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\section*{Ordered fields (Taylor and Mann, \S 2.1--2.2)}

\begin{definition}
A \emph{field} is a set $F$ with operations ``addition'' and
``multiplication'' satisfying the following:
\begin{itemize}
\item If $a$ and $b$ are in $F$, then so are $a+b$ and $ab$.
\item Addition and multiplication are each associative and
commutative, and together they are distributive: $a(b+c) = ab + ac$
for all $a, b, c \in F$.
\item $F$ contains elements called 0 and 1 satisfying:
\[
a+0 = a \quad \text{and} \quad a \cdot 1 = a
\]
for all $a \in F$.
\item For all $a \in F$, there exists an element $b \in F$ so that
$a+b=0$.
\item For all $a \in F$ with $a \neq 0$, there exists an element $b
\in F$ so that $ab=1$.
\end{itemize}
\end{definition}

\begin{example}
The set of rational numbers $\Q$, the set of real numbers $\R$, and
the set of complex numbers $\C$ each form fields.  The integers $\Z$
and the non-negative integers $\N$ do not.
\end{example}

\begin{definition}
A field $F$ is \emph{ordered} if it has an ordering $<$ so that:
\begin{itemize}
\item For all $a, b \in F$, exactly one of these holds:
\[
a<b, \quad a=b, \quad a>b.
\]
\item For all $a, b, c \in F$, if $a<b$, then $a+c<b+c$.
\item For all $a, b \in F$, if $a>0$ and $b>0$, then $ab>0$.
\end{itemize}
\end{definition}

For example, $\Q$ and $\R$ are ordered fields, while $\C$ is not.

\begin{definition}
Suppose that $F$ is an ordered field and $S$ is a subset of $F$.  An
\emph{upper bound} for $S$ is any element $M$ of $F$ so that $M \geq
x$ for all $x \in S$.  A \emph{least upper bound} for $S$ is any
element $L$ which is an upper bound for $S$ and which also has the
property that every $a<L$ is not an upper bound for $S$: $L$ is the
smallest upper bound for $S$.

An ordered field $F$ has the \emph{least upper bound property} if 
any nonempty subset $S \subseteq F$ with an upper bound has a least
upper bound.
\end{definition}

For example, $\Q$ does not have the least upper bound property.

\begin{theorem}
There is an ordered field $\R$, the \emph{field of real numbers},
which has the least upper bound property and contains $\Q$ as a
subfield.
\end{theorem}

We will not prove this theorem.  Instead, we will essentially use it
as our definition of the field of real numbers.





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