$U = \{1, 2, 3, 4, 5, 6\}$, $A = \{2, 3\}$ and $B = \{3, 4, 5\}$. Find $A'$, $B'$, $A' \cap B'$, $A \cup B$ and hence show that $(A \cup B)' = A' \cap B'$.

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Accepted Answer

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> [!TLDR]
> Both $(A\cup B)'$ and $A'\cap B'$ equal $\{1,6\}$, verifying De Morgan's law.
>
> With $U=\{1,2,3,4,5,6\}$, $A=\{2,3\}$, $B=\{3,4,5\}$:
> $$A'=\{1,4,5,6\},\qquad B'=\{1,2,6\}.$$
> $$A'\cap B'=\{1,6\}.$$
> $$A\cup B=\{2,3,4,5\}\Rightarrow (A\cup B)'=\{1,6\}.$$ …

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. $U = \{1, 2, 3, 4, 5, 6\}$ , $A = \{2, 3\}$ and $B = \{3, 4, 5\}$ . Find $A'$ , $B'$ , $A' \cap B'$ , $A \cup B$ and hence show that $(A \cup B)' = A' \cap B'$ .

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Q.U={1,2,3,4,5,6}U = \{1, 2, 3, 4, 5, 6\}U={1,2,3,4,5,6}, A={2,3}A = \{2, 3\}A={2,3} and B={3,4,5}B = \{3, 4, 5\}B={3,4,5}. Find A′A'A′, B′B'B′, A′∩B′A' \cap B'A′∩B′, A∪BA \cup BA∪B and hence show that (A∪B)′=A′∩B′(A \cup B)' = A' \cap B'(A∪B)′=A′∩B′.

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Q. $U = \{1, 2, 3, 4, 5, 6\}$ , $A = \{2, 3\}$ and $B = \{3, 4, 5\}$ . Find $A'$ , $B'$ , $A' \cap B'$ , $A \cup B$ and hence show that $(A \cup B)' = A' \cap B'$ .

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