Problem 10

Question

Problem 1.1.10 Prove the following distributive laws: $A \cap(B \cup C)=(A \cap B) \cup(A \cap C), \quad A \cup(B \cap C)=(A \cup B) \cap(A \cup C)$

Step-by-Step Solution

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Answer

The distributive laws are true and have been proven algebraically.

1Step 1: Understand the Distributive Laws

The distributive laws in set theory relate intersection and union operations, much like distributive laws in algebra. The expressions to prove are: $ A \cap (B \cup C) = (A \cap B) \cup (A \cap C) $ and $ A \cup (B \cap C) = (A \cup B) \cap (A \cup C) $.

2Step 2: Prove the First Distributive Law (Forward Direction)

Begin with $ A \cap (B \cup C) $. An element $ x $ is in this set if it is in $ A $ and also in $ B \cup C $. Therefore, $ x $ can be in $ A \cap B $ or $ A \cap C $ if it is in $ B $ or $ C $ while being in $ A $. Hence, $ x \in (A \cap B) \cup (A \cap C) $.

3Step 3: Prove the First Distributive Law (Backward Direction)

Now start with $ (A \cap B) \cup (A \cap C) $. An element $ x $ in this set must be in either $ A \cap B $ or $ A \cap C $. If in $ A \cap B $, $ x $ is in $ A $ and $ B $; likewise, if in $ A \cap C $, $ x $ is in $ A $ and $ C $. In both cases, $ x \in A \cap (B \cup C) $.

4Step 4: Prove the Second Distributive Law (Forward Direction)

Consider $ A \cup (B \cap C) $. An element $ x $ is in this set if it is either in $ A $ or in $ B \cap C $. Hence, $ x $ must be in either $ A \cup B $ or $ A \cup C $, meaning it is also in $ (A \cup B) \cap (A \cup C) $.

5Step 5: Prove the Second Distributive Law (Backward Direction)

Start with $ (A \cup B) \cap (A \cup C) $. Here, an element $ x $ is in both $ A \cup B $ and $ A \cup C $. Therefore, $ x $ is in $ A $ or in both $ B $ and $ C $. Thus, $ x \in A \cup (B \cap C) $.

6Step 6: Conclusion

In both cases, we have shown the forward and backward inclusion, leading to equality between the sets. Thus, the distributive laws are proven.

Key Concepts

Distributive Laws in Set TheoryIntersection and UnionProof in Mathematics

Distributive Laws in Set Theory

In set theory, distributive laws connect two fundamental operations: intersection and union. Understanding these laws helps in manipulating and simplifying complex set expressions much like the distributive property in arithmetic.
Here are the relevant distributive laws to know:

The first law states: $ A \cap (B \cup C) = (A \cap B) \cup (A \cap C) $. This emphasizes how the intersection distributes over the union, indicating that an element belongs to the intersection of a set $ A $ and the union of sets $ B $ and $ C $, if and only if, it belongs to the intersection of $ A $ with each individual set, $ B $ and $ C $.
The second law says: $ A \cup (B \cap C) = (A \cup B) \cap (A \cup C) $. This shows that a union distributes over an intersection, meaning an element is in the union of $ A $ and the intersection of $ B $ and $ C $, if and only if, it is in the intersection of the union of $ A $ with each individual set, $ B $ and $ C $.

Understanding these laws enhances your ability to handle complex set operations and proofs effectively.

Intersection and Union

Intersection and union are two primary operations in set theory that help define relations between sets.

**Intersection ($\cap$)**: This operation returns a set containing all elements that are common to all sets involved. For the intersection $ A \cap B $, it means finding every element that is in both sets $ A $ and $ B $.
**Union ($\cup$)**: Conversely, the union operation results in a set containing every element that exists in at least one of the sets. For the union $ A \cup B $, it means including every element from both sets $ A $ and $ B $, without duplicating any elements.

These operations are fundamental in set theory, forming the basis for more complex manipulations and expressions. Recognizing how they work together, especially through the distributive laws, provides greater insights into the relationships between different sets.

Proof in Mathematics

Proving mathematical properties like the distributive laws in set theory involves demonstrating that two expressions are equivalent by showing that each is a subset of the other.
Here's a simple approach to follow when constructing a proof:

**Forward Direction**: Begin with one expression and demonstrate that every element of this set also belongs to the other set. This establishes that one side is at least a subset of the other.
**Backward Direction**: Then, start from the other expression and show that every element in this set is also present in the original set. This confirms that the sets are indeed equal since they contain exactly the same elements.
**Conclusion**: Affirm the equality by noting that since each set is a subset of the other, their contents are identical, solidifying both expressions as equivalent.

This two-step proof technique is crucial for verifying set identities, ensuring we comprehend the full relationship between set expressions.

Problem 10

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