Set operations — union, intersection, complement, difference with Venn diagrams

Question

Let $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ , $A = \{1, 2, 3, 4, 5\}$ , and $B = \{3, 4, 5, 6, 7\}$ . Find: (a) $A \cup B$ , (b) $A \cap B$ , (c) $A - B$ , (d) $A'$ , (e) $(A \cup B)'$ , (f) Verify De Morgan’s law: $(A \cup B)' = A' \cap B'$ .

Solution — Step by Step

$A \cup B$ = all elements in A or B or both = $\{1, 2, 3, 4, 5, 6, 7\}$

$A \cap B$ = elements in both A and B = $\{3, 4, 5\}$

$A - B$ = elements in A but not in B = $\{1, 2\}$

$A'$ = elements in U but not in A = $\{6, 7, 8, 9, 10\}$

$B'$ = elements in U but not in B = $\{1, 2, 8, 9, 10\}$

$(A \cup B)' = U - (A \cup B) = \{8, 9, 10\}$

$A' \cap B' = \{6, 7, 8, 9, 10\} \cap \{1, 2, 8, 9, 10\} = \{8, 9, 10\}$

Since $(A \cup B)' = \{8, 9, 10\} = A' \cap B'$ , De Morgan’s law is verified.

Why This Works

graph TD
    A["Set Operation"] --> B["Union A ∪ B"]
    A --> C["Intersection A ∩ B"]
    A --> D["Difference A - B"]
    A --> E["Complement A'"]
    B --> F["Everything in either set"]
    C --> G["Only what is common"]
    D --> H["In A but not in B"]
    E --> I["Everything in U not in A"]
    A --> J["De Morgan's Laws"]
    J --> K["A ∪ B complement = A' ∩ B'"]
    J --> L["A ∩ B complement = A' ∪ B'"]

Set operations are the foundation of mathematical logic. Union corresponds to “OR,” intersection to “AND,” and complement to “NOT.” De Morgan’s laws connect these operations — they say “not(A or B) = (not A) and (not B)” and “not(A and B) = (not A) or (not B).” These laws appear everywhere — in probability, logic gates, database queries, and programming.

The cardinal number formula is essential for problem solving: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$ . We subtract the intersection because it was counted twice.

Alternative Method

For JEE problems involving three sets, use the inclusion-exclusion principle:

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

Draw a Venn diagram with three overlapping circles and fill in values from the innermost region outward. Start with $n(A \cap B \cap C)$ , then fill the two-set-only regions, then single-set-only regions. This approach prevents double-counting errors.

Common Mistake

Confusing $A - B$ with $B - A$ . These are NOT the same unless $A = B$ . Here, $A - B = \{1, 2\}$ but $B - A = \{6, 7\}$ . Difference is not commutative. Also, students often forget that $A' = U - A$ , not just “not A” — the complement depends on the universal set $U$ . Always check what $U$ is before computing complements.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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