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\title{General Mathematics and Computational Science I}
\author{Exercise 6}
\date{September 22, 2005}
\maketitle

\begin{enumerate}

\item (Bernoulli inequality.)  Use induction to prove that for every
$x \in \mathbb{Q}$ with $x \geq -1$ and every $n \in \mathbb{N}$,
\begin{equation*}
  (1+x)^n \geq 1 + n \, x \,.
\end{equation*}

\item Recall the definition of addition and multiplication on the
rationals.  For $x=[a/b]\in \mathbb{Q}$ and $y=[c/d]\in \mathbb{Q}$,
we define $x+y=[(ad+bc)/bd]$ and $x\cdot y = [ac/bd]$.

Show that multiplication is well-defined.  I.e., show that the
definition is independent of the choice of representative under the
equivalence relation $(a,b) \sim (a',b')$ if $a\cdot b'=a'\cdot b$.

\item We say a field $\mathbb{F}$ is \emph{ordered} if there exists a
relation $>$ which respects addition and multiplication in the
following way.  For any $x,y,z\in\mathbb{F}$,
\begin{itemize}
\item[(O1)] if $x>y$, then $x+z>y+z$;
\item[(O2)] if $x>y$ and $z>0$, then $xz>yz$.
\end{itemize}

Recall from class that the set of rational numbers $\mathbb{Q}$ can be
ordered as follows.  A rational number, i.e.\ an equivalence class
$[a/b]\in \mathbb{Q}$, is \emph{positive} if $ab>0$ and
\emph{negative} if $ab<0$.  Then, for rationals $x=[a/b]$ and
$y=[c/d]$, we define $x>y$ if $x-y>0$.

Show that this order relation satisfies (O1) and (O2).

\end{enumerate}
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