avewright/proofwiki-math huggingface dataset:
Theorem: Symmetric Difference with Union does not form Ring

Let $S$ be a set.
Let:
:$\symdif$ denote the symmetric difference operation
:$\cup$ denote the set union operation
:$\powerset S$ denote the power set of $S$.
Then $\struct {\powerset S, \symdif, \cup}$ does not form a ring.

Proof:
For $\struct {S, \symdif, \cup}$ to be a ring, it is a necessary condition that $\cup$ be distributive over $*$.
Also, the identity element for set union and symmetric difference must be different.
However:
:$(1): \quad$ the identity for union and symmetric difference is $\O$ for both operations
:$(2): \quad$ set union is not distributive over symmetric difference:
From Symmetric Difference of Unions:
:$\paren {R \cup T} \symdif \paren {S \cup T} = \paren {R \symdif S} \setminus T$
The result follows.
{{qed}}
Theorem: Set of Order m times n Matrices does not form Ring

Let $m, n \in \N_{>0}$ be non-zero natural numbers such that $m > n$.
Let $S$ be the set of all matrices of order $m \times n$.
Then the algebraic structure $\struct {S, +, \times}$ is not a ring.
Note that $\times$ denotes conventional matrix multiplication.

Proof:
For $\struct {S, +, \times}$ to be a ring, it is a necessary condition that $\struct {S, \times}$ is a semigroup.
For $\struct {S, \times}$ to be a semigroup, it is a necessary condition that $\struct {S, \times}$ is closed.
That is:
:$\forall x, y \in S: x \times y \in S$
Let $\mathbf A$ and $\mathbf B$ be elements of $S$.
The matrix multiplication operation is defined on $\mathbf A$ and $\mathbf B$ {{iff}} $\mathbf A$ is of order $m \times p$ and $\mathbf A$ is of order $p \times n$, for some $m, n, p \in \N_{>0}$.
That is, the second dimension of $\mathbf A$ must be equal to the first dimension of $\mathbf B$.
But in this case, the second dimension of $\mathbf A$ is $n$, and the first dimension of $\mathbf B$ is $m$.
As we have been given that $m \ne n$, it follows that matrix multiplication operation is not defined on $S$.
Hence the result.
{{qed}}
Theorem: Set of Order 3 Vectors under Cross Product does not form Ring

Let $S$ be the set of all vectors in a vector space of dimension $3$.
Let $\times$ denote the cross product operation.
Then the algebraic structure $\struct {S, +, \times}$ is not a ring.

Proof:
For $\struct {S, +, \times}$ to be a ring, it is a necessary condition that $\struct {S, \times}$ is a semigroup.
For $\struct {S, \times}$ to be a semigroup, it is a necessary condition that $\times$ is associative on $S$.
However, from Vector Cross Product is not Associative, this is not the case here.
The result follows.
{{qed}}
Theorem: Integers under Subtraction do not form Semigroup

Let $\struct {\Z, -}$ denote the algebraic structure formed by the set of integers under the operation of subtraction.
Then $\struct {\Z, -}$ is not a semigroup.

Proof:
It is to be demonstrated that $\struct {\Z, -}$ does not satisfy the semigroup axioms.
We then have Subtraction on Numbers is Not Associative.
So, for example:
:$3 - \paren {2 - 1} = 2 \ne \paren {3 - 2} - 1 = 0$
Thus it has been demonstrated that $\struct {\Z, -}$ does not satisfy {{Semigroup-axiom|1}}.
Hence the result.
{{qed}}
Theorem: Cardinality of Maximal Independent Subset Equals Rank of Set

Let $M = \struct{S, \mathscr I}$ be a matroid.
Let $A \subseteq S$.
Let $X$ be a maximal independent subset of $A$.
Then:
:$\card X = \map \rho A$
where $\rho$ is the rank function on $M$.

Proof:
From Independent Subset is Contained in Maximal Independent Subset:
:$\exists Y \in \mathscr I : X \subseteq Y \subseteq A : \card Y = \map \rho A$
By definition of a maximal independent Subset of $A$:
:$X = Y$
The result follows.
{{qed}}
Category:Matroid Independent Subsets
Category:Matroid Rank Functions
n60ritlugze101due1fljmp5t5kjm87
Theorem: Motion of Body with Constant Mass

Let $B$ be a body with constant mass $m$ undergoing a force $\mathbf F$.
Then the equation of motion of $B$ is given by:
:$\mathbf F = m \mathbf a$
where $\mathbf a$ is the acceleration of $B$.

Proof:
{{begin-eqn}}
{{eqn | l = \mathbf F
      | r = \map {\dfrac \d {\d t} } {m \mathbf v}
      | c = Newton's Second Law of Motion
}}
{{eqn | r = m \dfrac {\d \mathbf v} {\d t} + \mathbf v \dfrac {\d m} {\d t}
      | c = 
}}
{{eqn | r = m \dfrac {\d \mathbf v} {\d t} + 0
      | c = Derivative of Constant
}}
{{eqn | r = m \mathbf a
      | c = {{Defof|Acceleration}}
}}
{{end-eqn}}
{{qed}}
Theorem: Like Vector Quantities are Multiples of Each Other

Let $\mathbf a$ and $\mathbf b$ be like vector quantities.
Then:
:$\mathbf a = \dfrac {\size {\mathbf a} } {\size {\mathbf b} } \mathbf b$
where:
:$\size {\mathbf a}$ denotes the magnitude of $\mathbf a$
:$\dfrac {\size {\mathbf a} } {\size {\mathbf b} } \mathbf b$ denotes the scalar product of $\mathbf b$ by $\dfrac {\size {\mathbf a} } {\size {\mathbf b} }$.

Proof:
By the definition of like vector quantities:
:$\mathbf a$ and $\mathbf b$ are '''like vector quantities''' {{iff}} they have the same direction.
By definition of unit vector:
:$\dfrac {\mathbf a} {\size {\mathbf a} } = \dfrac {\mathbf b} {\size {\mathbf b} }$
as both are in the same direction, and both have length $1$.
By definition of scalar division:
:$\dfrac 1 {\size {\mathbf a} } \mathbf a = \dfrac 1 {\size {\mathbf b} } \mathbf b$
Hence, multiplying by $\size {\mathbf a}$:
:$\mathbf a = \dfrac {\size {\mathbf a} } {\size {\mathbf b} } \mathbf b$
{{qed}}
Theorem: Direct Product Norm is Norm

Let $\struct {X, \norm {\, \cdot \,}}$ and $\struct {Y, \norm {\, \cdot \,}}$ be normed vector spaces.
Let $V = X \times Y$ be a direct product of vector spaces $X$ and $Y$ together with induced component-wise operations.
Let $\norm {\tuple {x, y} }$ be the direct product norm.
Then $\norm {\tuple {x, y} }$ is a norm on $V$.

Proof:
=== Positive Definiteness ===
Let $\tuple {x , y} \in V$.
Then:
{{begin-eqn}}
{{eqn | l = \norm {\tuple {x, y} } 
      | r = \map \max {\norm x, \norm y}
      | c = {{defof|Direct Product Norm}}
}}
{{eqn | o = \ge
      | r =  0
      | c = Norm Axiom $N1$: Positive Definiteness
}}
{{end-eqn}}
Suppose $\norm {\tuple {x, y} } = 0$.
Then:
{{begin-eqn}}
{{eqn | l = 0 
      | o = \le
      | r = \norm x 
      | c = Norm Axiom $N1$: Positive Definiteness      
}}
{{eqn | o = \le 
      | r = \map \max {\norm x, \norm y} 
}}
{{eqn | r = \norm {\tuple {x, y} }
      | c = {{defof|Direct Product Norm}}
}}
{{eqn | r = 0
}}
{{end-eqn}}
Hence, $\norm x = 0$
{{qed|lemma}}
Theorem: Vector Quantity as Scalar Product of Unit Vector Quantity

Let $\mathbf a$ be a vector quantity.
Then:
:$\mathbf a = \size {\mathbf a} \mathbf {\hat a}$
where:
:$\size {\mathbf a}$ denotes the magnitude of $\mathbf a$
:$\mathbf {\hat a}$ denotes the unit vector in the direction $\mathbf a$.

Proof:
{{begin-eqn}}
{{eqn | l = \size {\mathbf {\hat a} }
      | r = 1
      | c = {{Defof|Unit Vector}}
}}
{{eqn | ll= \leadsto
      | l = \mathbf {\hat a} \times \size {\mathbf {\hat a} }
      | r = \mathbf {\hat a}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \size {\mathbf a} \times \size {\mathbf {\hat a} } \times \mathbf {\hat a}
      | r = \size {\mathbf a} \times \mathbf {\hat a}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \mathbf a
      | r = \size {\mathbf a} \mathbf {\hat a}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Scalar Product of Magnitude by Unit Vector Quantity

Let $\mathbf a$ be a vector quantity.
Let $m$ be a scalar quantity.
Then:
:$m \mathbf a = m \paren {\size {\mathbf a} \hat {\mathbf a} } = \paren {m \size {\mathbf a} } \hat {\mathbf a}$
where:
:$\size {\mathbf a}$ denotes the magnitude of $\mathbf a$
:$\hat {\mathbf a}$ denotes the unit vector in the direction $\mathbf a$.

Proof:
{{begin-eqn}}
{{eqn | l = \mathbf a
      | r = \size {\mathbf a} \hat {\mathbf a}
      | c = Vector Quantity as Scalar Product of Unit Vector Quantity
}}
{{eqn | ll= \leadsto
      | l = m \mathbf a
      | r = m \paren {\size {\mathbf a} \hat {\mathbf a} }
      | c = 
}}
{{end-eqn}}
Then:
{{finish|hard to prove something trivial}}
{{qed}}
Theorem: Like Unit Vectors are Equal

Let $\mathbf a$ and $\mathbf b$ be like vector quantities.
Then:
:$\mathbf {\hat a} = \mathbf {\hat b}$
where $\mathbf {\hat a}$ and $\mathbf {\hat b}$ denote the unit vectors in the direction of $\mathbf a$ and $\mathbf b$.

Proof:
By definition of like vector quantities, $\mathbf a$ and $\mathbf b$ have the same direction.
By definition of unit vector, $\mathbf {\hat a}$ and $\mathbf {\hat b}$ are both of magnitude $1$.
Hence the result, by Equality of Vector Quantities.
{{qed}}
Category:Vectors
4dc6u0f04r1xrbj90fu5p7eaj30d0uk
Theorem: Vector Quantity can be Expressed as Sum of 3 Non-Coplanar Vectors

Let $\mathbf r$ be a vector quantity embedded in space.
Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be non-coplanar.
Then $\mathbf r$ can be expressed uniquely as the resultant of $3$ vector quantities which are each parallel to one of $\mathbf a$, $\mathbf b$ and $\mathbf c$.

Proof:
400px
Let $\mathbf {\hat a}$, $\mathbf {\hat b}$ and $\mathbf {\hat c}$ be unit vectors in the directions of $\mathbf a$, $\mathbf b$ and $\mathbf c$ respectively.
Let $O$ be a point in space.
Take $\vec {OP} := \mathbf r$.
With $OP$ as its space diagonal, construct a parallelepiped with edges $OA$, $OB$ and $OC$ parallel to $\mathbf {\hat a}$, $\mathbf {\hat b}$ and $\mathbf {\hat c}$ respectively.
Only one such parallelepiped can be so constructed.
Let $x$, $y$ and $z$ be the length of the edges $OA$, $OB$ and $OC$ respectively.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf r
      | r = \vec {OA} + \vec {AF} + \vec {FP}
      | c = 
}}
{{eqn | r = \vec {OA} + \vec {OB} + \vec {OC}
      | c = 
}}
{{eqn | r = x \mathbf {\hat a} + y \mathbf {\hat b} + z \mathbf {\hat c}
      | c = 
}}
{{end-eqn}}
Thus $\mathbf r$ is the resultant of the $3$ components $x \mathbf {\hat a}$, $y \mathbf {\hat b}$ and $z \mathbf {\hat c}$ which, by construction, are parallel to $\mathbf a$, $\mathbf b$ and $\mathbf c$ respectively.
The fact that only one parallelepiped can be constructed in the above proves uniqueness.
{{qed}}
Theorem: Vectors are Equal iff Components are Equal

Two vector quantities are equal {{iff}} they have the same components.

Proof:
Let $\mathbf a$ and $\mathbf b$ be vector quantities.
Then by Vector Quantity can be Expressed as Sum of 3 Non-Coplanar Vectors, $\mathbf a$ and $\mathbf b$ can be expressed uniquely as components.
So if $\mathbf a$ and $\mathbf b$ then the components of $\mathbf a$ are the same as the components of $\mathbf b$
Suppose $\mathbf a$ and $\mathbf b$ have the same components: $\mathbf x$, $\mathbf y$ and $\mathbf z$.
Then by definition:
:$\mathbf a = \mathbf x + \mathbf y + \mathbf z$
and also:
:$\mathbf b = \mathbf x + \mathbf y + \mathbf z$
and trivially:
:$\mathbf a = \mathbf b$
{{qed}}
Theorem: Characteristics of Birkhoff-James Orthogonality

Let $\struct {V, \norm {\,\cdot\,} }$ be a normed linear space.
Let $x, y \in V$.
Then $x$ and $y$ are '''Birkhoff-James orthogonal''' {{iff}} either:
:$(1): \quad x = 0$
or:
:$(2): \quad$ there exists a continuous functional $ f$ on $\struct {V, \norm {\,\cdot\,} }$ such that:
::::$\norm f = 1$
::::$\map f x = \norm x$
::::$\map f y = 0$

Proof:
=== Necessary Condition ===
Let $x \perp_B y$.
Let $V' \subset V$ be the subspace spanned by $x$ and $y$.
Define $\overline f$ on $V'$ as:
:$\map {\overline f} {a x + b y} = a \norm x$
for $a$ and $b$ scalars.
Clearly, $\overline f$ is linear and:
:$\map {\overline f} x = \norm x$
:$\map {\overline f} y = 0$
Further:
{{begin-eqn}}
{{eqn | l = \norm {a x + b y}
      | r = \size a \norm {x + \dfrac b a y}
      | c = 
}}
{{eqn | o = \ge
      | r = \size a \norm x
      | c = 
}}
{{eqn | r = \size {\map {\overline f} {a x + b y} }
      | c = 
}}
{{end-eqn}}
proving that $\overline f$ is a bounded functional of norm $1$.
Now by Hahn-Banach Theorem, $\overline f$ can be extended to a functional $f$ on $V$ such that $\norm f = \norm {\overline f} = 1$
This proves the necessity.
{{qed|lemma}}
Theorem: Components of Vector in terms of Direction Cosines

Let $\mathbf r$ be a vector quantity embedded in a Cartesian $3$-space.
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 $\cos \alpha$, $\cos \beta$ and $\cos \gamma$ be the direction cosines of $\mathbf r$ with respect to the $x$-axis, $y$-axis and $z$-axis respectively.
Let $x$, $y$ and $z$ be the components of $\mathbf r$ in the $\mathbf i$, $\mathbf j$ and $\mathbf k$ directions respectively.
Let $r$ denote the magnitude of $\mathbf r$, that is:
:$r := \size {\mathbf r}$
Then:
{{begin-eqn}}
{{eqn | l = x
      | r = r \cos \alpha
}}
{{eqn | l = y
      | r = r \cos \beta
}}
{{eqn | l = z
      | r = r \cos \gamma
}}
{{end-eqn}}

Proof:
:480px
By definition, the direction cosines are the cosines of the angles that $\mathbf r$ makes with the coordinate axes.
By definition of the components of $\mathbf r$:
:$\mathbf r = x \mathbf i + y \mathbf j + z \mathbf k$
Thus:
:$\mathbf r = r \cos \alpha \mathbf i + r \cos \beta \mathbf j + r \cos \gamma \mathbf k$
and the result follows.
{{qed}}
Theorem: Magnitude of Vector Quantity in terms of Components

Let $\mathbf r$ be a vector quantity embedded in a Cartesian $3$-space.
Let $\mathbf r$ be expressed in terms of its components:
:$\mathbf r = x \mathbf i + y \mathbf j + z \mathbf k$
where $\mathbf i$, $\mathbf j$ and $\mathbf k$ denote the unit vectors in the positive directions of the $x$-axis, $y$-axis and $z$-axis respectively.
Then:
:$\size {\mathbf r} = \sqrt {x^2 + y^2 + z^2}$
where $\size {\mathbf r}$ denotes the magnitude of $\mathbf r$.

Proof:
Let the initial point of $\mathbf r$ be $\tuple {x_1, y_1, z_1}$.
Let the terminal point of $\mathbf r$ be $\tuple {x_2, y_2, z_2}$.
Thus, by definition of the components of $\mathbf r$, the magnitude of $\mathbf r$ equals the distance between $\tuple {x_1, y_1, z_1}$ and $\tuple {x_2, y_2, z_2}$.
The result follows from Distance Formula in 3 Dimensions.
{{qed}}
Theorem: Components of Zero Vector Quantity are Zero

Let $\mathbf r$ be a vector quantity embedded in a Cartesian $3$-space.
Let $\mathbf r$ be expressed in terms of its components:
:$\mathbf r = x \mathbf i + y \mathbf j + z \mathbf k$
Let $\mathbf r$ be the zero vector.
Then:
:$x = y = z = 0$

Proof:
By definition of the zero vector, the magnitude of $\mathbf r$ is equal to zero.
By Magnitude of Vector Quantity in terms of Components:
:$\size {\mathbf r} = \sqrt {x^2 + y^2 + z^2} = 0$
where $\size {\mathbf r}$ denotes the magnitude of $\mathbf r$.
As each of $x$, $y$ and $z$ are real numbers, each of $x^2$, $y^2$ and $z^2$ is non-negative.
so in order for $\sqrt {x^2 + y^2 + z^2} = 0$, it must follow that each of $x$, $y$ and $z$ is zero.
Hence the result.
{{qed}}
Theorem: Unit Vector in terms of Direction Cosines

Let $\mathbf r$ be a vector quantity embedded in a Cartesian $3$-space.
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 $\cos \alpha$, $\cos \beta$ and $\cos \gamma$ be the direction cosines of $\mathbf r$ with respect to the $x$-axis, $y$-axis and $z$-axis respectively.
Let $\mathbf {\hat r}$ denote the unit vector in the direction of $\mathbf r$.
Then:
:$\mathbf {\hat r} = \paren {\cos \alpha} \mathbf i + \paren {\cos \beta} \mathbf j + \paren {\cos \gamma} \mathbf k$

Proof:
From Components of Vector in terms of Direction Cosines:
:$(1): \quad \mathbf r = r \cos \alpha \mathbf i + r \cos \beta \mathbf j + r \cos \gamma \mathbf k$
where $r$ denotes the magnitude of $\mathbf r$, that is:
:$r := \size {\mathbf r}$
By Unit Vector in Direction of Vector:
:$\mathbf {\hat r} = \dfrac {\mathbf r} r$
The result follows by multiplication of both sides of $(1)$ by $\dfrac 1 r$.
{{qed}}
Theorem: Vectors from Sum and Difference

Let $\mathbf a$ and $\mathbf b$ be vector quantities.
Let $\mathbf c = \mathbf a + \mathbf b$ and $\mathbf d = \mathbf a - \mathbf b$ be given.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf a
      | r = \dfrac 1 2 \paren {\mathbf c + \mathbf d}
}}
{{eqn | l = \mathbf b
      | r = \dfrac 1 2 \paren {\mathbf c - \mathbf d}
}}
{{end-eqn}}

Proof:
{{begin-eqn}}
{{eqn | l = \dfrac 1 2 \paren {\mathbf c + \mathbf d}
      | r = \dfrac 1 2 \paren {\paren {\mathbf a + \mathbf b} + \paren {\mathbf a - \mathbf b} }
      | c = 
}}
{{eqn | r = \dfrac 1 2 \paren {\mathbf a + \mathbf b + \mathbf a - \mathbf b}
      | c = 
}}
{{eqn | r = \dfrac 1 2 \paren {2 \mathbf a}
      | c = 
}}
{{eqn | r = \mathbf a
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \dfrac 1 2 \paren {\mathbf c - \mathbf d}
      | r = \dfrac 1 2 \paren {\paren {\mathbf a + \mathbf b} - \paren {\mathbf a - \mathbf b} }
      | c = 
}}
{{eqn | r = \dfrac 1 2 \paren {\mathbf a + \mathbf b - \mathbf a + \mathbf b}
      | c = 
}}
{{eqn | r = \dfrac 1 2 \paren {2 \mathbf b}
      | c = 
}}
{{eqn | r = \mathbf b
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Direct Product of Banach Spaces is Banach Space

Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be normed vector spaces.
Let $V = X \times Y$ be a direct product of vector spaces $X$ and $Y$ together with induced component-wise operations.
Let $\norm {\, \cdot \,}_{X \times Y}$ be the direct product norm.
Suppose $X$ and $Y$ are Banach spaces.
Then $V$ is a Banach space.

Proof:
Let $\sequence {\tuple {x_n, y_n}}_{n \mathop \in \N}$ be a Cauchy sequence in $V$:
:$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n > N: \norm {\tuple {x_n, y_n} - \tuple {x_m, y_m} }_{X \times Y} < \epsilon$
We have that:
{{begin-eqn}}
{{eqn | l = \norm {x_n - x_m}_X
      | o = \le
      | r = \map \max {\norm {x_n - x_m}_X, \norm {y_n - y_m}_Y }
}}
{{eqn | r = \norm {\tuple {x_n - x_m, y_n - y_m} }_{X \times Y}
      | c = {{Defof|Direct Product Norm}}
}}
{{eqn | r =  \norm {\tuple {x_n, y_n} - \tuple {x_m, y_m} }_{X \times Y}
      | c = induced component-wise operations
}}
{{eqn | o = <
      | r = \epsilon
}}
{{end-eqn}}
Hence, $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy sequence in $X$.
By assumption, $X$ is a Banach space.
Therefore, $\sequence {x_n}_{n \mathop \in \N}$ converges to $x \in X$.
By analogous arguments, $\sequence {y_n}_{n \mathop \in \N}$ converges to $y \in Y$.
So, by Convergence in Direct Product Norm, we have: 
:$\sequence {\tuple {x_n, y_n}}_{n \mathop \in \N}$ converges to $\tuple {x, y}$.
By definition, $V$ is a Banach space.
{{qed}}
Theorem: Cartesian Plane Rotated with respect to Another

Let $\mathbf r$ be a position vector embedded in a Cartesian plane $\CC$ with origin $O$.
Let $\CC$ be rotated anticlockwise through an angle $\varphi$ about the axis of rotation $O$.
Let $\CC'$ denote the Cartesian plane in its new position.
Let $\mathbf r$ be kept fixed during this rotation.
Let $\tuple {x, y}$ denote the components of $\mathbf r$ {{WRT}} $\CC$.
Let $\tuple {x', y'}$ denote the components of $\mathbf r$ {{WRT}} $\CC'$.
Then:
{{begin-eqn}}
{{eqn | l = x'
      | r = x \cos \varphi + y \sin \varphi
}}
{{eqn | l = y'
      | r = -x \sin \varphi + y \cos \varphi
}}
{{end-eqn}}

Proof:
:400px
Let $\mathbf r$ be represented by a directed line segment whose initial point coincides with the origin $O$.
Let the terminal point of $\mathbf r$ be identified with the point $P$.
Let $\CC$ be rotated to $\CC'$ through an angle $\varphi$ as shown, keeping $P$ fixed.
We have that:
{{begin-eqn}}
{{eqn | l = x'
      | r = OA + BP
      | c = by inspection
}}
{{eqn | r = x \cos \varphi + y \sin \varphi
      | c = {{Defof|Cosine}}, {{Defof|Sine}}
}}
{{eqn | l = y'
      | r = xB - xA
      | c = by inspection
}}
{{eqn | r = y \cos \varphi - x \sin \varphi
      | c = {{Defof|Cosine}}, {{Defof|Sine}}
}}
{{end-eqn}}
hence the result.
{{qed}}
Theorem: Dot Product of Perpendicular Vectors

Let $\mathbf a$ and $\mathbf b$ be vector quantities such that $\mathbf a \ne \bszero$ and $\mathbf b \ne \bszero$.
Let $\mathbf a \cdot \mathbf b$ denote the dot product of $\mathbf a$ and $\mathbf b$.
Then:
:$\mathbf a \cdot \mathbf b = 0$
{{iff}}:
:$\mathbf a$ and $\mathbf b$ are perpendicular.

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 $\mathbf b$ be perpendicular, by definition $\theta = 90 \degrees$.
The result follows by Cosine of Right Angle, which gives that $\cos 90 \degrees = 0$.
{{qed}}
Theorem: Dot Product of Orthonormal Basis Vectors

Let $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ be an orthonormal basis of a vector space $V$.
Then:
:$\forall i, j \in \set {1, 2, \ldots, n}: \mathbf e_i \cdot \mathbf e_j = \delta_{i j}$
where:
:$\mathbf e_i \cdot \mathbf e_j$ denotes the dot product of $\mathbf e_i$ and $\mathbf e_j$
:$\delta_{i j}$ denotes the Kronecker delta.

Proof:
By definition of orthonormal basis:
:$(1): \quad \tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ is an orthogonal basis of $V$
:$(2): \quad \norm {\mathbf e_1} = \norm {\mathbf e_2} = \cdots = \norm {\mathbf e_1} = 1$
From $(1)$ we have by definition that $\mathbf e_i$ and $\mathbf e_j$ are perpendicular whenever $i \ne j$
Thus, from Dot Product of Perpendicular Vectors, for all $i$ and $j$, when $i \ne j$:
:$\mathbf e_i \cdot \mathbf e_j = 0$
From Dot Product of Vector with Itself, for all $i$ we have:
:$\mathbf e_i \cdot \mathbf e_i = \norm {\mathbf e_i}^2$
From $(2)$, we have that:
:$\norm {\mathbf e_i} = 1$
and so:
:$\mathbf e_i \cdot \mathbf e_i = 1$
Putting this together, we have:
:$\forall i, j \in \set {1, 2, \ldots, n}: \mathbf e_i \cdot \mathbf e_j = \begin {cases} 1 & : i = j \\ 0 & : i \ne j \end {cases}$
The result follows by definition of the Kronecker delta.
{{qed}}
Theorem: Partial Derivative wrt x of sin x y over cos (x + y)

:$\dfrac \partial {\partial x} \dfrac {\sin x y} {\map \cos {x + y} } = \dfrac {y \map \cos {x + y} \cos x y + \map \sin {x + y} \sin x y} {\map {\cos^2} {x + y} }$

Proof:
{{begin-eqn}}
{{eqn | l = \frac \partial {\partial x} \frac {\sin x y} {\map \cos {x + y} }
      | r = \frac {\map \cos {x + y} y \cos x y - \sin x y \paren {-\map \sin {x + y} } } {\map {\cos^2} {x + y} }
      | c = Quotient Rule for Derivatives, treating $y$ as a constant
}}
{{eqn | r = \dfrac {y \map \cos {x + y} \cos x y + \map \sin {x + y} \sin x y} {\map {\cos^2} {x + y} }
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Second Partial Derivative wrt r of ln (r^2 + s)

:$\dfrac {\partial^2} {\partial r^2} \map \ln {r^2 + s} = \dfrac {2 \paren {s - r^2} } {\paren {r^2 + s}^2}$

Proof:
{{begin-eqn}}
{{eqn | l = \dfrac {\partial^2} {\partial r^2} \map \ln {r^2 + s}
      | r = \map {\dfrac \partial {\partial r} } {\dfrac \partial {\partial r} \map \ln {r^2 + s} }
      | c = {{Defof|Second Partial Derivative}}
}}
{{eqn | r = \dfrac \partial {\partial r} \frac 1 {r^2 + s} 2 r
      | c = Derivative of Natural Logarithm, Chain Rule for Derivatives, treating $s$ as a constant
}}
{{eqn | r = \frac {2 \paren {r^2 + s} - 2 r \paren {2 r} } {\paren {r^2 + s}^2}
      | c = Quotient Rule for Derivatives, again treating $s$ as a constant
}}
{{eqn | r = \frac {2 \paren {s - r^2} } {\paren {r^2 + s}^2}
      | c = simplifying
}}
{{end-eqn}}
{{qed}}