If $U = \{1, 2, 3, 4, 5, 6\}$, $A = \{2, 3\}$ and $B = \{3, 4, 5\}$, then prove that $(A \cup B)' = A' \cap B'$.

**OR**

If set $A = \{3, 5, 7, 9, 11\}$, $B = \{7, 9, 11, 13\}$, $C = \{11, 13, 15\}$ and $D = \{15, 17\}$, then find the following: (i) $A \cap (B \cup D)$ (ii) $(A

Question

If $U = \{1, 2, 3, 4, 5, 6\}$, $A = \{2, 3\}$ and $B = \{3, 4, 5\}$, then prove that $(A \cup B)' = A' \cap B'$.

**OR**

If set $A = \{3, 5, 7, 9, 11\}$, $B = \{7, 9, 11, 13\}$, $C = \{11, 13, 15\}$ and $D = \{15, 17\}$, then find the following: (i) $A \cap (B \cup D)$ (ii) $(A \cap B) \cap (B \cup C)$ (iii) $(A \cup D) \cap (B \cup C)$

Accepted Answer

>

> [!TLDR]
> Direct computation from $U=\{1,\ldots,6\}$, $A=\{2,3\}$, $B=\{3,4,5\}$ shows both $(A\cup B)'$ and $A'\cap B'$ equal $\{1,6\}$, proving De Morgan's Law.
>
> Given $U = \{1,2,3,4,5,6\}$, $A = \{2,3\}$, $B = \{3,4,5\}$.
>
> **Left side:** $A \cup B = \{2,3,4,5\}$.
>
> $(A \cup B)' = U - (A \cup B) = \{1, 6\}$.
>
> **Right side:** $A' = U - A = \{1,4,5,6\}$.
>
> $B' = U - B = \{1,2,6\}$.
> …

Operation	What it asks	Bucket contains
Union ( $A \cup B$ )	In either?	Everything from $A$ or $B$ (or both)
Intersection ( $A \cap B$ )	In both?	Only the overlap
Difference ( $A \setminus B$ )	In $A$ but not $B$ ?	Just the part of $A$ that doesn't touch $B$
Complement ( $A^c$ )	Not in $A$ ?	Everything outside $A$ (inside $U$ )
Symmetric Difference ( $A \triangle B$ )	In exactly one?	The two crescent-shaped parts, excluding the overlap

Q.If $U = \{1, 2, 3, 4, 5, 6\}$ , $A = \{2, 3\}$ and $B = \{3, 4, 5\}$ , then prove that $(A \cup B)' = A' \cap B'$ . OR If set $A = \{3, 5, 7, 9, 11\}$ , $B = \{7, 9, 11, 13\}$ , $C = \{11, 13, 15\}$ and $D = \{15, 17\}$ , then find the following:

Concept understanding — Set Operations

The idea in plain words

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