Sets - Intersection of Two Sets-advanced

Given $A = \{x : x \text{ is a prime number } < 15\}$ and $B = \{x : x \text{ is a factor of } 30\}$. Find $A \cap B$.

First, list the elements of both sets in roster form: $A = \{2, 3, 5, 7, 11, 13\}$ $B = \{1, 2, 3, 5, 6, 10, 15, 30\}$ To find $A \cap B$, identify elements present in both sets: $A \cap B = \{2, 3, 5\}$

Intersection includes only the elements that satisfy both conditions: being a prime number less than 15 AND being a factor of 30.

In a class of 60 students, 40 students like Mathematics and 35 like Science. If every student likes at least one subject, find the number of students who like both Mathematics and Science.

Let $M$ be the set of students who like Mathematics and $S$ be the set of students who like Science. Given: $n(M \cup S) = 60$ $n(M) = 40$ $n(S) = 35$ Using the formula: $n(M \cup S) = n(M) + n(S) - n(M \cap S)$ $60 = 40 + 35 - n(M \cap S)$ $60 = 75 - n(M \cap S)$ $n(M \cap S) = 75 - 60 = 15$ So, 15 students like both subjects.

We use the Principle of Inclusion-Exclusion to find the overlap (intersection) between the two groups.

If $A = \{3, 6, 9, 12, 15, 18, 21\}$, $B = \{4, 8, 12, 16, 20\}$, and $C = \{2, 4, 6, 8, 10, 12\}$, find $(A \cap B) \cap C$.

Step 1: Find $(A \cap B)$ $A \cap B = \{12\}$ (since 12 is the only common multiple of 3 and 4 in the sets) Step 2: Find $(A \cap B) \cap C$ $(A \cap B) \cap C = \{12\} \cap \{2, 4, 6, 8, 10, 12\}$ $(A \cap B) \cap C = \{12\}$

The Associative law states that $(A \cap B) \cap C$ is the set of elements common to all three sets.