nyuuzyou/wikis huggingface dataset:
Rational Number Space is not Extremally Disconnected

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ does not form an extremally disconnected space.
== Proof ==
The definition of extremally disconnected space can be stated as:
:A $T_2$ (Hausdorff) topological space $T = \struct {S, \tau}$ is extremally disconnected the closure of every open set of $T$ is open.
Hence the existence will be demonstrated of an open set in $\struct {\Q, \tau_d}$ whose closure is not open.
First we establish that $\struct {\Q, \tau_d}$ is indeed a Hausdorff space.
Indeed, we have:
:Rational Numbers form Metric Space
:Metric Space is Hausdorff
We have that $\openint 0 1$ is open in $\struct {\R, \tau_d}$.
Thus $\openint 0 1 \cap \Q$ is open in $\struct {\Q, \tau_d}$.
We show that $0$ and $1$ are limit points of $\openint 0 1 \cap \Q$.
For any $\epsilon \in \R_{>0}$:
:$\openint {-\epsilon} \epsilon \cap \Q \cap \openint 0 1 = \openint 0 \epsilon \cap \Q$
From Between two Real Numbers exists Rational Number:
:$\openint 0 \epsilon \cap \Q \ne \O$
Similarly:
:$\openint {1 - \epsilon} 1 \cap \Q \ne \O$
Thus $0$ and $1$ are limit points of $\openint 0 1 \cap \Q$.
Hence:
:$0, 1 \in \paren {\openint 0 1 \cap \Q}'$
Thus:
:$\closedint 0 1 \cap \Q \subseteq \paren {\openint 0 1 \cap \Q}^-$
where $\paren {\openint 0 1 \cap \Q}^-$ denotes the closure of $\openint 0 1 \cap \Q$.
Now we show that for any $x \notin \closedint 0 1$, $x$ is not a limit point of $\openint 0 1 \cap \Q$.
Suppose $x < 0$.
Then $\dfrac x 2 < 0$.
Thus:
:$\openint {x + \dfrac x 2} {x - \dfrac x 2} \cap \Q \cap \openint 0 1 = \O$
Similarly for $y > 1$:
:$\openint {y - \dfrac {y - 1} 2} {y + \dfrac {y - 1} 2} \cap \Q \cap \openint 0 1 = \O$
Thus:
:$\map \complement {\closedint 0 1 \cap \Q} \subseteq \map \complement {\paren {\openint 0 1 \cap \Q}^-}$
Hence we have:
:$\closedint 0 1 \cap \Q = \paren {\openint 0 1 \cap \Q}^-$
From Between two Real Numbers exists Rational Number:
:$\openint {-\epsilon} \epsilon \cap \Q \cap \map \complement {\closedint 0 1} = \openint {-\epsilon} 0 \cap \Q \ne \O$
A neighborhood of $0$ must intersect $\map \complement {\closedint 0 1}$.
So $\closedint 0 1 \cap \Q$ is not open in $\struct {\Q, \tau_d}$.
Hence $\openint 0 1$ is that open set in $\struct {\Q, \tau}$ whose closure is not open of which we were to demonstrate the existence.
Thus the result follows from definition of extremally disconnected space.
Category:Rational Number Space
Category:Extremally Disconnected Spaces
Rational Number Space is not Locally Compact Hausdorff Space

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ is not a locally compact Hausdorff Space.
== Proof ==
For $\struct {\Q, \tau_d}$ to be a locally compact Hausdorff Space, it is required that every point of $\Q$ has a compact neighborhood.
Let $x \in \Q$.
Let $N \subseteq \Q$ be a neighborhood of $x$.
Then:
:$\exists U \in \tau: x \in U \subseteq N \subseteq \Q$.
 $N$ is compact.
By Compact Set of Rational Numbers is Nowhere Dense, $N$ is nowhere dense.
Thus $N^-$ contains no open set of $\Q$ which is non-empty.
But $U$ is a non-empty open set of $\Q$.
By Set is Subset of its Topological Closure, $N \subseteq N^-$.
Because Subset Relation is Transitive:
:$U \subseteq N^-$
This is a contradiction.
Hence $\struct {\Q, \tau_d}$ is not a locally compact Hausdorff Space.
== Also see ==
* Irrational Number Space is not Locally Compact Hausdorff Space
== Sources ==
* : Part $\text {II}$: Counterexamples: $30$. The Rational Numbers: $8$
Category:Rational Number Space
Category:Locally Compact Hausdorff Spaces
Rational Number Space is not Scattered

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ is not scattered.
== Proof ==
For a space to be scattered, it needs by definition to have no subset which is dense-in-itself.
From Rational Number Space is Dense-in-itself, $\struct {\Q, \tau_d}$ is dense-in-itself.
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $30$. The Rational Numbers: $9$
Category:Rational Number Space
Category:Scattered Spaces
Rational Number Space is not Weakly Sigma-Locally Compact

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ is not weakly $\sigma$-locally compact.
== Proof ==
From Rational Number Space is not Locally Compact Hausdorff Space, $\struct {\Q, \tau_d}$ is not weakly locally compact.
Hence the result from definition of weakly $\sigma$-locally compact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $30$. The Rational Numbers: $8$
Category:Rational Number Space
Category:Weakly Sigma-Locally Compact Spaces
Rational Number as Power of Number with Itself

== Theorem ==
Every rational number in the interval $\openint {\paren {\dfrac 1 e}^{\frac 1 e} }{+\infty}$ can be written either as:
: $a^a$ for some irrational number $a$
or as:
: $n^n$ for some natural number $n$.
== Proof ==
$\dfrac \d {\d x} x^x = \dfrac \d {\d x} e^{x \ln x} = e^{x \ln x} \paren {\ln x + 1}$
So we have $\dfrac \d {\d x} x^x > 0$ for every $x > \dfrac 1 e$.
Thus $x^x: \openint {\dfrac 1 e} {+\infty} \to \openint {\paren {\dfrac 1 e}^{\frac 1 e} } {+\infty}$ is bijective.
For each $y \in \openint {\paren {\dfrac 1 e}^{\frac 1 e} } {+\infty}$, we can find some $x \in \openint {\dfrac 1 e} {+\infty}$ such that $y = x^x$.
The result above is therefore equivalent to:
:$a^a$ is irrational if $a$ is rational and not a natural number.
Suppose $a^a$ is rational for some rational number $a > \dfrac 1 e$.
We express $a^a$ and $a$ in canonical form: 
:$\paren {\dfrac m n}^{\frac m n} = \dfrac b c$
where $m, n, b, c \in \N_{> 0}$ and $\map \gcd {b, c} = \map \gcd {m, n} = 1$.
This gives $\dfrac {m^m} {n^m} = \dfrac {b^n} {c^n}$.
By Canonical Form of Rational Number is Unique:
:$n^m = c^n$
If $n = 1$ then $a = \dfrac m n$ is a natural number, and $a^a$ is rational.
 $n \ge 1$.
Then there is a prime such that $p \divides n$.
This prime must also divide $c$.
Write $n = p^i k$, $c = p^j l$, where $i, j, k, l \in \N_{> 0}$ and $\map \gcd {p, k} = \map \gcd {p, l} = 1$.
Then $n^m = p^{i m} k^m = p^{j n} l^n = c^n$.
Since $\map \gcd {p, k^m} = \map \gcd {p, l^n} = 1$, we must have $p^{i m} = p^{j n}$.
Thus $i m = j n = j k p^i$.
Since $\map \gcd {m, n} = 1$, $p \nmid m$.
So we have $p^i \divides i$.
By Absolute Value of Integer is not less than Divisors, $p^i \le i$.
However by Bernoulli's Inequality:
:$p^i \ge 2^i \ge 1 + i > i$
This is a contradiction.
Therefore if $a$ is not a natural number, $a^a$ is irrational.
This proves the result.
== Sources ==
* 
Category:Number Theory
Rational Number can be Expressed as Simple Finite Continued Fraction

== Theorem == 
Let $q \in \Q$ be a rational number.
Then $q$ can be expressed as a simple finite continued fraction.
== Proof ==
Let $q = \dfrac a b$ be a rational number expressed in canonical form.
That is $b > 0$ and $a \perp b = 1$.
By the Euclidean Algorithm, we have:
Thus from the system of equations on the , we get:
This shows that $q$ has the SFCF $\sqbrk {q_1, q_2, q_3, \ldots, q_n}$.
== Note ==
It can be seen from this proof that there is a close connection between continued fractions and the Euclidean Algorithm.
== Sources ==
* 
Category:Simple Continued Fractions
Category:Rational Numbers
Rational Number is Algebraic

== Theorem ==
Let $r \in \Q$ be a rational number.
Then $r$ is also an algebraic number.
== Proof ==
Let $r$ be expressed in the form:
:$r = \dfrac p q$
Consider the linear function in $x$:
:$q x - p = 0$
which has the solution:
:$x = \dfrac p q$
Hence the result, by definition of algebraic number.
== Sources ==
* : Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental
* 
Category:Algebraic Numbers
Category:Rational Numbers
Rational Number is Algebraic of Degree 1

== Theorem ==
Let $r \in \Q$ be a rational number.
Then $r$ is an algebraic number of degree $1$.
== Proof ==
Let $r$ be expressed in the form:
:$r = \dfrac p q$
By Rational Number is Algebraic, $r$ can be expressed as the root of the linear function:
:$q x - p = 0$
A polynomial of degree zero is a constant polynomial, and has no roots.
Hence the result by definition of degree of algebraic number.
== Sources ==
* : Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental
Category:Algebraic Numbers
Category:Rational Numbers
Rational Number is Computable Real Number

== Theorem ==
Let $r \in \Q$ be a rational number.
Then, $r$ is a computable real number.
== Proof ==
Let $\sequence {x_n}$ be defined as:
:$x_n = r$
By Constant Sequence of Rational Number is Computable:
:$\sequence {x_n}$ is a computable rational sequence.
By Computable Rational Sequence is Computable Real Sequence:
:$\sequence {x_n}$ is a computable real sequence.
By Term of Computable Real Sequence is Computable:
:$x_0 = r$ is a computable real number.
Category:Computability Theory
Rational Number plus Irrational Number is Irrational

== Theorem ==
Rational number plus irrational number is irrational.
That is, let $x \in \Q$, $y \in \R \setminus \Q$ and $x + y = z$.
Then $z \in \R \setminus \Q$.
== Proof ==
 $z \in \Q$.
By definition of rational numbers:
:$\exists a, b \in \Z, b \ne 0: x = \dfrac a b$
:$\exists c, d \in \Z, d \ne 0: z = \dfrac c d$
Then:
This shows that $y$ is rational, which is a contradiction.
By Proof by Contradiction, $z$ is irrational.
Category:Rational Numbers
Category:Irrational Numbers
Rational Numbers Null Set under Lebesgue Measure

== Theorem ==
Let $\lambda$ be $1$-dimensional Lebesgue measure on $\R$.
Let $\Q$ be the set of rational numbers.
Then:
:$\map \lambda \Q = 0$
that is, $\Q$ is a $\lambda$-null set.
== Proof ==
We have that the Rational Numbers are Countably Infinite.
The result follows from Countable Set is Null Set under Lebesgue Measure.
== Sources ==
* : $\S 4$: Problem $11 \ \text{(ii)}$
Category:Lebesgue Measure
Rational Numbers are Countably Infinite

== Theorem ==
The set $\Q$ of rational numbers is countably infinite.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Sources ==
* : Chapter $1$: Mappings: $\S 15 \epsilon$
* : Chapter $1$: Preliminaries: $1.4$ Set Notation: Infinite sets
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $155$
* : Appendix $\text{A}.6$: Cardinality: Problem $\text{A}.6.2$
* 
* 
* : Chapter $1$: General Background: $\S 2$ Countable or uncountable?
* 
* 
Category:Rational Numbers
Category:Countable Sets
Category:Rational Numbers are Countably Infinite
Rational Numbers are Countably Infinite/Proof 1

== Theorem ==
== Proof ==
The rational numbers are arranged thus:
:$\dfrac 0 1, \dfrac 1 1, \dfrac {-1} 1, \dfrac 1 2, \dfrac {-1} 2, \dfrac 2 1, \dfrac {-2} 1, \dfrac 1 3, \dfrac 2 3, \dfrac {-1} 3, \dfrac {-2} 3, \dfrac 3 1, \dfrac 3 2, \dfrac {-3} 1, \dfrac {-3} 2, \dfrac 1 4, \dfrac 3 4, \dfrac {-1} 4, \dfrac {-3} 4, \dfrac 4 1, \dfrac 4 3, \dfrac {-4} 1, \dfrac {-4} 3 \ldots$
It is clear that every rational number will appear somewhere in this list.
Thus it is possible to set up a bijection between each rational number and its position in the list, which is an element of $\N$.
== Sources ==
* : Introduction to Set Theory: $2$. Set Theoretical Equivalence and Denumerability
* : $\S 2.2$: Countable sets: Example $4$
* : Chapter $2$: 'And you do addition?': $\S 2.1$: Functions: Example $2.1.6$
* : Chapter $1$: General Background: $\S 2$ Countable or uncountable?
Category:Rational Numbers are Countably Infinite
Rational Numbers are Countably Infinite/Proof 2

== Theorem ==
== Proof ==
Let us define the mapping $\phi: \Q \to \Z \times \N$ as follows:
:$\forall \dfrac p q \in \Q: \phi \left({\dfrac p q}\right) = \left({p, q}\right)$
where $\dfrac p q$ is in canonical form.
Then $\phi$ is clearly injective.
From Cartesian Product of Countable Sets is Countable‎, we have that $\Z \times \N$ is countably infinite.
The result follows directly from Domain of Injection to Countable Set is Countable‎.
Category:Rational Numbers are Countably Infinite
Rational Numbers are Countably Infinite/Proof 3

== Theorem ==
== Proof ==
For each $n \in \N$, define $S_n$ to be the set:
:$S_n := \set {\dfrac m n: m \in \Z}$
By Integers are Countably Infinite, each $S_n$ is countably infinite.
Because each rational number can be written down with a positive denominator, it follows that:
:$\forall q \in \Q: \exists n \in \N: q \in S_n$
which is to say:
:$\ds \bigcup_{n \mathop \in \N} S_n = \Q$
By Countable Union of Countable Sets is Countable, it follows that $\Q$ is countable.
Since $\Q$ is manifestly infinite, it is countably infinite.
== Sources ==
* : Appendix $\text{A}$: Set Theory: Countable Sets
Category:Rational Numbers are Countably Infinite
Rational Numbers are Countably Infinite/Proof 4

== Theorem ==
== Proof ==
Let $Q_\pm = \set {q \in \Q: \pm q > 0}$.
For every $q \in Q_+$, there exists at least one pair $\tuple {m, n} \in \N \times \N$ such that $q = \dfrac m n$.
Therefore, we can find an injection $i: Q_+ \to \N \times \N$.
By Cartesian Product of Natural Numbers with Itself is Countable, $\N \times \N$ is countable.
Hence $Q_+$ is countable, by Domain of Injection to Countable Set is Countable.
The map $-: q \mapsto -q$ provides a bijection from $Q_-$ to $Q_+$, hence $Q_-$ is also countable.
Hence $\Q$ is countable.
== Sources ==
* : $2.5 \ \text{(v)}$
Category:Rational Numbers are Countably Infinite
Rational Numbers are Dense Subfield of P-adic Numbers

== Theorem ==
Let $p$ be any prime number.
Let $\norm {\,\cdot\,}^{\Q}_p$ be the p-adic norm on the rational numbers $\Q$.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\phi: \Q \to \Q_p$ be the mapping defined by:
:$\map \phi r = \eqclass {r, r, r, \dotsc} {}$
where $\eqclass {r, r, r, \dotsc} {}$ is the left coset in $\Q_p$ that contains the constant sequence $\sequence {r, r, r, \dotsc}$.
Then:
:$\struct{\Q, \norm {\,\cdot\,}^{\Q}_p }$ is isometrically isomorphic to $\map \phi \Q$ which is a dense subfield of $\Q_p$.
That is, $\struct{\Q, \norm {\,\cdot\,}^{\Q}_p }$ can be identified as a dense subfield of $\struct {\Q_p, \norm {\,\cdot\,}_p}$ and $\norm {\,\cdot\,}_p$ as an extension of $\norm {\,\cdot\,}^\Q_p$.
== Proof ==
From P-adic Numbers form Completion of Rational Numbers with P-adic Norm:
:$\struct {\Q_p, \norm {\,\cdot\,}_p}$ is a completion of $\struct {\Q, \norm {\,\cdot\,}^\Q_p}$
From Embedding Division Ring into Quotient Ring of Cauchy Sequences:
:the mapping $\phi: \Q \to \Q_p$ is a distance-preserving monomorphism.
From Normed Division Ring is Dense Subring of Completion:
:$\struct {\Q, \norm {\, \cdot \,}^\Q_p }$ is isometrically isomorphic to $\struct {\map \phi \Q, \norm {\, \cdot \,}_p }$ which is a dense subfield of $\struct {\Q_p, \norm {\, \cdot \,}_p }$.
Category:P-adic Number Theory
Rational Numbers are Densely Ordered

== Theorem ==
Let $a, b \in \Q$ such that $a < b$.
Then $\exists c \in \Q: a < c < b$.
That is, the set of rational numbers is densely ordered.
== Proof ==
From the definition of rational numbers, we can express $a$ and $b$ as $a = \dfrac {p_1} {q_1}, b = \dfrac {p_2} {q_2}$.
Thus from Mediant is Between:
:$\dfrac {p_1} {q_1} < \dfrac {p_1 + p_2} {q_1 + q_2} < \dfrac {p_2} {q_2}$
From Rational Numbers form Field:
:$\dfrac {p_1 + p_2} {q_1 + q_2} \in \Q$
Hence $c = \dfrac {p_1 + p_2} {q_1 + q_2}$ is an element of $\Q$ strictly between $a$ and $b$.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations
Category:Rational Numbers
Category:Densely Ordered
Rational Numbers are F-Sigma Set in Real Line

== Theorem ==
Let $\struct {\R, \tau}$ be the real number line considered asa topological space with the usual (Euclidean) topology.
Then:
:$\Q$ is an $F_\sigma$ set in $\struct {\R, \tau}$.
== Proof ==
Define the set of subsets of $\R$ as:
:$\FF := \set {\set x: x \in \Q}$
By Closed Real Interval is Closed Set:
:$\forall x \in \Q: \closedint x x = \set x$ is closed (in topological sense)
Then:
:$\forall A \in \FF: A$ is closed
By Cardinality of Set of Singletons:
:$\card \FF = \card \Q$
where $\card \FF$ denotes the cardinality of $\FF$.
By Rational Numbers are Countably Infinite:
:$\Q$ is countable.
Therefore by Set is Countable if Cardinality equals Cardinality of Countable Set:
:$\FF$ is countable.
By Union of Set of Singletons:
:$\bigcup \FF = \Q$
Thus, by definition, $\Q$ is an $F_\sigma$ set.
== Sources ==
* 
* 
Category:F-Sigma Sets
Rational Numbers are Totally Disconnected

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ is a totally disconnected space.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $6.5$: Components: Examples $6.5.1$
Category:Rational Numbers
Category:Totally Disconnected Spaces
Category:Rational Numbers are Totally Disconnected
Rational Numbers are Totally Disconnected/Proof 1

== Theorem ==
== Proof ==
Let $x, y \in \Q$ such that $x \ne y$.
From Between two Rational Numbers exists Irrational Number, there exists $\alpha \in \R \setminus \Q$ such that $x < \alpha < y$.
From Rational Numbers are not Connected, it follows that $x$ and $y$ belong to different components of $\Q$.
As $x$ and $y$ are arbitrary, it follows that no rational number is in the same component as any other rational number.
That is, the components of $\Q$ are singetons.
Hence the result, by definition of a totally disconnected space.
Category:Rational Numbers are Totally Disconnected
Rational Numbers are Well-Orderable

== Theorem ==
The set $\Q$ of rational numbers is well-orderable.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering: Discussion
Category:Rational Numbers
Category:Well-Orderings
Category:Rational Numbers are Well-Orderable
Rational Numbers are Well-Orderable/Proof 1

== Theorem ==
== Proof ==
From Rational Numbers are Countably Infinite, $\Q$ is a countable set.
The result follows from Countable Set is Well-Orderable.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering: Discussion
Category:Rational Numbers are Well-Orderable
Rational Numbers are Well-Orderable/Proof 2

== Theorem ==
== Proof ==
The rational numbers are arranged thus:
:$\dfrac 0 1, \dfrac 1 1, \dfrac {-1} 1, \dfrac 1 2, \dfrac {-1} 2, \dfrac 2 1, \dfrac {-2} 1, \dfrac 1 3, \dfrac 2 3, \dfrac {-1} 3, \dfrac {-2} 3, \dfrac 3 1, \dfrac 3 2, \dfrac {-3} 1, \dfrac {-3} 2, \dfrac 1 4, \dfrac 3 4, \dfrac {-1} 4, \dfrac {-3} 4, \dfrac 4 1, \dfrac 4 3, \dfrac {-4} 1, \dfrac {-4} 3 \ldots$
It is clear that every rational number will appear somewhere in this list.
Thus it is possible to set up a bijection $f: \Q \to \N$ between each rational number and its position in the list, which is an element of $\N$.
We set up an ordering $\preccurlyeq$ defined as:
:$\forall x, y \in \Q: x \preccurlyeq y \iff \map f x \le \map f y$
where $\le$ is the usual ordering on $\N$.
It remains to demonstrate that $\preccurlyeq$ is a well-ordering.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering: Discussion
Category:Rational Numbers are Well-Orderable
Rational Numbers are not Connected

== Theorem ==
The set of rational numbers $\Q$ is not a connected topological space.
== Proof ==
Let $\alpha \in \R$ be an irrational number.
By definition, $\alpha \notin \Q$.
Consider the sets:
:$S := \Q \cap \left({-\infty \,.\,.\, \alpha}\right)$
:$T := \Q \cap \left({\alpha \,.\,.\, \infty}\right)$
Let $x \in S$.
Let $B_\epsilon \left({x}\right)$ be the open $\epsilon$-ball of $x$ in $\Q$.
Then:
:$\forall x \in S: \exists \epsilon \in \R_{>0}: B_\epsilon \left({x}\right) \subseteq S$
by setting $\epsilon = \alpha - x$.
Similarly:
:$\forall x \in T: \exists \epsilon \in \R_{>0}: B_\epsilon \left({x}\right) \subseteq T$
by setting $\epsilon = x - \alpha$.
Thus $S$ and $T$ are open sets of $\Q$.
Then:
:$S \cup T = \Q$
:$S \cap T = \O$
:$S, T \ne \O$
So $S$ and $T$ fulfil the conditions for $S \mid T$ to be a separation of $\Q$.
Hence the result by definition of connected topological space.
== Sources ==
* : $6.2$: Connectedness: Example $6.2.9$
Category:Connected Spaces
Category:Rational Numbers
Rational Numbers are not Discrete Space

== Theorem ==
The set of rational numbers $\Q$ does not form a discrete space.
== Proof ==
Follows from:
:Rational Number Space is not Extremally Disconnected
:Discrete Space is Extremally Disconnected.
== Sources ==
* : $6.5$: Components: Examples $6.5.1$
Category:Rational Numbers
Category:Totally Disconnected Spaces
Rational Numbers form F-Sigma Set in Reals

== Theorem ==
Let $\Q$ be the set of rational numbers.
Let $\struct {\R, \tau}$ be the real number line with the usual (Euclidean) topology.
Then $\Q$ is a $F_\sigma$ set in $\R$.
== Proof ==
Let $\alpha \in \Q$ be a rational number.
From Real Number is Closed in Real Number Line, $\set \alpha$ is a closed set of $\R$.
From Rational Numbers are Countably Infinite, $\ds \bigcup_{\alpha \mathop \in \Q} \set \alpha$ is a countable union.
Thus $\Q = \ds \bigcup_{\alpha \mathop \in \Q} \set \alpha$ is a countable union of closed sets of $\R$.
Hence the result by definition of $F_\sigma$ set.
== Sources ==
* : Part $\text {II}$: Counterexamples: $30$. The Rational Numbers: $2$
Category:Rational Numbers
Category:Real Number Line with Euclidean Topology
Category:F-Sigma Sets
Rational Numbers form Field

== Theorem ==
Consider the algebraic structure $\struct {\Q, +, \times}$, where:
:$\Q$ is the set of all rational numbers
:$+$ is the operation of rational addition
:$\times$ is the operation of rational multiplication.
Then $\struct {\Q, +, \times}$ forms a field.
== Proof ==
This is demonstrated in the formal definition of rational numbers.
== Also see ==
* Definition:Field of Rational Numbers
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 1$. Rings and Fields: Example $2$
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers
* : Chapter $4$: Fields: $\S 15$. Examples of Fields: Example $14$
* : Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 88$
* : $\S 4$: Number systems $\text{I}$: The rational numbers
* : Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Example $2$
* : $\S 19$: Properties of $\Z_m$ as an algebraic system
* : $\S 55$. Special types of ring and ring elements: $(4)$
* : $\S 2.4$: The rational numbers and some finite fields
* 
* 
Category:Examples of Fields
Category:Rational Numbers
Rational Numbers form Integral Domain

== Theorem ==
The set of rational numbers $\Q$ forms an integral domain under addition and multiplication: $\struct {\Q, +, \times}$.
== Proof ==
Recall that Rational Numbers form Field.
The result then follows directly from Field is Integral Domain.
== Sources ==
* : Chapter $1$: Integral Domains: $\S 3$. Definition of an Integral Domain
Category:Examples of Integral Domains
Category:Rational Numbers
Rational Numbers form Metric Space

== Theorem ==
Let $\Q$ be the set of all rational numbers.
Let $d: \Q \times \Q \to \R$ be defined as:
:$\map d {x_1, x_2} = \size {x_1 - x_2}$
where $\size x$ is the absolute value of $x$.
Then $d$ is a metric on $\Q$ and so $\struct {\Q, d}$ is a metric space.
== Proof ==
From the definition of absolute value:
:$\size {x_1 - x_2} = \sqrt {\paren {x_1 - x_2}^2}$
This is the same as the Euclidean metric.
This is shown in Euclidean Metric on Real Vector Space is Metric to be a metric.
From Rational Numbers form Vector Space, it follows that the set of all rational numbers is a 1-dimensional Euclidean space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $30$. The Rational Numbers: $4$
Category:Rational Numbers
Category:Rational Number Space