Question

if U={4,7,9,10,11,12,15,16} ,A={7,8,11,12} and B={4,8,12,15} then verify De Morgan's laws for complementation​

Answer 1

according to De Morgan's law for complementation.

if U is universal set of set A and set B, then,

• (i) $(A\cap B)'=A'\cup B'$
• (ii) $(A\cup B)'=A'\cap B'$

given, U = {4, 7, 9, 10, 11, 12, 15, 16}

A = {7, 8, 11, 12} and B = {4, 8, 12, 15}

so, $(A\cap B)$ = {7, 8, 11, 12}$\cap$ { 4, 8, 12, 15}

= {8, 12}

so, $(A\cap B)'=U-(A\cap B)$

= {4, 7, 9, 10, 11, 12, 15, 16} - {8, 12}

= {4, 7, 9, 10, 11, 15, 16}.....(1)

now, A' = U - A

= {4, 7, 9, 10, 11, 12, 15, 16} - {7, 8, 11, 12}

= {4, 9, 10, 15, 16}

B' = U - B

= {4, 7, 9, 10, 11, 12, 15, 16} - {4, 8, 12, 15}

= {7, 9 10, 11, 16}

so, A' U B' = {4, 9, 10, 15, 16} U {7, 9 10, 11, 16}

= {4, 7, 9, 10, 11, 15, 16} ....(2)

from equations (1) and (2),

we get, $(A\cap B)'=A'\cup B'$

[ note : similarly you can prove second one too. ]

Answer 2

Answer:

De Morgan's law for complementation.  

(i) $(A\cap B)'=A'\cup B'$

(ii) $(A\cup B)'=A'\cap B'$

Given:- U = {4, 7, 9, 10, 11, 12, 15, 16} , A = {7, 8, 11, 12} and B = {4, 8, 12, 15}  

first, calculate  $A\cap B$

In this we need to calculate common numbers in set A and B

$(A\cap B)=\{7,8,11,12\} \cap \{4,8,12,15\}$

$(A\cap B)=\{8,12\}$

So,  $(A\cap B)'=U-(A\cap B)$  

$\{4, 7, 9, 10, 11, 12, 15, 16\}-\{8,12\}$  

= {4, 7, 9, 10, 11, 15, 16}                           .....(1)

now, $A' = U - A$

$= \{4, 7, 9, 10, 11, 12, 15, 16\} - \{7, 8, 11, 12\}$

$= \{4, 9, 10, 15, 16\}$

$B' = U - B$

$= \{4, 7, 9, 10, 11, 12, 15, 16\} - \{4, 8, 12, 15\}$

$= \{7, 9, 10, 11, 16\}$  

then, $A' \cup B' =\{4, 9, 10, 15, 16\}\cup\{7, 9 10, 11, 16\}$

$A' \cup B' =\{4, 7, 9, 10, 11, 15, 16\}$

= {4, 7, 9, 10, 11, 15, 16}                          ......(2)

from equations (1) and (2),  

we get,  $(A\cap B)'=A' \cup B'$

Hence proved (i) $(A \cap B)'=A' \cup B'$

Now, we will prove (ii) $(A\cup B)'=A'\cap B'$

first, calculate $(A\cup B)=A\cup B$

$(A\cup B)=\{7,8,11,12\} \cup \{4,8,12,15\}$

$(A\cup B)=\{4, 7,8,11,12, 15 \}$

Now,  $(A\cup B)'=U-(A\cup B)$  

$=\{4,7,9,10,11,12,15,16 \}-\{4, 7,8,11,12, 15 \}$  

={9,10,16}                                       .......(3)

and $A' \cap B'=\{4, 9, 10, 15, 16\} \cap \{7, 9, 10, 11, 16\}$

={9,10,16}                                      .......(4)

from equations (3) and (4),  

we get,   $(A\cup B)'=A' \cap B'$

Hence proved (ii)  $(A\cup B)'=A' \cap B'$