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\documentclass{math215}

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\usepackage{times,txfonts}

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% in the header.
\doclabel{Math 215: Homework 8a}
\docauthor{Your name here}
\docdate{March 18, 2013}

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% by uncommenting the line below.  Feel free to make changes to that file.
\input{math215extras.tex}

\newtheorem*{definition}{Definition}

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\begin{document}


\begin{proposition}{4.5} For all $k\in\Ints_{\ge 0}$, $k!\in\Nats$.
\end{proposition}
\begin{pf}
Your proof here.
\end{pf}

\begin{proposition}{4.7 (i)}  For all $k\in\Nats$, $5^{2k}-1$ is divisible by $24$.
\end{proposition}
\begin{pf}
Your proof here.

\begin{proposition}{4.6(iii)} Let $b\in \Ints$ and $m,k\ge 0$. Then
  $(b^m)^k=b^{mk}$.
\end{proposition}
\begin{pf}
\end{pf}

\begin{proposition}{4.11} For all $k\in\Nats$,
\[
2\sum_{j=1}^k j = k(k+1).
\]
\end{proposition}
\begin{pf}
\end{pf}

\begin{proposition}{4.A} Suppose $a$ and $b$ are integers such that $a\neq 0$ and $a\mid b$. Then there exists a unique integer $j$ such that $b=aj$.
\end{proposition}
\begin{pf}
\end{pf}

\hrule

There is nothing more to prove on this homework.  The discussion below 
explains how to rewrite the result of Proposition 4.11 more naturally
using Proposition 4.A.

\begin{definition} Suppose that $a$ and $b$ are integers such that $a\neq 0$
  and $a\mid b$.  We define
$$
\frac{b}{a} = j  
$$
where $j$ is the unique integer such that $b=aj$
\end{definition}

If $a$, $b$, and $c$ are integers (with $a\neq 0$) and if we write 
$$
\frac{b}{a} = c
$$
we mean $a$ divides $b$ and $b=ca$.  To show that $\frac{b}{a}=c$
you simply show that $b=ac$.  With this definition in mind, Proposition
4.11 can be rephrased 
\[
\sum_{j=1}^k j = \frac{k(k+1)}{2}.
\]

\end{pf}

\end{document}
