Math Problem Statement

Prove A ∪ B = B ∪ A and A ∩ B = B ∩ A

Solution

We are tasked with proving the following two set identities:

  1. AB=BAA \cup B = B \cup A
  2. AB=BAA \cap B = B \cap A

1. Proving AB=BAA \cup B = B \cup A

To prove that AB=BAA \cup B = B \cup A, we need to show that every element in ABA \cup B is also in BAB \cup A, and vice versa.

Step 1: Prove ABBAA \cup B \subseteq B \cup A

  • Let xABx \in A \cup B. By definition of the union of sets, this means xAx \in A or xBx \in B.
    • If xAx \in A, then xBAx \in B \cup A (since xx is in AA, and BAB \cup A includes all elements of AA).
    • If xBx \in B, then xBAx \in B \cup A (since xx is in BB, and BAB \cup A includes all elements of BB).
  • Therefore, xBAx \in B \cup A, proving that ABBAA \cup B \subseteq B \cup A.

Step 2: Prove BAABB \cup A \subseteq A \cup B

  • Let xBAx \in B \cup A. By definition of the union of sets, this means xBx \in B or xAx \in A.
    • If xBx \in B, then xABx \in A \cup B (since xx is in BB, and ABA \cup B includes all elements of BB).
    • If xAx \in A, then xABx \in A \cup B (since xx is in AA, and ABA \cup B includes all elements of AA).
  • Therefore, xABx \in A \cup B, proving that BAABB \cup A \subseteq A \cup B.

Since both inclusions hold, we have shown that AB=BAA \cup B = B \cup A.

2. Proving AB=BAA \cap B = B \cap A

Next, we need to prove that AB=BAA \cap B = B \cap A. Again, we show that every element of ABA \cap B is in BAB \cap A, and vice versa.

Step 1: Prove ABBAA \cap B \subseteq B \cap A

  • Let xABx \in A \cap B. By the definition of intersection, this means that xAx \in A and xBx \in B.
  • Since xAx \in A and xBx \in B, it follows that xBAx \in B \cap A (because BAB \cap A includes all elements that are in both BB and AA).
  • Therefore, xBAx \in B \cap A, proving that ABBAA \cap B \subseteq B \cap A.

Step 2: Prove BAABB \cap A \subseteq A \cap B

  • Let xBAx \in B \cap A. By the definition of intersection, this means that xBx \in B and xAx \in A.
  • Since xBx \in B and xAx \in A, it follows that xABx \in A \cap B (because ABA \cap B includes all elements that are in both AA and BB).
  • Therefore, xABx \in A \cap B, proving that BAABB \cap A \subseteq A \cap B.

Since both inclusions hold, we have shown that AB=BAA \cap B = B \cap A.

Conclusion

We have proven both identities:

  1. AB=BAA \cup B = B \cup A
  2. AB=BAA \cap B = B \cap A

Would you like more details or further clarification on any part of the proof?

Here are 5 related questions:

  1. What are the properties of set unions and intersections in general?
  2. How can De Morgan's Laws be applied to set operations?
  3. What is the difference between the symmetric difference and union/intersection of sets?
  4. How can Venn diagrams help visualize set identities?
  5. How do set operations behave in the context of finite and infinite sets?

Tip: When proving set identities, it's helpful to break the problem into smaller inclusions (showing both subsets) to demonstrate equality.

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Math Problem Analysis

Mathematical Concepts

Set Theory
Union of Sets
Intersection of Sets
Set Identities

Formulas

A ∪ B = B ∪ A
A ∩ B = B ∩ A

Theorems

Commutative Property of Set Union
Commutative Property of Set Intersection

Suitable Grade Level

Grades 8-10