Given the following sets, for each set operation provide the elements of the resulting set in set notation or using a well-known set, i. e. N, the set of natural numbers

Z, the set of all integers
Z+, the set of all positive integers
Z-, the set of all negative integers
A, the set of all integers evenly divisible by 3
B={4, 5, 9, 10 }
C={12,4,11,14)
D={3,6,9}
E= {4,6, 16}

a. B U C
b. A ⋂ D
c. A⋂C
d. (B+D)+E
e. AUE
f. E-D
g. D-E
h. (Z- Z+)- Z

Check the explanation

Explanation:

Z -> the set of all integers

Z+ -> the set of all positive integers

Z- -> the set of all negative integers

A -> the set of all integers evenly divisible by 3

B -> {4,5,9,10}

C -> {2,4,11,14}

D -> {3,6,9}

E -> {4,6,16}

Questions:

a. B U C

    {2,4,5,9,10,11,14}

b. A ∩ D

    {3,6,9}

c. A ∩ C

    {} or null set or ∅

d. (B⊕D) ⊕ E

   A ⊕ B denotes the Symmetric difference between A and B.

   The result of the operation is the elements that in either of the sets but not in their intersection.

   For the question

   B -> {4,5,9,10}

   D -> {3,6,9}

   Elements that are not in the intersection of the two sets that are.

   {3,4,5,6,10} here 9 is omitted because it is present in both the sets.

   {3,4,5,6,10} ⊕ E

   E -> {4,6,16}

   The result will be {3,5,19,16} here 4,6 are the intersection of both sets.

   So the answer is {3,5,19,16}

e. A U D

    the set of all integers divisible evenly by 3.

f. E – D

    {4,16}

g. D – E

    {3,9}

h. (Z – Z+) – Z-

    (Z – Z+) means remove all the integers from Z that are present in     Z+ which results all the negative integers and 0.

   ({0,Z-} – Z- ) results in {0}.

   So the final answer is {0}

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