Sets - Set-advanced

If $A = \{1, 2, 3\}$, find the Power Set $P(A)$ and verify the number of elements.

The subsets of $A$ are:

1. $\emptyset$ (Empty set)
2. $\{1\}, \{2\}, \{3\}$ (Singletons)
3. $\{1, 2\}, \{2, 3\}, \{1, 3\}$ (Pairs)
4. $\{1, 2, 3\}$ (The set itself) Therefore, $P(A) = \{\emptyset, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{2, 3\}, \{1, 3\}, \{1, 2, 3\}\}$. Since $n(A) = 3$, the number of elements is $2^3 = 8$.

The power set includes every possible combination of elements from the original set. The formula $2^n$ correctly predicts the 8 subsets found.

In a group of 100 students, 50 play Football, 40 play Cricket, and 10 play both. Find the number of students who play neither game if the Universal Set consists of all 100 students.

Let $F$ be the set of students playing Football and $C$ be the set of students playing Cricket. $n(F) = 50$, $n(C) = 40$, $n(F \cap C) = 10$. Using the formula: $n(F \cup C) = n(F) + n(C) - n(F \cap C)$ $n(F \cup C) = 50 + 40 - 10 = 80$ The number of students playing neither game is given by $n(U) - n(F \cup C)$: $100 - 80 = 20$

First, we find the total number of students playing at least one game using the addition principle. Subtracting this from the total number of students gives those who play neither.

Given $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, $A = \{2, 4, 6, 8\}$, and $B = \{2, 3, 5, 7\}$, verify De Morgan's Law $(A \cup B)' = A' \cap B'$.

1. Find $A \cup B$: $A \cup B = \{2, 3, 4, 5, 6, 7, 8\}$
2. Find $(A \cup B)'$: $(A \cup B)' = U - \{2, 3, 4, 5, 6, 7, 8\} = \{1, 9\}$
3. Find $A'$ and $B'$: $A' = U - A = \{1, 3, 5, 7, 9\}$ $B' = U - B = \{1, 4, 6, 8, 9\}$
4. Find $A' \cap B'$: $A' \cap B' = \{1, 3, 5, 7, 9\} \cap \{1, 4, 6, 8, 9\} = \{1, 9\}$ Since $(A \cup B)' = \{1, 9\}$ and $A' \cap B' = \{1, 9\}$, the law is verified.

De Morgan's Law states that the complement of a union is equal to the intersection of the complements. We calculate both sides independently to show they result in the same set.