Question

Set $U=\{1,2,3,4,5,6,7,8,9\}$
$A=\{x:x\in N,30\le x^2\le 70\}$
$B=\{x:x\text{is a prime number}<10\}$
Find $A^{\prime },B^{\prime },(A\cup B),A^{\prime }\cap B^{\prime },{(A-B)}^{\prime }$

Answer 1

Given sets $U=\{1,2,3,4,5,6,7,8,9\}$,
$A=\{x:x\in N,30\le x^2\le 70\}=\{30\le 6^2\le 7^2\le 8^2\le 70\}$
$\therefore A=\{6,7,8\}$
Similar way, $B=\{x:x\text{is a prime number}<10\}=\{2,3,5,7\}$
Since, $A^{\prime }=U-A$
$=\{1,2,3,4,5,6,7,8,9\}-\{6,7,8\}$
$\therefore A^{\prime }=\{1,2,3,4,5,9\}$
$B^{\prime }=U-B$
$=\{1,2,3,4,5,6,7,8,9\}-\{2,3,5,7\}$
$\therefore B^{\prime }=\{1,4,6,8,9\}$

Consider $(A\cup B)=\{6,7,8\}\cup \{2,3,5,7\}$
$\therefore (A\cup B)=\{2,3,5,6,7,8\}$
Now to find, ${(A\cup B)}^{\prime }=U-(A\cup B)$
$=\{1,2,3,4,5,6,7,8,9\}-\{2,3,5,6,7,8\}$
$\therefore {(A\cup B)}^{\prime }=\{1,4,9\}$
$A^{\prime }\cap B^{\prime }=\{1,2,3,4,5,9\}\cap \{1,4,6,8,9\}$
$\therefore A^{\prime }\cap B^{\prime }=\{1,4,9\}$

$(A-B)=\{6,7,8\}-\{2,3,5,7\}$
$(A-B)=\{6,8\}$
Thus,$(A-B{)}^{\prime }=U-(A-B)$
$=(\{1,2,3,4,5,6,7,8,9\}-\{6,8\}$
$\therefore (A-B{)}^{\prime }=\{1,2,3,4,5,7,9\}$