Pregunta
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Let \( A= \) \{apple, prune, cantaloupe, pear\}, \( B= \) (peach, lemon\}, and \( C= \) \{peach, kiwi, berry\}. (a) Find (i) \( n(A \cup B) \), (ii) \( n(A \cup C) \), and (iii) \( n(B \cup C) \). (b) In which case is the number of elements in the union not the sum of the number of elements in the individual sets? (a) (i) \( n(A \cup B)=6 \) (ii) \( n(A \cup C)=7 \) (iii) \( n(B \cup C)=4 \) (b) In which case is the number of elements in the union not the sum of the number of elements in the individual sets? (iii) (i) (ii) (i) AInd (iii) (i) and (ii) (ii) and (iii) (i). (ii), and (iii) None of the cases

Ask by Wood Little. in the United States
Feb 03,2025

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- **Part (a):** - (i) \( n(A \cup B) = 6 \) - (ii) \( n(A \cup C) = 7 \) - (iii) \( n(B \cup C) = 4 \) - **Part (b):** - The union does not equal the sum of the individual set sizes only in case (iii): \( B \cup C \).

Solución

Let's analyze the given sets and solve each part step-by-step. **Given Sets:** - \( A = \{ \text{apple}, \text{prune}, \text{cantaloupe}, \text{pear} \} \) → \( |A| = 4 \) - \( B = \{ \text{peach}, \text{lemon} \} \) → \( |B| = 2 \) - \( C = \{ \text{peach}, \text{kiwi}, \text{berry} \} \) → \( |C| = 3 \) --- ### **Part (a): Calculate the Number of Elements in Unions** 1. **\( n(A \cup B) \):** - **Union of A and B:** \[ A \cup B = \{ \text{apple}, \text{prune}, \text{cantaloupe}, \text{pear}, \text{peach}, \text{lemon} \} \] - **Number of Elements:** \[ n(A \cup B) = 6 \] 2. **\( n(A \cup C) \):** - **Union of A and C:** \[ A \cup C = \{ \text{apple}, \text{prune}, \text{cantaloupe}, \text{pear}, \text{peach}, \text{kiwi}, \text{berry} \} \] - **Number of Elements:** \[ n(A \cup C) = 7 \] 3. **\( n(B \cup C) \):** - **Union of B and C:** \[ B \cup C = \{ \text{peach}, \text{lemon}, \text{kiwi}, \text{berry} \} \] - **Number of Elements:** \[ n(B \cup C) = 4 \] --- ### **Part (b): When is the Union Not the Sum of Individual Sets?** The number of elements in the union of two sets is **not** simply the sum of the number of elements in each set **if and only if** the sets have at least one element in common. This is because the overlapping elements are counted twice when simply adding the sizes. **Analyzing Each Case:** 1. **\( A \cup B \):** - **Common Elements:** None - **Sum of Sizes:** \( |A| + |B| = 4 + 2 = 6 \) - **Union Size:** \( n(A \cup B) = 6 \) - **Conclusion:** \( n(A \cup B) = |A| + |B| \) 2. **\( A \cup C \):** - **Common Elements:** None - **Sum of Sizes:** \( |A| + |C| = 4 + 3 = 7 \) - **Union Size:** \( n(A \cup C) = 7 \) - **Conclusion:** \( n(A \cup C) = |A| + |C| \) 3. **\( B \cup C \):** - **Common Elements:** \( \text{peach} \) - **Sum of Sizes:** \( |B| + |C| = 2 + 3 = 5 \) - **Union Size:** \( n(B \cup C) = 4 \) - **Conclusion:** \( n(B \cup C) < |B| + |C| \) **Therefore, the union does not equal the sum of the individual set sizes only in case (iii): \( B \cup C \).** --- **Final Answers:** - **Part (a):** - (i) \( n(A \cup B) = 6 \) - (ii) \( n(A \cup C) = 7 \) - (iii) \( n(B \cup C) = 4 \) - **Part (b):** - **Only in case (iii): \( B \cup C \)**

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Beyond the Answer

To find the number of elements in the unions of sets \( A \), \( B \), and \( C \), we first need to determine the elements contained in each. Set \( A = \{ \text{apple, prune, cantaloupe, pear} \} \) has 4 distinct elements. Set \( B = \{ \text{peach, lemon} \} \) has 2 distinct elements. Set \( C = \{ \text{peach, kiwi, berry} \} \) has 3 distinct elements. Now, let's compute the unions: (i) \( A \cup B = \{ \text{apple, prune, cantaloupe, pear, peach, lemon} \} \) Here, 'peach' from set \( B \) is already included in \( A \), so we have: \( n(A \cup B) = 6 \). (ii) \( A \cup C = \{ \text{apple, prune, cantaloupe, pear, peach, kiwi, berry} \} \) Again, 'peach' from \( C \) overlaps with \( A \): \( n(A \cup C) = 7 \). (iii) \( B \cup C = \{ \text{peach, lemon, kiwi, berry} \} \) This time, 'peach' is present in both \( B \) and \( C \): \( n(B \cup C) = 4 \). Now for part (b), here are the number of elements in each case: - In (i), we had 6 elements total in the union, not simply the sum of individual sets (4 + 2 = 6). - In (ii), we had 7 which also does follow this checking out with (4 + 3 - 1 for 'peach'). - In (iii), 4 elements, as 'peach' is counted in both sets B and C. Thus, the answer to part (b) is: None of the cases, because despite overlaps, we still correctly accounted for the total in all scenarios.

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