nyuuzyou/wikis huggingface dataset:
Identity of Monoid is Unique

== Theorem ==
Let $\struct {S, \circ}$ be a monoid.
Then $S$ has a unique identity.
== Proof ==
As $\struct {S, \circ}$ is an algebraic structure, the result Identity is Unique can be applied directly.
== Also see ==
* Inverse in Monoid is Unique
* Identity of Group is Unique
== Sources ==
* : Chapter $2$: The Definition of Group Structure: $\S 26 \beta$
* : $2$: Examples of Groups and Homomorphisms: $2.2$ Definitions $\text{(i)}$
* : $\S 31.1$ Identity element and inverses
* : Chapter $1$: Basic Definitions and Results
Category:Monoids
Identity of Points

== Theorem ==
Two points share the same position they are the same points.
== Proof ==
Let $a$ be a point with position $P_1$.
Let $b$ be a point with position $P_2$.
By hypothesis, $P_1 = P_2$.
By Leibniz's Law, two objects are the same object they share every property in common.
By the definition of point, the only property possessed by a point is position.
We have:
:$P_1 = P_2 \dashv \vdash a = b$
Hence the result.
 
== Also see ==
* 
Category:Euclidean Geometry
Identity of Power Set with Intersection

== Theorem ==
Let $S$ be a set and let $\powerset S$ be its power set.
Consider the algebraic structure $\struct {\powerset S, \cap}$, where $\cap$ denotes set intersection.
Then $S$ serves as the identity for $\struct {\powerset S, \cap}$.
== Proof ==
We note that by Set is Subset of Itself, $S \subseteq S$ and so $S \in \powerset S$ from the definition of the power set.
From Intersection with Subset is Subset‎, we have:
:$A \subseteq S \iff A \cap S = A = S \cap A$
By definition of power set:
:$A \subseteq S \iff A \in \powerset S$
So:
:$\forall A \in \powerset S: A \cap S = A = S \cap A$
Thus we see that $S$ acts as the identity element.
== Sources ==
* : Exercise $1.4: \ 9$
* : $\S 4.3$. Units and zeros: Example $75$
* : Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses
* : $\S 2$: Sets and functions: Operations
Category:Set Intersection
Category:Power Set
Category:Examples of Identity Elements
Identity of Power Set with Union

== Theorem ==
Let $S$ be a set and let $\powerset S$ be its power set.
Consider the algebraic structure $\struct {\powerset S, \cup}$, where $\cup$ denotes set union.
Then the empty set $\O$ serves as the identity for $\struct {\powerset S, \cup}$.
== Proof ==
From Empty Set is Element of Power Set:
:$\O \in \powerset S$
From Union with Empty Set:
:$\forall A \subseteq S: A \cup \O = A = \O \cup A$
By definition of power set:
:$A \subseteq S \iff A \in \powerset S$
So:
:$\forall A \in \powerset S: A \cup \O = A = \O \cup A$
Thus we see that $\O$ acts as the identity for $\struct {\powerset S, \cup}$.
== Sources ==
* : Exercise $1.4: \ 9$
* : $\S 4.3$. Units and zeros: Example $75$
* : Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses
Category:Set Union
Category:Power Set
Category:Empty Set
Category:Examples of Identity Elements
Identity of Subgroup

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $H$ be a subgroup of group $G$.
Then the identity of $H$ is also $e$.
== Proof ==
A group is a monoid.
From the Cancellation Laws, all of its elements are cancellable.
The result then follows from Identity of Cancellable Monoid is Identity of Submonoid.
== Also see ==
* Identity of Submonoid is not necessarily Identity of Monoid
== Sources ==
* : $\S 5.2$. Subgroups
* : Chapter $\text{II}$: Groups: Subgroups
* : $\S 36.2 \ \text{(i)}$: Subgroups
* : Chapter $4$: Subgroups: Proposition $4.2$: Remark $2$
* 
* 
Category:Subgroups
Category:Identity Elements
Identity of Subgroup of Dipper Semigroup is not Identity of Dipper

== Theorem ==
Let $m, n \in \Z$ be integers such that $m, n > 0$.
Let $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$ denote the dipper semigroup.
Let $\struct {H, +_{m, n} }$ be the subgroup of $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$ where $H = \set {k \in \N: m \le k < m + n}$
Then the identity of $\struct {H, +_{m, n} }$ is not the identity of $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$.
== Proof ==
This is demonstrated by Proof by Counterexample.
First we note that by Existence of Subgroup of Dipper Semigroup:
:$\struct {H, +_{m, n} }$ is indeed a subgroup of $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$
:the identity of $\struct {H, +_{m, n} }$ is $n$.
But we note that by definition of $+_{m, n}$:
:$0 +_{m, n} n = n - k n$
where $m \le n - k n$
As we have specified that $m > 0$, it follows that:
:$0 +_{m, n} n > 0$
Thus $n$ is not the identity of $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$.
== Examples ==
Category:Dipper Semigroups
Category:Examples of Identity Elements
Category:Identity of Subgroup of Dipper Semigroup is not Identity of Dipper
Identity of Subgroup of Dipper Semigroup is not Identity of Dipper/Examples/(m, n) = (3, 4)

== Examples of Use of Identity of Subgroup of Dipper Semigroup is not Identity of Dipper ==
Consider the dipper semigroup $\struct {N_{<7}, +_{3, 4} }$.
Let $H = \set {x \in \N: 3 \le x < 7} = \set {3, 4, 5, 6}$.
From Existence of Subgroup of Dipper Semigroup Example: $\struct {H, +_{3, 4} }$
:$\struct {H, +_{3, 4} }$ is a subgroup of $\struct {N_{<7}, +_{3, 4} }$
whose identity is $4$.
We have that:
:$0 +_{3, 4} 4 = 4$
and so $4$ is not the identity of $\struct {N_{<7}, +_{3, 4} }$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.4$
Category:Identity of Subgroup of Dipper Semigroup is not Identity of Dipper
Identity of Submagma containing Identity of Magma is Same Identity

== Theorem ==
Let $\struct {S, \circ}$ be a magma which has an identity $e$.
Let $\struct {T, \circ}$ be a submagma of $\struct {S, \circ}$ such that $e \in T$.
Then $e$ is the identity of $T$.
== Proof ==
From Identity is Unique, there can be only one identity $e$ of $\struct {S, \circ}$.
We have that:
That is, $e$ is the (unique} identity of $T$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets
Category:Identity Elements
Category:Magmas
Identity of Submonoid is not necessarily Identity of Monoid

== Theorem ==
Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$.
Let $\struct {T, \circ}$ be a submonoid of $\struct {S, \circ}$ whose identity is $e_T$.
Then it is not necessarily the case that $e_T = e_S$.
== Proof ==
Let $\struct {S, \times}$ be the semigroup formed by the set of order $2$ square matrices over the real numbers $R$ under (conventional) matrix multiplication.
Let $T$ be the subset of $S$ consisting of the matrices of the form $\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}$ for $x \in \R$.
From Matrices of the Form $\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}$ we have that $\struct {T, \times}$ is a subsemigroup of $\struct {S, \times}$.
From (some result somewhere) $\struct {S, \times}$ has an identity $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ which is not in $\struct {T, \times}$.
However, note that:
demonstrating that $\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ is the identity of $\struct {T, \times}$.
== Also see ==
* Identity of Cancellable Monoid is Identity of Submonoid
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets
* : $\S 32.2$ Identity element and inverses
Category:Submonoids
Identity of Subsemigroup of Group

== Definition ==
Let $\struct {G, \circ}$ be a group whose identity element is $e$.
Let $\struct {H, \circ}$ be a subsemigroup of $\struct {G, \circ}$.
If $\struct {H, \circ}$ has an identity element, then that identity is $e$.
== Proof ==
From the Cancellation Laws, every element of $\struct {G, \circ}$ is cancellable.
From Identity is only Idempotent Cancellable Element, there is only one element $x$ of $\struct {G, \circ}$ satisfying $x \circ x = x$, and that is $e$.
As $x \in H \implies x \in G$, the same applies to $\struct {H, \circ}$.
So if there is an element in $\struct {H, \circ}$ such that $x \circ x = x$, it must be $e$.
Hence the result.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Theorem $8.3$
Category:Subsemigroups
Category:Group Theory
Category:Identity Elements
If Compact Between then Way Below

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Let $x, k, y \in S$ such that:
:$x \preceq k$ and $k \preceq y$ and $k \in \map K L$
where $\map K L$ denotes the compact subset of $L$.
Then $x \ll y$
where $\ll$ denotes the way below relation.
== Proof ==
By definition of compact subset:
:$k$ is compact.
By definition of compact:
:$k \ll k$
Thus by Preceding and Way Below implies Way Below:
:$x \ll y$
== Sources ==
* 
* 
Category:Way Below Relation
If Definite Integral of a(x)h(x) vanishes for any C^0 h(x) then C^0 a(x) vanishes

== Theorem ==
Let $\map \alpha x$ be a continuous real function on the closed real interval $\closedint a b$.
Let $\ds \int_a^b \map \alpha x \map h x \rd x = 0$ for every real function $\map h x \in C^0 \closedint a b$ such that $\map h a = 0$ and $\map h b = 0$.
where $C^0 \closedint a b$ means continuous functions on $\closedint a b$.
Then $\map \alpha x = 0$ for all $x \in \closedint a b$.
== Proof ==
 the real function $\map \alpha x$ is nonzero at some point in $\closedint a b$ for some arbitrary $\map h x$. 
Due to belonging to $C^0$ it is also nonzero in some interval $\closedint {x_1} {x_2}$ contained in $\closedint a b$.
Let us choose $\map h x$ to be of a specific form, while still satisfying the requirements in the statement of the theorem:
:$\map h x = \begin {cases}
\map \sgn {\map \alpha x} \paren {x - x_1} \paren {x_2 - x} & : x \in \closedint {x_1} {x_2} \\
0 & : x \notin \closedint {x_1} {x_2}
\end {cases}$
Then:
:$\ds \int_a^b \map \alpha x \map h x \rd x = \int_{x_1}^{x_2} \size {\map \alpha x} \paren {x - x_1} \paren {x_2 - x} \rd x$
where we used the fact that:
:$\map \alpha x = \map \sgn {\map \alpha x} \size {\map \alpha x}$
as well as:
:$\map {\sgn^2} x = 1$ if $x \ne 0$ and $x \in \R$.
The integrand is positive for $x \in \closedint {x_1} {x_2}$, whence the integral is positive.
However, that contradicts the condition on the integral in the statement of the theorem.
Thus, with the provided assumption the condition for the integral does not hold for all $\map h x$ with aforementioned conditions.
Hence the result by Proof by Contradiction.
== Remark ==
The lemma still holds if we replace $C^0 \closedint a b$ by $C^n \closedint a b$.
To see this, modify $\map h x$:
:$\map h x = \begin {cases}
\map \sgn {\map \alpha x} \paren {\paren {x - x_1} \paren {x_2 - x} }^{n + 1} & : x \in \closedint {x_1} {x_2} \\
0 & : x \notin \closedint {x_1} {x_2}
\end {cases}$
then we can use the same proof
== Sources ==
* : $\S 1.3$: The Variation of a Functional. A Necessary Condition for an Extremum
Category:Calculus of Variations
If Double Integral of a(x, y)h(x, y) vanishes for any C^2 h(x, y) then C^0 a(x, y) vanishes

== Theorem ==
Let $\map \alpha {x, y}$, $\map h {x, y}$ be functions in $\R$.
Let $\alpha \in C^0$ in a closed region $R$ whose boundary is $\Gamma$.
Let $h \in C^2$ in $R$ and $h = 0$ on $\Gamma$.
Let:
:$\ds \iint_R \map \alpha {x, y} \map h {x, y} \rd x \rd y = 0$
Then $\map \alpha {x, y}$ vanishes everywhere in $R$.
== Proof ==
 $\map \alpha {x, y}$ is nonzero at some point in $R$.
Then $\map \alpha {x, y}$ is also nonzero in some disk $D$ such that:
:$\paren {x - x_0}^2 + \paren {y - y_0}^2 \le \epsilon^2$
Suppose:
:$\map h {x, y} = \map \sgn {\map \alpha {x, y} } \paren {\epsilon^2 - \paren {x - x_0}^2 + \paren {y - y_0}^2}^3$ 
in this disk and $0$ elsewhere.
Thus $\map h {x, y}$ satisfies conditions of the theorem.
However:
:$\ds \iint_R \map \alpha {x, y} \map h {x, y} \rd x \rd y = \iint_D \size {\map \alpha {x, y} } \paren {\epsilon^2 - \paren {x - x_0}^2 + \paren {y - y_0}^2}^3 \ge 0$
Hence the result, by Proof by Contradiction.
== Sources ==
* : $\S 1.5$: The Case of Several Variables
Category:Calculus of Variations
If Element Does Not Belong to Ideal then There Exists Prime Ideal Including Ideal and Excluding Element

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a distributive lattice.
Let $I$ be an ideal in $L$.
Let $x$ be an element of $S$.
Suppose $x \notin I$
Then there exists a prime ideal $P$ in $L$: $I \subseteq P$ and $x \notin P$
== Proof ==
By Upper Closure of Element is Filter:
:$x^\succeq$ is a filter.
We will prove that
:$I \cap x^\succeq = \O$
:
:$\exists y: y \in I \cap x^\succeq$
By definition of intersection:
:$y \in I$ and $y \in x^\succeq$
By definition of upper closure of element:
:$x \preceq y$
By definition of lower section:
:$x \in I$
This contradicts $x \notin I$
Thus by If Ideal and Filter are Disjoint then There Exists Prime Ideal Including Ideal and Disjoint from Filter:
:there exists a prime ideal $P$ in $L$: $I \subseteq P$ and $P \cap x^\succeq = \O$
By definition of reflexivity:
:$x \preceq x$
By definition of upper closure of element:
:$x \in x^\succeq$
Thus by definitions of intersection and empty set:
:$x \notin P$
== Sources ==
* 
* 
Category:Prime Ideals (Order Theory)
If Every Element Pseudoprime is Prime then Way Below Relation is Multiplicative

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below continuous distributive lattice.
Let every element $p \in S$: $p$ is pseudoprime $\implies p$ is prime.
Then $\ll$ is multiplicative
where $\ll$ denotes the way below relation of $L$.
== Proof ==
Let $a, x, y \in S$ such that:
:$a \ll x$ and $a \ll y$
:
:$a \not\ll x \wedge y$
We will prove that:
:$\forall z \in S: z \in \paren {x \wedge y}^\ll \implies z \notin a^\succeq$
Let $z \in S$ such that:
:$z \in \paren {x \wedge y}^\ll$
By definition of way below closure:
:$z \ll x \wedge y$
:
:$z \in a^\succeq$
By definition of upper closure of element:
:$a \preceq z$
By Preceding and Way Below implies Way Below:
:$a \ll x \wedge y$
This contradicts $a \not\ll x \wedge y$
By definitions of empty set and intersection:
:$\paren {x \wedge y}^\ll \cap a^\succeq = \O$
By Way Below Closure is Ideal in Bounded Below Join Semilattice:
:$\paren {x \wedge y}^\ll$ is ideal in $L$.
By Upper Closure of Element is Filter:
:$a^\succeq$ is filter on $L$.
By If Ideal and Filter are Disjoint then There Exists Prime Ideal Including Ideal and Disjoint from Filter:
:there exists prime ideal $P$ in $L$:
:$\paren {x \wedge y}^\ll \subseteq P$ and $P \cap a^\succeq = \O$
By definition of pseudoprime element:
:$\sup P$ is pseudoprime.
By assumption:
:$\sup P$ is prime.
By definition of reflexivity:
:$a \preceq a$
By definition of upper closure of element:
:$a \in a^\succeq$
By the axiom of approximation:
:$\sup \paren {x \wedge y}^\ll = x \wedge y$
By definition of up-complete:
:$\paren {x \wedge y}^\ll$ admits a supremum
and
:$P$ admits a supremum.
By Supremum of Subset:
:$x \wedge y \preceq \sup P$
By definition of prime element:
:$x \preceq \sup P$ or $y \preceq \sup P$
By Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal:
:$a \in P$
By definition of intersection:
:$a \in P \cap a^\succeq$
This contradicts $P \cap a^\succeq = \O$
== Sources ==
* 
* 
Category:Prime Elements
Category:Way Below Relation
If First of Four Numbers in Geometric Sequence is Cube then Fourth is Cube

== Theorem ==
Let $P = \tuple {a, b, c, d}$ be a geometric sequence of integers.
Let $a$ be a cube number.
Then $d$ is also a cube number.
== Proof ==
From Form of Geometric Sequence of Integers:
:$P = \tuple {k p^3, k p^2 q, k p q^2, k q^3}$
for some $k, p, q \in \Z$.
If $a = k p^3$ is a cube number it follows that $k$ is a cube number: $k = r^3$, say.
So:
:$P = \tuple {r^3 p^3, r^3 p^2 q, r^3 p q^2, r^3 q^3}$
and so $d = r^3 q^3 = \paren {r q}^3$.
== Sources ==
* : Book $\text{VIII}$. Propositions
Category:Geometric Sequences of Integers
If First of Three Numbers in Geometric Sequence is Square then Third is Square

== Theorem ==
Let $P = \tuple {a, b, c}$ be a geometric sequence of integers.
Let $a$ be a square number.
Then $c$ is also a square number.
== Proof ==
From Form of Geometric Sequence of Integers:
:$P = \tuple {k p^2, k p q, k q^2}$
for some $k, p, q \in \Z$.
If $a = k p^2$ is a square number it follows that $k$ is a square number: $k = r^2$, say.
So:
:$P = \tuple {r^2 p^2, r^2 p q, r^2 q^2}$
and so $c = r^2 q^2 = \paren {r q}^2$.
== Sources ==
* : Book $\text{VIII}$. Propositions
Category:Geometric Sequences of Integers
If Ideal and Filter are Disjoint then There Exists Prime Filter Including Filter and Disjoint from Ideal

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a distributive lattice.
Let $I$ be an ideal in $L$.
Let $F$ be a filter on $L$ such that
:$F \cap I = \O$
Then there exists a prime filter $P$ in $L$:
$F \subseteq P$ and $P \cap I = \O$
== Proof ==
By Dual Distributive Lattice is Distributive:
:$L^{-1}$ is a distributive lattice
where $L^{-1} = \struct {S, \succeq}$ denotes the dual of $L$.
By Filter is Ideal in Dual Ordered Set:
:$I' := F$ as an ideal in $L^{-1}$.
By Ideal is Filter in Dual Ordered Set:
:$F' := I$ as a filter on $L^{-1}$.
By assumption:
:$I' \cap F' = \O$
By If Ideal and Filter are Disjoint then There Exists Prime Ideal Including Ideal and Disjoint from Filter:
:there exists prime ideal $P'$ in $L^{-1}$: $I' \subseteq P'$ and $P' \cap F' = \O$
By Prime Filter is Prime Ideal in Dual Lattice:
:$P := P'$ as a prime filter on $L$.
Thus there exists prime filter $P$ on $L$: $F \subseteq P$ and $P \cap I = \O$
== Sources ==
* 
* 
Category:Prime Ideals (Order Theory)
If Infimum of Filtered Subset belongs to Element of Sub-Basis then Subset and Element Intersect implies Infimum of Subset belongs to Closure of Subset

== Theorem ==
Let $T = \struct {S, \preceq, \tau}$ be a complete topological lattice with lower topology.
Let $B$ be an analytic sub-basis of $T$.
Let $F$ be a filtered subset of $S$ such that
:$\forall A \in B: \inf F \in A \implies F \cap A \ne \O$
Then $\inf F \in F^-$
where $F^-$ denotes the topological closure of $F$.
== Proof ==
We will prove that
:$\forall A \in B, x \in F \cap A, y \in F: y \preceq x \implies y \in A$
Let $A \in B$, $x \in F \cap A$, $y \in F$.
By definition of sub-basis:
:$A$ is open.
By Open Subset is Lower Section in Lower Topology:
:$A$ is a lower section.
By definition of intersection:
:$x \in A$.
Thus by definition of lower section:
:$y \preceq x \implies y \in A$.
Define $\ds H := \set {\bigcap G: G \subseteq B, G \text { is finite} }$
By definitions of sub-basis and basis:
:$H$ is basis of $T$.
We will prove that
:$\forall A \in H: \inf F \in A \implies F \cap A \ne \O$
Let $A \in H$ such that
:$\inf F \in A$
By definition of $H$:
:$\exists G \subseteq B: A = \bigcap G \land G$ is finite.
By definition of intersection:
:$\forall C \in G: \inf F \in C \in B$
By assumption:
:$\forall C \in G: F \cap C \ne \O$
By definition of non-empty set:
:$\forall C \in G: \exists x: x \in F \cap C$
By Axiom of Choice:
:$\exists f: G \to F: \forall C \in G: \map f C \in F \cap C$
By Image of Mapping from Finite Set is Finite:
:$f \sqbrk G$ is finite.
By Filtered iff Finite Subsets have Lower Bounds:
:$\exists h \in F: \forall x \in f \sqbrk G: h \preceq x$
Then
:$\forall C \in G: h \in C$
By definition of intersection:
:$h \in \bigcap G$
Thus by definitions of intersection and non-empty set:
:$F \cap A \ne \O$
Thus by Characterization of Closure by Basis:
:$\inf F \in F^-$
== Sources ==
* 
Category:Topological Order Theory
If Ratio of Cube to Number is as between Two Cubes then Number is Cube

== Theorem ==
Let $a, b, c, d \in \Z$ be integers such that:
:$\dfrac a b = \dfrac {c^3} {d^3}$
Let $a$ be a cube number.
Then $b$ is also a cube number.
== Proof ==
From :
:$\left({c^3, c^2 d, c d^2, d^3}\right)$
is a geometric sequence.
From :
:$\left({a, m_1, m_2, b}\right)$
is a geometric sequence for some $m$.
We have that $a$ is a cube number.
From :
:$b$ is a cube number.
== Sources ==
* : Book $\text{VIII}$. Propositions
Category:Ratios
Category:Cube Numbers
If Ratio of Square to Number is as between Two Squares then Number is Square

== Theorem ==
Let $a, b, c, d \in \Z$ be integers such that:
:$\dfrac a b = \dfrac {c^2} {d^2}$
Let $a$ be a square number.
Then $b$ is also a square number.
== Proof ==
From :
:$\tuple {c^2, c d, d^2}$
is a geometric sequence.
From :
:$\tuple {a, m, b}$
is a geometric sequence for some $m$.
We have that $a$ is a square number.
From :
:$b$ is a square number.
== Sources ==
* : Book $\text{VIII}$. Propositions
Category:Ratios
Category:Square Numbers
If Set Exists then Empty Set Exists

== Theorem ==
If at least one set exists, then there exists an empty set.
== Proof ==
Let $S$ be a set.
By the axiom of class comprehension, there is an empty class:
:$\O = \set { x : x \ne x }$
Since $x \in \O$ is never true, it follows vacuously that:
:$x \in \O \implies x \in S$
By the subclass definition:
:$\O \subseteq S$
By Subclass of Set is Set, $\O$ is a set.
== Sources ==
* : Chapter $1$: Separation Schema
Category:Gödel-Bernays Class Theory
Category:Empty Set
If n is Triangular then so is (2m+1)^2 n + m(m+1)/2

== Theorem ==
Let $n$ be a triangular number.
Let $m \in \Z_{\ge 0}$ be a positive integer.
Then $\paren {2 m + 1}^2 n + \dfrac {m \paren {m + 1} } 2$ is also a triangular number.
== Proof ==
Let $n$ be a triangular number.
Then:
:$\exists k \in \Z: n = \dfrac {k \paren {k + 1} } 2$
So:
which is of the form:
:$\dfrac {t \paren {t + 1} } 2$
where $t = \paren {2 m + 1} k + m$.
Category:Triangular Numbers
If n is Triangular then so is 25n + 3

== Theorem ==
Let $n$ be a triangular number.
Then $25 n + 3$ is also triangular.
== Proof ==
Let $n$ be triangular.
Then:
:$\exists k \in \Z: n = \dfrac {k \paren {k + 1} } 2$
So:
which is triangular.
== Historical Note ==
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.3$ Early Number Theory: Problems $1.3$: $1 \ \text {(d)}$
Category:Triangular Numbers
If n is Triangular then so is 49n + 6

== Theorem ==
Let $n$ be a triangular number.
Then $49 n + 6$ is also triangular.
== Proof ==
Let $n$ be triangular.
Then:
:$\exists k \in \Z: n = \dfrac {k \paren {k + 1} } 2$
So:
which is triangular.
== Historical Note ==
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.3$ Early Number Theory: Problems $1.3$: $1 \ \text {(d)}$
Category:Triangular Numbers
If n is Triangular then so is 9n + 1

== Theorem ==
Let $n$ be a triangular number.
Then $9 n + 1$ is also triangular.
== Proof ==
Let $n$ be triangular.
Then:
:$\exists k \in \Z: n = \dfrac {k \paren {k + 1} } 2$
So:
which is triangular.
== Historical Note ==
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.3$ Early Number Theory: Problems $1.3$: $1 \ \text {(d)}$
Category:Triangular Numbers
If n is Triangular then so is 9n + 1/Historical Note

== Historical Note on If $n$ is Triangular then so is $9 n + 1$ ==
, in his of $1980$, reports that the result If $n$ is Triangular then so is $9 n + 1$ was published by in $1775$.
He published this along with If $n$ is Triangular then so is $25 n + 3$ and If $n$ is Triangular then so is $49 n + 6$.
There is an obvious pattern.
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.3$ Early Number Theory: Problems $1.3$: $1 \ \text {(d)}$
Category:Historical Notes
Ill-Conditioned Problem/Examples/Arbitrary Example 1

== Example of Ill-Conditioned Problem ==
Consider the zeroes of the polynomials:
The zeroes of $\map p x$ are all $0$.
However, the zeroes of $\map q x$ are all of modulus $0 \cdotp 1$.
Hence, while $\map q x$ is a tiny perturbation of $\map p x$, the zeroes differ by a modulus some $8$ orders of magnitude larger.
== Sources ==
* 
* 
Category:Examples of Ill-Conditioned Problems
Ill-Conditioned Problem/Examples/Arbitrary Example 2

== Example of Ill-Conditioned Problem ==
Consider the simultaneous equations:
These have the solution:
However, the simultaneous equations:
have the solution:
So a change in the $4$th decimal place of one coefficient leads to a completely different solution.
This can be explained by the fact that the matrix of coefficients is nearly singular.
== Sources ==
* 
* 
Category:Examples of Ill-Conditioned Problems