avewright/proofwiki-math huggingface dataset:
Theorem: Length of Element of Arc in Orthogonal Curvilinear Coordinates

Let $\tuple {q_1, q_2, q_3}$ denote a set of orthogonal curvilinear coordinates.
Let the relation between those orthogonal curvilinear coordinates and Cartesian coordinates be expressed as:
{{begin-eqn}}
{{eqn | l = x
      | r = \map x {q_1, q_2, q_3}
}}
{{eqn | l = y
      | r = \map y {q_1, q_2, q_3}
}}
{{eqn | l = z
      | r = \map z {q_1, q_2, q_3}
}}
{{end-eqn}}
where $\tuple {x, y, z}$ denotes the Cartesian coordinates.
Let $S$ be an infinitesimal arc.
Let $\d s$ be the length of $S$
Then:
{{begin-eqn}}
{{eqn | l = \d s^2
      | r = {h_1}^2 {\rd q_1}^2 + {h_2}^2 {\rd q_2}^2 + {h_3}^2 {\rd q_3}^2
      | c = 
}}
{{eqn | r = \sum_i {h_i}^2 {\rd q_i}^2
      | c = for $i \in \set {1, 2, 3}$
}}
{{end-eqn}}
where:
:$\d q_i$ is the projection of $S$ onto the unit normal to the curvilinear coordinate surface determined by $q_i$, for $i \in \set {1, 3}$
:${h_i}^2 = \paren {\dfrac {\partial x} {\partial q_i} }^2 + \paren {\dfrac {\partial y} {\partial q_i} }^2 + \paren {\dfrac {\partial z} {\partial q_i} }^2$

Proof:
By definition of the metric of $\tuple {q_1, q_2, q_3}$:
{{begin-eqn}}
{{eqn | l = \d s^2
      | r = \d x^2 + \d y^2 + \d z^2
      | c = 
}}
{{eqn | r = \sum_{i, j} {h_{i j} }^2 \rd q_i \rd q_j
      | c = for $i, j \in \set {1, 2, 3}$
}}
{{end-eqn}}
From Value of Curvilinear Coordinate Metric:
:$\forall i, j \in \set {1, 2, 3}: {h_{i j} }^2 = \dfrac {\partial x} {\partial q_i} \dfrac {\partial x} {\partial q_j} + \dfrac {\partial y} {\partial q_i} \dfrac {\partial y} {\partial q_j} + \dfrac {\partial z} {\partial q_i} \dfrac {\partial z} {\partial q_j}$
But we have that $\tuple {q_1, q_2, q_3}$ is orthogonal.
From Definition 1 of orthogonal curvilinear coordinates:
:$\dfrac {\partial x} {\partial q_i} \dfrac {\partial x} {\partial q_j} + \dfrac {\partial y} {\partial q_i} \dfrac {\partial y} {\partial q_j} + \dfrac {\partial z} {\partial q_i} \dfrac {\partial z} {\partial q_j} = 0$
wherever $i \ne j$.
Hence we have:
{{begin-eqn}}
{{eqn | l = \d s^2
      | r = {h_{1 1} }^2 {\rd q_1}^2 + {h_{2 2} }^2 {\rd q_2}^2 + {h_{3 3} }^2 {\rd q_3}^2
      | c = 
}}
{{eqn | r = \sum_i {h_{i i} }^2 {\rd q_i}^2
      | c = for $i \in \set {1, 2, 3}$
}}
{{end-eqn}}
Elements $h_{i i}$ are those elements of the metric which do not vanish when $\tuple {q_1, q_2, q_3}$ is orthogonal.
To streamline notation, we rename them $h_1$, $h_2$ and $h_3$.
Hence:
{{begin-eqn}}
{{eqn | l = \d s^2
      | r = {h_1}^2 {\rd q_1}^2 + {h_2}^2 {\rd q_2}^2 + {h_3}^2 {\rd q_3}^2
      | c = 
}}
{{eqn | r = \sum_i {h_i}^2 {\rd q_i}^2
      | c = for $i \in \set {1, 2, 3}$
}}
{{end-eqn}}
where:
:${h_i}^2 = \paren {\dfrac {\partial x} {\partial q_i} }^2 + \paren {\dfrac {\partial y} {\partial q_i} }^2 + \paren {\dfrac {\partial z} {\partial q_i} }^2$
{{qed}}
Theorem: Laplacian of Function in Orthogonal Curvilinear Coordinates

Let $\map \psi {q_1, q_2, q_3}$ denote a real-valued function embedded in an orthogonal curvilinear coordinate system.
Then the Laplacian of $\psi$ can be expressed as:
:$\nabla^2 \psi = \dfrac 1 {h_1 h_2 h_3} \paren {\map {\dfrac \partial {\partial q_1} } {\dfrac {h_2 h_3} {h_1} \dfrac {\partial \psi} {\partial q_1} } + \map {\dfrac \partial {\partial q_2} } {\dfrac {h_3 h_1} {h_2} \dfrac {\partial \psi} {\partial q_2} } + \map {\dfrac \partial {\partial q_3} } {\dfrac {h_1 h_2} {h_3} \dfrac {\partial \psi} {\partial q_3} } }$
where:
:${h_i}^2 = \paren {\dfrac {\partial x} {\partial q_i} }^2 + \paren {\dfrac {\partial y} {\partial q_i} }^2 + \paren {\dfrac {\partial z} {\partial q_i} }^2$

Proof:
{{ProofWanted|a coherent understanding of exactly what it means would be useful here}}
Theorem: Shortest Distance between Two Points is Straight Line

The shortest distance between $2$ points is a straight line.

Proof:
Let $s$ be the length of a curve between $2$ points $A$ and $B$.
The problem becomes one of finding the curve for which $\ds \int_a^B \rd s$ is a minimum.
{{ProofWanted|In due course as the work progresses}}
Hence such a curve has the equation:
:$y = m x + c$
which defines a straight line.
Theorem: Equivalence of Definitions of Matroid Circuit Axioms

Let $S$ be a finite set.
Let $\mathscr C$ be a non-empty set of subsets of $S$.
{{TFAE|axiom=Circuit Axioms (Matroid)|view=Matroid Circuit Axioms}}
=== Formulation 1 ===
$\mathscr C$ satisfies the circuit axioms:
{{:Axiom:Circuit Axioms (Matroid)/Formulation 1}}
=== Formulation 2 ===
$\mathscr C$ satisfies the circuit axioms:
{{:Axiom:Circuit Axioms (Matroid)/Formulation 2}}
=== Formulation 3 ===
$\mathscr C$ satisfies the circuit axioms:
{{:Axiom:Circuit Axioms (Matroid)/Formulation 3}}

Proof:
=== Formulation 1 implies Formulation 2 ===
{{:Equivalence of Definitions of Matroid Circuit Axioms/Formulation 1 Implies Formulation 2}}{{qed|lemma}}
Theorem: Rank of Matroid Circuit is One Less Than Cardinality

Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $C \subseteq S$ be a circuit of $M$.
Let $\rho: \powerset S \to \Z$ denote the rank function of $M$.
Then:
:$\map \rho C = \card C -1$

Proof:
By definition of a circuit:
:$C$ is dependent
By matroid axiom $(\text I 1)$:
:$C \ne \O$
Let $x \in C$.
Theorem: Bound for Cardinality of Matroid Circuit

Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $C \subseteq S$ be a circuit of $M$.
Let $\rho: \powerset S \to \Z$ denote the rank function of $M$.
Then:
:$\card C \le \map \rho S + 1$

Proof:
By definition of a circuit:
:$C$ is dependent
By matroid axiom $(\text I 1)$:
:$C \ne \O$
Let $x \in C$.
From Set Difference is Subset and Set Difference with Disjoint Set:
:$C \setminus \set x \subsetneq C$
From Proper Subset of Matroid Circuit is Independent and matroid axiom $(\text I 1)$:
:$C \setminus \set x \in \mathscr I$
We have:
{{begin-eqn}}
{{eqn | l = \map \rho S
      | o = \ge
      | r = \card {C \setminus \set x}
      | c = {{Defof|Rank Function (Matroid)|Rank Function}}
}}
{{eqn | r = \card C - \card {\set x}
      | c = Cardinality of Set Difference with Subset
}}
{{eqn | r = \card C - 1
      | c = Cardinality of Singleton
}}
{{eqn | ll = \leadsto
      | l = \map \rho S + 1
      | o = \ge
      | r = \card C
      | c = adding $1$ to both sides of the equation
}}
{{end-eqn}}
{{qed}}
Theorem: Matroid with No Circuits Has Single Base

Let $M = \struct {S, \mathscr I}$ be a matroid with no circuits.
Then:
:$S$ is the only base on $M$.

Proof:
From Dependent Subset Contains a Circuit:
:$M$ has no dependent subsets
By definition of dependent subsets:
:Every subset of $S$ is independent
In particular:
:$S \in \mathscr I$
By definition of independent subsets:
:$X \in \mathscr I \implies X \subseteq S$
Hence $S$ is a base on $M$ by definition.
Let $X$ be a base on $M$.
Then $X \subseteq S$.
By definition of a base on $M$:
:$X$ is a maximal independent subset of $M$
As $S \in \mathscr I$, by definition of a  maximal set:
:$X = S$
Hence:
:$S$ is the only base on $M$.
{{qed}}
Theorem: Proper Subset of Matroid Circuit is Independent

Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $C \subseteq S$ be a circuit of $M$.
Then every proper subset $A$ of $C$ is independent.

Proof:
By definition of a circuit of $M$:
:$C$ is a minimum dependent subset of $M$
By definition of the minimum dependent subset of $M$:
:every proper subset $A$ of $C$ is not a dependent subset.
By definition of a dependent subset:
:every proper subset $A$ of $C$ is an independent subset.
{{qed}}
Theorem: Supremum Norm is Norm/Space of Bounded Sequences

The supremum norm on the vector space of bounded sequences is a norm.

Proof:
=== {{NormAxiomVector|1|nolink}} ===
Let $x \in \ell^\infty$.
By definition of supremum norm:
:$\ds \norm {\mathbf x}_\infty = \sup_{n \mathop \in \N} \size {x_n}$
The complex modulus of $x_n$ is real and non-negative.
Hence, $\norm {\mathbf x}_\infty \ge 0$.
Suppose $\norm {\mathbf x}_\infty = 0$.
Then:
{{begin-eqn}}
{{eqn | l = \norm {\mathbf x}_\infty
      | r = 0
}}
{{eqn | ll= \leadsto
      | l = \sup_{n \mathop \in \N} \size {x_n}
      | r = 0
}}
{{eqn | ll= \leadsto
      | l = x_n
      | r = 0
      | c = Complex Modulus equals Zero iff Zero
}}
{{eqn | ll= \leadsto
      | l = \bf x
      | r = \sequence 0_{n \mathop \in \N}
}}
{{end-eqn}}
Thus {{NormAxiomVector|1}} is satisfied.
{{qed|lemma}}
Theorem: Parallelogram Law for Vector Subtraction

Let $\mathbf u$ and $\mathbf v$ be vectors.
Consider a parallelogram, two of whose adjacent sides represent $\mathbf y$ and $\mathbf v$ (in magnitude and direction).
:400px
Then the diagonal of the parallelogram connecting the terminal points of $\mathbf u$ and $\mathbf v$ represents the magnitude and direction of $\mathbf u - \mathbf v$, the difference of $\mathbf u$ and $\mathbf v$.

Proof:
We can construct a parallelogram as follows:
:400px
and the construction is apparent.
Theorem: Space of Bounded Sequences with Supremum Norm forms Normed Vector Space

The vector space of bounded sequences with the supremum norm forms a normed vector space.

Proof:
We have that:
:Space of Bounded Sequences with Pointwise Addition and Pointwise Scalar Multiplication on Ring of Sequences forms Vector Space
:Supremum norm on the space of bounded sequences is a norm
By definition, $\struct {\ell^\infty, \norm {\, \cdot \,}_\infty}$ is a normed vector space.
{{qed}}
Theorem: Riesz's Lemma

Let $X$ be a normed vector space.
Let $Y$ be a proper closed linear subspace of $X$. 
Let $\alpha \in \openint 0 1$.
Then there exists $x_\alpha \in X$ such that: 
:$\norm {x_\alpha} = 1$
with:
:$\norm {x_\alpha - y} > \alpha$
for all $y \in Y$.

Proof:
Since $Y < X$:
:$X \setminus Y$ is non-empty. 
Since $Y$ is closed:
:$X \setminus Y$ is open.
Let $x \in X \setminus Y$.
Then there exists $\epsilon > 0$ such that: 
:$\map {B_\epsilon} x \subset X \setminus Y$
So, for all $y \in Y$, we must have: 
:$\norm {x - y} \ge \epsilon$
That is: 
:$\inf \set {\norm {x - y} \colon y \in Y} \ge \epsilon$
For brevity, let:
:$d = \inf \set {\norm {x - y} \colon y \in Y}$
Since $\alpha^{-1} > 1$, there exists $z \in Y$ with: 
:$\norm {x - z} < d \alpha^{-1}$
otherwise the infimum would be at least $d \alpha^{-1} > d$, a contradiction.
Since $x \in X \setminus Y$ and $z \in Y$, we clearly have $x \ne z$.
So, we can set: 
:$ x_\alpha = \dfrac {x - z} {\norm {x - z} }$
Clearly:
:$\norm {x_\alpha} = \dfrac {\norm {x - z} } {\norm {x - z} } = 1$
Now, for any $y \in Y$ we have: 
{{begin-eqn}}
{{eqn | l = \norm {x_\alpha - y}
      | r = \norm {\frac {x - z} {\norm {x - z} } - y}
}}
{{eqn | r = \frac {\norm {x - \paren {z + \norm {x - z} y} } } {\norm {x - z} }
}}
{{end-eqn}}
We've already seen that: 
:$\norm {x - z} < d \alpha^{-1}$
Since $Y$ is closed under linear combination, we have:
:$z + \norm {x - z} y \in Y$
and so: 
:$\norm {x - \paren {z + \norm {x - z} y} } \ge d$
We conclude that: 
{{begin-eqn}}
{{eqn | l = \norm {x_\alpha - y}
      | r = \frac {\norm {x - \paren {z + \norm {x - z} y} } } {\norm {x - z} }
}}
{{eqn | o = >
      | r = \frac d {d \alpha^{-1} } 
}}
{{eqn | r = \alpha
}}
{{end-eqn}}
{{qed}}
Theorem: Riesz's Lemma

Let $X$ be a normed vector space.
Let $Y$ be a proper closed linear subspace of $X$. 
Let $\alpha \in \openint 0 1$.
Then there exists $x_\alpha \in X$ such that: 
:$\norm {x_\alpha} = 1$
with:
:$\norm {x_\alpha - y} > \alpha$
for all $y \in Y$.

Proof:
Consider the normed quotient vector space $X / Y$ with quotient mapping $\pi$.
From Operator Norm of Quotient Mapping in Quotient Normed Vector Space is 1, we have: 
:$\norm \pi_{\map B {X, X/Y} } = 1$ 
Since $\alpha \in \openint 0 1$, there exists $x_\alpha \in X$ with $\norm {x_\alpha} = 1$ and: 
:$\norm {\map \pi {x_\alpha} }_{X/Y} > \alpha$
by the definition of the norm on the space of bounded linear transformations.
That is, by the definition of the quotient norm:
:$\ds \inf_{z \mathop \in Y} \norm {x_\alpha - z} > \alpha$
So there exists $y \in Y$ such that: 
:$\norm {x_\alpha - y} > \alpha$
{{qed}}
Theorem: Normed Vector Space is Finite Dimensional iff Unit Sphere is Compact

Let $X$ be a normed vector space.
Let $\Bbb S = \map {\Bbb S_1} 0$ be the unit sphere centred at $0$ in $X$.
Then $X$ is finite dimensional {{iff}} $\Bbb S$ is compact.

Proof:
=== Necessary Condition ===
{{:Normed Vector Space is Finite Dimensional iff Unit Sphere is Compact/Necessary Condition}}{{qed|lemma}}
Theorem: Total Force on Point Charge from 2 Point Charges

Let $q_1$, $q_2$ and $q_3$ be point charges.
Let $\mathbf F_{ij}$ denote the electrostatic force exerted on $q_j$ by $q_i$.
Let $\mathbf F_i$ denote the electrostatic force exerted on $q_i$ by the combined action of the other two point charges.
Then the electrostatic force $\mathbf F_1$ exerted on $q_1$ by the combined action of $q_2$ and $q_3$ is given by:
{{begin-eqn}}
{{eqn | l = \mathbf F_1
      | r = \mathbf F_{21} + \mathbf F_{31}
      | c = 
}}
{{eqn | r = \dfrac {q_2 q_1} {4 \pi \varepsilon_0 {r_{2 1} }^3} \mathbf r_{2 1} + \dfrac {q_3 q_1} {4 \pi \varepsilon_0 {r_{3 1} }^3} \mathbf r_{3 1}
      | c = 
}}
{{end-eqn}}
where:
:$\mathbf F_{21} + \mathbf F_{31}$ denotes the vector sum of $\mathbf F_{21}$ and $\mathbf F_{31}$
:$\mathbf r_{ij}$ denotes the displacement from $q_i$ to $q_j$
:$r_{ij}$ denotes the distance between $q_i$ and $q_j$
:$\varepsilon_0$ denotes the vacuum permittivity.

Proof:
:500px
By definition, the electrostatic force $\mathbf F_{21}$ and $\mathbf F_{31}$ are vector quantities.
Hence their resultant can be found by using the Parallelogram Law.
The result follows from Coulomb's Law of Electrostatics.
{{qed}}
Theorem: Total Force on Point Charge from Multiple Point Charges

Let $q_1, q_2, \ldots, q_n$ be point charges.
For all $i$ in $\set {1, 2, \ldots, n}$ where $i \ne j$, let $\mathbf F_{i j}$ denote the force exerted on $q_j$ by $q_i$.
For all $i$ in $\set {1, 2, \ldots, n}$, let $\mathbf F_i$ denote the force exerted on $q_i$ by the combined action of all the other point charges.
Then the electrostatic force $\mathbf F_i$ exerted on $q_i$ by the combined action of all the other point charges is given by:
{{begin-eqn}}
{{eqn | l = \mathbf F_i
      | r = \sum_{\substack {1 \mathop \le j \mathop \le n \\ i \mathop \ne j} } \mathbf F_{j i}
      | c = 
}}
{{eqn | r = \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le n \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i}
      | c = 
}}
{{end-eqn}}
where:
:the summation denotes the vector sum of $\mathbf F_{ji}$
:$\mathbf r_{ij}$ denotes the displacement from $q_i$ to $q_j$
:$r_{ij}$ denotes the distance between $q_i$ and $q_j$
:$\varepsilon_0$ denotes the vacuum permittivity.

Proof:
The proof proceeds by induction.
For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:
:$\ds \mathbf F_i = \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le n \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i}$
$\map P 2$ is the case:
$\mathbf F_1 = \dfrac 1 {4 \pi \varepsilon_0} \dfrac {q_1 q_2} {r_{2 1}^3} \mathbf r_{2 1}$
which is Coulomb's Law of Electrostatics.
Thus $\map P 0$ is seen to hold.
Theorem: Scalar Multiplication by Zero gives Zero Vector

Let $\mathbf a$ be a vector quantity.
Let $0 \mathbf a$ denote the scalar product of $\mathbf a$ with $0$.
Then:
:$0 \mathbf a = \bszero$
where $\bszero$ denotes the zero vector.

Proof:
By definition of scalar product:
:$\size {0 \mathbf a} = 0 \size {\mathbf a}$
where $\size {\mathbf a}$ denotes the magnitude of $\mathbf a$.
Thus:
:$\size {0 \mathbf a} = 0$
That is: $0 \mathbf a$ is a vector quantity whose magnitude is zero.
Hence, by definition, $0 \mathbf a$ is the zero vector.
{{qed}}
Theorem: Space of Continuously Differentiable on Closed Interval Real-Valued Functions with C^1 Norm forms Normed Vector Space

Space of Continuously Differentiable on Closed Interval Real-Valued Functions with $C^1$ norm forms a normed vector space.

Proof:
Let $I := \closedint a b$ be a closed real interval.
We have that:
:Space of Continuously Differentiable on Closed Interval Real-Valued Functions with Pointwise Addition and Pointwise Scalar Multiplication forms Vector Space
:$\map {C^1} I$ norm on the space of continuously differentiable on closed interval real-valued functions is a norm
By definition, $\struct {\map {C^1} I, \norm {\, \cdot \,}_{1, \infty} }$ is a normed vector space.
{{qed}}
Theorem: Ring of Endomorphisms is Ring with Unity

Let $\struct {G, \oplus}$ be an abelian group.
Let $\mathbb G$ be the set of all group endomorphisms of $\struct {G, \oplus}$.
Let $\struct {\mathbb G, \oplus, *}$ denote the '''ring of endomorphisms''' on $\struct {G, \oplus}$.
Then $\struct {\mathbb G, \oplus, *}$ is a ring with unity.

Proof:
By Structure Induced by Group Operation is Group, $\struct {\mathbb G, \oplus}$ is an abelian group.
{{explain|The abelian nature of $\struct {\mathbb G, \oplus}$ needs to be demonstrated -- we have to invoke a result either that demonstrates that commutativity is preserved, or generate a similar result to the above but for an abelian group}}
By Set of Homomorphisms to Abelian Group is Subgroup of All Mappings, it follows that $\struct {\mathbb G, \oplus}$ is a subgroup of $\struct {G^G, \oplus}$.
Next, we establish that $*$ is associative.
By definition, $\forall u, v \in \mathbb G: u * v = u \circ v$ where $u \circ v$ is defined as composition of mappings.
Associativity of $*$ follows directly from Composition of Mappings is Associative.
Next, we establish that $*$ is distributive over $\oplus$.
Let $u, v, w \in \mathbb G$.
Then:
:$\paren {u \oplus v} * w = \paren {u \oplus v} \circ w$
:$u * \paren {v \oplus w} = u \circ \paren {v \oplus w}$
So let $x \in G$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\paren {\paren {u \oplus v} * w} } x
      | r = \map {\paren {\paren {u \oplus v} \circ w} } x
      | c = 
}}
{{eqn | r = \map {\paren {u \oplus v} } {\map w x}
      | c = 
}}
{{eqn | r = \map u {\map w x} \oplus \map v {\map w x}
      | c = 
}}
{{eqn | r = \map {\paren {u \circ w} } x \oplus \map {\paren {v \circ w} } x
      | c = 
}}
{{eqn | r = \map {\paren {u * w} } x \oplus \map {\paren {v * w} } x
      | c = 
}}
{{end-eqn}}
So $\paren {u \oplus v} * w = \paren {u * w} \oplus \paren {v * w}$.
Similarly:
{{begin-eqn}}
{{eqn | l = \map {\paren {u * \paren {v \oplus w} } } x
      | r = \map {\paren {u \circ \paren {v \oplus w} } } x
      | c = 
}}
{{eqn | r = \map u {\map {\paren {v \oplus w} } x}
      | c = 
}}
{{eqn | r = \map u {\map v x \oplus \map w x}
      | c = 
}}
{{eqn | r = \map u {\map v x} \oplus \map u {\map w x}
      | c = $u$ has the morphism property
}}
{{eqn | r = \map {\paren {u \circ v} } x \oplus \map {\paren {u \circ w} } x
      | c = 
}}
{{eqn | r = \map {\paren {u * v} } x \oplus \map {\paren {u * w} } x
      | c = 
}}
{{end-eqn}}
So:
:$u * \paren {v \oplus w} = \paren {u * v} \oplus \paren {u * w}$
Thus $*$ is distributive over $\oplus$.
The ring axioms are satisfied, and $\struct {\mathbb G, \oplus, *}$ is a ring.
The zero is easily checked to be the mapping which takes everything to the identity:
:$e: G \to \set {e_G}: \map e x = e_G$
The unity is easily checked to be the identity mapping, which is known to be an automorphism.
{{Qed}}
Theorem: Ring of Endomorphisms is not necessarily Commutative Ring

Let $\struct {G, \oplus}$ be an abelian group.
Let $\mathbb G$ be the set of all group endomorphisms of $\struct {G, \oplus}$.
Let $\struct {\mathbb G, \oplus, *}$ denote the '''ring of endomorphisms''' on $\struct {G, \oplus}$.
Then $\struct {\mathbb G, \oplus, *}$ is not necessarily a commutative ring with unity.

Proof:
From Ring of Endomorphisms is Ring with Unity, we have that $\struct {\mathbb G, \oplus, *}$ is a ring with unity.
It remains to show that the operation $*$ is not necessarily commutative.
{{ProofWanted}}
Category:Rings of Endomorphisms
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Theorem: Set of Endomorphisms of Non-Abelian Group is not Ring

Let $\struct {G, \oplus}$ be a group which is non-abelian.
Let $\mathbb G$ be the set of all group endomorphisms of $\struct {G, \oplus}$.
Let $*: \mathbb G \times \mathbb G \to \mathbb G$ be the operation defined as:
:$\forall u, v \in \mathbb G: u * v = u \circ v$
where $u \circ v$ is defined as composition of mappings.
Then the algebraic structure $\struct {\mathbb G, \oplus, *}$ is not a ring.

Proof:
In order to be a ring, it is necessary that the additive operation $\oplus$ is commutative.
However, as $\struct {G, \oplus}$ is specifically defined as being non-abelian, a fortiori $\oplus$ is not commutative.
Hence the result.
{{qed}}
Theorem: Set of Positive Integers does not form Ring

Let $\Z_{\ge 0}$ denote the set of positive integers.
Then the algebraic structure $\struct {\Z_{\ge 0}, +, \times}$ does not form a ring.

Proof:
For $\struct {\Z_{\ge 0}, +, \times}$ to be a ring, it is necessary for the algebraic structure $\struct {\Z_{\ge 0}, +}$ to form a group.
But from {{Corollary|Natural Numbers under Addition do not form Group}}:
:$\struct {\Z_{\ge 0}, +}$ is not a group.
{{qed}}
Theorem: Space of Continuously Differentiable on Closed Interval Real-Valued Functions with C^1 Norm is Banach Space

Let $I := \closedint a b$ be a closed real interval.
Let $\map C I$ be the space of real-valued functions continuous on $I$.
Let $\map {C^1} I$ be the space of real-valued functions, continuously differentiable on $I$.
Let $\norm {\, \cdot \,}_{1, \infty}$ be the $\CC^1$ norm.
$\struct {\map {C^1} I, \norm {\, \cdot \,}_{1, \infty} }$ be the normed space of real-valued functions, continuously differentiable on $I$.
Then $\struct {\map {C^1} I, \norm {\, \cdot \,}_{1, \infty} }$ is a Banach space.

Proof:
Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\struct {\map {C^1} I, \norm {\, \cdot \,}_{1, \infty} }$:
:$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n \ge N: \norm {x_n - x_m}_{1, \infty} < \epsilon$
Theorem: Rational Numbers whose Denominators are not Divisible by 4 do not form Ring

Let $S$ be the set defined as:
:$S = \set {\dfrac m n : m, n \in \Z, m \perp n, 4 \nmid n}$
That is, $S$ is defined as the set of rational numbers such that, when expressed in canonical form, their denominators are not divisible by $4$.
Then the algebraic structure $\struct {S, +, \times}$ is not a ring.

Proof:
For $\struct {S, +, \times}$ to be a ring, it is a necessary condition that $\struct {S, \times}$ is a semigroup.
For $\struct {S, \times}$ to be a semigroup, it is a necessary condition that $\struct {S, \times}$ is closed.
That is:
:$\forall x, y \in S: x \times y \in S$
Let $x = \dfrac 1 2$ and $y = \dfrac 3 2$.
Both $x$ and $y$ are in $S$, as both are rational numbers expressed in canonical form whose denominators are not divisible by $4$.
But then:
:$x \times y = \dfrac 1 2 \times \dfrac 3 2 = \dfrac 3 4$
which is a rational numbers expressed in canonical form whose denominator ''is'' divisible by $4$.
Hence $x \times y \notin S$ and so $\struct {S, \times}$ is not closed.
The result follows.
{{qed}}
Theorem: Order 2 Square Matrices with Zero Diagonals do not form Ring

Let $S$ be the set of square matrices of order $2$ whose diagonal elements are zero.
Then the algebraic structure $\struct {S, +, \times}$ is not a ring.
Note that $\times$ denotes conventional matrix multiplication.

Proof:
For $\struct {S, +, \times}$ to be a ring, it is a necessary condition that $\struct {S, \times}$ is a semigroup.
For $\struct {S, \times}$ to be a semigroup, it is a necessary condition that $\struct {S, \times}$ is closed.
That is:
:$\forall x, y \in S: x \times y \in S$
Let $x = \begin {pmatrix} 0 & 1 \\ 1 & 0 \end {pmatrix}$ and $y = \begin {pmatrix} 0 & 2 \\ 2 & 0 \end {pmatrix}$.
Both $x$ and $y$ are in $S$, as both are square matrices of order $2$ whose diagonal elements are zero.
But then:
:$x y = \begin {pmatrix} 0 & 1 \\ 1 & 0 \end {pmatrix} \begin {pmatrix} 0 & 2 \\ 2 & 0 \end {pmatrix} = \begin {pmatrix} 2 & 0 \\ 0 & 2 \end {pmatrix}$
which is a square matrix of order $2$ whose diagonal elements are ''not'' zero.
Hence $x y \notin S$ and so $\struct {S, \times}$ is not closed.
The result follows.
{{qed}}