(i) Set A has 3 elements, and set B has 6 elements. What can be the minimum number of elements in th

Q: Why can't the minimum number of elements be 0? If I have 3 elements in A and 6 in B, can't they be disjoint?

That is a fantastic question! If they were disjoint, the union would be $3 + 6 = 9$. But we are looking for the *minimum* possible union. To minimize the union, we need to maximize the overlap (the intersection). The maximum overlap is limited by the size of the smaller set, which is 3. So, we subtract 3 from 9 to get 6. If they were disjoint, that would actually give us the *maximum* possible union!

Q: Why does $A \subset B$ minimize the union?

Think of it this way: when $A$ is a subset of $B$, every single element of $A$ is already 'living' inside $B$. Therefore, when you form the union $A \cup B$, you aren't adding any *new* elements to the set $B$. The union is simply $B$ itself, which has 6 elements. This is the smallest possible size because you cannot have a union smaller than the largest set involved.

Q: Is the Cartesian product always distributive over intersection?

Yes, absolutely! The Cartesian product $A \times (B \cap C)$ is indeed equal to $(A \times B) \cap (A \times C)$. It behaves very much like multiplication over addition in algebra. This property is a cornerstone of set theory and is incredibly useful for proving identities like the one in this problem.

Question

(i) Set $A$ has 3 elements, and set $B$ has 6 elements. What can be the minimum number of elements in the set $A \cup B$?
(ii) $P, Q, R$ are subsets of a set $A$. Is the following equality true?
$R 	imes (P^c \cup Q^c)^c = (R 	imes P) \cap (R 	imes Q)$?

Sigma Percentile · Accepted Answer

### The Art of Set Theory: Minimization and Logic

Welcome, future engineers! Today, we are going to peel back the layers of a seemingly simple set theory problem. Often, in the heat of a JEE Advanced exam, we tend to rush through the basics, thinking they are 'too easy.'

But the beauty of mathematics lies in the elegance of its foundations. Let us walk through this problem, not just to find the answer, but to understand the logic that governs the behavior of sets.

### Part I: The Tug-of-War of Minimization

Imagine you have two sets, $A$ and $B$. Set $A$ is a small group of 3 elements, and set $B$ is a larger group of 6 elements. We are asked to find the **minimum** number of elements in their union, $n(A \cup B)$.

To solve this, we must invoke the Principle of Inclusion-Exclusion. This is our primary tool:

$$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$

Think of this equation as a balance. We know $n(A) = 3$ and $n(B) = 6$. Substituting these, we get:

$$n(A \cup B) = 3 + 6 - n(A \cap B) = 9 - n(A \cap B)$$

Here is the crucial insight: to make the union as small as possible, we must subtract the largest possible value from 9. In other words, we need to maximize the intersection, $n(A \cap B)$.

What is the largest possible overlap between a set of 3 and a set of 6? It is simply the size of the smaller set, which is 3. This occurs when $A$ is entirely contained within $B$ (i.e., $A \subset B$).

When this happens, the union is just $9 - 3 = 6$. We have successfully minimized the union!

### Part II: The Elegance of Set Identities

Now, let us shift gears to the second part of our journey. We are presented with a complex-looking equality involving Cartesian products and complements: $R 	imes (P^c \cup Q^c)^c = (R 	imes P) \cap (R 	imes Q)$.

Do not let the notation intimidate you. In JEE Advanced, complex expressions are often just simple concepts wearing a disguise. Let us break down the Left Hand Side (LHS) using De Morgan's Law.

The law states that the complement of a union is the intersection of the complements. Applying this to the term $(P^c \cup Q^c)^c$, we get:

$$(P^c \cup Q^c)^c = (P^c)^c \cap (Q^c)^c$$

Since the complement of a complement returns the original set, $(P^c)^c = P$ and $(Q^c)^c = Q$. Thus, our expression simplifies beautifully to $P \cap Q$. The LHS becomes $R 	imes (P \cap Q)$.

Now, look at the Right Hand Side (RHS): $(R 	imes P) \cap (R 	imes Q)$. This is a classic application of the distributive property of the Cartesian product over intersection.

Just as $a(b + c) = ab + ac$ in basic algebra, the Cartesian product distributes over the intersection of sets:

$$R 	imes (P \cap Q) = (R 	imes P) \cap (R 	imes Q)$$

### The Conclusion

When we compare our simplified LHS and our simplified RHS, we see they are identical. The equality holds true!

By systematically breaking down the problem—first by visualizing the constraints of the union, and then by applying the fundamental laws of set algebra—we have navigated through the complexity with clarity.

Remember, in physics and mathematics, the most complex problems are often just a series of simple, elegant steps waiting to be uncovered. Keep practicing, keep visualizing, and keep falling in love with the logic!

Visualized Solution

Visualizing Sets $A$ and $B$

The Principle of Inclusion-Exclusion

Substituting the Values

Maximizing the Intersection

The Subset Condition

Analyzing Part (ii): Set Equality

Applying De Morgan's Law

Simplifying the LHS

Distributive Property of Cartesian Product

Conclusion and Summary

The Sigma Insight: Types of Sets and Set Operations

The Art of Set Theory

Part I

Part II

The Conclusion

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