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\documentclass{math215}

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\doclabel{Math 215: Homework 1a Solutions}
\docauthor{David Maxwell}
\docdate{January 25, 2013}

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\begin{document}

\begin{proposition}{1.7} If $m$ is an integer, then $0+m=m$
and $1\cdot m=m$.
\end{proposition} 
\begin{pf}
Let $m\in\Ints$.  Then by additive commutivity
\[
0+m = m+0.
\]
But $m+0=m$ by Axiom 1.2.  So $0+m=m$.

On the other hand, by multiplicative commutivity,
\[
1\cdot m = m \cdot 1.
\]
Axiom 1.3 implies $m\cdot 1=m$, so $1\cdot m=m$.
\end{pf}

\begin{proposition}{1.8} If $m$ is an integer, then $(-m)+m=0$.
\end{proposition}
\begin{pf}
Let $m\in \Ints$.  Then additive commutativity implies
\[
(-m)+m = m+(-m).
\]
Since Axiom 1.4 implies $m+(-m)=0$, it follows that 
$(-m)+m=0$.
\end{pf}


\begin{proposition}{1.11(iii)} Let $m$, $n$ and $p$ be integers.
	Then $m+(n+p)=(p+m)+n$.
\end{proposition}
\begin{pf}
Suppose $m$, $n$, and $p$ are integers.
Then
\begin{align*}
        m+(n+p) &= m+(p+n) &\text{by additive commutativity}\\
        &= (m+p) + n &\text{by additive associativity} \\
        &= (p+m) +n &\text{by additive commutativity.}
\end{align*}
\end{pf}

\end{document}
