Finish updating notes

week4/03-02-2022/03-02-2022.tex

@@ -21,12 +21,60 @@
 
 \date{01/20/2022}
 \title{%
-        LectureName\\
+        Intro Set Theory\\
             \large MATH--190--01 : Discrete Mathematics for Computing}
 \author{Blizzard MacDougall}
 \begin{document}
 \maketitle
 \pagenumbering{arabic}
 
+\section{Sets}
+A set is an unordered list. A set is said to contain elements or objects.
+Notationally, $a\in S$ is \say{a is contained in S}. $a\notin S$ also exists.
+A set is typically an upper case letter, and lower case letters are elements.
+
+The roster method just lists all the elements in a list, with curly braces around them.
+If you're going to use an ellipses, you need to give enough info for it to be obvious.
+Sets don't \emph{have} to have any common properties, but its helpful if they do.
+
+There's also \emph{set builder notation}. That is, $S={x\in D | P(x)}$, read as
+\say{The set $S$, defined as all $x$ contained in $D$ such that $P(x)$}.
+Examples:
+\begin{equation}
+    \begin{split}
+    \mathbb{N}=\{0,1,2,3,...\}\\
+    \mathbb{Z}=\{...-3,-2,-1,0,1,2,3,...\}\\
+    \mathbb{Z}^+=\mathbb{Z}_+=\mathbb{N}+0\\
+    \mathbb{Q}=\{\frac{a}{b}|a,b\in \mathbb{Z}\ and\ b\neq0\}\\
+    \mathbb{R}=All\ real\ numbers\\
+    \mathbb{R}^+=\mathbb{R}_+=\{x\in \mathbb{R}| x> 0\}\\
+    \mathbb{C}=all\ complex\ numbers
+    \end{split}
+\end{equation}
+Note that natural numbers are kinda fuzzy. This course says zero is natural, others may not.
+Every set is a subset of itself.
+Subsets are noted as $A\subseteq B$. i.e. A is a subset of B.
+\begin{equation}
+    \mathbb{N}\subseteq\mathbb{Z}\subseteq\mathbb{R}\subseteq\mathbb{Q}\subseteq\mathbb{C}
+\end{equation}
+
+Every set has two difinitive subsets, itself, and the empty set (or the null set).$\emptyset$
+A subset that is strictly smaller than the superset is called a proper subset.
+You can check for this with $\subset$.
+Note that there is only one empty set.
+
+Two sets are equal if they contain the same elements. Alternatively, sets are equal
+if they contain each other. To show something's unique, show that if there are two of the thing,
+then prove they're equal.
+
+\section{Cardinality}
+Note that we're sticking with finite sets for now.
+
+If we have a set with exactly $n$ distinct elements, (where $n\in \mathbb{N}$), then
+the cardinality of a set is $n$. Cardinality is notationally written as $\#S=|S|=n$.
+If $S$ is not finite, its infinite. Cardinality of infinte sets are written with $\aleph_n$.
+
+Notably, $|\emptyset|=0$.
+$\wp$
 
 \end{document}

week5/10-02-2022.tex

@@ -0,0 +1,60 @@
+%        File: 20-01-2022.tex
+%     Created: 12:28:35 Thu, 20 Jan 2022 EST
+% Last Change: 12:28:35 Thu, 20 Jan 2022 EST
+%
+\documentclass[letterpaper]{article}
+\usepackage{amsmath}
+\usepackage{graphicx}
+\usepackage{cancel}
+\usepackage{amssymb}
+\usepackage{listings}
+\usepackage[shortlabels]{enumitem}
+\usepackage{soul}
+%\usepackage[smartEllipses,hashEnumerators,hybrid]{markdown}
+\usepackage{geometry}
+\usepackage{dirtytalk}
+\usepackage{lplfitch}
+
+\geometry{portrait, margin=1in}
+
+%\begin{minted}[linenos,bgcolor=LightGray]{[language]}
+
+\date{01/20/2022}
+\title{%
+            Set Theory: Set Operations\\
+            \large MATH--190--01 : Discrete Mathematics for Computing}
+\author{Blizzard MacDougall}
+\begin{document}
+\maketitle
+\pagenumbering{arabic}
+Union $\cup$: Set equivalency to \verb|OR|
+
+Intersection $\cap$: Set equivalency to \verb|AND|
+
+Disjoint: When the intersection is the empty set
+
+Cardinality for union is $|A|+|B|-|A+B|$
+
+The difference between sets ($A-B$) is the objects only in $A$. This is also denoted
+$A\setminus B$. Remember, order matters.
+
+$\overline A = U-A$ (U is the universe)
+
+Generalized unionization is $\underset{i=1}{\overset{n}{\cup}}$
+Generalized intersection is similar.
+
+What is $|A\cup B\cup C|$?
+Because they're overlapping circles, its
+\begin{equation}
+    |A|+|B|+|C|-|A\cap B|-|A\cap C|-B\cap C|+|A\cap B\cap C|
+\end{equation}
+
+Four sets?
+\begin{equation}
+    \begin{split}
+    |A|+|B|+|C|+|D|-(|A\cap B|+...+|A\cap D|)+(|A\cap B\cap C| +...+|B\cap C\cap D|) -|A\cap B\cap C\cap D|
+    \end{split}
+\end{equation}
+
+In short, subtract even-subsets, add odd subsets
+\end{document}