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\doclabel{Math 215: Homework 12}
\docauthor{Your name here}
\docdate{April 20, 2012}

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\begin{document}

\begin{proposition}{9.18}  For all $m,n\in\Ints$, 
\[
e(m\cdot n) = e(m)\cdot e(n).
\]
\end{proposition}
\begin{proof}
Your proof goes here.  Attempt maximal laziness.
\end{proof}

\begin{proposition}{9.19}  If $n,m\in\Ints$ then
\[
n<m\quad\text{if and only if}\quad e(n)<e(m).
\]
\end{proposition}
\begin{proof}
For full credit, you must be maximally lazy.  Use Lemma 9.E proved in class,
and use a string of if and only ifs.
\end{proof}

\begin{proposition}{11.3} If $x,y,z\in\Reals$ with $y\neq 0$ and $z\neq 0$,
  then 
\[
\frac{xz}{yz}=\frac{x}{y}.
\]
\end{proposition}
\begin{proof}
Your proof goes here.
\end{proof}

\begin{lemma}{11.A} Suppose $a,b\in\Ints$ and $b\neq 0$. Let $g=\gcd(a,b)$,
so $g\in\Nats$, $g\mid a$ and $g\mid b$.  Then
\[
\gcd\left(\frac{a}{g},\frac{b}{g}\right)=1.
\]
\end{lemma}
\begin{proof}
Recall that if $c\mid d$, and $c\neq 0$, then $\frac{c}{d}$ is
the unique integer $j$ such that $c=jd$.

Feel free to use Proposition 6.30; I'll give a proof in class.
\end{proof}

\begin{proposition}{11.4} Let $x\in\Rats$.  Then there exist
  integers $a$ and $b$ such that $b>0$, $\gcd(a,b)=1$, and 
  $x=a/b$.
\end{proposition}
\begin{proof}
The proof is not long if you take advantage of Proposition 11.3 and Lemma 11.A.  Ignore the hint in the book; it will likely lead to a hand-wavy proof.
\end{proof}

\hrule
The following proposition might take you some work.  Please start
on it now. It will not be due until the next week, however.
\begin{proposition}{9.G} Suppose $f:\Ints\rightarrow\Reals$
  and that for all $n,m\in\Ints$, $f(n+m)=f(n)+f(m)$ and $f(n\cdot m)=f(n)\cdot f(m)$.  Then either $f(n)=0_{\Reals}$ for all $n\in\Ints$
or $f=e$.
\end{proposition}
\begin{proof}
Your proof here.  Hint: either $f(1_\Ints)=0_\Reals$ or not.
\end{proof}
In chapter 9 we constructed
a function $e$ from $\Ints$ to $\Reals$ that obeyed nice rules with
respect to arithmetic.  The point of proposition 9.G is that there
was only one good way to do this.


\end{document}