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\documentclass{math215}

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\doclabel{Math 215: Homework 13b}
\docauthor{Your name here}
\docdate{April 29, 2013}

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\begin{document}

\begin{proposition}{HW13.1} 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.


\begin{proposition}{11.17}  Suppose $x,y\in\Reals$ and $x<y$.  Then
  there exists an irrational number $z$ such that $x<z<y$.
\end{proposition}
\begin{proof}
Your proof goes here.
\end{proof}

\begin{proposition}{10.B} Suppose $x\in\Reals$ and $M\ge 0$.  Then
$|x|\le M$ if and only if $-M\le x\le M$.
\end{proposition}
\begin{proof}
Your proof goes here.
\end{proof}

\begin{proposition}{HW 13.2}  $\Nats\subseteq\Reals$ is not bounded above.
\end{proposition}
\begin{proof}
Your proof goes here.  Hint:  If $\Nats$ were bounded above, it would have a supremum $x$.  I bet there's a natural number larger than $x-\frac{1}{2}$ then...
\end{proof}


\end{document}