Find the following

Question

Let U be the set of counting numbers between 11 and 20, inclusive. 

A = {x element U : x is divisible by 2}

B = {11,12,13,14,15}

C = {x element U : x is odd}

 

Find (i) [(B intersect A) \ C] x (A union B)'

 

(ii) P((B intersect C) \ A)

 

 

You don't know how happy am I when someone answers this. :)

Answer 1

First figure out the elements in U, A, B, and C.  Everything will follow once you have these lists.

 

U = 11, 12, 13, 14, 15, 16, 17, 18, 19, 20

 

A = the set of all the elements of U that are divisible by 2 = {12, 14, 16, 18, 20}

 

B = {11, 12, 13, 14, 15}

 

C = the set of all the elements of U that are odd = {11, 13, 15, 17, 19}

 

So first let's do i:

 

[(B intersect A) \ C] x (A union B)'

 

(B intersect A) is the set of all numbers that are in both B and A = {12, 14}

(B intersect A) \ C is the set of numbers in {12, 14} that are NOT in C.  Neither 12 nor 14 is in C, so this again gives us {12, 14}

(A union B) is the set of all numbers in either A or B = {11, 12, 13, 14, 15, 16, 18, 20}

 

QUESTION:  Is the ' symbol after "(A union B)" intentional or is it a typo?  It it intended to mean the complement of (A union B)?  If so, then (A union B)' is the set of all numbers in U that are NOT in (A union B), which is {17, 19}.

 

Finally (B intersect A) \ C x (A union B) is the set of ordered pairs where the first number in the pair is in {12, 14} and the second number is in {11, 12, 13, 14, 15, 16, 18, 20} -- or {17, 19} depending on the answer to my question about the ' .  Whichever the case, just form the list of ordered pairs as I described and put them together as a set.

 

ii.  (B intersect C) is the set of all elements in both B and C.  That's {11, 13, 15}

 

(B intersect C) \ A) is the set of all elements in both B and C that are NOT in A.  Since none of the elements in {11, 13, 15} are even, we still have {11, 13, 15}.

 

Finally P((B intersect C) \ A) means the power set of {11, 13, 15}.  That's the set of all subsets of {11, 13, 15}.  The subsets of {11, 13, 15} are {}, {11}, {13}, {15}, {11, 13}, {11, 15}, {13, 15}, {11, 13, 15}.  So the final answer should be written as:

 

{ {}, {11}, {13}, {15}, {11, 13}, {11, 15}, {13, 15}, {11, 13, 15} }