Question

If U= {2,3,5,7,9} is the universal set and A = {3,7} , B = {2,5,7,9}, then prove that :

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

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

Answer 1

(i) U = {2,3,5,7,9} is the universal set

A = {3,7} , B = {2,5,7,9}

$A\cup B$ = { $x:xϵAorxϵB$}

= {2,3,5,7,9}

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

= {2,3,5,7,9}

= U - $A\cup B$

= $\varphi$

RHS = $A^{\prime }\cap B^{\prime }$

A' = {$xϵU:x\text{/}ϵA$}

= {2,5, 9}

B' = {$xϵU:x\text{/}ϵB$ }

={3}

$\therefore A^{\prime }\cap B^{\prime }$ = {2, 5, 9} $\cap$ {3}

= $\varphi$ [$\therefore$ the two sets are disjoint]

LHS = RHS Proved.

(ii) LHS = $(A\cap B{)}^{\prime }$

Now,

$A\cap B$ = {$x|xϵaandxϵB$}

= {7}

$\therefore (A\cap B{)}^{\prime }$ = {7}

= {x~ \epsilon~ U : x~ \not {\epsilon } ~7\)}

= {2,3,5,9}

RHS = A${}^{\prime }\cup$ B'

Now, A' = {2,5,9} [from (i)]

and B' = {3} [from (i)]

$\therefore A^{\prime }\cup B^{\prime }$= {2,3,5,9}

Hence, LHS = RHS Proved.