nyuuzyou/wikis huggingface dataset:
Alexandroff Extension of Rational Number Space is Sequentially Compact

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Let $p$ be a new element not in $\Q$.
Let $\Q^* := \Q \cup \set p$.
Let $T^* = \struct {\Q^*, \tau^*}$ be the Alexandroff extension on $\struct {\Q, \tau_d}$.
Then $T^*$ is a sequentially compact space.
== Proof ==
The strategy here is to demonstrate that every sequence in $T^*$ is either contained in a compact subspace of $T^*$, or must contain a subsequence which converges to $p$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $35$. One Point Compactification Topology: $6$
Category:Alexandroff Extensions
Category:Sequentially Compact Spaces
Category:Rational Number Space
Alexandroff Extension of Rational Number Space is T1 Space

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Let $p$ be a new element not in $\Q$.
Let $\Q^* := \Q \cup \set p$.
Let $T^* = \struct {\Q^*, \tau^*}$ be the Alexandroff extension on $\struct {\Q, \tau_d}$.
Then $T^*$ is a $T_1$ (Fréchet) space.
== Proof ==
From Condition for Alexandroff Extension to be $T_1$ Space, $T^*$ is a $T_1$ space $\struct {\Q, \tau_d}$ is also a $T_1$ space.
From Rational Numbers form Metric Space, $\struct {\Q, d}$ is a metric space.
From Metric Space fulfils all Separation Axioms, $\struct {\Q, \tau_d}$ is a $T_1$ space.
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $35$. One Point Compactification Topology: $4$
Category:Alexandroff Extensions
Category:T1 Spaces
Category:Rational Number Space
Alexandroff Extension of Rational Number Space is not Hausdorff

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Let $p$ be a new element not in $\Q$.
Let $\Q^* := \Q \cup \set p$.
Let $T^* = \struct {\Q^*, \tau^*}$ be the Alexandroff extension on $\struct {\Q, \tau_d}$.
Then $T^*$ is not a $T_2$ (Hausdorff) space.
== Proof ==
From Condition for Alexandroff Extension to be $T_2$ Space, $T^*$ is a $T_2$ space $\struct {\Q, \tau_d}$ is a locally compact Hausdorff Space.
But from Rational Number Space is not Locally Compact Hausdorff Space, $\struct {\Q, \tau_d}$ is not a locally compact Hausdorff Space.
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $35$. One Point Compactification Topology: $4$
Category:Alexandroff Extensions
Category:Hausdorff Spaces
Category:Rational Number Space
Alexandroff Extension which is T2 Space is also T4 Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a non-empty topological space.
Let $p$ be a new element not in $S$.
Let $S^* := S \cup \set p$.
Let $T^* = \struct {S^*, \tau^*}$ be the Alexandroff extension on $S$.
Let $T^*$ be a $T_2$ (Hausdorff) space.
Then $T^*$ is a $T_4$ space.
== Proof ==
We have:
:Alexandroff Extension is Compact
:Compact Hausdorff Space is $T_4$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $34$. One Point Compactification Topology: $3$
Category:Alexandroff Extensions
Category:Hausdorff Spaces
Category:T4 Spaces
Algebra Defined by Ring Homomorphism is Algebra

== Theorem ==
Let $R$ be a commutative ring.
Let $\struct {S, +, *}$ be a ring.
Let $f : R \to S$ be a ring homomorphism.
Let the image of $f$ be a subset of the center of $S$.
Let $\struct {S_R, *}$ be the algebra defined by the ring homomorphism $f$.
Then $\struct {S_R, *}$ is an algebra over $R$.
== Proof ==
== Also see ==
* Algebra Defined by Ring Homomorphism is Associative
* Algebra Defined by Ring Homomorphism on Ring with Unity is Unitary
* Algebra Defined by Ring Homomorphism on Commutative Ring is Commutative
Category:Algebras
Algebra Defined by Ring Homomorphism is Associative

== Theorem ==
Let $R$ be a commutative ring.
Let $\struct {S, +, *}$ be a ring with unity.
Let $f: R \to S$ be a ring homomorphism.
Let the image of $f$ be a subset of the center of $S$.
Let $\struct {S_R, *}$ be the algebra defined by the ring homomorphism $f$.
Then $\struct {S_R, *}$ is an associative algebra.
== Proof ==
By definition, the multiplication of $\struct {S_R, *}$ is the ring product of $S$.
Thus it follows immediately from the fact that $S$ is a ring, that $\struct {S_R, *}$ is an associative algebra.
== Also see ==
* Algebra Defined by Ring Homomorphism is Algebra
* Algebra Defined by Ring Homomorphism on Ring with Unity is Unitary
* Algebra Defined by Ring Homomorphism on Commutative Ring is Commutative
Category:Algebras
Category:Associative Algebras
Algebra Defined by Ring Homomorphism on Commutative Ring is Commutative

== Theorem ==
Let $R$ be a commutative ring.
Let $\struct {S, +, *}$ be a commutative ring.
Let $f: R \to S$ be a ring homomorphism.
Let $\struct {S_R, *}$ be the algebra defined by the ring homomorphism $f$.
Then $\struct {S_R, *}$ is a commutative algebra.
== Proof ==
Note that by Center of Commutative Ring, the image of $f$ is indeed a subset of the center of $S$.
By definition, the multiplication of $\struct {S_R, *}$ is the ring product of $S$.
Thus it follows immediately from the fact that $S$ is a ring, that $\struct {S_R, *}$ is a commutative algebra.
== Also see ==
* Algebra Defined by Ring Homomorphism is Algebra
* Algebra Defined by Ring Homomorphism is Associative
* Algebra Defined by Ring Homomorphism on Ring with Unity is Unitary
Category:Algebras
Category:Commutative Algebras
Algebra Defined by Ring Homomorphism on Ring with Unity is Unitary

== Theorem ==
Let $R$ be a commutative ring.
Let $\struct {S, +, *}$ be a ring with unity.
Let $f: R \to S$ be a ring homomorphism.
Let $\struct {S_R, *}$ be the algebra defined by the ring homomorphism $f$.
Then $\struct {S_R, *}$ is a unitary algebra.
== Proof ==
By definition, the multiplication of $\struct {S_R, *}$ is the ring product of $S$.
Thus it follows immediately from the fact that $S$ is a ring with unity, that $\struct {S_R, *}$ is a unitary algebra.
Category:Algebras
Category:Unital Algebras
Algebra Loop/Examples/Order 3

== Example of Algebra Loop ==
The following is the Cayley table of the only operation $\circ$ on $S = \set {e, a, b}$ such that $\struct {S, \circ}$ is an algebra loop whose identity is $e$:
:$\begin{array}{r|rrr}
\circ & e & a & b \\
\hline
e & e & a & b
\\
a & a & b & e
\\
b & b & e & a
\\
\end{array}$
Hence, up to isomorphism, there is only one algebra loop with $3$ elements.
This is isomorphic to the additive group of integers modulo $3$.
== Proof ==
The initial specification allows us to populate the first few elements of the Cayley table:
:$\begin{array}{r|rrr}
\circ & e & a & b \\
\hline
e & e & a & b
\\
a & a & & 
\\
b & b & & 
\\
\end{array}$
Then we note that:
because both $a$ and $b$ already appear in the row and column of those cells.
Hence we have:
:$\begin{array}{r|rrr}
\circ & e & a & b \\
\hline
e & e & a & b
\\
a & a & & e
\\
b & b & e & 
\\
\end{array}$
and the remaining cells are likewise forced.
The isomorphism $\phi$ to the additive group of integers modulo $3$ can be established as:
The fact that this is indeed an isomorphism follows by inspection.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.8 \ \text {(a)}$
Category:Examples of Algebra Loops
Category:Examples of Cayley Tables
Category:Groups of Order 3
Algebra Loop/Examples/Order 5

== Example of Algebra Loop ==
The following is the Cayley table of an operation $\circ$ on $S = \set {e, a, b, c, d}$ such that $\struct {S, \circ}$ is an algebra loop whose identity is $e$, but which is not actually a group:
:$\begin{array}{r|rrr}
\circ & e & a & b & c & d \\
\hline
e & e & a & b & c & d
\\
a & a & e & d & b & c
\\
b & b & c & e & d & a
\\
c & c & d & a & e & b
\\
d & d & b & c & a & e
\\
\end{array}$
== Proof ==
We note that:
demonstrating that $\circ$ is not associative.
Hence $\struct {S, \circ}$ is not a group.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.8 \ \text {(c)}$
Category:Examples of Algebra Loops
Category:Examples of Cayley Tables
Algebra from Cayley-Dickson Construction is not Real Star-Algebra

== Theorem ==
Let $A$ be a $*$-algebra.
Let $A'$ be constructed from $A$ using the Cayley-Dickson construction.
Then $A'$ is not a real $*$-algebra.
== Proof ==
Let the conjugation operator on $A$ be $*$.
 $A'$ is a real $*$-algebra whose conjugation operator is $*'$.
Then by definition:
:$\forall a \in A': {a^*}' = a$
Let $a = \tuple {x, y} \in A'$.
Then by definition of the Cayley-Dickson construction:
:$x, y \in A$
By definition of the conjugation operator:
:${\tuple {x, y}^*}' = \tuple {x^*, -y}$
So for ${a^*}' = a$ it must be that $x = x^*$ and $y = -y$ from definition of Equality of Ordered Pairs.
That is, $y = 0$.
But $A'$ is a $*$-algebra, which is by definition a unitary division algebra.
So $\exists y \in A: y = 1 \ne 0$.
From that contradiction, it follows that $A'$ can not be a real $*$-algebra.
== Sources ==
* : The Octonions (2002): 2.2 The Cayley-Dickson Construction: Proposition $1$
Category:Real Star-Algebras
Category:Cayley-Dickson Construction
Algebra obtains Norm Structure through Algebra Isomorphism

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $A$ and $B$ be algebras over $\GF$.
Let $\norm {\, \cdot \,}_B$ be an algebra norm on $B$.
Let $\phi : A \to B$ be an algebra isomorphism.
Define $\norm {\, \cdot \,}_A : A \to \R$ by:
:$\norm a_A = \norm {\map \phi a}_B$
Then $\norm {\, \cdot \,}_A$ is an algebra norm on $A$.
== Proof ==
From Vector Space obtains Norm Structure through Linear Isomorphism:
:$\norm {\, \cdot \,}$ is a vector space norm on $A$.
We want to show that:
:$\norm {a b}_A \le \norm a_A \norm b_A$
for each $a, b \in A$.
Let $a, b \in A$.
We have:
So $\norm {\, \cdot \,}_A$ is an algebra norm.
Category:Normed Algebras
Algebra of Sets contains Empty Set

== Theorem ==
Let $S$ be a set.
Let $\RR$ be an algebra of sets on $S$.
Then:
:$\O \in \RR$
where $\O$ denotes the empty set.
== Proof 1 ==
== Proof 2 ==
Category:Algebra of Sets contains Empty Set
Category:Algebras of Sets
Category:Empty Set
Algebra of Sets is Boolean Algebra

== Theorem ==
An algebra of sets is a Boolean algebra.
== Proof ==
Let $\RR \subseteq \powerset S$ be a set $S$ upon which an algebra of sets has been constructed.
We identify:
We demonstrate that the Boolean algebra axioms (formulation $2$) are fulfilled.
;:
From :
:$\forall a, b \in \RR: a \cup b \in \RR$
From :
:$\forall a, b \in \RR: \relcomp S a \in \RR$
Then:
Hence:
:$\forall a, b \in \RR: a \cap b \in \RR$
and $\RR$ fulfils .
;:
We have:
and $\RR$ fulfils .
;:
We have:
and $\RR$ fulfils .
;:
We have:
and $\RR$ fulfils .
;:
We have:
and $\RR$ fulfils .
;:
and $\RR$ fulfils .
All of the Boolean algebra axioms (formulation $2$) are fulfilled.
Hence the result.
== Sources ==
* 
* 
Category:Algebras of Sets
Category:Boolean Algebras
Algebra of Sets is Closed under Intersection

== Theorem ==
Let $S$ be a set.
Let $\RR$ be an algebra of sets on $S$.
Then:
:$\forall A, B \in S: A \cap B \in \RR$
== Proof 1 ==
== Proof 2 ==
Category:Algebra of Sets is Closed under Intersection
Category:Algebras of Sets
Category:Set Intersection
Algebra of Sets is Closed under Set Difference

== Theorem ==
Let $S$ be a set.
Let $\RR$ be an algebra of sets on $S$.
Then:
:$\forall A, B \in S: A \setminus B \in \RR$
== Proof 1 ==
== Proof 2 ==
Category:Algebra of Sets is Closed under Set Difference
Category:Algebras of Sets
Category:Set Difference
Algebra of all Mappings is Algebra

== Theorem ==
Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {A, +_A, \circ_A, \ast_A}$ be an $R$-algebra. 
Let $S$ be a set.
Let $\struct {A^S, +, \circ, \ast}$ the $R$-algebra of all mappings from $S$ to $A$.
Then $\struct {A^S, +, \circ, \ast}$ is an $R$-algebra. 
== Proof ==
From Module of All Mappings is Module, $\struct {A^S, +, \circ}$ is a module. 
It remains to show that $\ast$ is $R$-bilinear.
Let $f, g, h \in A^S$ and $\lambda \in R$. 
We have, for $x \in S$: 
Similarly:
So $\struct {A^S, +, \circ, \ast}$ is an $R$-algebra. 
Category:Algebras over Rings
Algebra over Field/Examples/Vectors in 3-Space with Vector Product

== Examples of Algebras over Fields ==
Let $V$ be the vector space formed of the set of all vectors in space.
Then $\struct {V, \times}$ forms an algebra over the field of vectors in space where $\times$ is the vector cross product.
== Proof ==
== Sources ==
* 
Category:Examples of Algebras over Fields
Algebra over Field Embeds into Unitization as Ideal

== Theorem ==
Let $K$ be a field.
Let $A$ be an algebra over $K$. 
Let $A_+$ be the unitization of $A$.
Let:
:$A_0 = \set {\tuple {x, 0_K} : x \in A} \subseteq A_+$.
Then $A_0$ is an ideal in $A_+$. 
== Proof ==
This follows from Ideal of Algebra over Field Embeds into Unitization as Ideal, since $A$ is an ideal of itself.
Category:Unitizations of Algebras over Fields
Algebra over Field Embeds into Unitization as Vector Subspace

== Theorem ==
Let $K$ be a field.
Let $A$ be an algebra over $K$ that is not unital. 
Let $A_+$ be the unitization of $A$.
Let:
:$A_0 = \set {\tuple {x, 0_K} : x \in A} \subseteq A_+$.
Then $A_0$ is a vector subspace of $A_+$.
== Proof ==
This follows from Vector Subspace of Algebra over Field Embeds into Unitization as Vector Subspace.
Category:Unitizations of Algebras over Fields
Algebraic Closure of Rational Number Field is Algebraic Number Field

== Theorem ==
Let $\Q$ denote the field of rational numbers.
Let $\Bbb A$ denote the field of algebraic numbers.
The algebraic closure of $\Q$ is $\Bbb A$.
== Proof ==
== Sources ==
* 
Category:Algebraic Closures
Category:Rational Numbers
Category:Algebraic Numbers
Algebraic Curve/Examples/Circle

== Example of Algebraic Curve ==
The circle is an example of an algebraic curve.
From Cartesian Equation of Circle, a circle can be expressed by an equation in the form:
:$\paren {x - a}^2 + \paren {y - b}^2 = R^2$
For example:
:$x^2 + y^2 = 4$
is the equation defining a circle whose center is at $\tuple {0, 0}$ and whose radius is $2$.
== Sources ==
* 
Category:Examples of Algebraic Curves
Algebraic Curve/Examples/Conic Section

== Example of Algebraic Curve ==
The conic section is an example of an algebraic curve.
From Equation of Conic Section, a conic section can be expressed by an equation in the form:
:$a x^2 + b x y + c y^2 + d x + e y + f = 0$
for some $a, b, c, d, e, f \in \R$.
== Sources ==
* 
Category:Examples of Algebraic Curves
Algebraic Curve/Examples/Elliptic Curve

== Example of Algebraic Curve ==
The elliptic curve is an example of an algebraic curve.
By definition of the standard form of the elliptic curve, an elliptic curve can be expressed by an equation in the form:
:$y^3 = x^3 + a x + b$
for some $a, b \in \R$.
== Sources ==
* 
Category:Examples of Algebraic Curves
Algebraic Element of Degree 3 is not Element of Field Extension of Degree Power of 2

== Theorem ==
Let $K / F$ be a finite field extension of degree $2^m$.
Let $\alpha \in K$ be algebraic over $F$ with degree $3$.
Then $\alpha \notin K$.
== Proof ==
 $\alpha \in K$.
From Degree of Element of Finite Field Extension divides Degree of Extension:
:$\map \deg \alpha \divides \map \deg {K / F}$
But:
:$3 \nmid 2^m$
From this contradiction, it follows that $\alpha \notin K$.
== Sources ==
* : Chapter $8$: Field Extensions: $\S 38$. Simple Algebraic Extensions: Theorem $74 \ \text {(ii)}$
Category:Field Extensions
Algebraic Element of Ring Extension/Examples/Root 2

== Example of Algebraic Element of Ring Extension ==
:$\sqrt 2$ is an algebraic element over the integers $\Z$.
== Proof ==
We have that:
Thus $\sqrt 2$ is a root of $x^2 - 2$.
Hence the result by definition of algebraic element of $\Z$.
== Sources ==
* : $\S 64$. Polynomial rings over an integral domain: Definition $(2)$
Category:Examples of Algebraic Numbers
Category:Square Root of 2
Algebraic Expansion/Examples/(x + 1)^2/Proof 1

== Example of Algebraic Expansion ==
== Proof ==
Use Square of Sum:
:$\forall x, y \in \R: \paren {x + y}^2 = x^2 + 2 x y + y^2$
substituting $1$ for $y$.
Category:Examples of Algebraic Expansions
Algebraic Expansion/Examples/(x + 1)^3

== Example of Algebraic Expansion ==
The following is an algebraic expansion:
:$\paren {x + 1}^3 = x^3 + 3 x^2 + 3 x + 1$ 
== Proof 1 ==
== Proof 2 ==
== Sources ==
* 
* 
Category:Examples of Algebraic Expansions
Algebraic Function/Examples/Square Root

== Examples of Algebraic Functions ==
Let $f: \C \to \C$ be the complex function:
:$\forall z \in \C: \map f z = z^{1/2}$
Then $f$ is an algebraic function.
== Proof ==
We have that $w = z^{1/2}$ is a solution to the polynomial equation:
:$w^2 - z = 0$
The result follows.
== Sources ==
* : $2$: Functions, Limits and Continuity: The Elementary Functions: $10$
Category:Examples of Algebraic Functions
Algebraic Invariants for Group of Permutations of Variables

== Theorem ==
Let $S = \set {x_1, x_2, \ldots,x_n}$ be a set of algebraic variables.
The algebraic invariants for the group of permutations of $S$ are those generated by the elementary symmetric polynomials:
== Proof ==
== Sources ==
* 
* 
Category:Invariants
Category:Elementary Symmetric Polynomials
Algebraic Irrational Number/Examples/Root 5

== Example of Algebraic Irrational Number ==
The number $\sqrt 5$ is an example of an algebraic irrational number.
== Proof ==
We have that $\sqrt 5$ is the root of the polynomial equation $x^2 - 5 = 0$.
Hence by definition $\sqrt 5$ is an algebraic number.
From Square Root of Prime is Irrational, $\sqrt 5$ is also irrational.
Hence the result, by definition of algebraic irrational number.
== Sources ==
* 
* 
Category:Examples of Algebraic Irrational Numbers
Algebraic Number/Examples

== Examples of Algebraic Numbers ==
The algebraic numbers include the following:
=== Rational Number is Algebraic ===
=== $\sqrt 2$ is Algebraic ===
=== $\sqrt {2 + \sqrt 3}$ is Algebraic ===
=== $\sqrt [3] {2 + \sqrt 2}$ is Algebraic ===
=== $\sqrt [3] 2 + \sqrt 3$ is Algebraic ===
=== $\sqrt 3 + \sqrt 2$ is Algebraic ===
=== $2 - \sqrt 2 i$ is Algebraic ===
=== $\sqrt [3] 4 - 2 i$ is Algebraic ===
=== Golden Mean is Algebraic ===
=== Imaginary Unit is Algebraic ===
=== $-7$ is Algebraic ===
=== $\frac 5 2$ is Algebraic ===
=== $3 - i$ is Algebraic ===
=== $\sqrt [3] 6$ is Algebraic ===
=== $\frac 1 3 \paren {1 + i \sqrt 2}$ is Algebraic ===
Category:Examples of Algebraic Numbers
Algebraic Number/Examples/2 - Root 2 i

== Example of Algebraic Number ==
:$2 - \sqrt 2 i$ is an algebraic number.
== Proof ==
Let $x = 2 - \sqrt 2 i$.
We have that:
Thus $2 - \sqrt 2 i$ is a root of $x^2 - 4 x + 6 = 0$.
Hence the result by definition of algebraic number.
</onlyinclude>
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $158 \ \text{(b)}$
Category:Examples of Algebraic Numbers
Algebraic Number/Examples/5 over 2

== Example of Algebraic Number ==
:$\dfrac 5 2$ is an algebraic number.
== Proof ==
$\dfrac 5 2$ is a root of $2 x - 5 = 0$.
Hence the result by definition of algebraic number.
</onlyinclude>
== Sources ==
* 
* 
Category:Examples of Algebraic Numbers
Algebraic Number/Examples/Cube Root of (2 plus Root 2)

== Example of Algebraic Number ==
:$\sqrt [3] {2 + \sqrt 2}$ is an algebraic number.
== Proof ==
Let $x = \sqrt [3] {2 + \sqrt 2}$.
We have that:
Thus $\sqrt [3] {2 + \sqrt 2}$ is a root of $x^6 - 4 x^3 + 2 = 0$.
Hence the result by definition of algebraic number.
</onlyinclude>
== Sources ==
* : Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental
Category:Examples of Algebraic Numbers
Algebraic Number/Examples/Cube Root of 2 plus Root 3

== Example of Algebraic Number ==
:$\sqrt [3] 2 + \sqrt 3$ is an algebraic number.
== Proof ==
Let $x = \sqrt [3] 2 + \sqrt 3$.
We have that:
Thus $\sqrt [3] 2 + \sqrt 3$ is a root of a non-zero polynomial with rational coefficients, namely $x^6 - 9 x^4 - 4 x^3 + 27 x^2 - 36 x - 23 = 0$.
Hence the result by definition of algebraic number.
</onlyinclude>
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $158 \ \text{(a)}$
Category:Examples of Algebraic Numbers
Algebraic Number/Examples/Cube Root of 4 minus 2 i

== Example of Algebraic Number ==
:$\sqrt [3] 4 - 2 i$ is an algebraic number.
== Proof ==
Let $x = \sqrt [3] 4 - 2 i$.
We have that:
Thus $\sqrt [3] 4 - 2 i$ is a root of $x^6 + 12 x^4 - 8 x^3 + 48 x^2 + 96 x + 80 = 0$.
Hence the result by definition of algebraic number.
</onlyinclude>
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Miscellaneous Problems: $47 \ \text {(b)}$
Category:Examples of Algebraic Numbers
Algebraic Number/Examples/Golden Mean

== Example of Algebraic Number ==
:The golden mean $\phi = \dfrac {1 + \sqrt 5} 2$ is an algebraic number.
== Proof ==
From Golden Mean as Root of Quadratic:
:$\phi$ is a root of $x^2 - x - 1 = 0$.
Hence the result by definition of algebraic number.
</onlyinclude>
== Sources ==
* 
* 
Category:Examples of Algebraic Numbers
Category:Golden Mean
Algebraic Number/Examples/Imaginary Unit

== Example of Algebraic Number ==
:The imaginary unit $i$ is an algebraic number.
== Proof ==
We have by definition of imaginary unit that $i^2 = -1$.
Therefore $i$ is a root of $x^2 + 1 = 0$.
Hence the result by definition of algebraic number.
</onlyinclude>
== Sources ==
* : Chapter $8$: Field Extensions: $\S 38$. Simple Algebraic Extensions: Example $76$
Category:Examples of Algebraic Numbers
Algebraic Number/Examples/Root 2

== Example of Algebraic Number ==
:$\sqrt 2$ is an algebraic number.
== Proof ==
We have that:
Thus $\sqrt 2$ is a root of $x^2 - 2$.
Hence the result by definition of algebraic number.
== Sources ==
* : Chapter $8$: Field Extensions: $\S 38$. Simple Algebraic Extensions: Example $76$
* : Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental
* 
Category:Examples of Algebraic Numbers
Category:Square Root of 2
Algebraic Number/Examples/Root 3 plus Root 2

== Example of Algebraic Number ==
:$\sqrt 3 + \sqrt 2$ is an algebraic number.
== Proof ==
Let $x = \sqrt 3 + \sqrt 2$.
We have that:
Thus $\sqrt 3 + \sqrt 2$ is a root of $x^4 - 10 x^2 + 1 = 0$.
Hence the result by definition of algebraic number.
</onlyinclude>
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Miscellaneous Problems: $47 \ \text {(a)}$
Category:Examples of Algebraic Numbers
Algebraic Number/Examples/Root of (2 plus Root 3)

== Example of Algebraic Number ==
:$\sqrt {2 + \sqrt 3}$ is an algebraic number.
== Proof ==
Let $x = \sqrt {2 + \sqrt 3}$.
We have that:
Thus $\sqrt {2 + \sqrt 3}$ is a root of $x^4 - 4 x^2 + 1 = 0$.
Hence the result by definition of algebraic number.
</onlyinclude>
== Sources ==
* : Chapter $8$: Field Extensions: $\S 38$. Simple Algebraic Extensions: Example $76$
Category:Examples of Algebraic Numbers
Algebraic Numbers are Countable

== Theorem ==
The set $\Bbb A$ of algebraic numbers is countable.
== Proof ==
By definition, $\Bbb A$ is the subset of the complex numbers which consists of roots of polynomials with coefficients in $\Q$.
We can prove the theorem by a cardinality argument, counting the number of such polynomials and roots.
By Set of Polynomials over Infinite Set has Same Cardinality, the set $\Q \sqbrk x$ of polynomials over $\Q$ is countable.
Next, note that $\Bbb A$ can be written as the union of the set of roots of each polynomial.
That is:
:$\Bbb A = \ds \bigcup_{\map p x \mathop \in \Q \sqbrk x} \set {x \in \C: \map p x = 0}$
From Polynomial over Field has Finitely Many Roots, $\Bbb A$ is a union of countably many finite sets.
From Countable Union of Countable Sets is Countable it follows that $\Bbb A$ is countable.
== Sources ==
* : Introduction to Set Theory: $2$. Set Theoretical Equivalence and Denumerability
* : $\S 2$: Equivalence of Sets. The Power of a Set: Problem $4$
Category:Countable Sets
Category:Polynomial Theory
Category:Algebraic Numbers
Algebraic Numbers form Field

== Theorem ==
Let $\Bbb A$ denote the set of algebraic numbers in $\C$.
Then the algebraic structure $\struct {\Bbb A, +, \times}$ is a field.
In the above, $+$ and $\times$ denote under complex addition and complex multiplication respectively.
== Proof ==
Category:Examples of Fields
Category:Algebraic Number Theory