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For each Multiple Choice Question (MCQ), four options are given. One of them is the correct answer. Make your choice (1,2,3 or 4). Write your answers in the brackets provided..

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1)  

 Given that U = {g,t,x,o,l,u,o,g}, A = {g,t,u,o}, and B = {g,t,x}, verify De Morgans laws of complementation. (A∩B)‘ = A‘∪B‘     (must use , )

A‘∪B‘ = 


Answer:_______________




2)  

 Given that U = {p,z,r,u,i,v,n,y}, A = {p,z,v,n}, and B = {p,z,r}, verify De Morgans laws of complementation. (A∪B)‘ = A‘∩B‘     (must use , )

A‘∩B‘ =   



Answer:_______________




3)  

 Given that U = {g,x,w,r,f,c,g,x}, A = {g,x,c,g}, and B = {g,x,w}, verify De Morgans laws of complementation. (A∩B)‘ = A‘∪B‘     (must use , )

A‘∪B‘ = 


Answer:_______________




4)  

 Given that U = {o,d,h,u,m,z,h,f}, A = {o,d,z,h}, and B = {o,d,h}, verify De Morgans laws of complementation. (A∪B)‘ = A‘∩B‘     (must use , )

A‘∩B‘ =   



Answer:_______________




5)  

 Given that U = {e,o,j,s,d,k,w,x}, A = {e,o,k,w}, and B = {e,o,j}, verify De Morgans laws of complementation. (A∩B)‘ = A‘∪B‘     (must use , )

A‘∪B‘ = 


Answer:_______________




6)  

 Given that U = {h,w,p,a,d,z,a,t}, A = {h,w,z,a}, and B = {h,w,p}, verify De Morgans laws of complementation. (A∪B)‘ = A‘∩B‘     (must use , )

A‘∩B‘ =   



Answer:_______________




7)  

 Given that U = {s,y,h,g,v,d,n,i}, A = {s,y,d,n}, and B = {s,y,h}, verify De Morgans laws of complementation. (A∩B)‘ = A‘∪B‘     (must use , )

A‘∪B‘ = 


Answer:_______________




8)  

 Given that U = {p,s,q,l,e,e,a,o}, A = {p,s,e,a}, and B = {p,s,q}, verify De Morgans laws of complementation. (A∪B)‘ = A‘∩B‘     (must use , )

A‘∩B‘ =   



Answer:_______________




9)  

 Given that U = {t,r,i,i,z,y,w,d}, A = {t,r,y,w}, and B = {t,r,i}, verify De Morgans laws of complementation. (A∩B)‘ = A‘∪B‘     (must use , )

A‘∪B‘ = 


Answer:_______________




10)  

 Given that U = {y,l,h,u,g,p,x,a}, A = {y,l,p,x}, and B = {y,l,h}, verify De Morgans laws of complementation. (A∪B)‘ = A‘∩B‘     (must use , )

A‘∩B‘ =   



Answer:_______________




 

1)  

 Given that U = {t,c,e,z,k,x,s,f}, A = {t,c,x,s}, and B = {t,c,e}, verify De Morgans laws of complementation. (A∩B)‘ = A‘∪B‘     (must use , )

A‘∪B‘ =  Answer: e,z,k,x,s,f


SOLUTION 1 :

verify De Morgan‘s laws of complementation.

(i)  (A∩B)‘ = A‘∪B‘

(1)  to verify : (A∩B)‘ = A‘∪B‘

A∩B = {t,c,x,s}∪{t,c,e}  

        =  {t,c,e,x,s}

       (A∩B)‘  = U∖(A∩B)

        =   {t,c,e,z,k,x,s,f}∖{t,c}

        (A∩B)‘ =  {e,z,k,x,s,f}   ......      (i)

      A‘  = U∖A

            =  {t,c,e,z,k,x,s,f}∖{t,c,x,s}

         A‘ = {e,z,k,f}

   B‘  = U∖B

       =  {t,c,e,z,k,x,s,f}∖{t,c,e}

              B‘ =  {z,k,x,s,f}

A‘∪B‘ = {e,z,k,f}∖ {z,k,x,s,f}

A‘∪B‘ = {e,z,k,x,s,f}   .......      (2)

From (i) and (2), we have (A∩B)‘ = A‘∪B‘




2)  

 Given that U = {y,q,d,n,l,y,h,p}, A = {y,q,y,h}, and B = {y,q,d}, verify De Morgans laws of complementation. (A∪B)‘ = A‘∩B‘     (must use , )

A‘∩B‘ =  Answer: n,l,p 



SOLUTION 1 :

verify De Morgan‘s laws of complementation.

(i)  (A∪B)‘ = A‘∩B‘

(ii)  (A∩B)‘ = A‘∪B‘

(1)  to verify : (A∪B)‘ = A‘∩B‘

A∪B = {y,q,y,h}∪{y,q,d}  

        =  {y,q,d,y,h}

       (A∪B)‘  = U∖(A∪B)

        =   {y,q,d,n,l,y,h,p}∖{y,q,d,y,h}

        (A∪B)‘ =  {n,l,p}   ......      (i)

      A‘  = U∖A

            =  {y,q,d,n,l,y,h,p}∖{y,q,y,h}

         A‘ = {d,n,l,p}

   B‘  = U∖B

       =  {y,q,d,n,l,y,h,p}∖{y,q,d}

              B‘ =  {n,l,y,h,p}

A‘∩B‘ = {d,n,l,p}∖ {n,l,y,h,p}

A‘∩B‘ = {n,l,p}   .......      (2)

From (i) and (2), we have (A∪B)‘ = A‘∩B‘




3)  

 Given that U = {r,x,n,o,b,s,i,g}, A = {r,x,s,i}, and B = {r,x,n}, verify De Morgans laws of complementation. (A∩B)‘ = A‘∪B‘     (must use , )

A‘∪B‘ =  Answer: n,o,b,s,i,g


SOLUTION 1 :

verify De Morgan‘s laws of complementation.

(i)  (A∩B)‘ = A‘∪B‘

(1)  to verify : (A∩B)‘ = A‘∪B‘

A∩B = {r,x,s,i}∪{r,x,n}  

        =  {r,x,n,s,i}

       (A∩B)‘  = U∖(A∩B)

        =   {r,x,n,o,b,s,i,g}∖{r,x}

        (A∩B)‘ =  {n,o,b,s,i,g}   ......      (i)

      A‘  = U∖A

            =  {r,x,n,o,b,s,i,g}∖{r,x,s,i}

         A‘ = {n,o,b,g}

   B‘  = U∖B

       =  {r,x,n,o,b,s,i,g}∖{r,x,n}

              B‘ =  {o,b,s,i,g}

A‘∪B‘ = {n,o,b,g}∖ {o,b,s,i,g}

A‘∪B‘ = {n,o,b,s,i,g}   .......      (2)

From (i) and (2), we have (A∩B)‘ = A‘∪B‘




4)  

 Given that U = {x,c,l,q,f,r,i,k}, A = {x,c,r,i}, and B = {x,c,l}, verify De Morgans laws of complementation. (A∪B)‘ = A‘∩B‘     (must use , )

A‘∩B‘ =  Answer: q,f,k 



SOLUTION 1 :

verify De Morgan‘s laws of complementation.

(i)  (A∪B)‘ = A‘∩B‘

(ii)  (A∩B)‘ = A‘∪B‘

(1)  to verify : (A∪B)‘ = A‘∩B‘

A∪B = {x,c,r,i}∪{x,c,l}  

        =  {x,c,l,r,i}

       (A∪B)‘  = U∖(A∪B)

        =   {x,c,l,q,f,r,i,k}∖{x,c,l,r,i}

        (A∪B)‘ =  {q,f,k}   ......      (i)

      A‘  = U∖A

            =  {x,c,l,q,f,r,i,k}∖{x,c,r,i}

         A‘ = {l,q,f,k}

   B‘  = U∖B

       =  {x,c,l,q,f,r,i,k}∖{x,c,l}

              B‘ =  {q,f,r,i,k}

A‘∩B‘ = {l,q,f,k}∖ {q,f,r,i,k}

A‘∩B‘ = {q,f,k}   .......      (2)

From (i) and (2), we have (A∪B)‘ = A‘∩B‘




5)  

 Given that U = {h,k,v,l,y,i,a,f}, A = {h,k,i,a}, and B = {h,k,v}, verify De Morgans laws of complementation. (A∩B)‘ = A‘∪B‘     (must use , )

A‘∪B‘ =  Answer: v,l,y,i,a,f


SOLUTION 1 :

verify De Morgan‘s laws of complementation.

(i)  (A∩B)‘ = A‘∪B‘

(1)  to verify : (A∩B)‘ = A‘∪B‘

A∩B = {h,k,i,a}∪{h,k,v}  

        =  {h,k,v,i,a}

       (A∩B)‘  = U∖(A∩B)

        =   {h,k,v,l,y,i,a,f}∖{h,k}

        (A∩B)‘ =  {v,l,y,i,a,f}   ......      (i)

      A‘  = U∖A

            =  {h,k,v,l,y,i,a,f}∖{h,k,i,a}

         A‘ = {v,l,y,f}

   B‘  = U∖B

       =  {h,k,v,l,y,i,a,f}∖{h,k,v}

              B‘ =  {l,y,i,a,f}

A‘∪B‘ = {v,l,y,f}∖ {l,y,i,a,f}

A‘∪B‘ = {v,l,y,i,a,f}   .......      (2)

From (i) and (2), we have (A∩B)‘ = A‘∪B‘




6)  

 Given that U = {n,b,e,b,u,v,t,i}, A = {n,b,v,t}, and B = {n,b,e}, verify De Morgans laws of complementation. (A∪B)‘ = A‘∩B‘     (must use , )

A‘∩B‘ =  Answer: b,u,i 



SOLUTION 1 :

verify De Morgan‘s laws of complementation.

(i)  (A∪B)‘ = A‘∩B‘

(ii)  (A∩B)‘ = A‘∪B‘

(1)  to verify : (A∪B)‘ = A‘∩B‘

A∪B = {n,b,v,t}∪{n,b,e}  

        =  {n,b,e,v,t}

       (A∪B)‘  = U∖(A∪B)

        =   {n,b,e,b,u,v,t,i}∖{n,b,e,v,t}

        (A∪B)‘ =  {b,u,i}   ......      (i)

      A‘  = U∖A

            =  {n,b,e,b,u,v,t,i}∖{n,b,v,t}

         A‘ = {e,b,u,i}

   B‘  = U∖B

       =  {n,b,e,b,u,v,t,i}∖{n,b,e}

              B‘ =  {b,u,v,t,i}

A‘∩B‘ = {e,b,u,i}∖ {b,u,v,t,i}

A‘∩B‘ = {b,u,i}   .......      (2)

From (i) and (2), we have (A∪B)‘ = A‘∩B‘




7)  

 Given that U = {l,k,b,t,v,i,u,j}, A = {l,k,i,u}, and B = {l,k,b}, verify De Morgans laws of complementation. (A∩B)‘ = A‘∪B‘     (must use , )

A‘∪B‘ =  Answer: b,t,v,i,u,j


SOLUTION 1 :

verify De Morgan‘s laws of complementation.

(i)  (A∩B)‘ = A‘∪B‘

(1)  to verify : (A∩B)‘ = A‘∪B‘

A∩B = {l,k,i,u}∪{l,k,b}  

        =  {l,k,b,i,u}

       (A∩B)‘  = U∖(A∩B)

        =   {l,k,b,t,v,i,u,j}∖{l,k}

        (A∩B)‘ =  {b,t,v,i,u,j}   ......      (i)

      A‘  = U∖A

            =  {l,k,b,t,v,i,u,j}∖{l,k,i,u}

         A‘ = {b,t,v,j}

   B‘  = U∖B

       =  {l,k,b,t,v,i,u,j}∖{l,k,b}

              B‘ =  {t,v,i,u,j}

A‘∪B‘ = {b,t,v,j}∖ {t,v,i,u,j}

A‘∪B‘ = {b,t,v,i,u,j}   .......      (2)

From (i) and (2), we have (A∩B)‘ = A‘∪B‘




8)  

 Given that U = {t,z,p,n,b,a,w,k}, A = {t,z,a,w}, and B = {t,z,p}, verify De Morgans laws of complementation. (A∪B)‘ = A‘∩B‘     (must use , )

A‘∩B‘ =  Answer: n,b,k 



SOLUTION 1 :

verify De Morgan‘s laws of complementation.

(i)  (A∪B)‘ = A‘∩B‘

(ii)  (A∩B)‘ = A‘∪B‘

(1)  to verify : (A∪B)‘ = A‘∩B‘

A∪B = {t,z,a,w}∪{t,z,p}  

        =  {t,z,p,a,w}

       (A∪B)‘  = U∖(A∪B)

        =   {t,z,p,n,b,a,w,k}∖{t,z,p,a,w}

        (A∪B)‘ =  {n,b,k}   ......      (i)

      A‘  = U∖A

            =  {t,z,p,n,b,a,w,k}∖{t,z,a,w}

         A‘ = {p,n,b,k}

   B‘  = U∖B

       =  {t,z,p,n,b,a,w,k}∖{t,z,p}

              B‘ =  {n,b,a,w,k}

A‘∩B‘ = {p,n,b,k}∖ {n,b,a,w,k}

A‘∩B‘ = {n,b,k}   .......      (2)

From (i) and (2), we have (A∪B)‘ = A‘∩B‘




9)  

 Given that U = {i,s,x,d,t,m,w,z}, A = {i,s,m,w}, and B = {i,s,x}, verify De Morgans laws of complementation. (A∩B)‘ = A‘∪B‘     (must use , )

A‘∪B‘ =  Answer: x,d,t,m,w,z


SOLUTION 1 :

verify De Morgan‘s laws of complementation.

(i)  (A∩B)‘ = A‘∪B‘

(1)  to verify : (A∩B)‘ = A‘∪B‘

A∩B = {i,s,m,w}∪{i,s,x}  

        =  {i,s,x,m,w}

       (A∩B)‘  = U∖(A∩B)

        =   {i,s,x,d,t,m,w,z}∖{i,s}

        (A∩B)‘ =  {x,d,t,m,w,z}   ......      (i)

      A‘  = U∖A

            =  {i,s,x,d,t,m,w,z}∖{i,s,m,w}

         A‘ = {x,d,t,z}

   B‘  = U∖B

       =  {i,s,x,d,t,m,w,z}∖{i,s,x}

              B‘ =  {d,t,m,w,z}

A‘∪B‘ = {x,d,t,z}∖ {d,t,m,w,z}

A‘∪B‘ = {x,d,t,m,w,z}   .......      (2)

From (i) and (2), we have (A∩B)‘ = A‘∪B‘




10)  

 Given that U = {x,a,s,n,v,y,j,b}, A = {x,a,y,j}, and B = {x,a,s}, verify De Morgans laws of complementation. (A∪B)‘ = A‘∩B‘     (must use , )

A‘∩B‘ =  Answer: n,v,b 



SOLUTION 1 :

verify De Morgan‘s laws of complementation.

(i)  (A∪B)‘ = A‘∩B‘

(ii)  (A∩B)‘ = A‘∪B‘

(1)  to verify : (A∪B)‘ = A‘∩B‘

A∪B = {x,a,y,j}∪{x,a,s}  

        =  {x,a,s,y,j}

       (A∪B)‘  = U∖(A∪B)

        =   {x,a,s,n,v,y,j,b}∖{x,a,s,y,j}

        (A∪B)‘ =  {n,v,b}   ......      (i)

      A‘  = U∖A

            =  {x,a,s,n,v,y,j,b}∖{x,a,y,j}

         A‘ = {s,n,v,b}

   B‘  = U∖B

       =  {x,a,s,n,v,y,j,b}∖{x,a,s}

              B‘ =  {n,v,y,j,b}

A‘∩B‘ = {s,n,v,b}∖ {n,v,y,j,b}

A‘∩B‘ = {n,v,b}   .......      (2)

From (i) and (2), we have (A∪B)‘ = A‘∩B‘