nyuuzyou/wikis huggingface dataset:
Ratio of Successive Local Maxima for Underdamped Free Vibration

== Theorem ==
Consider a physical system $S$ whose behaviour can be described with the second order ODE in the form:
:$(1): \quad \dfrac {\d^2 x} {\d t^2} + 2 b \dfrac {\d x} {\d t} + a^2 x = 0$
for $a, b \in \R_{>0}$.
Let $b < a$, so as to make $S$ underdamped.
:
Let $T$ be the period of oscillation of $S$.
Let $x_1$ and $x_2$ be successive local maxima of $x$.
Then:
:$\dfrac {x_1} {x_2} = e^{b T}$
== Proof ==
Let the position of $S$ be described in the canonical form:
:$(1): \quad x = \dfrac {x_0 \, a} \alpha e^{-b t} \map \cos {\alpha t - \theta}$
where:
:$\alpha = \sqrt {a^2 - b^2}$
:$\theta = \map \arctan {\dfrac b \alpha}$
From Period of Oscillation of Underdamped System is Regular, the period of oscillation $T$ is given by:
:$T = \dfrac {2 \pi} {a^2 - b^2}$
From Interval between Local Maxima for Underdamped Free Vibration, the local maxima of $x$ occur at:
:$t = 0, T, 2 T, \ldots$
Let $x_1$ occur at $n T$.
Thus $x_2$ occurs at $\paren {n + 1} T$.
Differentiating $t$:
As $T = \dfrac {2 \pi} {a^2 - b^2} = \dfrac {2 \pi} \alpha$, we have:
Hence:
as required.
== Sources ==
* : $\S 3.20$: Problem $(2)$
Category:Mathematical Physics
Ratio of Volumes of Regular Dodecahedron and Regular Icosahedron in Same Sphere

== Theorem ==
=== Lemma ===
== Proof ==
Let a regular dodecahedron, a regular icosahedron and a cube be inscribed in a given sphere.
From :
:the circle which circumscribes the regular pentagon which is the face of the regular dodecahedron is the same size as the circle which circumscribes the equilateral triangle which is the face of the regular icosahedron.
In a sphere, equal sections are equally distant from the center.
Thus the perpendiculars from the center to the faces of the regular icosahedron and regular dodecahedron are equal.
So the pyramids whose apices are the center of the sphere and whose bases are the faces of these polyhedra are of the same height.
Therefore from :
:the ratios of these pyramids are to one another as their bases.
Thus:
Therefore from :
== Sources ==
* : The So-Called Book $\text{XIV}$, by Hypsicles
* : $12$
* : $12$
Category:Cubes
Category:Regular Dodecahedra
Category:Regular Icosahedra
Ratio of Volumes of Regular Dodecahedron and Regular Icosahedron in Same Sphere/Lemma

== Lemma to Ratio of Volumes of Regular Dodecahedron and Regular Icosahedron in Same Sphere ==
== Proof ==
:
Let $AB$ be cut at $C$ in extreme and mean ratio where $AC$ is the greater segment.
Let $DE$ be cut at $F$ in extreme and mean ratio where $DF$ is the greater segment.
It is to be demonstrated that:
:$AB : AC = DE : DF$
Thus:
== Sources ==
* : The So-Called Book $\text{XIV}$, by Hypsicles
Category:Platonic Solids
Rational Addition Identity is Zero

== Theorem ==
The identity of rational number addition is $0$:
:$\exists 0 \in \Q: \forall a \in \Q: a + 0 = a = 0 + a$
== Proof ==
From the definition, the field $\struct {\Q, +, \times}$ of rational numbers is the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
From Zero of Inverse Completion of Integral Domain, for any $k \in \Z^*$, the element $\dfrac {0_D} k$ of $\Q$ serves as the zero of $\struct {\Q, +, \times}$.
Hence $\dfrac 0 k$ is the identity for $\left({\Q, +}\right)$:
Similarly for $\dfrac 0 k + \dfrac a b$.
Next we note that it is a zero:
Hence we define the zero of $\struct {\Q, +, \times}$ as $0$ and identify it with the set of all elements of $\Q$ of the form $\dfrac 0 k$ where $k \in \Z^*$.
Category:Rational Addition
Rational Addition is Associative

== Theorem ==
The operation of addition on the set of rational numbers $\Q$ is associative:
:$\forall x, y, z \in \Q: x + \paren {y + z} = \paren {x + y} + z$
== Proof ==
Follows directly from the definition of rational numbers as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
So $\struct {\Q, +, \times}$ is a field, and therefore a fortiori $+$ is associative on $\Q$.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Introduction
Category:Rational Addition
Category:Examples of Associative Operations‎
Category:Associative Law of Addition
Rational Addition is Closed

== Theorem ==
The operation of addition on the set of rational numbers $\Q$ is well-defined and closed:
:$\forall x, y \in \Q: x + y \in \Q$
== Proof ==
Follows directly from the definition of rational numbers as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
So $\struct {\Q, +, \times}$ is a field, and therefore a fortiori $+$ is well-defined and closed on $\Q$.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Introduction
* : Chapter $1$: Integral Domains: $\S 2$. Operations: Example $1$
* : Chapter $1$: Real Numbers and Functions of a Real Variable: $1.1$ Real Numbers
* : $1$: Complex Numbers: The Real Number System: $3$
Category:Rational Addition
Category:Algebraic Closure
Rational Addition is Commutative

== Theorem ==
The operation of addition on the set of rational numbers $\Q$ is commutative:
:$\forall x, y \in \Q: x + y = y + x$
== Proof ==
Follows directly from the definition of rational numbers as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
So $\struct {\Q, +, \times}$ is a field, and therefore a fortiori $+$ is commutative on $\Q$.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Introduction
Category:Rational Addition
Category:Examples of Commutative Operations
Category:Commutative Law of Addition
Rational Cut has Smallest Upper Number

== Theorem ==
Let $r \in \Q$ be rational.
Let $\alpha$ be the rational cut consisting of all rational numbers $p$ such that $p < r$.
Then $\alpha$ is indeed a cut, and has a smallest upper number that is $r$.
== Proof ==
That $\alpha$ fulfils conditions $(1)$ and $(2)$ of the definition of a cut follows directly from that definition.
Then it is noted that if $p \in \alpha$ then from Mediant is Between:
:$p < \dfrac {p + r} 2 < r$
and so $p$ cannot be the greatest element of $\alpha$.
Hence it is seen that $\alpha$ fulfils all conditions to be a cut.
Because it is not the case that $r < r$, it follows that $r \notin \alpha$.
Thus $r$ is an upper number of $\alpha$.
 $q \in \Q$ is an upper number of $\alpha$ such that $q < r$.
But then by definition of a cut, $q \in \alpha$.
That is, $q$ is not an upper number of $\alpha$.
Hence by Proof by Contradiction it follows that there is no upper number of $\alpha$ which is smaller than $r$.
That is, $r$ is the smallest upper number of $\alpha$.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.6$. Theorem
Category:Cuts
Rational Division is Closed

== Theorem ==
The set of rational numbers less zero is closed under division:
:$\forall a, b \in \Q_{\ne 0}: a / b \in \Q_{\ne 0}$
== Proof ==
From the definition of division:
:$a / b := a \times \paren {b^{-1} }$
where $b^{-1}$ is the inverse for rational multiplication.
From Non-Zero Rational Numbers under Multiplication form Infinite Abelian Group, the algebraic structure $\struct {\Q_{\ne 0}, \times}$ is a group.
From it follows that every $b \in \Q_{\ne 0}$ has an inverse element $b^{-1} \in \Q$ under multiplication.
From it follows that $\Q_{\ne 0}$ is closed under multiplication.
Hence the result:
:$\forall a, b \in \Q_{\ne 0}: a / b \in \Q_{\ne 0}$
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Introduction
* : Chapter $1$: Real Numbers and Functions of a Real Variable: $1.1$ Real Numbers
* : $1$: Complex Numbers: The Real Number System: $3$
Category:Rational Division
Category:Algebraic Closure
Rational Function/Examples/Arbitrary Example 1

== Example of Rational Functions == 
The function defined as:
:$\map f x = \dfrac {x^3 + 2 x + 3} {x + 6}$
is a rational function which is not defined at $x = 6$.
== Sources ==
* : Chapter $1$: Real Numbers and Functions of a Real Variable: $1.3$ Functions of a Real Variable: $\text {(e)}$ Rational Functions $(6)$
Category:Examples of Rational Functions
Rational Function/Examples/Arbitrary Example 2

== Example of Rational Functions == 
The function defined as:
:$\map f x = \dfrac {2 x^2 + 3 x + 4} {x^3 + 2}$
is a rational function which is not defined at $x = -\sqrt [3] 2$.
== Sources ==
* 
* 
Category:Examples of Rational Functions
Rational Multiplication Distributes over Addition

== Theorem ==
The operation of multiplication on the set of rational numbers $\Q$ is distributive over addition:
:$\forall x, y, z \in \Q: x \times \paren {y + z} = \paren {x \times y} + \paren {x \times z}$
:$\forall x, y, z \in \Q: \paren {y + z} \times x = \paren {y \times x} + \paren {z \times x}$
== Proof ==
Follows directly from the definition of rational numbers as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
So $\struct {\Q, +, \times}$ is a field, and therefore $\times$ is distributive over $+$ on $\Q$.
Category:Rational Addition
Category:Rational Multiplication
Category:Examples of Distributive Operations
Rational Multiplication Identity is One

== Theorem ==
The identity of rational number multiplication is $1$:
:$\exists 1 \in \Q: \forall a \in \Q: a \times 1 = a = 1 \times a$
== Proof ==
From the definition, the field $\struct {\Q, +, \times}$ of rational numbers is the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
From the properties of the quotient structure, elements of $\Q$ of the form $\dfrac p p$ where $p \ne 0$ act as the identity for multiplication.
From Equal Elements of Field of Quotients, we have that:
:$\dfrac p p = \dfrac {1 \times p} {1 \times p} = \dfrac 1 1$
Hence $\dfrac p p$ is the identity for $\struct {\Q, \times}$:
Similarly for $\dfrac p p \times \dfrac a b$.
Hence we define the unity of $\struct {\Q, +, \times}$ as $1$ and identify it with the set of all elements of $\Q$ of the form $\dfrac p p$ where $ \in \Z^*$.
Category:Rational Multiplication
Rational Multiplication is Associative

== Theorem ==
The operation of multiplication on the set of rational numbers $\Q$ is associative:
:$\forall x, y, z \in \Q: x \times \paren {y \times z} = \paren {x \times y} \times z$
== Proof ==
Follows directly from the definition of rational numbers as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
So $\struct {\Q, +, \times}$ is a field, and therefore a fortiori $\times$ is associative on $\Q$.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Introduction
Category:Rational Multiplication
Category:Examples of Associative Operations‎
Category:Associative Law of Multiplication
Rational Multiplication is Closed

== Theorem ==
The operation of multiplication on the set of rational numbers $\Q$ is well-defined and closed:
:$\forall x, y \in \Q: x \times y \in \Q$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Introduction
* : Chapter $1$: Integral Domains: $\S 2$. Operations: Example $1$
* : Chapter $1$: Real Numbers and Functions of a Real Variable: $1.1$ Real Numbers
* : $1$: Complex Numbers: The Real Number System: $3$
Category:Rational Multiplication
Category:Algebraic Closure
Category:Rational Multiplication is Closed
Rational Multiplication is Closed/Proof 1

== Theorem ==
== Proof ==
Follows directly from the definition of rational numbers as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
So $\struct {\Q, +, \times}$ is a field, and therefore a fortiori $\times$ is well-defined and closed on $\Q$.
Category:Rational Multiplication is Closed
Rational Multiplication is Closed/Proof 2

== Theorem ==
== Proof ==
From the definition of rational numbers, there exists four integers $p$, $q$, $r$, $s$, where:
:$q \ne 0$
:$s \ne 0$
:$\dfrac p q = x$
:$\dfrac r s = y$
We have that:
:$p \times r \in \Z$
:$q \times s \in \Z$
Since $q \ne 0$ and $s \ne 0$, we have that:
:$q \times s \ne 0$
Therefore, by the definition of rational numbers:
:$x \times y = \dfrac {p \times r} {q \times s} \in \Q$
Hence the result.
Category:Rational Multiplication is Closed
Rational Multiplication is Commutative

== Theorem ==
The operation of multiplication on the set of rational numbers $\Q$ is commutative:
:$\forall x, y \in \Q: x \times y = y \times x$
== Proof ==
Follows directly from the definition of rational numbers as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
So $\struct {\Q, +, \times}$ is a field, and therefore a fortiori $\times$ is commutative on $\Q$.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Introduction
Category:Rational Multiplication
Category:Examples of Commutative Operations
Category:Commutative Law of Multiplication
Rational Number Expressible as Sum of Reciprocals of Distinct Squares

== Theorem ==
Let $x$ be a rational number such that $0 < x < \dfrac {\pi^2} 6 - 1$.
Then $x$ can be expressed as the sum of a finite number of reciprocals of distinct squares.
== Proof ==
That no rational number such that $x \ge \dfrac {\pi^2} 6 - 1$ can be so expressed follows from Riemann Zeta Function of 2:
:$\ds \sum_{n \mathop = 1}^n \dfrac 1 {n^2} = 1 + \dfrac 1 {2^2} + \dfrac 1 {3^2} + \dotsb = \dfrac {\pi^2} 6$
That is, using all the reciprocals of distinct squares, you can never get as high as $\dfrac {\pi^2} 6 - 1$.
It remains to be shown that for all rational numbers $x$ less than $\dfrac {\pi^2} 6 - 1$, you can make $x$ with a subset of them.
== Examples ==
== Sources ==
* 
* : $0,64493 34067 \ldots$
* : Egyptian Fractions
Category:Square Numbers
Category:Reciprocals
Category:Unit Fractions
Category:Rational Number Expressible as Sum of Reciprocals of Distinct Squares
Rational Number Expressible as Sum of Reciprocals of Distinct Squares/Examples/One Half

== Examples of Rational Number Expressible as Sum of Reciprocals of Distinct Squares ==
$\dfrac 1 2$ can be expressed as the sum of a finite number of reciprocals of distinct squares as follows:
:$\dfrac 1 2 = \dfrac 1 {2^2} + \dfrac 1 {3^2} + \dfrac 1 {4^2} + \dfrac 1 {5^2} + \dfrac 1 {7^2} + \dfrac 1 {12^2} + \dfrac 1 {15^2} + \dfrac 1 {20^2} + \dfrac 1 {28^2} + \dfrac 1 {35^2}$
== Sources ==
* 
* : Egyptian Fractions
Category:Rational Number Expressible as Sum of Reciprocals of Distinct Squares
Rational Number Expressible as Sum of Reciprocals of Distinct Squares/Mistake

== Source Work ==
: 
:The Puzzles:
::Egyptian Fractions
== Mistake ==
:The sum of the series $1 + 1 / 2^2 + 1 / 3^2 + 1 / 4^2 \ldots = \pi^2 / 6$, so the sum of different Egyptian fractions whose denominators are squares cannot exceed $\pi^2 / 6$, but might equal, for example, $\frac 1 2$.
== Correction ==
It is implicit that $1$ is not included in the set of Egyptian fractions.
We have that:
Hence the sentence should end:
:... cannot exceed $\pi^2 / 6 - 1$, but might equal, for example, $\frac 1 2$.
== Sources ==
* : Egyptian Fractions
Category:Mistakes/Curious and Interesting Puzzles
Rational Number Not in Cut is Greater than Element of Cut

== Theorem ==
Let $\alpha$ be a cut.
Let $p \in \alpha$.
Let $q \in \Q$ such that $q \notin \alpha$.
Then $q > p$.
== Proof ==
We have been given that $p \in \alpha$ while $q \notin \alpha$.
 $q \le p$.
If $q = p$ then $q \in \alpha$ immediately.
If $q < p$ then $q \in \alpha$ by property $(2)$ of the definition of cut.
In either case $q \in \alpha$ which contradicts $q \notin \alpha$.
Hence by Proof by Contradiction it follows that $q > p$.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.5$. Theorem
Category:Cuts
Rational Number Space is Completely Normal

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ is a completely normal space.
== Proof ==
From Euclidean Space is Complete Metric Space, a Euclidean space is a metric space.
From Metric Space fulfils all Separation Axioms it follows that $\struct {\Q, \tau_d}$ is a completely normal space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $30$. The Rational Numbers: $4$
Category:Rational Number Space
Category:Completely Normal Spaces
Rational Number Space is Dense-in-itself

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ is dense-in-itself.
== Proof ==
Let $x \in \Q$.
Let $U \subseteq \R$ be an open set of $\struct {\Q, \tau_d}$ such that $x \in U$.
From Basis for Euclidean Topology on Real Number Line, the set of all open real intervals of $\R$ form a basis for $\struct {\R, \tau_d}$.
By Basis for Topological Subspace, the set of all intersections of $\Q$ and open real intervals of $\R$ form a basis for $\struct {\Q, \tau_d}$.
So there exists $V = \Q \cap \openint {x - \epsilon} {x + \epsilon} \subseteq U$ for some $\epsilon > 0$ such that $x \in V$.
By Between two Real Numbers exists Rational Number we have:
:$\exists r \in \Q: x < r < x + \epsilon$.
Then $r \in \Q \cap \openint {x - \epsilon} {x + \epsilon}$.
Therefore $x$ is not isolated.
Hence $\struct {\Q, \tau_d}$ is dense-in-itself.
== Also see ==
* Irrational Number Space is Dense-in-itself
== Sources ==
* : Part $\text {II}$: Counterexamples: $30$. The Rational Numbers: $9$
Category:Rational Number Space
Category:Denseness
Rational Number Space is Meager

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ is meager.
== Proof ==
From Rational Numbers are Countably Infinite, $\Q$ is a countable union of singleton subsets.
From Singleton Set is Nowhere Dense in Rational Space, each of those singleton subsets is nowhere dense in $\struct {\Q, \tau_d}$.
The result follows from definition of meager.
== Sources ==
* : Part $\text {II}$: Counterexamples: $30$. The Rational Numbers: $6$
* 
* 
Category:Rational Number Space
Category:Meager Spaces
Rational Number Space is Paracompact

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ is paracompact.
== Proof ==
From Euclidean Space is Complete Metric Space, a Euclidean space is a metric space.
The result follows from Metric Space is Paracompact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $30$. The Rational Numbers: $4$
Category:Rational Number Space
Category:Paracompact Spaces
Rational Number Space is Second-Countable

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ is second-countable.
== Proof ==
From Rational Numbers form Metric Space, $\struct {\Q, \tau_d}$ is a metric space.
From Rational Number Space is Separable, $\struct {\Q, \tau_d}$ is a separable space.
The result follows from Separable Metric Space is Second-Countable.
== Sources ==
* : Part $\text {II}$: Counterexamples: $30$. The Rational Numbers: $7$
Category:Rational Number Space
Category:Second-Countable Spaces
Rational Number Space is Separable

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ is separable.
== Proof ==
From Rational Numbers are Countably Infinite, $\Q$ is itself countable.
The result follows by Countable Space is Separable.
== Sources ==
* : Part $\text {II}$: Counterexamples: $30$. The Rational Numbers: $7$
Category:Rational Number Space
Category:Separable Spaces
Rational Number Space is Sigma-Compact

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ is $\sigma$-compact.
== Proof ==
From Rational Numbers are Countably Infinite, $\Q$ is countable.
Hence the result from definition of Countable Space is $\sigma$-Compact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $30$. The Rational Numbers: $8$
Category:Rational Number Space
Category:Sigma-Compact Spaces
Rational Number Space is Topological Space

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space formed by the rational numbers $\Q$ under the usual (Euclidean) topology $\tau_d$.
Then $\tau_d$ forms a topology.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $30$. The Rational Numbers
Category:Rational Number Space
Category:Rational Number Space is Topological Space
Rational Number Space is Topological Space/Proof 2

== Theorem ==
== Proof ==
Let $\left({\R, \tau_d}\right)$ be the real number space $\R$ under the Euclidean topology $\tau_d$.
By definition of rational numbers, $\Q \subseteq \R$.
From Topological Subspace is Topological Space we have that $\left({\Q, \tau_d}\right)$ is a topology.
Category:Rational Number Space
Rational Number Space is Totally Separated

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ is totally separated.
== Proof ==
Let $x, y \in \Q$.
From Between two Rational Numbers exists Irrational Number:
:$\exists \alpha \in \R \setminus \Q: x < \alpha < y$
Consider the unbounded open real intervals:
:$A := \openint \gets \alpha$, $B := \openint \alpha \to$
Let:
:$U := A \cap \Q, V := B \cap \Q$
Let $\beta \in \Q$.
Then either:
:$(1): \quad \beta < \alpha$
in which case:
:$\beta \in U$
or:
:$(2): \quad \beta > \alpha$
in which case:
:$\beta \in V$
Thus $U \cup V = \Q$.
Let $a \in U$.
Then $a < \alpha$ and so $a \notin V$.
Similarly, let $b \in V$.
Then $b > \alpha$ and so $b \notin U$.
Then note that $x \in U$ and $y \in V$.
Thus $U \ne \O$ and $V \ne \O$.
Thus, by definition, $U$ and $V$ constitute a separation of $\Q$ such that $x \in U$ and $y \in V$.
Hence the result by definition of totally separated.
== Sources ==
* : Part $\text {II}$: Counterexamples: $30$. The Rational Numbers: $9$
Category:Rational Number Space
Category:Totally Separated Spaces
Rational Number Space is Zero Dimensional

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ is zero dimensional.
== Proof ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $30$. The Rational Numbers: $9$
Category:Rational Number Space
Category:Zero Dimensional Spaces
Rational Number Space is not Complete Metric Space

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean metric $d$.
Then $\struct {\Q, \tau_d}$ is not a complete metric space.
== Proof ==
Proof by Counterexample:
First note that Rational Numbers form Metric Space.
It remains to be shown that $\struct {\Q, \tau_d}$ is not complete.
Consider the sequence $\sequence {a_n}$ given by:
:$a_n := \dfrac {f_{n + 1} } {f_n}$
where $\sequence {f_n}$ is the sequence of Fibonacci numbers.
By Ratio of Consecutive Fibonacci Numbers:
:$\ds \lim_{n \mathop \to \infty} a_n = \phi := \dfrac {1 + \sqrt 5} 2$
But Square Root of Prime is Irrational.
That means $\dfrac {1 + \sqrt 5} 2$ is likewise irrational.
Thus $\sequence {a_n}$ is a Cauchy sequence of rational numbers which converges to a number which is not in $\Q$.
The result follows by definition of complete metric space.
In fact, any sequence of rational numbers that converges to an irrational number (in the metric space $\R$) is a Cauchy sequence that does not converge in $\Q$.
== Also see ==
* Normed Vector Space of Rational Numbers is not Banach Space
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Introduction: $1.2$. Remark
* 
* 
* 
Category:Rational Number Space
Category:Complete Metric Spaces