nyuuzyou/wikis huggingface dataset:
Invertible Continuous Linear Operator has Unique Inverse

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,} }$ be the normed vector space.
Let $\map {CL} X := \map {CL} {X, X}$ be a continuous linear transformation space.
Let $I \in \map {CL} X$ be the identity element.
Suppose $A \in \map {CL} X$ is invertible.
Then there is a unique $B \in \map {CL} X$ such that $A \circ B = B \circ A = I$.
== Proof ==
Let $B_1, B_2 \in \map {CL} X$.
Suppose:
:$A \circ B_1 = I = B_1 \circ A$
:$A \circ B_2 = I = B_2 \circ A$
Then:
== Sources ==
* : Chapter $\S 2.4$: Composition of continuous linear transformations
Category:Continuous Linear Transformations
Category:Inverse Mappings
Invertible Continuous Linear Operator is Bijective

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,} }$ be the normed vector space.
Let $\map {CL} X := \map {CL} {X, X}$ be a continuous linear transformation space.
Let $I \in \map {CL} X$ be the identity element.
Suppose $A \in \map {CL} X$ is invertible.
Then $A$ is bijective.
== Proof ==
=== $A$ is injective ===
Let $x, y \in X$ be such that $\map A x = \map A y$.
Then:
:$A^{-1} \circ \map A x = A^{-1} \circ \map A y$
where $A^{-1}$ is the inverse of $A$.
By definition:
:$A^{-1} \circ A = I$
Hence:
:$x = y$
By definition, $A$ is injective.
=== $A$ is surjective ===
Let $y \in X$
Then:
:$x := \map {A^{-1} } y \in X$
Moreover:
Hence:
:$\forall y \in X : \exists x \in X : \map A x = y$
By definition, $A$ is surjective.
By definition, $A$ is bijective.
== Sources ==
* : Chapter $\S 2.4$: Composition of continuous linear transformations
Category:Continuous Linear Transformations
Category:Bijections
Invertible Element containing Identity in Power Structure

== Theorem ==
Let $\struct {S, \circ}$ be a magma.
Let identity element $e \in S$ be an identity element of $\struct {S, \circ}$.
Let $\struct {\powerset S, \circ_\PP}$ be the power structure of $\struct {S, \circ}$.
Let $X \subseteq S$ such that:
:$e \in X$
:$X$ is invertible for $\circ_PP$.
Then $X = \set e$.
== Proof ==
Let $X \subseteq S$ be invertible for $\circ_PP$ and such that $e \in X$.
From Identity Element for Power Structure, $\struct {\powerset S, \circ_\PP}$ has an identity element $J = \set e$.
We have:
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets: Exercise $9.7 \ \text {(a)}$
Category:Power Structures
Category:Inverse Elements
Invertible Element of Associative Structure is Cancellable

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure where $\circ$ is associative.
Let $\struct {S, \circ}$ have an identity element $e_S$.
An element of $\struct {S, \circ}$ which is invertible is also cancellable.
=== Corollary ===
== Proof ==
Let $a \in S$ be invertible.
Suppose $a \circ x = a \circ y$.
Then:
A similar argument shows that $x \circ a = y \circ a \implies x = y$.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Theorem $7.1$
Category:Associativity
Category:Inverse Elements
Category:Cancellability
Category:Invertible Element of Associative Structure is Cancellable
Invertible Element of Associative Structure is Cancellable/Corollary

== Theorem ==
Let $\struct {S, \circ}$ be a monoid whose identity element is $e_S$.
An element of $\struct {S, \circ}$ which is invertible is also cancellable.
== Proof ==
By definition, a monoid is an associative algebraic structure with an identity element.
The result follows from Invertible Element of Associative Structure is Cancellable.
Category:Invertible Element of Associative Structure is Cancellable
Invertible Elements of Monoid form Subgroup

== Theorem ==
Let $\struct {S, \circ}$ be a monoid whose identity element is $e$.
Let $U \subseteq S$ be the subset of $S$ consisting of the invertible elements of $S$.
Then $\struct {U, \circ}$ forms a subgroup of $S$.
== Proof ==
We have from Inverse of Identity Element is Itself that $e$ is invertible.
Hence $e \in U$ and so $U \ne \O$.
Let $x, y \in U$.
As $x$ and $y$ are invertible, it follows that $x^{-1}$ and $y^{-1}$ both exist in $S$.
Both $x^{-1}$ and $y^{-1}$ also have inverses $x$ and $y$ respectively, and so themselves are invertible.
Hence both $x^{-1} \in U$ and $y^{-1} \in U$.
Then we have:
and similarly for $\paren {y^{-1} \circ x^{-1} } \circ \paren {x \circ y}$
Hence $x \circ y$ is invertible.
The result follows by the Two-Step Subgroup Test.
== Sources ==
* : $2$: Examples of Groups and Homomorphisms: $2.3$
Category:Monoids
Category:Subgroups
Category:Inverse Elements
Invertible Elements of Monoid form Subgroup of Cancellable Elements

== Theorem ==
Let $\struct {S, \circ}$ be an monoid whose identity is $e_S$.
Let $C$ be the set of all cancellable elements of $S$.
Let $T$ be the set of all invertible elements of $S$.
Then $\struct {T, \circ}$ is a subgroup of $\struct {C, \circ}$.
== Proof ==
From Cancellable Elements of Monoid form Submonoid, $\struct {C, \circ}$ is a submonoid of $\struct {S, \circ}$.
Let its identity be $e_C$ (which may or may not be the same as $e_S$).
Let $T$ be the set of all invertible elements of $S$.
From Invertible Element of Monoid is Cancellable, all the invertible elements of $S$ are also all cancellable, so $T \subseteq C$.
Let $x, y \in T$.
Clearly $x^{-1}, y^{-1} \in T$, as if $x, y$ are invertible, then so are their inverses.
Taking the group axioms in turn:
=== ===
By Inverse of Product, $x^{-1} \circ y^{-1} \in T$.
Thus $\struct {T, \circ}$ is closed.
=== ===
This is inherited from $S$, by Subsemigroup Closure Test.
Thus $\struct {T, \circ}$ is associative.
=== ===
All the elements of $\struct {C, \circ}$ are by definition cancellable
So, by Identity of Cancellable Monoid is Identity of Submonoid:
:$e_C \in T$
Thus $\struct {T, \circ}$ has an identity element.
=== ===
By Inverse of Inverse in Monoid:
:$\struct {x^{-1} \circ y^{-1} }^{-1} \in T$
Thus every element of $\struct {T, \circ}$ has an inverse.
All the group axioms are thus seen to be fulfilled, and so $\struct {T, \circ}$ is a group.
== Sources ==
* : Chapter $4$. Groups: Exercise $13$
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Theorem $8.5$
Category:Group Theory
Category:Monoids
Category:Cancellability
Invertible Elements under Natural Number Multiplication

== Theorem ==
Let $\N$ be the natural numbers.
Let $\times$ denote multiplication.
Then the only invertible element of $\N$ for $\times$ is $1$.
== Proof ==
$m \in \N$ is invertible for $\times$.
Let $n \in \N: m \times n = 1$.
Then from Natural Numbers have No Proper Zero Divisors:
:$m \ne 0$ and $n \ne 0$
Thus, $1 \le m$ and $1 \le n$.
If $1 \le m$ then from Ordering on Natural Numbers is Compatible with Multiplication:
:$1 \le n < m \times n$
This contradicts $m \times n = 1$.
The result follows.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.13$: Corollary
Category:Natural Number Multiplication
Invertible Integers under Multiplication

== Theorem ==
The only invertible elements of $\Z$ for multiplication (that is, units of $\Z$) are $1$ and $-1$.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
Let $x > 0$ and $x y > 0$.
 first that $y \le 0$.
Then from Multiplicative Ordering on Integers and Ring Product with Zero:
:$x y \le x \, 0 = 0$
From this contradiction we deduce that $y > 0$.
Let $x > 0$ and $x y = 1$.
Then:
:$y > 0$
and by Natural Numbers are Non-Negative Integers:
:$y \in \N$
Hence by Invertible Elements under Natural Number Multiplication:
:$x = 1$
Thus $1$ is the only element of $\N$ that is invertible for multiplication.
Therefore by Natural Numbers are Non-Negative Integers and Product with Ring Negative, the result follows.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 1$. Rings and Fields: Example $1$
* : Chapter $4$. Groups: Exercise $13$
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets
* : Chapter $\text {IV}$: Rings and Fields: $20$. The Integers: Theorem $20.10$
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 26$. Divisibility: Example $47$
* : $\S 55$. Special types of ring and ring elements: $(6)$
* : Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers
Category:Integer Multiplication
Category:Units of Rings
Invertible Integers under Multiplication/Corollary 1

== Corollary to Invertible Integers under Multiplication ==
Let $a, b \in \Z$ such that $a b = 1$.
Then $a = b = \pm 1$.
== Proof ==
Let $a b = 1$.
From Integer Multiplication Identity is One, $1$ is the identity element for $\Z$.
Then, by definition, both $a$ and $b$ are invertible elements of $\Z$ for multiplication.
So by Invertible Integers under Multiplication, $a = \pm 1$ and $b = \pm 1$.
As $-1 \times 1 = -1$, it follows that either:
:$a = b = 1$
or
:$a = b = -1$.
== Sources ==
* : Chapter $2$: Integers and natural numbers: $\S 2.2$: Divisibility and factorization in $\mathbf Z$
Category:Integer Multiplication
Invertible Matrix Corresponds with Change of Basis

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $G$ be an $n$-dimensional unitary $R$-module.
Let $\sequence {a_n}$ be an ordered basis of $G$.
Let $\mathbf P = \sqbrk \alpha_n$ be a square matrix of order $n$ over $R$.
Let $\ds \forall j \in \closedint 1 n: b_j = \sum_{i \mathop = 1}^n \alpha_{i j} a_i$.
Then $\sequence {b_n}$ is an ordered basis of $G$ $\mathbf P$ is nonsingular.
== Proof ==
From Change of Basis Matrix is Nonsingular, if $\sequence {b_n}$ is an ordered basis of $G$ then $\mathbf P$ is nonsingular.
Now let $\mathbf P$ be nonsingular.
Then by the corollary to Set of Linear Transformations is Isomorphic to Matrix Space, there is an automorphism $u$ of $G$ which satisfies $\mathbf P = \sqbrk {u; \sequence {a_n} }$.
Therefore, as $\forall j \in \closedint 1 n: b_j = \map u {a_j}$, it follows that $\sequence {b_n}$ is also an ordered basis of $G$.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices: Theorem $29.3$
Category:Linear Algebra
Category:Matrix Algebra
Involution is Permutation

== Theorem ==
Let $S$ be a set.
Let $f: S \to S$ be an involution.
Then $f$ is a permutation.
== Proof ==
By definition, a permutation is a bijection from a set to itself.
Thus it is sufficient to show that $f$ is a bijection.
By definition of involution, for each $x \in S$:
:$\map f {\map f x} = x$
By Equality of Mappings:
:$f \circ f = I_S$
where $I_S$ is the identity mapping on $S$.
Thus $f$ is both a left inverse and a right inverse of itself.
The result follows from Bijection iff Left and Right Inverse.
Category:Involutions
Category:Permutations
Irrational Number/Examples/Cube Root of 2

== Example of Irrational Number ==
$\sqrt [3] 2$ is irrational.
== Proof ==
 $\sqrt [3] 2 = \dfrac m n$ for integers $m$ and $n$ such that:
:$m \perp n$
where $\perp$ denotes coprimality.
Then:
:$m^3 = 2 n^2$
Thus $2 \divides m^3$ and so $2 \divides m$.
Hence:
:$m = 2 k$
for some $k \in \Z$.
Then:
:$8 k^2 = 2 n^2$
and so $2 \divides n$.
But then we have $2 \divides m$ and $2 \divides n$
Hence $m$ and $n$ are not coprime after all.
From this contradiction the result follows.
== Sources ==
* : $\S 1$: Real Numbers: Exercise $\S 1.20 \ (6)$
Category:Examples of Irrational Numbers
Irrational Number/Examples/Square Root of 3

== Example of Irrational Number ==
$\sqrt 3$ is irrational.
== Proof ==
 $\sqrt 3 = \dfrac m n$ for integers $m$ and $n$ such that:
:$m \perp n$
where $\perp$ denotes coprimality.
Then:
:$m^2 = 3 n^2$
Thus $3 \divides m^2$ and so $3 \divides m$.
Hence:
:$m = 3 k$
for some $k \in \Z$.
Then:
:$9 k^2 = 3 n^2$
and so $3 \divides n$.
But then we have $3 \divides m$ and $3 \divides n$
Hence $m$ and $n$ are not coprime after all.
From this contradiction the result follows.
== Sources ==
* : $\S 1$: Real Numbers: Exercise $\S 1.20 \ (6)$
Category:Examples of Irrational Numbers
Irrational Number Space is Complete Metric Space

== Theorem ==
Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space under the Euclidean topology $\tau_d$.
Then $\struct {\R \setminus \Q, \tau_d}$ is completely metrizable.
== Proof ==
Let $\set {r_i}$ be an enumeration of rational numbers.
Let $d: \R \times \R \to \R$ be the Euclidean plus metric:
:$\ds \map d {x, y} := \size {x - y} + \sum_{i \mathop = 1}^\infty 2^{-i} \map \inf {1, \size {\max_{j \mathop \le i} \frac 1 {\size {x - r_j} } - \max_{j \mathop \le i} \frac 1 {\size {y - r_j} } } }$
Let $\sequence {x_n}$ converge in $\struct {\R \setminus \Q, \tau_d}$ to a rational number.
For each $x_i$ in $\sequence {x_n}$, there exists $j > i$ such that $\map d {x_i, x_j} \ge \size {x_i - x_j} + 2^{-k}$ for some $k$.
Hence $\sequence {x_n}$ is not a Cauchy sequence.
Thus if $\sequence {x_n}$ is a Cauchy sequence it converges to an irrational number.
== Sources ==
* : Part $\text {II}$: Counterexamples: $31$. The Irrational Numbers: $5$
Category:Irrational Number Space
Category:Complete Metric Spaces
Irrational Number Space is Completely Normal

== Theorem ==
Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space under the Euclidean topology $\tau_d$.
Then $\struct {\R \setminus \Q, \tau_d}$ is a completely normal space.
== Proof ==
From Euclidean Space is Complete Metric Space, a Euclidean space is a metric space.
Hence in particular $\struct {\R, \tau_d}$ is a metric space.
From Subspace of Metric Space is Metric Space, it follows that $\struct {\R \setminus \Q, \tau_d}$ is likewise a metric space.
From Metric Space fulfils all Separation Axioms it follows that $\struct {\R \setminus \Q, \tau_d}$ is a completely normal space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $31$. The Irrational Numbers: $4$
Category:Irrational Number Space
Category:Completely Normal Spaces
Irrational Number Space is Dense-in-itself

== Theorem ==
Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space under the Euclidean topology $\tau_d$.
Then $\struct {\R \setminus \Q, \tau_d}$ is dense-in-itself.
== Proof ==
Let $x \in \R \setminus \Q$.
Let $U \subseteq \R$ be an open set of $\struct {\R \setminus \Q, \tau_d}$ 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 $\R \setminus \Q$ and open real intervals of $\R$ form a basis for $\struct {\R \setminus \Q, \tau_d}$.
So there exists $V = \paren {\R \setminus \Q} \cap \openint {x - \epsilon} {x + \epsilon} \subseteq U$ for some $\epsilon > 0$ such that $x \in V$.
By Between two Real Numbers exists Irrational Number we have:
:$\exists r \in \R \setminus \Q: x < r < x + \epsilon$.
Then $r \in \paren {\R \setminus \Q} \cap \openint {x - \epsilon} {x + \epsilon}$.
Therefore $x$ is not isolated.
Hence $\struct {\R \setminus \Q, \tau_d}$ is dense-in-itself.
== Also see ==
* Rational Number Space is Dense-in-itself
== Sources ==
* : Part $\text {II}$: Counterexamples: $31$. The Irrational Numbers: $9$
Category:Irrational Number Space
Category:Denseness
Irrational Number Space is Non-Meager

== Theorem ==
Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space under the Euclidean topology $\tau_d$.
Then $\struct {\R \setminus \Q, \tau_d}$ is non-meager.
== Proof ==
From Irrational Number Space is Complete Metric Space, $\struct {\R \setminus \Q, d}$ is a complete metric space.
From the Baire Category Theorem, a complete metric space is also a Baire space.
The result follows from Baire Space is Non-Meager.
== Sources ==
* : Part $\text {II}$: Counterexamples: $31$. The Irrational Numbers: $6$
Category:Irrational Number Space
Category:Non-Meager Spaces
Irrational Number Space is Paracompact

== Theorem ==
Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space under the Euclidean topology $\tau_d$.
Then $\struct {\R \setminus \Q, \tau_d}$ 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: $31$. The Irrational Numbers: $4$
Category:Irrational Number Space
Category:Paracompact Spaces
Irrational Number Space is Second-Countable

== Theorem ==
Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space under the Euclidean topology $\tau_d$.
Then $\struct {\R \setminus \Q, \tau_d}$ is second-countable.
== Proof ==
From Irrational Numbers form Metric Space, $\struct {\R \setminus \Q, \tau_d}$ is a metric space.
From Irrational Number Space is Separable, $\struct {\R \setminus \Q, \tau_d}$ is a separable space.
The result follows from Separable Metric Space is Second-Countable.
== Sources ==
* : Part $\text {II}$: Counterexamples: $31$. The Irrational Numbers: $7$
Category:Irrational Number Space
Category:Second-Countable Spaces
Irrational Number Space is Separable

== Theorem ==
Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space under the Euclidean topology $\tau_d$.
Then $\struct {\R \setminus \Q, \tau_d}$ is separable.
== Proof ==
Let $S$ be the set defined as:
:$S = \set {\pi + q: q \in \Q}$
From Rational Numbers are Countably Infinite, $\Q$ is countable.
Therefore $S$ is also countable.
From $\pi$ is Irrational:
:$\pi \in \R \setminus \Q$
It follows from Rational Addition is Closed that:
:$\forall q \in \Q: \pi + q \in \R \setminus \Q$
and so:
:$S \subseteq \R \setminus \Q$
From Rationals plus Irrational are Everywhere Dense in Irrationals, it follows that $S$ is everywhere dense in $\R \setminus \Q$.
Thus we have constructed a countable subset $S$ of $\R \setminus \Q$ which is everywhere dense in $\R \setminus \Q$.
The result follows by definition of separable space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $31$. The Irrational Numbers: $7$
Category:Irrational Number Space
Category:Separable Spaces
Irrational Number Space is Topological Space

== Theorem ==
Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space formed by the irrational numbers $\R \setminus \Q$ under the usual (Euclidean) topology $\tau_d$.
Then $\tau_d$ forms a topology.
== Proof ==
Let $\struct {\R, \tau_d}$ be the real number space $\R$ under the Euclidean topology $\tau_d$.
By definition of irrational numbers, $\R \setminus \Q \subseteq \R$.
From Topological Subspace is Topological Space we have that $\struct {\R \setminus \Q, \tau_d}$ is a topology.
== Sources ==
* : Part $\text {II}$: Counterexamples: $31$. The Irrational Numbers
Category:Irrational Number Space
Irrational Number Space is Totally Separated

== Theorem ==
Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space under the Euclidean topology $\tau_d$.
Then $\struct {\R \setminus \Q, \tau_d}$ is totally separated.
== Proof ==
Let $x, y \in \R \setminus \Q$.
From Between two Real Numbers exists Rational Number:
:$\exists \alpha \in \Q: x < \alpha < y$
Consider the unbounded open real intervals:
:$A := \openint \gets \alpha$, $B := \openint \alpha \to$
Let:
:$U := A \cap \paren {\R \setminus \Q}$, $V := B \cap \paren {\R \setminus \Q}$
Let $\beta \in \R \setminus \Q$.
Then either:
:$(1): \quad \beta < \alpha$
in which case:
:$\beta \in U$
or:
:$(2): \quad \beta > \alpha$
in which case:
:$\beta \in V$
Thus $U \cup V = \R \setminus \Q$.
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 $\R \setminus \Q$ such that $x \in U$ and $y \in V$.
Hence the result by definition of totally separated.
== Sources ==
* : Part $\text {II}$: Counterexamples: $31$. The Irrational Numbers: $9$
Category:Irrational Number Space
Category:Totally Separated Spaces
Irrational Number Space is Zero Dimensional

== Theorem ==
Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space under the Euclidean topology $\tau_d$.
Then $\struct {\R \setminus \Q, \tau_d}$ is zero dimensional.
== Proof ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $31$. The Irrational Numbers: $9$
Category:Irrational Number Space
Category:Zero Dimensional Spaces
Irrational Number Space is not Locally Compact Hausdorff Space

== Theorem ==
Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space under the Euclidean topology $\tau_d$.
Then $\struct {\R \setminus \Q, \tau_d}$ is not a locally compact Hausdorff Space.
== Proof ==
For $\struct {\R \setminus \Q, \tau_d}$ to be a locally compact Hausdorff Space, it is required that every point of $\R \setminus \Q$ has a compact neighborhood.
Let $x \in \R \setminus \Q$.
Let $N \subseteq \R \setminus \Q$ be a neighborhood of $x$.
Then:
:$\exists U \in \tau: x \in U \subseteq N \subseteq \R \setminus \Q$.
 $N$ is compact.
By Compact Set of Irrational Numbers is Nowhere Dense, $N$ is nowhere dense.
Thus $N^-$ contains no open set of $\R \setminus \Q$ which is non-empty.
But $U$ is a non-empty open set of $\R \setminus \Q$.
By Set is Subset of its Topological Closure, $N \subseteq N^-$.
Because Subset Relation is Transitive:
:$U \subseteq N^-$
This is a contradiction.
Hence $\struct {\R \setminus \Q, \tau_d}$ is not a locally compact Hausdorff Space.
== Also see ==
* Rational Number Space is not Locally Compact Hausdorff Space
== Sources ==
* : Part $\text {II}$: Counterexamples: $31$. The Irrational Numbers: $8$
Category:Irrational Number Space
Category:Locally Compact Hausdorff Spaces
Irrational Number Space is not Scattered

== Theorem ==
Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space under the Euclidean topology $\tau_d$.
Then $\struct {\R \setminus \Q, \tau_d}$ is not scattered.
== Proof ==
For a space to be scattered, it needs by definition to have no subset which is dense-in-itself.
From Irrational Number Space is Dense-in-itself, $\struct {\R \setminus \Q, \tau_d}$ is dense-in-itself.
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $31$. The Irrational Numbers: $9$
Category:Irrational Number Space
Category:Scattered Spaces
Irrational Number Space is not Weakly Sigma-Locally Compact

== Theorem ==
Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space under the Euclidean topology $\tau_d$.
Then $\struct {\R \setminus \Q, \tau_d}$ is not weakly $\sigma$-locally compact.
== Proof ==
From Irrational Number Space is not Locally Compact Hausdorff Space, $\struct {\R \setminus \Q, \tau_d}$ is not weakly locally compact.
Hence the result from definition of weakly $\sigma$-locally compact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $31$. The Irrational Numbers: $8$
Category:Irrational Number Space
Category:Weakly Sigma-Locally Compact Spaces
Irrational Number divided by Rational Number is Irrational

== Theorem ==
Let $x$ be a irrational number. 
Let $y$ be a non-zero rational number.
Then: 
:$\dfrac x y$
is irrational.
== Proof ==
 $\dfrac x y$ is rational number.
Then there exists an integer $p_1$ and a natural number $q_1$ such that: 
:$\dfrac x y = \dfrac {p_1} {q_1}$
That is: 
:$x = \dfrac {p_1} {q_1} y$
From the fact that $y$ is rational, we similarly have that there exists an integer $p_2$ and a natural number $q_2$ such that:
:$y = \dfrac {p_2} {q_2}$
Then: 
:$x = \dfrac {p_1 p_2} {q_1 q_2}$
From Integer Multiplication is Closed, we have that $p_1 p_2$ is an integer. 
From Natural Number Multiplication is Closed, we have that $q_1 q_2$ is a natural number.
Let $p_3 = p_1 p_2$ and $q_3 = q_1 q_2$.
Then $x$ is expressible in the form: 
:$x = \dfrac {p_3} {q_3}$
where $p_3$ is an integer and $q_3$ is a natural number.
This, however, implies that $x$ is rational, which is a contradiction.
By Proof by Contradiction, we conclude that $\dfrac x y$ is irrational.
Category:Rational Numbers
Category:Irrational Numbers
Irrational Number has Periodic Continued Fraction iff Quadratic

== Theorem ==
Let $x$ be an irrational number.
Then $x$ is a quadratic irrational its continued fraction expansion is periodic.
== Proof ==
Follows from:
* Limit of Simple Infinite Periodic Continued Fraction is Quadratic Irrational
* Quadratic Irrational has Periodic Continued Fraction Expansion
Category:Simple Continued Fractions
Category:Quadratic Irrationals