Problem 59 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Let $$ \begin{array}{l} U=\\{1,2,3,4,5,6,7,8,9,10\\} \\ A=\\{1,3,5,7,9\\} \\ B=\\{1,2,4,7,8\\} \\ C=\\{2,4,6,8\\} \end{array} $$ Verify each equation by direct computation. a. \(A \cup(B \cup C)=(A \cup B) \cup C\) b. \(A \cap(B \cap C)=(A \cap B) \cap C\)

Step-by-Step Solution

Answer

We have verified both equations by direct computation: a. \(A \cup (B \cup C) = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}\) and \((A \cup B) \cup C = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}\) \\ So, \(A \cup (B \cup C) = (A \cup B) \cup C\) b. \(A \cap (B \cap C) = \emptyset\) and \((A \cap B) \cap C = \emptyset\) \\ So, \(A \cap (B \cap C) = (A \cap B) \cap C\)

Step 1: Calculate \(B \cup C\)

To calculate the union of two sets, we take all elements that belong to either B or C or both. There are no duplicates in the union, so if an element appears in both sets, we only include it once. In this case: \(B = \{1, 2, 4, 7, 8\}\) and \(C = \{2, 4, 6, 8\}\) So, \(B \cup C = \{1, 2, 4, 6, 7, 8\}\)

Step 2: Calculate \(A \cup (B \cup C)\)

Now that we know the result of \(B \cup C\), we can calculate the union of A and this result. Again, we take all elements that belong to either A or \(B \cup C\) or both: \(A = \{1, 3, 5, 7, 9\}\) and \(B \cup C = \{1, 2, 4, 6, 7, 8\}\) So, \(A \cup (B \cup C) = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}\)

Step 3: Calculate \(A \cup B\)

HNow, let's calculate the union of A and B directly: \(A = \{1, 3, 5, 7, 9\}\) and \(B = \{1, 2, 4, 7, 8\}\) So, \(A \cup B = \{1, 2, 3, 4, 5, 7, 8, 9\}\)

Step 4: Calculate \((A \cup B) \cup C\)

We will now compute the union of the result we obtained in step 3 with set C: \(A \cup B = \{1, 2, 3, 4, 5, 7, 8, 9\}\) and \(C = \{2, 4, 6, 8\}\) So, \((A \cup B) \cup C = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}\) Since the resulting sets from Step 2 and Step 4 are identical, the first equation is verified: \(A \cup (B \cup C) = (A \cup B) \cup C\)

Step 5: Calculate \(B \cap C\)

To calculate the intersection of two sets, we take the elements that appear in both sets: \(B = \{1, 2, 4, 7, 8\}\) and \(C = \{2, 4, 6, 8\}\) So, \(B \cap C = \{2, 4, 8\}\)

Step 6: Calculate \(A \cap (B \cap C)\)

Now that we know the result of \(B \cap C\), we can calculate the intersection of A and this result: \(A = \{1, 3, 5, 7, 9\}\) and \(B \cap C = \{2, 4, 8\}\) So, \(A \cap (B \cap C) = \emptyset\)

Step 7: Calculate \(A \cap B\)

Now, let's calculate the intersection of A and B directly: \(A = \{1, 3, 5, 7, 9\}\) and \(B = \{1, 2, 4, 7, 8\}\) So, \(A \cap B = \{1, 7\}\)

Step 8: Calculate \((A \cap B) \cap C\)

We will now compute the intersection of the result we obtained in step 7 with set C: \(A \cap B = \{1, 7\}\) and \(C = \{2, 4, 6, 8\}\) So, \((A \cap B) \cap C = \emptyset\) Since the resulting sets from Step 6 and Step 8 are identical, the second equation is verified: \(A \cap (B \cap C) = (A \cap B) \cap C\) We have now verified both equations by direct computation.

Key Concepts

Union of Sets

In set theory, the union of sets is a fundamental operation that unites elements from multiple sets. The union combines all elements from each set, ensuring each element appears only once.
To perform a union, list all distinct elements from the involved sets. For instance, consider sets:

• \(A = \{1, 3, 5, 7, 9\}\)
• \(B = \{1, 2, 4, 7, 8\}\)
• \(C = \{2, 4, 6, 8\}\)

The union of \(B \cup C\) means we gather all elements from both \(B\) and \(C\), so \(B \cup C = \{1, 2, 4, 6, 7, 8\}\).
When performing a full union with \(A\), we find that \(A \cup (B \cup C) = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}\). Notice there's no repetition, each element is listed once. This approach confirms that the order in which unions are performed does not alter the result, reflecting the associative property.

Intersection of Sets

Intersection of sets involves identifying elements common to multiple sets, listing only shared elements in the intersection result. For example, consider the same sets:

• \(A = \{1, 3, 5, 7, 9\}\)
• \(B = \{1, 2, 4, 7, 8\}\)
• \(C = \{2, 4, 6, 8\}\)

Applying the intersection operation, \(B \cap C\) identifies elements present in both \(B\) and \(C\), which are \(\{2, 4, 8\}\).
Performing \(A \cap (B \cap C)\), no common elements exist between \(A\) and the result of \(B \cap C\). Thus, we have \(A \cap (B \cap C) = \emptyset\), indicating an empty set. Similarly, \((A \cap B) \cap C\) results in \(\emptyset\). Hence, the intersection is associative; the order of operations doesn't affect the final outcome when intersecting three sets.

Associative Property

The associative property is a key concept in mathematics, applicable to various operations, including set theory. It demonstrates that the grouping of elements doesn't affect the result for union and intersection operations.
For the union of sets, consider the expression \((A \cup B) \cup C\). First, combining \(A\) and \(B\) yields \(\{1, 2, 3, 4, 5, 7, 8, 9\}\), and uniting with \(C\) gives \(\{1, 2, 3, 4, 5, 6, 7, 8, 9\}\). Similarly, \(A \cup (B \cup C)\) directly achieves the same set, proving the associative law via a common result.
For intersections, \((A \cap B) \cap C\) and \(A \cap (B \cap C)\) both lead to the empty set \(\emptyset\).
Therefore, associative property confirms that regardless of how elements are grouped in these operations, the end result remains consistent, simplifying complex set expressions.