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+\documentclass[a4paper]{article}
+
+% Imports
+\usepackage{amssymb}
+\usepackage{amsmath}
+\usepackage{multicol}
+\usepackage{ragged2e}
+\usepackage{blindtext}
+\usepackage[english]{babel} %this is the dictionary you will use
+\usepackage{graphicx,mathdots,chemarr,fancyvrb,comment} %some more packages
+\usepackage{tikz} %some more packages
+%the packages from here on will help with creating a graph
+%tikzpicture
+\usepackage[many]{tcolorbox}
+\usepackage{wrapfig}
+\usepackage{scalerel}
+\usepackage{pict2e}
+\usepackage{tkz-euclide}
+\usepackage{scalerel}
+\usepackage{pict2e}
+\usepackage{tkz-euclide}
+\usepackage{listings}
+\usepackage{color}
+\usepackage{xifthen}
+\usepackage{hyperref}
+\graphicspath{ {/home/jliu/Documents/texassets/} }
+\definecolor{dkgreen}{rgb}{0,0.6,0}
+%\definecolor{gray}{rgb}{0.5,0.5,0.5}
+\definecolor{red}{HTML}{ffb3b3}
+\definecolor{redbar}{HTML}{ff0000}
+\definecolor{mauve}{rgb}{0.58,0,0.82}
+\definecolor{cyanbar}{HTML}{00bfff}
+\definecolor{cyan}{HTML}{b3f0ff}
+\definecolor{greenbar}{HTML}{00ff00}
+\definecolor{green}{HTML}{8cd98c}
+\definecolor{main}{HTML}{5989cf}    % setting main color to be used
+\definecolor{sub}{HTML}{cde4ff}     % setting sub color to be used
+\tcbset{
+	sharp corners,
+	colback = white,
+	before skip = 0.2cm,    % add extra space before the box
+	after skip = 0.5cm      % add extra space after the box
+}                           % setting global options for tcolorbox
+\makeatletter
+
+
+%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 $n > N  > {#3}$}}} 
+}
+\newcommand{\infobox}[2]{\begin{InfoBox}
+		\smash{\raisebox{-5pt}{\includegraphics[width=0.77cm,height=0.68cm]{information}}}{\bf #1}\newline\newline
+		{#2}
+\end{InfoBox}}
+\newcommand{\warningbox}[2]{\begin{WarningBox}
+		\smash{\raisebox{-6pt}{\includegraphics[width=0.70cm,height=0.70cm]{warning}}}
+		{\bf #1}\newline\newline
+		{#2}
+\end{WarningBox}}
+\newcommand{\theorybox}[2]{\begin{TheoryBox}
+		\smash{\raisebox{-6pt}{\includegraphics[width=0.70cm,height=0.70cm]{theorem}}}
+		{\bf #1}\newline\newline
+		{#2}
+\end{TheoryBox}}
+\newcommand{\notebox}[2]{\begin{NoteBox}
+		\smash{\raisebox{-6pt}{\includegraphics[width=0.55cm,height=0.70cm]{reminder}}}
+		{\bf #1}\newline\newline
+		{#2}
+\end{NoteBox}}
+\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
+	\hskip -\arraycolsep
+	\let\@ifnextchar\new@ifnextchar
+	\array{#1}}
+\newtcolorbox{InfoBox}{
+	colback = sub, 
+	colframe = main, 
+	boxrule = 0pt, 
+	leftrule = 6pt % left rule weight
+}
+\newtcolorbox{WarningBox}{
+	colback = red, 
+	colframe = redbar, 
+	boxrule = 0pt, 
+	leftrule = 6pt % left rule weight
+}
+\newtcolorbox{TheoryBox}{
+	colback = cyan, 
+	colframe = cyanbar, 
+	boxrule = 0pt, 
+	leftrule = 6pt % left rule weight
+}
+\newtcolorbox{NoteBox}{
+	colback = green, 
+	colframe = greenbar, 
+	boxrule = 0pt, 
+	leftrule = 6pt % left rule weight
+}
+\newtcolorbox{TitleBox}{
+	boxrule = 2pt,
+	rounded corners
+}
+\makeatother
+\usepackage[letterpaper,left=6mm,includemp=true,marginparwidth=12mm,marginparsep=1mm,reversemarginpar,right=19mm,
+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}