Answer:
a) {{}, {a}}.
b) {{}, {a}, {b}, {a, b}}.
c) {{}, {1}, {2}, {1, 2}, {3}, {1, 3}, {2, 3}, {1, 2, 3}, {4}, {1, 4}, {2, 4}, {1, 2, 4}, {3, 4}, {1, 3, 4}, {2, 3, 4}, {1, 2, 3, 4}}.
Step-by-step explanation:
The power set of a set is the set of all subset of the set in question. The number of power sets (including the empty set) of a set with [tex]n[/tex] (where [tex]n \in \mathbb{Z}[/tex]) unique elements is [tex]2^{n}[/tex].
In other words, there shall be
This explanation shows how to find the power set using binary numbers (only 0 and 1.) (Credit: Mathsisfun.)
List all the binary numbers that are equivalent to decimals ranging from 0 to [tex]2 - 1 = 1[/tex].
[tex]\begin{array}{l|l}\text{Decimal}&\text{Binary}\\ 0 & 0 \\ 1 & 1\end{array}[/tex].
Reverse the original set. Each digit in the binary number corresponds to a member of the original set (i.e. a letter in a) and b) or a number in c).) 0 means that the element is absent in the subset and 1 means that the element is present.
[tex]\begin{array}{c|l}a & \text{Element of the Power Set}\\ 0 & \{\}\\ 1 & \{a\}\end{array}[/tex].
The power set of a) thus contains:
Similarly, list all the binary numbers that are equivalent to decimals ranging from 0 to [tex]4 - 1 = 3[/tex].
[tex]\begin{array}{l|l}\text{Decimal}&\text{Binary}\\ 0 & 00 \\ 1 & 01 \\ 2 & 10 \\ 3 & 11\end{array}[/tex].
[tex]\begin{array}{cc|l}b & a & \text{Element of the Power Set}\\ 0 & 0 & \{\}\\ 0 & 1 & \{a\}\\ 1 & 0 & \{b\} \\ 1 & 1 & \{a, b\}\end{array}[/tex].
The power set of b) thus contains:
Similarly, list all the binary numbers that are equivalent to decimals ranging from 0 to [tex]16 - 1 = 15[/tex].
[tex]\begin{array}{l|l}\text{Decimal}&\text{Binary}\\ 0 & 0000 \\ 1 & 0001 \\ 2 & 0010 \\ 3 & 0011\\4 & 0100 \\ 5 & 0101\\ 6 & 0110 \\ 7 & 0111\\ 8 & 1000\\ 9 & 1001\\ 10 & 1010\\ 11 & 1011\\ 12& 1100 \\13 & 1101 \\ 14 & 1110\\ 15 & 1111 \end{array}[/tex].
[tex]\begin{array}{cccc|l}4 & 3 & 2 &1& \text{Element of the Power Set}\\ 0 & 0 & 0 & 0 &\{\}\\ 0 & 0 & 0 & 1 & \{1\}\\ 0 & 0 & 1 & 0 & \{2\} \\ 0 & 0 &1 & 1 & \{1, 2\} \\ 0 & 1 & 0 & 0 & \{3\} \\ 0 & 1 & 0 & 1& \{1, 3\}\\ 0 & 1 & 1 & 0& \{2, 3\}\\ 0 & 1 & 1 & 1 & \{1, 2, 3\} \\ 1 & 0 & 0 & 0 & \{4\} \\ 1 & 0 & 0 & 1 & \{1, 4\}\\ 1& 0 & 1 &0&\{2, 4\}\\ 1 & 0 & 1 & 1 &\{1, 2, 4\}\\ 1 & 1 & 0 & 0 & \{3, 4\} \\ 1 & 1 & 0 & 1 & \{1, 3, 4\} \\ 1 & 1 & 1 & 0 & \{2, 3, 4\} \\ 1 & 1 & 1 & 1 & \{1, 2, 3, 4\}\end{array}[/tex].
The power set of c) thus contains:
