avewright/proofwiki-math huggingface dataset:
Theorem: Construction of Direct Product of Fields

Let $\struct {F, +_F, \times_F}$ be a field whose zero is $0$ and whose multiplicative identity is $1$.
Let $E = F \times F$ be the Cartesian product of $F$ with itself.
Let addition be defined on $E$ by:
:$\forall a, b, c, d \in F: \tuple {a, b} +_E \tuple {c, d} := \tuple {a +_F c, b +_F d}$
Let multiplication be defined on $E$ by:
:$\forall a, b, c, d \in F: \tuple {a, b} \times_E \tuple {c, d} := \tuple {\paren {a \times_F c} +_F \paren {-\paren {b \times_F d} }, \paren {a \times_F d} +_F \paren {b \times_F c} }$
Then $\struct {E, +_E, \times_E}$ is a field.

Proof:
In order to define the structure rigorously, each of the field addition and field multiplication operations were explicitly stated in the above.
However, in order to simplify presentation, the operations will be denoted in the following as:
{{begin-eqn}}
{{eqn | q = \forall a, b, c, d \in F
      | l = \tuple {a, b} + \tuple {c, d}
      | r = \tuple {a + c, b + d}
      | c = 
}}
{{eqn | l = \tuple {a, b} \tuple {c, d}
      | r = \tuple {a c - b d, a d + b c}
      | c = 
}}
{{end-eqn}}
when it is clear from the context which operation is implied.
We check the criteria for $E$ to be a field.
First we note that $\struct {E, +_E}$ is the external group direct product $\struct {F, +_F} \times \struct {F, +_F}$, where $\struct {F, +_F}$ is an Abelian group.
Hence from External Direct Product of Abelian Groups is Abelian Group we have that $\struct {E, +_E}$ is an Abelian group whose identity is $\tuple {0, 0}$.
Now we consider the algebraic structure $\tuple {E, \times_E}$.
We specifically wish to demonstrate that $\tuple {E^*, \times_E}$, where $E^* := E \setminus \tuple {0, 0}$, is an Abelian group.
Taking the group axioms in turn:
Theorem: Field Homomorphism is either Trivial or Injection

Let $\struct {E, +_E, \times_E}$ and $\struct {F, +_F, \times_F}$ be fields.
Let $\phi: E \to F$ be a (field) homomorphism.
Then $\phi$ is either an injection or the trivial homomorphism.

Proof:
This is an instance of Ring Homomorphism from Field is Monomorphism or Zero Homomorphism.
{{qed}}
Theorem: Surjective Field Homomorphism is Field Isomorphism

Let $E$ and $F$ be fields.
Let $\phi: E \to F$ be a (field) homomorphism.
Let $\phi$ be a surjection.
Then $\phi$ is an isomorphism.

Proof:
As asserted, let $\phi$ be a surjection.
From Field Homomorphism is either Trivial or Injection, $\phi$ is either an injection or the trivial homomorphism.
If $\phi$ is an injection, then, by definition, $\phi$ is a bijection.
Hence, again by definition, $\phi$ is an isomorphism.
If $\phi$ is not an injection, then $\phi$ is the trivial homomorphism.
But from Field Contains at least 2 Elements, $\Img \phi$ cannot in that case be a field.
Hence if $\phi$ is not an injection, then $\phi$ cannot be a surjection.
Hence the result.
{{qed}}
Theorem: Field Contains at least 2 Elements

Let $\struct {F, +, \times}$ be a field.
Then $E$ contains at least $2$ elements.

Proof:
By definition, $\struct {F, +, \times}$ is an algebraic structure such that:
:$\struct {F, +}$ is an abelian group whose identity element is $0 \in F$
:$\struct {F^*, \times}$ is also an abelian group, where $F^* = F \setminus \set 0$.
From Group is not Empty, $\struct {F^*, \times}$ cannot be the empty set.
So there are at least $2$ elements in $F$: 
:the identity element $0$ of $\struct {F, +}$
:the identity element $1$ of $\struct {F^*, \times}$.
{{qed}}
Category:Field Theory
kutt8kj3a6zfm6cdam3o7qfw05j7tjd
Theorem: Subfield Test/Four-Step

$\struct {K, +, \circ}$ is a subfield of $\struct {F, +, \times}$ {{iff}} these all hold:
:$(1): \quad K^* \ne \O$
:$(2): \quad \forall x, y \in K: x + \paren {-y} \in K$
:$(3): \quad \forall x, y \in K: x \times y \in K$
:$(4): \quad x \in K^* \implies x^{-1} \in K^*$
where $K^*$ denotes $K \setminus \set {0_F}$.

Proof:
=== Necessary Condition ===
Let $\struct {K, +, \times}$ be a subfield of $\struct {F, +, \circ}$.
Then the conditions $(1)$ to $(4)$ all hold by virtue of the field axioms.
{{qed|lemma}}
Theorem: Subfield Test/Three-Step

$\struct {K, +, \times}$ is a subfield of $\struct {F, +, \times}$ {{iff}} these all hold:
:$(1): \quad K^* \ne \O$
:$(2): \quad \forall x, y \in K: x - y \in K$
:$(3): \quad \forall x \in K: \forall y \in K^*: \dfrac x y \in K$
where $K^*$ denotes $K \setminus \set {0_F}$.

Proof:
=== Necessary Condition ===
Let $\struct {K, +, \times}$ be a subfield of $\struct {F, +, \times}$.
Then the conditions $(1)$ to $(3)$ all hold by virtue of the field axioms.
{{qed|lemma}}
Theorem: Complex Numbers of Type Rational a plus b root 2 form Field

Let $\Q \sqbrk {\sqrt 2}$ denote the set:
:$\Q \sqbrk {\sqrt 2} := \set {x \in \C: x = a + b \sqrt 2: a, b \in \Q}$
that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are rational numbers.
Then the algebraic structure:
:$\struct {\Q \sqbrk {\sqrt 2}, +, \times}$
where $+$ and $\times$ are conventional addition and multiplication on real numbers, is a number field.

Proof:
From Real Numbers of Type Rational a plus b root 2 form Field, the set:
:$\Q \sqbrk {\sqrt 2} := \set {x \in \R: x = a + b \sqrt 2: a, b \in \Q}$
forms a field.
As $\Q \sqbrk {\sqrt 2} \subseteq \R$ and $\R \subseteq \C$ it follows that $\struct {\Q \sqbrk {\sqrt 2}, +, \times}$ is a subfield of $\C$.
Hence the result by definition of number field.
{{qed}}
Theorem: Number Field has Rational Numbers as Subfield

Let $F$ be a number field.
The field of rational numbers $\struct {\Q, +, \times}$ is a subfield of $F$.

Proof:
Let $F$ be a number field.
By definition, $F$ is a subfield of the field of complex numbers $\struct {\C, +, \times}$
We have that Rational Numbers form Subfield of Complex Numbers.
Let $G = \Q \cap F$ be the intersection of $\Q$ with $F$.
By Intersection of Subfields is Subfield, $G$ is a subfield of $\C$ and so also a number field.
From Intersection is Subset we have that $G \subseteq \Q$.
So $G$ is a field which is also a subset of $\Q$.
Hence by definition $G$ is a subfield of $\Q$.
But from Rational Numbers form Prime Field, $\struct {\Q, +, \times}$ is a prime field.
That is, the only subfield of $\Q$ is $\Q$ itself.
Hence:
:$G = \Q$
and from Intersection with Subset is Subset:
:$\Q \subseteq F$
Hence $\Q$ is a subset of $F$ which is also a field.
That is, $\Q$ is a subfield of $F$.
{{qed}}
Theorem: Element to Power of Positive Characteristic of Field is Zero

Let $\struct {F, +, \times}$ be a field whose zero is $0$ and whose unity is $1$.
Let the characteristic of $F$ be $n$ such that $n > 0$.
Then:
:$n \cdot a = 0$
where $n \cdot a$ denotes the power of $a$ in the context of the additive group $\struct {F, +}$:
:$n \cdot a = \begin {cases}
0 & : n = 0 \\
\paren {\paren {n - 1} \cdot a} + a & : n > 0
\end {cases}$

Proof:
By definition, the characteristic of $\struct {F, +, \times}$ is the order of the additive group $\struct {F, +}$.
By Element to Power of Group Order is Identity it follows directly that:
:$n \cdot a = 0$
{{qed}}
Theorem: Field has Characteristic of Zero iff exists Monomorphism from Rationals

Let $F$ be a field.
Then:
:there exists a field monomorphism $\phi: \Q \to F$ from the field of rational numbers $\Q$ and $F$.
{{iff}}:
:$\Char F = 0$
where $\Char F$ denotes the characteristic of $F$.

Proof:
=== Necessary Condition ===
Let $\Char F = 0$.
Then from Field of Characteristic Zero has Unique Prime Subfield, $F$ has a unique prime subfield $K$ such that:
:$K \cong \Q$
where $\cong$ denotes isomorphism.
Thus there exists an isomorphism from $\Q$ to a subfield of $F$.
From Injection to Image is Bijection, it follows that $\phi$ is an injection into $F$.
Hence the result by definition of field monomorphism.
{{qed}}
Theorem: Field has Prime Characteristic p iff exists Monomorphism from Field of Integers Modulo p

Let $F$ be a field.
Then:
:there exists some prime number $p$ such that $\Char F = p$
{{iff}}:
:there exists a field monomorphism $\phi: \Z_p \to F$
where:
:$\Char F$ denotes the characteristic of $F$.
:$\Z_p$ denotes the field of integers modulo $p$.

Proof:
Let $\struct {F, +, \times}$ be a field whose zero is $0$ and whose unity is $1$.
Theorem: Group is Cancellable Monoid

Let $\struct {G, \circ}$ be a group.
Then $\struct {G, \circ}$ is a cancellable monoid.

Proof:
By definition, a group is {{afortiori}} a monoid.
From Group Operation is Cancellable, $\circ$ is a cancellable operation in $G$.
Hence the result by definition of cancellable monoid.
{{qed}}
Theorem: Intersection of Submonoids with Monoid Identity is Submonoid

Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$.
Let $I$ be an indexing set.
Let $\family {S_\alpha}_{\alpha \mathop \in I}$ be a family of submonoids of $S$.
For each $S_\alpha \in \family {S_\alpha}_{\alpha \mathop \in I}$, let $e_S \in S_\alpha$.
Let $\ds \bigcap_{\alpha \mathop \in I} S_\alpha$ denote the intersection of $\family {S_\alpha}$
Then $\ds \bigcap_{\alpha \mathop \in I} S_\alpha$ is a submonoid of $S$.

Proof:
First we show that $\struct {\ds \bigcap_{\alpha \mathop \in I} S_\alpha, \circ}$ is a semigroup:
Theorem: Generated Submonoid is Intersection of Submonoids containing Generator

Let $\struct {M, \circ}$ be a monoid whose identity is $e_M$.
Let $S \subseteq M$.
Let $\struct {H, \circ}$ be the '''submonoid of $\struct {M, \circ}$ generated by $S$'''.
Then $\struct {H, \circ}$ is the intersection of all submonoids of $\struct {M, \circ}$ containing $S \cup \set {e_M}$.

Proof:
Let $\struct {H, \circ}$ be the '''submonoid of $\struct {M, \circ}$ generated by $S$'''.
Then by definition $H$ is the smallest (with respect to set inclusion) submonoid of $\struct {M, \circ}$ containing $S \cup \set {e_M}$.
Let $\mathbb S$ be the set of submonoids of $\struct {M, \circ}$ containing $S \cup \set {e_M}$.
We need to show that $H = \ds \bigcap \mathbb S$.
Because $H$ is a submonoid of $\struct {M, \circ}$ containing $S \cup \set {e_M}$:
:$H \in \mathbb S$
By Intersection is Subset:
:$\ds \bigcap \mathbb S \subseteq H$
{{qed|lemma}}
By Intersection of Submonoids with Monoid Identity is Submonoid:
:$\ds \bigcap \mathbb S$ is a submonoid of $\struct {M, \circ}$ containing $S \cup \set {e_M}$.
Because $H$ is the smallest (with respect to set inclusion) submonoid of $\struct {M, \circ}$ containing $S \cup \set {e_M}$:
:$H \subseteq \ds \bigcap \mathbb S$
By definition of set equality:
:$H = \ds \bigcap \mathbb S$
Hence the result.
{{qed}}
Theorem: Fundamental Solution to Reduced Linear First Order ODE with Constant Coefficients

Let $H$ be the Heaviside step function.
Let $\lambda \in \R$.
Let $\map f x = \map H x \map \exp {\lambda x}$.
Let $T_f$ be the Schwartz distribution associated with $f$.
Then, in the distributional sense, $T_f$ is the fundamental solution of
:$\paren {\dfrac \d {\d x} - \lambda} T_f = \delta$

Proof:
$x \stackrel f {\longrightarrow} \map H x \map \exp {\lambda x}$ is a continuously differentiable real function on $\R \setminus \set 0$ and possibly has a discontinuity at $x = 0$.
By Differentiable Function as Distribution we have that:
:$T'_f = T_{f'}$
Moreover:
:$x < 0 \implies \paren {\map H x \map \exp {\lambda x} }' = 0$
:$x > 0 \implies \paren { {\map H x} \map \exp { \lambda x} }' = \lambda \map \exp {\lambda x}$
Altogether:
:$\forall x \in \R \setminus \set 0 : \paren { {\map H x} \map \sin x}' = \lambda \map H x \map \exp {\lambda x}$
Furthermore:
:$\map f {0^+} - \map f {0^-} = \map H {0^+} \map \exp {0^+} - \map H {0^-} \map \exp {0^-} = 1$
By the Jump Rule:
:$T'_{\map H x \map \exp {\lambda x} } = T_{\lambda \map H x \map \exp {\lambda x} } + \delta$
Rearranging the terms and using the linearity of Schwartz distribution gives:
:$T'_{\map H x \map \exp {\lambda x} } - \lambda T_{\map H x \map \exp {\lambda x} } = \delta$
{{qed}}
Theorem: Fundamental Solution to y'' + w y

Let $H$ be the Heaviside step function.
Let $\omega \in \R : \omega \ne 0$.
Let $\map f x = \map H x \dfrac {\map \sin {\omega x} } \omega$.
Let $T_f$ be the Schwartz distribution associated with $f$.
Then, in the distributional sense, $T_f$ is the fundamental solution of
:$\paren {\dfrac {\d^2} {\d x^2} + \omega^2} T_f = \delta$

Proof:
$x \stackrel f {\longrightarrow} \map H x \dfrac {\map \sin {\omega x} } \omega$ is a continuously differentiable real function on $\R \setminus \set 0$ and possibly has a discontinuity at $x = 0$.
By Differentiable Function as Distribution we have that $T'_f = T_{f'}$.
Moreover:
:$x < 0 \implies \paren {\map H x \dfrac {\map \sin {\omega x} } {\omega} }' = 0$
:$x > 0 \implies \paren {\map H x \dfrac {\map \sin {\omega x} } {\omega} }' = \map \cos {\omega x}$
Altogether:
:$\forall x \in \R \setminus \set 0 : \paren { {\map H x} \dfrac {\map \sin {\omega x} } \omega}' = \map H x \map \cos {\omega x}$
Furthermore:
:$\map f {0^+} - \map f {0^-} = \map H {0^+} \dfrac {\map \sin {\omega \cdot 0^+} } \omega - \map H {0^-} \dfrac {\map \sin {\omega \cdot 0^-} } \omega = 0$
By the Jump Rule:
:$T'_{\map H x \frac {\map \sin {\omega x} } \omega} = T_{\map H x \map \cos {\omega x} }$
$x \stackrel g {\longrightarrow} \map H x \map \cos {\omega x}$ is a continuously differentiable real function on $\R \setminus \set 0$ and possibly has a discontinuity at $x = 0$.
By Differentiable Function as Distribution we have that:
:$T'_g = T_{g'}$
Moreover:
:$x < 0 \implies \paren {\map H x \map \cos \omega}' = 0$
:$x > 0 \implies \paren {\map H x \map \cos \omega}' = -\omega \map \sin {\omega x}$
Altogether:
:$\forall x \in \R \setminus \set 0 : \paren {\map H x \map \cos \omega}' = - \map H x \omega \map \sin {\omega x}$
Furthermore:
:$\map g {0^+} - \map g {0^-} = \map H {0^+} \map \cos {\omega \cdot 0^+} - \map H {0^-} \map \cos {\omega \cdot 0^-} = 1$
By the Jump Rule:
:$T'_{\map H x \map \cos {\omega x}} = T_{- \omega \map H x \map \sin {\omega x} } + \delta$
Hence:
:$T' '_{\map H x \frac {\map \sin {\omega x} } \omega} = T_{- \omega^2 \map H x \frac {\map \sin {\omega x} } \omega} + \delta$
Rearranging the terms and using the linearity of Schwartz distribution gives:
:$T' '_{\map H x \frac {\map \sin {\omega x} } \omega} + \omega^2 T_{\map H x \frac {\map \sin {\omega x} } \omega} = \delta$
{{qed}}
Theorem: Order of Floor Function

Let $\floor x$ denote the floor function of $x$. 
Then:
:$\floor x = x + \map \OO 1$
where $\OO$ is big-O notation.

Proof:
From Floor is between Number and One Less: 
:$\floor x \le x < \floor x + 1$
so:
:$0 \le x - \floor x < 1$
By the definition of the absolute value function, we have: 
:$\size {\floor x - x} < 1$
so by the definition of Big-O notation, we have:
:$\floor x - x = \map \OO 1$
We can conclude that:
:$\floor x = x + \map \OO 1$
{{qed}}
Category:Analytic Number Theory
Category:Floor Function
91l6pse7wyc2dsq116m1c1rco063run
Theorem: Fundamental Solution to Nth Derivative

Let $H$ be the Heaviside step function.
Let $n \in \N_{>0}$.
Let $\map {f_n} x = \map H x \dfrac {x^{n - 1} } {\paren {n - 1}!}$.
Let $T_{f_n}$ be the Schwartz distribution associated with $f_n$.
Then, in the distributional sense, $T_{f_n}$ is the fundamental solution of
:$\dfrac {\rd^n} {\rd x^n} T_{f_n} = \delta$

Proof:
Proof by Principle of Mathematical Induction:
Theorem: Coordinate Representation of Divergence

Let $\struct {M, g}$ be a Riemannian manifold.
Let $U \subseteq M$ be an open set.
Let $\tuple {x^i}$ be local smooth coordinates.
Let $X$ be a smooth vector field on $M$.
Let $\operatorname {div}$ be the divergence operator.
Then:
:$\map {\operatorname {div} } {X^i \dfrac \partial {\partial x^i}} = \dfrac 1 {\sqrt g} \dfrac \partial {\partial x^i} \paren {X^i \sqrt g}$
{{explain|BEWARE: The specification of this result has been changed: in the last expression, $\paren {X^i \sqrt {\det g} }$ has been changed to $\paren {X^i \sqrt g}$ THIS HAS GOT TO BE REVIEWED.}}
{{MissingLinks}}

Proof:
Let $\omega$ be the Riemannian volume form on $M$.
Recall that the divergence of $X$ is defined by the relation:
:$\map \d {\omega \rfloor X} = \map {\operatorname {div} } X \omega$, where $\rfloor$ denotes the interior product.
In local coordinates we have:
:$\omega = \sqrt g \rd x_1 \wedge \dots \wedge \d x_n$
Thus:
:$\omega \rfloor X = \paren {-1}^{i - 1} \sqrt g X^i \rd x_1 \wedge \cdots \wedge \widehat {\d x_i} \wedge \cdots \d x_n$
Thus:
:$\d \omega = \map {\dfrac \partial {\partial x_i} } {\sqrt g X^i} \rd x_1 \wedge \cdots \wedge \d x_n$
Thus:
:$\map {\operatorname {div} } X = \dfrac 1 {\sqrt g} \dfrac \partial {\partial x_i} \paren {X^i \sqrt g}$
{{Proofread}}
Theorem: Transplanting Theorem/Corollary

Let $\struct {S, \circ}$ be an algebraic structure.
Let $f: S \to S$ be an automorphism on $\struct {S, \circ}$.
Then the transplant of $\circ$ under $f$ is $\circ$ itself.

Proof:
From the Transplanting Theorem there exists one and only one operation $\circ$ such that $f: \struct {S, \circ} \to \struct {S, \circ}$ is an automorphism.
{{qed}}
Theorem: Invertible Element of Associative Structure is Cancellable/Corollary

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.
{{Qed}}
Category:Invertible Element of Associative Structure is Cancellable
3z54y864c0xutnlpqvr3ewiqiqcirwl
Theorem: Smooth Real Function times Derivative of Dirac Delta Distribution

Let $\alpha \in \map {C^\infty} \R$ be a smooth real function.
Let $\delta \in \map {\DD'} \R$ be the Dirac delta distribution.
Then in the Schwartz distributional sense it holds that:
:$\alpha \cdot \delta' = \map \alpha 0 \delta' - \map {\alpha'} 0 \delta$

Proof:
Let $\phi \in \map \DD \R$ be a test function.
{{begin-eqn}}
{{eqn | l = \map {\alpha \cdot \delta'} \phi
      | r = \map {\delta'} {\alpha \phi}
      | c = {{Defof|Multiplication of Schwartz Distribution by Smooth Function}}
}}
{{eqn | r = - \map \delta {\paren {\alpha \phi}'}
      | c = {{Defof|Distributional Derivative}}
}}
{{eqn | r = - \map \delta {\alpha' \phi + \alpha \phi'}
}}
{{eqn | r = - \map {\alpha'} 0 \map \phi 0 - \map \alpha 0 \map {\phi'} 0
      | c = {{Defof|Dirac Delta Distribution}}
}}
{{eqn | r = -\map {\alpha'} 0 \map \delta \phi - \map \alpha 0 \map \delta {\phi'}
      | c = {{Defof|Dirac Delta Distribution}}
}}
{{eqn | r = -\map {\alpha'} 0 \map \delta \phi + \map \alpha 0 \map {\delta'} \phi
      | c = {{Defof|Distributional Derivative}}
}}
{{end-eqn}}
{{qed}}
Theorem: Parity Group is Only Group with 2 Elements

Let $\struct {G, \circ}$ be a group with exactly $2$ elements.
Then $\struct {G, \circ}$ is isomorphic to the parity group, which can be exemplified $\struct {\Z_2, +_2}$.
That is, the additive group of integers modulo $2$.

Proof:
We have that $2$ is a prime number.
Hence $\struct {\Z_2, +_2}$ is a prime group.
The result follows from Prime Groups of Same Order are Isomorphic.
Theorem: Identity of Submagma containing Identity of Magma is Same Identity

Let $\struct {S, \circ}$ be a magma which has an identity $e$.
Let $\struct {T, \circ}$ be a submagma of $\struct {S, \circ}$ such that $e \in T$.
Then $e$ is the identity of $T$.

Proof:
From Identity is Unique, there can be only one identity $e$ of $\struct {S, \circ}$.
We have that:
{{begin-eqn}}
{{eqn | l = x
      | o = \in
      | r = T
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | o = \in
      | r = S
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x \circ e
      | r = x = e \circ x
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \forall x \in T
      | l = x \circ e
      | r = x = e \circ x
      | c = 
}}
{{end-eqn}}
That is, $e$ is the (unique} identity of $T$.
{{qed}}
Theorem: Subsemigroup of Monoid is not necessarily Monoid

Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$.
Let $\struct {T, \circ}$ be a subsemigroup of $\struct {S, \circ}$
Then it is not necessarily the case that $\struct {T, \circ}$ has an identity.

Proof:
Consider the set of integers under multiplication $\struct {\Z, \times}$.
From Integers under Multiplication form Monoid, $\struct {\Z, \times}$ is a monoid.
Let $n \in \Z$ such that $n > 1$.
Let $n \Z$ be the set of integer multiples of $n$:
:$\set {x \in \Z: n \divides x}$
where $\divides$ denotes divisibility.
From Integer Multiples under Multiplication form Semigroup, $\struct {n \Z, \times}$ is a semigroup which has no identity.
By construction, $n \Z$ is a subset of $\Z$.
Hence $\struct {n \Z, \times}$ is a subsemigroup of $\struct {\Z, \times}$ which has no identity.
{{qed}}
Theorem: Image of P-adic Norm

Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime number $p$.
Then the image of $\norm {\,\cdot\,}_p$ is:
:$\Img {\norm {\,\cdot\,}_p} = \set {p^n : n \in \Z} \cup \set 0$

Proof:
This follows immediately from:
:{{Defof|P-adic Norm}}
:{{Defof|P-adic Valuation}}
{{qed}}
Category:P-adic Number Theory
loyf7o0bnhfz37u556teza8xvny0slx
Theorem: Left Operation is Closed for All Subsets

:$\struct {T, \leftarrow}$ is a subsemigroup of $\struct {S, \leftarrow}$.

Proof:
From Structure under Left Operation is Semigroup we have that $\struct {S, \leftarrow}$ is a semigroup, whatever the nature of $S$.
Let $T \in \powerset S$.
Then:
:From Structure under Left Operation is Semigroup, $\struct {T, \leftarrow}$ is a semigroup, and therefore a subsemigroup of $\struct {S, \leftarrow}$.
This applies whatever $S$ is and whatever the subset $T$ is.
{{qed}}
Theorem: Right Operation is Closed for All Subsets

:$\struct {T, \rightarrow}$  is a subsemigroup of $\struct {S, \rightarrow}$.

Proof:
From Structure under Right Operation is Semigroup we have that $\struct {S, \rightarrow}$ is a semigroup, whatever the nature of $S$.
Let $T \in \powerset S$.
Then:
:From Structure under Right Operation is Semigroup, $\struct {T, \rightarrow}$ is a semigroup, and therefore a subsemigroup of $\struct {S, \rightarrow}$.
This applies whatever $S$ is and whatever the subset $T$ is.
{{qed}}
Theorem: Equivalence Class Equivalent Statements/1 iff 2

Let $\RR$ be an equivalence relation on $S$.
Let $x, y \in S$.
{{TFAE}}
:$(1): \quad x$ and $y$ are in the same $\RR$-class
:$(2): \quad \eqclass x \RR = \eqclass y \RR$

Proof:
By Equivalence Class is Unique:
:$\eqclass x \RR$ is the unique $\RR$-class to which $x$ belongs
and:
:$\eqclass y \RR$ is the unique $\RR$-class to which $y$ belongs.
As these are unique for each, they must be the same set.
{{qed}}
Theorem: Equivalence Class Equivalent Statements/2 iff 3

Let $\RR$ be an equivalence relation on $S$.
Let $x, y \in S$.
{{TFAE}}
:$\eqclass x \RR = \eqclass y \RR$
:$x \mathrel \RR y$

Proof:
By Equivalence Class holds Equivalent Elements:
{{:Equivalence Class holds Equivalent Elements}}
Theorem: Equivalence Class Equivalent Statements/3 iff 4

Let $\RR$ be an equivalence relation on $S$.
Let $x, y \in S$.
{{TFAE}}
:$x \mathrel \RR y$
:$x \in \eqclass y \RR$

Proof:
This follows directly by the definition of equivalence class.
{{qed}}
Theorem: Equivalence Class Equivalent Statements/3 iff 5

Let $\RR$ be an equivalence relation on $S$.
Let $x, y \in S$.
{{TFAE}}
:$x \mathrel \RR y$
:$y \in \eqclass x \RR$

Proof:
This follows through dint of the symmetry of $\RR$ and the definition of Equivalence Class.
{{qed}}
Theorem: Equivalence Class Equivalent Statements/3 iff 6

Let $\RR$ be an equivalence relation on $S$.
Let $x, y \in S$.
{{TFAE}}
:$x \mathrel \RR y$
:$\eqclass x \RR \cap \eqclass y \RR \ne \O$

Proof:
Follows directly from Equivalence Classes are Disjoint.
{{qed}}