avewright/proofwiki-math huggingface dataset:
Theorem: Median of Normal Distribution

Let $X \sim \Gaussian \mu {\sigma^2}$ for some $\mu \in \R, \sigma \in \R_{> 0}$, where $N$ is the normal distribution.
Then the median of $X$ is equal to $\mu$.

Proof:
From the definition of the normal distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac 1 {\sigma \sqrt {2 \pi} } \map \exp {-\dfrac {\paren {x - \mu}^2} {2 \sigma^2} }$
Note that $f_X$ is non-zero, sufficient to ensure a unique median.
By the definition of a median, to prove that $\mu$ is the median of $X$ we must verify: 
:$\ds \map \Pr {X < \mu} = \int_{-\infty}^\mu \map {f_X} x \rd x = \frac 1 2$
We have:
{{begin-eqn}}
{{eqn | l = \int_{-\infty}^\mu \map {f_X} x \rd x
      | r = \frac 1 {\sigma \sqrt {2 \pi} } \int_{-\infty}^\mu \map \exp {-\frac {\paren {x - \mu}^2} {2 \sigma^2} } \rd x
}}
{{eqn | r = \frac {\sqrt 2 \sigma} {\sigma \sqrt {2 \pi} } \int_{-\infty}^{\frac {\mu - \mu} {\sqrt 2 \sigma} } \map \exp {-t^2} \rd t
      | c = substituting $t = \dfrac {x - \mu} {\sqrt 2 \sigma}$
}}
{{eqn | r = \frac 1 {\sqrt \pi} \int_{-\infty}^0 \map \exp {-t^2} \rd t
}}
{{eqn | r = \frac 1 {2 \sqrt \pi} \int_{-\infty}^\infty \map \exp {-t^2} \rd t
      | c = Definite Integral of Even Function
}}
{{eqn | r = \frac {\sqrt \pi} {2 \sqrt \pi}
      | c = Gaussian Integral
}}
{{eqn | r = \frac 1 2
}}
{{end-eqn}}
{{qed}}
Category:Normal Distribution
Category:Medians
sydbbti8z38vo1o6z5vzqt61ti18kuc
Theorem: Expectation of Gamma Distribution

Let $X \sim \map \Gamma {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.
The expectation of $X$ is given by:
:$\expect X = \dfrac \alpha \beta$

Proof:
From the definition of the Gamma distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac {\beta^\alpha x^{\alpha - 1} e^{-\beta x} } {\map \Gamma \alpha}$
From the definition of the expected value of a continuous random variable: 
:$\ds \expect X = \int_0^\infty x \map {f_X} x \rd x$
So:
{{begin-eqn}}
{{eqn	| l = \expect X
	| r = \frac {\beta^\alpha} {\map \Gamma \alpha} \int_0^\infty x^\alpha e^{-\beta x} \rd x
}}
{{eqn	| r = \frac {\beta^\alpha} {\map \Gamma \alpha} \int_0^\infty \left({\frac t \beta}\right)^\alpha e^{-t} \frac {\rd t} \beta
	| c = substituting $t = \beta x$
}}
{{eqn	| r = \frac {\beta^\alpha} {\beta^{\alpha + 1} \map \Gamma \alpha} \int_0^\infty t^\alpha e^{-t} \rd t
}}
{{eqn	| r = \frac {\map \Gamma {\alpha + 1} } {\beta \map \Gamma \alpha} 
	| c = {{Defof|Gamma Function}}
}}
{{eqn	| r = \frac {\alpha \map \Gamma \alpha} {\beta \map \Gamma \alpha} 
	| c = Gamma Difference Equation
}}
{{eqn	| r = \frac \alpha \beta
}}
{{end-eqn}}
{{qed}}
Theorem: Expectation of Gamma Distribution

Let $X \sim \map \Gamma {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.
The expectation of $X$ is given by:
:$\expect X = \dfrac \alpha \beta$

Proof:
By Moment Generating Function of Gamma Distribution, the moment generating function of $X$ is given by: 
:$\map {M_X} t = \paren {1 - \dfrac t \beta}^{-\alpha}$
for $t < \beta$. 
From Moment in terms of Moment Generating Function: 
:$\expect X = \map { {M_X}'} 0$
From Moment Generating Function of Gamma Distribution: First Moment: 
:$\map { {M_X}'} t = \dfrac {\beta^\alpha \alpha} {\paren {\beta - t}^{\alpha + 1} }$
Hence setting $t = 0$:
{{begin-eqn}}
{{eqn	| l = \map { {M_X}'} 0
	| r = \frac {\beta^\alpha \alpha} {\paren {\beta - 0}^{\alpha + 1} }
}}
{{eqn	| r = \frac {\beta^\alpha \alpha} {\beta^{\alpha + 1} }
}}
{{eqn	| r = \frac \alpha \beta
}}
{{end-eqn}}
{{qed}}
Theorem: Variance of Gamma Distribution

Let $X \sim \map \Gamma {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.
The variance of $X$ is given by:
:$\var X = \dfrac \alpha {\beta^2}$

Proof:
From the definition of the Gamma distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac {\beta^\alpha x^{\alpha - 1} e^{-\beta x} } {\map \Gamma \alpha}$
From Variance as Expectation of Square minus Square of Expectation: 
:$\ds \var X = \int_0^\infty x^2 \map {f_X} x \rd x - \paren {\expect X}^2$
So:
{{begin-eqn}}
{{eqn	| l = \var X
	| r = \frac {\beta^\alpha} {\map \Gamma \alpha} \int_0^\infty x^{\alpha + 1} e^{-\beta x} \rd x - \paren {\frac \alpha \beta}^2
	| c = Expectation of Gamma Distribution
}}
{{eqn	| r = \frac {\beta^\alpha} {\map \Gamma \alpha} \int_0^\infty \paren {\frac t \beta}^{\alpha + 1} e^{-t} \frac {\d t} \beta - \frac {\alpha^2} {\beta^2}
	| c = substituting $t = \beta x$
}}
{{eqn	| r = \frac {\beta^\alpha} {\beta^{\alpha + 2} \map \Gamma \alpha} \int_0^\infty t^{\alpha + 1} e^{-t} \rd t - \frac {\alpha^2} {\beta^2}
}}
{{eqn	| r = \frac {\map \Gamma {\alpha + 2} } {\beta^2 \map \Gamma \alpha} - \frac {\alpha^2} {\beta^2}
	| c = {{Defof|Gamma Function}}
}}
{{eqn	| r = \frac {\map \Gamma {\alpha + 2} - \alpha^2 \map \Gamma \alpha} {\beta^2 \map \Gamma \alpha}
}}
{{eqn	| r = \frac {\alpha \paren {\alpha + 1} \map \Gamma \alpha - \alpha^2 \map \Gamma \alpha} {\beta^2 \map \Gamma \alpha}
	| c = Gamma Difference Equation
}}
{{eqn	| r = \frac {\alpha \map \Gamma \alpha \paren {\alpha + 1 - \alpha} } {\beta^2 \map \Gamma \alpha}
}}
{{eqn	| r = \frac \alpha {\beta^2}
}}
{{end-eqn}}
{{qed}}
Theorem: Variance of Gamma Distribution

Let $X \sim \map \Gamma {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.
The variance of $X$ is given by:
:$\var X = \dfrac \alpha {\beta^2}$

Proof:
By Moment Generating Function of Gamma Distribution, the moment generating function of $X$ is given by: 
:$\map {M_X} t = \paren {1 - \dfrac t \beta}^{-\alpha}$
for $t < \beta$. 
From Variance as Expectation of Square minus Square of Expectation: 
:$\var X = \expect {X^2} - \paren {\expect X}^2$
From Expectation of Gamma Distribution:
:$\expect X = \dfrac \alpha \beta$
From Moment Generating Function of Gamma Distribution: Second Moment: 
:$\map { {M_X}''} t = \dfrac {\beta^\alpha \alpha \paren {\alpha + 1} } {\paren {\beta - t}^{\alpha + 2} }$
From Moment in terms of Moment Generating Function, we also have: 
:$\expect {X^2} = \map { {M_X}''} 0$
Setting $t = 0$, we obtain the second moment: 
{{begin-eqn}}
{{eqn | l = \map {M''_X} 0 
      | r = \expect {X^2}
}}
{{eqn | r = \frac {\beta^\alpha \alpha \paren {\alpha + 1} } {\paren {\beta - 0}^{\alpha + 2} }
}}
{{eqn | r = \frac {\beta^\alpha \alpha \paren {\alpha + 1} } {\beta^{\alpha + 2} }
}}
{{eqn | r = \frac {\alpha \paren {\alpha + 1} } {\beta^2}
}}
{{end-eqn}}
So: 
{{begin-eqn}}
{{eqn | l = \var X
      | r = \frac {\alpha \paren {\alpha + 1} } {\beta^2} - \frac {\alpha^2} {\beta^2}
}}
{{eqn | r = \frac {\alpha^2 + \alpha - \alpha^2} {\beta^2}
}}
{{eqn | r = \frac \alpha {\beta^2}
}}
{{end-eqn}}
{{qed}}
Theorem: Sum of Bernoulli Numbers by Power of Two and Binomial Coefficient

Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \dbinom {2 n + 1} {2 k} 2^{2 k} B_{2 k}
      | r = \binom {2 n + 1} 2 2^2 B_2 + \binom {2 n + 1} 4 2^4 B_4 + \binom {2 n + 1} 6 2^6 B_6 + \cdots
      | c = 
}}
{{eqn | r = 2 n
      | c = 
}}
{{end-eqn}}
where $B_n$ denotes the $n$th Bernoulli number.

Proof:
Let $B_k$ denote the $k$th Bernoulli number
Let $\map {B_k} x$ denote the $k$th Bernoulli polynomial
By Value of Odd Bernoulli Polynomial at One Half:
{{begin-eqn}}
{{eqn | l = \map {B_{2 n + 1} } {\frac 1 2}
      | r = 0
}}
{{eqn | ll= \leadsto
      | l = \sum_{k \mathop = 0}^{2 n + 1} \binom {2 n + 1} k \map {B_k} {\frac 1 2}^{2 n + 1 - k}
      | r = 0
}}
{{eqn | ll= \leadsto
      | l = 2^{2 n + 1} \sum_{k \mathop = 0}^{2 n + 1} \binom {2 n + 1} k \map {B_k} {\frac 1 2}^{2 n + 1 - k}
      | r = 0
}}
{{eqn | ll= \leadsto
      | l = \sum_{k \mathop = 0}^{2 n + 1} \binom {2 n + 1} k \map {B_k} {\frac 1 2}^{-k}
      | r = 0
}}
{{eqn | ll= \leadsto
      | l = \sum_{k \mathop = 0}^{2 n + 1} \binom {2 n + 1} k B_k 2^k
      | r = 0
}}
{{end-eqn}}
By Odd Bernoulli Numbers Vanish:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^{2 n + 1} \binom {2 n + 1} k B_k 2^k
      | r = \paren {2 n + 1} 2 B_1 + \sum_{k \mathop = 0}^n \binom {2 n + 1} {2 k} B_{2 k} 2^{2 k}
}}
{{eqn | r = -\paren {2 n + 1} + \sum_{k \mathop = 0}^n \binom {2 n + 1 } {2 k} B_{2 k} 2^{2 k}
      | c = $B_1 = -\dfrac 1 2$
}}
{{eqn | ll= \leadsto
      | l = \sum_{k \mathop = 0}^n \binom {2 n + 1} {2 k} B_{2 k} 2^{2 k}
      | r = 2 n + 1
}}
{{eqn | ll = \leadsto
      | l = \sum_{k \mathop = 1}^n \binom {2 n + 1} {2 k} B_{2 k} 2^{2 k}
      | r = 2 n + 1 - B_0
}}
{{eqn | r = 2 n
      | c = $B_0 = 1$
}}
{{end-eqn}}
{{qed}}
Theorem: Sum of Bernoulli Numbers by Power of Two and Binomial Coefficient

Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \dbinom {2 n + 1} {2 k} 2^{2 k} B_{2 k}
      | r = \binom {2 n + 1} 2 2^2 B_2 + \binom {2 n + 1} 4 2^4 B_4 + \binom {2 n + 1} 6 2^6 B_6 + \cdots
      | c = 
}}
{{eqn | r = 2 n
      | c = 
}}
{{end-eqn}}
where $B_n$ denotes the $n$th Bernoulli number.

Proof:
The proof proceeds by induction.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{k \mathop = 1}^n \dbinom {2 n + 1} {2 k} 2^{2 k} B_{2 k} = 2 n$
=== Basis for the Induction ===
$\map P 1$ is the case:
{{begin-eqn}}
{{eqn | l = \binom {2 \times 1 + 1} 2 2^2 B_2
      | r = \frac {2 \times 3} 2 \times 2^2 \times \frac 1 6
      | c = Binomial Coefficient with Two, {{Defof|Bernoulli Numbers}}
}}
{{eqn | r = 2
      | c = after simplification
}}
{{eqn | r = 2 \times 1
      | c = 
}}
{{end-eqn}}
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P r$ is true, where $r \ge 2$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
:$\ds \sum_{k \mathop = 1}^r \dbinom {2 r + 1} {2 k} 2^{2 k} B_{2 k} = 2 r$
from which it is to be shown that:
:$\ds \sum_{k \mathop = 1}^{r + 1} \dbinom {2 \paren {r + 1} + 1} {2 k} 2^{2 k} B_{2 k} = 2 \paren {r + 1}$
=== Induction Step ===
This is the induction step:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^{r + 1} \dbinom {2 \paren {r + 1} + 1} {2 k} 2^{2 k} B_{2 k}
      | r = \sum_{k \mathop = 1}^r \dbinom {2 \paren {r + 1} + 1} {2 k} 2^{2 k} B_{2 k} + \dbinom {2 \paren {r + 1} + 1} {2 \paren {r + 1} } 2^{2 \paren {r + 1} } B_{2 \paren {r + 1} }
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 1}^r \dbinom {2 r + 3} {2 k} 2^{2 k} B_{2 k} + \paren {2 r + 3} 2^{2 r + 2} B_{2 r + 2}
      | c = Binomial Coefficient with Self minus One and simplification
}}
{{end-eqn}}
{{finish|This approach may not be workable. May be easier / better to start with Definition:Bernoulli Numbers/Recurrence Relation and take it from there.}}
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{>0}: \sum_{k \mathop = 1}^n \dbinom {2 n + 1} {2 k} 2^{2 k} B_{2 k} = 2 n$
Theorem: Moment in terms of Moment Generating Function

Let $X$ be a random variable. 
Let $M_X$ be the moment generating function of $X$.
Then: 
:$\expect {X^n} = \map { {M_X}^{\paren n} } 0$
where:
:$n$ is a non-negative integer
:${M_X}^{\paren n}$ denotes the $n$th derivative of $M_X$
:$\expect {X^n}$ denotes the expectation of $X^n$.

Proof:
{{begin-eqn}}
{{eqn | l = \map { {M_X}^{\paren n} } t
      | r = \frac {\d^n} {\d t^n} \expect {e^{t X} }
      | c = {{Defof|Moment Generating Function}}
}}
{{eqn | r = \frac {\d^n} {\d t^n} \expect {\sum_{m \mathop = 0}^\infty \frac {t^m X^m} {m!} }
      | c = Power Series Expansion for Exponential Function
}}
{{eqn | r = \frac {\d^n} {\d t^n} \sum_{m \mathop = 0}^\infty \expect {\frac {t^m X^m} {m!} }
      | c = Expectation is Linear
}}
{{eqn | r = \sum_{m \mathop = 0}^\infty \frac {\d^n} {\d t^n} \paren {\frac {t^m} {m!} } \expect {X^m}
      | c = Expectation is Linear, Power Series is Termwise Differentiable within Radius of Convergence
}}
{{eqn | r = \sum_{m \mathop = n}^\infty \frac {m^{\underline n} t^{m - n} } {m!} \expect {X^m}
      | c = Nth Derivative of Mth Power
}}
{{eqn | r = \sum_{m \mathop = n}^\infty \frac { m! t^{m - n} } {m! \paren {m - n}!} \expect {X^m}
      | c = Falling Factorial as Quotient of Factorials
}}
{{eqn | r = \frac {t^{n - n} } {\paren {n - n}!} \expect {X^n} + \sum_{m \mathop = n + 1}^\infty \frac {t^{m - n} } {\paren {m - n}!} \expect {X^m}
}}
{{eqn | r = \expect {X^n} + \sum_{m \mathop = n + 1}^\infty \frac {t^{m - n} } {\paren {m - n}!} \expect {X^m}
}}
{{end-eqn}}
Setting $t = 0$ yields the result. 
{{qed}}
Theorem: Moment Generating Function of Continuous Uniform Distribution

Let $X \sim \ContinuousUniform a b$ for some $a, b \in \R$ denote the continuous uniform distribution on the interval $\closedint a b$.
Then the moment generating function of $X$ is given by: 
:$\map {M_X} t = \begin {cases} \dfrac {e^{t b} - e^{t a} } {t \paren {b - a} } & t \ne 0 \\ 1 & t = 0 \end{cases}$

Proof:
From the definition of the continuous uniform distribution, $X$ has probability density function: 
:$\map {f_X} x = \begin{cases} \dfrac 1 {b - a} & a \le x \le b \\ 0 & \text{otherwise} \end{cases}$
From the definition of a moment generating function:
:$\ds \map {M_X} t = \expect {e^{t X} } = \int_{-\infty}^\infty e^{t x} \map {f_X} x \rd x$
where $\expect \cdot$ denotes expectation.
First, consider the case $t \ne 0$. 
Then: 
{{begin-eqn}}
{{eqn | l = \map {M_X} t
      | r = \int_{-\infty}^a 0 e^{t x} \rd x + \int_a^b \frac{ e^{t x} } {b - a} \rd x + \int_b^\infty 0 e^{t x} \rd x
}}
{{eqn | r = \intlimits {\frac {e^{t x} } {t \paren {b - a} } } a b
      | c = Primitive of $e^{a x}$, Fundamental Theorem of Calculus
}}
{{eqn | r = \frac {e^{t b} - e^{t a} } {t \paren {b - a} }
}}
{{end-eqn}}
In the case $t = 0$, we have $\expect {X^0} = \expect 1 = 1$. 
{{qed}}
Theorem: Moment Generating Function of Normal Distribution

Let $X \sim \Gaussian \mu {\sigma^2}$ for some $\mu \in \R, \sigma \in \R_{> 0}$, where $N$ is the normal distribution.
Then the moment generating function $M_X$ of $X$ is given by: 
:$\map {M_X} t = \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2}$

Proof:
From the definition of the normal distribution, $X$ has probability density function:
:$\map {f_X} x = \dfrac 1 {\sigma \sqrt{2 \pi} } \map \exp {-\dfrac {\paren {x - \mu}^2} {2 \sigma^2} }$
From the definition of a moment generating function:
:$\ds \map {M_X} t = \expect { e^{t X} } = \int_{-\infty}^\infty e^{t x} \map {f_X} x \rd x$
So: 
{{begin-eqn}}
{{eqn | l = \map {M_X} t
      | r = \frac 1 {\sigma \sqrt{2 \pi} } \int_{-\infty}^\infty \map \exp {t x - \frac {\paren {x - \mu}^2} {2 \sigma^2} } \rd x
}}
{{eqn | r = \frac {\sqrt 2 \sigma} {\sigma \sqrt {2 \pi} } \int_{-\infty}^\infty \map \exp {\paren {\sqrt 2 \sigma u + \mu} t - u^2} \rd u
      | c = substituting $u = \dfrac {x - \mu} {\sqrt 2 \sigma}$
}}
{{eqn | r = \frac {\exp \mu t} {\sqrt \pi} \int_{-\infty}^\infty \map \exp {-\paren {u^2 - \sqrt 2 \sigma u t} } \rd u
}}
{{eqn | r = \frac {\exp \mu t} {\sqrt \pi} \int_{-\infty}^\infty \map \exp {-\paren {u - \frac {\sqrt 2} 2 \sigma t}^2 + \frac 1 2 \sigma^2 t^2} \rd u
}}
{{eqn | r = \frac {\map \exp {\mu t + \frac 1 2 \sigma^2 t^2} } {\sqrt \pi} \int_{-\infty}^\infty \map \exp {-v^2} \rd v
      | c = substituting $v = u - \dfrac {\sqrt 2} 2 \sigma t$
}}
{{eqn | r = \frac {\sqrt \pi \map \exp {\mu t + \frac 1 2 \sigma^2 t^2} } {\sqrt \pi}
      | c = Gaussian Integral
}}
{{eqn | r = \map \exp {\mu t + \dfrac 1 2 \sigma^2 t^2}
}}
{{end-eqn}}
{{qed}}
Theorem: Equality of Vector Quantities

Two vector quantities are equal {{iff}} they have the same magnitude and direction.
That is:
:$\mathbf a = \mathbf b \iff \paren {\size {\mathbf a} = \size {\mathbf b} \land \hat {\mathbf a} = \hat {\mathbf b} }$
where:
:$\hat {\mathbf a}$ denotes the unit vector in the direction of $\mathbf a$
:$\size {\mathbf a}$ denotes the magnitude of $\mathbf a$.

Proof:
Let $\mathbf a$ and $\mathbf b$ be expressed in component form:
{{begin-eqn}}
{{eqn | l = \mathbf a
      | r = a_1 \mathbf e_1 + a_2 \mathbf e_2 + \cdots + a_n \mathbf e_n
      | c = 
}}
{{eqn | l = \mathbf b
      | r = b_1 \mathbf e_1 + b_2 \mathbf e_2 + \cdots + b_n \mathbf e_n
      | c = 
}}
{{end-eqn}}
where $\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n$ denote the unit vectors in the positive directions of the coordinate axes of the Cartesian coordinate space into which $\mathbf a$ has been embedded.
Thus $\mathbf a$ and $\mathbf b$ can be expressed as:
{{begin-eqn}}
{{eqn | l = \mathbf a
      | r = \tuple {a_1, a_2, \ldots, a_n}
      | c = 
}}
{{eqn | l = \mathbf b
      | r = \tuple {b_1, b_2, \ldots, b_n}
      | c = 
}}
{{end-eqn}}
By definition of vector length, we have that:
{{begin-eqn}}
{{eqn | l = \size {\mathbf a}
      | r = \size {\tuple {a_1, a_2, \ldots, a_n} }
      | c = 
}}
{{eqn | r = \sqrt {\paren {a_1^2 + a_2^2 + \ldots + a_n^2} }
      | c = 
}}
{{end-eqn}}
and similarly:
{{begin-eqn}}
{{eqn | l = \size {\mathbf b}
      | r = \size {\tuple {b_1, b_2, \ldots, b_n} }
      | c = 
}}
{{eqn | r = \sqrt {\paren {b_1^2 + b_2^2 + \ldots + b_n^2} }
      | c = 
}}
{{end-eqn}}
From Vector Quantity as Scalar Product of Unit Vector Quantity, it follows that:
{{begin-eqn}}
{{eqn | l = \hat {\mathbf a}
      | r = \widehat {\tuple {a_1, a_2, \ldots, a_n} }
      | c = 
}}
{{eqn | r = \dfrac 1 {\sqrt {\paren {a_1^2 + a_2^2 + \ldots + a_n^2} } } \mathbf a
      | c = 
}}
{{end-eqn}}
and similarly:
{{begin-eqn}}
{{eqn | l = \hat {\mathbf b}
      | r = \widehat {\tuple {b_1, b_2, \ldots, b_n} }
      | c = 
}}
{{eqn | r = \dfrac 1 {\sqrt {\paren {b_1^2 + b_2^2 + \ldots + b_n^2} } } \mathbf b
      | c = 
}}
{{end-eqn}}
Theorem: Vector Addition is Commutative

Let $\mathbf a, \mathbf b$ be vector quantities.
Then:
:$\mathbf a + \mathbf b = \mathbf b + \mathbf a$

Proof:
From the Parallelogram Law:
:350px
{{finish}}
Theorem: Vector Addition is Associative

Let $\mathbf a, \mathbf b, \mathbf c$ be vectors.
Then:
:$\mathbf a + \paren {\mathbf b + \mathbf c} = \paren {\mathbf a + \mathbf b} + \mathbf c$
where $+$ denotes vector addition.

Proof:
:420px
Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be positioned in space so they are end to end as in the above diagram.
Let $\mathbf v$ be a vector representing the closing side of the polygon whose other $3$ sides are represented by $\mathbf a$, $\mathbf b$ and $\mathbf c$.
By the Parallelogram Law we can add any pair of vectors, and add a third vector to their resultant.
Hence we have:
{{begin-eqn}}
{{eqn | l = \mathbf v
      | r = \mathbf a + \mathbf b + \mathbf c
      | c = 
}}
{{eqn | r = \paren {\mathbf a + \mathbf b} + \mathbf c
      | c = 
}}
{{eqn | r = \mathbf a + \paren {\mathbf b + \mathbf c}
      | c = 
}}
{{eqn | r = \paren {\mathbf a + \mathbf c} + \mathbf b
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Vector Addition is Associative

Let $\mathbf a, \mathbf b, \mathbf c$ be vectors.
Then:
:$\mathbf a + \paren {\mathbf b + \mathbf c} = \paren {\mathbf a + \mathbf b} + \mathbf c$
where $+$ denotes vector addition.

Proof:
Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be expressed in component form:
{{begin-eqn}}
{{eqn | l = \mathbf a
      | r = a_1 \mathbf e_1 + a_2 \mathbf e_2 + \dotsb + a_n \mathbf e_n
}}
{{eqn | l = \mathbf b
      | r = b_1 \mathbf e_1 + b_2 \mathbf e_2 + \dotsb + b_n \mathbf e_n
}}
{{eqn | l = \mathbf c
      | r = c_1 \mathbf e_1 + c_2 \mathbf e_2 + \dotsb + c_n \mathbf e_n
}}
{{eqn | ll= \leadsto
      | l = \paren {\mathbf a + \mathbf b} + \mathbf c
      | r = \paren {\sum_{j \mathop = 1}^n a_j \mathbf e_j + \sum_{j \mathop = 1}^n b_j \mathbf e_j} + \sum_{j \mathop = 1}^n c_j \mathbf e_j
      | c = 
}}
{{eqn | r = \sum_{j \mathop = 1}^n \paren {a_j + b_j} \mathbf e_j + \sum_{j \mathop = 1}^n c_j \mathbf e_j
      | c = Scalar Multiplication of Vectors is Distributive over Vector Addition
}}
{{eqn | r = \sum_{j \mathop = 1}^n \paren {\paren {a_j + b_j} + c_j} \mathbf e_j
      | c = Scalar Multiplication of Vectors is Distributive over Vector Addition
}}
{{eqn | r = \sum_{j \mathop = 1}^n \paren {a_j + \paren {b_j + c_j} } \mathbf e_j
      | c = Associative Law of Addition
}}
{{eqn | r = \sum_{j \mathop = 1}^n a_j \mathbf e_j + \sum_{j \mathop = 1}^n \paren {b_j + c_j} \mathbf e_j
      | c = Scalar Multiplication of Vectors is Distributive over Vector Addition
}}
{{eqn | r = \sum_{j \mathop = 1}^n a_j \mathbf e_j + \paren {\sum_{j \mathop = 1}^n b_j + \sum_{j \mathop = 1}^n c_j \mathbf e_j}
      | c = Scalar Multiplication of Vectors is Distributive over Vector Addition
}}
{{eqn | r = \mathbf a + \paren {\mathbf b + \mathbf c}
      | c = Scalar Multiplication of Vectors is Distributive over Vector Addition
}}
{{end-eqn}}
{{qed}}
Theorem: Scalar Multiplication of Vectors is Associative

Let $\mathbf a$ be a vector quantity.
Let $m, n$ be scalar quantities.
Then:
:$m \paren {n \mathbf a} = \paren {m n} \mathbf a = n \paren {m \mathbf a}$

Proof:
{{ProofWanted|Need to consider which definition you start from}}
Theorem: Scalar Multiplication of Vectors is Distributive over Scalar Addition

Let $\mathbf a$ be a vector quantity.
Let $m, n$ be scalar quantities.
Then:
:$\paren {m + n} \mathbf a = m \mathbf a + n \mathbf a$

Proof:
{{ProofWanted|Need to consider which definition you start from}}
Theorem: Scalar Multiplication of Vectors is Distributive over Vector Addition

Let $\mathbf a, \mathbf b$ be a vector quantities.
Let $m$ be a scalar quantity.
Then:
:$m \paren {\mathbf a + \mathbf b} = m \mathbf a + m \mathbf b$

Proof:
:400px
Let $\mathbf a = \vec {OP}$ and $\mathbf b = \vec {PQ}$.
Then:
:$\vec {OQ} = \mathbf a + \mathbf b$
Let $P'$ and $Q'$ be points on $OP$ and $OQ$ respectively so that:
:$OP' : OP = OQ' : OQ = m$
Then $P'Q'$ is parallel to $PQ$ and $m$ times it in length.
Thus:
:$\vec {P'Q'} = m \mathbf b$
which shows that:
{{begin-eqn}}
{{eqn | l = m \paren {\mathbf a + \mathbf b}
      | r = \vec {OQ'}
      | c = 
}}
{{eqn | r = \vec {OP} + \vec {OP'}
      | c = 
}}
{{eqn | r = m \mathbf a + m \mathbf b
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Magnitude of Vector Cross Product equals Area of Parallelogram Contained by Vectors

Let $\mathbf a$ and $\mathbf b$ be vectors in a vector space of $3$ dimensions:
Let $\mathbf a \times \mathbf b$ denote the vector cross product of $\mathbf a$ with $\mathbf b$.
Then $\norm {\mathbf a \times \mathbf b}$ equals the area of the parallelogram two of whose sides are $\mathbf a$ and $\mathbf b$.

Proof:
By definition of vector cross product:
:$\mathbf a \times \mathbf b = \norm {\mathbf a} \norm {\mathbf b} \sin \theta \, \mathbf {\hat n}$
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
:$\mathbf {\hat n}$ is the unit vector perpendicular to both $\mathbf a$ and $\mathbf b$ in the direction according to the right-hand rule.
As $\mathbf {\hat n}$ is the unit vector:
:$\norm {\paren {\norm {\mathbf a} \norm {\mathbf b} \sin \theta \, \hat {\mathbf n} } } = \norm {\mathbf a} \norm {\mathbf b} \sin \theta$
By Area of Parallelogram, the area of the parallelogram equals the product of one of its bases and the associated altitude.
Let $\mathbf a$ denote the base of the parallelogram.
Then its altitude is $\norm {\mathbf b} \sin \theta$.
The result follows.
{{qed}}