avewright/proofwiki-math huggingface dataset:
Theorem: Image of Submodule under Linear Transformation is Submodule

Let $M$ be a submodule of $G$.
Then $\phi \sqbrk M$ is a submodule of $H$.

Proof:
Let $N = \phi \sqbrk M$ be the image set of $M$ under $\phi$.
By definition, a linear transformation $\phi: G \to H$ is, in particular, a (group) homomorphism from the group $\struct {G, +_G}$ to the group $\struct {H, +_H}$.
We have {{hypothesis}} that $M$ is a submodule of $G$.
So from Elements of Submodule form Subgroup, $M$ forms a subgroup of $G$.
From Group Homomorphism Preserves Subgroups, $N$ is therefore a subgroup of $H$.
{{qed|lemma}}
It remains to be shown that $N$ is closed for scalar product:
:$\forall \lambda \in R, x \in N: \lambda \circ_H x \in N$
As $M$ is a submodule of $G$, $M$ itself is closed for scalar product:
:$\forall \lambda \in R, y \in N: \lambda \circ_G y \in M$
We have:
{{begin-eqn}}
{{eqn | q = \forall n \in N: \exists y \in M
      | l = x
      | r = \map \phi y
      | c = Definition of $\phi$
}}
{{eqn | ll= \leadsto
      | q = \forall \lambda \in R
      | l = \lambda \circ_H x
      | r = \lambda \circ_H \map \phi y
      | c = 
}}
{{eqn | r = \map \phi {\lambda \circ_G y}
      | c = {{Defof|Morphism Property}} applied to $\phi$
}}
{{eqn | ll= \leadsto
      | l = \lambda \circ_H x
      | o = \in
      | r = \phi \sqbrk M
      | c = as $\lambda \circ_G y \in M$
}}
{{eqn | ll= \leadsto
      | l = \lambda \circ_H x
      | o = \in
      | r = N
      | c = Definition of $N$
}}
{{end-eqn}}
{{qed}}
Theorem: Preimage of Submodule under Linear Transformation is Submodule

Let $N$ be a submodule of $H$.
Then $\phi^{-1} \sqbrk N$ is a submodule of $G$.

Proof:
Let $M = \phi^{-1} \sqbrk N$ be the preimage of $N$ under $\phi$.
{{AimForCont}} $M$ is not a submodule of $G$.
This means that $M$ does not fulfil all the module axioms.
First suppose that:
:$(1): \quad \struct {M, +_G}$ is not a subgroup of $G$.
Then $M$ is not a group.
Then by the One-Step Subgroup Test:
:$\exists x, y \in M: x +_G \paren {-y} \notin M$
Then:
{{begin-eqn}}
{{eqn | q = \exists x, y \in M
      | l = x +_G \paren {-y}
      | o = \notin
      | r = M
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \phi {x +_G \paren {-y} }
      | o = \notin
      | r = N
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \phi x +_H \map \phi {-y}
      | o = \notin
      | r = N
      | c = 
}}
{{end-eqn}}
But by definition of $\phi$, both $\map \phi x \in N$ and $\map \phi {-y} \in N$.
Hence $\struct {N, +_G}$ is not closed.
By {{Group-axiom|0}} as applied to $N$, it follows that $N$ is not a group.
Hence $N$ is not a subgroup of $H$.
From this contradiction it follows that $(1)$ is false.
Thus $\struct {M, +_G}$ is a subgroup of $G$.
{{qed|lemma}}
We have {{hypothesis}} that $M$ is not a submodule of $G$.
We have shown that $\struct {M, +_G}$ is a subgroup of $G$.
Hence it must be the case that $M$ is not closed for scalar product:
{{begin-eqn}}
{{eqn | n = 2
      | q = \exists \lambda \in R, x \in M
      | l = \lambda \circ_G x
      | o = \notin
      | r = M
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \phi {\lambda \circ_G x}
      | o = \notin
      | r = N
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \lambda \circ_H {\map \phi x}
      | o = \notin
      | r = N
      | c = 
}}
{{end-eqn}}
But by definition of $\phi$, both $\map \phi x \in N$.
Hence $N$ is not closed for scalar product.
From this contradiction it follows that $(2)$ is false.
Thus $\struct {M, +_G}$ is closed for scalar product.
{{qed|lemma}}
We have shown that $M$ is a subgroup of $G$ which is closed for scalar product.
That is, $M$ is a submodule of $G$.
{{Qed}}
Theorem: Test Function Space with Pointwise Addition and Pointwise Scalar Multiplication forms Vector Space

Let $\map \DD {\R^d}$ be the test function space.
Let $\struct {\C, +_\C, \times_\C}$ be the field of complex numbers.
Let $\paren +$ be the pointwise addition of test functions.
Let $\paren {\, \cdot \,}$ be the pointwise scalar multiplication of test functions over $\C$.
Then $\struct {\map \DD {\R^d}, +, \, \cdot \,}_\C$ is a vector space.

Proof:
Let $f, g, h \in \map \DD {\R^d}$ be test functions with the compact support $K$.
Let $\lambda, \mu \in \C$.
Let $\map 0 x$ be a real-valued function such that:
:$\map 0 x : \R^d \to 0$.
Let us use real number addition and multiplication.
$\forall x \in \R^d$ define pointwise addition as:
:$\map {\paren {f + g}} x := \map f x +_\C \map g x$.
Define pointwise scalar multiplication as:
:$\map {\paren {\lambda \cdot f}} x := \lambda \times_\C \map f x$
Let $\map {\paren {-f} } x := -\map f x$.
Theorem: Image of Linear Transformation is Submodule

Let $\Img \phi$ denote the image set of $\phi$.
Then $\Img \phi$ is a submodule of $H$.

Proof:
By Module is Submodule of Itself, $\struct {G, +_G, \circ_G}_R$ is a submodule of $\struct {G, +_G, \circ_G}_R$.
The result follows from Image of Submodule under Linear Transformation is Submodule.
{{Qed}}
Theorem: Kernel of Linear Transformation is Submodule

Let $\map \ker \phi$ denote the kernel of $\phi$.
Then $\map \ker \phi$ is a submodule of $G$.

Proof:
By definition, the kernel of $\phi$ is defined as:
:$\map \ker \phi := \phi^{-1} \sqbrk {\set {e_H} }$
where $e_H$ is the identity of $\struct {H, +_H}$.
where $\phi^{-1} \sqbrk S$ denotes the preimage of $S$ under $\phi$.
From Null Module is Module:
:$\struct {\set {e_H}, +_H, \circ_H}_R$ is a module
where $\struct {\set {e_H}, +_H, \circ_H}_R$ is the null module.
We have that $\struct {\set {e_H}, +_H, \circ_H}_R$ is a module which is a subset of $H$.
It follows that $\struct {\set {e_H}, +_H, \circ_H}_R$ is a submodule of $H$.
The result follows from Preimage of Submodule under Linear Transformation is Submodule.
{{qed}}
Theorem: Locally Integrable Function defines Distribution

Let $f \in \map {L^1_{loc}} {\R^d}$ be a locally integrable function.
Let $\map \DD {\R^d}$ be the test function space.
Let $T_f : \map \DD {\R^d} \to \C$ be a mapping.
Then $T_f$ is a Schwartz distribution.

Proof:
=== Existence ===
Let $\phi \in \map \DD {\R^d}$ be a test function.
Let $T_f$ be defined as
:$\ds T_f = \int_{\R^d} \map f {\mathbf x} \map \phi {\mathbf x} \rd \mathbf x$
By definition, $\phi$ has the compact support.
Together with the properties of $f$ we have that $T_f$ is bounded with respect to any compact range of integration.
{{qed|lemma}}
Theorem: Equivalence of Definitions of P-adic Valuation on P-adic Numbers

Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.
{{TFAE|def = P-adic Valuation on P-adic Numbers|view = the $p$-adic valuation on $\struct {\Q_p, \norm {\,\cdot\,}_p}$}}

Proof:
Let $x \in \Q_p \setminus \set 0$.
Let $l$ be the index of the first non-zero coefficient in the $p$-adic expansion:
:$l = \min \set {i: i \ge m \land d_i \ne 0}$
From P-adic Norm of P-adic Expansion is determined by First Nonzero Coefficient:
:$\norm x_p = p^{-l}$
By definition of real general logarithm:
:$-\log_p \norm x_p = l$
The result follows.
{{qed}}
Category:P-adic Valuation on P-adic Numbers
9fajue1pqrizpvygykshgqy5sq1fwcu
Theorem: Similarity Mapping is Automorphism

Let $G$ be a vector space over a field $\struct {K, +, \times}$.
Let $\beta \in K$.
Let $s_\beta: G \to G$ be the similarity on $G$ defined as:
:$\forall \mathbf x \in G: \map {s_\beta} {\mathbf x} = \beta \mathbf x$
If $\beta \ne 0$ then $s_\beta$ is an automorphism of $G$.

Proof:
By definition, a vector space automorphism on $G$ is a vector space isomorphism from $G$ to $G$ itself.
To prove that $s_\beta$ is a '''automorphism''' it is sufficient to demonstrate that:
By definition, a vector space isomorphism is a mapping $s_\beta: G \to G$ such that:
:$(1): \quad s_\beta$ is a bijection
:$(2): \quad \forall \mathbf x, \mathbf y \in G: \map {s_\beta} {\mathbf x + \mathbf y} = \map {s_\beta} {\mathbf x} + \map {s_\beta} {\mathbf y}$
:$(3): \quad \forall \mathbf x \in G: \forall \lambda \in K: \map {s_\beta} {\lambda \mathbf x} = \lambda \map {s_\beta} {\mathbf x}$
It has been established in Similarity Mapping is Linear Operator that $s_\beta$ is a linear operator on $G$.
Hence $(2)$ and $(3)$ follow by definition of linear operator.
It remains to prove bijectivity.
That is, that $s_\beta$ is both injective and surjective.
Let $1_K$ denote the multiplicative identity of $K$.
We have:
{{begin-eqn}}
{{eqn | q = \forall \mathbf x, \mathbf y \in G
      | l = \map {s_\beta} {\mathbf x}
      | r = \map {s_\beta} {\mathbf y}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \beta \, \mathbf x
      | r = \beta \, \mathbf y
      | c = Definition of $s_\beta$
}}
{{eqn | ll= \leadsto
      | l = \beta^{-1} \beta \, \mathbf x
      | r = \beta^{-1} \beta \, \mathbf y
      | c = {{Field-axiom|M4}}
}}
{{eqn | ll= \leadsto
      | l = 1_K \, \mathbf x
      | r = 1_K \, \mathbf y
      | c = {{Field-axiom|M3}}
}}
{{eqn | ll= \leadsto
      | l = \mathbf x
      | r = \mathbf y
      | c = {{Vector-space-axiom|4}}
}}
{{end-eqn}}
Hence it has been demonstrated that $s_\beta$ is injective.
Let $\mathbf y \in G$.
Consider $\beta^{-1} \in K$ defined such that $\beta \beta^{-1} = 1_K$.
By {{Field-axiom|M4}}, $\beta^{-1}$ always exists.
Then:
{{begin-eqn}}
{{eqn | q = \forall \mathbf y \in G: \exists \mathbf x \in G
      | l = \mathbf x
      | r = \beta^{-1} \mathbf y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \beta \, \mathbf x
      | r = \beta \beta^{-1} \mathbf y
      | c = 
}}
{{eqn | r = 1_K \mathbf y
      | c = 
}}
{{eqn | r = \mathbf y
      | c = {{Field-axiom|M3}}
}}
{{eqn | ll= \leadsto
      | q = \forall \mathbf y \in G: \exists \mathbf x \in G
      | l = \map {s_\beta} {\mathbf x}
      | r = \mathbf y
      | c = Definition of $s_\beta$
}}
{{end-eqn}}
Hence it has been demonstrated that $s_\beta$ is surjective.
Hence the result.
{{qed}}
Theorem: Inverse of Similarity Mapping

Let $G$ be a vector space over a field $K$.
Let $\beta \in K$ such that $\beta \ne 0$.
Let $s_\beta: G \to G$ be the similarity on $G$ defined as:
:$\forall \mathbf x \in G: \map {s_\beta} {\mathbf x} = \beta \mathbf x$
Let $\paren {s_\beta}^{-1}$ denote the inverse of $s_\beta$.
Then:
:$\paren {s_\beta}^{-1} = s_{\beta^{-1} }$
where $\beta^{-1}$ is the multiplicative inverse in $K$ of $\beta$.

Proof:
From Similarity Mapping is Automorphism, $s_\beta$ is an automorphism of $G$.
Hence $s_\beta$ is an vector space isomorphism from $G$ to $G$ itself.
So by definition $s_\beta$ is a bijection.
Hence the existence of this inverse $\paren {s_\beta}^{-1}$ follows from Bijection iff Left and Right Inverse.
By {{Field-axiom|M4}}, we have that there exists a multiplicative inverse $\beta^{-1}$ for $\beta$ such that:
:$\beta \beta^{-1} = 1_K = \beta^{-1} \beta$
where $1_K$ is the multiplicative identity of $K$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\paren {s_\beta}^{-1} } {\map {s_\beta} {\beta^{-1} \, \mathbf x} }
      | r = \beta^{-1} \mathbf x
      | c = {{Defof|Inverse Mapping}}
}}
{{eqn | ll= \leadsto
      | l = \map {\paren {s_\beta}^{-1} } {\beta \beta^{-1} \, \mathbf x}
      | r = \beta^{-1} \, \mathbf x
      | c = Definition of $s_\beta$
}}
{{eqn | ll= \leadsto
      | l = \map {\paren {s_\beta}^{-1} } {1_K \, \mathbf x}
      | r = \beta^{-1} \, \mathbf x
      | c = {{Field-axiom|M4}}
}}
{{eqn | ll= \leadsto
      | l = \map {\paren {s_\beta}^{-1} } {\mathbf x}
      | r = \beta^{-1} \, \mathbf x
      | c = {{Field-axiom|M3}}
}}
{{eqn | ll= \leadsto
      | l = \map {\paren {s_\beta}^{-1} } {\mathbf x}
      | r = \map {s_{\beta^{-1} } } {\mathbf x}
      | c = Definition of $s_{\beta^{-1} }$
}}
{{end-eqn}}
{{qed}}
Theorem: Existence of Orthonormal Frames

Let $\struct{M, g}$ be a $n$-dimensional Riemannian manifold with or without boundary. 
Let $p \in M$ be a point.
Let $TM$ be the tangent bundle of $M$.
Let $U \subseteq M$ be an open subset.
Suppose $\tuple {X_j}$ is a smooth local frame for $TM$ over $U$.
Then for all $p \in M$ there is a smooth orthonormal frame $\tuple {E_j}$ over $U$ such that:
:$\forall k \in \N : 1 \le k \le n : \map \span {\bigvalueat {E_1} p, \ldots, \bigvalueat {E_n} p} = \map \span {\bigvalueat {X_1} p, \ldots, \bigvalueat {X_n} p}$
In particular, for every $p \in M$ there is a smooth orthonormal frame $\tuple {E_j}$ defined on some neighborhood of $p$.

Proof:
{{ProofWanted|Use Gram-Schmidt on vectors $X_j$; obtain orthonormal vector fields $E_j$; denominators are nonvanishing so $E_j$ are smooth; apply this to any smooth local frame}}
Theorem: Value of Vandermonde Determinant/Formulation 1

Let $V_n$ be the '''Vandermonde determinant of order $n$''' defined as the following formulation:
{{:Definition:Vandermonde Determinant/Formulation 1}}
Its value is given by:
:$\ds V_n = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_j - x_i}$

Proof:
Let:
:<nowiki>$V_n = \begin {vmatrix}
1      & x_1    & {x_1}^2  & \cdots & {x_1}^{n - 2} & {x_1}^{n - 1} \\
1      & x_2    & {x_2}^2  & \cdots & {x_2}^{n - 2} & {x_2}^{n - 1} \\
\vdots & \vdots & \vdots   & \ddots & \vdots        & \vdots \\
1      & x_n    & {x_n}^2  & \cdots & {x_n}^{n - 2} & {x_n}^{n - 1}
\end {vmatrix}$</nowiki>
Let $\map f x$ be an arbitrary monic polynomial of degree $n - 1$:
:$\ds \map f x = x^{n - 1} + \sum_{i \mathop = 0}^{n - 2} a_i x^i$
Apply elementary column operations to $V_n$ repeatedly to show:
:$V_n = W$
where:
:<nowiki>$W = \begin {vmatrix}
1      & x_1    & {x_1}^2  & \cdots & {x_1}^{n - 2} & \map f {x_1} \\
1      & x_2    & {x_2}^2  & \cdots & {x_2}^{n - 2} & \map f {x_2} \\
\vdots & \vdots & \vdots   & \ddots & \vdots        & \vdots \\
1      & x_n    & {x_n}^2  & \cdots & {x_n}^{n - 2} & \map f {x_n}
\end {vmatrix}$</nowiki>
Select a specific degree $n - 1$ monic polynomial:
:$\ds \map f x = \prod_{k \mathop = 1}^{n - 1} \paren {x - x_k}$
The selected polynomial is zero at all values $x_1, \ldots, x_{n - 1}$.
Then the last column of $W$ is all zeros except the entry $\map f {x_n}$.
Expand $\map \det W$ by cofactors along the last column to prove:
{{begin-eqn}}
{{eqn | n = 1
      | l = V_n
      | r = \map f {x_n} V_{n - 1}
      | c = Laplace Expansion Theorem for Determinants for columns
}}
{{eqn | r = V_{n - 1} \prod_{k \mathop = 1}^{n - 1} \paren {x_n - x_k}
}}
{{end-eqn}}
For $n \ge 2$, let $\map P n$ be the statement:
:$\ds V_n = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_j - x_i}$
Mathematical induction will be applied.
=== Basis for the Induction ===
By definition, determinant $V_1 = 1$.
To prove $\map P 2$ is true, use equation $(1)$ with $n = 2$:
:$V_2 = \paren {x_2 - x_1} V_1$
This is the basis for the induction.
=== Induction Step ===
This is the induction step:
Let $\map P n$ is be assumed true.
We are to prove that $\map P {n + 1}$ is true.
As follows:
{{begin-eqn}}
{{eqn | l = V_{n + 1}
      | r = V_n \prod_{k \mathop = 1}^n \paren {x_{n + 1} - x_k}
      | c = from $(1)$, setting $n \to n + 1$
}}
{{eqn | r = \prod_{1 \mathop \le m \mathop < k \mathop \le n} \paren {x_k - x_m} \prod_{k \mathop = 1}^n \paren {x_{n + 1} - x_k}
      | c = Induction hypothesis with new indexing symbols: $i \to m, j \to k$
}}
{{eqn | r = \prod_{1 \mathop \le i \mathop < j \mathop \le n + 1} \paren {x_j - x_i}
      | c = simplifying
}}
{{end-eqn}}
Thus $\map P {n + 1}$ has been shown to be true.
The induction is complete.
{{qed}}
Theorem: Value of Vandermonde Determinant/Formulation 2

Let $V_n$ be the '''Vandermonde determinant of order $n$''' defined as the following formulation:
{{:Definition:Vandermonde Determinant/Formulation 2}}
Its value is given by:
:$\ds V_n = \prod_{1 \mathop \le j \mathop \le n} x_j \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_j - x_i}$

Proof:
The proof follows directly from that for Value of Vandermonde Determinant/Formulation 1 and the result Determinant with Row Multiplied by Constant.
{{finish}}
{{Namedfor|Alexandre-Théophile Vandermonde}}
Theorem: Quantity of Positive Integers Divisible by Particular Integer

Let $d$ be a positive integer. 
Let $x \ge 1$ be a real number.
Then: 
:$\ds \sum_{n \le x, \, d \divides n} 1 = \floor {\frac x d}$
That is: 
:there are $\floor {\dfrac x d}$ natural numbers less than or equal to $x$ that are divisible by $d$.

Proof:
Consider the sum: 
:$\ds \sum_{n \le x, \, d \divides n} 1$
Note that a natural number $n \le x$ is divisible by $d$ {{iff}}:
:there exists a natural number $k$ such that $n = d k$.
So we are counting the natural numbers $k$ such that $d k \le x$. 
That is, the natural numbers $k$ such that: 
:$k \le \dfrac x d$ 
So:
{{begin-eqn}}
{{eqn	| l = \sum_{n \le x, \, d \divides n} 1
	| r = \sum_{k \le x/d} 1
}}
{{eqn	| r = \floor {\frac x d}
}}
{{end-eqn}}
{{qed}}
Category:Analytic Number Theory
38sh5omu3ayezzbj583do34oebk6qql
Theorem: Equivalence of Definitions of P-adic Integer

Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.
{{TFAE|def=P-adic Integer}}
=== Definition 1 ===
{{:Definition:P-adic Integer/Definition 1}}
=== Definition 2 ===
{{:Definition:P-adic Integer/Definition 2}}

Proof:
=== Definition 1 implies Definition 2 ===
{{:Equivalence of Definitions of P-adic Integer/Definition 1 Implies Definition 2}}{{qed|lemma}}
Theorem: Order of Sum of Reciprocal of Primes

:$\ds \sum_{p \mathop \le x} \frac 1 p = \map \ln {\ln x} + \map \OO 1$

Proof:
We have: 
{{begin-eqn}}
{{eqn	| l = \int_p^x \frac 1 {t \ln^2 t} \rd t
	| r = \int_{\ln p}^{\ln x} \frac {e^u} {e^u u^2} \rd u
	| c = substituting $t \mapsto e^u$
}}
{{eqn	| r = \int_{\ln p}^{\ln x} \frac 1 {u^2} \rd u
}}
{{eqn	| r = \intlimits {-\frac 1 u} {\ln p} {\ln x}
	| c = Primitive of Power, Fundamental Theorem of Calculus
}}
{{eqn	| r = \frac 1 {\ln p} - \frac 1 {\ln x}
}}
{{end-eqn}}
So we can write: 
{{begin-eqn}}
{{eqn	| l = \sum_{p \mathop \le x} \frac 1 p
	| r = \sum_{p \mathop \le x} \paren {\frac {\ln p} p \times \frac 1 {\ln p} }
	| c = since $p \ge 2$, $\ln p \ne 0$
}}
{{eqn	| r = \sum_{p \mathop \le x} \frac {\ln p} p \paren {\int_p^x \frac 1 {t \ln^2 t} \rd t + \frac 1 {\ln x} }
}}
{{eqn	| r = \sum_{p \mathop \le x} \frac {\ln p} p \paren {\int_p^x \frac 1 {t \ln^2 t} \rd t} + \frac 1 {\ln x} \sum_{p \mathop \le x} \frac {\ln p} p
}}
{{end-eqn}}
We look to interchange summation and integral in our first term.
Theorem: Similarity Mapping on Plane with Scale Factor Minus 1

Let $s_{-1}: \R^2 \to \R^2$ be a similarity mapping on $\R^2$ whose scale factor is $-1$.
Then $s_{-1}$ is the same as the rotation $r_\pi$ of the plane about the origin one half turn.

Proof:
Let $P = \tuple {x, y} \in \R^2$ be an aribtrary point in the plane.
Then:
{{begin-eqn}}
{{eqn | l = \map {r_\pi} P
      | r = \tuple {\paren {\cos \pi - \sin \pi} x, \paren {\sin \pi + \cos \pi} y}
      | c = Rotation of Plane about Origin is Linear Operator
}}
{{eqn | r = \tuple {\paren {\paren {-1} - 0} x, \paren {0 + \paren {-1} } y}
      | c = Cosine of Straight Angle, Sine of Straight Angle
}}
{{eqn | r = \tuple {-x, -y}
      | c = 
}}
{{eqn | r = \paren {-1} \tuple {x, y}
      | c = 
}}
{{eqn | r = \map {s_{-1} } P
      | c = Definition of $s_{-1}$
}}
{{end-eqn}}
{{qed}}
Theorem: Similarity Mapping on Plane with Negative Parameter

Let $\beta \in \R_{<0}$ be a (strictly) negative real number.
Let $s_\beta: \R^2 \to \R^2$ be the similarity mapping on $\R^2$ whose scale factor is $\beta$.
Then $s_\beta$ is a stretching or contraction followed by a rotation one half turn.

Proof:
Let $\beta = -\gamma$ where $\gamma \in \R_{>0}$.
Let $P = \tuple {x, y} \in \R^2$ be an aribtrary point in the plane.
Then:
{{begin-eqn}}
{{eqn | l = \map {s_\beta} P
      | r = \tuple {\paren {-\gamma} x, \paren {-\gamma} y}
      | c = Definition of $\beta$
}}
{{eqn | r = \paren {-1} \tuple {\gamma x, \gamma y}
      | c = 
}}
{{eqn | r = \paren {-1} \map {s_\gamma} P
      | c = Definition of $s_\gamma$
}}
{{eqn | r = \map {s_{-1} } {\map {s_\gamma} P}
      | c = Definition of $s_{-1}$
}}
{{end-eqn}}
Because $\gamma > 0$ we have by definition that $s_\gamma$ is a stretching or contraction.
From Similarity Mapping on Plane with Scale Factor Minus 1, $s_{-1}$ is the plane rotation of the plane about the angle $\pi$.
Hence, by definition of half turn:
:$s_\beta$ is a stretching or contraction followed by a rotation one half turn.
{{qed}}
Theorem: Similarity Mapping on Plane Commutes with Half Turn about Origin

Let $\beta \in \R_{>0}$ be a (strictly) positive real number.
Let $s_{-\beta}: \R^2 \to \R^2$ be the similarity mapping on $\R^2$ whose scale factor is $-\beta$.
Then $s_{-\beta}$ is the same as:
:a stretching or contraction of scale factor $\beta$ followed by a rotation one half turn
and:
:a rotation one half turn followed by a stretching or contraction of scale factor $\beta$.

Proof:
Let $P = \tuple {x, y} \in \R^2$ be an aribtrary point in the plane.
From Similarity Mapping on Plane with Negative Parameter, $s_{-\beta}$ is a stretching or contraction of scale factor $\beta$ followed by a rotation one half turn.
Thus:
{{begin-eqn}}
{{eqn | l = \map {s_{-\beta} } P
      | r = \map {s_{-1} } {\map {s_\beta} P}
      | c = Similarity Mapping on Plane with Negative Parameter
}}
{{eqn | r = \paren {-1} \map {s_\beta} P
      | c = Definition of $s_{-1}$
}}
{{eqn | r = \paren {-1} \tuple {\beta x, \beta y}
      | c = Definition of $s_\beta$
}}
{{eqn | r = \tuple {-\beta x, -\beta y}
      | c = 
}}
{{eqn | r = \beta \tuple {-x, -y}
      | c = 
}}
{{eqn | r = \beta \map {s_{-1} } P
      | c = Definition of $s_{-1}$
}}
{{eqn | r = \map {s_\beta} {\map {s_{-1} } P}
      | c = Definition of $s_\beta$
}}
{{end-eqn}}
That is:
:$s_\beta$ is a rotation one half turn followed by a stretching or contraction of scale factor $\beta$.
{{qed}}
Theorem: P-adic Integers is Valuation Ring Induced by P-adic Norm

Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.
Then:
:the $p$-adic integers, $\Z_p$, is the valuation ring induced by the non-Archimedean norm $\norm {\,\cdot\,}_p$

Proof:
By the definition of the $p$-adic integers:
:$\Z_p = \set {x \in \Q_p : \norm x_p \le 1}$
From P-adic Numbers form Non-Archimedean Valued Field:
:$\struct {\Q_p, \norm {\,\cdot\,}_p}$ is a non-Archimedean valued field.
By definition of the valuation ring induced by a non-Archimedean norm:
:$\Z_p$ is the valuation ring induced by $\norm {\,\cdot\,}_p$
{{qed}}
Theorem: Equivalence of Definitions of P-adic Integer/Definition 1 Implies Definition 2

Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.
Let $x \in \Q_p$ such that $\norm x_p \le 1$.
Then the canonical expansion of $x$ contains only positive powers of $p$.

Proof:
Let $x \in \Q_p$ such that $\norm x_p \le 1$.
From P-adic Integer is Limit of Unique P-adic Expansion, there exists a $p$-adic expansion of the form:
:$\ds \sum_{n \mathop = 0}^\infty d_n p^n$
By definition of the canonical expansion:
:$\ds \sum_{n \mathop = 0}^\infty d_n p^n$ is the canonical expansion of $x$
It follows that the canonical expansion of $x$ contains only positive powers of $p$.