avewright/proofwiki-math huggingface dataset:
Theorem: Right Limit Function is Right-Continuous

Let $I$ be an open interval.
Let $f : I \to \R$ be a real function such that:
:for each $x \in I$, the right limit:
::$\ds \lim_{y \mathop \to x^+} \map f y$ 
:exists.
Define $f_\leftarrow : I \to \R$ by:
:$\ds \map {f_\leftarrow} x = \lim_{y \mathop \to x^+} \map f y$
for each $x \in I$.
Then $f_\leftarrow$ is right-continuous.
Further, if $f$ is right-continuous at $x$, then $\map {f_\leftarrow} x = \map f x$.

Proof:
Let $\epsilon > 0$. 
Let $x \in I$. 
By the definition of the right limit at $x$, there exists $\delta > 0$ such that for all $y \in I$ satisfying $x < y < x + \delta$, we have:
:$\ds \size {\map f y - \map {f_\leftarrow} x} < \frac \epsilon 3$
Take $y \in I$ such that $x < y < x + \delta$. 
Let $\sequence {x_n}_{n \in \N}$ be a sequence with $x < x_n < \delta$ and $x_n \to x$. 
For example, we may take:
:$\ds x_n = x + \frac \delta {n + 1}$
for each $n \ge 1$. 
Also let $\sequence {y_n}_{n \in \N}$ be a sequence with $y < y_n < x + \delta$ and $y_n \to y$. 
For example, we may take:
:$\ds y_n = y + \frac {\paren {x + \delta} - y} {n + 1}$
for each $n \ge 1$.
By Limit of Function by Convergent Sequences: Corollary, we have:
:$\map f {x_n} \to \map {f_\leftarrow} x$
and:
:$\map f {y_n} \to \map {f_\leftarrow} y$
as $n \to \infty$. 
We can now write:
{{begin-eqn}}
{{eqn	| l = \size {\map f {x_n} - \map f {y_n} }
		| r = \size {\map f {x_n} - \map f {y_n} - \map {f_\leftarrow} x + \map {f_\leftarrow} x}
}}
{{eqn	| o = \le 
		| r = \size {\map f {x_n} - \map {f_\leftarrow} x} + \size {\map {f_\leftarrow} x - \map f {y_n} }
}}
{{eqn	| o = <
		| r = \frac {2 \epsilon} 3
		| c = since $x_n \in \openint x {x + \delta}$ and $y_n \in \openint x {x + \delta}$
}}
{{end-eqn}}
Taking $n \to \infty$ and using the Squeeze Theorem, we have:
:$\ds \size {\map {f_\leftarrow} x - \map {f_\leftarrow} y} \le \frac {2 \epsilon} 3 < \epsilon$
for all $y \in \openint x {x + \delta}$. 
Hence we have shown that if $x \in I$, then there exists $\delta > 0$ such that whenever $y \in \openint x {x + \delta}$ we have:
:$\size {\map {f_\leftarrow} x - \map {f_\leftarrow} y} < \epsilon$
Finally, if right-continuous at $x$, then:
:$\ds \lim_{y \mathop \to x^+} \map f y = \map f x$
That is:
:$\map {f_\leftarrow} x = \map f x$
by the definition of $f_\leftarrow$.
{{qed}}
Category:Right-Continuous Functions
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Theorem: Derived Subgroup of Abelian Group is Trivial

Let $G$ be a abelian group with identity $e$.
Let $G'$ be the derived subgroup of $G$.
Then $G'$ is the trivial group $\set e$.

Proof:
For $g, h \in G$, let $\sqbrk {g, h}$ denote the commutator of $g$ and $h$:
:$\sqbrk {g, h} := g^{-1} \circ h^{-1} \circ g \circ h$
Let $C$ be the set of the commutators of $G$:
:$C = \set {\sqbrk {g, h}: g, h \in G}$
From Commutators are Identity iff Group is Abelian, $C = \set e$.
From Trivial Subgroup is Subgroup, $C = \set e$ is a subgroup of $G$.
From Subgroup Generated by Subgroup:
:$\gen C = C$
where $\gen C$ denotes the subgroup generated by $C$.
From the definition of derived subgroup:
:$G' := \gen C$
Hence the result:
:$G' = \set e$
{{qed}}
Theorem: Biholomorphic Function from Open Unit Disk to Right Half-Plane

Let $\mathbb D = \set {z : \cmod z < 1}$ be the open unit disk. 
Let $\mathbb H_r = \set {z : \map \Re z > 0}$ be the right half-plane.
Define $f : \mathbb D \to \C$ by:
:$\ds \map f z = \frac {1 - z} {1 + z}$
for each $z \in \mathbb D$.
Then $f$ is a biholomorphic function $\mathbb D \to \mathbb H_r$.

Proof:
From Möbius Transformation is Bijection, defining the extended map $f_\ast : \overline \C \to \overline \C$ by:
:<nowiki>$\map {f_\ast} z = \begin {cases} \dfrac {1 - z} {1 + z} & : z \ne -1 \\
\infty & : z = -1 \\
-1 & : z = \infty \end{cases}$</nowiki>
we obtain a bijection.
Hence $f : \mathbb D \to \C$ is an injection.
To show that $f$ is a surjection $\mathbb D \to \mathbb H_r$ it is enough to show that if $\map {f_\ast} z \in \mathbb H_r$, then $\cmod z < 1$.
We first demonstrate that this is the case.
Suppose that it is true that if $\map {f_\ast} z \in \mathbb H_r$ for $z \in \overline \C$, then $\cmod z < 1$.
For each $w \in \mathbb H_r$, there will exist $z \in \overline \C$ such that $\map {f_\ast} z = w$.
We will then have $z \in \mathbb D$, and hence $w = \map {f_\ast} z = \map f z$.
Hence we will have shown that $f$ is surjective.
Now, we have, for $z \not \in \set {-1, \infty}$:
{{begin-eqn}}
{{eqn	| l = \map {f_\ast} z
		| r = \frac {1 - z} {1 + z}
}}
{{eqn	| r = \frac {\paren {1 - z} \paren {1 + \overline z} } {\paren {1 + z} \overline {1 + z} }
		| c = Sum of Complex Conjugates
}}
{{eqn	| r = \frac {1 - \paren {z - \overline z} - z \overline z} {\cmod {1 + z}^2}
		| c = Product of Complex Number with Conjugate
}}
{{eqn	| r = \frac {1 - 2 i \map \Im z - \cmod z^2} {\cmod {1 + z}^2}
		| c = Difference of Complex Number with Conjugate, Product of Complex Number with Conjugate
}}
{{eqn	| r = \paren {\frac {1 - \cmod z^2} {\cmod {1 + z}^2} } + 2 i \paren {\frac {\map \Im z} {\cmod {1 + z}^2} }
}}
{{end-eqn}}
If $z = \infty$, then $\map {f_\ast} z = -1 \not \in \mathbb H_r$. 
Similarly $\map {f_\ast} {-1} = \infty \not \in \mathbb H_r$. 
Hence $\map \Re {\map {f_\ast} z} > 0$ {{iff}}:
:$\ds \frac {1 - \cmod z^2} {\cmod {1 + z}^2} > 0$
which is the case {{iff}} $1 - \cmod z^2 > 0$, which is the case {{iff}} $\cmod z < 1$.
Hence, due to the prior discussion, $f$ is surjective.
Since we have $\map \Re {\map f z} > 0$ {{iff}} $\cmod z < 1$, we can consider $f$ as a mapping $\mathbb H_r \to \mathbb D$.
Denote this mapping as $f^{-1} : \mathbb H_r \to \mathbb D$. 
From Combination Theorem for Complex Derivatives: Quotient Rule, $f : \mathbb D \to \mathbb H_r$ is holomorphic.
We have, for $z \in \mathbb D$:
{{begin-eqn}}
{{eqn	| l = \map {\paren {f^{-1} \circ f} } z
		| r = \frac {1 - \frac {1 - z} {1 + z} } {1 + \frac {1 - z} {1 + z} }
}}
{{eqn	| r = \frac {\paren {1 + z} - \paren {1 - z} } {1 + z + \paren {1 - z} }
}}
{{eqn	| r = \frac {2 z} 2
}}
{{eqn	| r = z
}}
{{end-eqn}}
Hence $f^{-1}$ is the inverse of $f$. 
From Combination Theorem for Complex Derivatives: Quotient Rule again, $f^{-1}$ is holomorphic.
$f$ is therefore biholomorphic as required.
{{qed}}
Category:Biholomorphic Functions
Category:Möbius Transformations
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Theorem: Derived Subgroup is Trivial iff Group is Abelian

Let $G$ be a group with identity $e$.
Let $G'$ be the derived subgroup of $G$.
Then:
:$G'$ is the trivial group $\set e$
{{iff}}:
:$G$ is abelian.

Proof:
=== Sufficient Condition ===
This is proved in Derived Subgroup of Abelian Group is Trivial.
{{qed|lemma}}
Theorem: Skew Lines Cannot Exist in Euclidean Plane

Let $\PP$ be a Euclidean plane.
Let $\LL_1$ and $\LL_2$ be straight lines embedded in $\PP$.
Then it is not possible for $\LL_1$ and $\LL_2$ to be '''skew'''

Proof:
{{Recall|Skew Lines}}
{{:Definition:Skew Lines}}
The proof then follows directly from {{EuclidPostulateLink|Fifth}}:
{{:Axiom:Euclid's Fifth Postulate}}
That is, if $\LL_1$ and $\LL_2$ do not intersect, they are parallel.
{{qed}}
Theorem: Intersection of Plane with Sphere is Great Circle iff Passing through Center

Let $S$ be a sphere.
Let $P$ be a plane which intersects $S$ but is not tangent to $S$.
Let $C$ be the circle formed by the intersection between $P$ and $S$.
Then $C$ is a '''great circle''' of $S$ {{iff}} $P$ passes through the center of $S$.
Otherwise $C$ is a '''small circle''' of $S$.

Proof:
From Intersection of Plane with Sphere is Circle it is confirmed that $C$ is indeed a circle.
{{Recall|Great Circle}}
{{:Definition:Great Circle}}
{{Recall|Small Circle}}
{{:Definition:Small Circle/Definition 1}}
The result follows directly.
{{qed}}
Theorem: Measure of Total Solid Angle around Point

The total solid angle around a point has a measure of $4 \pi$ steradians.

Proof:
Let $Q$ denote this total solid angle
{{Recall|steradian}}
{{:Definition:Steradian}}
Hence $Q$ is equal to the total surface area of $S$ divided by $r^2$.
That is:
{{begin-eqn}}
{{eqn | l = Q
      | r = \dfrac A {r^2}
      | c = where $A$ is the total surface area of $S$
}}
{{eqn | r = \dfrac {4 \pi r^2} {r^2}
      | c = Surface Area of Sphere
}}
{{eqn | r = 4 \pi
      | c = simplification
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Measure of Trihedral Angle formed by 3 Mutually Perpendicular Half-Lines

Let $Q$ be the trihedral angle formed by $3$ half-lines each of which is perpendicular to the other two half-lines.
Then $Q$ measures $\dfrac \pi 2$ steradians.

Proof:
{{ProofWanted|Prove that $Q$ subtends $1/8$ of a sphere then use Measure of Total Solid Angle around Point}}
Theorem: Symmetry Group of Equilateral Triangle is Solvable

The symmetry group $D_3$ of the equilateral triangle is a solvable group.

Proof:
From Symmetry Group of Equilateral Triangle is Symmetric Group, $D_3$ is the symmetric group on $3$ letters $S_3$.
The result follows from Examples of Composition Series: Symmetric Group $S_3$.
{{qed}}
Theorem: Condition for Group to be Solvable

Let $G$ be a group.
Let $G$ be such that:
:$G$ has a normal subgroup $H$ which is solvable
:the quotient group $G / H$ is cyclic.
Then $G$ is a solvable group.

Proof:
{{Recall|Solvable Group}}
{{:Definition:Solvable Group}}
Let $\set e = H_0 \lhd H_1 \lhd \cdots \lhd H_n = H$ be a composition series in which each factor is a cyclic group.
By assumption, the above composition series could be extended to $\set e = H_0 \lhd H_1 \lhd \cdots \lhd H_n \lhd G$ in which each factor is still a cyclic group.
Hence the result.
{{qed}}
Theorem: Stone Space of Boolean Lattice is Hausdorff

Let $B = \struct {S, \preceq}$ be a Boolean lattice.
Let $\map S B = \struct {U, \tau}$ be the Stone space of $B$.
Then, $\map S B$ is a Hausdorff space.

Proof:
Let $x, y \in U$ be arbitrary, with $x \ne y$.
By definition of set equality, there is some $a \in S$ such that:
:$\neg \paren {a \in x \iff a \in y}$
{{WLOG}}, assume that $a \in x$ and $a \notin y$.
Now, let:
:$X = \set {z \in U : a \in z}$
:$Y = \set {z \in U : \paren {\neg a} \in z}$
By definition of the Stone space of $B$, each of $X$ and $Y$ is open in $\map S B$.
From the above, it is immediate that $x \in X$.
As $a \notin y$, we have by Ultrafilter on Boolean Lattice Contains Element or Complement that $\paren {\neg a} \in y$.
Hence, $y \in Y$.
If there existed some $z \in X \cap Y$, then $\set {a, \neg a} \subseteq z$.
But then, we would have $\bot = a \wedge \neg a \in z$ by definition of a filter.
Thus, it would hold that $z = S$, since $\bot \preceq c$ for every $c \in S$.
As that contradicts the definition of an ultrafilter, we must have $X \cap Y = \O$.
We have constructed two disjoint open sets $X, Y$ such that:
:$x \in X$
:$y \in Y$
As $x \ne y$ were arbitrary, it follows that $\map S B$ is Hausdorff.
{{qed}}
Category:Boolean Lattices
Category:Hausdorff Spaces
Category:Stone Spaces
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Theorem: Ambiguous Case for Spherical Triangle/Side-Side-Angle

Let the sides $a$ and $b$ be known.
Let the angle $\sphericalangle B$ also be known.
Then it may not be possible to know the value of $\sphericalangle A$.

Proof:
From the Spherical Law of Sines, we have:
:$\dfrac {\sin a} {\sin A} = \dfrac {\sin b} {\sin B} = \dfrac {\sin c} {\sin C}$
from which:
:$\sin A = \dfrac {\sin a \sin A} {\sin b}$
We find that $0 < \sin A \le 1$.
We have that:
:$\sin A = \map \sin {\pi - A}$
and so unless $\sin A = 1$ and so $A = \dfrac \pi 2$, it is not possible to tell which of $A$ or $\pi - A$ provides the correct solution.
{{qed}}
Theorem: Ambiguous Case for Spherical Triangle/Angle-Angle-Side

Let the angles $\sphericalangle A$ and $\sphericalangle B$ be known.
Let the side $b$ also be known.
Then it may not be possible to know the value of $a$.

Proof:
From the Spherical Law of Sines, we have:
:$\dfrac {\sin a} {\sin A} = \dfrac {\sin b} {\sin B} = \dfrac {\sin c} {\sin C}$
from which:
:$\sin a = \dfrac {\sin b \sin A} {\sin B}$
We find that $0 < \sin a \le 1$.
We have that:
:$\sin a = \map \sin {\pi - a}$
and so unless $\sin a = 1$ and so $a = \dfrac \pi 2$, it is not possible to tell which of $a$ or $\pi - a$ provides the correct solution.
{{qed}}
Theorem: Group Automorphism is Endomorphism

Let $G$ a group.
Let $\phi: G \to G$ be a (group) automorphism on $G$.
Then $\phi$ is a group endomorphism.

Proof:
{{Recall|Group Automorphism}}
{{:Definition:Group Automorphism}}
{{Recall|Group Isomorphism}}
{{:Definition:Group Isomorphism}}
{{Recall|Group Endomorphism}}
{{:Definition:Group Endomorphism}}
We have been given that $\phi$ is a group automorphism.
Hence {{afortiori}} $\phi$ is a group homomorphism from $G$ to $G$.
Hence $\phi$ is a group endomorphism by definition.
{{qed}}
Theorem: Fully Characteristic Subgroup is Characteristic

Let $G$ a group.
Let $H$ a fully characteristic subgroup of $G$.
Then $H$ is a characteristic subgroup of $G$.

Proof:
By definition of a fully characteristic subgroup,
$H$ is invariant under every group endomorphism of $G$.
By Group Automorphism is Endomorphism, every group automorphism is a group endomorphism.
Hence, invariance over every group endomorphism implies invariance over every group automorphism.
And by definition of a characteristic subgroup,
$H$ is '''characteristic'''.
{{qed}}
Theorem: Proper Ideal is Prime iff Contains Element or Complement

Let $\struct {S, \preceq}$ be a boolean lattice.
Let $I$ be a proper ideal on $S$.
Then:
:$I$ is a prime ideal
{{iff}}:
:for every $a \in S$, either $a \in I$ or $\neg a \in I$.

Proof:
=== Necessary Condition ===
Suppose that $I$ is a prime ideal.
Then, by definition:
:$S \setminus I$ is a filter
{{AimForCont}} $a, \neg a \notin I$.
Then, $a, \neg \in S \setminus I$.
Hence, by Filter is Closed under Meet:
:$\neg a \wedge a \in S \setminus I$
but then by definition of complement:
:$\bot = \neg \wedge a \notin I$
which contradicts Bottom in Ideal.
{{qed|lemma}}
Theorem: Group whose Derived Subgroup is Trivial is Abelian

Let $G$ be a group with identity $e$.
Let $G'$ be the derived subgroup of $G$.
Let $G'$ be the trivial group $\set e$.
Then $G$ is abelian.

Proof:
We assume {{hypothesis}} that the derived subgroup of $G$ is $\set e$.
{{AimForCont}} $G$ is non-abelian.
Hence by definition:
:$\exists g, h \in G: g h \ne h g$
Multiplying both sides by $g^{-1} h^{-1}$:
:$\exists g, h \in G: g^{-1} h^{-1} g h \ne e$
But from the definition of the derived subgroup:
:$G' := \set {g^{-1} h^{-1} g h : g, h \in G}$
Hence:
:$G' \ne \set e$
So by definition, $G'$ is not a trivial group.
By Proof by Contraposition it follows that $G$ is abelian.
{{qed}}
Theorem: Boolean Prime Ideal Theorem/Element Extension Lemma

Let $\struct {B, \vee, \wedge, \le}$ be a Boolean lattice.
Let $F \subseteq B$ be a filter on $B$.
Let $a, x \in B$ such that:
:$a \notin F$
Then either:
:$a \vee x \notin F$
or:
:$a \vee \neg x \notin F$

Proof:
{{AimForCont}} that both:
:$a \vee x \in F$
and:
:$a \vee \neg x \in F$
Then:
{{begin-eqn}}
{{eqn | l = F
      | o = \ni
      | r = \paren {a \vee x} \wedge \paren {a \vee \neg x}
      | c = Filter is Closed under Meet
}}
{{eqn | r = \paren {a \wedge \paren {a \vee \neg x} } \vee \paren {x \wedge \paren {a \vee \neg x} }
      | c = Distributive Lattice Axiom $\paren {1'}$
}}
{{eqn | r = a \vee \paren {\paren {x \wedge a} \vee \paren {x \wedge \neg x} }
      | c = Meet Absorbs Join, Distributive Lattice Axiom $\paren 1$
}}
{{eqn | r = a \vee \paren {\paren {a \wedge x} \vee \bot}
      | c = Meet is Commutative, {{Defof|Complement (Lattice Theory)}}
}}
{{eqn | r = a
      | c = {{Defof|Bottom of Lattice|index = 2}}, Join Absorbs Meet
}}
{{end-eqn}}
which contradicts the hypothesis.
{{qed}}
Category:Boolean Lattices
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Theorem: Interior of Complement equals Complement of Closure

Let $T$ be a topological space.
Let $H \subseteq T$.
Let $H^-$ denote the closure of $H$ and $H^\circ$ denote the interior of $H$.
Let $\map \complement H$ be the complement of $H$ in $T$:
:$\map \complement H = T \setminus H$
Then:
:$\paren {\map \complement H}^\circ = \map \complement {H^-}$
This can alternatively be written:
:$\paren {T \setminus H}^\circ = T \setminus H^-$
which, it can be argued, is easier to follow.

Proof:
{{begin-eqn}}
{{eqn | l = H^\circ
      | r = T \setminus \paren {T \setminus H^\circ}
      | c = Relative Complement of Relative Complement
}}
{{eqn | r = T \setminus \paren {\paren {T \setminus H}^-}
      | c = Complement of Interior equals Closure of Complement
}}
{{end-eqn}}
and so:
{{begin-eqn}}
{{eqn | l = \paren {T \setminus H}^\circ
      | r = T \setminus \paren {\paren {T \setminus \paren {T \setminus H} }^-}
      | c = Complement of Interior equals Closure of Complement
}}
{{eqn | r = T \setminus H^-
      | c = Relative Complement of Relative Complement
}}
{{end-eqn}}
{{qed}}
Theorem: Vector Subspace contains Zero Vector

Let $V$ be a vector space over a division ring $K$ whose unity is $1_K$.
Let $U \subseteq V$ be a subspace of $V$.
Then $U$ contains the zero vector of $V$.

Proof:
Demonstrated during the course of the One-Step Vector Subspace Test.
{{qed}}
Theorem: Singleton of Greatest Element in Two is Completely Prime Filter

Let $\struct{\mathbf 2, \vee, \wedge, \preceq}$ denote the (Boolean Lattice) $\mathbf 2$.
Then:
:$\set \top$ is a completely prime filter

Proof:
By definition of (Boolean Lattice) $\mathbf 2$:
:$\mathbf 2 := \set {\bot, \top}$
endowed with the logical operations $\lor$, $\land$ defined by the following Cayley tables:
:<nowiki>$\begin{array}{c|cc}
 \lor & \bot & \top \\ \hline
 \bot & \bot & \top \\
 \top & \top & \top
\end{array} \qquad \begin{array}{c|cc}
 \land & \bot & \top \\ \hline
 \bot  & \bot & \bot \\
 \top  & \bot & \top
\end{array}$
</nowiki>
and ordering $\preceq$ defined by:
:$\preceq = \set{ \tuple{\bot, \bot}, \tuple{\bot, \top}, \tuple{\top, \top}}$.
From Two is a Locale:
:$\struct{\mathbf 2, \vee, \wedge, \preceq}$ is a locale
By definition of locale:
:$\struct{\mathbf 2, \vee, \wedge, \preceq}$ is a complete lattice
From Characterization of Completely Prime Filter in Complete Lattice, it is sufficient to show:
:$(1)\quad\forall A \subseteq \mathbf 2 : \bigvee A \in \set \top \iff \paren{\exists a \in A : a \in \set \top}$
:$(2)\quad\forall $ finite $A \subseteq \mathbf 2 : \bigwedge A \in \set \top \iff \paren{\forall a \in A : a \in \set \top}$
where:
:* $\bigvee A$ denotes the supremum of $A$ in $\mathbf 2$
:* $\bigwedge A$ denotes the infimum of $A$ in $\mathbf 2$
From Power Set of Doubleton:
:$\powerset {\mathbf 2} = \set {\O, \set \bot, \set \top, \set {\bot, \top} }$
Theorem: Supplemental Chords are Perpendicular

Let $\CC$ be a circle.
Let $K_1$ and $K_2$ be a pair of supplemental chords of $\CC$.
Then $K_1$ and $K_2$ are perpendicular.

Proof:
This is merely a restatement of Thales' Theorem stated in terms of supplemental chords.
{{qed}}
Theorem: Squares which are Sum of Two Cubes/Formulation 2

{{begin-eqn}}
{{eqn | l = A
      | r = \dfrac 1 4 \paren {3 n^3 + 6 n^2 - n}
}}
{{eqn | l = B
      | r = \dfrac 1 4 \paren {-3 n^3 + 6 n^2 + n}
}}
{{eqn | l = C
      | r = \dfrac 1 8 \paren {6 n^2 \paren {3 n^2 + 1} }
}}
{{end-eqn}}
where $n \in \C$.

Proof:
{{begin-eqn}}
{{eqn	| l = \paren {\frac A 4}^3 + \paren {\frac B 4}^3
	    | r = \frac 1 {64} \paren {A^3 + B^3}
	    | c = 
}}
{{eqn	| r = \paren {\frac C 8}^2
	    | c = 
}}
{{end-eqn}}
From Squares which are Sum of Two Cubes: Formulation 1, we have:
{{:Squares which are Sum of Two Cubes/Formulation 1}}
Therefore:
{{begin-eqn}}
{{eqn | l = A
      | r = \dfrac 1 4 \paren {3 n^3 + 6 n^2 - n}
}}
{{eqn | l = B
      | r = \dfrac 1 4 \paren {-3 n^3 + 6 n^2 + n}
}}
{{eqn | l = C
      | r = \dfrac 1 8 \paren {6 n^2 \paren {3 n^2 + 1} }
}}
{{end-eqn}}
{{qed}}
Category:Sums of Cubes
3bqd83fimzfyneysrtdnq5ash02gw8p
Theorem: Equivalence of Definitions of Symmetric Matrices

{{TFAE|def = Symmetric Matrix}}
Let $\mathbf A$ be a square matrix over a set $S$.

Proof:
Place the elements of $\mathbf A$ onto the $x y$ plane such that element $a_{i j}$ is at coordinate $\tuple {j, -i}$.
The indices are reversed because the vertical coordinate--the row index--is listed first in matrix element notation.
The negative sign on the $i$ index is because the rows of a matrix are counted from the top.
Thus, by Definition 1, where $a_{i j} = a_{j i}$, a symmetric matrix is one in which the elements found at coordinate $\tuple {j, -i}$ is equal to the one at $\tuple {i, -j}$.
The main diagonal of $\mathbf A$ consists of elements $a_{i i}$, which are at coordinates $\tuple {i, -i}$.
These elements lie along the line $L$ given by:
:$y = -x$.
Since $L$ passes through the origin and Reflection of Plane in Line through Origin is Linear Operator, the plane reflection about $L$ is a linear operator and can be represented by a matrix.
Let the matrix $\mathbf R$ represent the plane reflection about $L$.
From Matrix Equation of Plane Reflection with $\alpha = -45 \degrees$, we have:
:$\mathbf R = \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}$
The reflected matrix element lies at:
{{begin-eqn}}
{{eqn | l = \mathbf R \begin{bmatrix} j \\ -i \end{bmatrix}
      | r = \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} j \\ -i \end{bmatrix}
}}
{{eqn | r = \begin{bmatrix} i \\ -j \end{bmatrix}
}}
{{end-eqn}}
Thus, by Definition 2, where matrix elements that map onto each other after a reflection about the main diagonal are equal, a symmetric matrix is one in which the elements found at coordinate $\tuple {j, -i}$ is equal to the one at $\mathbf R \tuple {j, -i} = \tuple {i, -j}$.
Since both definitions give the same coordinates for the equal matrix element, the definitions are equivalent.
{{qed}}
Theorem: Squares which are Sum of Two Cubes/Formulation 3

{{begin-eqn}}
{{eqn | l = A
      | r = k^2 \paren {3 n^3 + 6 n^2 - n}
}}
{{eqn | l = B
      | r = k^2 \paren {-3 n^3 + 6 n^2 + n}
}}
{{eqn | l = C
      | r = k^3 \paren {6 n^2 \paren {3 n^2 + 1} }
}}
{{end-eqn}}
where $n, k \in \C$.

Proof:
{{begin-eqn}}
{{eqn	| l = \paren {k^2 A}^3 + \paren {k^2 B}^3
	    | r = k^6 \paren {A^3 + B^3}
	    | c = 
}}
{{eqn	| r = \paren {k^3 C}^2
	    | c = 
}}
{{end-eqn}}
From Squares which are Sum of Two Cubes: Formulation 1, we have:
{{:Squares which are Sum of Two Cubes/Formulation 1}}
Therefore:
{{begin-eqn}}
{{eqn | l = A
      | r = k^2 \paren {3 n^3 + 6 n^2 - n}
}}
{{eqn | l = B
      | r = k^2 \paren {-3 n^3 + 6 n^2 + n}
}}
{{eqn | l = C
      | r = k^3 \paren {6 n^2 \paren {3 n^2 + 1} }
}}
{{end-eqn}}
{{qed}}
Category:Sums of Cubes
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Theorem: Inclusion Functor is Functor

Let $\mathbf D$ be a metacategory, and let $\mathbf C$ be a subcategory of $\mathbf D$.
Let $\Iota_{\mathbf C}: \mathbf C \to \mathbf D$ be the inclusion functor on $\mathbf C$
Then:
:$\Iota_{\mathbf C}$ is a (covariant) functor

Proof:
Let $f, g$ be morphisms of $\mathbf C$ such that $g \circ f$ is defined.
Then:
{{begin-eqn}}
{{eqn | l = \map {\Iota_{\mathbf C} } {g \circ f}
      | r = g \circ f
      | c = {{Defof|Inclusion Functor}}
}}
{{eqn | r = \map {\Iota_{\mathbf C} } g \circ \map {\Iota_{\mathbf C} } f
      | c = {{Defof|Inclusion Functor}}
}}
{{end-eqn}}
Also, for any object $C$ of $\mathbf C$:
{{begin-eqn}}
{{eqn | l = \map {\Iota_{\mathbf C} } {i_C}
      | r = i_C
      | c = {{Defof|Inclusion Functor}}
}}
{{eqn | r = i_{\map {\Iota_{\mathbf C} } C}
      | c = {{Defof|Inclusion Functor}}
}}
{{end-eqn}}
Hence $\Iota_{\mathbf C}$ is shown to be a covariant functor.
{{qed}}