9. Find the result of the following binary subtraction operations:
(i) $(1010111)_{2}-(11001)_{2}-(11110)_{2}=$ ?
(ii) $(10110001)_{2}-(1101100)_{2}-(11110)_{2}=$ ?
9.
(i) Solution: We first use the conventional subtraction operation, to get

Therefore, we have $(1010111)_{2}-(11001)_{2}-(11110)_{2}=(100000)_{2}$.
Now we will use the method of finding complements to perform the operation. First, we find the complement of $(11001)_{2}$ to get $(00111)_{2}=(111)_{2}$, then add the complement to the minuend $(1010111)_{2}$ to get

Then subtract $(100000)_{2}$, to obtain the answer for $(1010111)_{2}-(11001)_{2}$
$$\begin{array}{r}
(10111110)_{2} \\
-\quad(1000000)_{2} \\
\hline
\end{array}$$

Now we find $(111110)_{2}-(11110)_{2}$. First, we find the complement of $(11110)_{2}$ to get $(00010)_{2}=(10)_{2}$, then add the complement to the minuend $(111110)_{2}$ to get
$$\begin{array}{r}
(111110)_{2} \\
+\quad(10)_{2} \\
\hline(1000000)_{2}
\end{array}$$

Then subtract $(100000)_{2}$, to obtain the answer for $(111110)_{2}-(11110)_{2}$
$$\begin{array}{r}
(1000000)_{2} \\
-\quad(1000000)_{2} \\
\hline(100000)_{2}
\end{array}$$

Therefore, we have $(1010111)_{2}-(11001)_{2}-(11110)_{2}=(100000)_{2}$.
(ii) Solution: We first use the conventional subtraction operation to get

Therefore, we have
$(10110001)_{2}-(1101100)_{2}-(11110)_{2}=(100111)_{2}$.
Now we will use the method of finding complements to perform the calculation. First, we find the complement of $(1101100)_{2}$ to get $(0010100)_{2}=(10100)_{2}$, then add the complement $(10100)_{2}$ to the minuend $(10110001)_{2}$ to get
$$\begin{array}{r}
(10110001)_{2} \\
+\quad \begin{array}{r}
(10100)_{2}
\end{array} \\
\hline
\end{array}$$

Then subtract $(10000000)_{2}$, to obtain the answer for $(10110001)_{2}-(1101100)_{2}$
$$\begin{array}{r}
(110000101)_{2} \\
-\frac{(10000000)_{2}}{(1000101)_{2}}
\end{array}$$

Now we find $(1000101)_{2}-(11110)_{2}$. First, we find the complement of $(11110)_{2}$ to get $(00010)_{2}=(10)_{2}$, then add the complement $(10)_{2}$ to the minuend $(1000101)_{2}$ to get
$$\begin{array}{r}
(1000101)_{2} \\
+\quad(10)_{2} \\
\hline(1000111)_{2}
\end{array}$$

Then subtract $(100000)_{2}$, to obtain the answer for $(1000101)_{2}-(11110)_{2}$
$$\begin{array}{r}
(1000111)_{2} \\
-\quad(100000)_{2} \\
\hline
\end{array}$$

Therefore, we have
$$\left(101100(1)_{2}-(1101100)_{2}-(11110)_{2}=(100111)_{2}\right.$$
$$\begin{array}{l}
(1010111)_{2} \\
\left(\begin{array}{llllll}
1 & 1 & 1 & 1 & 1 & 0
\end{array}\right)_{2} \\
\frac{-\quad\left(\begin{array}{lllll}
1 & 1 & 0 & 0 & 1
\end{array}\right)_{2}}{\left(\begin{array}{llllll}
1 & 1 & 1 & 1 & 1 & 0
\end{array}\right)_{2}} \\
\frac{-\left(\begin{array}{lllll}
1 & 1 & 1 & 1 & 0
\end{array}\right)_{2}}{\left(\begin{array}{llllll}
1 & 0 & 0 & 0 & 0 & 0
\end{array}\right)_{2}},
\end{array}$$
$$\begin{array}{l}
\text { (1 } 0 \\
+\frac{(111)_{2}}{(1011110)_{2}},
\end{array}$$
$$\begin{array}{l}
\text { ( } 10 \\
\left(\begin{array}{lllllll}
1 & 0 & 0 & 0 & 1 & 0 & 1
\end{array}\right)_{2} \\
\frac{-(1101100)_{2}}{(1000101)_{2}}, \\
\frac{-(11110)_{2}}{(100111)_{2}},
\end{array}$$