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\section{Extensions of $\mathbb{R}$}

\definition{}
An \textit{ordered field} consists of a set $S$, the operations $+$ and $\times$, and the relation $<$. \par
An ordered field must satisfy the following properties:

\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}

\definition{}
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


\problem{}
Which of the following are ordered fields?
\begin{itemize}[itemsep=2mm]
	\item $\mathbb{R}$ with the usual definitions of $+$, $\times$, $<$
	\item $\mathbb{R}$ with the usual definitions of $+$, $\times$, $\leq$ \par
	\note{Note that our relation here is $\leq$, not $<$}
	\item $\mathbb{Z}$ with the usual definitions of $+$, $\times$, $<$
	\item $\mathbb{Q}$ with the usual definitions of $+$, $\times$, $<$
	\item $\mathbb{C}$ with the usual definitions of $+$, $\times$, \par
	and with $(a + bi) < (c + di)$ iff $a < c$.
\end{itemize}

\vfill


\problem{}
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 $0 \times x = 0$
	\item $(-x)(-y) = xy$
	\item if $0 < x < y$, then $x^{-1} > y^{-1}$
\end{enumerate}


%\begin{solution}
%	\textbf{Part A:}
%	We know that $x^{-1} \times (x^{-1})^{-1} = 1$ \par
%	Thus $x \times (x^{-1} \times (x^{-1})^{-1}) = x \times 1 = x$ \par
%	We can rewrite this as $(x \times x^{-1}) \times (x^{-1})^{-1} = x$ \par
%	When then becomes $1 \times (x^{-1})^{-1} = x$ \par
%	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








\definition{}
In an ordered field, the \textit{magnitude} of a number x is defined as follows: \par
\begin{equation*}
	|x| =
	\begin{cases}
		x & \text{\tab} x \geq 0 \\
		-x & \text{\tab otherwise}
	\end{cases}
\end{equation*}

\definition{}
We say an element $\delta$ of an ordered field is \textit{infinitesimal} if $|nd| < 1$ for all $n \in \mathbb{Z^+}$. \par
\note{Note that $\mathbb{Z}^+$ is a subset of any nonarchimedian extension of $\mathbb{R}$.} \par

\vspace{2mm}

Likewise, we say $x$ is \textit{limited} if $|x| < n$ for some $n \in \mathbb{Z}^+$. \par
Elements that are not limited are \textit{unlimited}.

\definition{}
We say an element $x$ of a field is \textit{positive} if $x > 0$. \par
We say $x$ is \textit{negative} if $x < 0$. \par


\problem{}
Show that a positive $\delta$ is infinitesimal if and only if $\delta < x$ for all $x \in \mathbb{R}^+$. \par
Then, show that a negative $\delta$ is infinitesimal if and only if $\delta > x$ for every $x \in \mathbb{R}^-$.

\vfill


\problem{}
Prove the following statements: \par
\begin{itemize}
	\item If $\delta$ and $\varepsilon$ are infinitesimal, then $\delta + \varepsilon$ is infinitesimal.
	\item If $\delta$ is infinitesimal and $x$ is limited, then $x\delta$ is infinitesimal.
	\item If $x$ and $y$ are limited, $xy$ and $x+y$ are too.
	\item A nonzero $\delta$ is infinitesimal iff $\delta^{-1}$ is unlimited.
\end{itemize}

\vfill
\pagebreak


\problem{}
Let $\delta$ be a positive infinitesimal. Which is greater?
\begin{itemize}
	\item $\delta$ or $\delta^2$
	\item $(1 - \delta)$ or $(1 + \delta^2)^{-1}$
	\item $\frac{1 + \delta}{1 + \delta^2}$ or $\frac{2 + \delta^2}{2 + \delta^3}$ \par
	\note[Note]{we define $\frac{1}{x}$ as $x^{-1}$, and thus $\frac{a}{b} = a \times b^{-1}$}
\end{itemize}

\vfill


\definition{}<stpart>
We say two elements of an ordered field are \textit{infinitely close} if $x - y$ is infinitesimal. \par
We say that $x_0 \in \mathbb{R}$ is the \textit{standard part} of $x$ if it is infinitely close to $x$. \par

We will denote the standard part of $x$ as $\text{st}(x)$. \par
You may assume that $\text{st}(x)$ exists and is unique for limited $x$. \par

%\problem{}
%Let $H$ be positive unlimited. Determine which of the following are limited. \par

\problem{}
Show that $\text{st}(x + y) = \text{st}(x) + \text{st}(y)$ and $\text{st}(xy) = \text{st}(x) \text{st}(y)$. \par

\vfill
\pagebreak
