Answer :
To prove that [tex]\((A \cup B)^{\prime} = A^{\prime} \cap B^{\prime}\)[/tex], we'll use set theory and logical reasoning.
First, let us define some terms and notations:
- [tex]\( U \)[/tex] is the universal set, which contains all possible elements under consideration.
- [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are subsets of [tex]\( U \)[/tex].
- [tex]\( A' \)[/tex] (or [tex]\( \overline{A} \)[/tex]) represents the complement of [tex]\( A \)[/tex] in [tex]\( U \)[/tex]. Thus, [tex]\( A' = U - A \)[/tex].
- Similarly, [tex]\( B' \)[/tex] (or [tex]\( \overline{B} \)[/tex]) represents the complement of [tex]\( B \)[/tex] in [tex]\( U \)[/tex]. Thus, [tex]\( B' = U - B \)[/tex].
- [tex]\( (A \cup B)' \)[/tex] represents the complement of the union of [tex]\( A \)[/tex] and [tex]\( B \)[/tex] in [tex]\( U \)[/tex].
We need to prove that [tex]\( (A \cup B)' = A' \cap B' \)[/tex].
### Step-by-Step Proof:
1. Understand [tex]\( (A \cup B)^{\prime} \)[/tex]:
- The set [tex]\( A \cup B \)[/tex] includes all elements that are in [tex]\( A \)[/tex], in [tex]\( B \)[/tex], or in both.
- The complement [tex]\( (A \cup B)' \)[/tex] contains all elements that are not in [tex]\( A \cup B \)[/tex].
- Mathematically, [tex]\( (A \cup B)' = \{ x \in U \mid x \notin (A \cup B) \} \)[/tex].
2. Understand [tex]\( A^{\prime} \cap B^{\prime} \)[/tex]:
- The set [tex]\( A' \)[/tex] contains all elements that are not in [tex]\( A \)[/tex].
- The set [tex]\( B' \)[/tex] contains all elements that are not in [tex]\( B \)[/tex].
- Their intersection [tex]\( A' \cap B' \)[/tex] contains all elements that are not in [tex]\( A \)[/tex] and not in [tex]\( B \)[/tex].
- Mathematically, [tex]\( A' \cap B' = \{ x \in U \mid x \notin A \text{ and } x \notin B \} \)[/tex].
3. Show [tex]\( (A \cup B)' \subseteq A' \cap B' \)[/tex]:
- Take any element [tex]\( x \in (A \cup B)' \)[/tex].
- By definition, [tex]\( x \notin A \cup B \)[/tex].
- This means [tex]\( x \)[/tex] is neither in [tex]\( A \)[/tex] nor in [tex]\( B \)[/tex].
- Therefore, [tex]\( x \in A' \)[/tex] and [tex]\( x \in B' \)[/tex].
- Hence, [tex]\( x \in A' \cap B' \)[/tex].
4. Show [tex]\( A' \cap B' \subseteq (A \cup B)' \)[/tex]:
- Take any element [tex]\( x \in A' \cap B' \)[/tex].
- By definition, [tex]\( x \in A' \)[/tex] and [tex]\( x \in B' \)[/tex].
- This means [tex]\( x \notin A \)[/tex] and [tex]\( x \notin B \)[/tex].
- Therefore, [tex]\( x \notin A \cup B \)[/tex].
- Hence, [tex]\( x \in (A \cup B)' \)[/tex].
Since we have shown both [tex]\( (A \cup B)' \subseteq A' \cap B' \)[/tex] and [tex]\( A' \cap B' \subseteq (A \cup B)' \)[/tex], we can conclude that:
[tex]\[ (A \cup B)' = A' \cap B' \][/tex]
Thus, we have proven that the complement of the union of two sets [tex]\( A \)[/tex] and [tex]\( B \)[/tex] is equal to the intersection of their complements.
First, let us define some terms and notations:
- [tex]\( U \)[/tex] is the universal set, which contains all possible elements under consideration.
- [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are subsets of [tex]\( U \)[/tex].
- [tex]\( A' \)[/tex] (or [tex]\( \overline{A} \)[/tex]) represents the complement of [tex]\( A \)[/tex] in [tex]\( U \)[/tex]. Thus, [tex]\( A' = U - A \)[/tex].
- Similarly, [tex]\( B' \)[/tex] (or [tex]\( \overline{B} \)[/tex]) represents the complement of [tex]\( B \)[/tex] in [tex]\( U \)[/tex]. Thus, [tex]\( B' = U - B \)[/tex].
- [tex]\( (A \cup B)' \)[/tex] represents the complement of the union of [tex]\( A \)[/tex] and [tex]\( B \)[/tex] in [tex]\( U \)[/tex].
We need to prove that [tex]\( (A \cup B)' = A' \cap B' \)[/tex].
### Step-by-Step Proof:
1. Understand [tex]\( (A \cup B)^{\prime} \)[/tex]:
- The set [tex]\( A \cup B \)[/tex] includes all elements that are in [tex]\( A \)[/tex], in [tex]\( B \)[/tex], or in both.
- The complement [tex]\( (A \cup B)' \)[/tex] contains all elements that are not in [tex]\( A \cup B \)[/tex].
- Mathematically, [tex]\( (A \cup B)' = \{ x \in U \mid x \notin (A \cup B) \} \)[/tex].
2. Understand [tex]\( A^{\prime} \cap B^{\prime} \)[/tex]:
- The set [tex]\( A' \)[/tex] contains all elements that are not in [tex]\( A \)[/tex].
- The set [tex]\( B' \)[/tex] contains all elements that are not in [tex]\( B \)[/tex].
- Their intersection [tex]\( A' \cap B' \)[/tex] contains all elements that are not in [tex]\( A \)[/tex] and not in [tex]\( B \)[/tex].
- Mathematically, [tex]\( A' \cap B' = \{ x \in U \mid x \notin A \text{ and } x \notin B \} \)[/tex].
3. Show [tex]\( (A \cup B)' \subseteq A' \cap B' \)[/tex]:
- Take any element [tex]\( x \in (A \cup B)' \)[/tex].
- By definition, [tex]\( x \notin A \cup B \)[/tex].
- This means [tex]\( x \)[/tex] is neither in [tex]\( A \)[/tex] nor in [tex]\( B \)[/tex].
- Therefore, [tex]\( x \in A' \)[/tex] and [tex]\( x \in B' \)[/tex].
- Hence, [tex]\( x \in A' \cap B' \)[/tex].
4. Show [tex]\( A' \cap B' \subseteq (A \cup B)' \)[/tex]:
- Take any element [tex]\( x \in A' \cap B' \)[/tex].
- By definition, [tex]\( x \in A' \)[/tex] and [tex]\( x \in B' \)[/tex].
- This means [tex]\( x \notin A \)[/tex] and [tex]\( x \notin B \)[/tex].
- Therefore, [tex]\( x \notin A \cup B \)[/tex].
- Hence, [tex]\( x \in (A \cup B)' \)[/tex].
Since we have shown both [tex]\( (A \cup B)' \subseteq A' \cap B' \)[/tex] and [tex]\( A' \cap B' \subseteq (A \cup B)' \)[/tex], we can conclude that:
[tex]\[ (A \cup B)' = A' \cap B' \][/tex]
Thus, we have proven that the complement of the union of two sets [tex]\( A \)[/tex] and [tex]\( B \)[/tex] is equal to the intersection of their complements.
