Add last lecture notes (Yes there's some missing)

week3/27-01-2022/27-01-2022.tex

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+%        File: 27-01-2022.tex
+%     Created: 12:36:49 Thu, 27 Jan 2022 EST
+% Last Change: 12:36:49 Thu, 27 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{minted}
+\usepackage{geometry}
+\usepackage{dirtytalk}
+\usepackage{lplfitch}
+
+\geometry{portrait, margin=1in}
+
+%\begin{minted}[linenos,bgcolor=LightGray]{[language]}
+
+\date{01/27/2022}
+\title{%
+        Starting Proofs\\
+            \large MATH--190--01: Discrete Mathematics for Computing}
+\author{Blizzard MacDougall}
+\begin{document}
+\maketitle
+\pagenumbering{arabic}
+
+We have the following rules so far:
+\begin{itemize}
+    \item $A=A$
+    \item $A=B\to B=A$
+    \item $[(A=B)\land (B=C)] \to (A=C)$
+    \item $\mathbb{Z}$ is all whole integers.
+    \item Integers are closed under $+,\ -,\ \times$ but not $\div$
+\end{itemize}
+
+There are two ways to prove statements.
+\begin{itemize}
+    \item \textit{Constructive proofs} - Find the thing, or show how to.
+    \begin{itemize}
+        \item Perfect numbers exist (There are numbers which is the sum of its proper divisors.) 6
+        \item Suppose $r,s\in \mathbb{Z}$. Prove that there exists an integer $k$ such that $24r+4s=2k$
+    \end{itemize}
+    \item \textit{Nonconstructive proofs} - Show that it has to exist, but don't say how or what it is.
+    \begin{itemize}
+        \item Given any set of 2 consecutive integers, one of them is even.
+        \item Roth's theorem. If $\alpha$ is any algebraic irrational number, then there are
+        only finitely many rational numbers such that
+        $\left|\alpha - \frac{p}{q}\right| <\frac{1}{q^{2+\epsilon}}$
+    \end{itemize}
+\end{itemize}
+
+The easiest way to disprove a universal statement is that you have to show the inverse.
+That is, you have to find one instance where it isn't true.
+
+\emph{Proving} a universal statement is significantly harder. If $\mathbb{D}$ is finite,
+the method of exhaustion is easiest. However, the best technique is to generalize from a generic
+parameter. That is, pick a general element, then show that it satisfies the statement.
+
+Generally $x$ is reserved for a real number, and $n$ is reserved for integer use.
+
+To prove a $\forall x\in \mathbb{D}$:
+\begin{enumerate}
+    \item Express the statement to be proved in the form:
+    \begin{equation}
+        \forall x\in \mathbb{D} P(x)\to Q(x)
+    \end{equation}
+    \item "Suppose $x\in \mathbb{D} \land P(x)$"
+    \item Prove $Q(x)=\top$
+\end{enumerate}
+
+
+Example proof:\\
+
+For all integers $m$ and $n$, if $m$ and $n$ are odd, then $m\times n$ is odd.\\
+Proof: Let $m$ and $n$ be odd integers.\\
+Since $m$ is odd, there is an integer $r$ such that $m=2r+1$.\\
+Similarly, there is an integer $s$ such that $n=2s+1$.\\
+Since $m\times n = (2r+1)(2s+1) = 4rs+2r+2s+1 = 2(2rs+r+s) + 1$\\
+Let $k=2rs+r+s$. Note that $k$ is an integer.\\
+$\therefore m\times n=2k+1$, so $m\times n$ is odd. QED
+\end{document}

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

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+%        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{%
+        LectureName\\
+            \large MATH--190--01 : Discrete Mathematics for Computing}
+\author{Blizzard MacDougall}
+\begin{document}
+\maketitle
+\pagenumbering{arabic}
+
+
+\end{document}