Example 19 Find $(101101)_{2} \div(1111)_{2}=$ ?
First, find the complement of the divisor $(1111)_2$, which is $(0001)_2 = (1)_2$. Then add the complement $(1)_2$ to the dividend $(101101)_2$ and subtract $(10000)_2$ to obtain the remainder after the first subtraction of the divisor, i.e.,

Then add the complement $(1)_2$ to the remainder $(11110)_2$ and subtract $(10000)_2$ to obtain the remainder after the second subtraction of the divisor, i.e.,

Then add the complement $(1)_2$ to the remainder $(1111)_2$ and subtract $(10000)_2$ to obtain the remainder after the third subtraction of the divisor, i.e.,
$$\begin{array}{r}
\begin{array}{r}
11111)_2 \\
+\quad(1)_2
\end{array} \\
\hline
\end{array}$$

It is not hard to see that since it was subtracted three times and just completed, the quotient is 3. Converting the decimal number 3 to binary yields $(11)_2$, so
$$(101101)_2 \div (1111)_2 = (11)_2$$
$$\begin{array}{l}
\left(\begin{array}{llllll}
1 & 0 & 1 & 1 & 0 & 1
\end{array}\right)_2 \\
\frac{+(1)_2}{\left(\begin{array}{lllll}
1 & 0 & 1 & 1 & 10
\end{array}\right)_2} \\
\frac{-\left(\begin{array}{lllll}
1 & 0 & 0 & 0 & 0
\end{array}\right)_2}{\left(\begin{array}{llllll}
1 & 1 & 1 & 1 & 0
\end{array}\right)_2},
\end{array}$$
$$\begin{array}{l}
\left(\begin{array}{lllll}
1 & 1 & 1 & 1 & 0
\end{array}\right)_2 \\
\frac{+\quad(1)_2}{\left(\begin{array}{llll}
1 & 1 & 1 & 1
\end{array}\right)_2} \\
\frac{-(10000)_2}{(1111)_2},
\end{array}$$