Question

Let U = {1,2,3,4,5,6,7,8,9}, A = {1,2,3,4}, B = {2,4,6,8} and C = {3,4,5,6}. Find :

(i) $A^{\prime }$ (ii) $B^{\prime }$ (iii) $(A\cap C{)}^{\prime }$

(iv) $(A\cup B{)}^{\prime }$ (v) ${\left(A^{\prime }\right)}^{\prime }$ (vi) $(B-C{)}^{\prime }$

Answer 1

(i) U= {1,2,3,4,5,6,7,8,9}, A = {1,2,3,4}, B = {2,4,6,8}, C = {3,4,5,6}

By the complement of a set A, which respect to the universal set U, denoted by $A^{\prime }$ or $A^C$ or U- A, we mean {$xϵU:x\text{/}ϵA$ }.

Hence, $A^{\prime }$ = { $xϵU:x\text{/}ϵA$ } = {5,6,7,8,9}

(ii) $B^{\prime }$ = { $xϵU:x\text{/}ϵB$ } = {1,3,5,7,9}

(iii) $(A\cap C{)}^{\prime }$ = { $xϵU:x\text{/}ϵA\cap C$}

Now,

$A\cap C$ = {$x:xϵAandxϵC$} = {3,4}

$\therefore$ $(A\cap C{)}^{\prime }$ = {1,2,5,6,7,8,9}

(iv) $(A\cup B{)}^{\prime }$

$(A\cup B{)}^{\prime }$ = {$x:UϵA\cap B$ }

Now, $A\cap B$ = {$x:xϵAandxϵB$}

= {1,2,3,4,5,6,8}

$(A\cup B{)}^{\prime }$ = {5,7,9}

(v) ${\left(A^{\prime }\right)}^{\prime }$

${\left(A^{\prime }\right)}^{\prime }$ = {$xϵU:xϵA^{\prime }$ }

= {$xϵU:xϵA$} [$\because x\text{/}ϵA^{\prime }meansxepsilonA$ ]

= A [$\therefore A\subset U$]

(vi) $(B-C{)}^{\prime }$

$B-C^{\prime }$ = [ $xϵB:x\text{/}ϵC$ ] = {2,8}

Hence, $B-C^{\prime }$ = {$xϵU:x\text{/}ϵB-C$ }

= {$xϵU:x\text{/}ϵ2,8$}

= {1,3,4,5,6,7,9}