Question

If $U=\{4,8,12,16,20,24,28\},A=\{8,16,24\},B=\{4,16,20,28\}$.

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

Answer 1

The given sets are $U=\{4,8,12,16,20,24,28\}$, $A=\{8,16,24\}$ and $B=\{4,16,20,28\}$

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

Consider the following:

${(A\cup B)}^{\prime }=U-(A\cup B)=\{4,8,12,16,20,24,28\}-\left[\right\{8,16,24\}\cup \{4,16,20,28\left\}\right]$

$=\{4,8,12,16,20,24,28\}-\{4,8,16,20,24,28\}=\left\{12\right\}......\left(1\right)$

$A^{\prime }\cap B^{\prime }=[U-A]\cap [U-B]=\left[\right\{4,8,12,16,20,24,28\}-\{8,16,24\left\}\right]\cap \left[\right\{4,8,12,16,20,24,28\}-\{4,16,20,28\left\}\right]$

$=\{4,12,20,28\}\cap \{8,12,24\}=\left\{12\right\}......\left(2\right)$

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

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