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1 %$Header$
3 \chapter[Solutions: \ccilzeroxrefcomma{}Chapter \ref{ccil0}]
4 {Solutions: \ccilzeroxrefcomma{}Chapter \ref{ccil0}, \ccilzerolongtitle{}}
6 \label{ccis0}
8 \vworkexercisechapterheader{}
9 \begin{vworkexercisesolution}{\ref{exe:ccil0:sexe0:01}}
10 We can show this result in two ways. The first way, based on bit patterns, is to note
11 that adding an $m$-bit number, $u$, to its one's complement will result in a bit pattern
12 containing all 1's, i.e. $\forall i$, $u_{[i]} = 1$. Adding 1 to this bit pattern will
13 always produce $\forall i$, $u_{[i]} = 0$ with a carry out. Since the order of addition
14 does not matter, this establishes that adding $u$ to $\sim{}u+1$ will produce 0, thus showing
15 that $u$ and $\sim{}u+1$ are additive inverses. This method, although valid, does not
16 establish that $u$ and $\sim{}u+1$ actually represent additive inverses. For example, if
17 $u=-2^{m-1}$, $u=\sim{}u+1$, and clearly a non-zero number cannot be an additive inverse of
18 itself. Thus, it would be more comforting to show this result in a way that demonstrates the
19 actual values of the integers represented.
21 We present a second method now. Assume that $u \neq -2^{m-1}$, since
22 $-2^{m-1}$ cannot have an additive inverse in an $m$-bit signed integer.
23 If $u=0$, $\sim{}u+1=0$, so the relationship is clearly met. If $u<0$, then
24 $u_{[m-1]}=1$, and by
25 (\ccilzeroxrefhyphen\ref{eq:ccil0:sroi0:sros0:00}),
28 \end{vworkexercisesolution}
29 \vworkexercisechapterfooter
31 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
32 \vfill
33 \noindent\begin{figure}[!b]
34 \noindent\rule[-0.25in]{\textwidth}{1pt}
35 \begin{tiny}
36 \begin{verbatim}
37 $HeadURL$
38 $Revision$
39 $Date$
40 $Author$
41 \end{verbatim}
42 \end{tiny}
43 \noindent\rule[0.25in]{\textwidth}{1pt}
44 \end{figure}
45 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
46 %
47 %End of file C_CIS0.TEX


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