If $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, $A = \{2, 4, 6, 8\}$ and $B = \{2, 3, 5, 7\}$ then prove that $(A \cup B)' = A' \cap B'$.

Question

Accepted Answer

>

> [!TLDR]
> Direct computation shows $(A\cup B)' = \{1,9\} = A'\cap B'$, verifying De Morgan's Law for this example.
>
> Given: $U=\{1,2,3,4,5,6,7,8,9\}$, $A=\{2,4,6,8\}$, $B=\{2,3,5,7\}$.
>
> **Step 1 — find $A\cup B$:**
> $$A\cup B = \{2,3,4,5,6,7,8\}$$
>
> **Step 2 — find $(A\cup B)'$:** (elements of $U$ not in $A\cup B$)
> $$(A\cup B)' = \{1,9\}$$
>
> **Step 3 — find $A'$ and $B'$:**
> $$A' = U-A = \{1,3,5,7,9\}$$
> $$B' = U-B = \{1,4,6,8,9\}$$
> …

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 $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ , $A = \{2, 4, 6, 8\}$ and $B = \{2, 3, 5, 7\}$ then prove that $(A \cup B)' = A' \cap B'$ .

Concept understanding — Set Operations

The idea in plain words

Why this works

Step by step

A common slip …

Q.If U={1,2,3,4,5,6,7,8,9}U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}U={1,2,3,4,5,6,7,8,9}, A={2,4,6,8}A = \{2, 4, 6, 8\}A={2,4,6,8} and B={2,3,5,7}B = \{2, 3, 5, 7\}B={2,3,5,7} then prove that (A∪B)′=A′∩B′(A \cup B)' = A' \cap B'(A∪B)′=A′∩B′.

🎓Unlock everything free for 14 days

Q.If $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ , $A = \{2, 4, 6, 8\}$ and $B = \{2, 3, 5, 7\}$ then prove that $(A \cup B)' = A' \cap B'$ .

Unlock everything free for 14 days