Question

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

Answer 1

Step 1: Finding $A^{\prime }$
$U=\{1,2,3,4,5,6\},A=\{2,3\}$
$B=\{3,4,5\}$
So, $A^{\prime }=U-A$
$=\{1,2,3,4,5,6\}-\{2,3\}$
$=\{1,4,5,6\}$

Step 2: Finding $B^{\prime }$.
$B^{\prime }=U-B$
$B^{\prime }=\{1,2,3,4,5,6\}-\{3,4,5\}$
$=\{1,2,6\}$

Step 3: Finding $A^{\prime }\cap B^{\prime }$
$\because A^{\prime }=\{1,4,5,6\}$ and $B^{\prime }=\{1,2,6\}$
$\therefore A^{\prime }\cap B^{\prime }=\{1,6\}$

Step 4: Finding $A\cup B$
$A=\{2,3\},B=\{3,4,5\}$
$\therefore A\cap B=\{2,3,4,5\}$

Step 5: Finding $(A\cap B^{\prime })$
$\because A\cup B=\{2,3,4,5\}$
$\therefore (A\cup B{)}^{\prime }=U-(A\cup B)$
$=\{1,2,3,4,5,6\}-\{2,3,4,5\}$
$=\{1,6\}$

Step 6: Solve for $(A\cup B{)}^{\prime }=A^{\prime }\cap B^{\prime }$
$\because A^{\prime }\cap B^{\prime }=\{1,6\}$
And $(A\cup B{)}^{\prime }=\{1,6\}$
$\therefore (A\cup B{)}^{\prime }=A^{\prime }\cap B^{\prime }$
Hence proved.