Question

Convert the decimal expansion of each of these integers to a binaryexpansion.

1. \(321\)
2. \(1023\)
3. \(100632\)

Short answer

1. \(101000001\)
2. \(1111111111\)
3. \(11000100100011000\)

Step-by-step solution

Step 1(a): Converting the decimal expansion of each of the integer \(321\) to a binary expansion

You obtain the binary expansion of an integer by consecutively dividing the integer by 2 until you obtain 0.

\(\begin{aligned}{c}321 = 2 \cdot 160 + 1\\160 = 2 \cdot 80 + 0\\80 = 2 \cdot 40 + 0\end{aligned}\)

\(\begin{aligned}{c}40 = 2 \cdot 20 + 0\\20 = 2 \cdot 10 + 0\\10 = 2 \cdot 5 + 0\end{aligned}\)

\(\begin{aligned}{c}5 = 2 \cdot 2 + 1\\2 = 2 \cdot 1 + 0\\1 = 2 \cdot 0 + 1\end{aligned}\)

The successive remainders of each division represent the binary expansion from right to left.

Hence the solution is\(101000001\).

Step 2(b): Converting the decimal expansion of each of the integer \(1023\) to a binary expansion

You obtain the binary expansion of an integer by consecutively dividing the integer by 2 until you obtain0.

\(\begin{aligned}{c}1023 = 2 \cdot 511 + 1\\511 = 2 \cdot 255 + 1\\255 = 2 \cdot 127 + 1\end{aligned}\)

\(\begin{aligned}{c}127 = 2 \cdot 63 + 1\\63 = 2 \cdot 31 + 1\\31 = 2 \cdot 15 + 1\end{aligned}\)

\(\begin{aligned}{c}15 = 2 \cdot 7 + 1\\7 = 2 \cdot 3 + 1\\3 = 2 \cdot 1 + 1\\1 = 2 \cdot 0 + 1\end{aligned}\)

The successive remainders of each division represent the binary expansion from right to left.

Hence the solution is\(1111111111\).

Step 3(c): Converting the decimal expansion of each of the integer \(100632\) to a binary expansion

You obtain the binary expansion of an integer by consecutively dividing the integer by2 until you obtain 0.

\(\begin{aligned}{c}100632 = 2 \cdot 50316 + 0\\50316 = 2 \cdot 25158 + 0\\25158 = 2 \cdot 12579 + 0\end{aligned}\)

\(\begin{aligned}{c}12579 = 2 \cdot 6289 + 1\\6289 = 2 \cdot 3144 + 1\\3144 = 2 \cdot 1572 + 0\end{aligned}\)

\(\begin{aligned}{c}1672 = 2 \cdot 786 + 0\\786 = 2 \cdot 393 + 0\\393 = 2 \cdot 196 + 1\end{aligned}\)

\(\begin{aligned}{c}196 = 2 \cdot 98 + 0\\98 = 2 \cdot 49 + 0\\49 = 2 \cdot 24 + 1\end{aligned}\)

\(\begin{aligned}{c}24 = 2 \cdot 12 + 0\\12 = 2 \cdot 6 + 0\\6 = 2 \cdot 3 + 0\end{aligned}\)

\(\begin{aligned}{c}3 = 2 \cdot 1 + 1\\1 = 2 \cdot 0 + 1\end{aligned}\)

The successive remainders of each division represent the binary expansion from right to left.

Hence the solution is\(11000100100011000\).