Math Problem Statement

Prove that \(\overline{A \cup B} = \overline{A} \cap \overline{B}\).

Solution

The image contains a mathematical statement that needs to be proven. The statement is:

"Prove that AB=AB\overline{A \cup B} = \overline{A} \cap \overline{B}."

This is a well-known identity in set theory called De Morgan's Law.

Proof:

We need to prove that: AB=AB\overline{A \cup B} = \overline{A} \cap \overline{B} where:

  • AA and BB are any two sets.
  • A\overline{A} denotes the complement of AA, i.e., the set of all elements that are not in AA.
  • ABA \cup B is the union of sets AA and BB, meaning all elements that are in AA or BB or both.

1. Left-Hand Side (AB\overline{A \cup B}):

The complement of the union ABA \cup B consists of all elements that are not in ABA \cup B. This means: xAB    xABx \in \overline{A \cup B} \implies x \notin A \cup B So, xx is not in AA and is not in BB. This can be written as: xA and xBx \notin A \text{ and } x \notin B Thus: xA and xBx \in \overline{A} \text{ and } x \in \overline{B} Therefore: xABx \in \overline{A} \cap \overline{B} So, we have shown that: ABAB\overline{A \cup B} \subseteq \overline{A} \cap \overline{B}

2. Right-Hand Side (AB\overline{A} \cap \overline{B}):

Now, consider the right-hand side, which is AB\overline{A} \cap \overline{B}.

An element xx that belongs to AB\overline{A} \cap \overline{B} must be in both A\overline{A} and B\overline{B}. This means: xA and xBx \in \overline{A} \text{ and } x \in \overline{B} Which implies: xA and xBx \notin A \text{ and } x \notin B Therefore, xABx \notin A \cup B, which means: xABx \in \overline{A \cup B} So, we have shown that: ABAB\overline{A} \cap \overline{B} \subseteq \overline{A \cup B}

Conclusion:

Since both ABAB\overline{A \cup B} \subseteq \overline{A} \cap \overline{B} and ABAB\overline{A} \cap \overline{B} \subseteq \overline{A \cup B}, we conclude that: AB=AB\overline{A \cup B} = \overline{A} \cap \overline{B}

This completes the proof.


Would you like any further details or have any questions?

Here are some related questions to explore:

  1. How do De Morgan's laws extend to more than two sets?
  2. Can De Morgan's laws be applied to logical statements, and if so, how?
  3. What is the significance of set complements in probability theory?
  4. How does the intersection operation differ from the union in set theory?
  5. Can you provide an example where De Morgan's laws simplify a set expression?

Tip: De Morgan's laws are fundamental in both set theory and logic, helping simplify complex expressions involving unions, intersections, and complements.

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Math Problem Analysis

Mathematical Concepts

Set Theory
De Morgan's Laws

Formulas

\(\overline{A \cup B} = \overline{A} \cap \overline{B}\)

Theorems

De Morgan's Laws

Suitable Grade Level

Grades 11-12