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Mark 2024-02-05 08:55:55 -08:00
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@ -32,13 +32,18 @@
}
% This handout is pretty difficult. Make sure you can all solve all the problems yourself,
% and remember that each SECTION was a two-hour lesson with a smart class.
% From experience, the following holds:
% supremum is a better lesson than dual numbers, which is better than extensions.
\begin{document}
\maketitle
\input{parts/1 extensions}
\input{parts/2 dual}
\input{parts/x supremum}
\input{parts/supremum}
\input{parts/dual}
\input{parts/extensions}
\end{document}

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@ -21,8 +21,7 @@ In the problems below, let $\varepsilon$ a positive infinitesimal so that $\vare
\note{Note that $\varepsilon \neq 0$.}
\definition{}
The set of \textit{dual numbers} is a nonarchimedian extension of $\mathbb{R}$ \par
that consists of elements of the form $a + b\varepsilon$, where $a, b \in \mathbb{R}$.
The set of \textit{dual numbers} consists of elements of the form $a + b\varepsilon$, where $a, b \in \mathbb{R}$.
\problem{}
Compute $(a + b\varepsilon) \times (c + d\varepsilon)$.
@ -33,11 +32,7 @@ Compute $(a + b\varepsilon) \times (c + d\varepsilon)$.
\definition{}
Let $f(x)$ be an algebraic function $\mathbb{R} \to \mathbb{R}$. \par
(that is, a function we can write using the operators $+-\times\div$, powers, and roots) \par
\note[Note]{
Why this condition? These are the only operations we have in an ordered field! \\
Powers, roots, and division aren't directly available, but are fairly easy to define.
}
(that is, a function we can write using the operators $+-\times\div$ and integer powers) \par
\vspace{2mm}

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@ -13,7 +13,7 @@
% If you edit this, please give credit!
% Quality handouts take time to make.
\section{Nonarchimedian Extensions}
\section{Extensions of $\mathbb{R}$}
\definition{}
An \textit{ordered field} consists of a set $S$, the operations $+$ and $\times$, and the relation $<$. \par
@ -49,7 +49,16 @@ An ordered field must satisfy the following properties:
\end{itemize}
\definition{}
An ordered field that contains $\mathbb{R}$ is called a \textit{nonarchimedian extension} of $\mathbb{R}$.
An ordered field that contains $\mathbb{R}$ is called an \textit{extension} of $\mathbb{R}$.
\definition{}
The \textit{Archimedian property} states the following: \par
For all positive $x, y$, there exists an $n$ so that $nx \geq y$.
\theorem{}
All extensions of $\mathbb{R}$ are nonarchemedian. \par
Proving this is difficult.
\vfill
\pagebreak
@ -71,7 +80,8 @@ Which of the following are ordered fields?
\problem{}
Show that each of the following is true in any ordered field.
Show that each of the following is true in any ordered field. \par
The list of field axioms is provided below, for convenience.
\begin{enumerate}
\item if $x \neq 0$ then $(x^{-1})^{-1} = x$
\item $0 \times x = 0$
@ -89,8 +99,36 @@ Show that each of the following is true in any ordered field.
% And thus $(x^{-1})^{-1} = x$
%\end{solution}
\vfill
\begin{itemize}
\item \textbf{Properties of $+$:}
\begin{itemize}
\item Commutativity: $a + b = b + a$
\item Associativity: $a + (b + c) = (a + b) + c$
\item Identity: there exists an element $0$ so that $a + 0 = a$ for all $a \in S$
\item Inverse: for every $a$, there exists a $-a$ so that $a + (-a) = 0$
\end{itemize}
\item \textbf{Properties of $\times$:}
\begin{itemize}
\item Commutativity
\item Associativity
\item Identity (which we label $1$)
\item For every $a \neq 0$, there exists an inverse $a^{-1}$ so that $aa^{-1} = 1$
\item Distributivity: $a(b + c) = ab + ac$
\end{itemize}
\item \textbf{Properties of $<$:}
\begin{itemize}
\item Non-reflexive: $x < x$ is always false
\item Transitive: $x < y$ and $y < z$ imply $x < z$
\item Connected: for all $x, y \in S$, either $x < y$, $x > y$, or $x = y$.
\item If $x < y$ then $x + z < y + z$
\item If $x < y$ and $z > 0$, then $xz < yz$
\item $0 < 1$
\end{itemize}
\end{itemize}
\pagebreak

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@ -14,7 +14,7 @@
% Quality handouts take time to make.
\section*{Bonus: The supremum \& infimum}
\section*{The supremum \& the infimum}
\definition{}
In this section, we'll define a \say{real number} as a decimal, infinite or finite.