avewright/proofwiki-math huggingface dataset:
Theorem: Factorization of Natural Numbers within 4 n + 1 not Unique

Let:
:$S = \set {4 n + 1: n \in \N} = \set {1, 5, 9, 13, 17, \ldots}$
be the set of natural numbers of the form $4 n + 1$.
Then not all elements of $S$ have a complete factorization by other elements of $S$ which is unique.

Proof:
Proof by Counterexample:
Consider the number:
:$m = 693 = 3^2 \times 7 \times 11$
Thus:
:$m = 9 \times 77 = 21 \times 33$
We have that:
{{begin-eqn}}
{{eqn | l = 9
      | r = 4 \times 2 + 1
      | rr= \in S
}}
{{eqn | l = 77
      | r = 4 \times 19 + 1
      | rr= \in S
}}
{{eqn | l = 21
      | r = 4 \times 5 + 1
      | rr= \in S
}}
{{eqn | l = 33
      | r = 4 \times 8 + 1
      | rr= \in S
}}
{{end-eqn}}
The divisors of these numbers are as follows:
{{begin-eqn}}
{{eqn | l = 9
      | r = 3^2
      | c = where $3 \notin S$
}}
{{eqn | l = 77
      | r = 7 \times 11
      | c = where $7 \notin S$ and $11 \notin S$
}}
{{eqn | l = 21
      | r = 3 \times 7
      | c = where $3 \notin S$ and $7 \notin S$
}}
{{eqn | l = 33
      | r = 3 \times 11
      | c = where $3 \notin S$ and $11 \notin S$
}}
{{end-eqn}}
Thus $693$ has two different complete factorizations into elements of $S$.
Hence the result.
{{qed}}
Theorem: Solutions of Pythagorean Equation/Primitive

The set of all primitive Pythagorean triples is generated by:
:$\tuple {2 m n, m^2 - n^2, m^2 + n^2}$
where:
:$m, n \in \Z_{>0}$ are (strictly) positive integers
:$m \perp n$, that is, $m$ and $n$ are coprime
:$m$ and $n$ are of opposite parity
:$m > n$

Proof:
First we show that $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ is a Pythagorean triple:
{{begin-eqn}}
{{eqn | l = \paren {2 m n}^2 + \paren {m^2 - n^2}^2
      | r = 4 m^2 n^2 + m^4 - 2 m^2 n^2 + n^4
      | c =
}}
{{eqn | r = m^4 + 2 m^2 n^2 + n^4
      | c =
}}
{{eqn | r = \paren {m^2 + n^2}^2
      | c =
}}
{{end-eqn}}
So $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ is indeed a Pythagorean triple.
Now we establish that $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ is primitive:
{{AimForCont}} to the contrary, that $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ is not primitive.
So there is a prime divisor $p$ of both $2 m n$ and $m^2 - n^2$.
That is:
:$p \in \mathbb P: p \divides \paren {2 m n}, p \divides \paren {m^2 - n^2}$
Then from Prime Divides Power:
:$p \divides \paren {2 m n}^2$ and $p \divides \paren {m^2 - n^2}^2$
Hence from Common Divisor Divides Integer Combination:
:$p \divides \paren {m^2 + n^2}^2$
and from Prime Divides Power again:
:$p \divides \paren {m^2 + n^2}$
So from Common Divisor Divides Integer Combination:
:$p \divides \paren {m^2 + n^2} + \paren {m^2 - n^2} = 2 m^2$
:$p \divides \paren {m^2 + n^2} - \paren {m^2 - n^2} = 2 n^2$
But $p \ne 2$ as, because $m$ and $n$ are of opposite parity, $m^2 - n^2$ must be odd.
So $p \divides n^2$ and $p \divides m^2$ and so from Prime Divides Power, $p \divides n$ and $p \divides m$.
But as we specified that $m \perp n$, this is a contradiction.
Therefore $\paren {2 m n, m^2 - n^2, m^2 + n^2}$ is primitive.
{{qed|lemma}}
Now we need to show that every primitive Pythagorean triple is of this form:
So, suppose that $\tuple {x, y, z}$ is ''any'' primitive Pythagorean triple given in canonical form.
From Parity of Smaller Elements of Primitive Pythagorean Triple, $x$ and $y$ are of opposite parity.
By definition of canonical form $x$ is even and $y$ and $z$ are both odd.
As $y$ and $z$ are both odd, their sum and difference are both even.
Hence we can define:
:$s, t \in Z: s = \dfrac {z + y} 2, t = \dfrac {z - y} 2$.
Note that $s \perp t$ as any common divisor would also divide $s + t = z$ and $s - t = y$, and we know that $z \perp y$ from Elements of Primitive Pythagorean Triple are Pairwise Coprime.
Then from the Pythagorean equation:
:$x^2 = z^2 - y^2 = \paren {z + y} \paren {z - y} = 4 s t$
Hence:
:$\paren {\dfrac x 2}^2 = s t$
As $x$ is even, $\dfrac x 2$ is an integer and so $s t$ is a square.
So each of $s$ and $t$ must be square as they are coprime.
Now, we write $s = m^2$ and $t = n^2$ and substitute back:
:$x^2 = 4 s t = 4 m^2 n^2$ and so $x = 2 m n$
:$y = s - t = m^2 - n^2$
:$z = m^2 + n^2$
Finally, note that:
:$m \perp n$ from $s \perp t$ and Prime Divides Power
:$m$ and $n$ have opposite parity otherwise $y$ and $z$ would be even.
Thus, our primitive Pythagorean triple is of the form $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$.
{{qed}}
Theorem: Solutions of Pythagorean Equation/General

Let $x, y, z$ be a solution to the Pythagorean equation.
Then $x = k x', y = k y', z = k z'$, where:
:$\tuple {x', y', z'}$ is a primitive Pythagorean triple
:$k \in \Z: k \ge 1$

Proof:
Let $\tuple {x, y, z}$ be non-primitive solution to the Pythagorean equation.
Let:
:$\exists k \in \Z: k \ge 2, k \divides x, k \divides y$
such that $x \perp y$.
Then we can express $x$ and $y$ as $x = k x', y = k y'$.
Thus:
:$z^2 = k^2 x'^2 + k^2 y'^2 = k^2 z'^2$
for some $z' \in \Z$.
Let:
:$\exists k \in \Z: k \ge 2, k \divides x, k \divides z$
such that $x \perp z$
Then we can express $x$ and $z$ as $x = k x', z = k z'$.
Thus:
:$y^2 = k^2 z'^2 - k^2 x'^2 = k^2 y'^2$
for some $y' \in \Z$.
Similarly for any common divisor of $y$ and $z$.
Thus any common divisor of any pair of $x, y, z$ has to be a common divisor of Integers of the other.
Hence any non-primitive solution to the Pythagorean equation is a constant multiple of some primitive solution.
{{qed}}
Theorem: Goldbach Conjecture implies Goldbach's Marginal Conjecture

Suppose the Goldbach Conjecture holds:
:Every even integer greater than $2$ is the sum of two primes.
Then Goldbach's Marginal Conjecture follows:
:Every integer greater than $5$ can be written as the sum of three primes.

Proof:
Suppose the Goldbach Conjecture holds.
Let $n \in \Z$ such that $n > 5$.
Let $n$ be an odd integer.
Then $n - 3$ is an even integer greater than $2$.
By the Goldbach Conjecture:
:$n - 3 = p_1 + p_2$
where $p_1$ and $p_2$ are both primes.
Then:
:$n = p_1 + p_2 + 3$
As $3$ is prime, the result follows.
Let $n$ be an even integer.
Then $n - 2$ is an even integer greater than $3$ and so greater than $2$.
By the Goldbach Conjecture:
:$n - 2 = p_1 + p_2$
where $p_1$ and $p_2$ are both primes.
Then:
:$n = p_1 + p_2 + 2$
As $2$ is prime, the result follows.
{{qed}}
Category:Prime Numbers
Category:Goldbach Conjecture
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Theorem: Congruent Integers are of same Quadratic Character

Let $p$ be an odd prime.
Let $a \in \Z$ be an integer such that $a \not \equiv 0 \pmod p$.
Let $a \equiv b \pmod p$.
Then $a$ and $b$ have the same quadratic character.

Proof:
Let $a \equiv b \pmod p$.
Then by Congruence of Powers:
:$a^2 \equiv b^2 \pmod p$
Hence:
:$x^2 \equiv a \pmod p$ has a solution {{iff}} $x^2 \equiv b \pmod p$.
Hence the result.
{{qed}}
Category:Quadratic Residues
rqg7tzzvmszbgwmoy21zvr5o7qdyw94
Theorem: Weight of Sorgenfrey Line is Continuum

Let $T = \struct {\R, \tau}$ be the Sorgenfrey line.
Then $\map w T = \mathfrak c$
where
:$\map w T$ denotes the weight of $T$
:$\mathfrak c$ denotes continuum, the cardinality of real numbers.

Proof:
By definition of Sorgenfrey line, the set:
:$\BB = \set {\hointr x y: x, y \in \R \land x < y}$
is a basis of $T$.
By definition of weight:
:$\map w T \le \card \BB$
where $\card \BB$ denotes the cardinality of $\BB$.
By Cardinality of Basis of Sorgenfrey Line not greater than Continuum:
:$\card \BB \le \mathfrak c$
Thus
:$\map w T \le \mathfrak c$
It remains to show that:
:$\mathfrak c \le \map w T$
{{AimForCont}}
:$\mathfrak c \not \le \map w T$
Then:
:$\map w T < \mathfrak c$
By definition of weight, there exists a basis $\BB_0$ of $T$:
:$\map w T = \card {\BB_0}$
Then by Set of Subset of Reals with Cardinality less than Continuum has not Interval in Union Closure:
:$\exists x, y \in \R: x < y \land \hointr x y \notin \set {\bigcup A: A \subseteq \BB_0} = \tau$
By definition of $\BB$:
:$\hointr x y \in \BB \subseteq \tau$
Thus this contradicts by definition of subset with:
:$\hointr x y \notin \tau$
{{qed}}
Theorem: Construction of Regular Heptadecagon

It is possible to construct a regular hepadecagon (that is, a regular polygon with $17$ sides) using a compass and straightedge construction.

Proof:
It remains to be demonstrated that the line segment $NM$ is the side of a regular hepadecagon inscribed in circle $ACB$.
This will be done by demonstrating that $\angle NOM$ is equal to $\dfrac {2 \pi} {17}$ radians, that is, $\dfrac 1 {17}$ of the full circle $ACB$.
For convenience, let the radius $OA$ be equal to $4 a$.
By Pythagoras's Theorem, $AD = a \sqrt {17}$.
By definition of tangent, $OE = a \map \arctan {\dfrac {\angle ODA} 4}$.
By construction, $\angle EDF = \dfrac \pi 4$ radians.
Thus:
{{begin-eqn}}
{{eqn | l = \frac {\tan \angle ODE + \tan \angle ODF} {1 - \tan \angle ODE \tan \angle ODF}
      | r = \tan \angle EDF
      | c = Tangent of Sum
}}
{{eqn | r = 1
      | c = Tangent of $\dfrac \pi 4$
}}
{{end-eqn}}
{{finish|finish off -- the algebra gets complicated from here on in.}}
Theorem: Set of Subset of Reals with Cardinality less than Continuum has not Interval in Union Closure

Let $\BB$ be a set of subsets of $\R$, the set of all real numbers.
Let:
:$\card \BB < \mathfrak c$
where
:$\card \BB$ denotes the cardinality of $\BB$
:$\mathfrak c = \card \R$ denotes continuum.
Let $\FF = \set {\bigcup \GG: \GG \subseteq \BB}$.
Then:
:$\exists x, y \in \R: x < y \land \hointr x y \notin \FF$

Proof:
Define:
:$ Z = \leftset {x \in \R: \exists U \in \FF: x}$ is local minimum in $\rightset U$
By Set of Subsets of Reals with Cardinality less than Continuum Cardinality of Local Minimums of Union Closure less than Continuum:
:$\card Z < \mathfrak c$
Then by Cardinalities form Inequality implies Difference is Nonempty:
:$\R \setminus Z \ne \O$
Hence by definition of empty set:
:$\exists z: z \in \R \setminus Z$
By definition of difference:
:$z \in \R \land z \notin Z$
Thus $z < z + 1$.
We will show that $z$ is local minimum in $\hointr z {z + 1}$.
Thus:
:$z \in \hointr z {z + 1}$
Hence:
:$z - 1 < z$
Thus:
:$\openint {z - 1} z \cap \hointr z {z + 1} = \O$
Then by definition $z$ is a local minimum in $\hointr z {z + 1}$.
Because $z \notin Z$:
:$\hointr z {z + 1} \notin \FF$
{{qed}}
Theorem: Cardinality of Basis of Sorgenfrey Line not greater than Continuum

Let $T = \struct {\R, \tau}$ be the Sorgenfrey line.
Let
:$\BB = \set {\hointr x y: x, y \in \R \land x < y}$
be the basis of $T$.
Then $\card \BB \le \mathfrak c$
where
:$\card \BB$ denotes the cardinality of $\BB$
:$\mathfrak c = \card \R$ denotes the continuum.

Proof:
Define a mapping $f: \BB \to \R \times \R$:
:$\forall I \in \BB: \map f I = \tuple {\min I, \sup I}$
That is:
:$\map f {\hointr x y} = \tuple {x, y} \forall x, y \in \R: x < y$
We will show that $f$ is an injection by definition.
Let $I_1, I_2 \in \BB$ such that:
:$\map f {I_1} = \map f {I_2}$
{{begin-eqn}}
{{eqn | l = I_1
      | r = \hointr {\min I_1} {\sup I_1}
      | c = {{Defof|Half-Open Real Interval}}
}}
{{eqn | r = \hointr {\min I_2} {\sup I_2}
      | c = by $\map f {I_1} = \map f {I_2}$
}}
{{eqn | r = I_2
      | c = {{Defof|Half-Open Real Interval}}
}}
{{end-eqn}}
So:
:$I_1 = I_2$
Thus $f$ is an injection.
By Injection implies Cardinal Inequality:
:$\card \BB \le \card {\R \times \R}$
By Cardinal Product Equal to Maximum:
:$\card {\R \times \R} = \map \max {\mathfrak c, \mathfrak c}$
Thus:
:$\card \BB \le \mathfrak c$
{{qed}}
Theorem: Cardinalities form Inequality implies Difference is Nonempty

Let $X, Y$ be sets.
Let
:$\card X < \card Y$
where $\card X$ denotes the cardinality of $X$.
Then:
:$Y \setminus X \ne \O$

Proof:
{{AimForCont}} that:
:$Y \setminus X = \O$
Then by Set Difference with Superset is Empty Set:
:$Y \subseteq X$
Hence by Subset implies Cardinal Inequality:
:$\card Y \le \card X$
This contradicts:
:$\card X < \card Y$
Hence the result.
{{qed}}
Theorem: Slope of Secant

Let $f: \R \to \R$ be a real function.
Let the graph of $f$ be depicted on a Cartesian plane.
:400px
Let $AB$ be a secant of $f$ where:
:$A = \tuple {x, \map f x}$
:$A = \tuple {x + h, \map f {x + h} }$
Then the slope of $AB$ is given by:
:$\dfrac {\map f {x + h} - \map f x} h$

Proof:
The slope of $AB$ is defined as the change in $y$ divided by the change in $x$.
Between $A$ and $B$:
:the change in $x$ is $\paren {x + h} - x = h$
:the change in $y$ is $\map f {x + h} - \map f x$.
Hence the result.
{{qed}}
Theorem: Derivative of Curve at Point

Let $f: \R \to \R$ be a real function.
Let the graph $G$ of $f$ be depicted on a Cartesian plane.
Then the derivative of $f$ at $x = \xi$ is equal to the tangent to $G$ at $x = \xi$.

Proof:
Let $f: \R \to \R$ be a real function.
:400px
Let the graph $G$ of $f$ be depicted on a Cartesian plane.
Let $A = \tuple {\xi, \map f \xi}$ be a point on $G$.
Consider the secant $AB$ to $G$ where $B = \tuple {\xi + h, \map f {\xi + h} }$.
From Slope of Secant, the slope of $AB$ is given by:
:$\dfrac {\map f {x + h} - \map f x} h$
By taking $h$ smaller and smaller, the secant approaches more and more closely the tangent to $G$ at $A$:
:$\map {f'} x = \ds \lim_{h \mathop \to 0} \dfrac {\map f {x + h} - \map f x} h$
where $\lim$ denotes the limit as $h \to 0$.
{{qed}}
Theorem: Derivative of Square Function

Let $f: \R \to \R$ be the square function:
:$\forall x \in \R: \map f x = x^2$
Then the derivative of $f$ is given by:
:$\map {f'} x = 2 x$

Proof:
{{begin-eqn}}
{{eqn | l = \map {f'} x
      | r = \lim_{h \mathop \to 0} \frac {\map f {x + h} - \map f x} h
      | c = {{Defof|Derivative of Real Function at Point}}
}}
{{eqn | r = \lim_{h \mathop \to 0} \frac {\paren {x + h}^2 - x^2} h
      | c = 
}}
{{eqn | r = \lim_{h \mathop \to 0} \frac {x^2 + 2 x h + h^2 - x^2} h
      | c = 
}}
{{eqn | r = \lim_{h \mathop \to 0} \frac {2 x h + h^2} h
      | c = 
}}
{{eqn | r = \lim_{h \mathop \to 0} 2 x + h
      | c = 
}}
{{eqn | r = 2 x
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Derivative of Square Function

Let $f: \R \to \R$ be the square function:
:$\forall x \in \R: \map f x = x^2$
Then the derivative of $f$ is given by:
:$\map {f'} x = 2 x$

Proof:
From Power Rule for Derivatives:
:$\map {\dfrac \d {\d x} } {x^n} = n x^{n - 1}$
The result follows by setting $n = 2$.
{{qed}}
Theorem: Countable iff Cardinality not greater than Aleph Zero

Let $X$ be set.
$X$ is countable {{iff}}:
$\card X \le \aleph_0$
where:
:$\card X$ denotes the cardinality of $X$
:$\aleph_0 = \card \N$ by Aleph Zero equals Cardinality of Naturals.

Proof:
:$X$ is countable 
{{iff}}:
:there exists an injection $f: X \to \N$ by definition of countable set
{{iff}}:
:$\card X \le \card \N$ by Injection iff Cardinal Inequality
{{iff}}:
:$\card X \le \aleph_0$
{{qed}}
Theorem: Aleph Zero equals Cardinality of Naturals

$\aleph_0 = \card \N$
where
:$\aleph$ denotes the aleph mapping,
:$\card \N$ denotes the cardinality of $\N$.

Proof:
{{begin-eqn}}
{{eqn | l = \aleph_0
      | r = \card {\aleph_0}
      | c = Cardinal of Cardinal Equal to Cardinal
}}
{{eqn | r = \card {\omega}
      | c = {{Defof|Aleph Mapping}}
}}
{{eqn | r = \card {\N}
      | c = {{Defof|Natural Numbers|subdef = Construction}}
}}
{{end-eqn}}
{{qed}}
Theorem: Distance Moved by Body from Rest under Constant Acceleration

Let a body $B$ be stationary.
Let $B$ be subject to a constant acceleration.
Then the distance travelled by $B$ is proportional to the square of the length of time $B$ is under the acceleration.

Proof:
From Equations of Motion with Constant Acceleration: Distance after Time:
:$\mathbf s = \mathbf u t + \dfrac {\mathbf a t^2} 2$
where:
:$\mathbf s$ is the displacement of $B$ at time $t$ from its initial position at time $t$
:$\mathbf u$ is the velocity at time $t = 0$
:$\mathbf a$ is the constant acceleration $t$
In this scenario, $\mathbf u = \mathbf 0$.
Thus:
:$\mathbf s = \dfrac {\mathbf a} 2 t^2$
and so by taking the magnitudes of the vector quantities:
:$s = \dfrac a 2 t^2$
Hence the result, by definition of proportional.
{{qed}}
Theorem: Equations of Motion with Constant Acceleration/Velocity after Time

:$\mathbf v = \mathbf u + \mathbf a t$

Proof:
By definition of acceleration:
:$\dfrac {\d \mathbf v} {\d t} = \mathbf a$
By Solution to Linear First Order Ordinary Differential Equation:
:$\mathbf v = \mathbf c + \mathbf a t$
where $\mathbf c$ is a constant vector.
We are given the initial condition:
:$\bigvalueat {\mathbf v} {t \mathop = 0} = \mathbf u$
from which it follows immediately that:
:$\mathbf v = \mathbf u + \mathbf a t$
{{qed}}
Theorem: Equations of Motion with Constant Acceleration/Distance after Time

:$\mathbf s = \mathbf u t + \dfrac {\mathbf a t^2} 2$

Proof:
From Equations of Motion with Constant Acceleration: Velocity after Time:
:$\mathbf v = \mathbf u + \mathbf a t$
By definition of velocity, this can be expressed as:
:$\dfrac {\d \mathbf s} {\d t} = \mathbf u + \mathbf a t$
where both $\mathbf u$ and $\mathbf a$ are constant.
By Solution to Linear First Order Ordinary Differential Equation:
:$\mathbf s = \mathbf c + \mathbf u t + \dfrac {\mathbf a t^2} 2$
where $\mathbf c$ is a constant vector.
We are (implicitly) given the initial condition:
:$\bigvalueat {\mathbf s} {t \mathop = 0} = \mathbf 0$
from which it follows immediately that:
:$\mathbf s = \mathbf u t + \dfrac {\mathbf a t^2} 2$
{{qed}}
Theorem: Equations of Motion with Constant Acceleration/Velocity after Distance

:$\mathbf v \cdot \mathbf v = \mathbf u \cdot \mathbf u + 2 \mathbf a \cdot \mathbf s$

Proof:
From Equations of Motion with Constant Acceleration: Velocity after Time
:$\mathbf v = \mathbf u + \mathbf a t$
Then:
{{begin-eqn}}
{{eqn | l = \mathbf v \cdot \mathbf v
      | r = \paren {\mathbf u + \mathbf a t} \cdot \paren {\mathbf u + \mathbf a t}
      | c = 
}}
{{eqn | r = \mathbf u \cdot \mathbf u + \mathbf u \cdot \mathbf a t + \mathbf a t \cdot \mathbf u + \paren {\mathbf a t} \cdot \paren {\mathbf a t}
      | c = Dot Product Distributes over Addition
}}
{{eqn | r = \mathbf u \cdot \mathbf u + 2 \mathbf u \cdot \mathbf a t + \paren {\mathbf a t} \cdot \paren {\mathbf a t}
      | c = Dot Product Operator is Commutative
}}
{{eqn | r = \mathbf u \cdot \mathbf u + 2 \mathbf u \cdot \mathbf a t + \mathbf a \cdot \mathbf a t^2
      | c = Dot Product Associates with Scalar Multiplication
}}
{{eqn | n = 1
      | r = \mathbf u \cdot \mathbf u + \mathbf a \cdot \paren {2 \mathbf u t + \mathbf a t^2}
      | c = Dot Product Distributes over Addition
}}
{{end-eqn}}
From Equations of Motion with Constant Acceleration: Distance after Time:
:$\mathbf s = \mathbf u t + \dfrac {\mathbf a t^2} 2$
Substituting for $\mathbf s$ in $(1)$ gives:
:$\mathbf v \cdot \mathbf v = \mathbf u \cdot \mathbf u + 2 \mathbf a \cdot \mathbf s$
and the proof is complete.
{{qed}}
Theorem: Aleph Zero is less than Cardinality of Continuum

$\aleph_0 < \mathfrak c$
where
:$\aleph$ denotes the aleph mapping,
:$\mathfrak c$ denotes the cardinality of the continuum.

Proof:
By Power Set of Natural Numbers has Cardinality of Continuum:
:$\mathfrak c = \card {\powerset \N}$
where:
:$\powerset \N$ denotes the power set of $\N$
:$\card {\powerset \N}$ denotes the cardinality of $\powerset \N$.
By Cardinality of Set less than Cardinality of Power Set:
:$\card \N < \card {\powerset \N}$ 
Thus by Aleph Zero equals Cardinality of Naturals:
:$\aleph_0 < \mathfrak c$
{{qed}}
Theorem: Cardinality of Set less than Cardinality of Power Set

Let $X$ be a set.
Then:
:$\card X < \card {\powerset X}$
where
:$\card X$ denotes the cardinality of $X$,
:$\powerset X$ denotes the power set of $X$.

Proof:
By No Bijection from Set to its Power Set:
: there exist no bijections $X \to \powerset X$
Then by definition of set equivalence:
:$X \not\sim \powerset X$
Hence by definition of cardinality:
:$(1): \quad \card X \ne \card {\powerset X}$
By Cardinality of Set of Singletons:
:$(2): \quad \card {\set {\set {x}: x \in X} } = \card X$
By definition of subset:
:$\forall x \in X: \set {x} \subseteq X$
Then by definition of power set:
:$\forall x \in X: \set {x} \in \powerset X$
Hence by definition of subset:
:$\set {\set {x}: x \in X} \subseteq \powerset X$
Then by Subset implies Cardinal Inequality and $(2)$:
:$\card X \le \card {\powerset X}$
Thus by $(1)$:
:$\card X < \card {\powerset X}$
{{qed}}
Theorem: Volume of Solid of Revolution

Let $f: \R \to \R$ be a real function which is integrable on the interval $\closedint a b$.
Let the points be defined:
:$A = \tuple {a, \map f a}$
:$B = \tuple {b, \map f b}$
:$C = \tuple {b, 0}$
:$D = \tuple {a, 0}$
Let the figure $ABCD$ be defined as being bounded by the straight lines $y = 0$, $x = a$, $x = b$ and the curve defined by $\set {\map f x: a \le x \le b}$.
Let the solid of revolution $S$ be generated by rotating $ABCD$ around the $x$-axis (that is, $y = 0$).
Then the volume $V$ of $S$ is given by:
:$\ds V = \pi \int_a^b \paren {\map f x}^2 \rd x$

Proof:
:500px
Consider a rectangle bounded by the lines:
:$y = 0$
:$x = \xi$
:$x = \xi + \delta x$
:$y = \map f x$
Consider the right circular cylinder generated by revolving it about the $x$-axis.
By Volume of Right Circular Cylinder, the volume of this cylinder is:
:$V_\xi = \pi \paren {\map f x}^2 \delta x$
{{finish|Needs finishing off, needs a rigorous treatment.}}
Theorem: Acceleration is Second Derivative of Displacement with respect to Time

The '''acceleration''' $\mathbf a$ of a body $M$ is the second derivative of the displacement $\mathbf s$ of $M$ from a given point of reference {{WRT|Differentiation}} time $t$:
:$\mathbf a = \dfrac {\d^2 \mathbf s} {\d t^2}$

Proof:
By definition, the acceleration of a body $M$ is defined as the first derivative of the velocity $\mathbf v$ of $M$ relative to a given point of reference {{WRT|Differentiation}} time:
:$\mathbf a = \dfrac {\d \mathbf v} {\d t}$
Also by definition, the velocity of $M$ is defined as the first derivative of the displacement of $M$ from a given point of reference {{WRT|Differentiation}} time:
:$\mathbf v = \dfrac {\d \mathbf s} {\d t}$
That is:
:$\mathbf a = \map {\dfrac \d {\d t} } {\dfrac {\d \mathbf s} {\d t} }$
Hence the result by definition of the second derivative.
{{qed}}
Theorem: Equation of Catenary/Cartesian/Formulation 2

The '''catenary''' is described by the equation:
:$y = \dfrac a 2 \paren {e^{x / a} + e^{-x / a} } = a \cosh \dfrac x a$
where $a$ is a constant.
The lowest point of the chain is at $\tuple {0, a}$.

Proof:
Take the equation of the catenary according to Formulation 1:
:$y = \dfrac {e^{ax} + e^{-ax}} {2 a}$
Put this in a form which uses the hyperbolic cosine:
:$y = \dfrac {\cosh a x} a$
Replace $a$ with $\dfrac 1 a$:
:$y = a \cosh \dfrac x a$
Hence the result.
{{qed}}
Theorem: Cardinality of Power Set is Invariant

Let $X, Y$ be sets.
Let $\card X = \card Y$
where $\card X$ denotes the cardinality of $X$.
Then:
:$\card {\powerset X} = \card {\powerset Y}$
where $\powerset X$ denotes the power set of $X$.

Proof:
By definition of cardinality:
:$X \sim Y$
where $\sim$ denotes the set equivalence.
Then by definition of set equivalence:
: there exists a bijection $f: X \to Y$
By definition of bijection
:$f$ is an injection and a surjection.
By Mapping is Injection iff Direct Image Mapping is Injection:
:the direct image mapping $\map {f^\to}: \powerset X \to \powerset Y$ is an injection.
By Direct Image Mapping of Surjection is Surjection:
:$f^\to$ is a surjection.
Then by definition of bijection:
:$f^\to: \powerset X \to \powerset Y$ is a bijection.
Hence by definition of set equivalence:
:$\powerset X \sim \powerset Y$
Thus the result by definition of cardinality:
:$\card {\powerset X} = \card {\powerset Y}$
{{qed}}