%$Header$ \chapter[Solutions: \ccilzeroxrefcomma{}Chapter \ref{ccil0}] {Solutions: \ccilzeroxrefcomma{}Chapter \ref{ccil0}, \ccilzerolongtitle{}} \label{ccis0} \vworkexercisechapterheader{} \begin{vworkexercisesolution}{\ref{exe:ccil0:sexe0:01}} We can show this result in two ways. The first way, based on bit patterns, is to note that adding an $m$-bit number, $u$, to its one's complement will result in a bit pattern containing all 1's, i.e. $\forall i$, $u_{[i]} = 1$. Adding 1 to this bit pattern will always produce $\forall i$, $u_{[i]} = 0$ with a carry out. Since the order of addition does not matter, this establishes that adding $u$ to $\sim{}u+1$ will produce 0, thus showing that $u$ and $\sim{}u+1$ are additive inverses. This method, although valid, does not establish that $u$ and $\sim{}u+1$ actually represent additive inverses. For example, if $u=-2^{m-1}$, $u=\sim{}u+1$, and clearly a non-zero number cannot be an additive inverse of itself. Thus, it would be more comforting to show this result in a way that demonstrates the actual values of the integers represented. We present a second method now. Assume that $u \neq -2^{m-1}$, since $-2^{m-1}$ cannot have an additive inverse in an $m$-bit signed integer. If $u=0$, $\sim{}u+1=0$, so the relationship is clearly met. If $u<0$, then $u_{[m-1]}=1$, and by (\ccilzeroxrefhyphen\ref{eq:ccil0:sroi0:sros0:00}), \end{vworkexercisesolution} \vworkexercisechapterfooter %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \vfill \noindent\begin{figure}[!b] \noindent\rule[-0.25in]{\textwidth}{1pt} \begin{tiny} \begin{verbatim} $HeadURL$ $Revision$ $Date$ $Author$ \end{verbatim} \end{tiny} \noindent\rule[0.25in]{\textwidth}{1pt} \end{figure} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % %End of file C_CIS0.TEX