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\doclabel{Math 215: Homework 3}
\docauthor{Your name here.}
\docdate{February 10, 2012}

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\begin{document}

\noindent For this homework, the rules are as follows:

\begin{itemize}
  \item
For questions from Chapter 1, you {\bf do not} need to write down
justifications involving Axioms 1.1-1.4.  But you must write your
steps so that only one thing is being used at a time.

\item
For questions from Chapter 2, you may write your arguments using 
familiar rules of arithmetic {\bf so long as} you know how you
would convert your argument into a formal one using the 
techniques of Chapter 1.  If in doubt, write out the argument, and I
can let you know if the argument is correct.
\end{itemize}

\begin{proposition}{1.22}
\noindent\begin{itemize}
    \item[(i)] For all $m\in\Ints$, $-(-m)=m$.
    \item[(ii)] $-0=0$.
  \end{itemize}
\end{proposition}
\begin{pf}
Your proof goes here.
\end{pf}

\begin{proposition}{1.25(iii)} For all $m,n\in\Ints$,
  \[
  (-m)\cdot n = m\cdot(-n) = -(m\cdot n).
  \]
\end{proposition}
\begin{pf}
Your proof goes here.
\end{pf}
For full credit on this one, you should base your proof on 
Proposition 1.25(ii).

\begin{proposition}{2.3} $1\in\Nats$.
\end{proposition}
\begin{pf}
Your proof goes here.
\end{pf}
Hint: Use proof by contradiction. The first line of the proof will be:

``Suppose to the contrary that $1\not\in\Nats$.''

Take advantage of Proposition 2.2 and show that this leads to a contradiction.

\begin{proposition}{2.5} For each $n\in\Nats$ there exists $m\in\Nats$ such
  that $m>n$.
\end{proposition}
\begin{pf}
Your proof goes here.
\end{pf}
We'll define the symbol $<$ on Monday, but you can read about it on page 15.

\begin{proposition}{2.7(iv)} Let $m$, $n$, and $p\in\Ints$.  If $m<n$ and
$p<0$ then
\[
np < mp.
\]
\end{proposition}
\begin{pf}
Your proof goes here.
\end{pf}

\begin{proposition}{2.9} Let $m\in\Ints$.  If $m\neq 0$ then $m^2\in\Nats$.
\end{proposition}
\begin{pf}
Your proof goes here.
\end{pf}

\begin{proposition}{2.10} The equation $x^2=-1$ has no solution in $\Ints$.
\end{proposition}
\begin{pf}
Your proof goes here.
\end{pf}


\end{document}