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@@ -254,11 +254,11 @@ $+$ and $\times$ on $\Z/k$:
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\section*{Countable Sets}
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A set $X$ is finite if $\exists n \geq 0. $ a bijection $\{1, ...n\} \to X$ \\
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Pigeonhole Principle: for finite $X$, any injective $f: X \to X$ is also surjective. \\
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$\N$ is infinite. Proof: $f: \N \to \N$ is trivially injective, and $\neg\exists x.~f(x) = 0$, and so not surjective. \\
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$\N$ is infinite. Proof: $f: \N \to \N $ defined by $ f(x) = x + 1$ is trivially injective, and $\neg\exists x.~f(x) = 0$, and so not surjective. \\
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By the inverse of the Pigeonhole Principle, $\N$ is infinite.\\
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A set $X$ is \emph{countably infinite} iff there exists a bijection $\N \to X$. \\
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A set is \emph{countable} iff it is finite or countably infinite. \\
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Any subset of $\N$ is countable. Proof: Let $X \in \N$.\\
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Any subset of $\N$ is countable. Proof: Let $X \subseteq \N$.\\
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If $X$ is finite, it's trivially countable.
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Otherwise, $X$ is infinite and it must be shown that $X$ is countably infinite. \\
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For $k \in \N$, $X_{>k} = \{ n \in X | n > k \}$. Then $X_{>k} \not= \varnothing$, as $X$ would be a subset of $\{1..k\}$ and would be finite. \\
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1
MA2008/2024-09-24
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1
MA2008/2024-09-24
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76
MA2008/decls.tex
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76
MA2008/decls.tex
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\documentclass[fleqn]{article}
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\usepackage{amsmath,amssymb,amsthm}
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\usepackage[margin=0.25in]{geometry}
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\usepackage{enumitem}
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\usepackage{systeme}
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\usepackage{mathtools}
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\date{}
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\author{}
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\newcommand{\paren}[1]{\left(#1\right)}
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\newcommand{\powerset}[1]{\mathcal{P}\paren{#1}}
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\renewcommand{\Re}[1]{\operatorname{\mathbb{R}e}\paren{#1}}
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\renewcommand{\Im}[1]{\operatorname{\mathbb{{I}}m}\paren{#1}}
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\newcommand{\C}{\mathbb{C}}
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\newcommand{\N}{\mathbb{N}}
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\newcommand{\Z}{\mathbb{Z}}
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\newcommand{\Q}{\mathbb{Q}}
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\newcommand{\R}{\mathbb{R}}
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\newcommand{\conj}[1]{\overline{#1}}
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\renewcommand{\mod}[1]{\left|#1\right|}
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\newcommand{\abs}[1]{\left|#1\right|}
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\newcommand{\polar}[2]{#1\paren{\cos{\paren{#2}} + i\sin{\paren{#2}}}}
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\newcommand{\adj}[1]{\operatorname{adj}#1}
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\newcommand{\card}[1]{\left|#1\right|}
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\newcommand{\littletaller}{\mathchoice{\vphantom{\big|}}{}{}{}}
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\newcommand{\restr}[2]{{% we make the whole thing an ordinary symbol
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\left.\kern-\nulldelimiterspace % automatically resize the bar with \right
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#1 % the function
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\littletaller % pretend it's a little taller at normal size
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\right|_{#2} % this is the delimiter
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}}
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\makeatletter
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\renewcommand*{\env@matrix}[1][*\c@MaxMatrixCols c]{%
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\hskip -\arraycolsep
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\let\@ifnextchar\new@ifnextchar
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\array{#1}}
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\makeatother
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% https://gitlab.com/jim.hefferon/linear-algebra/-/blob/master/src/sty/linalgjh.sty
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\newlength{\grsteplength}
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\setlength{\grsteplength}{1.5ex plus .1ex minus .1ex}
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\newcommand{\grstep}[2][\relax]{%
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\ensuremath{\mathrel{
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\hspace{\grsteplength}\mathop{\longrightarrow}\limits^{#2\mathstrut}_{
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\begin{subarray}{l} #1 \end{subarray}}\hspace{\grsteplength}}}}
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\newcommand{\repeatedgrstep}[2][\relax]{\hspace{-\grsteplength}\grstep[#1]{#2}}
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\newcommand{\swap}{\leftrightarrow}
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% https://tex.stackexchange.com/a/198806
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\makeatletter
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\newcommand{\subalign}[1]{%
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\vcenter{%
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\Let@ \restore@math@cr \default@tag
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\baselineskip\fontdimen10 \scriptfont\tw@
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\advance\baselineskip\fontdimen12 \scriptfont\tw@
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\lineskip\thr@@\fontdimen8 \scriptfont\thr@@
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\lineskiplimit\lineskip
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\ialign{\hfil$\m@th\scriptstyle##$&$\m@th\scriptstyle{}##$\hfil\crcr
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#1\crcr
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}%
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}%
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}
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\makeatother
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\theoremstyle{definition}
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\newtheorem*{theorem}{Theorem}
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\newtheorem*{lemma}{Lemma}
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\newtheorem*{corollary}{Corollary}
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\theoremstyle{remark}
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\newtheorem*{note}{Note}
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27
MA2008/linear-transforms.tex
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27
MA2008/linear-transforms.tex
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\input{decls.tex}
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\title{Vector Spaces and Linear Transformations}
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\begin{document}
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\maketitle
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\begin{description}
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\item[Linear Transformation] A function $\phi: V \to W$ between vector spaces $V$ and $W$ (over some field $K$), such that
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\begin{align*}
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\phi(v + w) & \equiv \phi(v) + \phi(w) \\
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\phi(x \cdot v) & \equiv x \cdot \phi(v) \tag{For $x \in K$}
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\end{align*}
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\end{description}
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Differentiation is a linear transformation.
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Solutions to $f'' + f = 0$ for function $f$ are a vector space.
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\begin{theorem}
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For any scalars $\lambda, \mu \in \R$, there is a unique solution such that $f(0) = \mu$ and $f'(0) = \lambda$
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\end{theorem}
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The vector space is then two-dimensional, with basis $sin(x), cos(x)$
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\subsection*{}
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Vector spaces are used over finite fields in \emph{Algebraic Coding Theory}. The field is $\mathbb{F}_2 = \{0, 1\}$ - the integers mod 2.
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Binary strings of length $n$ are then a vector space over $\mathbb{F}_2^n$.
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ECC can be based on vector subspaces of $F_2^n$. (Vector subspaces are closed subsets of a vector space).
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\end{document}
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6
MA2008/tmpl.tex
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6
MA2008/tmpl.tex
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\input{decls.tex}
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\title{}
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\begin{document}
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\maketitle
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\end{document}
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