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.
