Math Problem Statement

) If A ∩ B = ∅, then |(A ∪ B)^n| ≥ |A^n| + |B^n| for every n ≥ 1

Solution

Given two sets AA and BB with the property that their intersection is empty (AB=A \cap B = \emptyset), we want to analyze the expression (AB)n|(A \cup B)^n| and determine if it is always greater than or equal to An+Bn|A^n| + |B^n| for every integer n1n \geq 1.

Step-by-Step Analysis

  1. Understanding (AB)n(A \cup B)^n:

    • The set (AB)n(A \cup B)^n represents the set of all nn-tuples where each element in the tuple is chosen from ABA \cup B.
    • Since AB=A \cap B = \emptyset, each element in any nn-tuple in (AB)n(A \cup B)^n can independently come from either AA or BB, but not both simultaneously.
  2. Cardinality of (AB)n(A \cup B)^n:

    • If we denote A=a|A| = a and B=b|B| = b, then AB=a+b|A \cup B| = a + b (because AA and BB are disjoint).
    • The set (AB)n(A \cup B)^n then contains (a+b)n(a + b)^n elements, as each position in an nn-tuple has a+ba + b choices.
  3. Cardinality of AnA^n and BnB^n:

    • The set AnA^n consists of all nn-tuples where each element is from AA only. Thus, An=an|A^n| = a^n.
    • Similarly, BnB^n consists of all nn-tuples where each element is from BB only. Therefore, Bn=bn|B^n| = b^n.
  4. Inequality Check:

    • We now compare (AB)n=(a+b)n|(A \cup B)^n| = (a + b)^n with An+Bn=an+bn|A^n| + |B^n| = a^n + b^n.
    • By the binomial theorem, we have: (a+b)n=an+bn+(positive terms involving products of a and b).(a + b)^n = a^n + b^n + \text{(positive terms involving products of } a \text{ and } b\text{)}.
    • Since each of the additional terms in the binomial expansion is positive for n1n \geq 1, it follows that: (a+b)nan+bn.(a + b)^n \geq a^n + b^n.
  5. Conclusion:

    • Therefore, for every integer n1n \geq 1, the inequality (AB)nAn+Bn|(A \cup B)^n| \geq |A^n| + |B^n| holds true.

This completes the proof.

Would you like a deeper explanation of any part?


  1. What if ABA \cap B \neq \emptyset? How does it affect the result?
  2. What happens to the inequality if n=0n = 0?
  3. Can this approach be extended if AA and BB are not finite?
  4. How does this relate to combinatorial counting principles?
  5. 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.

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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