MATH 350 - DISCRETE
MATHEMATICS - Operations on sets
Question Bank - Set 2
Liberty University
Question 1
Question
Let A={x∈Z|x2≤9}and B={x∈Z|x > −3}. Find the set C=A∩B.
Solution
Step 1: Determine the elements of set A. Since x2≤9, the possible values for
xare −3,−2,−1,0,1,2,3. Thus, A={−3,−2,−1,0,1,2,3}.
Step 2: Determine the elements of set B. Since x > −3, the possible values
for xare −2,−1,0,1,2,3, . . .. Thus, B={−2,−1,0,1,2,3, . . .}.
Step 3: Find the intersection of sets Aand Bto get set C.C=A∩B=
{−2,−1,0,1,2,3}.
Therefore, C={−2,−1,0,1,2,3}.
Question 2
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|5≤x≤15}. Find the
intersection, union, and relative complement of sets Aand B.
Solution
We are given:
A={x∈Z|2≤x≤10}
B={x∈Z|5≤x≤15}
Step 1: Find the intersection of sets Aand BTo find the intersection,
we need to determine the elements that are common to both sets Aand B. The
intersection of sets Aand B, denoted by A∩B, is given by:
A∩B={x∈Z|5≤x≤10}
Step 2: Find the union of sets Aand BTo find the union, we need to
combine all elements in both sets Aand B, without repetitions. The union of
sets Aand B, denoted by A∪B, is given by:
A∪B={x∈Z|2≤x≤15}
Step 3: Find the relative complement of set Ain set BThe relative
complement of set Ain set B, denoted by B\A, represents the elements in B
that are not in A. The relative complement of set Ain set Bis given by:
B\A={x∈Z|11 ≤x≤15}
Therefore, the intersection of sets Aand Bis {x∈Z|5≤x≤10}, the
union of sets Aand Bis {x∈Z|2≤x≤15}, and the relative complement of
set Ain set Bis {x∈Z|11 ≤x≤15}.
Question 3
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Compute the
following sets: a) A∩Bb) A∪Bc) A\Bd) B\A
Solution
a) To find A∩B, we need to determine the elements that are common in both
Aand B.
A∩B={x∈Z|5≤x≤10}
b) To find A∪B, we need to combine all the elements in sets Aand B
without repetitions.
A∪B={x∈Z|1≤x≤15}
c) To find A\B, we need to determine the elements in Athat are not in B.
A\B={x∈Z|1≤x < 5}
d) To find B\A, we need to determine the elements in Bthat are not in A.
B\A={x∈Z|11 ≤x≤15}
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Question 4
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Define set C={x|x∈Aor x∈
Bbut not both}. Find set C.
Solution
Step 1: Determine which elements are in set C.Ccontains elements that are
in Aor in B, but not in both. We can list out the elements of Cby examining
the elements of Aand B:A={1,2,3,4,5}B={3,4,5,6,7}
Step 2: Identify the elements that are in Aor Bbut not both. The elements
that are in Aor B, but not both, are: 1 (only in A) 2 (only in A) 6 (only in B)
7 (only in B)
Step 3: Write the set C. Therefore, the set Cis:
C={1,2,6,7}
Question 5
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Determine the set (A∪B)∩
(A∩B).
Solution
Step 1: Find A∪B
The union of two sets Aand Bis the set of all elements that are in Aor in B.
A∪B={1,2,3,4,5,6,7}.
Step 2: Find A∩B
The intersection of two sets Aand Bis the set of all elements that are in both
Aand B.
A∩B={3,4,5}.
Step 3: Find (A∪B)∩(A∩B)
To find the intersection of two sets (A∪B) and (A∩B), we take only the
elements that are common to both sets.
(A∪B)∩(A∩B) = {3,4,5}.
Therefore, (A∪B)∩(A∩B) = {3,4,5}.
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Question 6
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|5≤x≤15}. Find the set
(A∩B)∪(A∩Bc), where Bcdenotes the complement of set B.
Solution
Step 1: Find A∩BSince both Aand Bconsist of integers, their intersection
will be the set of integers that are common to both sets. A∩B={x∈Z|5≤
x≤10}={5,6,7,8,9,10}
Step 2: Find BcThe complement of set Bwill consist of integers that are
not in set B.Bc={x∈Z|x < 5 or x > 15}
Step 3: Find A∩BcSince Aand Bcconsist of integers, their intersection
will be the set of integers that are common to both sets. A∩Bc={x∈Z|2≤
x < 5 or x > 15}={2,3,4,16,17,18, ...}
Step 4: Find (A∩B)∪(A∩Bc) The union of two sets is the set of elements
that are in at least one of the sets. (A∩B)∪(A∩Bc) = {5,6,7,8,9,10} ∪
{2,3,4,16,17,18, ...}(A∩B)∪(A∩Bc) = {2,3,4,5,6,7,8,9,10,16,17,18, ...}
Therefore, (A∩B)∪(A∩Bc) = {2,3,4,5,6,7,8,9,10,16,17,18, ...}.
Question 7
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Determine the
set A∪B.
Solution
To find the set A∪B, we need to take the union of sets Aand B, which includes
all elements that are in A, in B, or in both Aand B.
Step 1: Write out the elements in set Aand set B.
A={1,2,3,4,5,6,7,8,9,10}
B={5,6,7,8,9,10,11,12,13,14,15}
Step 2: Combine the elements of sets Aand B, but remove any duplicates.
A∪B={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
Therefore, the set A∪Bconsists of the elements {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
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Question 8
Question
Let A={1,2,3},B={2,3,4}, and C={3,4,5}. Find (A∩B)∪(B∩C).
Solution
Step 1: Find A∩B. We need to find the intersection of sets Aand B.
A∩B={2,3}
Step 2: Find B∩C. We need to find the intersection of sets Band C.
B∩C={3,4}
Step 3: Find (A∩B)∪(B∩C). To find the union of two sets, we combine
all the elements from both sets, removing any duplicates.
(A∩B)∪(B∩C) = {2,3}∪{3,4}={2,3,4}
Therefore, (A∩B)∪(B∩C) = {2,3,4}.
Question 9
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|6≤x≤15}. Find the
cardinality of the set A∪B.
Solution
Step 1: Write down the elements of sets Aand B.
A={1,2,3,4,5,6,7,8,9,10}
B={6,7,8,9,10,11,12,13,14,15}
Step 2: Find the union of sets Aand B, denoted as A∪B.
A∪B={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
Step 3: Calculate the cardinality (number of elements) of the set A∪B.
|A∪B|= 15
Therefore, the cardinality of the set A∪Bis 15.
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Question 10
Question
Let A={x∈Z| −3≤x≤4}and B={x∈Z| −2≤x≤3}. Determine the
set (A∪B)∩(A∩B).
Solution
Step 1: We first find A∪B. This set contains all elements that are in set Aor
set B(including elements that are in both sets). Since set Aincludes integers
from −3 to 4 and set Bincludes integers from −2 to 3,
A∪B={x∈Z| −3≤x≤4 or −2≤x≤3}={x∈Z| −3≤x≤4}.
Step 2: Now we find A∩B. This set contains all elements that are in both
set Aand set B.
A∩B={x∈Z| −2≤x≤3} ∩ {x∈Z| −3≤x≤4}={x∈Z| −2≤x≤3}.
Step 3: Finally, we find (A∪B)∩(A∩B). This set contains all elements
that are in both sets A∪Band A∩B.
(A∪B)∩(A∩B) = {x∈Z| −3≤x≤4}∩{x∈Z| −2≤x≤3}={x∈Z| −2≤x≤3}.
Therefore, (A∪B)∩(A∩B) = {x∈Z| −2≤x≤3}.
Question 11
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Find the set
(A∪B)∩(A∩B).
Solution
Step 1: First, we find A∪B.
A∪B={x∈Z|1≤x≤10}∪{x∈Z|5≤x≤15}
={x∈Z|1≤x≤15}
Step 2: Next, we find A∩B.
A∩B={x∈Z|1≤x≤10}∩{x∈Z|5≤x≤15}
={x∈Z|5≤x≤10}
Step 3: Finally, we find (A∪B)∩(A∩B).
(A∪B)∩(A∩B) = {x∈Z|1≤x≤15}∩{x∈Z|5≤x≤10}
={5,6,7,8,9,10}
Therefore, (A∪B)∩(A∩B) = {5,6,7,8,9,10}.
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Question 12
Question
Let A={x∈Z| −3< x ≤5}and B={x∈Z| −2≤x < 4}. Find the set
A∩B.
Solution
To find the intersection of sets Aand B, we need to find the set of elements
that are common to both sets Aand B.
Step 1: Write the set Aexplicitly.
A={−2,−1,0,1,2,3,4,5}
Step 2: Write the set Bexplicitly.
B={−2,−1,0,1,2,3}
Step 3: Find the intersection of sets Aand B.
A∩B={−2,−1,0,1,2,3}
Therefore, the intersection of sets Aand Bis the set {−2,−1,0,1,2,3}.
Question 13
Question
Let A={x∈Z|0<x<10}and B={x∈Z|6<x<15}. Find the set
A∪Band express the result in set-builder notation.
Solution
Step 1: We first find the union of sets Aand B.
A∪B={x∈Z|0<x<10}∪{x∈Z|6<x<15}
Step 2: We combine the conditions for xto be in set Aor B. To find the
union, we take all elements that are in Aor in B.
A∪B={x∈Z|0<x<10}∪{x∈Z|6<x<15}={x∈Z|0<x<10 or 6 <x<15}
Step 3: We simplify the conditions for x.
A∪B={x∈Z|0<x<10 or 6 <x<15}={x∈Z|0<x<15}
Step 4: Therefore, the set A∪Bin set-builder notation is {x∈Z|0< x <
15}.
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Question 14
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Determine the set (A∪B)∩(A∩B).
Solution
Step 1: Find A∪B.
A∪B={1,2,3,4,5}∪{3,4,5,6,7}
={1,2,3,4,5,6,7}
Step 2: Find A∩B.
A∩B={1,2,3,4,5}∩{3,4,5,6,7}
={3,4,5}
Step 3: Find (A∪B)∩(A∩B).
(A∪B)∩(A∩B) = {1,2,3,4,5,6,7}∩{3,4,5}
={3,4,5}
Therefore, the set (A∪B)∩(A∩B) is {3,4,5}.
Question 15
Question
Let A={x∈R:x2<4}and B={x∈R:x2≥1}. Find the set A∩B.
Solution
Step 1: We first find the elements in set A:A={x∈R:x2<4}This implies
that −2<x<2, so A= (−2,2).
Step 2: Next, we find the elements in set B:B={x∈R:x2≥1}This
implies that x≤ −1 or x≥1, so B= (−∞,−1] ∪[1,∞).
Step 3: Finding the intersection of sets Aand B:A∩B= (−2,2) ∩
((−∞,−1] ∪[1,∞))
Step 4: To calculate the intersection, we first look for the intersection points:
Let −2<x<2 and x≤ −1. Since xcan’t be both less than 2 and less than
−1, this part of the intersection will be empty. Similarly, let −2< x < 2 and
x≥1. The common area here is 1 ≤x < 2.
Step 5: Putting it all together, A∩B= [1,2).
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Question 16
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|5≤x≤15}. Find the set
A∩B.
Solution
Step 1: Write the elements of Aand B.
A={2,3,4,5,6,7,8,9,10}
B={5,6,7,8,9,10,11,12,13,14,15}
Step 2: Find the intersection of Aand B.
A∩B={x∈Z|5≤x≤10}={5,6,7,8,9,10}
Therefore, A∩B={5,6,7,8,9,10}.
Question 17
Question
Let A={x∈N:x≤10}and B={x∈N:xis a prime number}. Find A∩B,
A∪B, and A−B.
Solution
Step 1: Find A∩B(the intersection of sets Aand B):
Since Bconsists of prime numbers, we have B={2,3,5,7}.
Thus, A∩B={2,3,5,7}.
Step 2: Find A∪B(the union of sets Aand B):
Since Aconsists of natural numbers less than or equal to 10 and Bconsists of
prime numbers, A∪Bwill be all numbers less than or equal to 10 and the prime
numbers up to 10.
Thus, A∪B={1,2,3,4,5,6,7,8,9,10}.
Step 3: Find A−B(the set difference of Aminus B):
A−Bwill include all the natural numbers less than or equal to 10 that are not
prime numbers.
Thus, A−B={1,4,6,8,9,10}.
Question 18
Question
Let A={x∈Z|4≤x≤10}and B={x∈Z|8≤x≤12}. Find the set
A∩B.
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Solution
Step 1: Write out the sets Aand Bexplicitly.
A={4,5,6,7,8,9,10}
B={8,9,10,11,12}
Step 2: Find the intersection of sets Aand B, denoted as A∩B, which
consists of elements that are in both sets.
A∩B={x|x∈Aand x∈B}
={8,9,10}
Therefore, A∩B={8,9,10}.
Question 19
Question
Let A={n∈Z|0≤n≤10}and B={n∈Z|5≤n≤15}. Find the set
A∪B.
Solution
Step 1: To find A∪B, we need to find all the elements that are in Aor in B
(or both).
Step 2: Set Acontains the elements {0,1,2,3,4,5,6,7,8,9,10}.
Step 3: Set Bcontains the elements {5,6,7,8,9,10,11,12,13,14,15}.
Step 4: The union of sets Aand Bwill contain all the unique elements from
both sets.
Step 5: Therefore, A∪B={0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 20
Question
Let A={x|xis a prime number less than 10}and B={x|xis a multiple of 3}.
Find each of the following:
1. A∩B
2. A∪B
3. A−B
4. (A∪B)′
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Solution
We first list out the elements of sets Aand B:
A={2,3,5,7}
B={3,6,9}
1. A∩B: This represents the intersection of sets Aand B, which includes
elements that are in both Aand B.
A∩B={3}
2. A∪B: This represents the union of sets Aand B, which includes all
unique elements from both sets.
A∪B={2,3,5,6,7,9}
3. A−B: This represents the set of elements that are in set Abut not in set
B.
A−B={2,5,7}
4. (A∪B)′: This represents the complement of the union of sets Aand B,
which includes all elements not in the union.
(A∪B)′={1,4,8}
Question 21
Question
Let A={x∈R|1<x<5}and B={x∈R|3<x<7}. Determine the set
A∩B.
Solution
Step 1: Find the intersection of the two sets by identifying the elements common
to both sets Aand B. Step 2: We have A={x∈R|1<x<5}and B={x∈
R|3<x<7}. Step 3: The intersection of Aand B, denoted by A∩B, consists
of elements that are in both sets. Therefore, A∩B={x∈R|3<x<5}. Step
4: Therefore, A∩B={x∈R|3<x<5}. This set represents the interval
where both sets Aand Boverlap.
Question 22
Question
Let A={2,4,6,8,10}and B={3,6,9}. Determine the set (A∪B)∩(A∩B).
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Solution
Step 1: First, we find A∪B.
A∪B={2,4,6,8,10}∪{3,6,9}
={2,3,4,6,8,9,10}
Step 2: Next, we find A∩B.
A∩B={2,4,6,8,10}∩{3,6,9}
={6}
Step 3: We now find (A∪B)∩(A∩B).
(A∪B)∩(A∩B) = {2,3,4,6,8,9,10}∩{6}
={6}
Therefore, the set (A∪B)∩(A∩B) is {6}.
Question 23
Question
Let A={x∈R|x2−4x+ 4 >0}and B={x∈R|x2−2x−3>0}. Find
A∩B.
Solution
Step 1: To find A={x∈R|x2−4x+ 4 >0}, we need to determine the values
of xthat satisfy x2−4x+ 4 >0. This can be factored as (x−2)2>0. The
only way the square of a real number can be greater than zero is if the number
itself is not equal to zero. Thus, the solution to Ais A={x∈R|x= 2}.
Step 2: To find B={x∈R|x2−2x−3>0}, we need to determine the
values of xthat satisfy x2−2x−3>0. This can be factored as (x−3)(x+1) >0.
We then construct a sign table to find the solution:
x < −1−1<x<3x > 3
(x−3)(x+ 1) −+ +
From the sign table, we see that the solution to Bis B={x∈R|x <
−1 or x > 3}.
Step 3: Now, we find the intersection of sets Aand B, denoted as A∩B,
which represents the set of all elements that are in both Aand B. So, A∩B=
{x∈R|x= 2 and (x < −1 or x > 3)}.
Therefore, the set A∩Bis A∩B={x∈R|x < −1 or x > 3}.
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Question 24
Question
Let Aand Bbe sets such that |A∩B|= 8, |A\B|= 5, and |B\A|= 7. Find
|A∪B|.
Solution
Step 1: We can use the formula for the cardinality of the union of two sets:
|A∪B|=|A|+|B|−|A∩B|
Step 2: We are given that |A∩B|= 8, |A\B|= 5, and |B\A|= 7. We can
express |A|,|B|, and |A∩B|in terms of these values:
|A|=|A∩B|+|A\B|= 8 + 5 = 13
|B|=|B∩A|+|B\A|=|A∩B|+|B\A|= 8 + 7 = 15
Step 3: Substitute the values of |A|,|B|, and |A∩B|into the formula for
the cardinality of the union:
|A∪B|=|A|+|B|−|A∩B|= 13 + 15 −8 = 20
Therefore, |A∪B|= 20.
Question 25
Question
Let A,B, and Cbe sets such that A⊆Band C∩A=∅. Prove or disprove:
A⊆(B∪C).
Solution
To prove or disprove A⊆(B∪C), we need to consider two cases:
Case 1: A⊆(B∪C) is true.
Case 2: A⊆(B∪C) is false.
Case 1: Assume A⊆(B∪C) is true. This means that every element of
Ais also an element of (B∪C). Since A⊆B, every element of Ais also an
element of B. Additionally, we are given that C∩A=∅, which means that
there are no elements in common between Cand A. Therefore, if A⊆Band
C∩A=∅, it must be the case that A⊆B∪C.
Case 2: Assume A⊆(B∪C) is false. This means that there exists an
element x∈Asuch that x /∈(B∪C). Since A⊆B, any element in Amust
also be in B. However, if x /∈(B∪C), this implies that x /∈Band x /∈C. This
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contradicts the fact that A⊆Band C∩A=∅, which implies that Aand C
have no elements in common.
Since both cases lead to a contradiction, we can conclude that A⊆(B∪C)
is false.
Question 26
Question
Let A={x∈Z|3≤x≤13, x is odd}and B={x∈Z|5≤x≤
15, x is prime}. Find A∩B.
Solution
Step 1: Determine the elements in set A. Since Aconsists of odd integers
between 3 and 13 inclusive, we have A={3,5,7,9,11,13}.
Step 2: Determine the elements in set B. Since Bconsists of prime numbers
between 5 and 15 inclusive, we have B={5,7,11,13}.
Step 3: Find the intersection of sets Aand Bto get A∩B. The intersection
of Aand Bis the set of elements that are in both Aand B, which is A∩B=
{5,7,11,13}. Thus, A∩B={5,7,11,13}.
Question 27
Question
Let A={x∈Z: 1 ≤x≤10}and B={x∈Z: 5 ≤x≤15}. Determine the
set (A∩B)∪(A∩Bc).
Solution
Step 1: First, let’s determine A∩B, the intersection of sets Aand B.
A∩B={x∈Z: 1 ≤x≤10}∩{x∈Z: 5 ≤x≤15}
This means that A∩Bcontains the integers that are both in set Aand in set
B. So, A∩Bwill be the set of integers from 5 to 10.
A∩B={5,6,7,8,9,10}
Step 2: Next, let’s determine A∩Bc, the intersection of set Aand the
complement of set B. To find Bc, we need to consider all integers that are not
in set B. In other words, Bcconsists of integers outside the range of 5 to 15.
Bc={x∈Z:x < 5 or x > 15}
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Taking the complement of set B, we have Bc={x∈Z:x < 5 or x > 15}.
Now, find the intersection of set Awith Bc.
A∩Bc={x∈Z: 1 ≤x≤10}∩{x∈Z:x < 5 or x > 15}
This means that A∩Bccontains the integers that are in set Abut not in set
B. So, A∩Bcwill be the set of integers from 1 to 4 and 11 to 15.
A∩Bc={1,2,3,4,11,12,13,14,15}
Step 3: Finally, we will determine (A∩B)∪(A∩Bc), the union of A∩B
and A∩Bc.
(A∩B)∪(A∩Bc) = {5,6,7,8,9,10}∪{1,2,3,4,11,12,13,14,15}
This union will contain all distinct elements from both sets A∩Band A∩Bc.
(A∩B)∪(A∩Bc) = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
Therefore, the set (A∩B)∪(A∩Bc) is {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 28
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Find the set
A∪B.
Solution
Step 1: First, let’s list down the elements of sets Aand B.
A={1,2,3,4,5,6,7,8,9,10}
B={5,6,7,8,9,10,11,12,13,14,15}
Step 2: Now, we find the union of sets Aand B, denoted by A∪B, which
consists of all elements present in either set A, set B, or in both sets.
Therefore, A∪B={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Thus, the set A∪Bis {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 29
Question
Let Aand Bbe two sets such that |A∪B|= 20, |A∩B|= 8, and |A\B|= 10.
Find |B\A|.
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Step 1: Find the intersection of sets Aand BTo find the intersection,
we need to determine the elements that are common to both sets Aand B. The
intersection of sets Aand B, denoted by A∩B, is given by:
A∩B={x∈Z|5≤x≤10}
Step 2: Find the union of sets Aand BTo find the union, we need to
combine all elements in both sets Aand B, without repetitions. The union of
sets Aand B, denoted by A∪B, is given by:
A∪B={x∈Z|2≤x≤15}
Step 3: Find the relative complement of set Ain set BThe relative
complement of set Ain set B, denoted by B\A, represents the elements in B
that are not in A. The relative complement of set Ain set Bis given by:
B\A={x∈Z|11 ≤x≤15}
Therefore, the intersection of sets Aand Bis {x∈Z|5≤x≤10}, the
union of sets Aand Bis {x∈Z|2≤x≤15}, and the relative complement of
set Ain set Bis {x∈Z|11 ≤x≤15}.
Question 3
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Compute the
following sets: a) A∩Bb) A∪Bc) A\Bd) B\A
Solution
a) To find A∩B, we need to determine the elements that are common in both
Aand B.
A∩B={x∈Z|5≤x≤10}
b) To find A∪B, we need to combine all the elements in sets Aand B
without repetitions.
A∪B={x∈Z|1≤x≤15}
c) To find A\B, we need to determine the elements in Athat are not in B.
A\B={x∈Z|1≤x < 5}
d) To find B\A, we need to determine the elements in Bthat are not in A.
B\A={x∈Z|11 ≤x≤15}
2
Question 4
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Define set C={x|x∈Aor x∈
Bbut not both}. Find set C.
Solution
Step 1: Determine which elements are in set C.Ccontains elements that are
in Aor in B, but not in both. We can list out the elements of Cby examining
the elements of Aand B:A={1,2,3,4,5}B={3,4,5,6,7}
Step 2: Identify the elements that are in Aor Bbut not both. The elements
that are in Aor B, but not both, are: 1 (only in A) 2 (only in A) 6 (only in B)
7 (only in B)
Step 3: Write the set C. Therefore, the set Cis:
C={1,2,6,7}
Question 5
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Determine the set (A∪B)∩
(A∩B).
Solution
Step 1: Find A∪B
The union of two sets Aand Bis the set of all elements that are in Aor in B.
A∪B={1,2,3,4,5,6,7}.
Step 2: Find A∩B
The intersection of two sets Aand Bis the set of all elements that are in both
Aand B.
A∩B={3,4,5}.
Step 3: Find (A∪B)∩(A∩B)
To find the intersection of two sets (A∪B) and (A∩B), we take only the
elements that are common to both sets.
(A∪B)∩(A∩B) = {3,4,5}.
Therefore, (A∪B)∩(A∩B) = {3,4,5}.
3
Question 6
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|5≤x≤15}. Find the set
(A∩B)∪(A∩Bc), where Bcdenotes the complement of set B.
Solution
Step 1: Find A∩BSince both Aand Bconsist of integers, their intersection
will be the set of integers that are common to both sets. A∩B={x∈Z|5≤
x≤10}={5,6,7,8,9,10}
Step 2: Find BcThe complement of set Bwill consist of integers that are
not in set B.Bc={x∈Z|x < 5 or x > 15}
Step 3: Find A∩BcSince Aand Bcconsist of integers, their intersection
will be the set of integers that are common to both sets. A∩Bc={x∈Z|2≤
x < 5 or x > 15}={2,3,4,16,17,18, ...}
Step 4: Find (A∩B)∪(A∩Bc) The union of two sets is the set of elements
that are in at least one of the sets. (A∩B)∪(A∩Bc) = {5,6,7,8,9,10} ∪
{2,3,4,16,17,18, ...}(A∩B)∪(A∩Bc) = {2,3,4,5,6,7,8,9,10,16,17,18, ...}
Therefore, (A∩B)∪(A∩Bc) = {2,3,4,5,6,7,8,9,10,16,17,18, ...}.
Question 7
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Determine the
set A∪B.
Solution
To find the set A∪B, we need to take the union of sets Aand B, which includes
all elements that are in A, in B, or in both Aand B.
Step 1: Write out the elements in set Aand set B.
A={1,2,3,4,5,6,7,8,9,10}
B={5,6,7,8,9,10,11,12,13,14,15}
Step 2: Combine the elements of sets Aand B, but remove any duplicates.
A∪B={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
Therefore, the set A∪Bconsists of the elements {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
4
Question 8
Question
Let A={1,2,3},B={2,3,4}, and C={3,4,5}. Find (A∩B)∪(B∩C).
Solution
Step 1: Find A∩B. We need to find the intersection of sets Aand B.
A∩B={2,3}
Step 2: Find B∩C. We need to find the intersection of sets Band C.
B∩C={3,4}
Step 3: Find (A∩B)∪(B∩C). To find the union of two sets, we combine
all the elements from both sets, removing any duplicates.
(A∩B)∪(B∩C) = {2,3}∪{3,4}={2,3,4}
Therefore, (A∩B)∪(B∩C) = {2,3,4}.
Question 9
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|6≤x≤15}. Find the
cardinality of the set A∪B.
Solution
Step 1: Write down the elements of sets Aand B.
A={1,2,3,4,5,6,7,8,9,10}
B={6,7,8,9,10,11,12,13,14,15}
Step 2: Find the union of sets Aand B, denoted as A∪B.
A∪B={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
Step 3: Calculate the cardinality (number of elements) of the set A∪B.
|A∪B|= 15
Therefore, the cardinality of the set A∪Bis 15.
5
Question 10
Question
Let A={x∈Z| −3≤x≤4}and B={x∈Z| −2≤x≤3}. Determine the
set (A∪B)∩(A∩B).
Solution
Step 1: We first find A∪B. This set contains all elements that are in set Aor
set B(including elements that are in both sets). Since set Aincludes integers
from −3 to 4 and set Bincludes integers from −2 to 3,
A∪B={x∈Z| −3≤x≤4 or −2≤x≤3}={x∈Z| −3≤x≤4}.
Step 2: Now we find A∩B. This set contains all elements that are in both
set Aand set B.
A∩B={x∈Z| −2≤x≤3} ∩ {x∈Z| −3≤x≤4}={x∈Z| −2≤x≤3}.
Step 3: Finally, we find (A∪B)∩(A∩B). This set contains all elements
that are in both sets A∪Band A∩B.
(A∪B)∩(A∩B) = {x∈Z| −3≤x≤4}∩{x∈Z| −2≤x≤3}={x∈Z| −2≤x≤3}.
Therefore, (A∪B)∩(A∩B) = {x∈Z| −2≤x≤3}.
Question 11
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Find the set
(A∪B)∩(A∩B).
Solution
Step 1: First, we find A∪B.
A∪B={x∈Z|1≤x≤10}∪{x∈Z|5≤x≤15}
={x∈Z|1≤x≤15}
Step 2: Next, we find A∩B.
A∩B={x∈Z|1≤x≤10}∩{x∈Z|5≤x≤15}
={x∈Z|5≤x≤10}
Step 3: Finally, we find (A∪B)∩(A∩B).
(A∪B)∩(A∩B) = {x∈Z|1≤x≤15}∩{x∈Z|5≤x≤10}
={5,6,7,8,9,10}
Therefore, (A∪B)∩(A∩B) = {5,6,7,8,9,10}.
6
Question 12
Question
Let A={x∈Z| −3< x ≤5}and B={x∈Z| −2≤x < 4}. Find the set
A∩B.
Solution
To find the intersection of sets Aand B, we need to find the set of elements
that are common to both sets Aand B.
Step 1: Write the set Aexplicitly.
A={−2,−1,0,1,2,3,4,5}
Step 2: Write the set Bexplicitly.
B={−2,−1,0,1,2,3}
Step 3: Find the intersection of sets Aand B.
A∩B={−2,−1,0,1,2,3}
Therefore, the intersection of sets Aand Bis the set {−2,−1,0,1,2,3}.
Question 13
Question
Let A={x∈Z|0<x<10}and B={x∈Z|6<x<15}. Find the set
A∪Band express the result in set-builder notation.
Solution
Step 1: We first find the union of sets Aand B.
A∪B={x∈Z|0<x<10}∪{x∈Z|6<x<15}
Step 2: We combine the conditions for xto be in set Aor B. To find the
union, we take all elements that are in Aor in B.
A∪B={x∈Z|0<x<10}∪{x∈Z|6<x<15}={x∈Z|0<x<10 or 6 <x<15}
Step 3: We simplify the conditions for x.
A∪B={x∈Z|0<x<10 or 6 <x<15}={x∈Z|0<x<15}
Step 4: Therefore, the set A∪Bin set-builder notation is {x∈Z|0< x <
15}.
7
Question 14
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Determine the set (A∪B)∩(A∩B).
Solution
Step 1: Find A∪B.
A∪B={1,2,3,4,5}∪{3,4,5,6,7}
={1,2,3,4,5,6,7}
Step 2: Find A∩B.
A∩B={1,2,3,4,5}∩{3,4,5,6,7}
={3,4,5}
Step 3: Find (A∪B)∩(A∩B).
(A∪B)∩(A∩B) = {1,2,3,4,5,6,7}∩{3,4,5}
={3,4,5}
Therefore, the set (A∪B)∩(A∩B) is {3,4,5}.
Question 15
Question
Let A={x∈R:x2<4}and B={x∈R:x2≥1}. Find the set A∩B.
Solution
Step 1: We first find the elements in set A:A={x∈R:x2<4}This implies
that −2<x<2, so A= (−2,2).
Step 2: Next, we find the elements in set B:B={x∈R:x2≥1}This
implies that x≤ −1 or x≥1, so B= (−∞,−1] ∪[1,∞).
Step 3: Finding the intersection of sets Aand B:A∩B= (−2,2) ∩
((−∞,−1] ∪[1,∞))
Step 4: To calculate the intersection, we first look for the intersection points:
Let −2<x<2 and x≤ −1. Since xcan’t be both less than 2 and less than
−1, this part of the intersection will be empty. Similarly, let −2< x < 2 and
x≥1. The common area here is 1 ≤x < 2.
Step 5: Putting it all together, A∩B= [1,2).
8
Question 16
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|5≤x≤15}. Find the set
A∩B.
Solution
Step 1: Write the elements of Aand B.
A={2,3,4,5,6,7,8,9,10}
B={5,6,7,8,9,10,11,12,13,14,15}
Step 2: Find the intersection of Aand B.
A∩B={x∈Z|5≤x≤10}={5,6,7,8,9,10}
Therefore, A∩B={5,6,7,8,9,10}.
Question 17
Question
Let A={x∈N:x≤10}and B={x∈N:xis a prime number}. Find A∩B,
A∪B, and A−B.
Solution
Step 1: Find A∩B(the intersection of sets Aand B):
Since Bconsists of prime numbers, we have B={2,3,5,7}.
Thus, A∩B={2,3,5,7}.
Step 2: Find A∪B(the union of sets Aand B):
Since Aconsists of natural numbers less than or equal to 10 and Bconsists of
prime numbers, A∪Bwill be all numbers less than or equal to 10 and the prime
numbers up to 10.
Thus, A∪B={1,2,3,4,5,6,7,8,9,10}.
Step 3: Find A−B(the set difference of Aminus B):
A−Bwill include all the natural numbers less than or equal to 10 that are not
prime numbers.
Thus, A−B={1,4,6,8,9,10}.
Question 18
Question
Let A={x∈Z|4≤x≤10}and B={x∈Z|8≤x≤12}. Find the set
A∩B.
9
Solution
Step 1: Write out the sets Aand Bexplicitly.
A={4,5,6,7,8,9,10}
B={8,9,10,11,12}
Step 2: Find the intersection of sets Aand B, denoted as A∩B, which
consists of elements that are in both sets.
A∩B={x|x∈Aand x∈B}
={8,9,10}
Therefore, A∩B={8,9,10}.
Question 19
Question
Let A={n∈Z|0≤n≤10}and B={n∈Z|5≤n≤15}. Find the set
A∪B.
Solution
Step 1: To find A∪B, we need to find all the elements that are in Aor in B
(or both).
Step 2: Set Acontains the elements {0,1,2,3,4,5,6,7,8,9,10}.
Step 3: Set Bcontains the elements {5,6,7,8,9,10,11,12,13,14,15}.
Step 4: The union of sets Aand Bwill contain all the unique elements from
both sets.
Step 5: Therefore, A∪B={0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 20
Question
Let A={x|xis a prime number less than 10}and B={x|xis a multiple of 3}.
Find each of the following:
1. A∩B
2. A∪B
3. A−B
4. (A∪B)′
10
Solution
We first list out the elements of sets Aand B:
A={2,3,5,7}
B={3,6,9}
1. A∩B: This represents the intersection of sets Aand B, which includes
elements that are in both Aand B.
A∩B={3}
2. A∪B: This represents the union of sets Aand B, which includes all
unique elements from both sets.
A∪B={2,3,5,6,7,9}
3. A−B: This represents the set of elements that are in set Abut not in set
B.
A−B={2,5,7}
4. (A∪B)′: This represents the complement of the union of sets Aand B,
which includes all elements not in the union.
(A∪B)′={1,4,8}
Question 21
Question
Let A={x∈R|1<x<5}and B={x∈R|3<x<7}. Determine the set
A∩B.
Solution
Step 1: Find the intersection of the two sets by identifying the elements common
to both sets Aand B. Step 2: We have A={x∈R|1<x<5}and B={x∈
R|3<x<7}. Step 3: The intersection of Aand B, denoted by A∩B, consists
of elements that are in both sets. Therefore, A∩B={x∈R|3<x<5}. Step
4: Therefore, A∩B={x∈R|3<x<5}. This set represents the interval
where both sets Aand Boverlap.
Question 22
Question
Let A={2,4,6,8,10}and B={3,6,9}. Determine the set (A∪B)∩(A∩B).
11
Solution
Step 1: First, we find A∪B.
A∪B={2,4,6,8,10}∪{3,6,9}
={2,3,4,6,8,9,10}
Step 2: Next, we find A∩B.
A∩B={2,4,6,8,10}∩{3,6,9}
={6}
Step 3: We now find (A∪B)∩(A∩B).
(A∪B)∩(A∩B) = {2,3,4,6,8,9,10}∩{6}
={6}
Therefore, the set (A∪B)∩(A∩B) is {6}.
Question 23
Question
Let A={x∈R|x2−4x+ 4 >0}and B={x∈R|x2−2x−3>0}. Find
A∩B.
Solution
Step 1: To find A={x∈R|x2−4x+ 4 >0}, we need to determine the values
of xthat satisfy x2−4x+ 4 >0. This can be factored as (x−2)2>0. The
only way the square of a real number can be greater than zero is if the number
itself is not equal to zero. Thus, the solution to Ais A={x∈R|x= 2}.
Step 2: To find B={x∈R|x2−2x−3>0}, we need to determine the
values of xthat satisfy x2−2x−3>0. This can be factored as (x−3)(x+1) >0.
We then construct a sign table to find the solution:
x < −1−1<x<3x > 3
(x−3)(x+ 1) −+ +
From the sign table, we see that the solution to Bis B={x∈R|x <
−1 or x > 3}.
Step 3: Now, we find the intersection of sets Aand B, denoted as A∩B,
which represents the set of all elements that are in both Aand B. So, A∩B=
{x∈R|x= 2 and (x < −1 or x > 3)}.
Therefore, the set A∩Bis A∩B={x∈R|x < −1 or x > 3}.
12
Question 24
Question
Let Aand Bbe sets such that |A∩B|= 8, |A\B|= 5, and |B\A|= 7. Find
|A∪B|.
Solution
Step 1: We can use the formula for the cardinality of the union of two sets:
|A∪B|=|A|+|B|−|A∩B|
Step 2: We are given that |A∩B|= 8, |A\B|= 5, and |B\A|= 7. We can
express |A|,|B|, and |A∩B|in terms of these values:
|A|=|A∩B|+|A\B|= 8 + 5 = 13
|B|=|B∩A|+|B\A|=|A∩B|+|B\A|= 8 + 7 = 15
Step 3: Substitute the values of |A|,|B|, and |A∩B|into the formula for
the cardinality of the union:
|A∪B|=|A|+|B|−|A∩B|= 13 + 15 −8 = 20
Therefore, |A∪B|= 20.
Question 25
Question
Let A,B, and Cbe sets such that A⊆Band C∩A=∅. Prove or disprove:
A⊆(B∪C).
Solution
To prove or disprove A⊆(B∪C), we need to consider two cases:
Case 1: A⊆(B∪C) is true.
Case 2: A⊆(B∪C) is false.
Case 1: Assume A⊆(B∪C) is true. This means that every element of
Ais also an element of (B∪C). Since A⊆B, every element of Ais also an
element of B. Additionally, we are given that C∩A=∅, which means that
there are no elements in common between Cand A. Therefore, if A⊆Band
C∩A=∅, it must be the case that A⊆B∪C.
Case 2: Assume A⊆(B∪C) is false. This means that there exists an
element x∈Asuch that x /∈(B∪C). Since A⊆B, any element in Amust
also be in B. However, if x /∈(B∪C), this implies that x /∈Band x /∈C. This
13
contradicts the fact that A⊆Band C∩A=∅, which implies that Aand C
have no elements in common.
Since both cases lead to a contradiction, we can conclude that A⊆(B∪C)
is false.
Question 26
Question
Let A={x∈Z|3≤x≤13, x is odd}and B={x∈Z|5≤x≤
15, x is prime}. Find A∩B.
Solution
Step 1: Determine the elements in set A. Since Aconsists of odd integers
between 3 and 13 inclusive, we have A={3,5,7,9,11,13}.
Step 2: Determine the elements in set B. Since Bconsists of prime numbers
between 5 and 15 inclusive, we have B={5,7,11,13}.
Step 3: Find the intersection of sets Aand Bto get A∩B. The intersection
of Aand Bis the set of elements that are in both Aand B, which is A∩B=
{5,7,11,13}. Thus, A∩B={5,7,11,13}.
Question 27
Question
Let A={x∈Z: 1 ≤x≤10}and B={x∈Z: 5 ≤x≤15}. Determine the
set (A∩B)∪(A∩Bc).
Solution
Step 1: First, let’s determine A∩B, the intersection of sets Aand B.
A∩B={x∈Z: 1 ≤x≤10}∩{x∈Z: 5 ≤x≤15}
This means that A∩Bcontains the integers that are both in set Aand in set
B. So, A∩Bwill be the set of integers from 5 to 10.
A∩B={5,6,7,8,9,10}
Step 2: Next, let’s determine A∩Bc, the intersection of set Aand the
complement of set B. To find Bc, we need to consider all integers that are not
in set B. In other words, Bcconsists of integers outside the range of 5 to 15.
Bc={x∈Z:x < 5 or x > 15}
14
Taking the complement of set B, we have Bc={x∈Z:x < 5 or x > 15}.
Now, find the intersection of set Awith Bc.
A∩Bc={x∈Z: 1 ≤x≤10}∩{x∈Z:x < 5 or x > 15}
This means that A∩Bccontains the integers that are in set Abut not in set
B. So, A∩Bcwill be the set of integers from 1 to 4 and 11 to 15.
A∩Bc={1,2,3,4,11,12,13,14,15}
Step 3: Finally, we will determine (A∩B)∪(A∩Bc), the union of A∩B
and A∩Bc.
(A∩B)∪(A∩Bc) = {5,6,7,8,9,10}∪{1,2,3,4,11,12,13,14,15}
This union will contain all distinct elements from both sets A∩Band A∩Bc.
(A∩B)∪(A∩Bc) = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
Therefore, the set (A∩B)∪(A∩Bc) is {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 28
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Find the set
A∪B.
Solution
Step 1: First, let’s list down the elements of sets Aand B.
A={1,2,3,4,5,6,7,8,9,10}
B={5,6,7,8,9,10,11,12,13,14,15}
Step 2: Now, we find the union of sets Aand B, denoted by A∪B, which
consists of all elements present in either set A, set B, or in both sets.
Therefore, A∪B={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Thus, the set A∪Bis {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 29
Question
Let Aand Bbe two sets such that |A∪B|= 20, |A∩B|= 8, and |A\B|= 10.
Find |B\A|.
15
Step 1: Find the intersection of sets Aand BTo find the intersection,
we need to determine the elements that are common to both sets Aand B. The
intersection of sets Aand B, denoted by A∩B, is given by:
A∩B={x∈Z|5≤x≤10}
Step 2: Find the union of sets Aand BTo find the union, we need to
combine all elements in both sets Aand B, without repetitions. The union of
sets Aand B, denoted by A∪B, is given by:
A∪B={x∈Z|2≤x≤15}
Step 3: Find the relative complement of set Ain set BThe relative
complement of set Ain set B, denoted by B\A, represents the elements in B
that are not in A. The relative complement of set Ain set Bis given by:
B\A={x∈Z|11 ≤x≤15}
Therefore, the intersection of sets Aand Bis {x∈Z|5≤x≤10}, the
union of sets Aand Bis {x∈Z|2≤x≤15}, and the relative complement of
set Ain set Bis {x∈Z|11 ≤x≤15}.
Question 3
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Compute the
following sets: a) A∩Bb) A∪Bc) A\Bd) B\A
Solution
a) To find A∩B, we need to determine the elements that are common in both
Aand B.
A∩B={x∈Z|5≤x≤10}
b) To find A∪B, we need to combine all the elements in sets Aand B
without repetitions.
A∪B={x∈Z|1≤x≤15}
c) To find A\B, we need to determine the elements in Athat are not in B.
A\B={x∈Z|1≤x < 5}
d) To find B\A, we need to determine the elements in Bthat are not in A.
B\A={x∈Z|11 ≤x≤15}
2
Question 4
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Define set C={x|x∈Aor x∈
Bbut not both}. Find set C.
Solution
Step 1: Determine which elements are in set C.Ccontains elements that are
in Aor in B, but not in both. We can list out the elements of Cby examining
the elements of Aand B:A={1,2,3,4,5}B={3,4,5,6,7}
Step 2: Identify the elements that are in Aor Bbut not both. The elements
that are in Aor B, but not both, are: 1 (only in A) 2 (only in A) 6 (only in B)
7 (only in B)
Step 3: Write the set C. Therefore, the set Cis:
C={1,2,6,7}
Question 5
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Determine the set (A∪B)∩
(A∩B).
Solution
Step 1: Find A∪B
The union of two sets Aand Bis the set of all elements that are in Aor in B.
A∪B={1,2,3,4,5,6,7}.
Step 2: Find A∩B
The intersection of two sets Aand Bis the set of all elements that are in both
Aand B.
A∩B={3,4,5}.
Step 3: Find (A∪B)∩(A∩B)
To find the intersection of two sets (A∪B) and (A∩B), we take only the
elements that are common to both sets.
(A∪B)∩(A∩B) = {3,4,5}.
Therefore, (A∪B)∩(A∩B) = {3,4,5}.
3
Question 6
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|5≤x≤15}. Find the set
(A∩B)∪(A∩Bc), where Bcdenotes the complement of set B.
Solution
Step 1: Find A∩BSince both Aand Bconsist of integers, their intersection
will be the set of integers that are common to both sets. A∩B={x∈Z|5≤
x≤10}={5,6,7,8,9,10}
Step 2: Find BcThe complement of set Bwill consist of integers that are
not in set B.Bc={x∈Z|x < 5 or x > 15}
Step 3: Find A∩BcSince Aand Bcconsist of integers, their intersection
will be the set of integers that are common to both sets. A∩Bc={x∈Z|2≤
x < 5 or x > 15}={2,3,4,16,17,18, ...}
Step 4: Find (A∩B)∪(A∩Bc) The union of two sets is the set of elements
that are in at least one of the sets. (A∩B)∪(A∩Bc) = {5,6,7,8,9,10} ∪
{2,3,4,16,17,18, ...}(A∩B)∪(A∩Bc) = {2,3,4,5,6,7,8,9,10,16,17,18, ...}
Therefore, (A∩B)∪(A∩Bc) = {2,3,4,5,6,7,8,9,10,16,17,18, ...}.
Question 7
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Determine the
set A∪B.
Solution
To find the set A∪B, we need to take the union of sets Aand B, which includes
all elements that are in A, in B, or in both Aand B.
Step 1: Write out the elements in set Aand set B.
A={1,2,3,4,5,6,7,8,9,10}
B={5,6,7,8,9,10,11,12,13,14,15}
Step 2: Combine the elements of sets Aand B, but remove any duplicates.
A∪B={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
Therefore, the set A∪Bconsists of the elements {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
4
Question 8
Question
Let A={1,2,3},B={2,3,4}, and C={3,4,5}. Find (A∩B)∪(B∩C).
Solution
Step 1: Find A∩B. We need to find the intersection of sets Aand B.
A∩B={2,3}
Step 2: Find B∩C. We need to find the intersection of sets Band C.
B∩C={3,4}
Step 3: Find (A∩B)∪(B∩C). To find the union of two sets, we combine
all the elements from both sets, removing any duplicates.
(A∩B)∪(B∩C) = {2,3}∪{3,4}={2,3,4}
Therefore, (A∩B)∪(B∩C) = {2,3,4}.
Question 9
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|6≤x≤15}. Find the
cardinality of the set A∪B.
Solution
Step 1: Write down the elements of sets Aand B.
A={1,2,3,4,5,6,7,8,9,10}
B={6,7,8,9,10,11,12,13,14,15}
Step 2: Find the union of sets Aand B, denoted as A∪B.
A∪B={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
Step 3: Calculate the cardinality (number of elements) of the set A∪B.
|A∪B|= 15
Therefore, the cardinality of the set A∪Bis 15.
5
Question 10
Question
Let A={x∈Z| −3≤x≤4}and B={x∈Z| −2≤x≤3}. Determine the
set (A∪B)∩(A∩B).
Solution
Step 1: We first find A∪B. This set contains all elements that are in set Aor
set B(including elements that are in both sets). Since set Aincludes integers
from −3 to 4 and set Bincludes integers from −2 to 3,
A∪B={x∈Z| −3≤x≤4 or −2≤x≤3}={x∈Z| −3≤x≤4}.
Step 2: Now we find A∩B. This set contains all elements that are in both
set Aand set B.
A∩B={x∈Z| −2≤x≤3} ∩ {x∈Z| −3≤x≤4}={x∈Z| −2≤x≤3}.
Step 3: Finally, we find (A∪B)∩(A∩B). This set contains all elements
that are in both sets A∪Band A∩B.
(A∪B)∩(A∩B) = {x∈Z| −3≤x≤4}∩{x∈Z| −2≤x≤3}={x∈Z| −2≤x≤3}.
Therefore, (A∪B)∩(A∩B) = {x∈Z| −2≤x≤3}.
Question 11
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Find the set
(A∪B)∩(A∩B).
Solution
Step 1: First, we find A∪B.
A∪B={x∈Z|1≤x≤10}∪{x∈Z|5≤x≤15}
={x∈Z|1≤x≤15}
Step 2: Next, we find A∩B.
A∩B={x∈Z|1≤x≤10}∩{x∈Z|5≤x≤15}
={x∈Z|5≤x≤10}
Step 3: Finally, we find (A∪B)∩(A∩B).
(A∪B)∩(A∩B) = {x∈Z|1≤x≤15}∩{x∈Z|5≤x≤10}
={5,6,7,8,9,10}
Therefore, (A∪B)∩(A∩B) = {5,6,7,8,9,10}.
6
Question 12
Question
Let A={x∈Z| −3< x ≤5}and B={x∈Z| −2≤x < 4}. Find the set
A∩B.
Solution
To find the intersection of sets Aand B, we need to find the set of elements
that are common to both sets Aand B.
Step 1: Write the set Aexplicitly.
A={−2,−1,0,1,2,3,4,5}
Step 2: Write the set Bexplicitly.
B={−2,−1,0,1,2,3}
Step 3: Find the intersection of sets Aand B.
A∩B={−2,−1,0,1,2,3}
Therefore, the intersection of sets Aand Bis the set {−2,−1,0,1,2,3}.
Question 13
Question
Let A={x∈Z|0<x<10}and B={x∈Z|6<x<15}. Find the set
A∪Band express the result in set-builder notation.
Solution
Step 1: We first find the union of sets Aand B.
A∪B={x∈Z|0<x<10}∪{x∈Z|6<x<15}
Step 2: We combine the conditions for xto be in set Aor B. To find the
union, we take all elements that are in Aor in B.
A∪B={x∈Z|0<x<10}∪{x∈Z|6<x<15}={x∈Z|0<x<10 or 6 <x<15}
Step 3: We simplify the conditions for x.
A∪B={x∈Z|0<x<10 or 6 <x<15}={x∈Z|0<x<15}
Step 4: Therefore, the set A∪Bin set-builder notation is {x∈Z|0< x <
15}.
7
Question 14
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Determine the set (A∪B)∩(A∩B).
Solution
Step 1: Find A∪B.
A∪B={1,2,3,4,5}∪{3,4,5,6,7}
={1,2,3,4,5,6,7}
Step 2: Find A∩B.
A∩B={1,2,3,4,5}∩{3,4,5,6,7}
={3,4,5}
Step 3: Find (A∪B)∩(A∩B).
(A∪B)∩(A∩B) = {1,2,3,4,5,6,7}∩{3,4,5}
={3,4,5}
Therefore, the set (A∪B)∩(A∩B) is {3,4,5}.
Question 15
Question
Let A={x∈R:x2<4}and B={x∈R:x2≥1}. Find the set A∩B.
Solution
Step 1: We first find the elements in set A:A={x∈R:x2<4}This implies
that −2<x<2, so A= (−2,2).
Step 2: Next, we find the elements in set B:B={x∈R:x2≥1}This
implies that x≤ −1 or x≥1, so B= (−∞,−1] ∪[1,∞).
Step 3: Finding the intersection of sets Aand B:A∩B= (−2,2) ∩
((−∞,−1] ∪[1,∞))
Step 4: To calculate the intersection, we first look for the intersection points:
Let −2<x<2 and x≤ −1. Since xcan’t be both less than 2 and less than
−1, this part of the intersection will be empty. Similarly, let −2< x < 2 and
x≥1. The common area here is 1 ≤x < 2.
Step 5: Putting it all together, A∩B= [1,2).
8
Question 16
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|5≤x≤15}. Find the set
A∩B.
Solution
Step 1: Write the elements of Aand B.
A={2,3,4,5,6,7,8,9,10}
B={5,6,7,8,9,10,11,12,13,14,15}
Step 2: Find the intersection of Aand B.
A∩B={x∈Z|5≤x≤10}={5,6,7,8,9,10}
Therefore, A∩B={5,6,7,8,9,10}.
Question 17
Question
Let A={x∈N:x≤10}and B={x∈N:xis a prime number}. Find A∩B,
A∪B, and A−B.
Solution
Step 1: Find A∩B(the intersection of sets Aand B):
Since Bconsists of prime numbers, we have B={2,3,5,7}.
Thus, A∩B={2,3,5,7}.
Step 2: Find A∪B(the union of sets Aand B):
Since Aconsists of natural numbers less than or equal to 10 and Bconsists of
prime numbers, A∪Bwill be all numbers less than or equal to 10 and the prime
numbers up to 10.
Thus, A∪B={1,2,3,4,5,6,7,8,9,10}.
Step 3: Find A−B(the set difference of Aminus B):
A−Bwill include all the natural numbers less than or equal to 10 that are not
prime numbers.
Thus, A−B={1,4,6,8,9,10}.
Question 18
Question
Let A={x∈Z|4≤x≤10}and B={x∈Z|8≤x≤12}. Find the set
A∩B.
9
Solution
Step 1: Write out the sets Aand Bexplicitly.
A={4,5,6,7,8,9,10}
B={8,9,10,11,12}
Step 2: Find the intersection of sets Aand B, denoted as A∩B, which
consists of elements that are in both sets.
A∩B={x|x∈Aand x∈B}
={8,9,10}
Therefore, A∩B={8,9,10}.
Question 19
Question
Let A={n∈Z|0≤n≤10}and B={n∈Z|5≤n≤15}. Find the set
A∪B.
Solution
Step 1: To find A∪B, we need to find all the elements that are in Aor in B
(or both).
Step 2: Set Acontains the elements {0,1,2,3,4,5,6,7,8,9,10}.
Step 3: Set Bcontains the elements {5,6,7,8,9,10,11,12,13,14,15}.
Step 4: The union of sets Aand Bwill contain all the unique elements from
both sets.
Step 5: Therefore, A∪B={0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 20
Question
Let A={x|xis a prime number less than 10}and B={x|xis a multiple of 3}.
Find each of the following:
1. A∩B
2. A∪B
3. A−B
4. (A∪B)′
10
Solution
We first list out the elements of sets Aand B:
A={2,3,5,7}
B={3,6,9}
1. A∩B: This represents the intersection of sets Aand B, which includes
elements that are in both Aand B.
A∩B={3}
2. A∪B: This represents the union of sets Aand B, which includes all
unique elements from both sets.
A∪B={2,3,5,6,7,9}
3. A−B: This represents the set of elements that are in set Abut not in set
B.
A−B={2,5,7}
4. (A∪B)′: This represents the complement of the union of sets Aand B,
which includes all elements not in the union.
(A∪B)′={1,4,8}
Question 21
Question
Let A={x∈R|1<x<5}and B={x∈R|3<x<7}. Determine the set
A∩B.
Solution
Step 1: Find the intersection of the two sets by identifying the elements common
to both sets Aand B. Step 2: We have A={x∈R|1<x<5}and B={x∈
R|3<x<7}. Step 3: The intersection of Aand B, denoted by A∩B, consists
of elements that are in both sets. Therefore, A∩B={x∈R|3<x<5}. Step
4: Therefore, A∩B={x∈R|3<x<5}. This set represents the interval
where both sets Aand Boverlap.
Question 22
Question
Let A={2,4,6,8,10}and B={3,6,9}. Determine the set (A∪B)∩(A∩B).
11
Solution
Step 1: First, we find A∪B.
A∪B={2,4,6,8,10}∪{3,6,9}
={2,3,4,6,8,9,10}
Step 2: Next, we find A∩B.
A∩B={2,4,6,8,10}∩{3,6,9}
={6}
Step 3: We now find (A∪B)∩(A∩B).
(A∪B)∩(A∩B) = {2,3,4,6,8,9,10}∩{6}
={6}
Therefore, the set (A∪B)∩(A∩B) is {6}.
Question 23
Question
Let A={x∈R|x2−4x+ 4 >0}and B={x∈R|x2−2x−3>0}. Find
A∩B.
Solution
Step 1: To find A={x∈R|x2−4x+ 4 >0}, we need to determine the values
of xthat satisfy x2−4x+ 4 >0. This can be factored as (x−2)2>0. The
only way the square of a real number can be greater than zero is if the number
itself is not equal to zero. Thus, the solution to Ais A={x∈R|x= 2}.
Step 2: To find B={x∈R|x2−2x−3>0}, we need to determine the
values of xthat satisfy x2−2x−3>0. This can be factored as (x−3)(x+1) >0.
We then construct a sign table to find the solution:
x < −1−1<x<3x > 3
(x−3)(x+ 1) −+ +
From the sign table, we see that the solution to Bis B={x∈R|x <
−1 or x > 3}.
Step 3: Now, we find the intersection of sets Aand B, denoted as A∩B,
which represents the set of all elements that are in both Aand B. So, A∩B=
{x∈R|x= 2 and (x < −1 or x > 3)}.
Therefore, the set A∩Bis A∩B={x∈R|x < −1 or x > 3}.
12
Question 24
Question
Let Aand Bbe sets such that |A∩B|= 8, |A\B|= 5, and |B\A|= 7. Find
|A∪B|.
Solution
Step 1: We can use the formula for the cardinality of the union of two sets:
|A∪B|=|A|+|B|−|A∩B|
Step 2: We are given that |A∩B|= 8, |A\B|= 5, and |B\A|= 7. We can
express |A|,|B|, and |A∩B|in terms of these values:
|A|=|A∩B|+|A\B|= 8 + 5 = 13
|B|=|B∩A|+|B\A|=|A∩B|+|B\A|= 8 + 7 = 15
Step 3: Substitute the values of |A|,|B|, and |A∩B|into the formula for
the cardinality of the union:
|A∪B|=|A|+|B|−|A∩B|= 13 + 15 −8 = 20
Therefore, |A∪B|= 20.
Question 25
Question
Let A,B, and Cbe sets such that A⊆Band C∩A=∅. Prove or disprove:
A⊆(B∪C).
Solution
To prove or disprove A⊆(B∪C), we need to consider two cases:
Case 1: A⊆(B∪C) is true.
Case 2: A⊆(B∪C) is false.
Case 1: Assume A⊆(B∪C) is true. This means that every element of
Ais also an element of (B∪C). Since A⊆B, every element of Ais also an
element of B. Additionally, we are given that C∩A=∅, which means that
there are no elements in common between Cand A. Therefore, if A⊆Band
C∩A=∅, it must be the case that A⊆B∪C.
Case 2: Assume A⊆(B∪C) is false. This means that there exists an
element x∈Asuch that x /∈(B∪C). Since A⊆B, any element in Amust
also be in B. However, if x /∈(B∪C), this implies that x /∈Band x /∈C. This
13
contradicts the fact that A⊆Band C∩A=∅, which implies that Aand C
have no elements in common.
Since both cases lead to a contradiction, we can conclude that A⊆(B∪C)
is false.
Question 26
Question
Let A={x∈Z|3≤x≤13, x is odd}and B={x∈Z|5≤x≤
15, x is prime}. Find A∩B.
Solution
Step 1: Determine the elements in set A. Since Aconsists of odd integers
between 3 and 13 inclusive, we have A={3,5,7,9,11,13}.
Step 2: Determine the elements in set B. Since Bconsists of prime numbers
between 5 and 15 inclusive, we have B={5,7,11,13}.
Step 3: Find the intersection of sets Aand Bto get A∩B. The intersection
of Aand Bis the set of elements that are in both Aand B, which is A∩B=
{5,7,11,13}. Thus, A∩B={5,7,11,13}.
Question 27
Question
Let A={x∈Z: 1 ≤x≤10}and B={x∈Z: 5 ≤x≤15}. Determine the
set (A∩B)∪(A∩Bc).
Solution
Step 1: First, let’s determine A∩B, the intersection of sets Aand B.
A∩B={x∈Z: 1 ≤x≤10}∩{x∈Z: 5 ≤x≤15}
This means that A∩Bcontains the integers that are both in set Aand in set
B. So, A∩Bwill be the set of integers from 5 to 10.
A∩B={5,6,7,8,9,10}
Step 2: Next, let’s determine A∩Bc, the intersection of set Aand the
complement of set B. To find Bc, we need to consider all integers that are not
in set B. In other words, Bcconsists of integers outside the range of 5 to 15.
Bc={x∈Z:x < 5 or x > 15}
14
Taking the complement of set B, we have Bc={x∈Z:x < 5 or x > 15}.
Now, find the intersection of set Awith Bc.
A∩Bc={x∈Z: 1 ≤x≤10}∩{x∈Z:x < 5 or x > 15}
This means that A∩Bccontains the integers that are in set Abut not in set
B. So, A∩Bcwill be the set of integers from 1 to 4 and 11 to 15.
A∩Bc={1,2,3,4,11,12,13,14,15}
Step 3: Finally, we will determine (A∩B)∪(A∩Bc), the union of A∩B
and A∩Bc.
(A∩B)∪(A∩Bc) = {5,6,7,8,9,10}∪{1,2,3,4,11,12,13,14,15}
This union will contain all distinct elements from both sets A∩Band A∩Bc.
(A∩B)∪(A∩Bc) = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
Therefore, the set (A∩B)∪(A∩Bc) is {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 28
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Find the set
A∪B.
Solution
Step 1: First, let’s list down the elements of sets Aand B.
A={1,2,3,4,5,6,7,8,9,10}
B={5,6,7,8,9,10,11,12,13,14,15}
Step 2: Now, we find the union of sets Aand B, denoted by A∪B, which
consists of all elements present in either set A, set B, or in both sets.
Therefore, A∪B={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Thus, the set A∪Bis {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 29
Question
Let Aand Bbe two sets such that |A∪B|= 20, |A∩B|= 8, and |A\B|= 10.
Find |B\A|.
15
Step 1: Find the intersection of sets Aand BTo find the intersection,
we need to determine the elements that are common to both sets Aand B. The
intersection of sets Aand B, denoted by A∩B, is given by:
A∩B={x∈Z|5≤x≤10}
Step 2: Find the union of sets Aand BTo find the union, we need to
combine all elements in both sets Aand B, without repetitions. The union of
sets Aand B, denoted by A∪B, is given by:
A∪B={x∈Z|2≤x≤15}
Step 3: Find the relative complement of set Ain set BThe relative
complement of set Ain set B, denoted by B\A, represents the elements in B
that are not in A. The relative complement of set Ain set Bis given by:
B\A={x∈Z|11 ≤x≤15}
Therefore, the intersection of sets Aand Bis {x∈Z|5≤x≤10}, the
union of sets Aand Bis {x∈Z|2≤x≤15}, and the relative complement of
set Ain set Bis {x∈Z|11 ≤x≤15}.
Question 3
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Compute the
following sets: a) A∩Bb) A∪Bc) A\Bd) B\A
Solution
a) To find A∩B, we need to determine the elements that are common in both
Aand B.
A∩B={x∈Z|5≤x≤10}
b) To find A∪B, we need to combine all the elements in sets Aand B
without repetitions.
A∪B={x∈Z|1≤x≤15}
c) To find A\B, we need to determine the elements in Athat are not in B.
A\B={x∈Z|1≤x < 5}
d) To find B\A, we need to determine the elements in Bthat are not in A.
B\A={x∈Z|11 ≤x≤15}
2
Question 4
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Define set C={x|x∈Aor x∈
Bbut not both}. Find set C.
Solution
Step 1: Determine which elements are in set C.Ccontains elements that are
in Aor in B, but not in both. We can list out the elements of Cby examining
the elements of Aand B:A={1,2,3,4,5}B={3,4,5,6,7}
Step 2: Identify the elements that are in Aor Bbut not both. The elements
that are in Aor B, but not both, are: 1 (only in A) 2 (only in A) 6 (only in B)
7 (only in B)
Step 3: Write the set C. Therefore, the set Cis:
C={1,2,6,7}
Question 5
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Determine the set (A∪B)∩
(A∩B).
Solution
Step 1: Find A∪B
The union of two sets Aand Bis the set of all elements that are in Aor in B.
A∪B={1,2,3,4,5,6,7}.
Step 2: Find A∩B
The intersection of two sets Aand Bis the set of all elements that are in both
Aand B.
A∩B={3,4,5}.
Step 3: Find (A∪B)∩(A∩B)
To find the intersection of two sets (A∪B) and (A∩B), we take only the
elements that are common to both sets.
(A∪B)∩(A∩B) = {3,4,5}.
Therefore, (A∪B)∩(A∩B) = {3,4,5}.
3
Question 6
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|5≤x≤15}. Find the set
(A∩B)∪(A∩Bc), where Bcdenotes the complement of set B.
Solution
Step 1: Find A∩BSince both Aand Bconsist of integers, their intersection
will be the set of integers that are common to both sets. A∩B={x∈Z|5≤
x≤10}={5,6,7,8,9,10}
Step 2: Find BcThe complement of set Bwill consist of integers that are
not in set B.Bc={x∈Z|x < 5 or x > 15}
Step 3: Find A∩BcSince Aand Bcconsist of integers, their intersection
will be the set of integers that are common to both sets. A∩Bc={x∈Z|2≤
x < 5 or x > 15}={2,3,4,16,17,18, ...}
Step 4: Find (A∩B)∪(A∩Bc) The union of two sets is the set of elements
that are in at least one of the sets. (A∩B)∪(A∩Bc) = {5,6,7,8,9,10} ∪
{2,3,4,16,17,18, ...}(A∩B)∪(A∩Bc) = {2,3,4,5,6,7,8,9,10,16,17,18, ...}
Therefore, (A∩B)∪(A∩Bc) = {2,3,4,5,6,7,8,9,10,16,17,18, ...}.
Question 7
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Determine the
set A∪B.
Solution
To find the set A∪B, we need to take the union of sets Aand B, which includes
all elements that are in A, in B, or in both Aand B.
Step 1: Write out the elements in set Aand set B.
A={1,2,3,4,5,6,7,8,9,10}
B={5,6,7,8,9,10,11,12,13,14,15}
Step 2: Combine the elements of sets Aand B, but remove any duplicates.
A∪B={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
Therefore, the set A∪Bconsists of the elements {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
4
Question 8
Question
Let A={1,2,3},B={2,3,4}, and C={3,4,5}. Find (A∩B)∪(B∩C).
Solution
Step 1: Find A∩B. We need to find the intersection of sets Aand B.
A∩B={2,3}
Step 2: Find B∩C. We need to find the intersection of sets Band C.
B∩C={3,4}
Step 3: Find (A∩B)∪(B∩C). To find the union of two sets, we combine
all the elements from both sets, removing any duplicates.
(A∩B)∪(B∩C) = {2,3}∪{3,4}={2,3,4}
Therefore, (A∩B)∪(B∩C) = {2,3,4}.
Question 9
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|6≤x≤15}. Find the
cardinality of the set A∪B.
Solution
Step 1: Write down the elements of sets Aand B.
A={1,2,3,4,5,6,7,8,9,10}
B={6,7,8,9,10,11,12,13,14,15}
Step 2: Find the union of sets Aand B, denoted as A∪B.
A∪B={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
Step 3: Calculate the cardinality (number of elements) of the set A∪B.
|A∪B|= 15
Therefore, the cardinality of the set A∪Bis 15.
5
Question 10
Question
Let A={x∈Z| −3≤x≤4}and B={x∈Z| −2≤x≤3}. Determine the
set (A∪B)∩(A∩B).
Solution
Step 1: We first find A∪B. This set contains all elements that are in set Aor
set B(including elements that are in both sets). Since set Aincludes integers
from −3 to 4 and set Bincludes integers from −2 to 3,
A∪B={x∈Z| −3≤x≤4 or −2≤x≤3}={x∈Z| −3≤x≤4}.
Step 2: Now we find A∩B. This set contains all elements that are in both
set Aand set B.
A∩B={x∈Z| −2≤x≤3} ∩ {x∈Z| −3≤x≤4}={x∈Z| −2≤x≤3}.
Step 3: Finally, we find (A∪B)∩(A∩B). This set contains all elements
that are in both sets A∪Band A∩B.
(A∪B)∩(A∩B) = {x∈Z| −3≤x≤4}∩{x∈Z| −2≤x≤3}={x∈Z| −2≤x≤3}.
Therefore, (A∪B)∩(A∩B) = {x∈Z| −2≤x≤3}.
Question 11
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Find the set
(A∪B)∩(A∩B).
Solution
Step 1: First, we find A∪B.
A∪B={x∈Z|1≤x≤10}∪{x∈Z|5≤x≤15}
={x∈Z|1≤x≤15}
Step 2: Next, we find A∩B.
A∩B={x∈Z|1≤x≤10}∩{x∈Z|5≤x≤15}
={x∈Z|5≤x≤10}
Step 3: Finally, we find (A∪B)∩(A∩B).
(A∪B)∩(A∩B) = {x∈Z|1≤x≤15}∩{x∈Z|5≤x≤10}
={5,6,7,8,9,10}
Therefore, (A∪B)∩(A∩B) = {5,6,7,8,9,10}.
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Question 12
Question
Let A={x∈Z| −3< x ≤5}and B={x∈Z| −2≤x < 4}. Find the set
A∩B.
Solution
To find the intersection of sets Aand B, we need to find the set of elements
that are common to both sets Aand B.
Step 1: Write the set Aexplicitly.
A={−2,−1,0,1,2,3,4,5}
Step 2: Write the set Bexplicitly.
B={−2,−1,0,1,2,3}
Step 3: Find the intersection of sets Aand B.
A∩B={−2,−1,0,1,2,3}
Therefore, the intersection of sets Aand Bis the set {−2,−1,0,1,2,3}.
Question 13
Question
Let A={x∈Z|0<x<10}and B={x∈Z|6<x<15}. Find the set
A∪Band express the result in set-builder notation.
Solution
Step 1: We first find the union of sets Aand B.
A∪B={x∈Z|0<x<10}∪{x∈Z|6<x<15}
Step 2: We combine the conditions for xto be in set Aor B. To find the
union, we take all elements that are in Aor in B.
A∪B={x∈Z|0<x<10}∪{x∈Z|6<x<15}={x∈Z|0<x<10 or 6 <x<15}
Step 3: We simplify the conditions for x.
A∪B={x∈Z|0<x<10 or 6 <x<15}={x∈Z|0<x<15}
Step 4: Therefore, the set A∪Bin set-builder notation is {x∈Z|0< x <
15}.
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Question 14
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Determine the set (A∪B)∩(A∩B).
Solution
Step 1: Find A∪B.
A∪B={1,2,3,4,5}∪{3,4,5,6,7}
={1,2,3,4,5,6,7}
Step 2: Find A∩B.
A∩B={1,2,3,4,5}∩{3,4,5,6,7}
={3,4,5}
Step 3: Find (A∪B)∩(A∩B).
(A∪B)∩(A∩B) = {1,2,3,4,5,6,7}∩{3,4,5}
={3,4,5}
Therefore, the set (A∪B)∩(A∩B) is {3,4,5}.
Question 15
Question
Let A={x∈R:x2<4}and B={x∈R:x2≥1}. Find the set A∩B.
Solution
Step 1: We first find the elements in set A:A={x∈R:x2<4}This implies
that −2<x<2, so A= (−2,2).
Step 2: Next, we find the elements in set B:B={x∈R:x2≥1}This
implies that x≤ −1 or x≥1, so B= (−∞,−1] ∪[1,∞).
Step 3: Finding the intersection of sets Aand B:A∩B= (−2,2) ∩
((−∞,−1] ∪[1,∞))
Step 4: To calculate the intersection, we first look for the intersection points:
Let −2<x<2 and x≤ −1. Since xcan’t be both less than 2 and less than
−1, this part of the intersection will be empty. Similarly, let −2< x < 2 and
x≥1. The common area here is 1 ≤x < 2.
Step 5: Putting it all together, A∩B= [1,2).
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Question 16
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|5≤x≤15}. Find the set
A∩B.
Solution
Step 1: Write the elements of Aand B.
A={2,3,4,5,6,7,8,9,10}
B={5,6,7,8,9,10,11,12,13,14,15}
Step 2: Find the intersection of Aand B.
A∩B={x∈Z|5≤x≤10}={5,6,7,8,9,10}
Therefore, A∩B={5,6,7,8,9,10}.
Question 17
Question
Let A={x∈N:x≤10}and B={x∈N:xis a prime number}. Find A∩B,
A∪B, and A−B.
Solution
Step 1: Find A∩B(the intersection of sets Aand B):
Since Bconsists of prime numbers, we have B={2,3,5,7}.
Thus, A∩B={2,3,5,7}.
Step 2: Find A∪B(the union of sets Aand B):
Since Aconsists of natural numbers less than or equal to 10 and Bconsists of
prime numbers, A∪Bwill be all numbers less than or equal to 10 and the prime
numbers up to 10.
Thus, A∪B={1,2,3,4,5,6,7,8,9,10}.
Step 3: Find A−B(the set difference of Aminus B):
A−Bwill include all the natural numbers less than or equal to 10 that are not
prime numbers.
Thus, A−B={1,4,6,8,9,10}.
Question 18
Question
Let A={x∈Z|4≤x≤10}and B={x∈Z|8≤x≤12}. Find the set
A∩B.
9
Solution
Step 1: Write out the sets Aand Bexplicitly.
A={4,5,6,7,8,9,10}
B={8,9,10,11,12}
Step 2: Find the intersection of sets Aand B, denoted as A∩B, which
consists of elements that are in both sets.
A∩B={x|x∈Aand x∈B}
={8,9,10}
Therefore, A∩B={8,9,10}.
Question 19
Question
Let A={n∈Z|0≤n≤10}and B={n∈Z|5≤n≤15}. Find the set
A∪B.
Solution
Step 1: To find A∪B, we need to find all the elements that are in Aor in B
(or both).
Step 2: Set Acontains the elements {0,1,2,3,4,5,6,7,8,9,10}.
Step 3: Set Bcontains the elements {5,6,7,8,9,10,11,12,13,14,15}.
Step 4: The union of sets Aand Bwill contain all the unique elements from
both sets.
Step 5: Therefore, A∪B={0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 20
Question
Let A={x|xis a prime number less than 10}and B={x|xis a multiple of 3}.
Find each of the following:
1. A∩B
2. A∪B
3. A−B
4. (A∪B)′
10
Solution
We first list out the elements of sets Aand B:
A={2,3,5,7}
B={3,6,9}
1. A∩B: This represents the intersection of sets Aand B, which includes
elements that are in both Aand B.
A∩B={3}
2. A∪B: This represents the union of sets Aand B, which includes all
unique elements from both sets.
A∪B={2,3,5,6,7,9}
3. A−B: This represents the set of elements that are in set Abut not in set
B.
A−B={2,5,7}
4. (A∪B)′: This represents the complement of the union of sets Aand B,
which includes all elements not in the union.
(A∪B)′={1,4,8}
Question 21
Question
Let A={x∈R|1<x<5}and B={x∈R|3<x<7}. Determine the set
A∩B.
Solution
Step 1: Find the intersection of the two sets by identifying the elements common
to both sets Aand B. Step 2: We have A={x∈R|1<x<5}and B={x∈
R|3<x<7}. Step 3: The intersection of Aand B, denoted by A∩B, consists
of elements that are in both sets. Therefore, A∩B={x∈R|3<x<5}. Step
4: Therefore, A∩B={x∈R|3<x<5}. This set represents the interval
where both sets Aand Boverlap.
Question 22
Question
Let A={2,4,6,8,10}and B={3,6,9}. Determine the set (A∪B)∩(A∩B).
11
Solution
Step 1: First, we find A∪B.
A∪B={2,4,6,8,10}∪{3,6,9}
={2,3,4,6,8,9,10}
Step 2: Next, we find A∩B.
A∩B={2,4,6,8,10}∩{3,6,9}
={6}
Step 3: We now find (A∪B)∩(A∩B).
(A∪B)∩(A∩B) = {2,3,4,6,8,9,10}∩{6}
={6}
Therefore, the set (A∪B)∩(A∩B) is {6}.
Question 23
Question
Let A={x∈R|x2−4x+ 4 >0}and B={x∈R|x2−2x−3>0}. Find
A∩B.
Solution
Step 1: To find A={x∈R|x2−4x+ 4 >0}, we need to determine the values
of xthat satisfy x2−4x+ 4 >0. This can be factored as (x−2)2>0. The
only way the square of a real number can be greater than zero is if the number
itself is not equal to zero. Thus, the solution to Ais A={x∈R|x= 2}.
Step 2: To find B={x∈R|x2−2x−3>0}, we need to determine the
values of xthat satisfy x2−2x−3>0. This can be factored as (x−3)(x+1) >0.
We then construct a sign table to find the solution:
x < −1−1<x<3x > 3
(x−3)(x+ 1) −+ +
From the sign table, we see that the solution to Bis B={x∈R|x <
−1 or x > 3}.
Step 3: Now, we find the intersection of sets Aand B, denoted as A∩B,
which represents the set of all elements that are in both Aand B. So, A∩B=
{x∈R|x= 2 and (x < −1 or x > 3)}.
Therefore, the set A∩Bis A∩B={x∈R|x < −1 or x > 3}.
12
Question 24
Question
Let Aand Bbe sets such that |A∩B|= 8, |A\B|= 5, and |B\A|= 7. Find
|A∪B|.
Solution
Step 1: We can use the formula for the cardinality of the union of two sets:
|A∪B|=|A|+|B|−|A∩B|
Step 2: We are given that |A∩B|= 8, |A\B|= 5, and |B\A|= 7. We can
express |A|,|B|, and |A∩B|in terms of these values:
|A|=|A∩B|+|A\B|= 8 + 5 = 13
|B|=|B∩A|+|B\A|=|A∩B|+|B\A|= 8 + 7 = 15
Step 3: Substitute the values of |A|,|B|, and |A∩B|into the formula for
the cardinality of the union:
|A∪B|=|A|+|B|−|A∩B|= 13 + 15 −8 = 20
Therefore, |A∪B|= 20.
Question 25
Question
Let A,B, and Cbe sets such that A⊆Band C∩A=∅. Prove or disprove:
A⊆(B∪C).
Solution
To prove or disprove A⊆(B∪C), we need to consider two cases:
Case 1: A⊆(B∪C) is true.
Case 2: A⊆(B∪C) is false.
Case 1: Assume A⊆(B∪C) is true. This means that every element of
Ais also an element of (B∪C). Since A⊆B, every element of Ais also an
element of B. Additionally, we are given that C∩A=∅, which means that
there are no elements in common between Cand A. Therefore, if A⊆Band
C∩A=∅, it must be the case that A⊆B∪C.
Case 2: Assume A⊆(B∪C) is false. This means that there exists an
element x∈Asuch that x /∈(B∪C). Since A⊆B, any element in Amust
also be in B. However, if x /∈(B∪C), this implies that x /∈Band x /∈C. This
13
contradicts the fact that A⊆Band C∩A=∅, which implies that Aand C
have no elements in common.
Since both cases lead to a contradiction, we can conclude that A⊆(B∪C)
is false.
Question 26
Question
Let A={x∈Z|3≤x≤13, x is odd}and B={x∈Z|5≤x≤
15, x is prime}. Find A∩B.
Solution
Step 1: Determine the elements in set A. Since Aconsists of odd integers
between 3 and 13 inclusive, we have A={3,5,7,9,11,13}.
Step 2: Determine the elements in set B. Since Bconsists of prime numbers
between 5 and 15 inclusive, we have B={5,7,11,13}.
Step 3: Find the intersection of sets Aand Bto get A∩B. The intersection
of Aand Bis the set of elements that are in both Aand B, which is A∩B=
{5,7,11,13}. Thus, A∩B={5,7,11,13}.
Question 27
Question
Let A={x∈Z: 1 ≤x≤10}and B={x∈Z: 5 ≤x≤15}. Determine the
set (A∩B)∪(A∩Bc).
Solution
Step 1: First, let’s determine A∩B, the intersection of sets Aand B.
A∩B={x∈Z: 1 ≤x≤10}∩{x∈Z: 5 ≤x≤15}
This means that A∩Bcontains the integers that are both in set Aand in set
B. So, A∩Bwill be the set of integers from 5 to 10.
A∩B={5,6,7,8,9,10}
Step 2: Next, let’s determine A∩Bc, the intersection of set Aand the
complement of set B. To find Bc, we need to consider all integers that are not
in set B. In other words, Bcconsists of integers outside the range of 5 to 15.
Bc={x∈Z:x < 5 or x > 15}
14
Taking the complement of set B, we have Bc={x∈Z:x < 5 or x > 15}.
Now, find the intersection of set Awith Bc.
A∩Bc={x∈Z: 1 ≤x≤10}∩{x∈Z:x < 5 or x > 15}
This means that A∩Bccontains the integers that are in set Abut not in set
B. So, A∩Bcwill be the set of integers from 1 to 4 and 11 to 15.
A∩Bc={1,2,3,4,11,12,13,14,15}
Step 3: Finally, we will determine (A∩B)∪(A∩Bc), the union of A∩B
and A∩Bc.
(A∩B)∪(A∩Bc) = {5,6,7,8,9,10}∪{1,2,3,4,11,12,13,14,15}
This union will contain all distinct elements from both sets A∩Band A∩Bc.
(A∩B)∪(A∩Bc) = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
Therefore, the set (A∩B)∪(A∩Bc) is {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 28
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Find the set
A∪B.
Solution
Step 1: First, let’s list down the elements of sets Aand B.
A={1,2,3,4,5,6,7,8,9,10}
B={5,6,7,8,9,10,11,12,13,14,15}
Step 2: Now, we find the union of sets Aand B, denoted by A∪B, which
consists of all elements present in either set A, set B, or in both sets.
Therefore, A∪B={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Thus, the set A∪Bis {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}.
Question 29
Question
Let Aand Bbe two sets such that |A∪B|= 20, |A∩B|= 8, and |A\B|= 10.
Find |B\A|.
15
Solution
Step 1: We know that |A∪B|=|A|+|B| − |A∩B|. Using the given values, we
can calculate |A|+|B| − 8 = 20.
Step 2: Since |A|+|B|= 20 + 8 = 28, we have |A|= 28 − |B|.
Step 3: We also have |A\B|=|A| − |A∩B|. Substituting the known values,
we get 10 = |A| − 8.
Step 4: By substituting |A|= 28 − |B|in the above equation, we get 10 =
28 − |B| − 8. Solving for |B|, we find |B|= 10.
Step 5: Now, to find |B\A|, we use the formula |B\A|=|B| − |A∩B|.
Substituting the known values, we get |B\A|= 10 −8 = 2.
Therefore, |B\A|= 2.
Question 30
Question
Let A,B, and Cbe sets such that A∪B=A∪C,A∩B=A∩C, and A=∅.
Prove or disprove that B=C.
Solution
To prove this statement, we can construct a counterexample.
Counterexample: Let A={1},B={1,2}, and C={1,3}.
We have A∪B={1,2}=A∪C.
Also, A∩B={1}=A∩C.
However, B={1,2} ={1,3}=C.
Therefore, the statement is disproved with the counterexample provided.
16
