Get expert insights and community support for your questions on IDNLearn.com. Discover the information you need from our experienced professionals who provide accurate and reliable answers to all your questions.
Sagot :
The given statement is
[tex]\[ n(A \cup B) = n(A) + n(B) \][/tex]
Let's analyze whether this statement is always true or not always true.
To do this, we need to understand what [tex]\( n(A \cup B) \)[/tex], [tex]\( n(A) \)[/tex], and [tex]\( n(B) \)[/tex] represent. Here:
- [tex]\( n(A) \)[/tex] denotes the number of elements in set [tex]\( A \)[/tex].
- [tex]\( n(B) \)[/tex] denotes the number of elements in set [tex]\( B \)[/tex].
- [tex]\( n(A \cup B) \)[/tex] denotes the number of elements in the union of sets [tex]\( A \)[/tex] and [tex]\( B \)[/tex].
The union of two sets [tex]\( A \)[/tex] and [tex]\( B \)[/tex], denoted by [tex]\( A \cup B \)[/tex], is the set of elements that are in [tex]\( A \)[/tex], or in [tex]\( B \)[/tex], or in both.
However, if there are elements that belong to both sets [tex]\( A \)[/tex] and [tex]\( B \)[/tex] (i.e., the intersection of [tex]\( A \)[/tex] and [tex]\( B \)[/tex], denoted as [tex]\( A \cap B \)[/tex], is not empty), then those elements would be counted twice if we simply added [tex]\( n(A) \)[/tex] and [tex]\( n(B) \)[/tex].
To correct this double-counting, the actual formula we use is:
[tex]\[ n(A \cup B) = n(A) + n(B) - n(A \cap B) \][/tex]
This formula accounts for the elements that are counted twice by subtracting the count of the elements in the intersection [tex]\( n(A \cap B) \)[/tex].
Thus, the given statement [tex]\( n(A \cup B) = n(A) + n(B) \)[/tex], which assumes no double-counting, does not hold true if there are elements common to both sets [tex]\( A \)[/tex] and [tex]\( B \)[/tex].
Therefore, the statement
[tex]\[ n(A \cup B) = n(A) + n(B) \][/tex]
is
[tex]\[ \boxed{\text{not always true}} \][/tex]
because it fails to account for the elements that are counted twice when sets [tex]\( A \)[/tex] and [tex]\( B \)[/tex] have common elements.
[tex]\[ n(A \cup B) = n(A) + n(B) \][/tex]
Let's analyze whether this statement is always true or not always true.
To do this, we need to understand what [tex]\( n(A \cup B) \)[/tex], [tex]\( n(A) \)[/tex], and [tex]\( n(B) \)[/tex] represent. Here:
- [tex]\( n(A) \)[/tex] denotes the number of elements in set [tex]\( A \)[/tex].
- [tex]\( n(B) \)[/tex] denotes the number of elements in set [tex]\( B \)[/tex].
- [tex]\( n(A \cup B) \)[/tex] denotes the number of elements in the union of sets [tex]\( A \)[/tex] and [tex]\( B \)[/tex].
The union of two sets [tex]\( A \)[/tex] and [tex]\( B \)[/tex], denoted by [tex]\( A \cup B \)[/tex], is the set of elements that are in [tex]\( A \)[/tex], or in [tex]\( B \)[/tex], or in both.
However, if there are elements that belong to both sets [tex]\( A \)[/tex] and [tex]\( B \)[/tex] (i.e., the intersection of [tex]\( A \)[/tex] and [tex]\( B \)[/tex], denoted as [tex]\( A \cap B \)[/tex], is not empty), then those elements would be counted twice if we simply added [tex]\( n(A) \)[/tex] and [tex]\( n(B) \)[/tex].
To correct this double-counting, the actual formula we use is:
[tex]\[ n(A \cup B) = n(A) + n(B) - n(A \cap B) \][/tex]
This formula accounts for the elements that are counted twice by subtracting the count of the elements in the intersection [tex]\( n(A \cap B) \)[/tex].
Thus, the given statement [tex]\( n(A \cup B) = n(A) + n(B) \)[/tex], which assumes no double-counting, does not hold true if there are elements common to both sets [tex]\( A \)[/tex] and [tex]\( B \)[/tex].
Therefore, the statement
[tex]\[ n(A \cup B) = n(A) + n(B) \][/tex]
is
[tex]\[ \boxed{\text{not always true}} \][/tex]
because it fails to account for the elements that are counted twice when sets [tex]\( A \)[/tex] and [tex]\( B \)[/tex] have common elements.
Thank you for participating in our discussion. We value every contribution. Keep sharing knowledge and helping others find the answers they need. Let's create a dynamic and informative learning environment together. For dependable and accurate answers, visit IDNLearn.com. Thanks for visiting, and see you next time for more helpful information.