$
Again, if $c$ is a prime number then the proof is complete.
Continuing in this manner, some prime number will be found which will divide the number before it.
This will be a divisor of $a$.
Because if not, then an infinite series of numbers will be divisors of $a$, each of which is less than the other.
This is impossible in numbers.
Hence the result.
== Sources ==
* : Book $\text{VII}$. Propositions
* : Chapter $7$: Patterns in Numbers: Euclid
Category:Prime Numbers
Composite Number has Prime Factor not Greater Than its Square Root

== Theorem ==
Let $n \in \N$ and $n = p_1 \times p_2 \times \cdots \times p_j$, $j \ge 2$, where $p_1, \ldots, p_j \in \Bbb P$ are prime factors of $n$.
Then $\exists p_i \in \Bbb P$ such that $p_i \le \sqrt n$.
That is, if $n \in \N$ is composite, then $n$ has a prime factor $p \le \sqrt n$. 
== Proof ==
Let $n$ be composite such that $n \ge 0$.
From Composite Number has Two Divisors Less Than It, we can write $n = a b$ where $a, b \in \Z$ and $1 < a, b < n$. 
, suppose that $a \le b$.
Let $a > \sqrt n$.
Then $b \ge a > \sqrt n$. 
However, if $b \ge a > \sqrt n$ is true, then:
:$n = a b > \sqrt n \sqrt n > n$
This is clearly a contradiction.
So:
:$a \le \sqrt n$
From Positive Integer Greater than 1 has Prime Divisor it follows that there is some prime $p$ which divides $a$.
From Absolute Value of Integer is not less than Divisors, we have that $p \le a$ and so:
:$p \le \sqrt n$
From Divisor Relation on Positive Integers is Partial Ordering:
: $p \divides n$
Hence the result.
== Comment ==
Suppose we are testing a number to see whether it is prime, or so as to find all its divisors.
One way to do this (which may not be the most efficient in all circumstances, but it gets the job done) is to divide it by successively larger primes until you find one that is a factor of the number.
Eventually you're bound to find a prime that is a factor, by Positive Integer Greater than 1 has Prime Divisor.
However, this result tells us that we don't need to go out that far.
If we've tested all the primes up to the square root of our target number without finding a divisor, we don't need to go any further because we know that our target number is prime after all.
== Sources ==
* : Chapter $\text{IV}$: The Prince of Amateurs
* : $\text I$: The Series of Primes: $1.2$ Prime numbers
* : Chapter $2$: Integers and natural numbers: $\S 2.2$: Divisibility and factorization in $\mathbf Z$: Exercise $2$
Category:Prime Numbers
Composite Number has Two Divisors Less Than It

== Theorem ==
Let $n \in \Z_{> 1}$ such that $n \notin \mathbb P$.
Then:
:$\exists a, b \in \Z: 1 < a < n, 1 < b < n: n = a b$
That is, a non-prime number greater than $1$ can be expressed as the product of two positive integers strictly greater than $1$ and less than $n$.
Note that these two numbers are not necessarily distinct.
== Proof ==
Since $n \notin \mathbb P$, it has a positive factor $a$ such that $a \ne 1$ and $a \ne n$.
Hence $\exists b \in \Z: n = a b$.
Thus by definition of factor:
:$a \divides n$
where $\divides$ denotes divisibility.
From Divisor Relation on Positive Integers is Partial Ordering:
:$a \le n$
As $a \ne n$, it follows that $a < n$.
From One Divides all Integers:
:$1 \divides a$
Thus from Divisor Relation on Positive Integers is Partial Ordering:
:$1 \le a$
Similarly, as $1 \ne a$ it follows that $1 < a$.
Since $a \ne n$, it follows that $b \ne 1$.
Similarly, since $a \ne 1$, it follows that $b \ne n$.
Thus:
:$b \divides n: 1 \ne b \ne n$
Arguing as above, we show that $1 < b < n$ and the result follows.
Note that we have not shown (and it is not necessarily the case) that $a \ne b$.
Category:Divisors
Composite Semilattice Homomorphisms is Semilattice Homomorphism

== Theorem ==
Let $S_1 = \struct{A_1, \circ_1}$, $S_2 = \struct{A_2, \circ_2}$ and $S_3 = \struct{A_3, \circ_3}$ be semilattices.
Let $\phi_1: S_1 \to S_2$ and $\phi_2: S_2 \to S_3$ be semilattice homomorphisms.
Let $\phi_2 \circ \phi_1 : A_1 \to A_3$ be the composite mapping of $\phi_1$ and $\phi_2$ 
Then:
:$\phi_2 \circ \phi_1$ is a semilattice homomorphism of $S_1$ to $S_3$
== Proof ==
Follows immediately from Composite of Homomorphisms is Homomorphism
Category:Semilattice Homomorphisms
Composite of Antisymmetric Relations is not necessarily Antisymmetric

== Theorem ==
Let $A$ be a set.
Let $\RR$ and $\SS$ be antisymmetric relations on $A$.
Then their composite $\RR \circ \SS$ is not necessarily also antisymmetric.
== Proof ==
;Proof by Counterexample:
Consider the ordering $\le$ on the natural numbers $\N$.
Consider its dual ordering $\ge$ also on $\N$.
Note that Dual Ordering is Ordering.
Both $\le$ and $\ge$ are antisymmetric relations.
We have:
:$1 \le 3$
:$3 \ge 2$
and similarly:
:$2 \le 3$
:$3 \ge 1$
Hence it follows that:
:$1 \le \circ \ge 2$
while at the same time:
:$2 \le \circ \ge 1$
and so while both $\le$ and $\ge$ are antisymmetric, their composite $\le \circ \ge$ is not.
Hence the result.
Category:Antisymmetric Relations
Category:Composite Relations