Problem 4

Question

Prove by means of set inclusion the following distributive law: $$ (A \cup B) \cap C=(A \cap C) \cup(B \cap C) $$

Step-by-Step Solution

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Answer

The distributive law $(A \cup B) \cap C = (A \cap C) \cup (B \cap C)$ is proven true by set inclusion.

1Step 1: Understand the Task

We need to prove that the distributive law $(A \cup B) \cap C = (A \cap C) \cup (B \cap C)$ holds. This can be done by showing that the left-hand side is a subset of the right-hand side and vice versa. So, our task involves proving two inclusions: $(A \cup B) \cap C \subseteq (A \cap C) \cup (B \cap C)$ and $(A \cap C) \cup (B \cap C) \subseteq (A \cup B) \cap C$.

2Step 2: Prove $(A \cup B) \cap C \subseteq (A \cap C) \cup (B \cap C)$

Assume $x \in (A \cup B) \cap C$. This means $x \in A \cup B$ and $x \in C$. From $x \in A \cup B$, we know that $x \in A$ or $x \in B$. If $x \in A$, then $x \in A \cap C$, and if $x \in B$, then $x \in B \cap C$. Thus, $x \in (A \cap C) \cup (B \cap C)$. Hence, $(A \cup B) \cap C \subseteq (A \cap C) \cup (B \cap C)$.

3Step 3: Prove $(A \cap C) \cup (B \cap C) \subseteq (A \cup B) \cap C$

Assume $x \in (A \cap C) \cup (B \cap C)$. Then $x \in A \cap C$ or $x \in B \cap C$. If $x \in A \cap C$, then $x \in A$ and $x \in C$. Similarly, if $x \in B \cap C$, then $x \in B$ and $x \in C$. Therefore, in either case, $x \in A \cup B$ and $x \in C$, implying $x \in (A \cup B) \cap C$. Hence, $(A \cap C) \cup (B \cap C) \subseteq (A \cup B) \cap C$.

4Step 4: Conclude the Proof

Since both $(A \cup B) \cap C \subseteq (A \cap C) \cup (B \cap C)$ and $(A \cap C) \cup (B \cap C) \subseteq (A \cup B) \cap C$, we conclude that $(A \cup B) \cap C = (A \cap C) \cup (B \cap C)$, proving the distributive law by set inclusion.

Key Concepts

Set InclusionSet TheorySubsetProof Methodology

Set Inclusion

Set inclusion is a fundamental concept in set theory. It involves expressing the relation between two sets, typically represented using the subset notation. If set A is a subset of set B, it means all elements of A are also elements of B, which is denoted as $A \subseteq B$. An important point to note is that set inclusion does not require equality between the sets. A can be entirely contained within B, but B might have additional elements not in A. In proofs, demonstrating set inclusion often requires showing the validity of this relationship in both directions: that one set is contained within the other and vice versa.

Set Theory

Set theory is the branch of mathematics that studies sets, which are collections of objects. It serves as a foundational system for mathematical logic and is primarily comprised of the concepts of union, intersection, complement, and inclusion. Understanding these concepts is key to solving problems involving sets.

Union ($A \cup B$): combines the elements of sets A and B.
Intersection ($A \cap B$): identifies common elements between sets A and B.
Complement: consists of elements not in the set.

These operations allow us to create different relationships between sets and are crucial for comprehending how distributive laws function in set theory.

Subset

In set theory, a subset is a set whose elements are all contained within another set. If A is a subset of B, it implies that every element in A is also in B, denoted as $A \subseteq B$. Understanding subsets is vital in proofs because to establish that one set is a subset of another, we need to demonstrate this point for every element:

If element $x$ belongs to A, then it must also belong to B.

This concept is particularly helpful when proving statements involving distributive laws, where we must show the subsets involved in both directions to prove equality.

Proof Methodology

In mathematical reasoning, proof methodology refers to the systematic approach used to demonstrate the truth of a statement. One common method in set theory is proof by set inclusion. This involves splitting the proof into two parts:

Proving that one set is a subset of another.
Proving the reverse inclusion, that the second set is a subset of the first.

If both inclusions are proven, we can conclude that the two sets are equal. This two-step process makes it possible to approach complex logical relationships with clarity and structure, particularly in verifying equations like the distributive law in set operations.

Problem 4

Problem 5

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Problem 5

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