Math Problem Statement

5 - Sets Again.pdf

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Solution

The document contains a proof assignment related to sets in mathematics. Here is a breakdown of the problems:

  1. Problem 1: Suppose AA, BB, and CC are sets. Prove the following: a. A(BC)=(AB)(AC)A \cap (B \cup C) = (A \cap B) \cup (A \cap C)
    b. A×(BC)=(A×B)(A×C)A \times (B \cap C) = (A \times B) \cap (A \times C)

These are two classic set-theoretic identities. Here’s a sketch of the proofs:

Proof for 1a:

Distributive Property of Intersection over Union

To prove:
A(BC)=(AB)(AC)A \cap (B \cup C) = (A \cap B) \cup (A \cap C)

Step-by-Step Proof:

  1. Start with the left-hand side A(BC)A \cap (B \cup C).

    • An element xA(BC)x \in A \cap (B \cup C) means that xAx \in A and xBCx \in B \cup C.
    • By definition of union, xBCx \in B \cup C means xBx \in B or xCx \in C.
    • Therefore, xABx \in A \cap B or xACx \in A \cap C, which implies x(AB)(AC)x \in (A \cap B) \cup (A \cap C).
  2. Now, the reverse direction: Take x(AB)(AC)x \in (A \cap B) \cup (A \cap C).

    • This means that xABx \in A \cap B or xACx \in A \cap C, so xAx \in A and (xB or xC)(x \in B \text{ or } x \in C).
    • Hence, xAx \in A and xBCx \in B \cup C, which implies xA(BC)x \in A \cap (B \cup C).

Since both directions hold, we have proved that A(BC)=(AB)(AC)A \cap (B \cup C) = (A \cap B) \cup (A \cap C).


Proof for 1b:

Distributive Property of Cartesian Product over Intersection

To prove:
A×(BC)=(A×B)(A×C)A \times (B \cap C) = (A \times B) \cap (A \times C)

Step-by-Step Proof:

  1. Start with the left-hand side A×(BC)A \times (B \cap C).

    • An element (a,x)A×(BC)(a, x) \in A \times (B \cap C) means that aAa \in A and xBCx \in B \cap C.
    • By the definition of intersection, xBCx \in B \cap C means xBx \in B and xCx \in C.
    • Therefore, (a,x)A×B(a, x) \in A \times B and (a,x)A×C(a, x) \in A \times C, which implies (a,x)(A×B)(A×C)(a, x) \in (A \times B) \cap (A \times C).
  2. Now, the reverse direction: Take (a,x)(A×B)(A×C)(a, x) \in (A \times B) \cap (A \times C).

    • This means that (a,x)A×B(a, x) \in A \times B and (a,x)A×C(a, x) \in A \times C, so aAa \in A, xBx \in B, and xCx \in C.
    • Hence, (a,x)A×(BC)(a, x) \in A \times (B \cap C).

Since both directions hold, we have proved that A×(BC)=(A×B)(A×C)A \times (B \cap C) = (A \times B) \cap (A \times C).


Would you like further details or clarifications? Here are some related questions to consider:

  1. What are the laws of set theory, such as De Morgan’s laws, that can help in these types of proofs?
  2. How do you represent sets graphically to understand the relationships between union, intersection, and Cartesian products?
  3. Can these distributive properties extend to larger collections of sets (e.g., more than two sets)?
  4. How do Cartesian products apply in fields like computer science or graph theory?
  5. What other operations in set theory, like complement or difference, can also be distributed?

Tip: When proving set identities, it's often helpful to work from both sides of the equation and verify that each element follows logically from the assumptions.

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Math Problem Analysis

Mathematical Concepts

Set Theory
Intersection
Union
Cartesian Product

Formulas

A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
A × (B ∩ C) = (A × B) ∩ (A × C)

Theorems

Distributive Property of Intersection over Union
Distributive Property of Cartesian Product over Intersection

Suitable Grade Level

Undergraduate