nyuuzyou/wikis huggingface dataset:
Null Sequences form Maximal Left and Right Ideal/Lemma 8

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\CC$ be the ring of Cauchy sequences over $R$
Let $\NN$ be the set of null sequences.
Then:
:There is no left ideal $\JJ$ of $\CC$ such that $\NN \subsetneq \JJ \subsetneq \CC$ 
== Proof ==
Let $\JJ$ be a left ideal of $\CC$ such that $\NN \subsetneq \JJ \subseteq \CC$.
It will be shown that $\JJ$ = $\CC$, from which the result will follow.
Let $\sequence {x_n} \in \JJ \setminus \NN$
By Inverse Rule for Cauchy sequences then 
:$\exists K \in \N: \sequence { \paren {x_{K + n} }^{-1} }_{n \mathop \in \N}$ is a Cauchy sequence.
Let $\sequence {y_n}$ be the sequence defined by:
:$y_n = \begin {cases} 0 & : n \le K \\ \paren {x_n}^{-1} & : n > K \end {cases}$
By Cauchy Sequence with Finite Elements Prepended is Cauchy Sequence then $\sequence {y_n} \in \CC$
By the definition of a left ideal the product $\sequence {y_n} \sequence {x_n} = \sequence {y_n x_n} \in \JJ$
By the definition of $\sequence {y_n}$ then:
:$y_n x_n = \begin {cases} 0 & : n \le K \\ 1 & : n > K \end {cases}$
Let $\mathcal 1 = \tuple {1, 1, 1, \dotsc}$ be the unity of $\CC$
Then $\mathcal 1 - \sequence {y_n} \sequence {x_n}$ is the sequence $\sequence {w_n}$ defined by
:$w_n = \begin{cases} 1 & : n \le K \\ 0 & : n > K \end {cases}$
By Convergent Sequence with Finite Elements Prepended is Convergent Sequence then $\sequence {w_n}$ is convergent to 0.
So $\sequence {w_n} \in \NN \subsetneq \JJ$
Since $\sequence {y_n} \sequence {x_n}, \sequence {w_n} \in \JJ$ by the definition of a ring ideal then:
:$\sequence {w_n} + \sequence {y_n} \sequence {x_n} = \mathcal 1 \in \JJ$
By the definition of a left ideal then:
:$\forall \sequence {a_n} \in \CC, \sequence {a_n} \circ \mathcal 1 = \sequence {a_n} \in \JJ$
Hence $\JJ = \CC$
== Sources ==
* : $\S 3.2$: Completions
Category:Null Sequences form Maximal Left and Right Ideal
Null Sets Closed under Countable Union

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\sequence {N_n}_{n \mathop \in \N}$ be a sequence of $\mu$-null sets.
Then $N := \ds \bigcup_{n \mathop \in \N} N_n$ is also a $\mu$-null set.
=== Signed Measure ===
=== Resolution of the Identity ===
== Proof ==
As $\mu$ is a measure:
:$\map \mu N \ge 0$
Also:
Hence necessarily:
:$\map \mu N = 0$
and $N$ is a $\mu$-null set.
== Sources ==
* : $\S 4$: Problem $10 \ \text{(iii)}$
Category:Null Sets Closed under Countable Union
Category:Null Sets
Category:Measure Theory
Null Sets Closed under Countable Union/Resolution of the Identity

== Theorem ==
Let $X$ be a topological space.
Let $\map \BB X$ be the Borel $\sigma$-algebra of $X$.
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\map B {\HH}$ be the space of bounded linear transformations on $\HH$.
Let $\EE : \map \BB X \to \map B {\HH}$ be a resolution of the identity.
Let $\set {A_j : j \in \N} \subseteq \map \BB X$ such that:
:$\map \EE {A_j} = 0$ for each $j \in \N$.
Let:
:$\ds A = \bigcup_{j \mathop = 1}^\infty A_j$
Then $\map \EE A = 0$.
== Proof ==
Let $x \in \HH$. 
Define $\EE_{x, x} : \map \BB X \to \C$ by:
:$\map {\EE_{x, x} } A = \innerprod {\map \EE A x} x$
for each $A \in \map \BB X$.
Since $\map \EE {A_j} = 0$ for each $j \in \N$, we have $\map {\EE_{x, x} } {A_j} = 0$ for each $j \in \N$. 
Since $\EE_{x, x}$ is a complex measure, we have:
:$\map {\EE_{x, x} } A = 0$ 
from countable additivity.
From Scalar-Valued Measure associated with Resolution of Identity is Positive Measure if Equal Vectors, we have:
:$\map {\EE_{x, x} } A = \norm {\map \EE A x}^2 = 0$ 
From , we have $\map \EE A x = 0$.
Further, $x \in \HH$ is arbitrary so we have $\map \EE A = 0$. 
== Sources ==
* : $12.19$: Proposition
Category:Resolutions of the Identity
Null Sets Closed under Countable Union/Signed Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$. 
Let $\sequence {N_i}_{i \mathop \in \N}$ be a sequence of $\mu$-null sets.
Then: 
:$\ds N = \bigcup_{i \mathop = 1}^\infty N_i$
is a $\mu$-null set.
 
== Proof ==
From Countable Union of Measurable Sets as Disjoint Union of Measurable Sets, there exists a sequence of pairwise disjoint $\Sigma$-measurable sets $\sequence {A_i}_{i \in \N}$ such that: 
:$\ds N = \bigcup_{i \mathop = 1}^\infty A_i$
We now show that if $E \in \Sigma$ has $E \subseteq N$, then $\map \mu E = 0$. 
Write: 
Since $\mu$ is countably additive, we have:
:$\ds \map \mu E = \sum_{i \mathop = 1}^\infty \map \mu {E \cap A_i}$
We have, from Intersection is Subset: 
:$E \cap A_i \subseteq A_i \subseteq N_i$
Since $N_i$ is a $\mu$-null set, we have that: 
:$\map \mu {E \cap A_i} = 0$ for each $i$. 
So:
:$\ds \map \mu E = \sum_{i \mathop = 1}^\infty \map \mu {E \cap A_i} = 0$
So:
:for each $E \in \Sigma$ with $E \subseteq N$, we have $\map \mu E = 0$.
So $N$ is a $\mu$-null set.
Category:Null Sets Closed under Countable Union
Category:Signed Measures
Null Sets Closed under Subset

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $N \in \Sigma$ be a $\mu$-null set, and let $M \in \Sigma$ be a subset of $N$.
Then $M$ is also a $\mu$-null set.
=== Resolution of the Identity ===
== Proof ==
As $\mu$ is a measure, $\map \mu M \ge 0$.
Also, by Measure is Monotone, $\map \mu M \le \map \mu N = 0$.
Hence $\map \mu M = 0$, and $M$ is a $\mu$-null set.
== Sources ==
* : $\S 4$: Problem $10 \ \text{(ii)}$
Category:Null Sets
Null Sets Closed under Subset/Resolution of the Identity

== Theorem ==
Let $X$ be a topological space.
Let $\map \BB X$ be the Borel $\sigma$-algebra of $X$.
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\map B {\HH}$ be the space of bounded linear transformations on $\HH$.
Let $\EE : \map \BB X \to \map B {\HH}$ be a resolution of the identity.
Let $E, E' \in \map \BB X$ be such that:
:$E \subseteq E'$ and $\map \EE {E'} = {\mathbf 0}_{\map B \HH}$
Then $\map \EE E = {\mathbf 0}_{\map B \HH}$.
== Proof 1 ==
 
== Proof 2 ==
Category:Resolutions of the Identity
Null Sets Closed under Subset/Resolution of the Identity/Proof 1

== Theorem ==
== Proof ==
For each $x, y \in \HH$, define $\EE_{x, y} : \map \BB X \to \C$ by:
:$\map {\EE_{x, y} } A = \innerprod {\map \EE A x} y$
for each $A \in \map \BB X$.
We then have:
:$\map {\EE_{x, y} } {E'} = 0$
for each $x, y \in \HH$.
From Null Sets Closed under Subset, we have $\map {\EE_{x, y} } E = 0$ for each $x, y \in \HH$.
That is, $\innerprod {\map \EE E x} y = 0$ for all $x, y \in \HH$.
From Linear Subspace Dense iff Zero Orthocomplement, we have $\HH^\bot = \set { {\mathbf 0}_\HH}$ and hence $\map \EE E x = {\mathbf 0}_\HH$ for each $x \in \HH$.
Hence $\map \EE E = {\mathbf 0}_{\map B \HH}$.
Category:Null Sets Closed under Subset
Null Sets Closed under Subset/Resolution of the Identity/Proof 2

== Theorem ==
== Proof ==
Let $\le_{\map B \HH}$ be the canonical preordering on $\map B \HH$.
From Measure is Monotone: Resolution of the Identity, we have:
:$\map \EE E \le_{\map B \HH} \map \EE {E'} = {\mathbf 0}_{\map B \HH}$
On the other hand from Bounds on Projection in Unital C*-Algebra, we have:
:${\mathbf 0}_{\map B \HH} \le_{\map B \HH} \map \EE E$
From Canonical Preordering of C*-Algebra is Antisymmetric, we then have:
:$\map \EE E = {\mathbf 0}_{\map B \HH}$
Category:Null Sets Closed under Subset
Null Space Closed under Scalar Multiplication

== Theorem ==
Let:
:$\map {\mathrm N} {\mathbf A} = \set {\mathbf x \in \R^n : \mathbf {A x} = \mathbf 0}$
be the null space of $\mathbf A$, where:
:$ \mathbf A_{m \times n} = \begin{bmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{m1} & a_{m2} & \cdots & a_{mn} \\
\end {bmatrix}$, $\mathbf x_{n \times 1} = \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix}$, $\mathbf 0_{m \times 1} = \begin {bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end {bmatrix}$
are matrices where each column is an element of a real vector space.
Then $\map {\mathrm N} {\mathbf A}$ is closed under scalar multiplication:
:$\forall \mathbf v \in \map {\mathrm N} {\mathbf A} ,\forall \lambda \in \R: \lambda \mathbf v \in \map {\mathrm N} {\mathbf A}$
== Proof ==
Let $\mathbf v \in \map {\mathrm N} {\mathbf A}$, $\lambda \in \R$.
By the definition of null space:
Observe that:
Hence the result, by the definition of null space.
== Also see ==
* Null Space Contains Zero Vector
* Null Space Closed under Vector Addition
* Null Space is Subspace
== Sources ==
* 
Category:Linear Algebra
Category:Null Spaces
Null Space Closed under Vector Addition

== Theorem ==
Let:
:$\map {\mathrm N} {\mathbf A} = \set {\mathbf x \in \R^n : \mathbf A \mathbf x = \mathbf 0}$
be the null space of $\mathbf A$, where:
:$\mathbf A_{m \times n} = \begin {bmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{m1} & a_{m2} & \cdots & a_{mn} \\
\end{bmatrix}$, $\mathbf x_{n \times 1} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}$ and $\mathbf 0_{m \times 1} = \begin {bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end {bmatrix}$ are matrices
:the column matrix $\mathbf x_{n \times 1}$ is interpreted as a vector in the real Euclidean space $\R^n$.
Then $\map {\mathrm N} {\mathbf A}$ is closed under vector addition:
:$\forall \mathbf v, \mathbf w \in \map {\mathrm N} {\mathbf A}: \mathbf v + \mathbf w \in \map {\mathrm N} {\mathbf A}$
== Proof ==
Let $\mathbf v, \mathbf w \in \map {\mathrm N} {\mathbf A}$.
By the definition of null space:
Next, observe that:
The order is correct, by hypothesis.
Hence the result, by the definition of null space.
== Also see ==
* Null Space Contains Zero Vector
* Null Space Closed under Scalar Multiplication
* Null Space is Subspace
== Sources ==
* 
Category:Linear Algebra
Category:Null Spaces
Null Space Contains Only Zero Vector iff Columns are Independent

== Theorem ==
Let:
be a matrix where:
:$\forall i: 1 \le i \le n: \mathbf a_i = \begin{bmatrix} a_{1i} \\ a_{2i} \\ \vdots \\ a_{mi} \end{bmatrix} \in \R^m$
are vectors.
Then:
:$\set {\mathbf a_1, \mathbf a_2, \cdots, \mathbf a_n}$ is a linearly independent set 
:
:$\map {\mathrm N} {\mathbf A} = \set {\mathbf 0_{n \times 1} }$
where $\map {\mathrm N} {\mathbf A}$ is the null space of $\mathbf A$.
== Proof ==
Let $\mathbf x = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \in \R^m$.
We have that:
=== Sufficient Condition ===
Let $\set {\mathbf a_1, \mathbf a_2, \cdots, \mathbf a_n}$ be linearly independent.
Then by definition:
:$\forall k: 1 \le k \le n: x_k = 0 \iff \mathbf x = \mathbf 0_{n \times 1}$
By the definition of null space:
:$\map {\mathrm N} {\mathbf A} = \set {\mathbf 0_{n \times 1} }$
=== Necessary Condition ===
Let $\map {\mathrm N} {\mathbf A} = \set {\mathbf 0_{n \times 1} }$.
Then by the definition of null space:
:$\mathbf x = \mathbf 0_{n \times 1}$
This means that:
:$\forall k: 1 \le k \le n: x_k = 0$
from which it follows that $\set {\mathbf a_1, \mathbf a_2, \cdots, \mathbf a_n}$ is linearly independent.
== Sources ==
* 
Category:Linear Independence
Category:Null Spaces
Null Space Contains Zero Vector

== Theorem ==
Let:
:$\map {\mathrm N} {\mathbf A} = \set {\mathbf x \in \R^n: \mathbf A \mathbf x = \mathbf 0}$
be the null space of $\mathbf A$, where:
$\quad \mathbf A_{m \times n} = \begin {bmatrix}
a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
a_{2 1} & a_{2 2} & \cdots & a_{2 n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{m 1} & a_{m 2} & \cdots & a_{m n} \\
\end {bmatrix}$ 
is a matrix in the matrix space $\map {\MM_\R} {m, n}$.
Then the null space of $\mathbf A$ contains the zero vector:
:$\mathbf 0 \in \map {\mathrm N} {\mathbf A}$
where:
:$\mathbf 0 = \mathbf 0_{m \times 1} = \begin {bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end {bmatrix}$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Null Space Closed under Scalar Multiplication
* Null Space Closed under Scalar Multiplication
* Null Space is Subspace
* Kernel of Linear Transformation contains Zero Vector
== Sources ==
* 
Category:Linear Algebra
Category:Null Spaces
Category:Null Space Contains Zero Vector
Null Space is Subspace

== Theorem ==
Let:
:$\map {\mathrm N} {\mathbf A} = \set {\mathbf x \in \R^n: \mathbf {A x} = \mathbf 0}$
be the null space of $\mathbf A$, where:
:$\mathbf A_{m \times n} = \begin {bmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{m1} & a_{m2} & \cdots & a_{mn} \\
\end {bmatrix}$, $\mathbf x_{n \times 1} = \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix}$, $\mathbf 0_{m \times 1} = \begin {bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end {bmatrix}$
are matrices.
Then $\map {\mathrm N} {\mathbf A}$ is a linear subspace of $\R^n$.
== Proof ==
$\map {\mathrm N} {\mathbf A} \subseteq \R^n$, by construction.
We have:
:$\mathbf 0 \in \map {\mathrm N} {\mathbf A}$, from Null Space Contains Zero Vector
:$\forall \mathbf v, \mathbf w \in \map {\mathrm N} {\mathbf A}: \mathbf v + \mathbf w \in \map {\mathrm N} {\mathbf A}$, from Null Space Closed under Vector Addition
:$\forall \mathbf v \in \map {\mathrm N} {\mathbf A}, \lambda \in \R: \lambda \mathbf v \in \map {\mathrm N} {\mathbf A}$, from Null Space Closed under Scalar Multiplication
The result follows from Vector Subspace of Real Vector Space.
== Sources ==
* 
Category:Linear Algebra
Category:Null Spaces
Null Space of Reduced Echelon Form

== Theorem ==
Let $\mathbf A$ be a matrix in the matrix space $\map {\MM_\R} {m, n}$ such that:
:$\mathbf A \mathbf x = \mathbf 0$
represents a homogeneous system of linear equations.
The null space of $\mathbf A$ is the same as that of the null space of the reduced row echelon form of $\mathbf A$:
:$\map {\mathrm N} {\mathbf A} = \map {\mathrm N} {\map {\mathrm {rref} } {\mathbf A} }$
== Proof ==
By the definition of null space:
:$\mathbf x \in \map {\mathrm N} {\mathbf A} \iff \mathbf A \mathbf x = \mathbf 0$
From the corollary to Row Equivalent Matrix for Homogeneous System has same Solutions:
:$\mathbf A \mathbf x = \mathbf 0 \iff \map {\mathrm {rref} } {\mathbf A} \mathbf x = \mathbf 0$
Hence the result, by the definition of set equality.
== Sources ==
* 
Category:Linear Algebra
Category:Null Spaces
Category:Echelon Matrices
Null String has Length Zero

== Definition ==
Let $\epsilon$ denote the null string.
Then:
:$\size \epsilon = 0$
where $\size \epsilon$ denotes the length of $\epsilon$.
== Proof ==
Follows directly from the definitions of null string and string length.
== Sources ==
* : Chapter $1$: Preliminaries: $1.1$ Strings, Alphabets and Languages
* : $\S 1.3$: Induction on Length of Wffs
Category:Collations
Null String is Identity Element for Concatenation Operator

== Theorem ==
Let $\AA$ be an alphabet of symbols.
Let $\WW$ denote the set of words in $\AA$.
Let $\epsilon$ denote the null string.
Let $C: \WW \times \WW \to \WW$ denote the concatenation operator on $\WW$:
:$\forall A, B \in \WW: \map C {A, B} := A B$
Then $\epsilon$ is the identity element for $C$.
== Proof ==
As $\epsilon$ is the null string:
:$\map C {\epsilon, A}$
Hence the result by definition of identity element.
== Sources ==
* : Chapter $1$: Preliminaries: $1.1$ Strings, Alphabets and Languages
Category:Concatenation (Formal Systems)
Null URM Program Computes Identity Function

== Theorem ==
The null URM program computes the identity function $I_\N: \N \to \N$, defined as:
:$\forall n \in \N: \map {I_\N} n = n$
== Proof ==
The null URM program by definition has no instructions.
Therefore, the contents of $R_1$ remain unchanged when "running" it.
Category:Null URM Program
Category:URM Programs
Category:Identity Mappings
Number Base Conversion

== Algorithm ==
The purpose of this algorithm is to convert a natural number from one base to another.
Let $x_0$ be the natural number in question.
Let $b$ be the number base into which it is to be converted, where $b \ge 2$.
:Step 0: Set $n := 0$.
:Step 1: From the Division Theorem, let $x_n = b x_{n+1} + r_n$ where $0 \le r_n < b$.
:Step 2: Set $n := n + 1$.
:Step 3: If $x_n < b$ go to step 4. Otherwise go to step 1.
:Step 4: Set $r_n = x_n$. The required output is $[r_n r_{n-1} \ldots r_1 r_0]_b$ where the notation is that used in the definition of number base.
== Analysis ==
=== Finiteness ===
Each cycle through the algorithm, step 1 is passed. In this step, $x_{n+1} \le \dfrac {x_n} b$ and so strictly decreases.
As $x_0$ is finite, the process is guaranteed to terminate in a finite number of steps.
=== Definiteness ===
Follows from the Division Theorem.
=== Inputs ===
The input is the natural number to be converted.
=== Outputs ===
The output is the base $b$ representation of the input.
=== Effectiveness ===
Follows from the Division Theorem.
Hence this process meets the criteria for being an algorithm, as required.
Category:Discrete Mathematics
Number Field/Examples/Rational Numbers

== Example of Number Field ==
The field of rational numbers $\Q$ is a number field.
== Proof ==
From Rational Numbers form Subfield of Complex Numbers, $\Q$ is a subfield of $\C$.
Hence the result by definition of number field.
== Sources ==
* : Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 88$
Category:Number Fields
Number Field/Examples/Real Numbers

== Example of Number Field ==
The field of real numbers $\R$ is a number field.
== Proof ==
From Real Numbers form Subfield of Complex Numbers, $\R$ is a subfield of $\C$.
Hence the result by definition of number field.
== Sources ==
* : Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 88$
Category:Number Fields
Number Field has Rational Numbers as Subfield

== Theorem ==
Let $F$ be a number field.
The field of rational numbers $\struct {\Q, +, \times}$ is a subfield of $F$.
== Proof ==
Let $F$ be a number field.
By definition, $F$ is a subfield of the field of complex numbers $\struct {\C, +, \times}$
We have that Rational Numbers form Subfield of Complex Numbers.
Let $G = \Q \cap F$ be the intersection of $\Q$ with $F$.
By Intersection of Subfields is Subfield, $G$ is a subfield of $\C$ and so also a number field.
From Intersection is Subset we have that $G \subseteq \Q$.
So $G$ is a field which is also a subset of $\Q$.
Hence by definition $G$ is a subfield of $\Q$.
But from Rational Numbers form Prime Field, $\struct {\Q, +, \times}$ is a prime field.
That is, the only subfield of $\Q$ is $\Q$ itself.
Hence:
:$G = \Q$
and from Intersection with Subset is Subset‎:
:$\Q \subseteq F$
Hence $\Q$ is a subset of $F$ which is also a field.
That is, $\Q$ is a subfield of $F$.
== Sources ==
* : Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 88 \delta$
Category:Number Fields
Category:Rational Numbers
Number Less One is Greater than Square Root

== Lemma ==
Let $n \in \N$ be an integer.
Then for $n > 2$:
:$n - 1 > \sqrt n$
== Proof ==
Let $n \in \N$ satisfy:
:$n - 1 > \sqrt n$
Then:
By definition, $n^2 - 3 n + 1$ is strictly increasing when the following inequality holds:
The solution to this is $n \ge 2$.
By inspection, $n = 3$ satisfies $n^2 - 3n + 1 > 0$.
As $n^2 - 3 n + 1$ is strictly increasing, the inequality holds for all $n \ge 3$.
Category:Inequalities
Number Plus One divides Power Plus One iff Odd

== Theorem ==
Let $q, n \in \Z_{>0}$.
Then:
:$\paren {q + 1} \divides \paren {q^n + 1}$
 $n$ is odd.
In the above, $\divides$ denotes divisibility.
== Proof ==
Let $n$ be odd.
Then from Sum of Odd Positive Powers:
:$\ds q^n + 1 = \paren {q + 1} \sum_{k \mathop = 1}^n \paren {-1}^k q^{k - 1}$
Let $n$ be even.
Consider the equation:
:$q^n + 1 = 0$
We have:
:$\paren {-1}^n + 1 = 2$
So $-1$ is not a root of $q^n + 1 = 0$.
By Polynomial Factor Theorem, $q + 1$ is not a divisor of $q^n + 1$ when $n$ is even.
== Sources ==
* : $\S 2.4$: The rational numbers and some finite fields: Further Exercises $8$
Category:Number Theory
Number Smaller than Lebesgue Number is also Lebesgue Number

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\epsilon \in \R_{>0}$ be a Lebesgue number for $M$.
Let $\epsilon' \in \R_{>0}: \epsilon' < \epsilon$.
Then $\epsilon'$ is also a Lebesgue number for $M$.
== Proof ==
By hypothesis, let $\epsilon \in \R_{>0}$ be a Lebesgue number for $M$.
Then by definition:
:$\forall x \in A: \exists \map U x \in \UU: \map {B_\epsilon} x \subseteq \map U x$
where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$ in $M$.
Let $\epsilon' \in \R_{>0}: \epsilon' < \epsilon$.
Let $y \in \map {B_{\epsilon'} } x$.
That is:
:$\map {B_{\epsilon'} } x \subseteq \map U x$
The result follows by definition of Lebesgue number.
== Sources ==
* : $7.2$: Sequential compactness
Category:Metric Spaces
Number Squared making Cube is itself Cube

== Theorem ==
Let $a \in \Z$ be an integer.
Let $a^2$ be a cube number.
Then $a$ is a cube number.
== Proof ==
$a^2$ and $a^3$ are both cube numbers.
From Between two Cubes exist two Mean Proportionals the sequence:
:$a^3, m_1, m_2, a^2$
is a geometric sequence of integers for some $m_1, m_2 \in \Z$.
By Geometric Sequences in Proportion have Same Number of Elements:
:$a^2, m_3, m_4, a$
is a geometric sequence of integers for some $m_3, m_4 \in \Z$.
From If First of Four Numbers in Geometric Sequence is Cube then Fourth is Cube it follows that $a$ is a cube number.
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Euclidean Number Theory
Number as Sum of Distinct Primes

== Theorem ==
For $n \ne 1, 4, 6$, $n$ can be expressed as the sum of distinct primes.
== Proof ==
Let $S = \set {s_n}_{n \mathop \in N}$ be the set of primes.
Then $S = \set {2, 3, 5, 7, 11, 13, \dots}$.
By Bertrand-Chebyshev Theorem:
:$s_{n + 1} \le 2 s_n$ for all $n \in \N$.
We observe that every integer $n$ where $6 < n \le 6 + s_6 = 19$ can be expressed as a sum of distinct elements in $\set {s_1, \dots, s_5} = \set {2, 3, 5, 7, 11}$.
Hence the result by Richert's Theorem.
Here is a demonstration of our claim:
Category:Prime Numbers
Number as Sum of Distinct Primes greater than 11

== Theorem ==
Every number greater than $45$ can be expressed as the sum of distinct primes greater than $11$.
== Proof ==
Let $S = \set {s_n}_{n \mathop \in N}$ be the set of primes greater than $11$ ordered by size.
Then $S = \set {11, 13, 17, 19, 23, 29, 31, 37, 41, 43, \dots}$.
By Bertrand-Chebyshev Theorem:
:$s_{n + 1} \le 2 s_n$ for all $n \in \N$.
We observe that every integer $n$ where $45 < n \le 45 + s_{11} = 92$ can be expressed as a sum of distinct elements in $\set {s_1, \dots, s_{10}} = \set {11, 13, 17, 19, 23, 29, 31, 37, 41, 43}$.
Hence the result by Richert's Theorem.
Here is a demonstration of our claim:
== Sources ==
* 
* : $45$
* : $45$
Category:Prime Numbers
Number divides Number iff Cube divides Cube

== Theorem ==
Let $a, b \in \Z$.
Then:
:$a^3 \divides b^3 \iff a \divides b$
where $\divides$ denotes integer divisibility.
== Proof ==
Let $a^3$ and $b^3$ be cube numbers.
From the corollary to Form of Geometric Sequence of Integers:
:$\tuple {a^3, a^2 b, a b^2, b^3}$
is a geometric sequence.
Let $a, b \in \Z$ such that $a^2 \divides b^2$.
Then from First Element of Geometric Sequence that divides Last also divides Second:
:$a^3 \divides a^2 b$
Thus:
Let $a \divides b$.
Then:
== Sources ==
* : Book $\text{VIII}$. Propositions
Category:Divisors
Category:Cube Numbers
Number divides Number iff Square divides Square

== Theorem ==
Let $a, b \in \Z$.
Then:
:$a^2 \divides b^2 \iff a \divides b$
where $\divides$ denotes integer divisibility.
== Proof ==
From Between two Squares exists one Mean Proportional:
:$\tuple {a^2, ab, b^2}$
is a geometric sequence.
Let $a, b \in \Z$ such that $a^2 \divides b^2$.
Then from First Element of Geometric Sequence that divides Last also divides Second:
:$a^2 \divides a b$
Thus:
Let $a \divides b$.
Then:
== Sources ==
* : Book $\text{VIII}$. Propositions
Category:Divisors
Category:Square Numbers