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\title{One's Complement}
\author{}

\begin{document}

\maketitle

Given an $n-$bit binary string, $I$, the leftmost bit indicates the sign of an integer in $1$s complement representation. In this left most position a $1$ indicates a negative value while a $0$ indicates a positive value. The representation for positive integers corresponds to unsigned representation where the leftmost bit must contain a $0$.

Negative integers are formed by reversing all bits to form the bitwise complement of the corresponding positive integer. If we represent $I$ by the $n-$bit binary sequence, $b_{n} \ldots b_1 $ then $-I$ in one's complement is given by $\overline{b_n } \ldots \overline{b_1}$ where $\overline{b_i}=1-b_i$ for all $i$.\bigskip

\noindent\textbf{Let's see what that looks like in Math speak}\bigskip

Let $I$ be a negative one's complement integer. The value of $I$ is obtained by forming its one's complement:
\begin{equation}
-I = \sum_{i=0}^{n-1}(1-b_i)\cdot2^i = \sum_{i=0}^{n-1}2^i - \sum_{i=0}^{n-1} b_i \cdot 2^i.
\end{equation}
Thus,

\begin{equation}
I = \sum_{i=0}^{n-1}b_i \cdot 2^i - (2^n - 1).
\end{equation}

Negative one's complement integers are formed by subtracting a bias of $2^n - 1$ from the positive integers. Taking into account the sign bit $bn$, the value for a positive or negative (n+1) bit one's complement integer is:
\begin{equation}
I = \sum_{i=0}^{n-1}b_i \cdot 2^i - b_n (2^n - 1).
\end{equation}

Recalling that the left most bit only represents the sign, the range of values for an $n-$bit one's complement integer is $-(2^{n-1}-1)$ to $2^{n-1}-1$.\bigskip

\noindent\textbf{Examples:}\bigskip

Since the complement of $0$ is $2^{n+1}-1$, there are different representations for $+0$ and $-0$ in one's complement. Examples of $8$-bit one's complement numbers:
\[
\begin{array}{cr}
Binary &amp; Decimal \\
00000000 &amp; 0 \\
11111111 &amp; -0 \\
00000011 &amp; 3 \\
11111100 &amp; -3 \\
\end{array}
\]

The range of $8-$bit one's complement integers is $-127$ to $+127$.

Addition of signed numbers in one's complement is performed using binary addition with end-around carry. If there is a carry out of the most significant bit of the sum, this bit must be added to the least significant bit of the sum.

To add decimal 17 to decimal -8 in 8-bit one's complement:\bigskip
\begin{center}
	\begin{tabular}{rrrrrr}
		&amp;  &amp; $0001$ &amp; $0001$ &amp;  &amp; $(17)$ \\
		$+$ &amp;  &amp; $1111$ &amp; $0111$ &amp;  &amp; $(-8)$ \\ \cline{1-4}\cline{6-6}
		&amp; $1$ &amp; $0000$ &amp; $1000$ &amp;  &amp;  \\
		&amp;  &amp; \multicolumn{1}{l}{$\hookrightarrow $} &amp; $+1$ &amp;  &amp;  \\ \cline{3-4}
		&amp;  &amp; $0000$ &amp; $1001$ &amp; $=$ &amp; $(9)$%
	\end{tabular}
\end{center}

\end{document}
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