Question

If $A=\{1,2,3\}$ and $B=\{2,3,4,5\}$ are the subsets of $U=\{1,2,3,4,5,6,7,8\}$

Verify $(A\cup B{)}^{\prime }=A^{\prime }\cap B^{\prime }$

Answer 1

We know that the compliment of set $A$ is $A^{\prime }=U-A$.

The given sets are $U=\{1,2,3,4,5,6,7,8\}$, $A=\{1,2,3\}$ and $B=\{2,3,4,5\}$

Consider the following:

${(A\cup B)}^{\prime }=U-(A\cup B)=\{1,2,3,4,5,6,7,8\}-\left[\right\{1,2,3\}\cup \{2,3,4,5\left\}\right]=\{1,2,3,4,5,6,7,8\}-\{1,2,3,4,5\}$

$=\{6,7,8\}......\left(1\right)$

$A^{\prime }\cap B^{\prime }=[U-A]\cap [U-B]=\left[\right\{1,2,3,4,5,6,7,8\}-\{1,2,3\left\}\right]\cap \left[\right\{1,2,3,4,5,6,7,8\}-\{2,3,4,5\left\}\right]$

$=\{4,5,6,7,8\}\cap \{1,6,7,8\}=\{6,7,8\}......\left(2\right)$

From equations 1 and 2, we get that ${(A\cup B)}^{\prime }=A^{\prime }\cap B^{\prime }$.

Hence, ${(A\cup B)}^{\prime }=A^{\prime }\cap B^{\prime }$.