159 lines
4.9 KiB
TeX
159 lines
4.9 KiB
TeX
% Copyright (C) 2023 <Mark (mark@betalupi.com)>
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%
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% This program is free software: you can redistribute it and/or modify
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% it under the terms of the GNU General Public License as published by
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% the Free Software Foundation, either version 3 of the License, or
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% (at your option) any later version.
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%
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% You may have received a copy of the GNU General Public License
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% along with this program. If not, see <https://www.gnu.org/licenses/>.
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%
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%
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%
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% If you edit this, please give credit!
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% Quality handouts take time to make.
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\section{Nonarchimedian Extensions}
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\definition{}
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An \textit{ordered field} consists of a set $S$, the operations $+$ and $\times$, and the relation $<$. \par
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An ordered field must satisfy the following properties:
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\begin{itemize}
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\item \textbf{Properties of $+$:}
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\begin{itemize}
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\item Commutativity: $a + b = b + a$
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\item Associativity: $a + (b + c) = (a + b) + c$
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\item Identity: there exists an element $0$ so that $a + 0 = a \forall a \in S$
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\item Inverse: for every $-a$, there exists a $-a$ so that $a + (-a) = 0$
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\end{itemize}
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\item \textbf{Properties of $\times$:}
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\begin{itemize}
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\item Commutativity
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\item Associativity
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\item Identity (which we label $1$)
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\item Inverse (which we label $a^{-1}$, and which doesn't exist for $0$)
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\item Distributivity: $a(b + c) = ab + ac$
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\end{itemize}
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\item \textbf{Properties of $<$:}
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\begin{itemize}
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\item Non-reflexive: $x < x$ is always false
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\item Transitive: $x < y$ and $y < z$ imply $x < z$
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\item Connected: for all $x, y \in S$, either $x < y$, $y > x$, or $x = y$.
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\item If $x < y$ then $x + z < y + z$
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\item If $x < y$ and $z > 0$, then $xz < yz$
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\item $0 < 1$
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\end{itemize}
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\end{itemize}
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\definition{}
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An ordered field that contains $\mathbb{R}$ is called a \textit{nonarchimedian extension} of $\mathbb{R}$.
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\problem{}
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Show that each of the following is true in any ordered field.
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\begin{enumerate}
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\item if $x \neq 0$ then $(x^{-1})^{-1} = x$
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\item $0 \times x = x$
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\item $(-x)(-y) = xy$
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\item if $0 < x < y$, then $x^{-1} > y^{-1}$
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\end{enumerate}
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%\begin{solution}
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% \textbf{Part A:}
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% We know that $x^{-1} \times (x^{-1})^{-1} = 1$ \par
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% Thus $x \times (x^{-1} \times (x^{-1})^{-1}) = x \times 1 = x$ \par
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% We can rewrite this as $(x \times x^{-1}) \times (x^{-1})^{-1} = x$ \par
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% When then becomes $1 \times (x^{-1})^{-1} = x$ \par
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% And thus $(x^{-1})^{-1} = x$
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%\end{solution}
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\vfill
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\pagebreak
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\definition{}
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In an ordered field, the \textit{magnitude} of a number x is defined as follows: \par
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\begin{equation*}
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|x| =
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\begin{cases}
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x & \text{\tab} x \geq 0 \\
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-x & \text{\tab otherwise}
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\end{cases}
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\end{equation*}
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\definition{}
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We say an element $\delta$ of an ordered field is \textit{infinitesimal} if $|nd| < 1$ for all $n \in \mathbb{Z^+}$. \par
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\note{Note that $\mathbb{Z}^+$ is a subset of any nonarchimedian extension.} \par
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\vspace{2mm}
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Likewise, we say $x$ is \textit{limited} if $|x| < n$ for some $n \in \mathbb{Z}^+$. \par
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Elements that are not limited are \textit{unlimited}.
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\definition{}
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We say an element $x$ of a field is \textit{positive} if $x > 0$. \par
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We say $x$ is \textit{negative} if $x < 0$. \par
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\problem{}
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Show that a positive $\delta$ is infinitesimal if and only if $\delta < x$ for all $x \in \mathbb{R}^+$. \par
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Then, show that a negative $\delta$ is infinitesimal if and only if it is bigger than every $x \in \mathbb{R}^-$.
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\vfill
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\problem{}
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Prove the following statements: \par
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\begin{itemize}
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\item If $\delta$ and $\varepsilon$ are infinitesimal, then $\delta + \varepsilon$ is infinitesimal.
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\item If $\delta$ is infinitesimal and $x$ is limited, then $a\delta$ is infinitesimal.
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\item If $x$ and $y$ are limited, $xy$ and $x+y$ are too.
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\item A nonzero $\delta$ is infinitesimal iff $\delta^{-1}$ is unlimited.
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\end{itemize}
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\vfill
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\problem{}
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Let $\delta$ be a positive infinitesimal. Which is greater?
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\begin{itemize}
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\item $\delta$ or $\delta^2$?
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\item $(1 - \delta)$ or $(1 + \delta^2)^{-1!}$?
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\item $\frac{1 + \delta}{1 + \delta^2}$ or $\frac{2 + \delta^2}{2 + \delta^3}$? \par
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\note[Note]{we define $\frac{1}{x}$ as $x^{-1}$, and thus $\frac{a}{b} = a \times b^{-1}$}
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\end{itemize}
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\vfill
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\pagebreak
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\definition{}
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We say two elements of an ordered field are \textit{infinitely close} if $x - y$ is infinitesimal. \par
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We say that $x_0 \in \mathbb{R}$ is a \textit{standard part} of $x$ if it is infinitely close to $x$. \par
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\problem{}
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We will denote the standard part of $x$ as $\text{st}(x)$. \par
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Show that $\text{st}(x)$ is well-defined for limited $x$. \par
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(In other words, Show that $x_0$ exists and is unique for limited $x$). \par
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\hint{To prove existance, consider $\text{sup}(\{a \in \mathbb{R} ~|~ a < x\}$)}
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\vfill
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%\problem{}
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%Let $H$ be positive unlimited. Determine which of the following are limited. \par
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\problem{}
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Show that $\text{st}(x + y) = \text{st}(x) + \text{st}(y)$ and $\text{st}(xy) = \text{st}(x) \text{st}(y)$. \par
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\vfill
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\pagebreak
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