avewright/proofwiki-math huggingface dataset:
Theorem: Curl Operator Distributes over Addition

Let $\map {\R^3} {x, y, z}$ denote the real Cartesian space of $3$ dimensions..
Let $\tuple {\mathbf i, \mathbf j, \mathbf k}$ be the standard ordered basis on $\R^3$.
Let $\mathbf f$ and $\mathbf g: \R^3 \to \R^3$ be vector-valued functions on $\R^3$:
:$\mathbf f := \tuple {\map {f_x} {\mathbf x}, \map {f_y} {\mathbf x}, \map {f_z} {\mathbf x} }$
:$\mathbf g := \tuple {\map {g_x} {\mathbf x}, \map {g_y} {\mathbf x}, \map {g_z} {\mathbf x} }$
Let $\nabla \times \mathbf f$ denote the curl of $f$.
Then:
:$\nabla \times \paren {\mathbf f + \mathbf g} = \nabla \times \mathbf f + \nabla \times \mathbf g$

Proof:
{{begin-eqn}}
{{eqn | l = \nabla \times \paren {\mathbf f + \mathbf g}
      | r = \begin {vmatrix} \mathbf i & \mathbf j & \mathbf k \\ \dfrac \partial {\partial x} & \dfrac \partial {\partial y} & \dfrac \partial {\partial z} \\ f_x + g_x & f_y + g_y & f_z + g_x \end {vmatrix}
      | c = {{Defof|Curl Operator}}
}}
{{eqn | r = \begin {vmatrix} \mathbf i & \mathbf j & \mathbf k \\ \dfrac \partial {\partial x} & \dfrac \partial {\partial y} & \dfrac \partial {\partial z} \\ f_x & f_y & f_z \end {vmatrix} + \begin {vmatrix} \mathbf i & \mathbf j & \mathbf k \\ \dfrac \partial {\partial x} & \dfrac \partial {\partial y} & \dfrac \partial {\partial z} \\ g_x & g_y & g_x \end {vmatrix}
      | c = Determinant as Sum of Determinants
}}
{{eqn | r = \nabla \times \mathbf f + \nabla \times \mathbf g
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Product Rule for Divergence

Let $\map {\mathbf V} {x_1, x_2, \ldots, x_n}$ be a vector space of $n$ dimensions.
Let $\mathbf A$ be a vector field over $\mathbf V$.
Let $U$ be a scalar field over $\mathbf V$.
Then:
:$\map {\operatorname {div} } {U \mathbf A} = \map U {\operatorname {div} \mathbf A} + \mathbf A \cdot \grad U$
where
:$\operatorname {div}$ denotes the divergence operator
:$\grad$ denotes the gradient operator
:$\cdot$ denotes dot product.

Proof:
From Divergence Operator on Vector Space is Dot Product of Del Operator and definition of the gradient operator:
{{begin-eqn}}
{{eqn | l = \operatorname {div} \mathbf V
      | r = \nabla \cdot \mathbf V
}}
{{eqn | l = \grad \mathbf U
      | r = \nabla U
}}
{{end-eqn}}
where $\nabla$ denotes the del operator.
Hence we are to demonstrate that:
:$\nabla \cdot \paren {U \, \mathbf A} = \map U {\nabla \cdot \mathbf A} + \paren {\nabla U} \cdot \mathbf A$
Let $\mathbf A$ be expressed as a vector-valued function on $\mathbf V$:
:$\mathbf A := \tuple {\map {A_1} {\mathbf r}, \map {A_2} {\mathbf r}, \ldots, \map {A_n} {\mathbf r} }$
where $\mathbf r = \tuple {x_1, x_2, \ldots, x_n}$ is an arbitrary element of $\mathbf V$.
Let $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ be the standard ordered basis of $\mathbf V$.
Then:
{{begin-eqn}}
{{eqn | l = \nabla \cdot \paren {U \mathbf A}
      | r = \sum_{k \mathop = 1}^n \frac {\map \partial {U A_k} } {\partial x_k}
      | c = {{Defof|Divergence Operator}}
}}
{{eqn | r = \sum_{k \mathop = 1}^n \paren {U \frac {\partial A_k} {\partial x_k} + \frac {\partial U} {\partial x_k} A_k}
      | c = Product Rule for Derivatives
}}
{{eqn | r = U \sum_{k \mathop = 1}^n \frac {\partial A_k} {\partial x_k} + \sum_{k \mathop = 1}^n \frac {\partial U} {\partial x_k} A_k
      | c = 
}}
{{eqn | r = \map U {\nabla \cdot \mathbf A} + \sum_{k \mathop = 1}^n \frac {\partial U} {\partial x_k} A_k
      | c = {{Defof|Divergence Operator}}
}}
{{eqn | r = \map U {\nabla \cdot \mathbf A} + \paren {\sum_{k \mathop = 1}^n \frac {\partial U} {\partial x_k} \mathbf e_k} \cdot \paren {\sum_{k \mathop = 1}^n A_k \mathbf e_k}
      | c = {{Defof|Dot Product}}
}}
{{eqn | r = \map U {\nabla \cdot \mathbf A} + \paren {\nabla U} \cdot \mathbf A
      | c = {{Defof|Gradient Operator}}, {{Defof|Vector (Linear Algebra)|Vector}}
}}
{{end-eqn}}
{{qed}}
Theorem: Moment Generating Function of Beta Distribution

Let $X \sim \BetaDist \alpha \beta$ denote the Beta distribution fior some $\alpha, \beta > 0$.
Then the moment generating function $M_X$ of $X$ is given by: 
:$\ds \map {M_X} t = 1 + \sum_{k \mathop = 1}^\infty \paren {\prod_{r \mathop = 0}^{k - 1} \frac {\alpha + r} {\alpha + \beta + r} } \frac {t^k} {k!}$

Proof:
From the definition of the Beta distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac {x^{\alpha - 1} \paren {1 - x}^{\beta - 1} } {\map \Beta {\alpha, \beta} }$
From the definition of a moment generating function:
:$\ds \map {M_X} t = \expect {e^{t X} } = \int_0^1 e^{t x} \map {f_X} x \rd x$
So:
{{begin-eqn}}
{{eqn	| l = \map {M_X} t
	| r = \frac 1 {\map \Beta {\alpha, \beta} } \int_0^1 e^{t x} x^{\alpha - 1} \paren {1 - x}^{\beta - 1} \rd x
}}
{{eqn	| r = \frac 1 {\map \Beta {\alpha, \beta} } \int_0^1 \paren {\sum_{k \mathop = 0}^\infty \frac {\paren {t x}^k} {k!} } x^{\alpha - 1} \paren {1 - x}^{\beta - 1} \rd x
	| c = Power Series Expansion for Exponential Function
}}
{{eqn	| r = \frac 1 {\map \Beta {\alpha, \beta} } \sum_{k \mathop = 0}^\infty \frac {t^k} {k!} \int_0^1 x^{\alpha + k - 1} \paren {1 - x}^{\beta - 1} \rd x
	| c = Power Series is Termwise Integrable within Radius of Convergence
}}
{{eqn	| r = \sum_{k \mathop = 0}^\infty \frac{t^k} {k!} \paren {\frac {\map \Beta {\alpha + k, \beta} } {\map \Beta {\alpha, \beta} } }
	| c = {{Defof|Beta Function/Definition 1|Beta Function}}
}}
{{eqn	| r = \frac {\map \Beta {\alpha, \beta} } {\map \Beta {\alpha, \beta} } \frac {t^0} {0!} + \sum_{k \mathop = 1}^\infty \frac{t^k} {k!} \paren {\frac {\map \Beta {\alpha + k, \beta} } {\map \Beta {\alpha, \beta} } }
}}
{{eqn	| r = 1 + \sum_{k \mathop = 1}^\infty \paren {\frac {\map \Gamma {\alpha + k} \map \Gamma \beta} {\map \Gamma {\alpha + \beta + k} } \cdot \frac {\map \Gamma {\alpha + \beta} } {\map \Gamma \alpha \map \Gamma \beta} } \frac{t^k} {k!}
	| c = {{Defof|Beta Function|index = 3|Beta Function}}
}}
{{eqn	| r = 1 + \sum_{k \mathop = 1}^\infty \paren {\frac {\map \Gamma {\alpha + k} } {\map \Gamma \alpha} \cdot \frac {\map \Gamma {\alpha + \beta} } {\map \Gamma {\alpha + \beta + k} } } \frac{t^k} {k!}
}}
{{eqn	| r = 1 + \sum_{k \mathop = 1}^\infty \paren {\frac {\map \Gamma \alpha \prod_{r \mathop = 0}^k \paren {\alpha + r} } {\map \Gamma \alpha} \cdot \frac {\map \Gamma {\alpha + \beta} } {\map \Gamma {\alpha + \beta} \prod_{r \mathop = 0}^k \paren {\alpha + \beta + r} } } \frac{t^k} {k!}
	| c = Gamma Difference Equation
}}
{{eqn	| r = 1 + \sum_{k \mathop = 1}^\infty \paren {\prod_{r \mathop = 0}^{k - 1} \frac {\alpha + r} {\alpha + \beta + r} } \frac{t^k} {k!}
	| c = Product of Products
}}
{{end-eqn}}
{{qed}}
Category:Moment Generating Functions
Category:Beta Distribution
e2vowzszpzcol258a2tcyykttsp0zgm
Theorem: Curl of Gradient is Zero

Let $\map {\R^3} {x, y, z}$ denote the real Cartesian space of $3$ dimensions..
Let $\map U {x, y, z}$ be a scalar field on $\R^3$.
Then:
:$\map \curl {\grad U} = \mathbf 0$
where:
:$\curl$ denotes the curl operator
:$\grad$ denotes the gradient operator.

Proof:
From Curl Operator on Vector Space is Cross Product of Del Operator and definition of the gradient operator:
{{begin-eqn}}
{{eqn | l = \grad \mathbf U
      | r = \nabla U
}}
{{eqn | l = \map \curl {\grad U}
      | r = \nabla \times \paren {\nabla U}
}}
{{end-eqn}}
where $\nabla$ denotes the del operator.
Hence we are to demonstrate that:
:$\nabla \times \paren {\nabla U} = \mathbf 0$
Let $\tuple {\mathbf i, \mathbf j, \mathbf k}$ be the standard ordered basis on $\R^3$.
Then:
{{begin-eqn}}
{{eqn | l = \nabla \times \paren {\nabla U}
      | r = \nabla \times \paren {\dfrac {\partial U} {\partial x} \mathbf i + \dfrac {\partial U} {\partial y} \mathbf j + \dfrac {\partial U} {\partial z} \mathbf k}
      | c = {{Defof|Gradient Operator}}
}}
{{eqn | r = \paren {\dfrac \partial {\partial y} \dfrac {\partial U} {\partial z} - \dfrac \partial {\partial z} \dfrac {\partial U} {\partial y} } \mathbf i + \paren {\dfrac \partial {\partial z} \dfrac {\partial U} {\partial x} - \dfrac \partial {\partial x} \dfrac {\partial U} {\partial z} } \mathbf j + \paren {\dfrac \partial {\partial x} \dfrac {\partial U} {\partial y} - \dfrac \partial {\partial y} \dfrac {\partial U} {\partial x} } \mathbf k
      | c = {{Defof|Curl Operator}}
}}
{{eqn | r = \paren {\dfrac {\partial^2 U} {\partial y \partial z} - \dfrac {\partial^2 U} {\partial z \partial y} } \mathbf i + \paren {\dfrac {\partial^2 U} {\partial z \partial x} - \dfrac {\partial^2 U} {\partial x \partial z} } \mathbf j + \paren {\dfrac {\partial^2 U} {\partial x \partial y} - \dfrac {\partial^2 U} {\partial y \partial x} } \mathbf k
      | c = 
}}
{{end-eqn}}
From Clairaut's Theorem:
:$\dfrac {\partial^2 U} {\partial x \partial y} = \dfrac {\partial^2 U} {\partial y \partial x}$
and the same {{mutatis}} for the other partial derivatives.
The result follows.
{{qed}}
Theorem: Raw Moment of Beta Distribution

Let $X \sim \BetaDist \alpha \beta$ for some $\alpha, \beta > 0$, where $\operatorname {Beta}$ is the Beta distribution. 
Then:
:$\ds \expect {X^n} = \prod_{r \mathop = 0}^{n - 1} \frac {\alpha + r} {\alpha + \beta + r}$
for positive integer $n$.

Proof:
By Moment Generating Function of Beta Distribution, the moment generating function, $M_X$, of $X$ is given by: 
:$\ds \map {M_X} t = \expect {e^{t X} } = 1 + \sum_{n \mathop = 1}^\infty \paren {\prod_{r \mathop = 0}^{n - 1} \frac {\alpha + r} {\alpha + \beta + r} } \frac {t^n} {n!}$
We also have: 
{{begin-eqn}}
{{eqn | l = \expect {e^{t X} }
      | r = \expect {\sum_{n \mathop = 0}^\infty \frac {t^n X^n} {n!} }
      | c = Power Series Expansion for Exponential Function
}}
{{eqn | r = \sum_{n \mathop = 0}^\infty \expect {\frac {t^n X^n} {n!} }
      | c = Expectation is Linear
}}
{{eqn | r = \sum_{n \mathop = 0}^\infty \expect {X^n} \frac {t^n} {n!}
      | c = Expectation is Linear
}}
{{eqn | r = \frac {t^0} {0!} \expect {X^0} + \sum_{n \mathop = 1}^\infty \expect {X^n} \frac {t^n} {n!}
}}
{{eqn | r = 1 + \sum_{n \mathop = 1}^\infty \expect {X^n} \frac {t^n} {n!}
      | c = Expectation of Constant
}}
{{end-eqn}}
Comparing coefficients yields the result. 
{{qed}}
Category:Beta Distribution
Category:Raw Moments
fk9kpglabwadtommkhid9g2svsosx76
Theorem: Expectation is Linear/Discrete

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ and $Y$ be random variables on $\struct {\Omega, \Sigma, \Pr}$.
Let $\expect X$ denote the expectation of $X$.
Then:
:$\forall \alpha, \beta \in \R: \expect {\alpha X + \beta Y} = \alpha \, \expect X + \beta \, \expect Y$

Proof:
Follows directly from Expectation of Function of Joint Probability Mass Distribution, thus:
{{begin-eqn}}
{{eqn | l = \expect {\alpha X + \beta Y}
      | r = \sum_x \sum_y \paren {\alpha x + \beta y} \, \map \Pr {X = x, Y = y}
      | c = Expectation of Function of Joint Probability Mass Distribution
}}
{{eqn | r = \alpha \sum_x x \sum_y \map \Pr {X = x, Y = y}
      | c = 
}}
{{eqn | o = +
      | r = \beta \sum_y y \sum_x \map \Pr {X = x, Y = y}
      | c = 
}}
{{eqn | r = \alpha \sum_x x \, \map \Pr {X = x} + \beta \sum_y y \, \map \Pr {Y = y}
      | c = {{Defof|Marginal Probability Mass Function}}
}}
{{eqn | r = \alpha \, \expect X + \beta \, \expect Y
      | c = {{Defof|Expectation}}
}}
{{end-eqn}}
{{qed}}
Theorem: Expectation is Linear/Continuous

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ and $Y$ be random variables on $\struct {\Omega, \Sigma, \Pr}$.
Let $E$ denote the expectation function.
Then:
:$\forall \alpha, \beta \in \R: \expect {\alpha X + \beta Y} = \alpha \expect X + \beta \expect Y$

Proof:
Let $\map \supp X$ and $\map \supp Y$ be the supports of $X$ and $Y$ respectively. 
Let $f_{X, Y} : \map \supp X \times \map \supp Y \to \R$ be the joint probability density function of $X$ and $Y$. 
Let $f_X$ and $f_Y$ be the marginal probability density functions of $X$ and $Y$. 
Then:
{{begin-eqn}}
{{eqn | l = \expect {\alpha X + \beta Y}
      | r = \int_{y \mathop \in \map \supp Y} \int_{x \mathop \in \map \supp X} \paren {\alpha x + \beta y} \map {f_{X, Y} } {x, y} \rd x \rd y
}}
{{eqn | r = \alpha \int_{y \mathop \in \map \supp Y} \int_{x \mathop \in \map \supp X} x \map {f_{X, Y} } {x, y} \rd x \rd y + \beta \int_{y \mathop \in \map \supp Y} \int_{x \mathop \in \map \supp X} y \map {f_{X, Y} } {x, y} \rd x \rd y
      | c = Linear Combination of Definite Integrals
}}
{{eqn | r = \alpha \int_{x \mathop \in \map \supp X} x \paren {\int_{y \mathop \in \map \supp Y} \map {f_{X , Y} } {x, y} \rd y} \rd x + \beta \int_{y \mathop \in \map \supp Y} y \paren {\int_{x \mathop \in \map \supp X} \map {f_{X, Y} } {x, y} \rd x} \rd y
      | c = rewriting
}}
{{eqn | r = \alpha \int_{x \mathop \in \map \supp X} x \map {f_X} x \rd x + \beta \int_{y \mathop \in \map \supp Y} y \map {f_Y} y \rd y
      | c = {{Defof|Marginal Probability Density Function}}
}}
{{eqn | r = \alpha \expect X + \beta \expect Y
      | c = {{Defof|Expectation of Continuous Random Variable}}
}}
{{end-eqn}}
{{qed}}
Theorem: Derivative of Constant Multiple/Real

Let $f$ be a real function which is differentiable on $\R$.
Let $c \in \R$ be a constant.
Then:
:$\map {\dfrac \d {\d x} } {c \map f x} = c \map {\dfrac \d {\d x} } {\map f x}$

Proof:
{{begin-eqn}}
{{eqn | l = \map {\dfrac \d {\d x} } {c \map f x}
      | r = c \map {\dfrac \d {\d x} } {\map f x} + \map f x \map {\dfrac \d {\d x} } c
      | c = Product Rule for Derivatives
}}
{{eqn | r = c \map {\dfrac \d {\d x} } {\map f x} + 0
      | c = Derivative of Constant
}}
{{eqn | r = c \map {\dfrac \d {\d x} } {\map f x}
      | c =
}}
{{end-eqn}}
{{qed}}
Theorem: Derivative of Constant Multiple/Complex

Let $D$ be an open subset of the set of complex numbers $\C$.
Let $f: D \to \C$ be a complex-differentiable function on $D$.
Let $c \in \C$ be a constant.
Then:
:$\forall z \in D : \map {D_z} {c \map f z} = c \map {D_z} {\map f z}$

Proof:
{{begin-eqn}}
{{eqn | l = \map {D_z} {c \map f z}
      | r = c \map {D_z} {\map f z} + \map f z \map {D_z} c
      | c = Product Rule for Complex Derivatives
}}
{{eqn | r = c \map {D_z} {\map f z} + 0
      | c = Complex Derivative of Constant
}}
{{eqn | r = c \map {D_z} {\map f z}
}}
{{end-eqn}}
{{qed}}
Category:Derivative of Constant Multiple
Category:Complex Analysis
lzqf5c0v7nygkina6g7nkejpcfqmrqe
Theorem: Bound on Complex Values of Gamma Function

Let $\map \Gamma z$ denote the Gamma function.
Then for any complex number $z = s + i t$, we have for $\size b \le \size t$:
:$\size {\map \Gamma {s + i t} } \le \dfrac {\size {s + i b} } {\size {s + i t} } \size {\map \Gamma {s + i b} }$

Proof:
From the Euler Form of the Gamma Function:
{{begin-eqn}}
{{eqn | l = \size {\map \Gamma {s + i t} }
      | r = \lim_{M \mathop \to \infty} \size {\dfrac 1 {s + i t} \prod_{n \mathop = 1}^M \dfrac {\paren {1 + \frac 1 n}^{s + i t} } {1 + \frac {s + i t} n} }
}}
{{eqn | r = \lim_{M \mathop \to \infty} \dfrac 1 {\size {s + i t} } \prod_{n \mathop = 1}^M \size {\dfrac {\paren {1 + \frac 1 n}^{s + i t} } {1 + \frac {s + i t} n} }
}}
{{eqn | r = \lim_{M \mathop \to \infty} \dfrac 1 {\size {s + i t} } \prod_{n \mathop = 1}^M \dfrac {\size {\paren {1 + \frac 1 n}^{s + i t} } } {\size {1 + \frac {s + i t} n} } }}
{{eqn | r = \lim_{M \mathop \to \infty} \dfrac 1 {\size {s + i t} } \prod_{n \mathop = 1}^M \dfrac {\size {\paren {1 + \frac 1 n}^s} } {\size {1 + \frac {s + i t} n} }
      | c = Modulus of Exponential of Imaginary Number is One
}}
{{end-eqn}}
Because $\size b \le \size t$, we have that:
{{begin-eqn}}
{{eqn | l = b^2
      | o = \le
      | r = t^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {\dfrac b n}^2
      | o = \le
      | r = \paren {\dfrac t n}^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {1 + \dfrac s n}^2 + \paren {\dfrac b n}^2
      | o = \le
      | r = \paren {1 + \dfrac s n}^2 + \paren {\dfrac t n}^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \size {1 + \frac {s + i b} n} 
      | o = \le
      | r = \size {1 + \frac {s + i t} n}
      | c = {{Defof|Complex Modulus}}
}}
{{eqn | ll= \leadsto
      | l = \dfrac 1 {\size {1 + \frac {s + i b} n} }
      | o = \ge 
      | r = \dfrac 1 {\size {1 + \frac {s + i t} n} }
      | c = 
}}
{{end-eqn}}
Using this we obtain:
{{begin-eqn}}
{{eqn | l = \size {\map \Gamma {s + i t} }
      | r = \lim_{M \mathop \to \infty} \dfrac 1 {\size {s + i t} } \prod_{n \mathop = 1}^M \dfrac {\size {\paren {1 + \frac 1 n}^s} } {\size {1 + \frac {s + i t} n} }
}}
{{eqn | o = \le
      | r = \lim_{M \mathop \to \infty} \dfrac 1 {\size {s + i t} } \prod_{n \mathop = 1}^M \dfrac {\size {\paren {1 + \frac 1 n}^s} } {\size {1 + \frac {s + i b} n} }
}}
{{eqn | r = \dfrac {\size {s + i b} } {\size {s + i t} } \lim_{M \mathop \to \infty} \dfrac 1 {\size {s + i b} } \prod_{n \mathop = 1}^M \dfrac {\size {\paren {1 + \frac 1 n}^s} } {\size {1 + \frac{s + i b} n} }
}}
{{eqn | r = \dfrac {\size {s + i b} } {\size {s + i t} } \lim_{M \mathop \to \infty} \dfrac 1 {\size {s + i b} } \size {\prod_{n \mathop = 1}^M \dfrac {\paren {1 + \frac 1 n}^{s + i b} } {1 + \frac{s + i b} n} }
}}
{{eqn | r = \dfrac {\size {s + i b} } {\size {s + i t} } \size {\map \Gamma {s + i b} }
}}
{{end-eqn}}
{{qed}}
Category:Gamma Function
pw70f4doua9xqrbrqlvo66jrr07gtth
Theorem: Set of Vectors defined by Directed Line Segments in Space forms Vector Space

Let $\R^3$ be a real cartesian space of $3$ dimensions.
Consider the set $S$ of directed line segments in $\R^3$.
Let the equivalence relation $\sim$ be applied to $\R^3$ such that:
:$\forall L_1, L_2 \in \R^3: L_1 \sim L_2$ {{iff}} there exists a translation $T$ such that $\map T {L_1} = L_2$
Let $\mathbb V$ denote the set of equivalence classes of $\sim$ on $S$.
The elements of $\mathbb V$ form a vector space where:
:$\forall \mathbf v_1, \mathbf v_2 \in \mathbb V: \mathbf v_1 + \mathbf v_2$ denotes the element of $\mathbb V$ exemplified by the result of the operation of joining of a representative element of element of $\mathbf v_2$ to the end of a similarly representative element of element of $\mathbf v_1$.
:$\forall \mathbf v \in \mathbb V, \lambda \in \R: \lambda \mathbf v$ denotes the element of $\mathbb V$ exemplified by the result of the operation of mulitplying a representative element of element of $\mathbf v$ by the scale factor $\lambda$.

Proof:
{{ProofWanted|This needs to be made considerably less clumsy}}
Theorem: Matrix is Nonsingular iff Determinant has Multiplicative Inverse/Sufficient Condition

Let $R$ be a commutative ring with unity.
Let $\mathbf A \in R^{n \times n}$ be a square matrix of order $n$.
Let  the determinant of $\mathbf A$ be invertible in $R$.
Then $\mathbf A$ is a nonsingular matrix.

Proof:
Let $\map \det {\mathbf A}$ be invertible in $R$.
From Matrix Product with Adjugate Matrix:
{{begin-eqn}}
{{eqn | l = \mathbf A \cdot \adj {\mathbf A}
      | r = \map \det {\mathbf A} \cdot \mathbf I_n
}}
{{eqn | l = \adj {\mathbf A} \cdot \mathbf A
      | r = \map \det {\mathbf A} \cdot \mathbf I_n
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \mathbf A \cdot \paren {\map \det {\mathbf A}^{-1} \cdot \adj {\mathbf A} }
      | r = \mathbf I_n
}}
{{eqn | l = \paren {\map \det {\mathbf A}^{-1} \cdot \adj {\mathbf A} } \cdot \mathbf A
      | r = \mathbf I_n
}}
{{end-eqn}}
Thus $\mathbf A$ is nonsingular, and:
:$\mathbf A^{-1} = \map \det {\mathbf A}^{-1} \cdot \adj {\mathbf A}$
{{qed}}
Theorem: Bienaymé-Chebyshev Inequality

Let $X$ be a random variable. 
Let $\expect X = \mu$ for some $\mu \in \R$. 
Let $\var X = \sigma^2$ for some $\sigma^2 \in \R_{> 0}$.
Then, for all $k > 0$: 
:$\map \Pr {\size {X - \mu} \ge k \sigma} \le \dfrac 1 {k^2}$

Proof:
Let $f$ be the function:
:$\map f x = \begin{cases} k^2 \sigma^2 & : \size {x - \mu} \ge k \sigma \\ 0 & : \text{otherwise} \end{cases}$
By construction:
:$\forall x \in \Dom f: \map f x \le \size {x - \mu}^2 = \paren {x - \mu}^2$
Hence from Expectation Preserves Inequality:
:$\expect {\map f X} \le \expect {\paren {X - \mu}^2}$
By definition of variance:
:$\expect {\paren {X - \mu}^2} = \var X = \sigma^2$
By definition of expectation of discrete random variable, we can show that:
{{begin-eqn}}
{{eqn | l = \expect {\map f X}
      | r = k^2 \sigma^2 \map \Pr {\size {X - \mu} \ge k \sigma} + 0 \cdot \map \Pr {\size {X - \mu} \le k \sigma}
      | c = 
}}
{{eqn | r = k^2 \sigma^2 \map \Pr {\size {X - \mu} \ge k \sigma}
      | c = 
}}
{{end-eqn}}
Putting this together, we have:
{{begin-eqn}}
{{eqn | l = \expect {\map f X}
      | o = \le
      | r = \expect {\paren {X - \mu}^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = k^2 \sigma^2 \map \Pr {\size {X - \mu} \ge k \sigma}
      | o = \le
      | r = \sigma^2
      | c = 
}}
{{end-eqn}}
By dividing both sides by $k^2 \sigma^2$, we get: 
:$\map \Pr {\size {X - \mu} \ge k \sigma} \le \dfrac 1 {k^2}$
{{qed}}
Theorem: Bienaymé-Chebyshev Inequality

Let $X$ be a random variable. 
Let $\expect X = \mu$ for some $\mu \in \R$. 
Let $\var X = \sigma^2$ for some $\sigma^2 \in \R_{> 0}$.
Then, for all $k > 0$: 
:$\map \Pr {\size {X - \mu} \ge k \sigma} \le \dfrac 1 {k^2}$

Proof:
Note that as $k > 0$ and $\sigma > 0$, we have $k \sigma > 0$.
We therefore have:
{{begin-eqn}}
{{eqn | l = \map \Pr {\size {X - \mu} \ge k \sigma}
      | r = \map \Pr {\paren {X - \mu}^2 \ge \paren {k \sigma}^2}
}}
{{eqn | o = \le
      | r = \frac {\expect {\paren {X - \mu}^2} } {\paren {k \sigma}^2}
      | c = as $k \sigma > 0$, we can apply {{Corollary|Markov's Inequality}}
}}
{{eqn | r = \frac {\sigma^2} {k^2 \sigma^2}
      | c = {{Defof|Variance}}
}}
{{eqn | r = \frac 1 {k^2}
}}
{{end-eqn}}
{{qed}}
Theorem: Sign of Composition of Permutations

Let $n \in \N$ be a natural number.
Let $N_n$ denote the set of natural numbers $\set {1, 2, \ldots, n}$.
Let $S_n$ denote the set of permutations on $N_n$.
Let $\map \sgn \pi$ denote the sign of $\pi$ of a permutation $\pi$ of $N_n$.
Let $\pi_1, \pi_2 \in S_n$.
Then:
:$\map \sgn {\pi_1} \map \sgn {\pi_2} = \map \sgn {\pi_1 \circ \pi_2}$
where $\pi_1 \circ \pi_2$ denotes the composite of $\pi_1$ and $\pi_2$.

Proof:
From Sign of Permutation on n Letters is Well-Defined, it is established that the sign each of $\pi_1$, $\pi_2$ and $\pi_1 \circ \pi_2$ is either $+1$ and $-1$.
By Existence and Uniqueness of Cycle Decomposition, each of $\pi_1$ and $\pi_2$ has a unique cycle decomposition.
Thus each of $\pi_1$ and $\pi_2$ can be expressed as the composite of $p_1$ and $p_2$ transpositions respectively.
Thus $\pi_1 \circ \pi_2$ can be expressed as the composite of $p_1 + p_2$ transpositions.
From Sum of Even Integers is Even, if $p_1$ and $p_2$ are both even then $p_1 + p_2$ is even.
In this case:
:$\map \sgn {\pi_1} = 1$
:$\map \sgn {\pi_2} = 1$
:$\map \sgn {\pi_1} \map \sgn {\pi_2} = 1 = 1 \times 1$
{{finish}}
{{improve}}
{{proofread}}
Theorem: Odd and Even Permutations of Set are Equivalent

Let $n \in \N_{> 0}$ be a natural number greater than $0$.
Let $S$ be a set of cardinality $n$.
Let $S_n$ denote the symmetric group on $S$ of order $n$.
Let $R_e$ and $R_o$ denote the subsets of $S_n$ consisting of even permutations and odd permutations respectively.
Then $R_e$ and $R_o$ are equivalent.

Proof:
From Symmetric Group on n Letters is Isomorphic to Symmetric Group, it is sufficient to investigate the symmetric group on n letters.
Let $\tau$ be a transposition.
By definition of sign:
:$\map \sgn \rho = -1$
By definition of odd permutation:
:$\tau \in R_o$
Moreover, also by definition of sign, for any $\rho \in R_e$ we have:
:$\rho \circ \tau \in R_o$
Define the mapping $\phi$:
:$\forall \rho \in R_e: \map \phi \sigma = \rho \circ \tau$
We claim that $\phi$ is injective.
Suppose:
:$\map \phi {\rho_1} = \map \phi {\rho_2}$
Then:
:$\rho_1 \circ \tau = \rho_2 \circ \tau$
Since each Transposition is Self-Inverse, it follows that:
:$\rho_1 = \rho_1 \circ \tau \circ \tau = \rho_2 \circ \tau \circ \tau = \rho_2$
Thus $\phi$ is indeed injective.
{{qed|lemma}}
We claim that $\phi$ is also surjective.
Let $\sigma \in R_o$. 
We need to show that there exists an element $\rho\in R_e$ such that:
:$\map \phi \rho = \sigma$
By definition of sign:
:$\sigma \circ \tau \in R_e$
Since each Transposition is Self-Inverse, it follows that:
:$\map \phi {\sigma \circ \tau} = \sigma \circ \tau \circ \tau = \sigma$
Thus $\phi$ is surjective.
{{qed|lemma}}
Since $\phi$ is injective and surjective, $\phi$ is the desired bijection.
Hence the result by definition of set equivalence.
{{qed}}