Let A and B be two subsets of the universal set U. Which of the following is equivalent to A ∪ (A ∩ B)?
Understand the Problem
The question is asking to determine which set expression is equivalent to the given expression involving sets A, B, and their relationships with the universal set U. We will analyze the expression A ∪ (A ∩ B) to find its equivalent.
Answer
$A$
Answer for screen readers
The expression $A \cup (A \cap B)$ is equivalent to $A$.
Steps to Solve
- Identify the Components of the Expression
We start with the expression $A \cup (A \cap B)$. In this expression, $A$ is a set, $B$ is another set, and we also need to remember that $\cup$ denotes the union and $\cap$ denotes the intersection of sets.
- Simplify the Intersection
The term $A \cap B$ represents the elements that are in both sets $A$ and $B$. Now we have:
$$ A \cup (A \cap B) $$
- Understand the Union
The union $A \cup (A \cap B)$ consists of all elements that are in set $A$ or in the intersection of sets $A$ and $B$. This means any element that is in $A$ is included in the union.
- Recognize Redundant Elements
Since the intersection $A \cap B$ consists of elements that are already in $A$, we can simplify the expression. Therefore, all elements in $A \cap B$ are accounted for in $A$, leading to:
$$ A \cup (A \cap B) = A $$
- Conclude with the Result
Now we have established that the given expression can be simplified to just set $A$.
The expression $A \cup (A \cap B)$ is equivalent to $A$.
More Information
This result illustrates a fundamental property of sets: the union of a set with its intersection with another set does not introduce any new elements. The only elements in the result are those that are already in the original set.
Tips
- Misunderstanding the operations: Confusing union and intersection can lead to incorrect interpretations. Ensure to remember that union combines all unique elements while intersection only takes the common elements.
- Forgetting about redundancy: It’s easy to overlook that the intersection $A \cap B$ does not add any new elements to the union if all are already within $A$.
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