Westonci.ca is the best place to get answers to your questions, provided by a community of experienced and knowledgeable experts. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
To solve the problem, let's carefully examine each statement given the condition [tex]\(A \cap B = \varnothing\)[/tex], which means that sets [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are disjoint.
### Statements to analyze:
1. [tex]\(A \cup B = 0\)[/tex]:
The union [tex]\(A \cup B\)[/tex] consists of all elements that are in [tex]\(A\)[/tex], in [tex]\(B\)[/tex], or in both (though there are no common elements since [tex]\(A \cap B = \varnothing\)[/tex]). This statement implies that the union of [tex]\(A\)[/tex] and [tex]\(B\)[/tex] is the empty set, which is not true unless both [tex]\(A\)[/tex] and [tex]\(B\)[/tex] themselves are empty. Therefore, this statement is false.
2. [tex]\(A\)[/tex] is equal to [tex]\(B\)[/tex] complement:
Given [tex]\(A \cap B = \varnothing\)[/tex], all elements in [tex]\(A\)[/tex] are not in [tex]\(B\)[/tex]. This can be rephrased as [tex]\(A \subseteq B'\)[/tex] where [tex]\(B'\)[/tex] is the complement of [tex]\(B\)[/tex] (the set of all elements not in [tex]\(B\)[/tex]). For [tex]\(A\)[/tex] to be exactly [tex]\(B'\)[/tex], each element in the universal set must be either in [tex]\(A\)[/tex] or in [tex]\(B\)[/tex]. Therefore, this statement is true because if [tex]\(A\)[/tex] and [tex]\(B\)[/tex] cover the entire universal set completely but do not overlap, [tex]\(A\)[/tex] is [tex]\(B\)[/tex] complement and vice versa.
3. [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are disjoint:
By definition, [tex]\(A\)[/tex] and [tex]\(B\)[/tex] being disjoint means [tex]\(A \cap B = \varnothing\)[/tex], which is given. Therefore, this statement is true.
4. [tex]\(A\)[/tex] is not a subset of [tex]\(B\)[/tex]:
Since [tex]\(A \cap B = \varnothing\)[/tex], sets [tex]\(A\)[/tex] and [tex]\(B\)[/tex] do not share any elements. This implies that [tex]\(A\)[/tex] cannot have any elements in common with [tex]\(B\)[/tex], and thus [tex]\(A\)[/tex] cannot be a subset of [tex]\(B\)[/tex] (unless [tex]\(A\)[/tex] is empty). Therefore, this statement is true.
### Determining the exception:
The only false statement in the given list is:
1. [tex]\(A \cup B = 0\)[/tex], which claims that the union of [tex]\(A\)[/tex] and [tex]\(B\)[/tex] is the empty set.
Based on the analysis, this is the false statement, and thus, all other statements are true.
So, the detailed solution shows that the statement that is not true is:
[tex]\(A \cup B = 0\)[/tex].
### Statements to analyze:
1. [tex]\(A \cup B = 0\)[/tex]:
The union [tex]\(A \cup B\)[/tex] consists of all elements that are in [tex]\(A\)[/tex], in [tex]\(B\)[/tex], or in both (though there are no common elements since [tex]\(A \cap B = \varnothing\)[/tex]). This statement implies that the union of [tex]\(A\)[/tex] and [tex]\(B\)[/tex] is the empty set, which is not true unless both [tex]\(A\)[/tex] and [tex]\(B\)[/tex] themselves are empty. Therefore, this statement is false.
2. [tex]\(A\)[/tex] is equal to [tex]\(B\)[/tex] complement:
Given [tex]\(A \cap B = \varnothing\)[/tex], all elements in [tex]\(A\)[/tex] are not in [tex]\(B\)[/tex]. This can be rephrased as [tex]\(A \subseteq B'\)[/tex] where [tex]\(B'\)[/tex] is the complement of [tex]\(B\)[/tex] (the set of all elements not in [tex]\(B\)[/tex]). For [tex]\(A\)[/tex] to be exactly [tex]\(B'\)[/tex], each element in the universal set must be either in [tex]\(A\)[/tex] or in [tex]\(B\)[/tex]. Therefore, this statement is true because if [tex]\(A\)[/tex] and [tex]\(B\)[/tex] cover the entire universal set completely but do not overlap, [tex]\(A\)[/tex] is [tex]\(B\)[/tex] complement and vice versa.
3. [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are disjoint:
By definition, [tex]\(A\)[/tex] and [tex]\(B\)[/tex] being disjoint means [tex]\(A \cap B = \varnothing\)[/tex], which is given. Therefore, this statement is true.
4. [tex]\(A\)[/tex] is not a subset of [tex]\(B\)[/tex]:
Since [tex]\(A \cap B = \varnothing\)[/tex], sets [tex]\(A\)[/tex] and [tex]\(B\)[/tex] do not share any elements. This implies that [tex]\(A\)[/tex] cannot have any elements in common with [tex]\(B\)[/tex], and thus [tex]\(A\)[/tex] cannot be a subset of [tex]\(B\)[/tex] (unless [tex]\(A\)[/tex] is empty). Therefore, this statement is true.
### Determining the exception:
The only false statement in the given list is:
1. [tex]\(A \cup B = 0\)[/tex], which claims that the union of [tex]\(A\)[/tex] and [tex]\(B\)[/tex] is the empty set.
Based on the analysis, this is the false statement, and thus, all other statements are true.
So, the detailed solution shows that the statement that is not true is:
[tex]\(A \cup B = 0\)[/tex].
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
