1. If sets [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are subsets of the universal set [tex]\(U\)[/tex].

[tex]\[U = \{x: x \in \mathbb{N} \text{ and } x \leq 12\}, \][/tex]
[tex]\[A = \{y: y \text{ is a prime number} \}, \][/tex]
[tex]\[B = \{z: z \text{ is a factor of 12}\}.\][/tex]

Answer the following questions:

(a) Define the intersection of sets with an example.

(b) Write the set operation defined by [tex]\(\{p: p \in A \text{ or } p \in B\}\)[/tex] and the cardinal number of set [tex]\(A\)[/tex].

(c) Prove that: [tex]\(\overline{A \cup B} = \bar{A} \cup \bar{B}\)[/tex].



Answer :

Let's solve the problem step by step.

1. Define the Necessary Sets:
- The universal set [tex]\( U \)[/tex] is all natural numbers less than or equal to 12:
[tex]\[ U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\} \][/tex]
- Set [tex]\( A \)[/tex] consists of prime numbers less than or equal to 12:
[tex]\[ A = \{2, 3, 5, 7, 11\} \][/tex]
- Set [tex]\( B \)[/tex] consists of the factors of 12:
[tex]\[ B = \{1, 2, 3, 4, 6, 12\} \][/tex]

2. Part (a): Define Intersection of Sets with Example:
- The intersection of sets [tex]\( A \)[/tex] and [tex]\( B \)[/tex] is the set of elements that are in both [tex]\( A \)[/tex] and [tex]\( B \)[/tex].
- Calculate [tex]\( A \cap B \)[/tex]:
[tex]\[ A \cap B = \{ x \mid x \in A \text{ and } x \in B \} \][/tex]
- From the definitions of [tex]\( A \)[/tex] and [tex]\( B \)[/tex], we find the common elements:
[tex]\[ A \cap B = \{2, 3\} \][/tex]

3. Part (b): Union of Sets and Cardinal Number of [tex]\( A \)[/tex]:
- The union of sets [tex]\( A \)[/tex] and [tex]\( B \)[/tex] is the set of elements that are in either [tex]\( A \)[/tex] or [tex]\( B \)[/tex] or both.
- Calculate [tex]\( A \cup B \)[/tex]:
[tex]\[ A \cup B = \{ x \mid x \in A \text{ or } x \in B \} \][/tex]
- From the definitions of [tex]\( A \)[/tex] and [tex]\( B \)[/tex], we combine the elements:
[tex]\[ A \cup B = \{1, 2, 3, 4, 5, 6, 7, 11, 12\} \][/tex]
- The cardinal number of set [tex]\( A \)[/tex], denoted [tex]\( n(A) \)[/tex], is the number of elements in [tex]\( A \)[/tex].
[tex]\[ n(A) = 5 \][/tex]

4. Part (c): Prove that [tex]\( \overline{A \cup B} = \overline{A} \cup \overline{B} \)[/tex]:
- The complement of [tex]\( A \cup B \)[/tex] with respect to [tex]\( U \)[/tex] is the set of elements in [tex]\( U \)[/tex] that are not in [tex]\( A \cup B \)[/tex].
- Calculate [tex]\( \overline{A \cup B} \)[/tex]:
[tex]\[ \overline{A \cup B} = U - (A \cup B) \][/tex]
- From the definitions, this is:
[tex]\[ \overline{A \cup B} = \{8, 9, 10\} \][/tex]
- The complement of [tex]\( A \)[/tex] with respect to [tex]\( U \)[/tex] is the set of elements in [tex]\( U \)[/tex] that are not in [tex]\( A \)[/tex]:
[tex]\[ \overline{A} = U - A = \{1, 4, 6, 8, 9, 10, 12\} \][/tex]
- The complement of [tex]\( B \)[/tex] with respect to [tex]\( U \)[/tex] is the set of elements in [tex]\( U \)[/tex] that are not in [tex]\( B \)[/tex]:
[tex]\[ \overline{B} = U - B = \{5, 7, 8, 9, 10, 11\} \][/tex]
- The union of the complements [tex]\( \overline{A} \)[/tex] and [tex]\( \overline{B} \)[/tex] is:
[tex]\[ \overline{A} \cup \overline{B} = \{ x \mid x \in \overline{A} \text{ or } x \in \overline{B} \} \][/tex]
- Combining the elements of [tex]\( \overline{A} \)[/tex] and [tex]\( \overline{B} \)[/tex]:
[tex]\[ \overline{A} \cup \overline{B} = \{1, 4, 5, 6, 7, 8, 9, 10, 11, 12\} \][/tex]
- However, note that we previously calculated:
[tex]\[ \overline{A \cup B} = \{8, 9, 10\} \][/tex]
- Therefore, [tex]\( \overline{A \cup B} \neq \overline{A} \cup \overline{B} \)[/tex] (as sets [tex]\(\overline{A} \cup \overline{B}\)[/tex] contain elements from [tex]\(A \cup B\)[/tex], which should not be there in [tex]\(\overline{A \cup B}\)[/tex]).

Hence, we have demonstrated the steps to determine the intersection, union, cardinality, and the relationship between the complements for given sets [tex]\( A \)[/tex] and [tex]\( B \)[/tex].