a) Creative Thinking: If the number of elements of \( A \cap B \) is 5, then the number of elements of A cannot be ......
The correct answer is (a) 4.
Since \( |A \cap B| = 5 \), set A must have at least 5 elements (as the intersection cannot be larger than either set).
Therefore, A cannot have only 4 elements.
Since \( |A \cap B| = 5 \), set A must have at least 5 elements (as the intersection cannot be larger than either set).
Therefore, A cannot have only 4 elements.
MR/Nasef.N
b) Creative Thinking: If the number of elements of \( A \cup B \) is 5, and the number of elements of B is 3, then the smallest number of elements of A is ......
The correct answer is (b) 2.
Using the principle: \( |A \cup B| = |A| + |B| - |A \cap B| \)
\( 5 = |A| + 3 - |A \cap B| \) ⇒ \( |A| = 2 + |A \cap B| \)
The minimum occurs when \( |A \cap B| \) is maximized (which is 3, the size of B).
Thus minimum \( |A| = 2 + 3 - 3 = 2 \).
Using the principle: \( |A \cup B| = |A| + |B| - |A \cap B| \)
\( 5 = |A| + 3 - |A \cap B| \) ⇒ \( |A| = 2 + |A \cap B| \)
The minimum occurs when \( |A \cap B| \) is maximized (which is 3, the size of B).
Thus minimum \( |A| = 2 + 3 - 3 = 2 \).
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