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{\bf Axioms for $\R$}
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\vspace*{.3in}
\noindent{\bf Field Axioms: }
There are operations $+$ (addition) and $\cdot$ (multiplication) on $\R$
with the following properties.
\begin{itemize}
\item $+$ and $\cdot$ are commutative: 
$$
x + y = y + x, \qquad x\cdot y = y\cdot x.
$$
\item $+$ and $\cdot$ are associative:
$$
x+(y+z) = (x+y) + z, \qquad x\cdot(y\cdot z) = (x\cdot y) \cdot z.
$$
\item There is an additive identity $0\in\R$ and a multiplicative identity
$1\in\R$:
$$
x+ 0 = x,\qquad x\cdot 1 = x.
$$
\item For each $x\in\R$ there is $y\in\R$ such that $x+y = 0$.  If
$x\neq 0$, then there is $z\in\R$ such that $xz = 1$.  (We write $y=-x$
and $z= 1/x$.)
\item Multiplication is distributive over addition:
$$
x\cdot(y+z) = x\cdot y + x\cdot z.
$$
\end{itemize}
\vspace*{.4in}

\noindent{\bf Order Axioms: } There is a relation $<$ on $\R$ with the
following properties
\begin{itemize}
\item (Trichotomy) For each $x,y\in\R$ exactly one of the following is true:
$x<y$, $y<x$, or $x=y$.
\item (Transitivity) If $x<y$ and $y<z$, then $x<z$.
\end{itemize}

\vspace*{.4in}
\noindent{\bf Axioms combining arithmetic and order:} 
\begin{itemize} 
\item If $x<y$ then $x+z<y+z$.
\item If $x,y>0$ then $x\cdot y > 0$.
\end{itemize}

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