avewright/proofwiki-math huggingface dataset:
Theorem: Cauchy-Bunyakovsky-Schwarz Inequality/Inner Product Spaces

Let $\mathbb K$ be a subfield of $\C$.
Let $V$ be a semi-inner product space over $\mathbb K$.
Let $x, y$ be vectors in $V$.
Then:
:$\size {\innerprod x y}^2 \le \innerprod x x \innerprod y y$

Proof:
Let $x, y \in V$.
Let $\lambda \in \mathbb K$.
Then:
{{begin-eqn}}
{{eqn | l = 0
      | o = \le
      | r = \innerprod {x - \lambda y} {x - \lambda y}
      | c = Semi-inner product axioms $\paren {3}$: Non-Negative Definiteness 
}}
{{eqn | r = \innerprod x x + \innerprod x {-\lambda y} + \innerprod {-\lambda y} x + \innerprod {-\lambda y} {-\lambda y}
      | c = Semi-inner product axioms $\paren {2}$:  	Sesquilinearity 
}}
{{eqn | r = \innerprod x x - \lambda^* \innerprod x y - \lambda \innerprod x y^* + \lambda \lambda^* \innerprod y y
      | c = Semi-inner product axioms $\paren {2}$ and $\paren {1}$ 
}}
{{end-eqn}}
where $\lambda^*$ denotes the complex conjugate of $\lambda$.
(If $\mathbb K$ is a subfield of $\R$ , then $\lambda^* = \lambda$.)
First, suppose $\innerprod y y \ne 0$.
Insert $\lambda = \innerprod x y \innerprod y y^{-1}$ in the inequality:
{{begin-eqn}}
{{eqn | l = 0
      | o = \le
      | r = \innerprod x x - \innerprod x y^* \paren {\innerprod y y^{-1} }^* \innerprod x y - \innerprod x y \innerprod y y^{-1} \innerprod x y^* + \innerprod x y \innerprod y y^{-1} \innerprod x y^* \paren {\innerprod y y^{-1} }^* \innerprod y y
      | c = Product of Complex Conjugates
}}
{{eqn | r = \innerprod x x - \size {\innerprod x y}^2 \paren {\innerprod y y^{-1} }^* - \size {\innerprod x y}^2 \innerprod y y^{-1} + \size {\innerprod x y}^2 \paren {\innerprod y y^{-1} }^* 
      | c = Modulus in Terms of Conjugate
}}
{{eqn | r = \innerprod x x - \size {\innerprod x y}^2 \innerprod y y^{-1} 
}}
{{end-eqn}}
Reorder the inequality to get:
:$\size {\innerprod x y}^2 \le \innerprod x x \innerprod y y$
{{qed|lemma}}
Next, suppose $\innerprod y y = 0$.
By Number Field has Rational Numbers as Subfield, $\Q \subset \mathbb K$.
Let $n \in \N \subset \Q \subset \mathbb K$ be a natural number.
Insert $\lambda = n \innerprod x y$ in the inequality:
{{begin-eqn}}
{{eqn | l = 0
      | o = \le
      | r = \innerprod x x - n^* \innerprod x y^* \innerprod x y - n \innerprod x y \innerprod x y^* + 0
      | c = Product of Complex Conjugates
}}
{{eqn | r = \innerprod x x - n \innerprod x y^* \innerprod x y - n \innerprod x y \innerprod x y^*
      | c = Complex Number equals Conjugate iff Wholly Real
}}
{{eqn | r = \innerprod x x - 2 n \size {\innerprod x y}^2
      | c = Modulus in Terms of Conjugate
}}
{{end-eqn}}
Rearrange the inequality to get:
:$\innerprod x x \ge 2 n \size {\innerprod x y}^2$
If $\size {\innerprod x y}^2 \ne 0$, then:
:$\dfrac {\innerprod x x} {2 \size {\innerprod x y}^2} \ge n$
for every $n \in \N$, contradicting the Axiom of Archimedes.
Therefore, by Proof by Contradiction:
:$\size {\innerprod x y}^2 = 0$.
Then:
:$\size {\innerprod x y}^2 \le 0 = \innerprod x x \innerprod y y$
{{qed}}
Theorem: Cauchy-Bunyakovsky-Schwarz Inequality/Inner Product Spaces

Let $\mathbb K$ be a subfield of $\C$.
Let $V$ be a semi-inner product space over $\mathbb K$.
Let $x, y$ be vectors in $V$.
Then:
:$\size {\innerprod x y}^2 \le \innerprod x x \innerprod y y$

Proof:
This proof assumes that $V$ is a semi-inner product space over $\R$.
Then for all $x, y \in V$, we have $\innerprod x y = \innerprod y x$ by property $(1')$ of semi-inner products.
Define $f_{x, y}: \R \to \R_{\ge 0}$ by:
:$\map {f_{x, y} } \lambda = \innerprod {x - \lambda y} {x - \lambda y}$
Then by property $(4)$ of semi-inner product:
:$\forall \lambda \in \R: \map {f_{x, y} } \lambda \ge 0$
For all $\lambda \in \R$, it follows that:
{{begin-eqn}}
{{eqn | l = \map {f_{x, y} } \lambda
      | r = \innerprod {x - \lambda y} {x - \lambda y}
}}
{{eqn | r = \innerprod x x + \innerprod x {-\lambda y} + \innerprod {-\lambda y} x + \innerprod {-\lambda y} {-\lambda y}
      | c = Property $(3)$ of Semi-Inner Product
}}
{{eqn | r = \innerprod x x - 2 \lambda \innerprod x y + \lambda^2 \innerprod y y
      | c = Properties $(1')$ and $(2)$ of Semi-Inner Product
}}
{{eqn | r = a \lambda^2 + b \lambda + c
}}
{{end-eqn}}
where we have put $a = \innerprod y y$, $b = -2 \innerprod x y$, and $c = \innerprod x x$.
Then $f_{x, y}$ is a quadratic polynomial which satisfies $\map {f_{x, y} } \lambda \ge 0$.
Hence $f_{x, y}$ has at most one distinct real root.
From Solution to Quadratic Equation, it follows that the discriminant $\Delta$ satisfies:
:$\Delta = b^2 - 4 a c \le 0$
Therefore:
:$4 \innerprod x y^2 - 4 \innerprod x x \innerprod y y \le 0$
which we rearrange as:
:$\size {\innerprod x y}^2 = \innerprod x y^2 \le \innerprod x x \innerprod y y$
{{qed}}
Theorem: Cauchy-Bunyakovsky-Schwarz Inequality/Definite Integrals

Let $f$ and $g$ be real functions which are continuous on the closed interval $\closedint a b$.
Then:
:$\ds \paren {\int_a^b \map f t \, \map g t \rd t}^2 \le \int_a^b \paren {\map f t}^2 \rd t \int_a^b \paren {\map g t}^2 \rd t$

Proof:
{{begin-eqn}}
{{eqn | q = \forall x \in \R
      | l = 0
      | o = \le
      | r = \paren {x \map f t + \map g t}^2
}}
{{eqn | l = 0
      | o = \le
      | r = \int_a^b \paren {x \map f t + \map g t}^2 \rd t
      | c = Relative Sizes of Definite Integrals
}}
{{eqn | r = x^2 \int_a^b \paren {\map f t}^2 \rd t + 2 x \int_a^b \map f t \, \map g t \rd t + \int_a^b \paren {\map g t}^2 \rd t
      | c = Linear Combination of Definite Integrals
}}
{{eqn | r = A x^2 + 2 B x + C
      | c = 
}}
{{end-eqn}}
where:
{{begin-eqn}}
{{eqn | l = A
      | r = \int_a^b \paren {\map f t}^2 \rd t
}}
{{eqn | l = B
      | r = \int_a^b \map f t \map g t \rd t
}}
{{eqn | l = C
      | r = \int_a^b \paren {\map g t}^2 \rd t
}}
{{end-eqn}}
The quadratic equation $A x^2 + 2 B x + C$ is non-negative for all $x$.
It follows (using the same reasoning as in Cauchy's Inequality) that the discriminant $\paren {2 B}^2 - 4 A C$ of this polynomial must be non-positive.
Thus:
:$B^2 \le A C$
and hence the result.
{{qed}}
Theorem: Cauchy-Schwarz Inequality/Complex Numbers

:$\ds \paren {\sum \cmod {w_i}^2} \paren {\sum \cmod {z_i}^2} \ge \cmod {\sum w_i z_i}^2$
where all of $w_i, z_i \in \C$.

Proof:
Let $w_1, w_2, \ldots, w_n$ and $z_1, z_2, \ldots, z_n$ be arbitrary complex numbers.
Take the Binet-Cauchy Identity:
:$\ds \paren {\sum_{i \mathop = 1}^n a_i c_i} \paren {\sum_{j \mathop = 1}^n b_j d_j} = \paren {\sum_{i \mathop = 1}^n a_i d_i} \paren {\sum_{j \mathop = 1}^n b_j c_j} + \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {a_i b_j - a_j b_i} \paren {c_i d_j - c_j d_i}$
and set $a_i = w_i, b_j = \overline {z_j}, c_i = \overline {w_i}, d_j = z_j $.
This gives us:
{{begin-eqn}}
{{eqn | l = \paren {\sum_{i \mathop = 1}^n w_i \overline {w_i} } \paren {\sum_{j \mathop = 1}^n \overline {z_j} z_j}
      | r = \paren {\sum_{i \mathop = 1}^n w_i z_i} \paren {\sum_{j \mathop = 1}^n \overline {z_j} \overline {w_j} }
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {w_i \overline {z_j} - w_j \overline {z_i} } \paren {\overline {w_i} z_j - \overline {w_j} z_i}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {\sum_{i \mathop = 1}^n w_i \overline {w_i} } \paren {\sum_{j \mathop = 1}^n \overline {z_j} z_j}
      | r = \paren {\sum_{i \mathop = 1}^n w_i z_i} \overline {\paren {\sum_{i \mathop = 1}^n w_i z_i} }
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {w_i \overline {z_j} - w_j \overline {z_i} } \overline {\paren {w_i \overline {z_j} - w_j \overline {z_i} } }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {\sum_{i \mathop = 1}^n \cmod {w_i}^2} \paren {\sum_{j \mathop = 1}^n \cmod {z_j}^2}
      | r = \cmod {\sum_{i \mathop = 1}^n w_i z_i}^2 + \sum_{1 \mathop \le i \mathop < j \mathop \le n} \cmod {w_i \overline {z_j} - w_j \overline {z_i} }^2
      | c = Modulus in Terms of Conjugate
}}
{{eqn | ll= \leadsto
      | l = \paren {\sum_{i \mathop = 1}^n \cmod {w_i}^2} \paren {\sum_{j \mathop = 1}^n \cmod {z_j}^2}
      | o = \ge
      | r = \cmod {\sum_{i \mathop = 1}^n w_i z_i}^2
      | c = Complex Modulus is Non-Negative
}}
{{end-eqn}}
Hence the result.
{{qed}}
{{Namedfor|Augustin Louis Cauchy|name2 = Karl Hermann Amandus Schwarz}}
Theorem: Ceiling of Floor is Floor

:$\ceiling {\floor x} = \floor x$

Proof:
Let $y = \floor x$.
By Floor Function is Integer, we have that $y \in \Z$.
Then from Real Number is Integer iff equals Ceiling, we have:
:$\ceiling y = y$
So:
:$\ceiling {\floor x} = \floor x$
{{qed}}
Theorem: Floor of Ceiling is Ceiling

:$\floor {\ceiling x} = \ceiling x$

Proof:
Let $y = \ceiling x$.
By Ceiling Function is Integer, we have that $y \in \Z$.
From Real Number is Integer iff equals Floor, we have:
:$\floor y = y$
So:
:$\floor {\ceiling x} = \ceiling x$
{{qed}}
Theorem: Floor equals Ceiling iff Integer

:<nowiki>$\floor x = \begin {cases}
\ceiling x & : x \in \Z \\
\ceiling x - 1 & : x \notin \Z \\
\end {cases}$</nowiki>
or equivalently:
:<nowiki>$\ceiling x = \begin {cases}
\floor x & : x \in \Z \\
\floor x + 1 & : x \notin \Z \\
\end {cases}$</nowiki>
where $\Z$ is the set of integers.

Proof:
From Real Number is Integer iff equals Floor:
:$x \in \Z \implies x = \floor x$
From Real Number is Integer iff equals Ceiling:
:$x \in \Z \implies x = \ceiling x$
So:
:$x \in \Z \implies \floor x = \ceiling x$
Now let $x \notin \Z$.
From the definition of the floor function:
:$\floor x = \map \sup {\set {m \in \Z: m \le x} }$
From the definition of the ceiling function:
:$\ceiling x = \map \inf {\set {m \in \Z: m \ge x} }$
Thus:
:$\floor x < x < \ceiling x$
Hence the result, from the definition of $\inf$ and $\sup$.
{{qed}}
Theorem: Ceiling of Negative equals Negative of Floor

:$\ceiling {-x} = -\floor x$

Proof:
From Integer equals Floor iff between Number and One Less we have:
:$x - 1 < \floor x \le x$
and so, by multiplying both sides by -1:
:$-x + 1 > -\floor x \ge -x$
From Integer equals Ceiling iff between Number and One More we have:
:$\ceiling x = n \iff x \le n < x + 1$
Hence:
:$-x \le -\floor x < -x + 1 \implies \ceiling {-x} = -\floor x$
{{Qed}}
Theorem: Floor of Negative equals Negative of Ceiling

:$\floor {-x} = -\ceiling x$

Proof:
From Integer equals Ceiling iff between Number and One More we have:
:$x \le \ceiling x < x + 1$
and so, by multiplying by -1:
:$-x \ge -\ceiling x > -x - 1$
From Integer equals Floor iff between Number and One Less we have:
:$\floor x = n \iff x - 1 < n \le x$
Hence:
:$-x - 1 < -\ceiling x \le -x \implies \floor {-x} = -\ceiling x$
{{qed}}
Theorem: Integer is Coprime to 1

Every integer is coprime to $1$.
That is:
:$\forall n \in \Z: n \perp 1$

Proof:
Follows from the definitions of coprime and greatest common divisor as follows.
When $n \in \Z: n \ne 0$ we have:
:$\gcd \set {n, 1} = 1$
Then by definition again:
:$\gcd \set {n, 0} = n$
and so when $n = 1$ we have:
:$\gcd \set {1, 0} = 1$
{{qed}}
Theorem: Integers Coprime to Zero

The only integers which are coprime to zero are $1$ and $-1$.
That is:
:$n \in \Z: n \perp 0 \iff n \in \set {1, -1}$
In particular, note that two integers which are coprime to each other cannot both be $0$.

Proof:
Let $n \in \Z$ such that $n \perp 0$.
First we note that from Coprime Integers cannot Both be Zero, it cannot be the case that $n = 0$.
From the definition of coprime, we have:
:$m \perp n \iff \gcd \set {m, n} = 1$
From the definition of greatest common divisor:
:$\gcd \set {n, 0} = \size n$
where $\size n$ is the absolute value of $n$.
Let $n \in \set {1, -1}$.
Then:
:$\gcd \set {n, 0} = \size n = 1$
and so $n \perp 0$.
{{qed}}
Theorem: Sum of Binomial Coefficients over Upper Index

{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^n \binom j m
      | r = \binom {n + 1} {m + 1}
      | c = 
}}
{{eqn | r = \dbinom 0 m + \dbinom 1 m + \dbinom 2 m + \cdots + \dbinom n m = \dbinom {n + 1} {m + 1}
      | c = 
}}
{{end-eqn}}

Proof:
Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
:$\ds \sum_{k \mathop = 0}^n \binom k m = \binom {n + 1} {m + 1}$
=== Basis for the Induction ===
$\map P 0$ says:
:$\dbinom 0 m = \dbinom 1 {m + 1}$
When $m = 0$ we have by definition:
:$\dbinom 0 0 = 1 = \dbinom 1 1$
When $m > 0$ we also have by definition:
:$\dbinom 0 m = 0 = \dbinom 1 {m + 1}$
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P r$ is true, where $r \ge 0$, then it logically follows that $\map P {r + 1}$ is true.
So this is our induction hypothesis:
:$\ds \sum_{k \mathop = 0}^r \binom k m = \binom {r + 1} {m + 1}$
Then we need to show:
:$\ds \sum_{k \mathop = 0}^{r + 1} \binom k m = \binom {r + 2} {m + 1}$
=== Induction Step ===
This is our induction step:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^{r + 1} \binom k m
      | r = \sum_{k \mathop = 0}^r \binom k m + \binom {r + 1} m
      | c = 
}}
{{eqn | r = \binom {r + 1} {m + 1} + \binom {r + 1} m
      | c = Induction Hypothesis
}}
{{eqn | r = \binom {r + 2} {m + 1}
      | c = Pascal's Rule
}}
{{end-eqn}}
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall m, n \in \Z, m \ge 0, n \ge 0: \sum_{k \mathop = 0}^n \binom k m = \binom {n + 1} {m + 1}$
{{qed}}
Theorem: Divisibility of Product of Consecutive Integers

The product of $n$ consecutive positive integers is divisible by the product of the ''first'' $n$ consecutive positive integers.
That is:
:$\ds \forall m, n \in \Z_{>0}: \exists r \in \Z: \prod_{k \mathop = 1}^n \paren {m + k} = r \prod_{k \mathop = 1}^n k$

Proof:
{{begin-eqn}}
{{eqn | l = \prod_{k \mathop = 1}^n \paren {m + k}
      | r = \frac {\paren {m + n}!} {m!}
      | c = 
}}
{{eqn | r = n! \frac {\paren {m + n}!} {m! \, n!}
      | c = 
}}
{{eqn | r = n! \binom {m + n} m
      | c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = \binom {m + n} m \prod_{k \mathop = 1}^n k
      | c = 
}}
{{end-eqn}}
Hence the result, and note that for a bonus we have identified exactly what the divisor is: 
:$\dbinom {m + n} m$
{{qed}}
Theorem: Equal Numbers are Congruent

:$\forall x, y, z \in \R: x = y \implies x \equiv y \pmod z$
where $x \equiv y \pmod z$ denotes congruence modulo $z$.

Proof:
{{begin-eqn}}
{{eqn | l = x
      | r = y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x - y
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x - y
      | r = 0 \cdot z
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | o = \equiv
      | r = y
      | rr= \pmod z
      | c = {{Defof|Congruence (Number Theory)|Congruence Modulo $z$}}
}}
{{end-eqn}}
{{qed}}
Theorem: Complex Roots of Unity in Exponential Form

Let $n \in \Z$ be an integer such that $n > 0$.
Let $z \in \C$ be a complex number such that $z^n = 1$.
Then:
:$U_n = \set {e^{2 i k \pi / n}: k \in \N_n}$
where $U_n$ is the set of $n$th roots of unity.
That is:
:$z \in \set {1, e^{2 i \pi / n}, e^{4 i \pi / n}, \ldots, e^{2 \paren {n - 1} i \pi / n} }$
Thus for every integer $n$, the number of $n$th roots of unity is $n$.
Setting $\omega := e^{2 i \pi / n}$, $U_n$ can then be written as:
:$U_n = \set {1, \omega, \omega^2, \ldots, \omega^{n - 1} }$

Proof:
Let $z \in \set {e^{2 i k \pi / n}: k \in \N_n}$.
Then:
:$z^n \in \set {e^{2 i k \pi}: k \in \N_n}$
Hence $z^n = 1$.
Now suppose $z^n = 1$.
We determine the possible values of $z$ using Roots of Complex Number.
Let $z = r e^{i \theta}$, where:
:$r$ is a positive real number
:$\theta$ is a real number.
Then $\cmod {z^n} = 1 \implies \cmod z = 1$ since $0 \le r$.
Similarly, we have $n \theta = 0 \bmod 2 \pi$.
So $\theta = \dfrac {2 k \pi} n$ for $k \in \Z$.
Hence the result.
{{qed}}
{{handwaving|This is a bit slapdash - anyone care to improve it?}}
Theorem: Roots of Unity under Multiplication form Cyclic Group

Let $n \in \Z$ be an integer such that $n > 0$.
The $n$th complex roots of unity under the operation of multiplication form the cyclic group which is isomorphic to $C_n$.

Proof:
From Complex Roots of Unity in Exponential Form:
:$U_n = \set {e^{2 i k \pi / n}: k \in \N_n}$
where $U_n$ is the set of complex $n$th roots of unity.
Let $\omega = e^{2 i \pi / n}$.
Then we have:
:$U_n = \set {\omega^k: k \in \N_n}$
that is:
:$U_n = \set {\omega^0, \omega^1, \omega^2, \ldots, \omega^{n - 1} }$
Let $\omega^a, \omega^b \in U_n$.
Then $\omega^a \omega^b = \omega^{a + b} \in U_n$.
Either $a + b < n$, in which case $\omega^{a + b} \in U_n$, or $a + b \ge n$, in which case:
{{begin-eqn}}
{{eqn | l = \omega^a \omega^b
      | r = \omega^{a + b}
      | c = 
}}
{{eqn | r = \omega^{n + t}
      | c = for some $t < n$
}}
{{eqn | r = \omega^n \omega^t
      | c = 
}}
{{eqn | r = \omega^t
      | c = as $\omega^n = 1$
}}
{{end-eqn}}
So $U_n$ is closed under multiplication.
We have that $\omega_0 = 1$ is the identity and that $\omega^{n - t}$ is the inverse of $\omega^t$.
Finally we note that $U_n$ is generated by $\omega$.
Hence the result, by definition of cyclic group, and from Cyclic Groups of Same Order are Isomorphic:
:$U_n = \gen \omega \cong C_n$.
{{qed}}
Theorem: Inverse Mapping is Unique

Let $f: S \to T$ be a mapping.
If $f$ has an inverse mapping, then that inverse mapping is unique.
That is, if:
:$f$ and $g$ are inverse mappings of each other
and
:$f$ and $h$ are inverse mappings of each other
then $g = h$.

Proof:
By the definition of inverse mapping:
{{begin-eqn}}
{{eqn | l = g \circ f
      | r = I_S
      | c = 
}}
{{eqn | r = h \circ f
      | c = 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = f \circ g
      | r = I_T
      | c = 
}}
{{eqn | r = f \circ h
      | c = 
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | l = h
      | r = h \circ I_T
      | c = 
}}
{{eqn | r = h \circ \paren {f \circ g}
      | c = 
}}
{{eqn | r = \paren {h \circ f} \circ g
      | c = Composition of Mappings is Associative
}}
{{eqn | r = I_S \circ g
      | c = 
}}
{{eqn | r = g
      | c = 
}}
{{end-eqn}}
So $g = h$ and the inverse is unique.
{{qed}}
Theorem: Inverse Mapping is Unique

Let $f: S \to T$ be a mapping.
If $f$ has an inverse mapping, then that inverse mapping is unique.
That is, if:
:$f$ and $g$ are inverse mappings of each other
and
:$f$ and $h$ are inverse mappings of each other
then $g = h$.

Proof:
We need to show that:
:$\forall t \in T: \map g t = \map h t$
So:
{{begin-eqn}}
{{eqn | l = \map f {\map g t}
      | r = t
      | c = {{Defof|Inverse Mapping}}
}}
{{eqn | ll= \leadsto
      | l = \map h t
      | r = \map h {\map f {\map g t} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map h t
      | r = \map g t
      | c = as $\forall s \in S: \map h {\map f s} = s$
}}
{{end-eqn}}
{{qed}}
Hence the result.
Theorem: Special Linear Group is Subgroup of General Linear Group

Let $K$ be a field whose zero is $0_K$ and unity is $1_K$.
Let $\SL {n, K}$ be the special linear group of order $n$ over $K$.
Then $\SL {n, K}$ is a subgroup of the general linear group $\GL {n, K}$.

Proof:
Because the determinants of the elements of $\SL {n, K}$ are not $0_K$, they are nonsingular.
So $\SL {n, K}$ is a subset of $\GL {n, K}$.
Now we need to show that $\SL {n, K}$ is a subgroup of $\GL {n, K}$.
Let $\mathbf A$ and $\mathbf B$ be elements of $\SL {n, K}$.
As $\mathbf A$ is nonsingular we have that it has an inverse  $\mathbf A^{-1} \in \GL {n, K}$.
From Determinant of Inverse Matrix:
:$\map \det {\mathbf A^{-1} } = \dfrac 1 {\map \det {\mathbf A} }$
and so:
:$\map \det {\mathbf A^{-1} } = 1$
So $\mathbf A^{-1} \in \SL {n, K}$.
Also, from Determinant of Matrix Product:
:$\map \det {\mathbf A \mathbf B} = \map \det {\mathbf A} \map \det {\mathbf B} = 1$
Hence the result from the Two-Step Subgroup Test.
{{qed}}
Theorem: Determinant of Unit Matrix

Let $R$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
The determinant of the unit matrix of order $n$ over $R$ is equal to $1_R$.

Proof:
Let $\mathbf I_n$ denote the unit matrix of order $n$ over $R$.
The proof proceeds by induction.
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:
:$\map \det {\mathbf I_n} = 1_R$
Theorem: Determinant of Inverse Matrix

Let $K$ be a field whose zero is $0_K$ and whose unity is $1_K$.
Let $\mathbf A$ be a nonsingular matrix of order $n$ over $K$.
Then the determinant of its inverse is given by:
:$\map \det {\mathbf A^{-1} } = \dfrac {1_K} {\map \det {\mathbf A} }$

Proof:
By definition of inverse matrix:
:$\mathbf A \mathbf A^{-1} = \mathbf I_n$
where $\mathbf I_n$ is the unit matrix.
By Determinant of Unit Matrix:
:$\map \det {\mathbf I_n} = 1_K$
By Determinant of Matrix Product:
:$\map \det {\mathbf A^{-1} } \map \det {\mathbf A} = \map \det {\mathbf A^{-1} \mathbf A}$
Hence the result.
{{qed}}
Theorem: Transposition is of Odd Parity

Let $S_n$ denote the set of permutations on $n$ letters.
Let $\pi \in S_n$ be a transposition.
Then $\pi$ is of odd parity.

Proof:
Let $\pi = \begin{pmatrix} 1 & 2 \end{pmatrix}$ be a transposition.
Let $\Delta_n$ be defined as Product of Differences.
Then $\forall n \in \N_{>0}: \pi \cdot \Delta_n$ produces only one sign change in $\Delta_n$, that is, the one occurring in the factor $\paren {x_1 - x_2}$.
Thus:
:$\begin{pmatrix} 1 & 2 \end{pmatrix} \cdot \Delta_n = - \Delta_n$
and thus $\begin{pmatrix} 1 & 2 \end{pmatrix}$ is odd.
From Conjugates of Transpositions:
:$\begin{pmatrix} 1 & k \end{pmatrix} = \begin{pmatrix} 2 & k \end{pmatrix} \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & k \end{pmatrix}$
Thus as $\begin{pmatrix} 2 & k \end{pmatrix}$ is self-inverse:
:$\begin{pmatrix} 1 & k \end{pmatrix} = \begin{pmatrix} 2 & k \end{pmatrix} \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & k \end{pmatrix}^{-1}$
But from Parity of Conjugate of Permutation:
:$\map \sgn {\begin{pmatrix} 2 & k \end{pmatrix} \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & k \end{pmatrix}^{-1} } = \sgn {\begin{pmatrix} 1 & 2 \end{pmatrix} }$
Thus:
:$\sgn {\begin{pmatrix} 1 & k \end{pmatrix} } = \sgn {\begin{pmatrix} 1 & 2 \end{pmatrix} }$
and thus $\begin{pmatrix} 1 & k \end{pmatrix}$ is odd.
Finally:
{{begin-eqn}}
{{eqn | l = \begin{pmatrix} h & k \end{pmatrix}
      | r = \begin{pmatrix} 1 & h \end{pmatrix} \begin{pmatrix} 1 & k \end{pmatrix} \begin{pmatrix} 1 & h \end{pmatrix}
      | c = Conjugates of Transpositions
}}
{{eqn | r = \begin{pmatrix} 1 & h \end{pmatrix} \begin{pmatrix} 1 & k \end{pmatrix} \begin{pmatrix} 1 & h \end{pmatrix}^{-1}
      | c = Transposition is Self-Inverse
}}
{{end-eqn}}
But from Parity of Conjugate of Permutation:
:$\map \sgn {\begin{pmatrix} 1 & h \end{pmatrix} \begin{pmatrix} 1 & k \end{pmatrix} \begin{pmatrix} 1 & h \end{pmatrix}^{-1} } = \sgn {\begin{pmatrix} 1 & k \end{pmatrix} }$
Thus:
:$\sgn {\begin{pmatrix} h & k \end{pmatrix} } = \sgn {\begin{pmatrix} 1 & k \end{pmatrix} }$
and thus $\begin{pmatrix} h & k \end{pmatrix}$ is odd.
{{qed}}
Theorem: Stabilizer in Group of Transformations

Let $X$ be any set with $n$ elements (where $n \in \Z_{>0}$).
Consider the symmetric group on $n$ letters $S_n$ as a group of transformations on $X$.
Let $x \in X$.
Then the stabilizer of $x$ is isomorphic to $S_{n - 1}$.

Proof:
Consider the initial segment of natural numbers $\N_n = \set {1, 2, \ldots, n}$.
By the definition of cardinality, $H$ is equivalent to $\N_n$.
{{WLOG}} we can consider $S_n$ acting directly on $\N_n$.
The stabilizer of $n$ in $\N_n$ is all the permutations of $S_n$ which fix $n$, which is clearly $S_{n - 1}$.
A permutation can be applied to $\N_n$ so that $i \to n$ for any $i$.
Thus one can build an isomorphism to show the result for a general $i$.
{{qed}}