De morgan's law for set theory - class-XI

We have the De Morgan's theorem,

$A\cup B=B-A+A-B+A\cap B\\implies B-A=A-B=\phi \implies A=B$

We know that $\displaystyle \left ( A\cup B \right )'=$ $\displaystyle A'\cap B'$ so we need to find $\displaystyle \left ( A\cup B \right ) $  first.
$n\displaystyle \left ( A\cup B \right ) =$ $n(A) + n(B) - n$$\displaystyle (A\cap B)$ $= 200 + 300 - 100 = 400$
$n\displaystyle \left ( A\cup B \right )' =$ $n(U) - n$$\displaystyle \left ( A\cup B \right )'=$$  700 - 400 = 300$
$\displaystyle \Rightarrow $ $n\displaystyle (A'\cap B')$$= 300$.

Since $A\subset B$, $A\cap B$ will be all the elements of A. So, the number of elements will be 5.
Since $A\subset B$, $A\cup B$ will be all the elements of A & B. So, the number of elements will be 8.

Students passed in Hindi $ = 70\% =n(H)$

Students passed in English $ = 80\% =n(E)$

Students passed in Both $ = 65\% =n(H\cap E)$

No. of students passed $=n(H \cup E)= n(H)+n(E)-n(H \cap E) = 70+80-65 =85\%$

No. of Students failed $= 100\% - n(H \cup E) = 100\% - 85\% = 15\%$

Fail: $n(M) =50\%$

$n(P) =50\%$

$n(M \cap P) =20\%$

$n(M\cup P) =n(M)+n(P)-n(M\cap P)$

$n(M\cup P) =50+50-20 = 80\%$

Pass:

n(Pass in both subjects) $=100\%$ -n(Fail in both the subjects)

$=100\% -80\% = 20\%$

$(A\cup B)' = A'\cap B'$ is called De Morgan's law.

n(Watch new on TV) $=65\%=n(T)$

n(Reads newspaper) $=40\%=n(N)$

n(Watch news on TV & read newspapers ) $=25\%=n(T\cap N)$

$n(N \cup T)= n(T)+n(N)-n(N\cap T)$

$n(N \cup T)= 65+40-25 = 80\%$

$n(N\cap T)^{c}=100-80=20\%$

Let $x\in A\cap (B - C)$

$n(A\cup R\cup P)= n(A) +n(R) +n(P)-n(A\cap R)-n(R\cap P) -n(P\cap A) + n(A\cap R \cap P)$

Using the principle of inclusion-exclusion and Venn diagram regions: n(A only) + n(B only) + n(C only) + n(A and B only) + n(B and C only) + n(A and C only) + n(A and B and C) = 20. Given n(A and B and C) = 1, n(A and B only) = 2, n(B and C only) = 4, and n(A and C only) = 4, we solve for the sum of 'only' regions.

Ans. $(c). n(A\cap B)=n(A\cup B)$
$=n(u)-n(A\cup B)$
$=n(u)-\left{ n\left( A \right) +n\left( B \right) -n\left( A\cap B \right)  \right}$
$=700-\left{ 200+300-100 \right} = 300$

By set theory definition, A - B = A intersect B complement. Therefore, A - (A - B) = A intersect (A intersect B complement) complement = A intersect (A complement union B) = (A intersect A complement) union (A intersect B) = empty union (A intersect B) = A intersect B.

Simplifying the expression: (A union B union C) intersect (A intersect B complement intersect C complement) complement intersect C complement. The term (A intersect B complement intersect C complement) complement is (A complement union B union C). Intersecting this with C complement leaves (A complement intersect C complement) union (B intersect C complement). Further intersection with (A union B union C) simplifies to B intersect C complement.

$Q={4,8};$     $P={2,3,5,7,11}$;     

Given 

Since, no elements of A and B are the same.
Therefore, the Intersection is empty.
i.e. $A \cap B = $ null
We know, $U = $  { $ N $ }
$A = $  { $ 5,6,7 $ }
$B = $  { $ 8,9,10 $ } 
($A \cup B$) $=$ {$5,6,7$) $ \cup $ {$8,9,10$}
               $=$ {$5,6,7,8,9,10$}
Therefore,  ($A \cup B$)'$= $ All natural numbers except $ {5,6,7,8,9,10} \dots (i)$
Now, $A'= $ All Natural numbers except ${5,6,7}$
and $B'=$ All Natural numbers except ${8,9,10}$
Therefore, $A' \cap B' = $ All Natural numbers except ${5,6,7,8,9,10}\dots (ii)$
By comparing $(i) $ and $(ii)$ we get,
$(A \cup B) = A' \cap B'$

$A = { x \in N : x < 6 }$

It is De Morgan's law that $(A\cup B)' = A'\cap B'$

$ { Y ' } = U - Y = { 3, 4, 7, 8, 9} $

$ X\cap Z = {0} $ as there is no element common in sets X and Z

$\Rightarrow (X\cap Z) ' = U - (X\cap Z) =  {3, 4, 5, 6, 7, 8, 9} $

Now, $Y' \cap (X\cap Z)'  = { 3, 4, 7, 8, 9} \cap {3, 4, 5, 6, 7, 8, 9} ={ 3, 4, 7, 8, 9} $

This is equal to $ X \cup Z = {3,4,7,8,9} $

Given, $U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}$, $A = {0, 3, 4, 7}$ ,$B = {1, 2, 8, 9}$

No. of players who played at least one game is:

De Morgan's law of intersection is $(A\cap B )' = A' \cup B' $

De Morgan's law of union is 

De Morgan's law states that the complement of a union is the intersection of the complements: (A union B)' = A' intersect B'.

Universal set includes ${-5,-4,-3,-2,-1,0,1,2,3,4,5,6}$

Given, $S={1,2,3,4,5,6,7}$ and let $A={2,5,7}$.

$A' -B' = B-A$