avewright/proofwiki-math huggingface dataset:
Theorem: Continuous Mappings preserve Convergent Sequences

Let $X, Y$ be normed vector spaces.
Let $c \in X$.
Let $f : X \to Y$ be a mapping.
Then $f$ is continuous at $c$ iff for every sequence $\sequence {x_n}_{n \mathop \in \N} \in X$ such that $\sequence {x_n}_{n \mathop \in \N}$ converges to $c$, $\sequence {\map f {x_n}}_{n \mathop \in \N}$ converges to $\map f c$.

Proof:
=== Necessary Condition ===
Let $f$ be continuous at $c$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$ such that $\sequence {x_n}_{n \mathop \in \N}$ converges to $c$.
Let $\epsilon \in \R_{\mathop > 0}$.
By definition of continuous mapping:
:$\exists \delta \in \R_{\mathop > 0} : \forall x \in X : \norm {x - c} < \delta \implies \norm {\map f x - \map f c} < \epsilon$
Since $\sequence {x_n}_{n \mathop \in \N}$ converges to $c$:
:$\exists N \in \N : \forall n \in \N : n > \N \implies \norm {x_n - c} < \delta$
By definition of continuity, and because $\sequence {x_n}_{n \mathop \in \N} \in X$:
:$\norm {\map f {x_n} - \map f c} < \epsilon$
Choice of $\epsilon$ was arbitrary.
Hence:
:$\forall \epsilon \in \R_{\mathop > 0} : \exists N \in \N : \forall n \in \N : n > \N \implies \norm {\map f {x_n} - \map f c} < \epsilon$
By definition, $\sequence {\map f {x_n}}_{n \mathop \in \N}$ converges to $\map f c$.
{{qed|lemma}}
Theorem: Zero Product of Numbers implies Factors are Zero

On all the number systems:
:natural numbers $\N$
:integers $\Z$
:rational numbers $\Q$
:real numbers $\R$
:complex numbers $\C$
the following holds.
Let $a \times b = 0$.
Then either $a = 0$ or $b = 0$.

Proof:
From Natural Numbers have No Proper Zero Divisors
:$\forall a, b \in \N: a \times b = 0 \implies a = 0 \text { or } b = 0$
We have:
:Integers form Integral Domain
:Rational Numbers form Integral Domain
:Real Numbers form Integral Domain
:Complex Numbers form Integral Domain
Hence by definition of integral domain:
:$a \times b = 0 \implies a = 0 \text { or } b = 0$
where $a, b \in \Z, \Q, \R, \C$.
{{qed}}
Theorem: Real Numbers form only Ordered Field which is Complete

The set of real numbers $\R$ is the only ordered field which also satisfies the Continuum Property.

Proof:
From Real Numbers form Totally Ordered Field we have that $\R$ forms an totally ordered field.
From the Continuum Property we have that $\R$ is complete.
It remains to be shown that any ordered field which also satisfies the Continuum Property is isomorphic to $\R$.
{{ProofWanted}}
Theorem: Sine of Integer Multiple of Argument/Formulation 4

{{begin-eqn}}
{{eqn | l = \map \sin {n \theta}
      | r = \paren {2 \cos \theta } \map \sin {\paren {n - 1 } \theta}  - \map \sin {\paren {n - 2 } \theta} 
}}
{{end-eqn}}

Proof:
{{begin-eqn}}
To proceed, we will require the following lemma:
Theorem: Continuous Mappings preserve Compact Subsets

Let $X, Y$ be normed vector spaces.
Let $K \subseteq X$ be a compact subset.
Suppose $f : X \to Y$ is a continuous mapping at each $x \in K$.
Then $\map f K$ is a compact subset of $Y$.

Proof:
Let $\sequence {y_n}_{n \mathop \in \N}$ be a sequence contained in $\map f K$.
Then:
:$\forall n \in \N : \exists x_n \in K : y_n = \map f {x_n}$
$K$ is compact.
By definition, there is a convergent subsequence $\sequence {x_{n_k}}_{k \mathop \in \N}$ convergent with the limit $L \in K$.
$f$ is a continuous mapping.
By Continuous Mappings preserve Convergent Sequences, $\sequence {\map f {x_{n_k}}}_{k \mathop \in \N} = \sequence {y_{n_k}}_{k \mathop \in \N}$ is convergent with limit $\map f L \in \map f K$.
By definition, $\map f K$ is compact.
{{qed}}
Theorem: Sine of Integer Multiple of Argument/Formulation 5

{{begin-eqn}}
{{eqn | l = \sin n \theta
      | r = \paren {\sin \frac {n \pi} 2} \paren {\sin \theta} + \paren {2 \cos \theta} \paren {\map \sin {\paren {n - 1} \theta} - \map \sin {\paren {n - 3} \theta} + \map \sin {\paren {n - 5} \theta} - \cdots}
      | c = 
}}
{{eqn | r = \paren {\sin \frac {n \pi} 2} \paren {\sin \theta} + 2 \cos \theta \paren {\sum_{k \mathop = 0}^n \paren {\sin \frac {k \pi} 2} \map \sin {\paren {n - k} \theta} }
      | c = 
}}
{{end-eqn}}

Proof:
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\ds \sin n \theta = \paren {\sin \frac {n \pi} 2} \paren {\sin \theta} + 2 \cos \theta \paren {\sum_{k \mathop = 0}^n \paren {\sin \frac {k \pi} 2} \map \sin {\paren {n - k} \theta} }$
Theorem: Equal Surfaces do not Intersect

Let $R$ be a region of space, which may be the interior of a body.
Let there exist a point-function $F$ on $R$ giving rise to a scalar field.
Let $S_1$ and $S_2$ be equal surfaces in $R$ upon which the value of $F$ on $S_1$ is different from the value of $F$ on $S_2$.
Then $S_1$ and $S_2$ do not intersect.

Proof:
Let:
:$\forall p \in S_1: \map F p = C_1$
:$\forall p \in S_2: \map F p = C_2$
By hypothesis, $C_1 \ne C_2$.
{{AimForCont}} there exists a point $P$ in $R$ such that both $P \in S_1$ and $P \in S_2$.
Then $\map F p = C_1$ and also $\map F p = C_2$.
This contradicts the fact that $F$ is a function.
Hence the result by Proof by Contradiction.
{{qed}}
Theorem: Dot Product with Zero Vector is Zero

:$\mathbf u \cdot \mathbf 0 = 0$
where $\mathbf 0$ denotes the zero vector.

Proof:
By definition of dot product:
{{begin-eqn}}
{{eqn | l = \mathbf u \cdot \mathbf 0
      | r = \norm {\mathbf u} \norm {\mathbf 0} \cos \theta
      | c = {{Defof|Dot Product|subdef = Real Euclidean Space}}
}}
{{eqn | r = \norm {\mathbf u} \times 0 \times \cos \theta
      | c = 
}}
{{eqn | r = 0
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Dot Product of Like Vectors

Let $\mathbf a$ and $\mathbf b$ be vector quantities such that $\mathbf a$ and $\mathbf b$ are like.
Let $\mathbf a \cdot \mathbf b$ denote the dot product of $\mathbf a$ and $\mathbf b$.
Then:
:$\mathbf a \cdot \mathbf b = \norm {\mathbf a} \norm {\mathbf b}$
where $\norm {\, \cdot \,}$ denotes the magnitude of a vector.

Proof:
By definition of dot product:
:$\mathbf a \cdot \mathbf b = \norm {\mathbf a} \norm {\mathbf b} \cos \theta$
where $\theta$ is the angle between $\mathbf a$ and $\mathbf b$.
When $\mathbf a$ and $\mathbf b$ are like, by definition $\theta = 0$.
The result follows by Cosine of Zero is One, which gives that $\cos 0 \degrees = 1$.
{{qed}}
Theorem: Dot Product of Elements of Standard Ordered Basis

Let $\tuple {\mathbf i, \mathbf j, \mathbf k}$ be the standard ordered basis of Cartesian $3$-space $S$.
Then:
:$\mathbf i \cdot \mathbf j = \mathbf j \cdot \mathbf k = \mathbf k \cdot \mathbf i = 0$
where $\cdot$ denotes the dot product.

Proof:
By definition, the Cartesian $3$-space is a frame of reference consisting of a rectangular coordinate system.
By definition of rectangular coordinate system, the coordinate axes are perpendicular to each other.
By definition of Component of Vector in $3$-space, the vectors $\mathbf i$, $\mathbf j$ and $\mathbf k$ are the unit vectors in the direction of the $x$-axis, $y$-axis and $z$-axis respectively.
Hence:
:$\mathbf i$ is perpendicular to $\mathbf j$
:$\mathbf j$ is perpendicular to $\mathbf k$
:$\mathbf k$ is perpendicular to $\mathbf i$
The result follows from Dot Product of Perpendicular Vectors.
{{qed}}
Theorem: Self-Product of Standard Ordered Basis Element equals 1

Let $\tuple {\mathbf i, \mathbf j, \mathbf k}$ be the standard ordered basis of Cartesian $3$-space $S$.
Then:
:$\mathbf i^2 = \mathbf j^2 = \mathbf k^2 = 1$
where $\mathbf i^2$ and so on denotes the square of a vector quantity:
:$\mathbf i^2 := \mathbf i \cdot \mathbf i$

Proof:
By definition, the Cartesian $3$-space is a frame of reference consisting of a rectangular coordinate system.
By definition of Component of Vector in $3$-space, the vectors $\mathbf i$, $\mathbf j$ and $\mathbf k$ are the unit vectors in the direction of the $x$-axis, $y$-axis and $z$-axis respectively.
Hence $\mathbf i^2$ is the square of a unit vector:
:$\mathbf i^2 = \norm {\mathbf i}^2 = 1^2 = 1$
and the same for $\mathbf j^2$ and $\mathbf k^2$.
{{qed}}
Theorem: Equation of Plane/Vector Form

Let $P$ be a plane which passes through a point $C$ whose position vector relative to the origin $O$ is $\mathbf c$.
Let $\mathbf p$ be the vector perpendicular to $P$ from $O$.
Let $\mathbf r$ be the position vector of an arbitrary point on $P$.
Then $P$ can be represented by the equation:
:$\mathbf p \cdot \paren {\mathbf r - \mathbf c} = 0$
where $\cdot$ denotes dot product.

Proof:
:600px
It is seen that $\mathbf r - \mathbf c$ lies entirely within the plane $P$.
As $P$ is perpendicular to $\mathbf p$, it follows that $\mathbf r - \mathbf c$ is perpendicular to $\mathbf p$.
Hence by Dot Product of Perpendicular Vectors:
:$\mathbf p \cdot \paren {\mathbf r - \mathbf c} = 0$
{{qed}}
Theorem: Resultant in Terms of Dot Product

Let $\mathbf a$ and $\mathbf b$ be vector quantities.
Let their resultant be $\mathbf v$:
:$\mathbf v = \mathbf a + \mathbf b$
Then:
:$\mathbf v^2 = \mathbf a^2 + 2 \mathbf a \cdot \mathbf b + \mathbf b^2$
where:
:$\mathbf v^2$ denotes the square of $\mathbf v$
:$\mathbf a \cdot \mathbf b$ denotes the dot product of $\mathbf a$ and $\mathbf b$

Proof:
{{begin-eqn}}
{{eqn | l = \mathbf v^2
      | r = \paren {\mathbf a + \mathbf b}^2
      | c = 
}}
{{eqn | r = \paren {\mathbf a + \mathbf b} \paren {\mathbf a + \mathbf b}
      | c = 
}}
{{eqn | r = \mathbf a \cdot \mathbf a + \mathbf a \cdot \mathbf b + \mathbf b \cdot \mathbf a + \mathbf b \cdot \mathbf b
      | c = 
}}
{{eqn | r = \mathbf a^2 + 2 \mathbf a \cdot \mathbf b + \mathbf b^2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Components of Vector in Terms of Dot Product

Let $\mathbf a$ be a vector quantity embedded in Cartesian $3$-space $S$.
Let $\mathbf i$, $\mathbf j$ and $\mathbf k$ be the unit vectors in the positive directions of the $x$-axis, $y$-axis and $z$-axis respectively.
Let $\mathbf a$ be expressed in component form:
:$\mathbf a = x \mathbf i + y \mathbf j + z \mathbf k$
Then:
{{begin-eqn}}
{{eqn | l = x
      | r = \mathbf a \cdot \mathbf i
      | c = 
}}
{{eqn | l = y
      | r = \mathbf a \cdot \mathbf j
      | c = 
}}
{{eqn | l = z
      | r = \mathbf a \cdot \mathbf k
      | c = 
}}
{{end-eqn}}

Proof:
Let $\mathbf a$ be a vector as described.
From the definition of cartesian space (by axes) and from the fact that $\mathbf i, \mathbf j, \mathbf k$ are unit vectors, it follows that $\tuple {\mathbf i, \mathbf j,\mathbf k}$  is an orthonormal basis.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf a \cdot \mathbf i
      | r = \paren {x \mathbf i + y \mathbf j + z \mathbf k} \cdot \mathbf i
}}
{{eqn | r = x \paren {\mathbf i \cdot \mathbf i} + y \paren {\mathbf j \cdot \mathbf i} + z \paren {\mathbf k \cdot \mathbf i}
      | c = Dot Product Operator is Bilinear
}}
{{eqn | r = x
      | c = {{Defof|Orthonormal Basis}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \mathbf a \cdot \mathbf j
      | r = \paren {x \mathbf i + y \mathbf j + z \mathbf k} \cdot \mathbf j
}}
{{eqn | r = x \paren {\mathbf i \cdot \mathbf j} + y \paren {\mathbf j \cdot \mathbf j} + z \paren {\mathbf k \cdot \mathbf j}
      | c = Dot Product Operator is Bilinear
}}
{{eqn | r = y
      | c = {{Defof|Orthonormal Basis}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \mathbf a \cdot \mathbf k
      | r = \paren {x \mathbf i + y \mathbf j + z \mathbf k} \cdot \mathbf k
}}
{{eqn | r = x \paren {\mathbf i \cdot \mathbf k} + y \paren {\mathbf j \cdot \mathbf k} + z \paren {\mathbf k \cdot \mathbf k}
      | c = Dot Product Operator is Bilinear
}}
{{eqn | r = z
      | c = {{Defof|Orthonormal Basis}}
}}
{{end-eqn}}
{{qed}}
Theorem: Electromotive Force in Closed Path in Electric Field is Zero

Let $\mathbf E$ be an electric field acting over a region of space $R$.
Let $\Gamma$ be a closed contour in $R$.
Then the electromotive force in $\Gamma$ is zero.

Proof:
In an electrostatic context, electric fields and magnetic fields are constant in time.
From Maxwell-Faraday Equation, we have
{{begin-eqn}}
{{eqn | l = \nabla \times \mathbf E
      | r = -\frac {\partial \mathbf B} {\partial t}
}}
{{eqn | r = 0
      | c = Magnetic field is static
}}
{{end-eqn}}
Therefore, by Definition 1, $\mathbf E$ is a conservative vector field.
Thus, by Definition 2, the integral of the electric field around a closed contour is
:$\ds \oint \mathbf E \cdot \rd \mathbf l = 0$
Since the electromotive force is defined as the line integral of the electric field, and the path of integration $\Gamma$ is closed, the result follows.
{{qed}}
Theorem: Extreme Value Theorem/Normed Vector Space

Let $X$ be a normed vector space.
Let $K \subseteq X$ be a compact subset.
Suppose $f : X \to \R$ is a continuous mapping at each $x \in K$.
Then:
:$\ds \exists c \in K : \map f c = \sup_{x \mathop \in K} \map f x = \max_{x \mathop \in K} \map f x$
:$\ds \exists d \in K : \map f d = \inf_{x \mathop \in K} \map f x = \min_{x \mathop \in K} \map f x$

Proof:
Let $K$ be compact.
By Continuous Mappings preserve Compact Subsets, $\map f K$ is compact.
By Compact Subset of Normed Vector Space is Closed and Bounded, $K$ is bounded.
Hence, $\map f K$ is bounded.
$K$ is nonempty, so $\map f K$ is non-empty.
By Characterizing Property of Supremum of Subset of Real Numbers, non-empty bounded subset of $\R$ has least upper bound.
Let $\ds M := \sup_{x \mathop \in K} \map f x$.
Then $M \in \R$.
Let $n \in \N$.
We have that:
$\ds M - \frac 1 n < M$
However, $M$ is the least upper bound.
Hence, $\ds M - \frac 1 n$ cannot be an upper bound for $\map f K$.
Therefore:
:$\ds \exists x_n \in K : \map f {x_n} > M - \frac 1 n$
Thus we have a sequence $\sequence {x_n}_{n \mathop \in \N}$ contained in $K$.
$K$ is compact.
By definition, there is a convergent subsequence $\sequence {x_{n_k}}_{k \mathop \in \N}$ with a limit $c \in K$.
$f$ is continuous.
By Continuous Mappings preserve Convergent Sequences, $\sequence {\map f {x_{n_k}}}_{k \mathop \in \N}$ converges with a limit $\map f c$.
But from $\ds \map f {x_n} > M - \frac 1 n$ it follows that $\map f c \ge M$.
On the other hand, by definition of $M$, $\map f c \le M$.
Therefore, $\map f c = M$.
Finally, since $c \in K$:
:$\ds \map f c \in \set {\map f x : x \in K}$.
Hence, $M$ is a maximum.
{{qed|lemma}}
For the second part, let $g : X \to \R$ be such that:
:$\forall x \in X : \map g x := - \map f x$
Then above arguments apply, if infima, lower bounds and minimum are replaced with suprema, upper bounds and maximum.
{{qed}}
Theorem: Sine of Integer Multiple of Argument/Formulation 7

{{begin-eqn}}
{{eqn | l = \sin n \theta
      | r = \paren {\sin^2 \frac {n \pi} 2} \paren {\sin \theta} + \paren {2 \sin \theta } \paren {\paren {0} \map \cos {\paren {n - 0} \theta} + \paren {1} \map \cos {\paren {n - 1} \theta} + \paren {0} \map \cos {\paren {n - 2} \theta} + \paren {1} \map \cos {\paren {n - 3} \theta} + \paren {0} \map \cos {\paren {n - 4} \theta} + \paren {1} \map \cos {\paren {n - 5} \theta} - \cdots}
      | c = 
}}
{{eqn | r = \paren {\sin^2 \frac {n \pi} 2} \paren {\sin \theta} + 2 \sin \theta \paren {\sum_{k \mathop = 0}^{n - 1} \paren {\sin^2 \frac {k \pi} 2} \map \cos {\paren {n - k} \theta} }
      | c = 
}}
{{end-eqn}}

Proof:
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\ds \sin n \theta = \paren {\sin^2 \frac {n \pi} 2} \paren {\sin \theta} + 2 \sin \theta \paren {\sum_{k \mathop = 0}^{n - 1} \paren {\sin^2 \frac {k \pi} 2} \map \cos {\paren {n - k} \theta} }$
Theorem: Cosine of Integer Multiple of Argument/Formulation 5

{{begin-eqn}}
{{eqn | l = \cos n \theta
      | r = \map \sin {\frac {\paren {n + 1} \pi} 2} + \paren {\sin \frac {n \pi} 2} \cos \theta + \paren {2 \cos \theta} \paren {\map \cos {\paren {n - 1} \theta} - \map \cos {\paren {n - 3} \theta} + \map \cos {\paren {n - 5} \theta} - \cdots}
      | c = 
}}
{{eqn | r = \map \sin {\frac {\paren {n + 1} \pi} 2} + \paren {\sin \frac {n \pi} 2} \cos \theta + 2 \cos \theta \paren {\sum_{k \mathop = 0}^{n - 1} \paren {\sin \frac {k \pi} 2} \map \cos {\paren {n - k} \theta} }
      | c = 
}}
{{end-eqn}}

Proof:
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\ds \cos n \theta = \map \sin {\frac {\paren {n + 1} \pi} 2} + \paren {\sin \frac {n \pi} 2} \cos \theta + 2 \cos \theta \paren {\sum_{k \mathop = 0}^{n - 1} \paren {\sin \frac {k \pi} 2} \map \cos {\paren {n - k} \theta} }$
Theorem: Cosine of Integer Multiple of Argument/Formulation 6

:$\map \cos {n \theta} = \paren {-2 \sin \theta} \map \sin {\paren {n - 1} \theta}  + \map \cos {\paren {n - 2} \theta}$

Proof:
{{begin-eqn}}
{{eqn | l = \map \cos {n \theta} 
      | r = \map \cos {\paren {n - 1} \theta + \theta}
      | c =
}}
{{eqn | r = -\sin \theta \map \sin {\paren {n - 1} \theta} + \cos \theta \map \cos {\paren {n - 1} \theta} 
      | c = Cosine of Sum
}}
{{eqn | r = - \sin \theta \map \sin {\paren {n - 1} \theta} + \cos \theta \paren {-\sin \theta \map \sin {\paren {n - 2} \theta} + \cos \theta \map \cos {\paren {n - 2} \theta} }
      | c = Cosine of Sum
}}
{{eqn | r = -\sin \theta \map \sin {\paren {n - 1} \theta} - \sin \theta \cos \theta \map \sin {\paren {n - 2} \theta} + \cos^2 \theta \map \cos {\paren {n - 2} \theta}
      | c = 
}}
{{eqn | r = -\sin \theta \map \sin {\paren {n - 1} \theta} - \sin \theta \cos \theta \map \sin {\paren {n - 2} \theta} + \paren {1 - \sin^2 \theta } \map \cos {\paren {n - 2} \theta}
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = -\sin \theta \map \sin {\paren {n - 1} \theta} - \sin \theta \paren {\cos \theta \map \sin {\paren {n - 2} \theta} + \sin \theta \map \cos {\paren {n - 2} \theta} } + \map \cos {\paren {n - 2} \theta}
      | c = 
}}
{{eqn | r = -\sin \theta \map \sin {\paren {n - 1} \theta} - \sin \theta \map \sin {\paren {n - 1} \theta} + \map \cos {\paren {n - 2} \theta}
      | c = Sine of Sum
}}
{{eqn | r = \paren {-2 \sin \theta} \map \sin {\paren {n - 1} \theta}  + \map \cos {\paren {n - 2} \theta} 
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Cosine of Integer Multiple of Argument
o7s77hxc0qmg9l2tw31bmdsv1a68wi0
Theorem: Cross Product of Perpendicular Vectors

Let $\mathbf a$ and $\mathbf b$ be vector quantities which are perpendicular.
Let $\mathbf a \times \mathbf b$ denote the cross product of $\mathbf a$ with $\mathbf b$.
Then:
:$\mathbf a \times \mathbf b = \norm {\mathbf a} \norm {\mathbf b} \mathbf {\hat n}$
where:
:$\norm {\mathbf a}$ denotes the length of $\mathbf a$
:$\hat {\mathbf n}$ is the unit vector perpendicular to both $\mathbf a$ and $\mathbf b$ in the direction according to the right-hand rule.

Proof:
By definition of cross product:
:$\norm {\mathbf a} \norm {\mathbf b} \sin \theta \, \mathbf {\hat n}$
where:
:$\norm {\mathbf a}$ denotes the length of $\mathbf a$
:$\theta$ denotes the angle from $\mathbf a$ to $\mathbf b$, measured in the positive direction
:$\hat {\mathbf n}$ is the unit vector perpendicular to both $\mathbf a$ and $\mathbf b$ in the direction according to the right-hand rule.
When $\mathbf a$ and $\mathbf b$ are perpendicular, $\theta = 90 \degrees$ by definition.
Hence from Sine of Right Angle, $\sin \theta = 1$.
Hence the result.
{{qed}}
Theorem: Cross Product of Elements of Standard Ordered Basis

Let $\tuple {\mathbf i, \mathbf j, \mathbf k}$ be the standard ordered basis of Cartesian $3$-space $S$.
Then:
:$\mathbf i \times \mathbf i = \mathbf j \times \mathbf j = \mathbf k \times \mathbf k = 0$
and:
{{begin-eqn}}
{{eqn | l = \mathbf i \times \mathbf j
      | m = \mathbf k
      | mo= =
      | r = -\mathbf j \times \mathbf i
}}
{{eqn | l = \mathbf j \times \mathbf k
      | m = \mathbf i
      | mo= =
      | r = -\mathbf k \times \mathbf j
}}
{{eqn | l = \mathbf k \times \mathbf i
      | m = \mathbf j
      | mo= =
      | r = -\mathbf i \times \mathbf k
}}
{{end-eqn}}
where $\times$ denotes the dot product.

Proof:
From Cross Product of Vector with Itself is Zero:
:$\mathbf i \times \mathbf i = \mathbf j \times \mathbf j = \mathbf k \times \mathbf k = 0$
Then we can take the definition of cross product:
:$\mathbf a \times \mathbf b = \begin {vmatrix}
\mathbf i & \mathbf j & \mathbf k \\
a_i & a_j & a_k \\
b_i & b_j & b_k \\
\end {vmatrix} = \mathbf a \times \mathbf b = \paren {a_j b_k - a_k b_j} \mathbf i - \paren {a_i b_k - a_k b_i} \mathbf j + \paren {a_i b_j - a_j b_i} \mathbf k$
and note that:
{{begin-eqn}}
{{eqn | l = \mathbf i 
      | r = 1 \mathbf i + 0 \mathbf j + 0 \mathbf k
}}
{{eqn | l = \mathbf j 
      | r = 0 \mathbf i + 1 \mathbf j + 0 \mathbf k
}}
{{eqn | l = \mathbf k 
      | r = 0 \mathbf i + 0 \mathbf j + 1 \mathbf k
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \mathbf i \times \mathbf j
      | r = \paren {0 \cdot 0 - 0 \cdot 1} \mathbf i - \paren {1 \cdot 0 - 0 \cdot 0} \mathbf j + \paren {1 \cdot 1 - 0 \cdot 0} \mathbf k
      | rr= = \mathbf k
}}
{{eqn | l = \mathbf j \times \mathbf k
      | r = \paren {1 \cdot 1 - 0 \cdot 0} \mathbf i - \paren {0 \cdot 1 - 0 \cdot 0} \mathbf j + \paren {0 \cdot 0 - 1 \cdot 0} \mathbf k
      | rr= = \mathbf i
}}
{{eqn | l = \mathbf k \times \mathbf i
      | r = \paren {0 \cdot 0 - 1 \cdot 0} \mathbf i - \paren {0 \cdot 0 - 1 \cdot 1} \mathbf j + \paren {0 \cdot 0 - 0 \cdot 1} \mathbf k
      | rr= = \mathbf j
}}
{{end-eqn}}
The remaining identities follow from Vector Cross Product is Anticommutative:
{{begin-eqn}}
{{eqn | l = \mathbf i \times \mathbf j
      | r = -\mathbf j \times \mathbf i
}}
{{eqn | l = \mathbf j \times \mathbf k
      | r = -\mathbf k \times \mathbf j
}}
{{eqn | l = \mathbf k \times \mathbf i
      | r = -\mathbf i \times \mathbf k
}}
{{end-eqn}}
{{qed}}
Theorem: Total Vector Area of Tetrahedron is Zero

Let $T$ be a tetrahedron whose faces have vector area $\mathbf S_1$, $\mathbf S_2$, $\mathbf S_3$ and $\mathbf S_4$.
Let the positive direction be defined as outward.
:300px
Then the total vector area is zero:
:$\mathbf S_1 + \mathbf S_2 + \mathbf S_3 + \mathbf S_4 = \mathbf 0$

Proof:
Let the vector areas be resolved upon the faces of $T$.
Some of the projections will be positive and some negative.
Let $T$ be imagined in a fluid which is in equilibrium under pressure.
Each face experiences a force which is normal to its plane and proportional to its area.
Because the fluid inside is in equilibrium with the fluid outside, the resultant of the forces on its faces will equal zero.
Hence the sum of all the vector areas is also zero, as the pressure is the same on all faces.
Hence the result.
{{qed}}
Theorem: Total Vector Area of Polyhedron is Zero

Let $P$ be a polyhedron.
Let the positive direction be defined as outward.
Let $\mathbf T$ be the total vector area of all the faces of $P$.
Then:
:$\mathbf T = \mathbf 0$

Proof:
$P$ can be geometrically divided into a finite number of tetrahedra.
Every face of these tetrahedra which are internal to $P$ appears twice: once with a positive vector area, and once with a negative normal.
Hence for any polyhedron, the total vector area is the sum of the vector areas of all the tetrahedra.
The result follows from Total Vector Area of Tetrahedron is Zero.
{{qed}}
Theorem: Total Vector Area of Closed Surface is Zero

Let $S$ be a closed surface.
Let $\d \mathbf S$ be an infinitesimal vector area around some point $P$ of $S$.
Then the total surface area of $S$ is given by:
:$\ds \iint_S \d \mathbf S = \mathbf 0$

Proof:
{{ProofWanted|The book waffles on about slicing $S$ into vanishingly small tetrahedra, but the argument is flimsy.}}
Theorem: Vectors are Coplanar iff Scalar Triple Product equals Zero

Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be vectors in a Cartesian $3$-space:
Let $\mathbf a \cdot \paren {\mathbf b \times \mathbf c}$ denote the scalar triple product of $\mathbf a$, $\mathbf b$ and $\mathbf c$.
Then:
:$\mathbf a \cdot \paren {\mathbf b \times \mathbf c} = 0$
{{iff}} $\mathbf a$, $\mathbf b$ and $\mathbf c$ are coplanar.

Proof:
From Magnitude of Scalar Triple Product equals Volume of Parallelepiped Contained by Vectors:
:$\size {\mathbf a \cdot \paren {\mathbf b \times \mathbf c} }$ equals the volume of the parallelepiped contained by $\mathbf a, \mathbf b, \mathbf c$.
The result follows.
{{qed}}