nyuuzyou/wikis huggingface dataset:
Gaussian Integers form Euclidean Domain/Proof 1

== Theorem ==
== Proof ==
We have by definition that $\Z \sqbrk i \subseteq \C$.
Let $a, b \in \Z \sqbrk i$.
We have from Modulus of Product that $\cmod a \cdot \cmod b = \cmod {a b}$.
From Complex Modulus is Non-Negative:
:$\forall a \in \C: \cmod a \ge 0$
and:
:$\cmod a = 0 \iff a = 0$
Let $a = x + i y$.
Suppose $a \in \Z \sqbrk i \ne 0$.
Then $x \ne 0$ or $y \ne 0$ and $x^2 \ge 1$ or $y^2 \ge 1$.
So:
:$\cmod a \ge 1$
Similarly, $b \in \Z \sqbrk i, b \ne 0 \implies \cmod b \ge 1$.
Thus it follows that $\cmod {a b} \ge \cmod a$ and so $\nu$ is a Euclidean valuation on $\Z \left[{i}\right]$.
Now, consider $x, y \in \Z \sqbrk i$.
We want to find $q, r \in \Z \sqbrk i$ such that $x = q y + r$.
Note that this means we want $r = y \paren {\dfrac x y - q}$ where $\dfrac x y$ is complex but not necessarily Gaussian.
We extend $\nu$ to the complex numbers and define $\nu: \C \to \C$ as:
:$\forall z \in \C: \map \nu z = \cmod z$
Then we have:
:$\map \nu r = \map \nu y \cdot \map \nu {\dfrac x y - q}$
Thus if $\map \nu {\dfrac x y - q} < 1$ we have:
:$\map \nu r < \map \nu y$
Consider the point $P = \dfrac x y$ as a point on the complex plane.
Let $Q$ lie at a point representing the Gaussian integer $q$ which lies closest to $P$.
The distance $P Q$ is at most half the length of a diagonal of a unit square in the complex plane.
Thus:
:$\map \nu {\dfrac x y - q} = \cmod {\dfrac x y - q} \le \dfrac {\sqrt 2} 2 = \dfrac 1 {\sqrt 2} < 1$
This element $q$ and the element $r$, where $\map \nu r < \map \nu y$, are the required values.
Thus $\Z \sqbrk i$ is a Euclidean domain.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 27$. Euclidean Rings: Example $52$
Category:Gaussian Integers form Euclidean Domain
Gaussian Integers form Euclidean Domain/Proof 2

== Theorem ==
== Proof ==
We have by definition that $\Z \sqbrk i \subseteq \C$.
Let $a, b \in \Z \sqbrk i$.
We have from Modulus of Product that:
:$\cmod a \cdot \cmod b = \cmod {a b}$
From Complex Modulus is Norm we have that:
:$\forall a \in \C: \cmod a \ge 0$
:$\cmod a = 0 \iff a = 0$
Suppose $a = x + i y \in \Z \sqbrk i$ and $a \ne 0$.
Then either $x \ne 0$ or $y \ne 0$, so either $x^2 \ge 1$ or $y^2 \ge 1$.
So $\cmod a^2 \ge 1$.
If also $b \in \Z \sqbrk i, b \ne 0$, then:
:$\map \nu {a b} = \cmod {a b}^2 \ge \cmod a^2 = \map \nu a$
Now, consider $x, y \in Z \sqbrk i$ with $y \ne 0$.
We want to find $q, r \in Z \sqbrk i$ such that $x = q y + r$ and $\map \nu r < \map \nu y$.
Note that this means we want $r = y \paren {\dfrac x y - q}$ where $\dfrac x y$ is complex but not necessarily Gaussian.
Consider the complex number $p = \dfrac x y = p_r + i p_i$.
We extend $\nu$ to the complex numbers and define $\nu: \C \to \C$ as:
:$\forall z \in \C: \map \nu z = \cmod z^2$
Let $q = q_r + i q_i$ be the Gaussian integer such that:
:$\map \nu {p - q} = \cmod {p - q}^2 = \paren {p_r - q_r}^2 + \paren {p_i - q_i}^2$
is minimal.
That is, $q_r$ is the nearest integer to $p_r$ and $q_i$ is the nearest integer to $p_i$.
A given real number can be at a distance at most $1/2$ from an integer, so it follows that:
:$(1): \quad \map \nu {p - q} \le \paren {\dfrac 1 2}^2 + \paren {\dfrac 1 2}^2 = \dfrac 1 2$
Now by Complex Modulus is Norm, for any two complex numbers $z_1, z_2$ we have:
:$\map \nu {z_1 z_2} = \map \nu {z_1} \, \map \nu {z_2}$
Thus we obtain:
On the other hand:
So letting $r = x - y q$ we have $\map \nu r < \map \nu y$.
Moreover we trivially have $x = q y + r$.
Thus $\Z \sqbrk i$ is a Euclidean domain.
Category:Gaussian Integers form Euclidean Domain
Gaussian Integers form Integral Domain

== Theorem ==
The ring of Gaussian integers $\struct {\Z \sqbrk i, +, \times}$ is an integral domain.
== Proof ==
The set of complex numbers $\C$ forms a field, which is by definition a division ring.
We have that $\Z \sqbrk i \subset \C$.
So from Cancellable Element is Cancellable in Subset, all non-zero elements of $\Z \sqbrk i$ are cancellable for complex multiplication.
The identity element for complex multiplication is $1 + 0 i$ which is in $\Z \sqbrk i$.
We also have that Complex Multiplication is Commutative.
From Identity of Cancellable Monoid is Identity of Submonoid, the identity element of $\struct {\Z \sqbrk i^*, \times}$ is the same as for $\struct {\C^*, \times}$.
So we have that:
:$\struct {\Z \sqbrk i, +, \times}$ is a commutative ring with unity
:All non-zero elements of $\struct {\Z \sqbrk i, +, \times}$ are cancellable.
Hence the result from definition of integral domain.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 26$. Divisibility: Example $50$
Category:Examples of Integral Domains
Category:Gaussian Integers
Gaussian Integers form Principal Ideal Domain

== Theorem ==
The ring of Gaussian integers:
:$\struct {\Z \sqbrk i, +, \times}$
forms a principal ideal domain.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $9$: Rings: Exercise $23$
Category:Gaussian Integers form Principal Ideal Domain
Category:Gaussian Integers
Category:Examples of Principal Ideal Domains
Gaussian Integers form Principal Ideal Domain/Proof 1

== Theorem ==
== Proof ==
From Gaussian Integers form Integral Domain, we have that $\struct {\Z \sqbrk i, +, \times}$ is an integral domain.
Let $a, d \in \Z \sqbrk i$ such that $d \ne 0$.
Suppose $\cmod a \ge \cmod d$.
Reference to an Argand diagram shows that one of:
:$a + d, a - d, a + i d, a - i d$
is closer to the origin than $a$ is.
So it is possible to subtract Gaussian integer multiples of $d$ from $a$ until the square of the modulus of the remainder drops below $\cmod d^2$.
That remainder can only take integer values.
Thus a Division Theorem result follows:
:$\exists q, r \in \Z \sqbrk i: a = q d + r$
where $\cmod r < \cmod d$.
Let $J$ be an arbitrary non-null ideal of $\Z \sqbrk i$.
Let $d$ be an element of minimum modulus in $J$.
Then the Division Theorem can be used to prove that $J = \ideal d$.
== Sources ==
* : Chapter $9$: Rings: Exercise $23$
Category:Gaussian Integers form Principal Ideal Domain
Gaussian Integers form Subgroup of Complex Numbers under Addition

== Theorem ==
The set of Gaussian integers $\Z \sqbrk i$, under the operation of complex addition, forms a subgroup of the set of additive group of complex numbers $\struct {\C, +}$.
== Proof ==
We will use the One-Step Subgroup Test.
This is valid, as the Gaussian integers are a subset of the complex numbers.
We note that $\Z \sqbrk i$ is not empty, as (for example) $0 + 0 i \in \Z \sqbrk i$.
Let $a + b i, c + d i \in \Z \sqbrk i$.
Then we have $-\paren {c + d i} = -c - d i$, and so:
We have that $a, b, c, d \in \Z$ and $\Z$ is an integral domain.
Therefore by definition $\Z$ is a ring.
So it follows that $a - c \in \Z$ and $b - d \in \Z$, and hence $\paren {a - c} + \paren {b - d} i \in \Z \sqbrk i$.
So by the One-Step Subgroup Test, $\Z \sqbrk i$ is a subgroup of $\struct {\C, +}$.
== Sources ==
* : Chapter $\text{II}$: Groups: Direct Products
Category:Subgroups
Category:Gaussian Integers
Gaussian Integers form Subring of Complex Numbers

== Theorem ==
The ring of Gaussian integers:
:$\struct {\Z \sqbrk i, +, \times}$
forms a subring of the set of complex numbers $\C$.
== Proof ==
From Complex Numbers form Field, $\C$ forms a field.
By definition, a field is a ring.
Thus it is possible to use the Subring Test.
We note that $\Z \sqbrk i$ is not empty, as (for example) $0 + 0 i \in \Z \sqbrk i$.
Let $a + b i, c + d i \in \Z \sqbrk i$.
Then we have $-\paren {c + d i} = -c - d i$, and so:
We have that $a, b, c, d \in \Z$ and $\Z$ is an integral domain, therefore by definition a ring.
So it follows that $a - c \in \Z$ and $b - d \in \Z$, and hence:
:$\paren {a - c} + \paren {b - d} i \in \Z \sqbrk i$
Now consider $\paren {a + b i} \paren {c + d i}$.
By the definition of complex multiplication, we have:
:$\paren {a + b i} \paren {c + d i} = \paren {a c - b d} + \paren {a d + b c} i$
As $a, b, c, d \in \Z$ and $\Z$ is a ring, it follows that:
:$a c - b d \in \Z$ and $ad + bc \in \Z$
So:
:$\paren {a + b i} \paren {c + d i} \in \Z \sqbrk i$
So by the Subring Test, $\Z \sqbrk i$ is a subring of $\C$.
== Also see ==
* Gaussian Integers form Integral Domain
* Gaussian Integers does not form Subfield of Complex Numbers
== Sources ==
* : Chapter $5$: Rings: $\S 19$. Subrings: Example $33$
* : Chapter $1$: Rings - Definitions and Examples: $2$: Some examples of rings: Ring Example $5$
* : Chapter $9$: Rings: Exercise $3$
Category:Subrings
Category:Gaussian Integers
Gaussian Isoperimetric Inequality

== Theorem ==
Let $A$ be a measurable subset of $\R^n$ endowed with the standard Gaussian measure $\gamma^n$ with the density $\dfrac {\map \exp {-\norm x^2 / 2} } {\paren {2 \pi}^{n/2} }$
Denote by:
:$A_\epsilon = \set {x \in \R^n: \map d {x, A} \le \epsilon}$
the $\epsilon$-extension of $A$.
The Gaussian isoperimetric inequality states that:
:$\ds \liminf_{\epsilon \mathop \to +0} \epsilon^{-1} \set {\map {\gamma^n} {A_\epsilon} - \map {\gamma^n} A} \ge \map \varphi {\map {\Phi^{-1} } {\map {\gamma^n} A } }$
where:
:$\map \varphi t = \dfrac {\map \exp {-t^2 / 2} } {\sqrt {2 \pi} }$
and:
:$\map \Phi t = \ds \int_{-\infty}^t \map \varphi s \rd s$
== Proof ==
Category:Measure Theory
Gaussian Rationals are Everywhere Dense in Complex Numbers

== Theorem ==
Let $\struct {\C, \cmod {\, \cdot \,}}$ be the normed vector space of complex numbers.
Let $\Q \sqbrk i = \set {a + i b: a, b \in \Q}$ be the set of Gaussian rational numbers.
Then $\Q \sqbrk i$ is everywhere dense in $\struct {\C, \cmod {\, \cdot \,}}$.
== Proof ==
Let $z = x + i y \in \C$ be a complex number.
Let $q = a + i b \in \Q \sqbrk i$ be a Gaussian rational number.
Then:
We have that rationals are everywhere dense in reals.
Then:
:$\forall x \in \R: \forall \epsilon_x \in \R_{> 0}: \exists a \in \Q: \size {x - a} < \epsilon_x$
:$\forall y \in \R: \forall \epsilon_y \in \R_{> 0}: \exists b \in \Q: \size {y - b} < \epsilon_y$
Let $\epsilon \in \R_{> 0}$.
Let $\epsilon_x, \epsilon_y \in \R_{> 0}$ be such that $\epsilon_x + \epsilon_y < \epsilon$.
Then:
:$\exists q \in \Q \sqbrk i: \cmod {z - q} < \epsilon$
$z$ and $\epsilon$ were arbitary.
Therefore:
:$\forall z \in \C: \forall \epsilon \in \R_{> 0}: \exists q \in \Q \sqbrk i: \cmod {z - q} < \epsilon$
By definition, $\Q \sqbrk i$ is everywhere dense in $\C$.
Category:Denseness
Category:Complex Numbers
Category:Gaussian Rationals
Gaussian Rationals form Number Field

== Theorem ==
The set of Gaussian rationals $\Q \sqbrk i$, under the operations of complex addition and complex multiplication, forms a number field.
== Proof ==
By definition, a number field is a subfield of the field of complex numbers $\C$.
Recall the definition of the Gaussian rationals:
:$\Q \sqbrk i = \set {z \in \C: z = a + b i: a, b \in \Q}$
From Complex Numbers form Field, $\C$ forms a field.
Thus it is possible to use the Subfield Test.
$\Q \sqbrk i$ is not empty, as (for example) $0 + 0 i \in \Q \sqbrk i$.
Let $a + b i, c + d i \in \Q \sqbrk i$.
Then we have $-\paren {c + d i} = -c - d i$, and so:
As $a, b, c, d \in \Q$ and $\Q$ is a field, it follows that $a - c \in \Q$ and $b - d \in \Q$, and hence $\paren {a - c} + \paren {b - d} i \in \Q \sqbrk i$.
Now consider $\paren {a + b i} \paren {c + d i}$.
By the definition of complex multiplication, we have:
:$\paren {a + b i} \paren {c + d i} = \paren {a c - b d} + \paren {a d + b c} i$
As $a, b, c, d \in \Q$ and $\Q$ is a field, it follows that $a c - b d \in \Q$ and $ad + bc \in \Q$ and so $\paren {a + b i} \paren {c + d i} \in \Q \sqbrk i$.
Finally, let $z = x + y i \in \Q \sqbrk i_{\ne 0} = \Q \sqbrk i \setminus \set {0 + 0 i}$, that is such that $x + y i \ne 0 + 0 i$.
Then, by Inverse for Complex Multiplication:
:$\dfrac 1 {x + y i} = \dfrac {x - y i} {x^2 + y^2}$
As $x$ and $y$ are not both zero, it follows that:
:$x^2 + y^2 \ne 0$ and so $x^2 + y^2 \in \Q_{\ne 0}$
Thus it follows that either $\dfrac x {x^2 + y^2} \in \Q_{\ne 0}$ or $\dfrac y {x^2 + y^2} \in \Q_{\ne 0}$ (or both, or course).
Thus $\dfrac 1 {x + y i} \in \Q \sqbrk i_{\ne 0}$.
So by the Subfield Test, $\Q \sqbrk i$ is a subfield of $\C$.
== Sources ==
* : Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 88 \gamma$
Category:Number Fields
Category:Gaussian Rationals
Gelfand's Spectral Radius Formula/Bounded Linear Operator

== Theorem ==
Let $\struct {X, \norm \cdot _X}$ be a Banach space over $\C$.
Let $\map B X$ be the set of bounded linear operators on $X$.
Let $\norm \cdot_{\map B X}$ denote the operator norm on $\map B X$.
Let $T \in \map B X$.
Let $\size {\map \sigma T}$ be the spectral radius of $T$.
Then:
:$\ds \size {\map \sigma T} = \lim_{n \mathop \to \infty} \paren {\norm {T^n}_{\map B X} }^{1/n} = \inf_{n \mathop \in \N_{>0} } \paren {\norm {T^n}_{\map B X} }^{1/n}$
== Proof ==
Let $z \in \C$ be such that:
:$\ds \cmod z > \inf_{n \mathop \in \N_{>0} } \paren {\norm {T^n}_{\map B X} }^{1/n}$
That is, there exists an $m \in \N_{>0}$ such that:
:$\ds \cmod z > \paren {\norm {T^m}_{\map B X} }^{1/m}$
Then:
exists, that is:
:$z \in \C \setminus \map \sigma T$
Therefore:
:$\forall z \in \C : \ds z \in \map \sigma T \implies \cmod z \le \inf_{n \mathop \in \N_{>0} } \paren {\norm{T^n}_{\map B X} }^{1/n}$
By definition of spectral radius we have:
:$\ds \size {\map \sigma T} \le \inf_{n \mathop \in \N_{>0} } \paren {\norm{T^n}_{\map B X} }^{1/n}$
It remains to show:
:$\ds \size{\map \sigma T} \ge \limsup_{n \mathop \to \infty} \paren {\norm {T^n}_{\map B X} }^{1/n}$
Let $\struct { {\map B X}^\ast, \norm {\cdot}_{ {\map B X}^\ast} }$ be the normed dual space of $\struct {\map B X, \norm \cdot_{\map B X} }$.
For $\ell \in {\map B X}^\ast$, we define a complex function $F_\ell : \C \setminus \map \sigma T \to \C$ by:
:$\map {F_\ell} z := \map \ell {\paren {T - z I}^{-1} }$
By Resolvent Mapping is Analytic, $F_\ell$ is analytic on $\C \setminus \map \sigma T$.
Moreover, if $\cmod z > \norm T_{\map B T}$, then:
Thus for all $\delta > 0$ and $m \in \N$:
By Existence of Support Functional, for each $m \in \N$ there is an $\ell_m \in {\map B X}^\ast$ such that:
:$\norm {\ell_m}_{ {\map B X}^\ast} = 1$
:$\map {\ell_m} {T^m} = \norm {T^m}_{\map B T}$
Thus:
Letting $m \to \infty$, we obtain:
:$\ds \limsup_{m \mathop \to \infty} \paren {\norm {T^m}_{\map B X} }^{1 / m} \le \size {\map \sigma T} + \delta$
The result follows by $\delta \to 0$.
== Sources ==
* : $17.1$: Normed Algebras
Category:Bounded Linear Operators
Category:Spectral Theory
Category:Gelfand's Spectral Radius Formula
Gelfand-Mazur Theorem

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a unital normed algebra over $\C$ where:
:$\map G A = A \setminus \set { {\mathbf 0}_A}$ 
where $\map G A$ is the group of units of $A$.
Then $A$ is isometrically algebra isomorphic to $\C$. 
== Proof ==
Define $\theta : \C \to A$ by: 
:$\map \theta \lambda = \lambda {\mathbf 1}_A$
for each $\lambda \in \C$.
For $z, w, \lambda \in \C$ we have:
so $\theta$ is linear. 
Further, for $\lambda, \mu \in \C$ we have:
:$\map \theta {\lambda \mu} = \paren {\lambda \mu} {\mathbf 1}_A = \paren {\lambda {\mathbf 1}_A} \paren {\mu {\mathbf 1}_A} = \map \theta \lambda \map \theta \lambda$
Hence $\theta$ is an algebra homomorphism.
For $\lambda \in \C$, we have: 
Hence $\theta$ is a isometry.
It remains to show that $\theta$ is surjective.
Let $x \in A$. 
From Spectrum of Element of Banach Algebra is Non-Empty: Corollary, we have $\map {\sigma_A} x \ne \O$ where $\map {\sigma_A} x$ is the spectrum of $x$ in $A$. 
Let $\lambda \in \map {\sigma_A} x$. 
Then $\lambda {\mathbf 1}_A - x \in A \setminus \map G A = \set { {\mathbf 0}_A}$. 
Hence we have $x = \lambda {\mathbf 1}_A = \map \theta \lambda$. 
That is, $x \in \Img \theta$. 
So $\theta$ is surjective.
So $\theta$ is an isometric isomorphism onto $\C$.
Hence $A$ is isometrically isomorphic to $\C$.
Category:Normed Algebras
Gelfand-Naimark Theorem/Commutative Case/Unital

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a commutative unital $\text C^\ast$-algebra.
Let $\Phi_A$ be the spectrum of $A$.
Let $\struct {\map \CC {\Phi_A}, \overline \cdot, \norm {\, \cdot \,} }$ be the $\text C^\ast$-algebra of continuous complex-valued functions vanishing at infinity.
Define $G : A \to \map \CC {\Phi_A}$ by:
:$\map G a = \hat a$
where $\hat a$ is the Gelfand transform of $a$.
Then $G$ is a isometric unital $\ast$-algebra isomorphism.
That is, the commutative unital $\text C^\ast$-algebras are precisely $\struct {\map \CC K, \overline \cdot, \norm {\, \cdot \,} }$ for compact Hausdorff spaces $K$, up to isometric $\ast$-algebra isomorphism. 
== Proof ==
From Complex-Valued Function on Compact Hausdorff Space Vanishes at Infinity, we have:
:$\map {\CC_0} K = \map \CC K$
for every compact Hausdorff space $K$.
Hence from Space of Continuous Functions Vanishing at Infinity is C*-Algebra, $\struct {\map \CC K, \overline \cdot, \norm {\, \cdot \,} }$ is indeed a $\text C^\ast$-algebra for each compact Hausdorff space $K$.
Finally, recall that $\Phi_A$ is a compact Hausdorff space from Spectrum of Banach Algebra is Weak-$\ast$ Locally Compact Hausdorff Space: Corollary.
Now from Gelfand-Naimark Theorem: Commutative Case, $G$ is a isometric $\ast$-algebra isomorphism.
Further, defining $\mathbf 1_{\map \CC K} : K \to \C$ by:
:$\map {\mathbf 1_{\map \CC K} } x = 1$
for each $x \in K$, we have:
:$\mathbf 1_{\map \CC K} \in \map \CC K$
and:
:$f \mathbf 1_{\map \CC K} = \mathbf 1_{\map \CC K} f = f$
for each $f \in \map \CC K$.
So $\struct {\map \CC K, \overline \cdot, \norm {\, \cdot \,} }$ is unital.
We now have:
for each $\phi \in \Phi_A$.
So $G$ is a isometric unital $\ast$-algebra isomorphism.
Hence every commutative unital $\text C^\ast$-algebra $A$ is isometrically $\ast$-algebra isomorphic to $\struct {\map \CC K, \overline \cdot, \norm {\, \cdot \,} }$ for some compact Hausdorff space $K$: in particular $K = \Phi_A$. 
Conversely, we have established that $\struct {\map \CC K, \overline \cdot, \norm {\, \cdot \,} }$ is a unital $\text C^\ast$-algebra for each compact Hausdorff space $K$.
Category:Gelfand-Naimark Theorem
Gelfand Representation Theorem

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a commutative Banach algebra over $\C$ such that:
:the spectrum $\Phi_A$ of $A$ is non-empty. 
Let $\hat a$ be the Gelfand transform of $a$.
Let $\struct {\map {\CC_0} {\Phi_A}, \norm {\, \cdot \,}_\infty}$ be the set of complex-valued functions vanishing at infinity with the supremum norm.
Define $G : A \to \map {\CC_0} {\Phi_A}$ by:
:$\map G a = \hat a$
for each $a \in A$.
Let $\map {\sigma_A} a$ be the spectrum of $a$ in $A$.
Let $r_A$ be the spectral radius of $a$ in $A$.
Then $G$ is an algebra homomorphism with:
:$(1) \quad \norm {\map G a}_\infty \le \norm a$ for each $a \in A$,
:$(2) \quad \map {r_A} a = \norm {\hat a}_\infty$ for each $a \in A$,
:$(3) \quad \map {\sigma_A} a = \hat a \sqbrk {\Phi_A}$ if $A$ is unital and $\map {\sigma_A} a = \hat a \sqbrk {\Phi_A} \cup \set 0$ otherwise. 
== Proof ==
=== Proof that $G$ is an algebra homomorphism ===
Let $a, b \in A$, $\lambda \in \C$ and $\phi \in \Phi_A$.
Then we have:
So we obtain:
:$\map G {a + \lambda b} = \map G a + \lambda \map G b$
So $G$ is linear.
Now let $a, b \in A$. 
We have:
So:
:$\map G {a b} = \map G a \map G b$
so $G$ is an algebra homomorphism.
=== Proof of $(2)$ ===
We have:
=== Proof of $(1)$ ===
We have:
=== Proof of $(3)$ ===
This is precisely:
:Spectrum of Element of Unital Commutative Banach Algebra in the case that $A$ is unital
: otherwise.
Category:Gelfand Transforms
Category:Commutative Banach Algebras
Gelfond's Constant is Transcendental

== Theorem ==
Gelfond's constant:
:$e^\pi$
is transcendental.
== Proof ==
From the Gelfond-Schneider Theorem:
If:
:$\alpha$ and $\beta$ are algebraic numbers such that $\alpha \notin \set {0, 1}$ 
:$\beta$ is either irrational or not wholly real
then $\alpha^\beta$ is transcendental.
We have that:
As:
:$i$ is algebraic
:$-2 i$ is algebraic and not wholly real
the conditions of the Gelfond-Schneider Theorem are fulfilled.
Hence the result.
== Historical Note ==
== Sources ==
* : $23 \cdotp 140 \, 692 \, 632 \, 7792 \, 69 \, 005 \, 729 \, 086 \ldots$
* : Chapter $\text {A}.29$: Liouville ()
* : Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental
* : $23 \cdotp 14069 \, 26327 \, 79269 \, 00572 \, 9086 \ldots$
Category:Exponential Function
Category:Gelfond's Constant
Gelfond's Constant is Transcendental/Historical Note

== Historical Note on Gelfond's Constant is Transcendental ==
The question of the transcendental nature of Gelfond's constant $e^\pi$ was raised in the context of the $7$th problem of the Hilbert $23$.
That Gelfond's Constant is Transcendental was initially established in $1929$ by .
It was since determined to be a special case of the Gelfond-Schneider Theorem, established .
== Sources ==
* : $2 \cdotp 665 \, 144 \ldots$
* : Chapter $\text {A}.29$: Liouville ()
* : Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental
* : $2 \cdotp 66514 \, 4 \ldots$
Category:Historical Notes
Gelfond's Constant minus Pi

== Theorem ==
Gelfond's constant minus $\pi$ is very close to $20$:
:$e^\pi - \pi \approx 20$
which is nothing more than an amusing coincidence.
== Proof ==
We have:
:$e^\pi \approx 23 \cdotp 14069 \, 26327 \, 79269 \ldots$
:$\pi \approx 3 \cdotp 14159 \, 26535 \, 89793$
Then:
 23.14069 26327 79269
- 3.14159 26535 89793
---------------------
 19.99909 99791 89486
---------------------
== Sources ==
* : $23 \cdotp 14069 \, 26327 \, 79269 \, 00572 \, 9086 \ldots$
Category:Pi
Category:Gelfond's Constant
Category:Approximate Relations between Pi and Euler's Number
Gelfond-Schneider Constant is Transcendental

== Theorem ==
The Gelfond-Schneider constant:
:$2^{\sqrt 2}$
is transcendental.
== Proof ==
From the Gelfond-Schneider Theorem:
If:
:$\alpha$ and $\beta$ are algebraic numbers such that $\alpha \notin \set {0, 1}$
:$\beta$ is either irrational or not wholly real
then $\alpha^\beta$ is transcendental.
From Rational Number is Algebraic:
:$2$ is algebraic.
From Square Root of 2 is Algebraic of Degree 2:
:$\sqrt 2$ is algebraic.
From Square Root of 2 is Irrational:
:$\sqrt 2$ is irrational.
Hence the result.
== Historical Note ==
== Sources ==
* : Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental
Category:Real Analysis
Gelfond-Schneider Constant is Transcendental/Historical Note

== Historical Note on Gelfond-Schneider Constant is Transcendental ==
The question of the transcendental nature of the Gelfond-Schneider constant $2^{\sqrt 2}$ was raised in the context of the $7$th problem of the Hilbert $23$.
That the Gelfond-Schneider Constant is Transcendental was established in $1930$ by .
It was since determined to be a special case of the Gelfond-Schneider Theorem, established .
== Sources ==
* : $2 \cdotp 665 \, 144 \ldots$
* : Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental
* : $2 \cdotp 66514 \, 4 \ldots$
Category:Historical Notes
Gelfond-Schneider Theorem/Examples/2 to Power of Root 2

== Example of Use of Gelfond-Schneider Theorem ==
$2^{\sqrt 2}$ is transcendental.
== Proof ==
We have that:
:$2$ is (trvially) rational, as $2 = \dfrac 2 1$
:$\sqrt 2$ is algebraic but irrational.
Hence by the Gelfond-Schneider Theorem, $2^{\sqrt 2}$ is transcendental.
== Sources ==
* 
Category:Examples of Use of Gelfond-Schneider Theorem
Gelfond-Schneider Theorem/Examples/3 to Power of i

== Example of Use of Gelfond-Schneider Theorem ==
$3^i$ is transcendental.
== Proof ==
We have that:
:$3$ is (trivially) rational, as $3 = \dfrac 3 1$
:$i$ is algebraic, but as it is (wholly) imaginary, it does not classify as being an element of the set of rational numbers $\Q$.
Hence by the Gelfond-Schneider Theorem, $3^i$ is transcendental.
== Sources ==
* 
Category:Examples of Use of Gelfond-Schneider Theorem