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%the packages from here on will help with creating a graph
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\usepackage{scalerel}
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\usepackage{tkz-euclide}
\usepackage{scalerel}
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\usepackage{tkz-euclide}
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%My Custom Commands
\newcommand{\mnewline}{\newline\newline\newline}
\newcommand{\mline}{\rule{0.5cm}{0.5pt}}
\newcommand{\proj}[1]{\text{Proj}_{#1}}
\newcommand{\st}{\ni:}
\newcommand{\evaline}[2]{\Big|^{#1}_{#2}}
\newcommand{\nulli}[1]{\text{Null }{#1}}
\newcommand{\ran}[1]{\text{ran }{#1}}
\newcommand{\col}[1]{\text{Col }({#1})}
\newcommand{\re}[1]{\text{Re}({#1})}
\newcommand{\im}[1]{\text{Im}({#1})}
\newcommand{\spa}[1]{\text{span}\{{#1}\}}
%\newcommand{\neproof}[3]{$\text{Let } \epsilon > {#2} \text{ be given}$\\\text{Choose $N = {#1}$\text{ Suppose $n > N  > {#3}$}}}
\newcommand{\neproof}[3]{ %The first one is without the 3rd argument and the second one is
	\ifthenelse{\isempty{#3}}{$\text{Let } \epsilon > {#2} \text{ be given}$\\\text{Choose $N = {#1}$\text{ Suppose $n > N$}}}
	{$\text{Let } \epsilon > {#2} \text{ be given}$\\\text{Choose $N = {#1}$\text{ Suppose $n > N  > {#3}$}}} 
}
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includefoot=true,top=19mm,nohead,footskip=12mm,bottom=6mm]{geometry}
% Here are the custom commands I have created. They are increadibly retarded
% mnewline: creates 3 newlines
% mline: Creates a horizontal line
% proj: Creates a Proj with a suitable subscript - Takes an argument
% st: creates a ni and a : as the 'such that'
% evaline: creates a vertical line for evaluated definite integrals. First argument is upper limit, second is lower. - Takes two arguments
% nulli: creates a Null (with a whitespace) - Takes an argument
% col: creates a Col (with a whitespace) - Takes an argument
% ran: creates a ran (with a whitespace) - Takes an argument
% re: creates a Re() - Takes an argument
% im: creates a Im() - Takes an argument
% sp: creates a span{} - Takes an argument
% neproof: Creates a cookie cutter N-epsilon proof. First argument set's N's value and second argument sets epsilon greater than value and the third (optional) argument sets the n > N > value. IF YOU DO NOT WANT THE THIRD ARGUMENT YOU NEED AN EMPTY CURLY BRACKET
\begin{document}
	\setlength{\parindent}{1cm}
	\begin{center}
		{\bf \Large Self-Study Linear Algebra}
	\end{center}
	\begin{TitleBox}
		\begin{center}
			{\bf bruh}
		\end{center}
	\end{TitleBox}
	\theorybox{The center of it all}{Linear algebra, at it's core, is the study of linear maps on finite-dimensional vector spaces.}
	\section{Foundational Knowledge}
	\subsubsection{Lists}
	\theorybox{Definition -- List -- Tuple}{A list is an ordered collection of non-negative {\bf finite} length of elements. Two lists are equal if and only if they contain the same elements in the same order and have the same length.}
	\notebox{Lists are fucking finite}{Lists are finite. If you have an infinite number of elements, you do not have a list/tuple.}
	\notebox{Notation}{Lists are surrounded by parenthesis, sets are surrounded by curly brackets.}
	\section{Fields}
	\theorybox{Definition -- Field}{A field is an algebric structure where the underlying set of objects contains at least two distinct elements, $0$ and $1$, along with two operations that satisfy seven total axioms. $0$ we usually refer to as zero or the additive identity, and $1$ we usually call one. The two operations, we usually refer to as addition and multiplication. Most people are familiar with them as $+$ and as $\times$. In this document, we shall refer the generalized addition as $\boxplus$ and generalized multiplication as $\boxdot$.\\
	The six axioms that must be satisfied for an algebraic structure to be considered a field are:
	\begin{enumerate}
	    \item Addition \& Multiplication must be associative
	    \item Addition \& Multiplication must be commutative
	    \item There must exist a unique additive identity such that the sum of any element with the additive identity results in the original element
	    \item There must exist a unique additive inverse for each element such that the sum of said element with it's inverse is the additive identity 
	    \item There must exist a multiplicative identity such the multiplication of any element with the identity is the original element
	    \item There must exist a unique multiplicative inverse for each element such that the multiplication of any element with the inverse gives the multiplicative identity
	    \item Multiplication must distribute over addition 
	\end{enumerate}
	Do note that often the identity axioms are squished together thus forming 6 axioms rather than 7, but just remember that these must be satisfied.\\\\
	Also remember that it is important for both operations to be closed. If they are not closed, then we do not have a field.
	}
	\subsubsection{Proof that Additive Inverses are unique for complex numbers}
	Let's assume towards a contradiction that additive inverses are not unique.\\
	Let $x \in \mathbb{C}$ and let $\alpha, \beta \in \mathbb{C}$ such that $\alpha \not= \beta$ and $x + \alpha = 0$ and $x + \beta = 0$.\\
	Let $x = a + bi$, $\alpha = c + di$ and $\beta = e + fi$ where $a, b, c, d, e, f \in \mathbb{R}$.\\
	We can observe that
	\begin{equation}
	    x + \alpha = x + \beta
	\end{equation}
	Then
	\begin{alignat*}{3}
	    &x + \alpha = x + \beta&\\
	    \Rightarrow&a + bi + c + di + -a = a + bi + e + fi + -a&\text{by additive inverse of $\mathbb{R}$}\\
	    \Rightarrow&bi + di + c - c = bi + fi + e + -c&\text{by additive inverse of $\mathbb{R}$}\\
	    \Rightarrow&bi + di = bi + fi + e - c&\\
	    \Rightarrow&di = fi + e - c&\text{(1)}\\
	    \Rightarrow&di + c = fi + e + -c + c&\\
	    \Rightarrow&di + c = fi + e&\text{As previously established, $-c$ is the additive inverse of $c$}\\
	    \Rightarrow&\alpha = \beta&\\
	\end{alignat*}
	And thus we have arrived at a contradiction, thus additive inverses are unique.\\\\
	Note that (1) can be justified by adding the additive inverse of $bi$ to both sides.
	\subsubsection{Proof that Multiplicative Inverses are unique for complex numbers}
	Let $x \in \mathbb{C}$ and let $\alpha, \beta \in \mathbb{C}$ such that $\alpha x = 1$ and $\beta x = 1$.\\
	Then
	\begin{alignat*}{3}
	    &\alpha&\qquad\\
	    =&1 \alpha&\qquad\text{by multiplicative identity}\\
	    =&\beta x \alpha&\qquad\text{by def}\\
	    =&\beta (x \alpha)&\qquad\text{since multiplication is associative}\\
	    =&\beta 1&\qquad\text{by def}\\
	    =&\beta&\qquad\\
	\end{alignat*}
	Thus $\alpha = \beta$, and thus multiplicative inverses are unique.
\end{document}