avewright/proofwiki-math huggingface dataset:
Theorem: Existence of Scalar for Vector Subspace Dimension One Less

Let $\sequence {\beta_n}$ be a sequence of scalars such that:
:$M = \set {\tuple {\lambda_1, \ldots, \lambda_n} \in K^n: \beta_1 \lambda_1 + \cdots + \beta_n \lambda_n = 0}$
Then there is a non-zero scalar $\gamma$ such that:
:$\forall k \in \closedint 1 n: \beta_k = \gamma \alpha_k$

Proof:
Let $\sequence { {e_n}'}$ be the ordered basis of $\paren {K^n}^*$ dual to the standard ordered basis of $K^n$.
Let $M = \map \ker \psi$, where $\ds \psi = \sum_{k \mathop = 1}^n \beta_k {e_k}'$.
From Equivalent Statements for Vector Subspace Dimension One Less:
:$\psi = M^\circ$
As $M^\circ$ is one-dimensional and since $\psi \ne 0$, it follows that:
:$\exists \gamma \ne 0: \psi = \gamma \phi$
Therefore:
:$\forall k \in \closedint 1 n: \beta_k = \gamma \alpha_k$
{{qed}}
Theorem: Kernel of Transpose of Linear Transformation is Annihilator of Image

Let $G$ and $H$ be $n$-dimensional vector spaces over a field.
Let $\map \LL {G, H}$ be the set of all linear transformations from $G$ to $H$.
Let $u \in \map \LL {G, H}$.
Let $u^\intercal$ be the transpose of $u$.
Then:
:$\map \ker {u^\intercal}$ is the annihilator of the image of $u$
where $\map \ker {u^\intercal}$ denotes the kernel of $u^\intercal$.

Proof:
From the definitions of:
:the transpose $u^\intercal$
:the annihilator $\paren {\map u G}^\circ$
it follows that:
:$\map {u^\intercal} y = 0 \iff y \in \paren {\map u G}^\circ$
Thus:
:$\map \ker {u^\intercal} = \paren {\map u G}^\circ$
{{qed}}
Theorem: Image of Transpose of Linear Transformation is Annihilator of Kernel

Let $G$ and $H$ be $n$-dimensional vector spaces over a field.
Let $\map \LL {G, H}$ be the set of all linear transformations from $G$ to $H$.
Let $u \in \map \LL {G, H}$.
Let $u^\intercal$ be the transpose of $u$.
Then:
:The image of $u^\intercal$ is the annihilator of $\map \ker u$.
where $\map \ker u$ denotes the kernel of $u$.

Proof:
Let $x \in \map \ker u$.
Let $H^*$ be the algebraic dual of $H$.
Let $\innerprod x t$ be the evaluation linear transformation.
Then:
:$\forall y \in H^*: \innerprod x {\map {u^\intercal} y} = \innerprod {\map u x} y = \innerprod 0 y = 0$
So:
:$\map {u^\intercal} {H^*} \subseteq \paren {\map \ker u}^\circ$
From Rank Plus Nullity Theorem and Results Concerning Annihilator of Vector Subspace:
{{begin-eqn}}
{{eqn | l = \map \dim {\map {u^\intercal} {H^*} }
      | r = n - \map \dim {\map \ker {u^\intercal} }
      | c = 
}}
{{eqn | r = n - \map \dim {\paren {\map u G}^\circ}
      | c = 
}}
{{eqn | r = \map \dim {\map u G}
      | c = 
}}
{{eqn | r = n - \map \dim {\map \ker u}
      | c = 
}}
{{eqn | r = \map \dim {\paren {\map \ker u}^\circ}
      | c = 
}}
{{end-eqn}}
Hence:
:$\map {u^\intercal} {H^*} = \paren {\map \ker u}^\circ$
{{qed}}
Theorem: Null Sequence Test for Convergence

Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring.
Let $\sequence {x_n}$ be a convergent sequence in $\struct {R, \norm {\,\cdot\,} }$ with limit $l$.
Let $\sequence {y_n}$ be a sequence.
Then:
:$\sequence {y_n}$ converges to the limit $l$ {{iff}} the sequence $\sequence {y_n - x_n}$ is a null sequence

Proof:
=== Necessary Condition ===
Let $\sequence {y_n}$ converge to the limit $l$.
From Difference Rule for Sequences in Normed Division Ring:
:$\ds \lim_{n \mathop \to \infty} y_n - x_n = l - l = 0$
Hence $\sequence {y_n - x_n}$ is a null sequence by definition.
{{qed|lemma}}
Theorem: Ring of Linear Operators is Ring

Let $R$ be a ring.
Let $\struct {G, +}$ be an abelian group..
Let $\struct {G, +, \circ}_R$ be an $R$-module.
Let $\struct {\map {\LL_R} G, +, \circ}$ be the ring of linear operators on $G$, where:
:$+$ denotes pointwise addition
:$\circ$ denotes composition of linear operators.
Then $\struct {\map {\LL_R} G, +, \circ}$ is a ring.

Proof:
Let $\phi$ and $\psi$ be elements of $\map {\LL_R} G$.
From Composite of R-Algebraic Structure Homomorphisms is Homomorphism, $\phi \circ \psi$ is also an element of $\map {\LL_R} G$.
That is, $\struct {\map {\LL_R} G, \circ}$ is closed.
From Set of Linear Transformations under Pointwise Addition forms Abelian Group, $\struct {\map {\LL_R} G, +}$ is an abelian group.
Hence by the One-Step Subgroup Test:
:$\forall \phi, \psi \in \map {\LL_R} G: \phi + \paren {-\psi} \in \map {\LL_R} G$
Then by the Subring Test:
:$\struct {\map {\LL_R} G, +, \circ}$ is a subring of the ring of endomorphisms of $\struct {G, +}$
Hence the result.
{{qed}}
Theorem: Distributional Derivative on Distributions is Linear Operator

The Distributional derivative on Schwartz distributions is a linear operator.

Proof:
Let $\phi, \psi \in \map \DD \R$ be test functions.
Let $\alpha \in \C$ be a complex number.
Let $T \in \map {\DD'} \R$ be a Schwartz distribution.
Then:
{{begin-eqn}}
{{eqn | l = \map {T'} {\phi + \psi}
      | r = - \map T {\paren {\phi + \psi}'}
      | c = {{Defof|Distributional Derivative}}
}}
{{eqn | r = - \map T {\phi' + \psi'}
      | c = Sum Rule for Derivatives
}}
{{eqn | r = - \paren {\map T {\phi'} + \map T {\psi'} }
      | c = {{Defof|Schwartz Distribution}}
}}
{{eqn | r = - \map T {\phi'} - \map T {\psi'}
}}
{{eqn | r = \map {T'} \phi + \map {T'} \psi
      | c = {{Defof|Distributional Derivative}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {T'} {\alpha \phi}
      | r = - \map T {\paren {\alpha \phi}'}
      | c = {{Defof|Distributional Derivative}}
}}
{{eqn | r = - \map T {\alpha \phi'}
}}
{{eqn | r = - \alpha \map T {\phi'}
      | c = {{Defof|Schwartz Distribution}}
}}
{{eqn | r = \alpha \map {T'} \phi
      | c = {{Defof|Distributional Derivative}}
}}
{{end-eqn}}
Thus the distributional derivative is a linear operator.
{{qed}}
Theorem: Distributional Derivative on Distributions is Continuous Operator

The distributional derivative on Schwartz distributions is a continuous operator.

Proof:
Let $\mathbf 0$ be the zero mapping.
Let $\sequence {\phi_n}_{n \mathop \in \N} \in \map \DD \R$ be a sequence of test functions.
Let $\sequence {\phi_n}_{n \mathop \in \N}$ converge to $\mathbf 0$ in the test function space:
:$\phi_n \stackrel \DD {\longrightarrow} \mathbf 0$
By definition, a test function is a smooth real function with a compact support $K$.
By differentiating a test function we get another smooth real function with a compact support not larger than $K$.
Thus, $\phi_n'$ is also a test function.
Since $\phi_n \stackrel \DD {\longrightarrow} \mathbf 0$, we have that for every multiindex $k$ the sequence $\sequence {D^k \phi_n}_{n \mathop \in \N}$ converges uniformly to $\mathbf 0$.
Obviously, all the sequences of $\sequence {D^k \phi_n}_{n \mathop \in \N}$ with $k \ge 1$ converge uniformly to $\mathbf 0$.
By definition, $\sequence {\phi_n'}_{n \mathop \in \N}$ converges to $\mathbf 0$ in the test function space:
:$\phi_n' \stackrel \DD {\longrightarrow} \mathbf 0$
Let $T \in \map {\DD'} \R$ be a Schwartz distribution.
Then:
:$\paren {\phi_n' \stackrel \DD {\longrightarrow} \mathbf 0} \implies \paren {\map T {\phi_n'} \to \map T {\mathbf 0} = 0}$
By definition of the distributional derivative:
:$\map T {\phi_n'} = - \map {T'} {\phi_n}$
Therefore:
:$\paren {\phi_n' \stackrel \DD {\longrightarrow} \mathbf 0} \implies \paren {\map {T'} {\phi_n} \to 0}$
By definition, the distributional derivative on Schwartz distributions is a continuous operator.
{{qed}}
Theorem: Composition of Linear Transformations is Isomorphic to Matrix Product

Let $R$ be a ring with unity.
Let $F$, $G$ and $H$ be free $R$-modules of finite dimension $p, n, m > 0$ respectively.
Let $\sequence {a_p}$, $\sequence {b_n}$ and $\sequence {c_m}$ be ordered bases 
Let $\map {\LL_R} {G, H}$ denote the set of all linear transformations from $G$ to $H$.
Let $\map {\MM_R} {m, n}$ be the $m \times n$ matrix space over $R$.
Let $\sqbrk {u; \sequence {c_m}, \sequence {b_n} }$ be the matrix of $u$ relative to $\sequence {b_n}$ and $\sequence {c_m}$.
Let $M: \map {\LL_R} {G, H} \to \map {\MM_R} {m, n}$ be defined as:
:$\forall u \in \map {\LL_R} {G, H}: \map M u = \sqbrk {u; \sequence {c_m}, \sequence {b_n} }$
Then:
:$\forall u \in \map {\LL_R} {F, G}, v \in \map {\LL_R} {G, H}: \sqbrk {v \circ u; \sequence {c_m}, \sequence {a_p} } = \sqbrk {v; \sequence {c_m}, \sequence {b_n} } \sqbrk {u; \sequence {b_n}, \sequence {a_p} }$

Proof:
Follows directly from Relative Matrix of Composition of Linear Transformations.
{{qed}}
Theorem: Slope of Normal is Minus Reciprocal of Tangent

Let $C$ be a curve defined by a real function which is differentiable.
Let $P$ be a point on $C$.
Let the curvature of $C$ at $P$ be non-zero.
Let $r$ be the slope of the tangent to $C$ at $P$.
Let $s$ be the slope of the normal to $C$ at $P$. 
Then:
:$r = -\dfrac 1 s$

Proof:
By definition, the normal to $C$ at $P$ is defined as being perpendicular to the tangent at $P$ and in the same plane as $P$.
The result follows from Condition for Straight Lines in Plane to be Perpendicular.
{{qed}}
Category:Normals to Curves
Category:Tangents
Category:Analytic Geometry
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Theorem: Change of Basis Matrix under Linear Transformation

Let $G$ and $H$ be free unitary $R$-modules of finite dimensions $n, m > 0$ respectively.
Let $\sequence {a_n}$ and $\sequence { {a_n}'}$ be ordered bases of $G$.
Let $\sequence {b_m}$ and $\sequence { {b_m}'}$ be ordered bases of $H$.
Let $u: G \to H$ be a linear transformation.
Let $\sqbrk {u; \sequence {b_m}, \sequence {a_n} }$ denote the matrix of $u$ relative to $\sequence {a_n}$ and $\sequence {b_m}$.
Let:
:$\mathbf A = \sqbrk {u; \sequence {b_m}, \sequence {a_n} }$
:$\mathbf B = \sqbrk {u; \sequence { {b_m}'}, \sequence { {a_n}'} }$
Then:
:$\mathbf B = \mathbf Q^{-1} \mathbf A \mathbf P$
where:
:$\mathbf P$ is the matrix corresponding to the change of basis from $\sequence {a_n}$ to $\sequence { {a_n}'}$
:$\mathbf Q$ is the matrix corresponding to the change of basis from $\sequence {b_m}$ to $\sequence { {b_m}'}$.

Proof:
We have $u = I_H \circ u \circ I_G$
and $\mathbf Q^{-1} = \sqbrk {I_H; \sequence { {b_m}'}, \sequence {b_m} }$.
Thus by Set of Linear Transformations is Isomorphic to Matrix Space:
{{begin-eqn}}
{{eqn | l = \mathbf Q^{-1} \mathbf A \mathbf P
      | r = \sqbrk {I_H; \sequence { {b_m}'}, \sequence {b_m} } \sqbrk {u; \sequence {b_m}, \sequence {a_n} } \sqbrk {I_G; \sequence {a_n}, \sequence { {a_n}'} }
      | c = 
}}
{{eqn | r = \sqbrk {I_H \circ u \circ I_G; \sequence { {b_m}'}, \sequence { {a_n}'} }
      | c = 
}}
{{eqn | r = \mathbf B
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Change of Basis Matrix under Linear Transformation/Converse

Let $G$ and $H$ be free unitary $R$-modules of finite dimensions $n, m > 0$ respectively.
Let $\sequence {a_n}$ be an ordered basis of $G$.
Let $\sequence {b_m}$ be an ordered basis of $H$.
Let $\mathbf A$ and $\mathbf B$ be $m \times n$ matrices over $R$.
Let there exist:
:a nonsingular matrix $\mathbf P$ of order $n$
:a nonsingular matrix $\mathbf Q$ of order $m$
such that:
:$\mathbf B = \mathbf Q^{-1} \mathbf A \mathbf P$
Then there exist:
:a linear transformation $u: G \to H$
:ordered bases $\sequence { {a_n}'}$ and $\sequence { {b_m}'}$ of $G$ and $H$ respectively
such that:
:$\mathbf A = \sqbrk {u; \sequence {b_m}, \sequence {a_n} }$
:$\mathbf B = \sqbrk {u; \sequence { {b_m}'}, \sequence { {a_n}'} }$
where $\sqbrk {u; \sequence {b_m}; \sequence {a_n} }$ denotes the matrix of $u$ relative to $\sequence {a_n}$ and $\sequence {b_m}$.

Proof:
Let:
:$\mathbf P = \sqbrk \alpha_n$
:$\mathbf Q = \sqbrk \beta_m$
Let:
:$\forall j \in \closedint 1 n: {a_j}' = \ds \sum_{i \mathop = 1}^n \alpha_{i j} a_i$
:$\forall j \in \closedint 1 m: {b_j}' = \ds \sum_{i \mathop = 1}^m \beta_{i j} b_i$
Then by Invertible Matrix Corresponds with Change of Basis:
:$\sequence { {a_n}'}$ and $\sequence { {b_m}'}$ are ordered bases of $G$ and $H$ respectively.
Also we have:
:$\mathbf P$ is the matrix corresponding to the change in basis from $\sequence {a_n}$ to $\sequence { {a_n}'}$
:$\mathbf Q$ is the matrix corresponding to the change in basis from $\sequence {b_m}$ to $\sequence { {b_m}'}$
:so $\mathbf Q^{-1}$ is the matrix corresponding to the change in basis from $\sequence { {b_m}'}$ to $\sequence {b_m}$
Let $\map {\LL_R} {G, H}$ be the set of all linear transformations from $G$ to $H$.
By Set of Linear Transformations is Isomorphic to Matrix Space, there exists $u \in \map {\LL_R} {G, H}$ such that:
:$\mathbf A = \sqbrk {u; \sequence {b_m}, \sequence {a_n} }$
But then, by Change of Basis Matrix under Linear Transformation:
:$\sqbrk {u; \sequence { {b_m}'}, \sequence { {a_n}'} } = \mathbf Q^{-1} \mathbf A \mathbf P = \mathbf B$
{{Qed}}
Theorem: P-adic Expansion of P-adic Unit

Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.
Let $\Z_p$ be the $p$-adic integers.
Let $a \in \Z_p$.
Let $\ldots a_n \ldots a_3 a_2 a_1 a_0$ be the canonical expansion of  $a$.
Then:
:$a$ is a $p$-adic unit {{iff}} $a_0 \ne 0$

Proof:
From P-adic Unit has Norm Equal to One:
:a is a $p$-adic unit {{iff}} $\norm a_p = 1 = p^0$
By definition of the canonical expansion:
:$a$ is the limit of the $p$-adic expansion $\ds \sum_{n \mathop = 0}^\infty a_n p^n$
From P-adic Norm of P-adic Expansion is determined by First Nonzero Coefficient:
:$\norm a_p = p^0$ {{iff}} $a_0 \ne 0$
The result follows.
{{qed}}
Theorem: Characterization of Rational P-adic Integer

Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.
Let $\Z_p$ be the $p$-adic integers for some prime $p$.
Let $\Q$ be the rational numbers.
Then:
:$\Z_p \cap \Q = \set{\dfrac a b \in \Q : p \nmid b}$

Proof:
Let $\norm{\,\cdot\,}^\Q _p$ denote the $p$-adic norm on the rational numbers.
We have:
{{begin-eqn}}
{{eqn | l = \Z_p \cap \Q
      | r = \set{\dfrac a b \in \Q : \norm {\dfrac a b}_p \le 1}
      | c = {{Defof|P-adic Integer|$p$-adic integers}}
}}
{{eqn | r = \set{\dfrac a b \in \Q : \norm{\dfrac a b}^\Q_p \le 1}
      | c = Rational Numbers are Dense Subfield of P-adic Numbers
}}
{{eqn | r = \set{\dfrac a b \in \Q : p \nmid b}
      | c = Valuation Ring of P-adic Norm on Rationals
}}
{{end-eqn}}
{{qed}}
Theorem: Characterization of Rational P-adic Unit

Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.
Let $\Z^\times_p$ be the $p$-adic units.
Let $\Q$ be the rational numbers.
Then:
:$\Z^\times_p \cap \Q = \set{\dfrac a b \in \Q : p \nmid ab}$

Proof:
Let $\norm{\,\cdot\,}^\Q _p$ denote the $p$-adic norm on the rational numbers.
We have:
{{begin-eqn}}
{{eqn | l = \Z^\times_p \cap \Q
      | r = \set{\dfrac a b \in \Q : \norm {\dfrac a b}_p = 1}
      | c = P-adic Unit has Norm Equal to One
}}
{{eqn | r = \set{\dfrac a b \in \Q : \norm {\dfrac a b}_p \le 1} \setminus \set{\dfrac a b \in \Q : \norm {\dfrac a b}_p < 1}
      | c = {{Defof|Set Difference}} 
}}
{{eqn | r = \set{\dfrac a b \in \Q : \norm{\dfrac a b}^\Q_p \le 1} \setminus \set{\dfrac a b \in \Q : \norm {\dfrac a b}^\Q_p < 1}
      | c = Rational Numbers are Dense Subfield of P-adic Numbers
}}
{{eqn | r = \set{\dfrac a b \in \Q : \norm{\dfrac a b}^\Q_p \le 1} \setminus \set{\dfrac a b \in \Q : p \nmid b, p \divides a}
      | c = Valuation Ideal of P-adic Norm on Rationals
}}
{{eqn | r = \set{\dfrac a b \in \Q : p \nmid b} \setminus \set{\dfrac a b \in \Q : p \nmid b, p \divides a}
      | c = Valuation Ring of P-adic Norm on Rationals
}}
{{eqn | r = \set{\dfrac a b \in \Q : p \nmid b, p \nmid a}
      | c = {{Defof|Set Difference}} 
}}
{{eqn | r = \set{\dfrac a b \in \Q : p \nmid ab}
      | c = Prime Divisor of Coprime Integers and Divisor Divides Multiple
}}
{{end-eqn}}
{{qed}}
Theorem: Equivalence of Definitions of Matrix Equivalence

Let $R$ be a ring with unity.
Let $\mathbf A, \mathbf B$ be $m \times n$ matrices over $R$.
{{TFAE|def = Matrix Equivalence}}

Proof:
This is specifically demonstrated in Change of Basis Matrix under Linear Transformation.
{{qed}}
Theorem: Equivalence of Definitions of Matrix Similarity

Let $R$ be a ring with unity.
Let $n \in \N_{>0}$ be a natural number.
Let $\mathbf A, \mathbf B$ be square matrices of order $n$ over $R$.
{{TFAE|def = Matrix Similarity}}

Proof:
This is specifically demonstrated in {{Corollary|Change of Basis Matrix under Linear Transformation}}.
{{qed}}
Theorem: Multiple Function on Ring is Homomorphism

:$g_a$ is a group homomorphism from $\struct {\Z, +}$ to $\struct {R, +}$.

Proof:
{{begin-eqn}}
{{eqn | l = \map {g_a} m + \map {g_a} n
      | r = m \cdot a + n \cdot a
      | c = {{Defof|Integral Multiple|subdef = Rings and Fields}}
}}
{{eqn | r = \paren {m + n} \cdot a
      | c = Integral Multiple Distributes over Ring Addition
}}
{{eqn | r = \map {g_a} {m + n}
      | c = {{Defof|Integral Multiple|subdef = Rings and Fields}}
}}
{{end-eqn}}
{{qed}}
Theorem: Principal Ideal of Characteristic of Ring is Subset of Kernel of Multiple Function

:$\ideal p \subseteq \map \ker {g_a}$
where:
:$\map \ker {g_a}$ is the kernel of $g_a$
:$\ideal p$ is the principal ideal of $\Z$ generated by $p$.

Proof:
We have from Multiple Function on Ring is Homomorphism that $g_a$ is a group homomorphism.
By definition of kernel:
:$x \in \map \ker {g_a} \iff \map {g_a} x = 0_R$
Hence to show that $\ideal p \subseteq \map \ker {g_a}$, we need to show that:
:$\forall x \in \ideal p: \map {g_a} x = 0_R$
By definition of '''characteristic''', $p \in \Z_{\ge 0}$ is such that $\ideal p$ is the kernel of $g_1$:
:$\map {g_1} n = n \cdot 1_R$
So:
{{begin-eqn}}
{{eqn | l = x
      | o = \in
      | r = \ideal p
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map {g_1} x
      | r = 0_R
      | c = {{Defof|Characteristic of Ring|index = 2}}: $\ideal p$ is the kernel of $g_1$
}}
{{eqn | n = 1
      | ll= \leadsto
      | l = x \cdot 1_R
      | r = 0_R
      | c = Definition of $g_a$ where here $a = 1$
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \map {g_a} x
      | r = x \cdot a
      | c = Definition of $g_a$
}}
{{eqn | r = x \cdot \paren {a \circ 1_R}
      | c = {{Defof|Multiplicative Identity}}
}}
{{eqn | r = a \circ \paren {x \cdot 1_R}
      | c = Multiple of Ring Product
}}
{{eqn | r = a \circ 0_R
      | c = from $(1)$
}}
{{eqn | r = 0_R
      | c = {{Defof|Ring Zero}}
}}
{{end-eqn}}
So:
:$x \in \map \ker {g_a}$
and so: 
:$\ideal p \subseteq \map \ker {g_a}$
{{qed}}
Theorem: Multiplication Function on Ring with Unity is Zero if Characteristic is Divisor

:$p \divides n \implies n \cdot a = 0_R$
where $p \divides n$ denotes that $p$ is a divisor of $n$.

Proof:
Let $p > 0$.
From Principal Ideal of Characteristic of Ring is Subset of Kernel of Multiple Function:
:$\ideal p \subseteq \map \ker {g_a}$
where:
:$\map \ker {g_a}$ is the kernel of $g_a$
:$\ideal p$ is the principal ideal of $\Z$ generated by $p$.
We have:
{{begin-eqn}}
{{eqn | q = 
      | l = p
      | o = \divides
      | r = n
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \exists k \in \Z
      | l = k p
      | r = n
      | c = 
}}
{{end-eqn}}
Then we have:
{{begin-eqn}}
{{eqn | q = 
      | l = kp = n
      | o = \in
      | r = \ideal p
      | c = Integral Ideal iff Set of Integer Multiples
}}
{{eqn | ll= \leadsto
      | l = n
      | o = \in
      | r = \map \ker {g_a}
      | c = Principal Ideal of Characteristic of Ring is Subset of Kernel of Multiple Function
}}
{{eqn | ll= \leadsto
      | l = \map {g_a} n
      | r = 0
      | c = {{Defof|Kernel of Group Homomorphism}}
}}
{{eqn | ll= \leadsto
      | l = n \cdot a
      | r = 0
      | c = Definition of $g_a$
}}
{{end-eqn}}
{{qed}}
Theorem: Kernel of Non-Zero Divisor Multiple Function is Primary Ideal of Characteristic

Let $a \in R$ such that $a$ is not a zero divisor of $R$.
Then:
:$\map \ker {g_a} = \ideal p$
where:
:$\map \ker {g_a}$ is the kernel of $g_a$
:$\ideal p$ is the principal ideal of $\Z$ generated by $p$.

Proof:
From Principal Ideal of Characteristic of Ring is Subset of Kernel of Multiple Function we have:
:$\ideal p \subseteq \map \ker {g_a}$
for all $a \in R$.
It remains to be shown that for all $a \in R$ such that $a$ is not a zero divisor of $R$:
:$\map \ker {g_a} \subseteq \ideal p$
So:
{{begin-eqn}}
{{eqn | l = n
      | o = \in
      | r = \map \ker {g_a}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = n \cdot a
      | r = 0_R
      | c = 
}}
{{eqn | ll= \leadsto
      | l = n \cdot \paren {a \circ 1_R}
      | r = 0_R
      | c = {{Defof|Unity of Ring}}
}}
{{eqn | ll= \leadsto
      | l = \paren {n \cdot a} \circ 1_R
      | r = 0_R
      | c = Multiple of Ring Product
}}
{{eqn | ll= \leadsto
      | l = a \circ \paren {n \cdot 1_R}
      | r = 0_R
      | c = Multiple of Ring Product
}}
{{eqn | ll= \leadsto
      | l = n \cdot 1_R
      | r = 0_R
      | c = as $a$ is not a zero divisor of $R$
}}
{{eqn | ll= \leadsto
      | l = n
      | o = \in
      | r = \ideal p
      | c = {{Defof|Characteristic of Ring|index = 2}}
}}
{{end-eqn}}
Hence the result by definition of subset.
{{qed}}
Theorem: Kernel of Multiple Function on Ring with Characteristic Zero is Trivial

Let $a \in R$ such that $a$ is not a zero divisor of $R$.
Let the characteristic of $R$ be $0$.
Then:
:$\map \ker {g_a} = \set {0_R}$
where $\ker$ denotes the kernel of $g_a$.

Proof:
For $n = 0$, we trivially have $n \cdot a = 0_R$.
{{AimForCont}} $\exists n \ne 0: n \cdot a = 0_R$.
Then:
{{begin-eqn}}
{{eqn | l = n
      | o = \in
      | r = \map \ker {g_a}
      | c = {{Defof|Kernel of Group Homomorphism}}
}}
{{eqn | ll= \leadsto
      | l = n
      | o = \in
      | r = \ideal p
      | c = Kernel of Non-Zero Divisor Multiple Function is Primary Ideal of Characteristic
}}
{{eqn-intertext|where $p$ is the characteristic of $R$}}
{{eqn | ll= \leadsto
      | l = p
      | o = \divides 
      | r = n
      | c = {{Defof|Integral Ideal}}
}}
{{eqn | ll= \leadsto
      | l = p
      | o = \ne
      | r = 0
      | c = 
}}
{{end-eqn}}
This contradicts our assertion that the characteristic of $R$ is $0$.
Hence by Proof by Contradiction there can be no such $n \ne 0$ such that $n \cdot a = 0_R$.
{{qed}}
Theorem: Multiple Function on Ring is Zero iff Characteristic is Divisor

Let $a \in R$ such that $a$ is not a zero divisor of $R$.
Then:
:$n \cdot a = 0_R$
{{iff}}:
:$p \divides n$

Proof:
Let $g_a: \Z \to R$ be the mapping from the integers into $R$ defined as:
:$\forall n \in \Z:\forall a \in R: \map {g_a} n = n \cdot a$
Then from Kernel of Non-Zero Divisor Multiple Function is Primary Ideal of Characteristic:
:$\map \ker {g_a} = \ideal p$
where:
:$\map \ker {g_a}$ is the kernel of $g_a$
:$\ideal p$ is the principal ideal of $\Z$ generated by $p$.
We have by definition of kernel:
:$n \in \map \ker {g_a} \iff n \cdot a = 0_R$
and by definition of principal ideal:
:$n \in \ideal p \iff p \divides n$
The result follows.
{{qed}}
Theorem: Canonical P-adic Expansion of Rational is Eventually Periodic

Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.
Let $x \in \Q_p$.
Then:
:$x$ is a rational number {{iff}} the canonical expansion of $x$ is eventually periodic.

Proof:
=== Necessary Condition ===
Let $x$ be a rational number.
{{:Canonical P-adic Expansion of Rational is Eventually Periodic/Necessary Condition}}{{qed|lemma}}
Theorem: Sum of Hyperbolic Sine and Cosine equals Exponential

Let $z \in \C$ be a complex number.
Then:
:$e^z = \cosh z + \sinh z$

Proof:
{{begin-eqn}}
{{eqn | l = \cosh z + \sinh z
      | r = \dfrac {e^z + e^{-z} } 2 + \dfrac {e^z - e^{-z} } 2
      | c = {{Defof|Hyperbolic Cosine}} and {{Defof|Hyperbolic Cosine}}
}}
{{eqn | r = \dfrac {e^z + e^z + e^{-z} - e^{-z} } 2
      | c = 
}}
{{eqn | r = e^z
      | c = 
}}
{{end-eqn}}
{{Qed}}
Theorem: Naturally Ordered Semigroup Axioms are Independent

Consider the naturally ordered semigroup axioms:
{{:Axiom:Naturally Ordered Semigroup Axioms}}
Each of the naturally ordered semigroup axioms is independent of all the others.
That is, you cannot drop any one of them and still uniquely define a naturally ordered semigroup.

Proof:
This will be proved by demonstrating that for each of the naturally ordered semigroup axioms, it is possible to create an algebraic structure which fulfils all the others, but is not a naturally ordered semigroup.