Math Problem Statement
) If A ∩ B = ∅, then |(A ∪ B)^n| ≥ |A^n| + |B^n| for every n ≥ 1
Solution
Given two sets and with the property that their intersection is empty (), we want to analyze the expression and determine if it is always greater than or equal to for every integer .
Step-by-Step Analysis
-
Understanding :
- The set represents the set of all -tuples where each element in the tuple is chosen from .
- Since , each element in any -tuple in can independently come from either or , but not both simultaneously.
-
Cardinality of :
- If we denote and , then (because and are disjoint).
- The set then contains elements, as each position in an -tuple has choices.
-
Cardinality of and :
- The set consists of all -tuples where each element is from only. Thus, .
- Similarly, consists of all -tuples where each element is from only. Therefore, .
-
Inequality Check:
- We now compare with .
- By the binomial theorem, we have:
- Since each of the additional terms in the binomial expansion is positive for , it follows that:
-
Conclusion:
- Therefore, for every integer , the inequality holds true.
This completes the proof.
Would you like a deeper explanation of any part?
- What if ? How does it affect the result?
- What happens to the inequality if ?
- Can this approach be extended if and are not finite?
- How does this relate to combinatorial counting principles?
- Could this inequality be applied to probability spaces as well?
Tip: When tackling inequalities, consider expanding terms using known theorems (like the binomial theorem) to reveal any hidden positive terms that might support your argument.
Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Set Theory
Cardinality
Binomial Theorem
Inequalities
Formulas
|(A ∪ B)^n| = (|A| + |B|)^n
|A^n| = |A|^n
|B^n| = |B|^n
Theorems
Binomial Theorem
Suitable Grade Level
Undergraduate Level
Related Recommendation
Inequality in Set Cardinality: Proving #(A ∪ B) ≤ #A + #B
Proof of |C| = |C ∩ A| + |(C ∩ B) \ A| + |C \ B| when A ⊆ B
Transforming Set Inequalities Using Inclusion-Exclusion: A∩C ≥ A∩B + B∩C − B
Counter-Example for Set Union |A+B| = |A| + |B|
Proof of Inequality for Intersection of Two Events in Probability