avewright/proofwiki-math huggingface dataset:
Theorem: Dandelin's Theorem

Let $\CC$ be a double napped right circular cone with apex $O$.
Let $\PP$ be a plane which intersects $\CC$ such that:
:$\PP$ does not pass through $O$
:$\PP$ is not perpendicular to the axis of $\CC$.
Let $\EE$ be the conic section arising as the intersection between $\PP$ and $\CC$.
Let $\SS$ and $\SS'$ be the Dandelin spheres with respect to $\PP$.

Proof:
Let $\SS$ and $\SS'$ be the Dandelin spheres with respect to $\PP$.
Let $P$ be a point on $\EE$.
Let $F$ and $F'$ be the points at which $\SS$ and $\SS'$ are tangent to $\PP$ respectively.
Let the generatrix of $\CC$ which passes through $P$ touch $\SS$ and $\SS'$ at $E$ and $E'$ respectively.
Because $PF$ and $PE$ are both tangent to $\SS$:
:$PF = PE$
and similarly:
:$PF' = PE'$
=== Proof for Ellipse ===
Let $\EE$ be an ellipse.
:540px
Then $E$ and $E'$ are on the same side of $O$, and so:
:$PF + PF' = PE + PE'$
But $PE + PE'$ is a constant.
Hence $\EE$ fulfils the equidistance property with respect to the points $F$ and $F'$.
Hence $F$ and $F'$ are the foci of $\EE$.
{{qed|lemma}}
=== Proof for Hyperbola ===
Let $\EE$ be a hyperbola.
:400px
Then $E$ and $E'$ are on the opposites sides of $O$, and so:
:$PF - PF' = PE - PE'$
But $PE - PE'$ is a constant.
Hence $\EE$ fulfils the equidistance property with respect to the points $F$ and $F'$.
Hence $F$ and $F'$ are the foci of $\EE$.
{{qed|lemma}}
=== Proof for Parabola ===
Let $\EE$ be a parabola.
:400px
Because $\PP$ is parallel to the generatrix of $\CC$, it follows that:
:$\triangle PHE = \triangle PHE'$
and so:
:$PF = PE'$
and it is seen that this is the focus-directrix property of the parabola.
{{qed}}
Theorem: Dandelin's Theorem/Foci/Proof

Let $\CC$ be a double napped right circular cone with apex $O$.
Let $\PP$ be a plane which intersects $\CC$ such that:
:$\PP$ does not pass through $O$
:$\PP$ is not perpendicular to the axis of $\CC$.
Let $\EE$ be the conic section arising as the intersection between $\PP$ and $\CC$.
Let $\SS$ and $\SS'$ be the Dandelin spheres with respect to $\PP$.
Then:
{{:Dandelin's Theorem/Foci}}

Proof:
Let $\SS$ and $\SS'$ be the Dandelin spheres with respect to $\PP$.
Let $P$ be a point on $\EE$.
Let $F$ and $F'$ be the points at which $\SS$ and $\SS'$ are tangent to $\PP$ respectively.
Let the generatrix of $\CC$ which passes through $P$ touch $\SS$ and $\SS'$ at $E$ and $E'$ respectively.
Because $PF$ and $PE$ are both tangent to $\SS$:
:$PF = PE$
and similarly:
:$PF' = PE'$
=== Proof for Ellipse ===
Let $\EE$ be an ellipse.
:540px
Then $E$ and $E'$ are on the same side of $O$, and so:
:$PF + PF' = PE + PE'$
But $PE + PE'$ is a constant.
Hence $\EE$ fulfils the equidistance property with respect to the points $F$ and $F'$.
Hence $F$ and $F'$ are the foci of $\EE$.
{{qed|lemma}}
=== Proof for Hyperbola ===
Let $\EE$ be a hyperbola.
:400px
Then $E$ and $E'$ are on the opposites sides of $O$, and so:
:$PF - PF' = PE - PE'$
But $PE - PE'$ is a constant.
Hence $\EE$ fulfils the equidistance property with respect to the points $F$ and $F'$.
Hence $F$ and $F'$ are the foci of $\EE$.
{{qed|lemma}}
=== Proof for Parabola ===
Let $\EE$ be a parabola.
:400px
Because $\PP$ is parallel to the generatrix of $\CC$, it follows that:
:$\triangle PHE = \triangle PHE'$
and so:
:$PF = PE'$
and it is seen that this is the focus-directrix property of the parabola.
{{qed}}
Theorem: Dandelin's Theorem/Foci

:$\SS$ and $\SS'$ are tangent to $\PP$ at the foci of $\EE$.

Proof:
Let $\SS$ and $\SS'$ be the Dandelin spheres with respect to $\PP$.
Let $P$ be a point on $\EE$.
Let $F$ and $F'$ be the points at which $\SS$ and $\SS'$ are tangent to $\PP$ respectively.
Let the generatrix of $\CC$ which passes through $P$ touch $\SS$ and $\SS'$ at $E$ and $E'$ respectively.
Because $PF$ and $PE$ are both tangent to $\SS$:
:$PF = PE$
and similarly:
:$PF' = PE'$
=== Proof for Ellipse ===
Let $\EE$ be an ellipse.
:540px
Then $E$ and $E'$ are on the same side of $O$, and so:
:$PF + PF' = PE + PE'$
But $PE + PE'$ is a constant.
Hence $\EE$ fulfils the equidistance property with respect to the points $F$ and $F'$.
Hence $F$ and $F'$ are the foci of $\EE$.
{{qed|lemma}}
=== Proof for Hyperbola ===
Let $\EE$ be a hyperbola.
:400px
Then $E$ and $E'$ are on the opposites sides of $O$, and so:
:$PF - PF' = PE - PE'$
But $PE - PE'$ is a constant.
Hence $\EE$ fulfils the equidistance property with respect to the points $F$ and $F'$.
Hence $F$ and $F'$ are the foci of $\EE$.
{{qed|lemma}}
=== Proof for Parabola ===
Let $\EE$ be a parabola.
:400px
Because $\PP$ is parallel to the generatrix of $\CC$, it follows that:
:$\triangle PHE = \triangle PHE'$
and so:
:$PF = PE'$
and it is seen that this is the focus-directrix property of the parabola.
{{qed}}
Theorem: Dandelin's Theorem/Directrices

Let $\KK$ and $\KK'$ be the planes in which the ring-contacts of $\CC$ with $\SS$ and $\SS'$ are embedded respectively.
:The intersections of $\KK$ and $\KK'$ with $\PP$ form the directrices of $\EE$.

Proof:
Let $\SS$ and $\SS'$ be the Dandelin spheres with respect to $\PP$.
Let $P$ be a point on $\EE$.
Let $F$ and $F'$ be the points at which $\SS$ and $\SS'$ are tangent to $\PP$ respectively.
Let the generatrix of $\CC$ which passes through $P$ touch $\SS$ and $\SS'$ at $E$ and $E'$ respectively.
Let $\theta$ be half the opening angle of $\CC$.
Let $\phi$ be the inclination of $\PP$ to the axis of $\CC$.
Let $\PP$ intersect $\KK$ in the straight line $NX$.
Let $PN$ be constructed perpendicular to $NX$.
Let $PK$ be constructed perpendicular to $\KK$.
Then:
:$PK = PN \cos \phi$
Also:
:$PK = PE \cos \theta = PF \cos \theta$
Hence:
:$\dfrac {PF} {PN} = \dfrac {\cos \phi} {\cos \theta}$
which is constant.
Hence $NX$ is a directrix of $\EE$ by definition.
A similar argument and construction applies with respect to $\SS'$ and $\KK$.
{{qed}}
Theorem: Dandelin's Theorem/Directrices/Proof

Let $\CC$ be a double napped right circular cone with apex $O$.
Let $\PP$ be a plane which intersects $\CC$ such that:
:$\PP$ does not pass through $O$
:$\PP$ is not perpendicular to the axis of $\CC$.
Let $\EE$ be the conic section arising as the intersection between $\PP$ and $\CC$.
Let $\SS$ and $\SS'$ be the Dandelin spheres with respect to $\PP$.
Then:
{{:Dandelin's Theorem/Directrices}}

Proof:
Let $\SS$ and $\SS'$ be the Dandelin spheres with respect to $\PP$.
Let $P$ be a point on $\EE$.
Let $F$ and $F'$ be the points at which $\SS$ and $\SS'$ are tangent to $\PP$ respectively.
Let the generatrix of $\CC$ which passes through $P$ touch $\SS$ and $\SS'$ at $E$ and $E'$ respectively.
Let $\theta$ be half the opening angle of $\CC$.
Let $\phi$ be the inclination of $\PP$ to the axis of $\CC$.
Let $\PP$ intersect $\KK$ in the straight line $NX$.
Let $PN$ be constructed perpendicular to $NX$.
Let $PK$ be constructed perpendicular to $\KK$.
Then:
:$PK = PN \cos \phi$
Also:
:$PK = PE \cos \theta = PF \cos \theta$
Hence:
:$\dfrac {PF} {PN} = \dfrac {\cos \phi} {\cos \theta}$
which is constant.
Hence $NX$ is a directrix of $\EE$ by definition.
A similar argument and construction applies with respect to $\SS'$ and $\KK$.
{{qed}}
Theorem: Conic Section is Curve of Second Degree

A '''conic section''' is a curve of second degree.

Proof:
{{ProofWanted|A rigorous approach to this needs more background}}
Theorem: Equation of Conic in Cartesian Coordinates is Quadratic

Let $\CC$ be a conic section.
Then $\CC$ can be expressed by an quadratic equation in $2$ variables.

Proof:
{{ProofWanted|Follows apparently from Conic Section is Curve of Second Degree. Sommerville is not rigorous about defining his terms.}}
Theorem: Intersecting Chord Theorem for Conic Sections

Consider a right circular cone $\CC$ with opening angle $2 \alpha$ whose apex is at $O$.
Consider a slicing plane $\PP$, not passing through $O$, at an angle $\beta$ to the axis of $\CC$.
Let the plane $OAA'$ through the axis of $\CC$ perpendicular to $\PP$ intersect $\PP$ in the line $AA'$.
Let $P$ be an arbitrary point on the intersection of $\PP$ with $\CC$.
Let $PM$ be constructed perpendicular to $AA'$.
Then:
:$PM^2 = k \cdot AM \cdot MA'$
where $k$ is the constant:
:$k = \dfrac {\map \sin {\beta + \alpha} \map \sin {\beta - \alpha} } {\cos \alpha}$

Proof:
{{ProofWanted|This is part of the muddled Construction of Conic Section in Sommerville. This collection of pages needs to be revisited with respect to a more coherent work than this one.}}
Theorem: P-adic Norm Characterisation of Divisibility by Power of p

Let $p \in \N$ be a prime.
Let $\Q$ denote the rational numbers.
Let $\norm{\,\cdot\,}$ denote the $p$-adic norm on $\Q$.
Then:
:$\forall a, b \in \Z: a \equiv b \pmod {p^n} \iff \norm {a - b}_p \le p^{-n}$

Proof:
Let $a, b \in \Z$.
We have:
{{begin-eqn}}
{{eqn | l = a \equiv b \pmod {p^n}
      | o = \iff
      | r = p^n \divides \paren{a - b}
      | c = {{defof|Congruence Modulo Integer}}
}}
{{eqn | o = \iff
      | r = n \le sup(m : p^m \divides \paren{a - b}
      | c = {{defof|Supremum of Set}}
}}
{{eqn | o = \iff
      | r = n \le \map {\nu_p} {a-b}
      | c = {{defof|P-adic Valuation}}
}}
{{eqn | o = \iff
      | r = p^n \le p^{\map {\nu_p} {a-b} }
      | c = Power Function on Base Greater than One is Strictly Increasing
}}
{{eqn | o = \iff
      | r = \dfrac 1 {p^{\map {\nu_p} {a-b} } } \le \dfrac 1 {p^n}
      | c = Inversion Mapping Reverses Ordering in Ordered Group
}}
{{eqn | o = \iff
      | r = \norm {a - b}_p \le p^{-n}
      | c = {{Defof|P-adic Norm}}
}}
{{end-eqn}}
{{qed}}
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Theorem: Continuously Differentiable Real Function at Removable Discontinuity

Let $f : \R \to \R$ be a real function.
Let $a \in \R$ be  real number.
Let $f$ be continuous on $\R$ and continuously differentiable in $\R \setminus \set a$.
Suppose that $a$ is a removable discontinuity of $f'$.
That is, suppose the limit $\ds \lim_{x \mathop \to a} \map {f'} x$ exists.
Then $f$ is continuously differentiable at $a$.

Proof:
By assumption, $\ds \lim_{x \mathop \to a} \map {f'} x = L$ exists.
By definition:
:$\forall \epsilon \in \R_{>0} : \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - a} < \delta \implies \size {\map {f'} x - L} < \epsilon$
Let $h := x - a$.
Suppose $0 < h < \delta$.
Consider a closed interval $\closedint 0 h$.
By assumption, $f$ is continuous on $\closedint 0 h$ and continuously differentiable in $\openint 0 h$.
By Mean Value Theorem:
:$\ds \exists \theta \in \openint 0 1 : \frac {\map f h - \map f 0} {h - 0} = \map {f'} {\theta h}$
Thus:
:$0 < \size {\theta h} \le \size h < \delta$ 
and:
{{begin-eqn}}
{{eqn | l = \size {\frac {\map f h - \map f 0} {h - 0} - L} 
      | r = \size {\map {f'} {\theta h} - L} 
}}
{{eqn | o = <
      | r = \epsilon
      | c = $\ds \lim_{x \mathop \to a} \map {f'} x = L$ exists
}}
{{end-eqn}}
Hence:
:$\ds \forall h \in \openint 0 \delta : \size {\frac {\map f h - \map f 0} {h - 0} - L} < \epsilon$
Consider a closed interval $\closedint {-h} 0$.
By assumption, $f$ is continuous on $\closedint {-h} 0$ and continuously differentiable in $\openint {-h} 0$.
By Mean Value Theorem:
:$\ds \exists \theta \in \openint 0 1 : \frac {\map f 0 - \map f {-h} } {0 - \paren {-h} } = \map {f'} {-\theta h}$
Thus:
$0 < \size {- \theta h} \le \size {-h} < \delta$
and:
{{begin-eqn}}
{{eqn | l = \size {\frac {\map f {- h} - \map f 0} {- h - 0} - L} 
      | r = \size {\map {f'} {- \theta h} - L} 
}}
{{eqn | o = < 
      | r = \epsilon
      | c = $\ds \lim_{x \mathop \to a} \map {f'} x = L$ exists
}}
{{end-eqn}}
Hence:
:$\ds \forall y \in \openint {-\delta} 0 : \size {\frac {\map f y - \map f 0} {y - 0} - L} < \epsilon$
Altogether, it holds that:
:$\ds \forall h \in \R : 0 < \size h < \delta : \size {\frac {\map f h - \map f 0} {h - 0} - L} < \epsilon$
Hence, $f$ is differentiable at $a$:
:$\ds \map {f'} a = L = \lim_{x \mathop \to a} \map {f'} x$.
By assumption, $f'$ is continuous on $\R \setminus \set a$.
Hence, $f$ is continuously differentiable on $\R$.
{{qed}}
Category:Continuity
Category:Differentiable Real Functions
cjoq1ug5uf41g83umciqsjlfm6zhz00
Theorem: Ellipse is Bounded in Plane

Let $E$ be an ellipse embedded in in a Euclidean plane.
Then $E$ is bounded.

Proof:
Let a Cartesian coordinate system be applied to the Euclidean plane in which $E$ is embedded.
Let $E$ be expressed in reduced form:
{{begin-eqn}}
{{eqn | l = \dfrac {x^2} {a^2} + \dfrac {y^2} {b^2}
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac x a
      | r = \sqrt {1 - \dfrac {y^2} {b^2} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac y b
      | r = \sqrt {1 - \dfrac {x^2} {a^2} }
      | c = 
}}
{{end-eqn}}
Hence:
:there are no real values of $x$ for $\size y > b$
:there are no real values of $y$ for $\size x > b$.
Hence $E$ exists entirely within the rectangle whose sides are $x = \pm a$ and $y = \pm b$.
The result follows.
{{Qed}}
Theorem: Circle is Ellipse with Equal Major and Minor Axes

Let $E$ be an ellipse whose major axis is equal to its minor axis.
Then $E$ is a circle.

Proof:
Let $E$ be embedded in a Cartesian plane in reduced form.
Then from Equation of Ellipse in Reduced Form $E$ can be expressed using the equation:
:$\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$
where the major axis and minor axis are $a$ and $b$ respectively.
Let $a = b$.
Then:
{{begin-eqn}}
{{eqn | l = \dfrac {x^2} {a^2} + \dfrac {y^2} {a^2}
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x^2 + y^2
      | r = a^2
      | c = 
}}
{{end-eqn}}
which by Equation of Circle center Origin is the equation of a circle whose radius is $a$.
{{qed}}
Theorem: Equation of Tangent to Ellipse in Reduced Form

Let $E$ be an ellipse embedded in a Cartesian plane in reduced form with the equation:
:$\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$
Let $P = \tuple {x_1, y_1}$ be a point on $E$.
The tangent to $E$ at $P$ is given by the equation:
:$\dfrac {x x_1} {a^2} + \dfrac {y y_1} {b^2} = 1$

Proof:
From the slope-intercept form of a line, the equation of a line passing through $P$ is:
:$y - y_1 = \mu \paren {x - x_1}$
If this line passes through another point $\tuple {x_2, y_2}$ on $E$, the slope of the line is given by:
:$\mu = \dfrac {y_2 - y_1} {x_2 - x_1}$
Because $P$ and $Q$ both lie on $E$, we have:
{{begin-eqn}}
{{eqn | l = \dfrac {x_1^2} {a^2} + \dfrac {y_1^2} {b^2}
      | r = 1 = \dfrac {x_2^2} {a^2} + \dfrac {y_2^2} {b^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {y_2^2} {b^2} - \dfrac {y_1^2} {b^2}
      | r = \dfrac {x_1^2} {a^2} - \dfrac {x_2^2} {a^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\paren {y_2 + y_1} \paren {y_2 - y_1} } {b^2}
      | r = \dfrac {\paren {x_1 + x_2} \paren {x_1 - x_2} } {a^2}
      | c = Difference of Two Squares
}}
{{eqn | ll= \leadsto
      | l = \dfrac {y_2 - y_1} {\paren {x_2 - x_1} }
      | r = -\dfrac {b^2 \paren {x_1 + x_2} } {a^2 \paren {y_1 + y_2} }
      | c = 
}}
{{end-eqn}}
As $Q$ approaches $P$, we have that $y_2 \to y_1$ and $x_2 \to x_1$.
The limit of the slope is therefore:
:$-\dfrac {2 b^2 x_1} {2 a^2 y_1} = -\dfrac {b^2 x_1} {a^2 y_1}$
The equation of the tangent $\TT$ to $\CC$ passing through $\tuple {x_1, y_1}$ is therefore:
{{begin-eqn}}
{{eqn | l = y - y_1
      | r = -\dfrac {b^2 x_1} {a^2 y_1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {y_1} {b^2} \paren {y - y_1}
      | r = -\dfrac {x_1} {a^2} \paren {x - x_1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {x x_1} {a^2} + \dfrac {y y_1} {b^2}
      | r = \dfrac {x_1^2} {a^2} + \dfrac {y_1^2} {b^2}
      | c = 
}}
{{eqn | r = 1
      | c = as $\tuple {x_1, y_1}$ is on $E$
}}
{{end-eqn}}
{{qed}}
Theorem: Metric on P-adic Numbers Extends Metric on Rationals

Let $p$ be any prime number.
Let $\struct{\Q, \norm {\,\cdot\,}^\Q_p}$ be the rational numbers $\Q$ with the $p$-adic norm $\norm {\,\cdot\,}^\Q_p$.
Let $d^{\Q}_p$ be the $p$-adic metric on the rational numbers.
Let $\struct{\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $d_p$ be the $p$-adic metric on the $p$-adic numbers.
Then $d_p$ is an extension of $d^{\Q}_p$.

Proof:
From Rational Numbers are Dense Subfield of P-adic Numbers:
:$\norm {\,\cdot\,}_p$ on $\Q_p$ is an extension of $\norm {\,\cdot\,}^Q_p$ on $\Q$.
By definition of $p$-adic metric on the rational numbers:
:$\forall x, y \in \Q : \map {d^\Q_p} {x, y} = \norm{x - y}^\Q_p$
By definition of $p$-adic metric on the $p$-adic numbers:
:$\forall x, y \in \Q : \map {d_p} {x, y} = \norm{x - y}_p$
It follows that the $p$-adic metric on $\Q_p$ is an extension of the $p$-adic metric on $\Q$.
{{qed}}
Category:P-adic Metrics
muxbupanq9430smsaokweh76b6am4uo
Theorem: Poisson Distribution Approximated by Hat-Check Distribution

Let $X$ be a discrete random variable which has the hat-check distribution with parameter $n$.
Then $X$ can be approximated by a Poisson distribution with parameter $\lambda = 1$.

Proof:
Let $X$ be as described.
Let $k \ge 0$ be fixed.
Then:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}
      | r = \lim_{n \mathop \to \infty} \dfrac 1 {\paren {n - \paren {n - k} }!} \sum_{s \mathop = 0}^{n - k} \dfrac {\paren {-1}^s} {s!}
      | c = setting $k = n - k$
}}
{{eqn | r = \frac {1^k} {k!}  e^{-1}
      | c = Taylor Series Expansion for Exponential Function and {{Defof|Poisson Distribution}}
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Simson Line Theorem

Let $\triangle ABC$ be a triangle.
Let $P$ be a point on the circumcircle of $\triangle ABC$.
Then the feet of the perpendiculars drawn from $P$ to each of the sides of $\triangle ABC$ are collinear.
:300px
This line is called the '''Wallace-Simson line'''.

Proof:
In the figure above, construct the lines $BP$ and $CP$.
:300px
By the converse of Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles, $EPDB$ is cyclic.
By the converse of Angles in Same Segment of Circle are Equal, $EPCF$ is cyclic.
{{WIP|Looking for the pages for the above converses and the (simple) unlinked theorem below. The theorem can be split trivially into Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles and Two Angles on Straight Line make Two Right Angles, but the same cannot be said for those converses. <br> Category:Cyclic Quadrilaterals is of no help}}
Therefore:
{{begin-eqn}}
{{eqn | l = \angle DEP
      | r = \angle DBP
      | c = Angles in Same Segment of Circle are Equal: $EPDB$ is cyclic
}}
{{eqn | r = \angle ACP
      | c = The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle: $ABPC$ is cyclic
}}
{{eqn | r = 180 \degrees - \angle PEF
      | c = Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles: $EPCF$ is cyclic
}}
{{end-eqn}}
This gives:
:$\angle DEP + \angle PEF = 180 \degrees$
Hence $DEF$ is a straight line.
{{qed}}
{{Namedfor|Robert Simson|cat = Simson}}
Theorem: Stone-Weierstrass Theorem/Compact Space

Let $T = \struct {X, \tau}$ be a compact topological space.
Let $\map C {X, \R}$ be the set of real-valued continuous functions on $T$.
Let $\times$ be the pointwise multiplication on $\map C {X, \R}$.
Let $\struct {\map C {X, \R}, \times}$ be the Banach algebra with respect to $\norm \cdot_\infty$.
{{ExtractTheorem|A definition page for Banach algebra $\map C {X, \R}$ and its verification?}}
{{explain|By the definition of Banach algebra, $\map C {X, \R}$ should be a commutative ring. Can it be confirmed that this is indeed the case?}}
Let $\AA$ be a unital subalgebra of $\map C {X, \R}$.
Suppose that $\AA$ separates points of $X$, that is:
:for distinct $p, q \in X$, there exists $h_{p q} \in \AA$ such that $\map {h_{p q} } p \ne \map {h_{p q} } q$.
Then the closure $\overline \AA$ of $\AA$ is equal to $\map C {X, \R}$.

Proof:
Let $\struct {C', \norm {\,\cdot\,}_{C'} }$ be the dual space of $\struct {\map C {X, \R}, \norm {\,\cdot\,}_\infty }$.
From Spanning Criterion of Normed Vector Space, it suffices to show that:
:$\forall \ell \in C' : \ell \restriction_\AA = 0 \implies \ell = 0$
Let $B' \subseteq C'$ be the closed unit ball, that is:
:$B' := \set {\ell \in C' : \norm \ell_{C'} \le 1}$
Let:
:$U := \set {\ell \in B' : \ell \restriction_\AA = 0}$
Let $\map E U$ be the set of extreme points of $U$.
Theorem: Integers with Metric Induced by P-adic Valuation

Let $p \in \N$ be a prime.
Let $d: \Z^2 \to \R_{\ge 0}$ be the mapping defined as:
:$\forall x, y \in \Z: \map d {x, y} = \begin {cases} 0 & : x = y \\ \dfrac 1 r & : x - y = p^{r - 1} k: r \in \N_{>0}, k \in \Z, p \nmid k \end {cases}$
Then $d$ is a metric on $\Z$.

Proof:
From Characterization of P-adic Valuation on Integers:
:$d$ is well-defined.
We prove the metric space axioms.
Theorem: Equation of Chord of Contact on Ellipse in Reduced Form

Let $\EE$ be an ellipse embedded in a Cartesian plane in reduced form with the equation:
:$\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$
Let $P = \tuple {x_0, y_0}$ be a point which is outside the boundary of $\EE$.
Let $UV$ be the chord of contact on $\EE$ with respect to $P$.
Then $UV$ can be defined by the equation:
:$\dfrac {x x_0} {a^2} + \dfrac {y y_0} {b^2} = 1$

Proof:
Let $\TT_1$ and $\TT_2$ be a tangents to $\EE$ passing through $P$.
Let:
:$\TT_1$ touch $\EE$ at $U = \tuple {x_1, y_1}$
:$\TT_2$ touch $\EE$ at $V = \tuple {x_2, y_2}$
Then the chord of contact on $\EE$ with respect to $P$ is defined as $UV$.
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From Equation of Tangent to Ellipse in Reduced Form, $\TT_1$ is expressed by the equation:
:$\dfrac {x x_1} {a^2} + \dfrac {y y_1} {b^2} = 1$
but as $\TT_1$ also passes through $\tuple {x_0, y_0}$ we also have:
:$\dfrac {x_0 x_1} {a^2} + \dfrac {y_0 y_1} {b^2} = 1$
This also expresses the condition that $U$ should lie on $\TT_1$:
:$\dfrac {x x_0} {a^2} + \dfrac {y y_0} {b^2} = 1$
Similarly, From Equation of Tangent to Circle Centered at Origin, $\TT_2$ is expressed by the equation:
:$\dfrac {x x_2} {a^2} + \dfrac {y y_2} {b^2} = 1$
but as $\TT_2$ also passes through $\tuple {x_0, y_0}$ we also have:
:$\dfrac {x_0 x_2} {a^2} + \dfrac {y_0 y_2} {b^2} = 1$
This also expresses the condition that $V$ should lie on $\TT_2$:
:$\dfrac {x x_0} {a^2} + \dfrac {y y_0} {b^2} = 1$
So both $U$ and $V$ lie on the straight line whose equation is:
:$\dfrac {x x_0} {a^2} + \dfrac {y y_0} {b^2} = 1$
and the result follows.
{{qed}}
Theorem: Polar is Locus of Harmonic Conjugates wrt Ellipse

Let $\EE$ be an ellipse embedded in the plane.
Let $P$ be an arbitrary point in the plane.
Let $\LL$ be the polar of $P$ {{WRT}} $\EE$.
Then $\LL$ is the locus of harmonic conjugates {{WRT}} $\EE$.

Proof:
{{ProofWanted|Need to craft the definition of harmonic conjugates {{WRT}} $\EE$. See Harmonic Property of Pole and Polar wrt Circle and explore what Sommerville says about this in chapter $\text {III}$ section $5$.}}
Theorem: Parabola has No Center

The parabola has no center.

Proof:
By definition, the center of a conic section is the midpoint of the straight line segment whose endpoints are the foci.
From Parabola has One Focus, it is not possible to construct such a line segment.
Hence the result.
{{qed}}
Category:Parabolas
Category:Centers of Conic Sections
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Theorem: Characterization of Derivative of Test Function

Let $\Phi, \phi \in \map \DD \R$ be test functions.
Let:
:$\ds X = \set {\Phi' : \Phi \in \map \DD \R} $
:$\ds Y = \set {\phi \in \map \DD \R : \int_{- \infty}^\infty \map \phi x \rd x = 0}$
Then $X = Y$.
Furthermore, for every $\phi \in Y$ there exists only one $\Phi \in \map \DD \R$ such that $\Phi' = \phi$.

Proof:
=== $X$ is a subset of $Y$ ===
By definition of test functions:
:$\paren {\Phi \in \map \DD \R} \implies \paren {\Phi' \in \map \DD \R}$
Moreover, a test function has compact support:
:$\ds \int_{-\infty}^\infty \map {\Phi'} x \rd x = \map \Phi \infty - \map \Phi {-\infty} = 0$
Hence:
:$X \subseteq Y$
{{qed|lemma}}