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X = 1.0000\ 1101 (269/256)

D = 1.0011\ 0110 (310/256)

Q = 0.1101\ 1110 (222/256)

D[1.3] = 1.001, so we use the ``1.001" column of chart 13.X. This means we select a quotient bit of 2 if the partial remainder is greater than or equal to 3.5, a quotient bit of 1 if the partial is greater or equal to than 1.0, a zero if the partial is greater than or equal to -1.5, -1 if the partial is greater than or equal to -3.75, and a -2 otherwise.

{\small
\begin{center}
\begin{tabular}{cccc}
    Initialization&amp;$D$&amp;$0001.0011\ 0110\ 00$&amp;\\
    &amp;$2D$&amp;$0010.0110\ 1100\ 00$&amp;\\
    &amp;$-D=\overline{D}+1$&amp;$1110.1100\ 1001\ 11$&amp;(+ 1 ulp)\\
    &amp;$-2D=\overline{2D}+1$&amp;$1101.1001\ 0011\ 11$&amp;(+ 1 ulp)\\
    &amp;&amp;&amp;\\
    &amp;$WS_{-1}=X$&amp;$0001.0000\ 1101\ 00$&amp;\\
    &amp;$WC_{-1}$&amp;$0000.0000\ 0000\ 00$&amp;\\
    \hdashline\\
    Step 0: &amp;$WS_{-1}   $&amp;$0001.0000\ 1101\ 00$&amp;\\
            &amp;$WC_{-1}   $&amp;$0000.0000\ 0000\ 0\mathbf{1}$&amp;($W_{msbs}=0001.000\ \text{so}\ q_0=1$)\\
            &amp;$-q_0D     $&amp;$1110.1100\ 1001\ 11$&amp;\\
    \cline{2-3}
            &amp;$sum       $&amp;$1111.1100\ 0100\ 10$&amp;$\ll2$\\
            &amp;$carry     $&amp;$0000.0001\ 0010\ 10$&amp;$\ll2$\\
    \hdashline\\
    Step 1: &amp;$WS_0      $&amp;$1111.0001\ 0010\ 00$&amp;\\
            &amp;$WC_0      $&amp;$0000.0100\ 1010\ 0\mathbf{0}$&amp;($W_{msbs}=1111.010\ \text{so}\ q_1=-1$)\\
            &amp;$-q_1D     $&amp;$0001.0011\ 0110\ 00$&amp;\\
    \cline{2-3}
            &amp;$sum       $&amp;$1110.0110\ 1110\ 00$&amp;$\ll2$\\
            &amp;$carry     $&amp;$0010.0010\ 0100\ 00$&amp;$\ll2$\\
    \hdashline\\
    Step 2: &amp;$WS_1      $&amp;$1001.1011\ 1000\ 00$&amp;\\
            &amp;$WC_1      $&amp;$1000.1001\ 0000\ 0\mathbf{1}$&amp;($W_{msbs}=0010.010\ \text{so}\ q_2=2$)\\
            &amp;$-q_2D     $&amp;$1101.1001\ 0011\ 11$\\
    \cline{2-3}
            &amp;$sum       $&amp;$1100.1011\ 1011\ 10$&amp;$\ll2$\\
            &amp;$carry     $&amp;$0011.0010\ 0000\ 10$&amp;$\ll2$\\
    \hdashline\\
    Step 3: &amp;$WS_2      $&amp;$0010.1110\ 1110\ 00$&amp;\\
            &amp;$WC_2      $&amp;$1100.1000\ 0010\ 0\mathbf{0}$&amp;($W_{msbs}=1111.011\ \text{so}\ q_3=-1$)\\
            &amp;$-q_3D     $&amp;$0001.0011\ 0110\ 00$\\
    \cline{2-3}
            &amp;$sum       $&amp;$1111.0101\ 1010\ 00$&amp;$\ll2$\\
            &amp;$carry     $&amp;$0001.0100\ 0100\ 00$&amp;$\ll2$\\
    \hdashline\\
    Step 4: &amp;$WS_3      $&amp;$1101.0110\ 1000\ 00$&amp;\\
            &amp;$WC_3      $&amp;$0101.0001\ 0000\ 0\mathbf{1}$&amp;($W_{msbs}=0010.011\ \text{so}\ q_4=2$)\\
            &amp;$-q_4D     $&amp;$1101.1001\ 0011\ 11$\\
    \cline{2-3}
            &amp;$sum       $&amp;$0101.1110\ 1011\ 10$&amp;$\ll2$\\
            &amp;$carry     $&amp;$1010.0010\ 0000\ 10$&amp;$\ll2$\\
    \hdashline\\
    Step 5: &amp;$WS_4      $&amp;$0111.1010\ 1110\ 00$&amp;\\
            &amp;$WC_4      $&amp;$1000.1000\ 0010\ 0\mathbf{0}$&amp;($W_{msbs}=0000.001\ \text{so}\ q_5=0$)\\
            &amp;$-q_5D     $&amp;$0000.0000\ 0000\ 00$\\
    \cline{2-3}
            &amp;$sum       $&amp;$1111.0010\ 1100\ 00$&amp;$\ll2$\\
            &amp;$carry     $&amp;$0001.0000\ 0100\ 00$&amp;$\ll2$\\
    \hdashline\\
    Terminate&amp;Quotient&amp;$00.11\ 01\ 11\ 10\ (00\ 1)$
\end{tabular}
\end{center}
}

\end{document}
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