nyuuzyou/wikis huggingface dataset:
Basel Problem as Infinite Product

== Theorem ==
:$\ds \dfrac {\pi^2} 6 = \prod_{p \mathop \in \mathbb P} \dfrac {p^2} {p^2 - 1}$
== Proof ==
From Sum of Reciprocals of Powers as Euler Product:
:$\ds \sum_{n \mathop \ge 1} \dfrac 1 {n^z} = \prod_p \frac 1 {1 - p^{-z} }$
for $z \in \C$ such that $\map \Re z > 1$.
Putting $z = 2$:
The result follows from Riemann Zeta Function of $2$.
== Sources ==
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
Category:Basel Problem
Bases of Finitely Generated Free Module have Equal Cardinality

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $M$ be a free $R$-module.
Let $M$ be finitely generated.
Let $B$ and $C$ be bases of $M$.
Then $B$ and $C$ are finite and have the same cardinality.
== Proof ==
== Also see ==
* Bases of Finitely Generated Vector Space have Equal Cardinality
Category:Module Theory
Category:Bases of Modules
Bases of Finitely Generated Vector Space have Equal Cardinality

== Theorem ==
Let $K$ be a division ring.
Let $G$ be a finitely generated $K$-vector space.
Then any two bases of $G$ are finite and equivalent.
== Proof ==
Since a basis is, by definition, both linearly independent and a generator, the result follows directly from Size of Linearly Independent Subset is at Most Size of Finite Generator.
== Also see ==
* Dimension Theorem for Vector Spaces
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.10$
* 
Category:Bases of Vector Spaces
Bases of Free Module have Equal Cardinality

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $M$ be a free $R$-module.
Let $B$ and $C$ be bases of $M$.
Then $B$ and $C$ are equivalent.
That is, they have the same cardinality.
== Proof ==
By definition, a basis is a generator.
By Basis of Free Module is No Greater than Generator, there exist:
:an injection $f : B \to C$
:an injection $g : C \to B$
By the Cantor-Bernstein-Schröder Theorem, $B$ and $C$ are equivalent.
== Also see ==
* Bases of Finitely Generated Free Module have Equal Cardinality
* Bases of Vector Space have Equal Cardinality
Category:Free Modules
Category:Bases of Modules
Bases of Vector Space have Equal Cardinality

== Theorem ==
Let $R$ be a division ring.
Let $V$ be a vector space over $R$.
Let $X$ and $Y$ be bases of $V$.
Then $X$ and $Y$ are equivalent.
That is, $X$ and $Y$ have the same cardinality.
== Proof ==
We will first prove that there is an injection from $X$ to $Y$.
Let $x \in X$.
By Expression of Vector as Linear Combination from Basis is Unique: General Result, there is a unique finite subset $C_x$ of $R \times Y$ such that:
:$\ds x = \sum_{\tuple {r, v} \mathop \in C_x} r \cdot v$ and
:$\forall \tuple {r, v} \in C_x: r \ne 0_R$
Define $\Phi: X \to \powerset Y$ by:
:$\map \Phi x := \Img {C_x}$
We next show that $\sequence {\map \Phi x}_{x \mathop \in X}$ satisfies the marriage condition. 
That is, for every finite subset $F$ of $X$:
:$\ds \card F \le \card {\bigcup \map \Phi F}$
Since $X$ is a basis, it is linearly independent.
By Subset of Linearly Independent Set is Linearly Independent, $F$ is also linearly independent.
By the definition of $\Phi$:
:$\ds F \subseteq \span \bigcup \map \Phi F$.
By Finite Union of Finite Sets is Finite, $\ds \bigcup \map \Phi F$ is finite.
Thus by Size of Linearly Independent Subset is at Most Size of Finite Generator:
:$\ds \size F \le \size {\bigcup \map \Phi F}$
By Hall's Marriage Theorem, there is an injection from $X$ into $Y$.
Precisely the same argument with $X$ and $Y$ interchanged shows that there is an injection from $Y$ into $X$ as well.
Thus by the Cantor-Bernstein-Schröder Theorem, $X$ and $Y$ are equivalent.
== Also known as ==
Some authors refer to this as the dimension theorem for vector spaces, but it should not be confused with e.g. the Rank Plus Nullity Theorem.
== Also see ==
* Bases of Finitely Generated Vector Space have Equal Cardinality
== Sources ==
* 
* 
* 
Category:Bases of Vector Spaces
Basic Inequality/One-Sided Shift Space of Finite Type

== Theorem ==
Let $\struct {X_\mathbf A ^+, \sigma}$ be a one-sided shift of finite type.
Let $F_\theta ^+$ be the space of Lipschitz functions on $X_\mathbf A ^+$.
Let $\size \cdot_\theta$ be the Lipschitz seminorm on $F_\theta ^+$.
Let $\norm \cdot_\infty$ be the supremum norm on $F_\theta ^+$.
Let $f \in F_\theta ^+$.
Let $u := \map \Re f$ be the real part of $f$.
Let $\LL_f$ and $\LL_u$ denote the transfer operators.
If $\LL_u$ is normalized, then there is a $C > 0$ such that:
:$\size {\LL_f ^n w}_\theta \le C \norm w_\infty + \theta ^n \size w_\theta$
for all $w \in F_\theta ^+$ and $n \in \N$.
== Proof ==
Let $x, y \in X_\mathbf A ^+$.
If $x_0 = y_0$, then:
where $N$ is the size of the alphabet of $X_\mathbf A ^+$.
If $x_0 \ne y_0$, then:
Thus:
:$\size {\LL_f w}_\theta \le \theta \size w_\theta + \tilde C \norm w_\infty$
where:
:$\tilde C := \max \set { N \size {e^w}_\theta \theta, 2 }$
In particular, for all $n \ge 1$:
Thus:
== Sources ==
* : Chapter $2$: The Ruelle Operator
Category:Ergodic Theory
Category:Functional Analysis
Basic Primitive Recursive Functions are Arithmetically Definable

== Theorem ==
Let $f: \N^k \to \N$ be a basic primitive recursive function.
Then there is a WFF $\map \phi {y, x_1, \dotsc, x_k}$ of $k + 1$ free variables and no quantifiers such that:
:$y = \map f {x_1, \dotsc, x_k}$
:$\N \models \map \phi {y \gets \sqbrk y, x_1 \gets \sqbrk {x_1}, \dotsc, x_k \gets \sqbrk {x_k} }$
where $\sqbrk a$ denotes the unary representation of $a \in \N$.
== Proof ==
=== Zero Function ===
Suppose $\map f x = 0$.
Then:
:$\map \phi {y, x} := y = 0$
Correctness is apparent.
=== Successor Function ===
Suppose $\map f x = \map s x$.
Then:
:$\map \phi {y, x} := y = \map s x$
Correctness is apparent.
=== Projection Function ===
Let $j, k \in \N$.
Suppose $\map f {x_1, \dotsc, x_k} = x_j$.
Then:
:$\map \phi {y, x_1, \dotsc, x_k} := y = x_j$
Correctness is apparent.
Category:Primitive Recursive Functions
Basic Properties of Complete Bipartite Graph

== Theorem ==
Let $G = \struct {A \mid B, E} =: K_{m, n}$ be the complete bipartite graph with $m$ vertices in $A$ and $n$ vertices in $B$.
Then $G$ has the following properties:
=== Indices of Complete Bipartite Graph Commute ===
=== Condition for Complete Bipartite Graph to be Edgeless ===
=== Null Graph is Complete Bipartite Graph ===
=== Complete Bipartite Graphs which are Trees ===
=== Complete Bipartite Graphs which are Regular ===
=== Complete Hamiltonian Bipartite Graph ===
=== Condition for Complete Bipartite Graph to be Semi-Hamiltonian ===
=== Complete Bipartite Graphs which are Complete Graphs ===
=== Complete Bipartite Graphs which are Path Graphs ===
=== Complete Bipartite Graphs which are Cycle Graphs ===
Category:Complete Bipartite Graphs
Basic Properties of Neighborhood in Metric Space

== Theorem ==
This page gathers together the basic properties of a neighborhood in a metric space.
=== $\text N 1$: Point in Metric Space has Neighborhood ===
=== $\text N 2$: Point in Metric Space is Element of its Neighborhood ===
=== $\text N 3$: Superset of Neighborhood in Metric Space is Neighborhood ===
=== $\text N 4$: Intersection of Neighborhoods in Metric Space is Neighborhood ===
=== $\text N 5$: Neighborhood in Metric Space has Subset Neighborhood ===
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Theorem $4.8$
Category:Neighborhoods
Basic Properties of Neighborhood in Topological Space

== Theorem ==
This page gathers together the basic properties of a neighborhood of a point in a topological space.
=== $\text N 1$: Point in Topological Space has Neighborhood ===
=== $\text N 2$: Point in Topological Space is Element of its Neighborhood ===
=== $\text N 3$: Superset of Neighborhood in Topological Space is Neighborhood ===
=== $\text N 4$: Intersection of Neighborhoods in Topological Space is Neighborhood ===
=== $\text N 5$: Neighborhood in Topological Space has Subset Neighborhood ===
== Sources ==
* : Chapter $3$: Topological Spaces: $\S 3$: Neighborhoods and Neighborhood Spaces: Theorem $3.1$
Category:Neighborhoods
Basic Results about Modules

== Theorem ==
Let $\struct {G, +_G}$ be an abelian group whose identity is $e$.
Let $\struct {R, +_R, \times_R}$ be a ring whose zero is $0_R$.
Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Let $x \in G, \lambda \in R, n \in \Z$.
Let $\sequence {x_m}$ be a sequence of elements of $G$.
Let $\sequence {\lambda_m}$ be a sequence of elements of $R$ that is, scalars.
Then:
=== Scalar Product with Identity ===
=== Scalar Product with Inverse ===
=== Scalar Product with Sum ===
=== Product with Sum of Scalar ===
=== Scalar Product with Product ===
== Also see ==
* Basic Results about Unitary Modules
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Theorem $26.2 \ (1) - (5)$
Category:Module Theory
Basic Results about Unitary Modules

== Theorem ==
Let $\struct {G, +_G}$ be an abelian group whose identity is $e$.
Let $\struct {R, +_R, \times_R}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $\struct {G, +_G, \circ}_R$ be an unitary $R$-module.
Let $x \in G, n \in \Z$.
Then:
=== Scalar Product with Inverse Unity ===
=== Scalar Product with Multiple of Unity ===
== Also see ==
* Basic Results about Modules
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Theorem $26.2 \ (6) - (7)$
Category:Unitary Modules
Basic Universe has Infinite Number of Elements

== Theorem ==
Let $V$ be a basic universe.
Then $V$ has an infinite number of elements.
== Proof ==
Let $a_n$ be the class defined as:
:$\forall n \in \N: a_n = \begin{cases} \O & : n = 0 \\ \set {a_{n - 1} } & : n > 0 \end {cases}$
It is shown by the Principle of Mathematical Induction that $a_n$ is a set for all $n \in \N$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$a_n$ is a set.
=== Basis for the Induction ===
$\map P 0$ is the case:
:$a_0$ is a set.
But $a_0 = \O$ by definition.
By the :
:$\O$ is a set.
:$\O \in V$
Thus $\map P 0$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$a_k$ is a set.
from which it is to be shown that:
:$a_{k + 1}$ is a set.
=== Induction Step ===
This is the induction step:
By the induction hypothesis we have that $a_k$ is a set.
From Singleton Class of Set is Set, its singleton class $\set {a_k}$ is also a set.
But by definition:
:$a_{k + 1} = \set {a_k}$
Hence $a_{k + 1}$ has been shown to be a set.
So $\map P k \implies \map P {k + 1}$ and so by the Principle of Mathematical Induction, $a_n$ is a set for all $n \in \N$.
It is now proved that $a_m \ne a_n$ when $m \ne n$.
, let $m > n$.
 $a_n = a_{m + n}$ for some $m > 0, n \ge 0$.
Let $n = 0$.
Then we would have:
:$a_m = \O$
But as $m > 0$ we have that $a_m = \set {a_{m - 1} }$ for some $m - 1 \in \N$.
But $\O \ne \set x$ for any set $x$.
Now suppose $n > 0$.
Then by Singleton Classes are Equal iff Sets are Equal:
:$a_{n - 1} = a_{n - 1 + m}$
We can repeat this step a total of $n$ times to reach:
:$a_0 = a_{0 + m}$
from which again it follows that $a_m = \O$ for some $m > 0$.
This has been shown to be false for all $m$.
Hence it cannot be the case that $a_n = a_{m + n}$ for some $m > 0, n \ge 0$.
Thus there exists an injection from $\N$ to $\set {a_n: n \in \N}$.
That is, $\set {a_n: n \in \N}$ is countably infinite by definition.
By definition of basic universe, $V$ is a universal class.
By definition of universal class, every set is an element of $V$.
Hence:
:$\forall n \in \N: a_n \in V$
and it follows that $V$ is infinite.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 4$ The pairing axiom: Discussion
Category:Basic Universe
Basic Universe is Inductive

== Theorem ==
Let $V$ be a basic universe.
Then $V$ is an inductive class.
== Proof ==
By definition of basic universe, $V$ is a class containing all sets as elements.
By the :
:the empty class $\O$ is a set.
Hence $\O$ is an element of $V$.
By the , if $x$ is a set, then $\powerset x$ is a set.
By definition of power set:
:$\set x \subseteq \powerset x$
By the , $\set x$ is a set.
By the , if $x$ is a set, then $\bigcup x$ is a set.
Thus $x \cup \set x$ is a set.
The result follows.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 2$ Definition of the Natural Numbers
Category:Basic Universe
Category:Inductive Classes
Basic Universe is Supercomplete

== Theorem ==
Let $V$ be a basic universe.
Then $V$ is supercomplete.
== Proof ==
By definition, a class $V$ is supercomplete $V$ is both transitive and swelled.
From the , $V$ is transitive.
From the , $V$ is swelled.
Hence the result.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 2$ Transitivity and supercompleteness
Category:Basic Universe
Category:Supercomplete Classes
Basic Universe is not Set

== Theorem ==
Let $V$ be a basic universe.
Then $V$ is not a set.
== Proof ==
 $V$ were a set.
Then by the the , $V$ is swelled.
That is, as $V$ is a set, every subclass of $V$ would also be a set.
From Class has Subclass which is not Element, $V$ has a subclass $S$ which is not an element of $V$.
That is:
:$\exists S \subseteq V: S \notin V$
But by definition of a basic universe, $V$ is a universal class.
That is:
:$S \in V$
This contradicts the deduction that $S \notin V$.
Hence the result by Proof by Contradiction.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 2$ Transitivity and supercompleteness
Category:Basic Universe
Basis (Hilbert Space)/Examples/L-2 Space over Interval of Zero to Two Pi

== Example of Basis (Hilbert Space) ==
Let $L^2_\C \closedint 0 {2 \pi}$ be the complex $L^2$ space over the closed interval $\closedint 0 {2 \pi}$.
For $n \in \Z$, let $e_n: \closedint 0 {2 \pi} \to \C$ be defined by:
:$\map {e_n} t = \paren{ 2 \pi }^{-1/2} \map \exp {i n t}$
Then $\set{ e_n : n \in \Z}$ is a basis for $L^2_\C \closedint 0 {2 \pi}$.
== Proof == 
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 4.$ Orthonormal Sets of Vectors and Bases
Category:Examples of Bases (Hilbert Space)
Category:Lp Spaces
Basis (Hilbert Space)/Examples/Real Vector Space

== Example of Basis (Hilbert Space) ==
Let $\R^d$ be the real vector space with $d$ dimensions.
Let $e_1, \ldots, e_d$ be the standard basis.
Then $\set{ e_1, \ldots, e_d }$ is a basis for the Hilbert space $\R^d$.
== Proof == 
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 4.$ Orthonormal Sets of Vectors and Bases: Example $4.4$
Category:Examples of Bases (Hilbert Space)
Basis (Hilbert Space)/Examples/Space of Square Summable Mappings

== Example of Basis (Hilbert Space) ==
Let $I$ be a set.
Let $\map {\ell^2} I$ be the space of square summable mappings over $I$.
For $i \in I$, define $e_i: I \to \GF$ as:
:$\map {e_i} j := \begin{cases}
1 &: i = j \\
0 &: i \ne j
\end{cases}$
Then $\set{ e_i : i \in I}$ is a basis for $\map {\ell^2} I$.
== Proof == 
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 4.$ Orthonormal Sets of Vectors and Bases: Example $4.5$
Category:Examples of Bases (Hilbert Space)
Category:Space of Square Summable Mappings
Basis Condition for Coarser Topology

== Theorem ==
Let $S$ be a set.
Let $\BB_1$ and $\BB_2$ be two bases on $S$.
Let $\tau_1$ and $\tau_2$ be the topologies generated by $\BB_1$ and $\BB_2$ respectively.
If $\BB_1$ and $\BB_2$ satisfy:
:$\forall U \in \BB_1 : \exists \AA \subseteq \BB_2: U = \bigcup \AA$
then $\tau_1$ is coarser than $\tau_2$.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
Let $\BB_1$ and $\BB_2$ satisfy:
:$\forall U \in \BB_1 : \exists \AA \subseteq \BB_2: U = \bigcup \AA$
Let $U \in \BB_1$.
Then there exists $\AA_2 \subseteq \BB_2$ such that $U = \bigcup \AA_2$.
By definition of the topology generated by $\BB_2$, it follows that $U \in \tau_2$.
Since $U$ was arbitrary, it follows that $\BB_1 \subseteq \tau_2$.
Let $W \in \tau_1$.
By definition of the topology generated by $\BB_1$:
:there exists $\AA_1 \subseteq \BB_1$ such that $W = \bigcup \AA_1$.
Since $\BB_1 \subseteq \tau_2$, it follows that $\AA_1 \subseteq \tau_2$.
By the open set axiom $(O1)$, then $W \in \tau_2$.
Since $W$ was arbitrary, it follows that $\tau_1 \subseteq \tau_2$.
The result follows.
Category:Topology
Basis Condition for Coarser Topology/Corollary 1

== Theorem ==
Let $S$ be a set.
Let $\BB_1$ and $\BB_2$ be two bases on $S$.
Let $\tau_1$ and $\tau_2$ be the topologies generated by $\BB_1$ and $\BB_2$ respectively.
If $\BB_1$ and $\BB_2$ satisfy:
:$\forall U \in \BB_1: \forall x \in U: \exists V \in \BB_2: x \in V \subseteq U$
then $\tau_1$ is coarser than $\tau_2$.
== Proof ==
Let $\BB_1$ and $\BB_2$ satisfy:
:$\forall U \in \BB_1: \forall x \in U: \exists V \in \BB_2: x \in V \subseteq U$
Let $U \in \BB_1$.
Let $\AA = \set {V \in \BB_2 : V \subseteq U}$
From Union of Family of Sets is Smallest Superset:
:$\bigcup \AA \subseteq V$
Let $x \in U$.
Then:
:$\exists V_x \in \BB_2 : x \in V_x \subseteq U$
Thus:
:$V_x \in \AA$
and
:$x \in V_x \subseteq \bigcup \AA$
Since $x$ was arbitrary, it follows that:
:$U \subseteq \bigcup \AA$
By the definition of set equality:
:$U = \bigcup \AA$
Since $U$ was arbitrary, it follows that $\BB_1$ and $\BB_2$ satisfy:
:$\forall U \in \BB_1 : \exists \AA \subseteq \BB_2: U = \bigcup \AA$
From Basis Condition for Coarser Topology, $\tau_1$ is coarser than $\tau_2$.
Category:Topology
Basis Condition for Coarser Topology/Corollary 2

== Theorem ==
Let $S$ be a set.
Let $\BB_1$ and $\BB_2$ be two bases on $S$.
Let $\tau_1$ and $\tau_2$ be the topologies generated by $\BB_1$ and $\BB_2$ respectively.
If $\BB_1 \subseteq \BB_2$ then $\tau_1$ is coarser than $\tau_2$.
== Proof ==
Let $\BB_1 \subseteq \BB_2$.
Let $U \in \BB_1$.
Let $\AA = \set U$.
Then:
:$\AA \subseteq \BB_2$
and
:$U = \bigcup \AA$
So:
:$\forall U \in \BB_1 : \exists \AA \subseteq \BB_2: U = \bigcup \AA$
From Basis Condition for Coarser Topology:
:$\tau_1$ is coarser than $\tau_2$
Category:Topology
Basis Element of Furstenberg Topology is Clopen

== Theorem ==
Let $\tau$ be the Furstenberg topology on the set of integers $\Z$.
Let $a, b \in \Z$ such that $a \ne 0$.
Then $a \Z + b$ is clopen in $\struct {\Z, \tau}$.
== Proof ==
$a \Z + b \in \tau$ by .
It remains to show:
:$\Z \setminus \paren {a \Z + b} \in \tau$
As $a \Z = \paren {-a} \Z$, we may assume $a > 0$.
If $a = 1$, then $\Z \setminus \Z = \O \in \tau$.
Thus we assume that $a \ge 2$.
Then:
That is:
:$\ds \Z \setminus \paren {a \Z + b} = \bigcup _{k \mathop \in \set {1, \ldots , a-1} } a \Z + k$
where $a \Z + k \in \tau$ for all $k \in \set {1, \ldots , a-1}$.
Therefore $\Z \setminus \paren {a \Z + b} \in \tau$.
Category:Furstenberg Topology
Basis Expansion of Irrational Number

== Theorem ==
A basis expansion of an irrational number never terminates and does not recur.
== Proof ==
We use a Proof by Contraposition.
Thus, we show that if a basis expansion of a (real) number terminates or recurs, then that number is rational.
 $x \in \R$ were to terminate in some number base $b$.
Then (using the notation of that definition):
:$\exists k \in \N: f_k = 0$
and so we can express $x$ precisely as:
:$x = \sqbrk {s . d_1 d_2 d_3 \ldots d_{k - 1} }_b$
This means:
:$x = s + \dfrac {d_1} b + \dfrac {d_2} {b^2} + \dfrac {d_3} {b^3} + \cdots + \dfrac {d_{k - 1} } {b^{k - 1} }$
This is the same as:
:$x = \dfrac {s b^{k - 1} + d_1 b^{k - 2} + d_2 b^{k - 3} + \cdots + d_{k - 1} } {b^{k - 1} }$
Both numerator and denominator are integers and so $x$ is a rational number.
Now suppose $x \in \R$ were to recur in some number base $b$.
Let $y$ be the recurring part of $x$.
That is, $y$ is obtained from $x$ by first subtracting $\floor x$ and then subtracting the non-recurring part of the basis expansion (if any).
Then we can express $y$ precisely as:
:$y = \sqbrk {0 . 0 0 \ldots 0 d_1 d_2 \ldots d_{k - 1} d_1 d_2 \ldots d_{k - 1} \ldots}_b$
for some number $m$ (possibly $m = 0$) of zeroes following the point.
We define:
:$z:= \sqbrk {0 . d_1 d_2 \ldots d_{k - 1} }_b = \dfrac {d_1} b + \dfrac {d_2} {b^2} + \dfrac {d_3} {b^3} + \cdots + \dfrac {d_{k - 1} } {b^{k - 1} }$
Then $z$ is a rational number.
Furthermore:
:$y = \dfrac z {b^m} + \dfrac z {b^{m + k} } + \dfrac z {b^{m + 2 k} } + \cdots = \dfrac z {b^m} \paren {1 + \dfrac 1 {b^k} + \dfrac 1 {b^{2 k} } + \cdots} = \dfrac z {b^m} \cdot \dfrac 1 {1 - \dfrac 1 {b^k} }$
so $y$ is a rational number as well.
Thus as $x - y$ is rational, $x$ is also rational.
== Sources ==
* : Chapter $1$: Real Numbers and Functions of a Real Variable: $1.1$ Real Numbers
* 
* 
* 
* 
Category:Irrational Numbers
Category:Basis Expansions
Category:Arithmetic
Basis Representation Theorem

== Theorem ==
Let $b \in \Z: b > 1$.
For every $n \in \Z_{> 0}$, there exists one and only one sequence $\sequence {r_j}_{0 \mathop \le j \mathop \le t}$ such that:
:$(1): \quad \ds n = \sum_{k \mathop = 0}^t r_k b^k$
:$(2): \quad \ds \forall k \in \closedint 0 t: r_k \in \N_b$
:$(3): \quad r_t \ne 0$
This unique sequence is called the representation of $n$ to the base $b$, or, informally, we can say $n$ is (written) in base $b$.
== Proof ==
Let $\map {s_b} n$ be the number of ways of representing $n$ to the base $b$.
We need to show that $\map {s_b} n = 1$ always.
Now, it is possible that some of the $r_k = 0$ in a particular representation.
So we may exclude these terms, and it won't affect the representation.
So, suppose:
:$n = r_k b^k + r_{k - 1} b^{k - 1} + \cdots + r_t b^t$
where $r_k \ne 0, r_t \ne 0$.
Then:
Note that we have already specified that $b > 1$.
So for each representation of $n$ to the base $b$, we can find a representation of $n - 1$.
If $n$ has another representation to the base $b$, then the same procedure will generate a new representation of $n - 1$.
Thus:
:$(1): \quad \map {s_b} n \le \map {s_b} {n - 1}$
Note that this holds even if $n$ has no representation at all, because if this is the case, then $\map {s_b} n = 0 \le \map {s_b} {n - 1}$.
So for $m, n \in \N$ such that $m \ge n$, the inequality $(1)$ implies the following:
:$\forall m, n: \map {s_b} m \le \map {s_b} {m - 1} \le \ldots \le \map {s_b} {n + 1} \le \map {s_b} n$
From N less than M to the N‎ and the fact that $b^n$ has at least one representation (itself), we see:
:$1 \le \map {s_b} {b^n} \le \map {s_b} n \le \map {s_b} 1 = 1$
The entries at either end of this inequality are $1$, so all the intermediate entries must also be $1$.
So $\map {s_b} n = 1$ and the theorem has been proved.
== Motivation ==
So, once we have chosen a base $b > 1$, we can express any positive integer $n$ uniquely as:
:$\ds n = \sum_{k \mathop = 0}^t {r_k b^k}: r_0, r_1, \ldots, r_t \in \set {0, 1, \ldots, b - 1}$
Then we can write $\ds n = \sum_{j \mathop = 0}^t {r_k b^k}$ as:
:$\sqbrk {r_t r_{t - 1} \ldots r_2 r_1 r_0}_b$
== Also see ==
* Definition:Number Base
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm: Theorem $24.2$
* : $\text {1-2}$ The Basis Representation Theorem: Theorem $\text {1-3}$
* : Chapter $1$: Properties of the Natural Numbers: $\S 21 \alpha$
Category:Number Bases
Category:Named Theorems