Math Problem Statement

State and prove De Morgan's laws for sets.

Solution

De Morgan's laws for sets describe how the complement of a union or intersection of sets can be expressed in terms of the complements of the individual sets. There are two main laws, which can be stated as follows:

  1. First Law (Union of Complements):
    (AB)c=AcBc(A \cup B)^c = A^c \cap B^c This means that the complement of the union of sets AA and BB is equal to the intersection of their complements.

  2. Second Law (Intersection of Complements):
    (AB)c=AcBc(A \cap B)^c = A^c \cup B^c This means that the complement of the intersection of sets AA and BB is equal to the union of their complements.

Proof of De Morgan's Laws

Proof of the First Law: (AB)c=AcBc(A \cup B)^c = A^c \cap B^c

  1. Let x(AB)cx \in (A \cup B)^c.

    • This implies that xx is not in ABA \cup B.
    • Therefore, xx is neither in AA nor in BB.
    • Hence, xAcx \in A^c and xBcx \in B^c.
    • This implies that xAcBcx \in A^c \cap B^c.

    Thus, (AB)cAcBc(A \cup B)^c \subseteq A^c \cap B^c.

  2. Conversely, let xAcBcx \in A^c \cap B^c.

    • This implies that xAcx \in A^c and xBcx \in B^c.
    • Therefore, xx is not in AA and not in BB.
    • Hence, xx is not in ABA \cup B.
    • This implies that x(AB)cx \in (A \cup B)^c.

    Thus, AcBc(AB)cA^c \cap B^c \subseteq (A \cup B)^c.

Since we have both (AB)cAcBc(A \cup B)^c \subseteq A^c \cap B^c and AcBc(AB)cA^c \cap B^c \subseteq (A \cup B)^c, we conclude that: (AB)c=AcBc(A \cup B)^c = A^c \cap B^c

Proof of the Second Law: (AB)c=AcBc(A \cap B)^c = A^c \cup B^c

  1. Let x(AB)cx \in (A \cap B)^c.

    • This implies that xx is not in ABA \cap B.
    • Therefore, xx is either not in AA or not in BB.
    • Thus, xAcx \in A^c or xBcx \in B^c.
    • This implies that xAcBcx \in A^c \cup B^c.

    Thus, (AB)cAcBc(A \cap B)^c \subseteq A^c \cup B^c.

  2. Conversely, let xAcBcx \in A^c \cup B^c.

    • This implies that xAcx \in A^c or xBcx \in B^c.
    • Therefore, xx is either not in AA or not in BB.
    • Hence, xx is not in ABA \cap B.
    • This implies that x(AB)cx \in (A \cap B)^c.

    Thus, AcBc(AB)cA^c \cup B^c \subseteq (A \cap B)^c.

Since we have both (AB)cAcBc(A \cap B)^c \subseteq A^c \cup B^c and AcBc(AB)cA^c \cup B^c \subseteq (A \cap B)^c, we conclude that: (AB)c=AcBc(A \cap B)^c = A^c \cup B^c

These are De Morgan's laws for sets.

Would you like further details on these proofs or have any questions?


Here are five related questions to explore:

  1. How do De Morgan's laws apply in Boolean algebra?
  2. Can De Morgan's laws be extended to three or more sets?
  3. What are the practical applications of De Morgan's laws in probability theory?
  4. How do De Morgan's laws relate to logic gates in digital circuits?
  5. How would you apply De Morgan's laws to solve complement problems in set theory?

Tip: De Morgan's laws are useful in simplifying expressions, especially in problems involving complements and unions or intersections.

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Math Problem Analysis

Mathematical Concepts

Set Theory
Set Complements
Logical Operations

Formulas

(A ∪ B)^c = A^c ∩ B^c
(A ∩ B)^c = A^c ∪ B^c

Theorems

De Morgan's Laws

Suitable Grade Level

Grades 10-12