Question

If S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} and A =

Answer 1

Answer:

a) = {0, 2, 3, 4, 5, 6, 8}, b = {0} empty set. c =  {0, 1, 6, 7, 8, 9} d =  {0, 1, 3, 5, 6, 7, 8, 9} e =  {0, 1, 5, 6, 7, 8, 9} f = {2,4} .

Step-by-step explanation:

Given:

S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

A =  {0, 2, 4, 6, 8}

B = {1, 3, 5, 7, 9}

C = {2, 3, 4, 5}

D = {1, 6, 7}

Solution:

a) A∪C = A + C - A∩C

            = {0, 2, 4, 6, 8} + {2, 3, 4, 5} - {2, 4}

            = {0, 2, 2, 3, 4, 4, 5, 6, 8} -  {2, 4}

            = {0, 2, 3, 4, 5, 6, 8}

b)  A∩B = ?

A∪B = A + B -   A∩B

A∪B +  A∩B = A + B

A∩B =  A + B - A∪B

        =   {0, 2, 4, 6, 8} +  {1, 3, 5, 7, 9} -   {0, 1, 2, 3, 4, 5, 6, 8, 9}

        = {0,1, 2, 3, 4, 5, 6,7, 8, 9} - {0, 1, 2, 3, 4, 5, 6, 8, 9}

A∩B = {0} empty set.

c) C′ = S - C

       = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} -  {2, 3, 4, 5}

       = {0, 1, 6, 7, 8, 9}

d) (C′∩D)∪B = ?

(C′∩D) = {0, 1, 6, 7, 8, 9} ∩ {1, 6, 7}

           = {1, 6, 7}

B =  {1, 3, 5, 7, 9}

(C′∩D)∪B = (C′∩D) + B - (C′∩D)∩B

                = {0, 1, 6, 7, 8, 9} +  {1, 3, 5, 7, 9} -  {1, 7, 9,}

                = {0, 1, 1, 3, 5, 6, 7, 7, 8, 9, 9} -  {1, 7, 9,}

                = {0, 1, 3, 5, 6, 7, 8, 9}

e) (S∩C)′ = S - (S∩C)′

(S∩C) =  {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} ∩ {2, 3, 4, 5}

              = {2, 3, 4, 5}

(S∩C)′ = S -  (S∩C)

           =  {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} -  {2, 3, 4, 5}

           =   {0, 1, 5, 6, 7, 8, 9}

f) A∩C∩D′ = ?

D′ = S - D

   = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 6, 7}

   = {0, 2, 3, 4, 5, 8, 9}

A∩C∩D′ = {0, 2, 4, 6, 8} ∩  {2, 3, 4, 5} ∩ {0, 2, 3, 4, 5, 8, 9}

              =  {2,4}