Westonci.ca is the premier destination for reliable answers to your questions, brought to you by a community of experts. Join our Q&A platform to get precise answers from experts in diverse fields and enhance your understanding. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
Alright, let's solve the problem step-by-step.
Given sets:
- Universal set [tex]\( U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \)[/tex]
- Set [tex]\( A = \{1, 2, 4, 7\} \)[/tex]
- Set [tex]\( B = \{1, 3, 5, 7, 9\} \)[/tex]
We need to use De Morgan's Law to find [tex]\((A \cap B)^{\prime}\)[/tex].
### Step 1: Find [tex]\( A \cap B \)[/tex]
First, we find the intersection of sets [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[ A \cap B = \{ x \mid x \in A \text{ and } x \in B \} \][/tex]
Elements in [tex]\( A \)[/tex] are: [tex]\( \{1, 2, 4, 7\} \)[/tex]
Elements in [tex]\( B \)[/tex] are: [tex]\( \{1, 3, 5, 7, 9\} \)[/tex]
The common elements in both sets are:
[tex]\[ A \cap B = \{1, 7\} \][/tex]
### Step 2: Find the complement of [tex]\( A \cap B \)[/tex]
The complement of [tex]\( A \cap B \)[/tex] with respect to the universal set [tex]\( U \)[/tex] is:
[tex]\[ (A \cap B)^{\prime} = U - (A \cap B) \][/tex]
[tex]\[ A \cap B = \{1, 7\} \][/tex]
So, the elements in the universal set [tex]\( U \)[/tex] that are not in [tex]\( A \cap B \)[/tex] are:
[tex]\[ (A \cap B)^{\prime} = \{2, 3, 4, 5, 6, 8, 9, 10\} \][/tex]
### Step 3: Verify Using De Morgan's Law
According to De Morgan's Law, [tex]\((A \cap B)^{\prime} = A^{\prime} \cup B^{\prime}\)[/tex].
First, find the complements of [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[ A^{\prime} = U - A \][/tex]
[tex]\[ A = \{1, 2, 4, 7\} \][/tex]
[tex]\[ A^{\prime} = \{3, 5, 6, 8, 9, 10\} \][/tex]
[tex]\[ B^{\prime} = U - B \][/tex]
[tex]\[ B = \{1, 3, 5, 7, 9\} \][/tex]
[tex]\[ B^{\prime} = \{2, 4, 6, 8, 10\} \][/tex]
Then, find the union of [tex]\( A^{\prime} \)[/tex] and [tex]\( B^{\prime} \)[/tex]:
[tex]\[ A^{\prime} \cup B^{\prime} = \{3, 5, 6, 8, 9, 10\} \cup \{2, 4, 6, 8, 10\} \][/tex]
[tex]\[ A^{\prime} \cup B^{\prime} = \{2, 3, 4, 5, 6, 8, 9, 10\} \][/tex]
### Conclusion
As expected from De Morgan's Law:
[tex]\[ (A \cap B)^{\prime} = \{2, 3, 4, 5, 6, 8, 9, 10\} \][/tex]
which verifies our original statement.
Thus:
[tex]\[ (A \cap B)^{\prime} = A^{\prime} \cup B^{\prime} = \{2, 3, 4, 5, 6, 8, 9, 10\} \][/tex]
So, the correct answer is:
[tex]\[ (A \cap B)^{\prime} = \{ 2, 3, 4, 5, 6, 8, 9, 10 \} \][/tex]
Given sets:
- Universal set [tex]\( U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \)[/tex]
- Set [tex]\( A = \{1, 2, 4, 7\} \)[/tex]
- Set [tex]\( B = \{1, 3, 5, 7, 9\} \)[/tex]
We need to use De Morgan's Law to find [tex]\((A \cap B)^{\prime}\)[/tex].
### Step 1: Find [tex]\( A \cap B \)[/tex]
First, we find the intersection of sets [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[ A \cap B = \{ x \mid x \in A \text{ and } x \in B \} \][/tex]
Elements in [tex]\( A \)[/tex] are: [tex]\( \{1, 2, 4, 7\} \)[/tex]
Elements in [tex]\( B \)[/tex] are: [tex]\( \{1, 3, 5, 7, 9\} \)[/tex]
The common elements in both sets are:
[tex]\[ A \cap B = \{1, 7\} \][/tex]
### Step 2: Find the complement of [tex]\( A \cap B \)[/tex]
The complement of [tex]\( A \cap B \)[/tex] with respect to the universal set [tex]\( U \)[/tex] is:
[tex]\[ (A \cap B)^{\prime} = U - (A \cap B) \][/tex]
[tex]\[ A \cap B = \{1, 7\} \][/tex]
So, the elements in the universal set [tex]\( U \)[/tex] that are not in [tex]\( A \cap B \)[/tex] are:
[tex]\[ (A \cap B)^{\prime} = \{2, 3, 4, 5, 6, 8, 9, 10\} \][/tex]
### Step 3: Verify Using De Morgan's Law
According to De Morgan's Law, [tex]\((A \cap B)^{\prime} = A^{\prime} \cup B^{\prime}\)[/tex].
First, find the complements of [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[ A^{\prime} = U - A \][/tex]
[tex]\[ A = \{1, 2, 4, 7\} \][/tex]
[tex]\[ A^{\prime} = \{3, 5, 6, 8, 9, 10\} \][/tex]
[tex]\[ B^{\prime} = U - B \][/tex]
[tex]\[ B = \{1, 3, 5, 7, 9\} \][/tex]
[tex]\[ B^{\prime} = \{2, 4, 6, 8, 10\} \][/tex]
Then, find the union of [tex]\( A^{\prime} \)[/tex] and [tex]\( B^{\prime} \)[/tex]:
[tex]\[ A^{\prime} \cup B^{\prime} = \{3, 5, 6, 8, 9, 10\} \cup \{2, 4, 6, 8, 10\} \][/tex]
[tex]\[ A^{\prime} \cup B^{\prime} = \{2, 3, 4, 5, 6, 8, 9, 10\} \][/tex]
### Conclusion
As expected from De Morgan's Law:
[tex]\[ (A \cap B)^{\prime} = \{2, 3, 4, 5, 6, 8, 9, 10\} \][/tex]
which verifies our original statement.
Thus:
[tex]\[ (A \cap B)^{\prime} = A^{\prime} \cup B^{\prime} = \{2, 3, 4, 5, 6, 8, 9, 10\} \][/tex]
So, the correct answer is:
[tex]\[ (A \cap B)^{\prime} = \{ 2, 3, 4, 5, 6, 8, 9, 10 \} \][/tex]
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.