MATH 350 - DISCRETE
MATHEMATICS - Operations on sets
Question Bank - Set 3
Liberty University
Question 1
Question
Let A={x∈Z|0< x ≤10}and B={x∈Z|5< x < 15}. Find
(A∪B)∩(A∩B).
Solution
Step 1: First, let’s find A∪Band A∩B.A∪B: This set contains all elements
that are in either Aor B.A∪B={x∈Z|0< x ≤10 or 5 < x < 15}
A∪B={x∈Z|(0 < x ≤10) or (5 <x<15)}A∪B={x∈Z|0<x<15}
Step 2: Now, let’s find A∩B.A∩B: This set contains all elements that
are common to both Aand B.A∩B={x∈Z|5< x ≤10}
Step 3: Finally, we find (A∪B)∩(A∩B). (A∪B)∩(A∩B) = {x∈Z|
(0 < x < 15) and (5 < x ≤10)}(A∪B)∩(A∩B) = {x∈Z|5< x ≤10}
(A∪B)∩(A∩B) = {6,7,8,9,10}
Therefore, (A∪B)∩(A∩B) = {6,7,8,9,10}.
Question 2
Question
Let A={x∈R|x2−4x+ 3 >0}and B={x∈R|x2−5x+ 6 >0}. Find
A∩B.
Solution
Step 1: We first find the solutions to the inequalities x2−4x+ 3 >0 and
x2−5x+ 6 >0.
For x2−4x+ 3 >0, we can factor this as (x−1)(x−3) >0. The solutions
are 1 <x<3.
For x2−5x+ 6 >0, we can factor this as (x−2)(x−3) >0. The solutions
are x < 2 or x > 3.
Step 2: Now, we find the intersection of Aand B.
The solutions to A∩Boccur when both inequalities are simultaneously
satisfied, so we look for the values of xwhich are in both 1 <x<3 and x < 2
or x > 3.
Taking the intersection of the intervals, we have 1 < x < 2 or 3 < x < 3,
which simplifies to 1 <x<2.
Therefore, A∩B={x∈R|1<x<2}.
Question 3
Question
Let A={x|1≤x≤5}and B={x|3≤x≤7}. Find the intersection of the
two sets A∩B.
Solution
Step 1: Write out the elements of sets Aand B.
A={1,2,3,4,5}, B ={3,4,5,6,7}
Step 2: Identify the common elements in sets Aand B.
A∩B={3,4,5}
Therefore, A∩B={3,4,5}.
Question 4
Question
Let A={x∈Z|0≤x≤10}and B={x∈Z|5≤x≤15}. Find the
intersection A∩Band the complement of A∪B.
Solution
Step 1: Find the intersection A∩B.
A∩B={x∈Z|0≤x≤10}∩{x∈Z|5≤x≤15}
={x∈Z|5≤x≤10}
2
Step 2: Find the complement of A∪B. To find the complement of A∪B,
we need to consider all the elements that are not in the set A∪B.
A∪B={x∈Z|0≤x≤10 or 5 ≤x≤15}
={0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
So, the complement of A∪Bis all integers outside the set {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}:
Z− {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
Question 5
Question
Let A={x∈R| −3≤x < 5}and B={x∈R| −1< x ≤4}. Find the sets
(A∩B)∪(A∩Bc) and (A∪B)∩(Ac∪B).
Solution
Let’s first find A∩Band A∩Bc:
A∩B: Elements that are in both Aand B.
A∩Bc: Elements that are in Abut not in B.
Step 1: Find A∩B.
A∩B={x∈R| −3≤x < 5 and −1< x ≤4}
A∩B={x∈R| −1<x<4}
Step 2: Find A∩Bc.
A∩Bc={x∈R| −3≤x < 5 and (x≤ −1 or x > 4)}
A∩Bc={x∈R| −3≤x < −1 or 4 <x<5}
Now, let’s find (A∩B)∪(A∩Bc):
(A∩B)∪(A∩Bc) = {x∈R| −1<x<4 or −3≤x < −1 or 4 <x<5}
(A∩B)∪(A∩Bc) = {x∈R| −3≤x < 5}=A
Next, let’s find (A∪B)∩(Ac∪B):
(A∪B)∩(Ac∪B) = {x∈R|(−3≤x < 5 or −1< x ≤4) and (x < −3 or x≥4)}
(A∪B)∩(Ac∪B) = {x∈R| −3≤x < −1 or 4 ≤x < 5}
Therefore,
(A∩B)∪(A∩Bc) = A
and
(A∪B)∩(Ac∪B) = {x∈R|(−3≤x < −1) or (4 ≤x < 5)}
.
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Question 6
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Find the
intersection A∩Band the complement of A∩B.
Solution
Step 1: Find A∩B.
A={1,2,3,4,5,6,7,8,9,10}
B={5,6,7,8,9,10,11,12,13,14,15}
A∩B={5,6,7,8,9,10}
Step 2: Find the complement of A∩B. The universal set in this case is
defined as {x∈Z|1≤x≤15}. The complement of A∩Bis given by:
({1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}−{5,6,7,8,9,10})
which simplifies to
{1,2,3,4,11,12,13,14,15}
Question 7
Question
Let A={x∈N|1≤x≤10}and B={x2|x∈N, x < 6}. Find the set A∩B.
Solution
Step 1: Find the elements of set B.
B={x2|x∈N, x < 6}
={12,22,32,42,52}
={1,4,9,16,25}
Step 2: Find the intersection of sets Aand B.
A∩B={x|x∈Aand x∈B}
={1,4,9}
Therefore, A∩B={1,4,9}.
Question 8
Question
Let A={x∈R|1≤x≤5}and B={x∈R|3≤x≤7}. Find the set A∩B.
4
Solution
To find the intersection of sets Aand B, we need to determine the elements
that are common to both sets.
A={x∈R|1≤x≤5}={1,2,3,4,5}
B={x∈R|3≤x≤7}={3,4,5,6,7}
Step 1: List the elements of set Aand set B.
A={1,2,3,4,5}
B={3,4,5,6,7}
Step 2: Identify the common elements in Aand B.
The intersection of sets Aand B, denoted A∩B, is the set containing
elements that are in both Aand B.
A∩B={3,4,5}
Therefore, the set A∩Bconsists of the elements {3,4,5}.
Question 9
Question
Let A={x∈Z:xis a prime number less than 10}and B={2,3,5,7}. Find
the following sets: 1. A∪B2. A∩B3. A\B4. B\A
Solution
1. To find A∪B, we need to find all the elements that are in either set Aor set
B(or both).
A∪B={2,3,5,7}∪{2,3,5,7}={2,3,5,7}
2. To find A∩B, we need to find all the elements that are in both set A
and set B.
A∩B={2,3,5,7}∩{2,3,5,7}={2,3,5,7}
3. To find A\B, we need to find all the elements that are in set Abut not
in set B.
A\B={2,3,5,7}\{2,3,5,7}={}
4. To find B\A, we need to find all the elements that are in set Bbut not
in set A.
B\A={2,3,5,7}\{2,3,5,7}={}
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Question 10
Question
Let A={x∈Z| −3≤x≤3}and B={x∈Z| −2≤x≤2}. Determine the
set (A∪B)∩(A∩B).
Solution
Step 1: First, find A∪B.
A∪B={−3,−2,−1,0,1,2,3} ∪ {−2,−1,0,1,2}
A∪B={−3,−2,−1,0,1,2,3}
Step 2: Next, find A∩B.
A∩B={−3,−2,−1,0,1,2,3} ∩ {−2,−1,0,1,2}
A∩B={−2,−1,0,1,2}
Step 3: Finally, find (A∪B)∩(A∩B).
(A∪B)∩(A∩B) = {−3,−2,−1,0,1,2,3} ∩ {−2,−1,0,1,2}
(A∪B)∩(A∩B) = {−2,−1,0,1,2}
Therefore, (A∪B)∩(A∩B) = {−2,−1,0,1,2}.
Question 11
Question
Let A={x∈R|1≤x≤5}and B={x∈R|3≤x≤7}. Find the set A∩B.
Solution
Step 1: Find the intersection of sets Aand Busing the given intervals.
A∩B={x∈R|1≤x≤5}∩{x∈R|3≤x≤7}
Step 2: Determine the overlapping interval between the two sets. The in-
tersection of the two intervals will be the set where they overlap, which is [3,5]
since 3 ≤x≤5.
Therefore, the set A∩Bis:
A∩B={x∈R|3≤x≤5}
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Question 12
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Find each of the
following sets:
a) A∪B
b) A∩B
c) A\B
d) B\A
Solution
a) To find A∪B, we need to take the union of sets Aand B, which includes all
elements that are in Aor in Bor in both.
A∪B={x∈Z|1≤x≤15}
A∪B={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
b) To find A∩B, we need to find the intersection of sets Aand B, which
includes all elements that are in both Aand B.
A∩B={x∈Z|5≤x≤10}
A∩B={5,6,7,8,9,10}
c) To find A\B, we need to find the elements that are in Abut not in B.
A\B={x∈Z|1≤x < 5}
A\B={1,2,3,4}
d) To find B\A, we need to find the elements that are in Bbut not in A.
B\A={x∈Z|11 ≤x≤15}
B\A={11,12,13,14,15}
Question 13
Question
Let A={1,2,3,4}and B={3,4,5,6}. Define the set C={x∈Z|x=
a−b, for some a∈Aand b∈B}. Find C.
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Solution
Step 1: To find the set C, we need to consider all possible differences between
elements of set Aand set B.
Step 2: Calculate the differences for all pairs of elements (a, b) where a∈A
and b∈B.
Step 3: Differences for (1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6),
(3,3), (3,4), (3,5), (3,6), (4,3), (4,4), (4,5), (4,6) are −2, −3, −4, −5, −1, −2,
−3, −4, 0, −1, −2, −3, 1, 0, −1, −2 respectively.
Step 4: Therefore, set C={−5,−4,−3,−2,−1,0,1}, where Cis the set of
all possible differences between elements of set Aand set B.
Question 14
Question
Let A={x∈Z| −3≤x≤3}and B={x∈Z|2≤x≤7}. Determine the
set A∪Band A∩B.
Solution
Step 1: To find the union of sets Aand B, we need to combine all elements that
are in either set.
A∪B={x∈Z| −3≤x≤7}
Step 2: To find the intersection of sets Aand B, we need to find the common
elements in both sets.
A∩B={x∈Z|2≤x≤3}
Therefore, we have A∪B={−3,−2,−1,0,1,2,3,4,5,6,7}and A∩B=
{2,3}.
Question 15
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|5≤x≤15}. Find the set
(A∩B)∪(A∩Bc).
Solution
Step 1: First, we find A∩B, which is the intersection of sets Aand B.
A∩B={x∈Z|5≤x≤10}={5,6,7,8,9,10}
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Step 2: Next, we find Bc, which is the complement of set B.
Bc={x∈Z|x < 5 or x > 15}={... −3,−2,−1,0,1,2,3,4,16,17,18, ...}
Step 3: Then, we find A∩Bc, which is the intersection of set Aand the
complement of set B.
A∩Bc={x∈Z|2≤x < 5}={2,3,4}
Step 4: Finally, we find (A∩B)∪(A∩Bc), which is the union of A∩Band
A∩Bc.
(A∩B)∪(A∩Bc) = {5,6,7,8,9,10}∪{2,3,4}={2,3,4,5,6,7,8,9,10}
Therefore, (A∩B)∪(A∩Bc) = {2,3,4,5,6,7,8,9,10}.
Question 16
Question
Let A={x∈R| −2≤x≤2}and B={x∈R|0< x ≤4}. Determine the
complement of A∪B.
Solution
To find the complement of A∪B, we first need to find A∪B.
Step 1: Find A∪B.
A∪B={x∈R| −2≤x≤2 or (0 < x ≤4)}
Therefore, A∪B={x∈R| −2≤x≤4}.
Step 2: Find the complement of A∪B. The complement of A∪Bis the
set of all elements that are in the universal set Rbut not in A∪B.
Therefore, the complement of A∪Bis {x∈R|x < −2 or x > 4}.
Question 17
Question
Let A={1,2,3,4}and B={3,4,5,6}. Find (A∪B)∩(A−B).
Solution
Step 1: Find A∪B.
A∪B={1,2,3,4}∪{3,4,5,6}
={1,2,3,4,5,6}
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Step 2: Find A−B.
A−B={1,2,3,4}−{3,4,5,6}
={1,2}
Step 3: Find (A∪B)∩(A−B).
(A∪B)∩(A−B) = {1,2,3,4,5,6}∩{1,2}
={1,2}
Therefore, (A∪B)∩(A−B) = {1,2}.
Question 18
Question
Let A={x∈Z|1≤x≤10},B={x∈Z|6≤x≤15}, and C={x∈Z|
8≤x≤12}. Find the set (A∩B)∪C.
Solution
Step 1: First, find the intersection of sets Aand B:
A∩B={x∈Z|6≤x≤10}
Step 2: Next, find the union of the intersection with set C:
(A∩B)∪C={x∈Z|6≤x≤10}∪{x∈Z|8≤x≤12}
={x∈Z|6≤x≤12}
Step 3: Therefore, the set (A∩B)∪Cis {x∈Z|6≤x≤12}.
Question 19
Question
Let A={x∈R|x2−4x+ 3 >0}and B={x∈R|x2−5x+ 6 <0}. Find
the intersection of sets Aand B.
Solution
To find the intersection of sets Aand B, we need to solve for xin the inequalities
defining each set and then find the values of xthat satisfy both inequalities.
Step 1: Find the values of xthat satisfy the inequality x2−4x+3 >0
in set A.
10
To solve x2−4x+ 3 >0, we first factor the quadratic expression if possible.
In this case, the expression factors as: (x−3)(x−1) >0. Next, we determine
the critical points by setting each factor equal to zero: x−3=0⇒x= 3 and
x−1=0⇒x= 1.
We can now test each interval created by these critical points to determine
when the inequality holds true. The solution is A= (1,3).
Step 2: Find the values of xthat satisfy the inequality x2−5x+6 <0
in set B.
To solve x2−5x+6 <0, we again factor the quadratic expression if possible.
In this case, the expression factors as: (x−2)(x−3) <0. Next, we determine
the critical points by setting each factor equal to zero: x−2=0⇒x= 2 and
x−3=0⇒x= 3.
We can now test each interval created by these critical points to determine
when the inequality holds true. The solution is B= (2,3).
Step 3: Find the intersection of sets Aand B.
To find the values of xthat satisfy both Aand B, we need to find the
intersection of the intervals (1,3) and (2,3). The only intersection point between
the two sets is the interval (2,3).
Therefore, the intersection of sets Aand Bis (2,3).
Question 20
Question
Let A={x|x∈Z,2≤x≤7}and B={x|x∈Z,5≤x≤10}. Determine
A∩Band A∪B.
Solution
Step 1: Determine A∩B. Since A={2,3,4,5,6,7}and B={5,6,7,8,9,10},
we have:
A∩B={x|x∈Aand x∈B}
Therefore, A∩B={5,6,7}.
Step 2: Determine A∪B. The union of sets Aand Bis denoted by A∪B
and is defined as:
A∪B={x|x∈Aor x∈B}
Thus, A∪B={2,3,4,5,6,7,8,9,10}.
Question 21
Question
Let A={x|xis a prime number less than 10}and B={3,5,7,9}. Find the
Cartesian product A×B.
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Solution
Step 1: A={2,3,5,7}since these are the prime numbers less than 10.
Step 2: A×B={(2,3),(2,5),(2,7),(2,9),(3,3),(3,5),(3,7),(3,9),
(5,3),(5,5),(5,7),(5,9),(7,3),(7,5),(7,7),(7,9)}.
Step 3: Therefore, A×B={(2,3),(2,5),(2,7),(2,9),(3,3),(3,5),(3,7),(3,9),
(5,3),(5,5),(5,7),(5,9),(7,3),(7,5),(7,7),(7,9)}.
Question 22
Question
Let A={x∈Z|4≤x≤10}and B={x∈Z|7≤x≤15}. Find the set
A∩B.
Solution
Step 1: First, let’s list the elements of sets Aand B.
A={4,5,6,7,8,9,10}
B={7,8,9,10,11,12,13,14,15}
Step 2: Now, we will find the intersection of sets Aand B, denoted by A∩B,
which contains elements that are in both Aand B.
A∩B={7,8,9,10}
Therefore, the set A∩Bis {7,8,9,10}.
Question 23
Question
Let A={x|x2−5x+ 6 = 0}and B={x|x2−3x−4 = 0}. Find the set
A∩B.
Solution
Step 1: Find the solutions for the equations in set Aand set B.
For set A: The equation x2−5x+6 = 0 can be factored as (x−2)(x−3) = 0.
So, the solutions are x= 2 and x= 3.
For set B: The equation x2−3x−4 = 0 can be factored as (x+1)(x−4) = 0.
So, the solutions are x=−1 and x= 4.
Step 2: Determine the intersection of sets Aand B. The intersection of two
sets Aand Bis the set of elements that are common to both sets. In this case,
A∩B={2,3} ∩ {−1,4}={}.
Therefore, the set A∩Bis the empty set ∅.
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Question 24
Question
Let A={n∈Z|3≤n≤10}and B={n∈Z|6≤n≤15}. Determine A∪B
and A∩B.
Solution
Step 1: To find A∪B, we need to determine all elements that are in either
set Aor set B(or both). Step 2: Set A={3,4,5,6,7,8,9,10}and set B=
{6,7,8,9,10,11,12,13,14,15}. Step 3: The union A∪Bincludes elements that
are in set Aor set B. Step 4: Therefore, A∪B={3,4,5,6,7,8,9,10,11,12,13,14,15}.
Step 5: To find A∩B, we need to determine all elements that are in both set
Aand set B.Step6 : T heintersectionA∩Bincludes elements that are in both
set Aand set B.Step7 : T heref ore,A∩B={6,7,8,9,10}.
Question 25
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Define the set C={x∈Z|1≤
x≤10}. Find (A∪B)∩C.
Solution
Step 1: First, let’s find A∪B, which is the union of sets Aand B.
A∪B={1,2,3,4,5}∪{3,4,5,6,7}
A∪B={1,2,3,4,5,6,7}
Step 2: Next, let’s find (A∪B)∩C, which is the intersection of the set we
found in Step 1 with set C.
(A∪B)∩C={1,2,3,4,5,6,7}∩{1,2,3,4,5,6,7,8,9,10}
(A∪B)∩C={1,2,3,4,5,6,7}
Therefore, (A∪B)∩C={1,2,3,4,5,6,7}.
Question 26
Question
Let A={x∈R|x2<4}and B={x∈R|x > 0}. Determine the set A∩B.
13
Solution
Step 1: We first determine the elements of set A. Since x2<4, we have
−2<x<2. Thus, A= (−2,2).
Step 2: Next, we determine the elements of set B. Since x > 0, the elements
of set Bare B= (0,∞).
Step 3: Now, we find the intersection of sets Aand B.
A∩B= (−2,2) ∩(0,∞)
Step 4: To find the intersection, we take the elements that are common in
both sets. Since 0 <2 and −2<0, the intersection is all the real numbers
greater than 0 but less than 2.
A∩B= (0,2)
Question 27
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|4≤x≤12}. Find the
intersection, union, and set difference A\B.
Solution
Step 1: Find the intersection A∩B.
A={1,2,3,4,5,6,7,8,9,10}
B={4,5,6,7,8,9,10,11,12}
A∩B={4,5,6,7,8,9,10}
Step 2: Find the union A∪B.
A∪B={1,2,3,4,5,6,7,8,9,10,11,12}
Step 3: Find the set difference A\B.
A\B={1,2,3}
Therefore, the intersection A∩B={4,5,6,7,8,9,10}, the union A∪B=
{1,2,3,4,5,6,7,8,9,10,11,12}, and the set difference A\B={1,2,3}.
Question 28
Question
Let A={1,2,3,4,5},B={3,4,5,6,7}, and C={2,4,6,8}. Compute the
following set: (A∩B)∪(A∩C).
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Solution
Step 1: Find A∩B.
A∩B={3,4,5}
Step 2: Find A∩C.
A∩C={2,4}
Step 3: Compute (A∩B)∪(A∩C).
(A∩B)∪(A∩C) = {3,4,5}∪{2,4}
Step 4: Simplify the union of the two sets.
(A∩B)∪(A∩C) = {2,3,4,5}
Therefore, (A∩B)∪(A∩C) = {2,3,4,5}.
Question 29
Question
Let A,B, and Cbe sets such that A⊆B. Prove or disprove the following
statement: (A∩C)⊆(B∩C).
Solution
To prove or disprove the statement (A∩C)⊆(B∩C), we will consider both
cases.
Case 1: Proving the Statement
We will prove that if A⊆B, then (A∩C)⊆(B∩C).
Step 1: Take an arbitrary element xfrom (A∩C). Since x∈(A∩C), we
have x∈Aand x∈C.
Step 2: Since A⊆B, the fact that x∈Aimplies x∈B.
Step 3: Now, we have shown that x∈Band x∈C, which implies x∈
(B∩C).
Step 4: Since our choice of xwas arbitrary, we have shown that if A⊆B,
then (A∩C)⊆(B∩C).
Therefore, the statement (A∩C)⊆(B∩C) is true if A⊆B.
Case 2: Disproving the Statement
We will now show a counterexample to disprove the statement for the general
case where A⊈B.
Step 1: Let A={1,2},B={1,2,3},C={2,3}.
Step 2: Then, (A∩C) = {2}and (B∩C) = {2,3}.
Step 3: It is clear that {2}⊈{2,3}, so the statement (A∩C)⊆(B∩C) is
not true when A⊈B.
Therefore, the statement (A∩C)⊆(B∩C) is true if A⊆B, and false in
general.
15
Solution
Step 1: We first find the solutions to the inequalities x2−4x+ 3 >0 and
x2−5x+ 6 >0.
For x2−4x+ 3 >0, we can factor this as (x−1)(x−3) >0. The solutions
are 1 <x<3.
For x2−5x+ 6 >0, we can factor this as (x−2)(x−3) >0. The solutions
are x < 2 or x > 3.
Step 2: Now, we find the intersection of Aand B.
The solutions to A∩Boccur when both inequalities are simultaneously
satisfied, so we look for the values of xwhich are in both 1 <x<3 and x < 2
or x > 3.
Taking the intersection of the intervals, we have 1 < x < 2 or 3 < x < 3,
which simplifies to 1 <x<2.
Therefore, A∩B={x∈R|1<x<2}.
Question 3
Question
Let A={x|1≤x≤5}and B={x|3≤x≤7}. Find the intersection of the
two sets A∩B.
Solution
Step 1: Write out the elements of sets Aand B.
A={1,2,3,4,5}, B ={3,4,5,6,7}
Step 2: Identify the common elements in sets Aand B.
A∩B={3,4,5}
Therefore, A∩B={3,4,5}.
Question 4
Question
Let A={x∈Z|0≤x≤10}and B={x∈Z|5≤x≤15}. Find the
intersection A∩Band the complement of A∪B.
Solution
Step 1: Find the intersection A∩B.
A∩B={x∈Z|0≤x≤10}∩{x∈Z|5≤x≤15}
={x∈Z|5≤x≤10}
2
Step 2: Find the complement of A∪B. To find the complement of A∪B,
we need to consider all the elements that are not in the set A∪B.
A∪B={x∈Z|0≤x≤10 or 5 ≤x≤15}
={0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
So, the complement of A∪Bis all integers outside the set {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}:
Z− {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
Question 5
Question
Let A={x∈R| −3≤x < 5}and B={x∈R| −1< x ≤4}. Find the sets
(A∩B)∪(A∩Bc) and (A∪B)∩(Ac∪B).
Solution
Let’s first find A∩Band A∩Bc:
A∩B: Elements that are in both Aand B.
A∩Bc: Elements that are in Abut not in B.
Step 1: Find A∩B.
A∩B={x∈R| −3≤x < 5 and −1< x ≤4}
A∩B={x∈R| −1<x<4}
Step 2: Find A∩Bc.
A∩Bc={x∈R| −3≤x < 5 and (x≤ −1 or x > 4)}
A∩Bc={x∈R| −3≤x < −1 or 4 <x<5}
Now, let’s find (A∩B)∪(A∩Bc):
(A∩B)∪(A∩Bc) = {x∈R| −1<x<4 or −3≤x < −1 or 4 <x<5}
(A∩B)∪(A∩Bc) = {x∈R| −3≤x < 5}=A
Next, let’s find (A∪B)∩(Ac∪B):
(A∪B)∩(Ac∪B) = {x∈R|(−3≤x < 5 or −1< x ≤4) and (x < −3 or x≥4)}
(A∪B)∩(Ac∪B) = {x∈R| −3≤x < −1 or 4 ≤x < 5}
Therefore,
(A∩B)∪(A∩Bc) = A
and
(A∪B)∩(Ac∪B) = {x∈R|(−3≤x < −1) or (4 ≤x < 5)}
.
3
Question 6
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Find the
intersection A∩Band the complement of A∩B.
Solution
Step 1: Find A∩B.
A={1,2,3,4,5,6,7,8,9,10}
B={5,6,7,8,9,10,11,12,13,14,15}
A∩B={5,6,7,8,9,10}
Step 2: Find the complement of A∩B. The universal set in this case is
defined as {x∈Z|1≤x≤15}. The complement of A∩Bis given by:
({1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}−{5,6,7,8,9,10})
which simplifies to
{1,2,3,4,11,12,13,14,15}
Question 7
Question
Let A={x∈N|1≤x≤10}and B={x2|x∈N, x < 6}. Find the set A∩B.
Solution
Step 1: Find the elements of set B.
B={x2|x∈N, x < 6}
={12,22,32,42,52}
={1,4,9,16,25}
Step 2: Find the intersection of sets Aand B.
A∩B={x|x∈Aand x∈B}
={1,4,9}
Therefore, A∩B={1,4,9}.
Question 8
Question
Let A={x∈R|1≤x≤5}and B={x∈R|3≤x≤7}. Find the set A∩B.
4
Solution
To find the intersection of sets Aand B, we need to determine the elements
that are common to both sets.
A={x∈R|1≤x≤5}={1,2,3,4,5}
B={x∈R|3≤x≤7}={3,4,5,6,7}
Step 1: List the elements of set Aand set B.
A={1,2,3,4,5}
B={3,4,5,6,7}
Step 2: Identify the common elements in Aand B.
The intersection of sets Aand B, denoted A∩B, is the set containing
elements that are in both Aand B.
A∩B={3,4,5}
Therefore, the set A∩Bconsists of the elements {3,4,5}.
Question 9
Question
Let A={x∈Z:xis a prime number less than 10}and B={2,3,5,7}. Find
the following sets: 1. A∪B2. A∩B3. A\B4. B\A
Solution
1. To find A∪B, we need to find all the elements that are in either set Aor set
B(or both).
A∪B={2,3,5,7}∪{2,3,5,7}={2,3,5,7}
2. To find A∩B, we need to find all the elements that are in both set A
and set B.
A∩B={2,3,5,7}∩{2,3,5,7}={2,3,5,7}
3. To find A\B, we need to find all the elements that are in set Abut not
in set B.
A\B={2,3,5,7}\{2,3,5,7}={}
4. To find B\A, we need to find all the elements that are in set Bbut not
in set A.
B\A={2,3,5,7}\{2,3,5,7}={}
5
Question 10
Question
Let A={x∈Z| −3≤x≤3}and B={x∈Z| −2≤x≤2}. Determine the
set (A∪B)∩(A∩B).
Solution
Step 1: First, find A∪B.
A∪B={−3,−2,−1,0,1,2,3} ∪ {−2,−1,0,1,2}
A∪B={−3,−2,−1,0,1,2,3}
Step 2: Next, find A∩B.
A∩B={−3,−2,−1,0,1,2,3} ∩ {−2,−1,0,1,2}
A∩B={−2,−1,0,1,2}
Step 3: Finally, find (A∪B)∩(A∩B).
(A∪B)∩(A∩B) = {−3,−2,−1,0,1,2,3} ∩ {−2,−1,0,1,2}
(A∪B)∩(A∩B) = {−2,−1,0,1,2}
Therefore, (A∪B)∩(A∩B) = {−2,−1,0,1,2}.
Question 11
Question
Let A={x∈R|1≤x≤5}and B={x∈R|3≤x≤7}. Find the set A∩B.
Solution
Step 1: Find the intersection of sets Aand Busing the given intervals.
A∩B={x∈R|1≤x≤5}∩{x∈R|3≤x≤7}
Step 2: Determine the overlapping interval between the two sets. The in-
tersection of the two intervals will be the set where they overlap, which is [3,5]
since 3 ≤x≤5.
Therefore, the set A∩Bis:
A∩B={x∈R|3≤x≤5}
6
Question 12
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Find each of the
following sets:
a) A∪B
b) A∩B
c) A\B
d) B\A
Solution
a) To find A∪B, we need to take the union of sets Aand B, which includes all
elements that are in Aor in Bor in both.
A∪B={x∈Z|1≤x≤15}
A∪B={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
b) To find A∩B, we need to find the intersection of sets Aand B, which
includes all elements that are in both Aand B.
A∩B={x∈Z|5≤x≤10}
A∩B={5,6,7,8,9,10}
c) To find A\B, we need to find the elements that are in Abut not in B.
A\B={x∈Z|1≤x < 5}
A\B={1,2,3,4}
d) To find B\A, we need to find the elements that are in Bbut not in A.
B\A={x∈Z|11 ≤x≤15}
B\A={11,12,13,14,15}
Question 13
Question
Let A={1,2,3,4}and B={3,4,5,6}. Define the set C={x∈Z|x=
a−b, for some a∈Aand b∈B}. Find C.
7
Solution
Step 1: To find the set C, we need to consider all possible differences between
elements of set Aand set B.
Step 2: Calculate the differences for all pairs of elements (a, b) where a∈A
and b∈B.
Step 3: Differences for (1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6),
(3,3), (3,4), (3,5), (3,6), (4,3), (4,4), (4,5), (4,6) are −2, −3, −4, −5, −1, −2,
−3, −4, 0, −1, −2, −3, 1, 0, −1, −2 respectively.
Step 4: Therefore, set C={−5,−4,−3,−2,−1,0,1}, where Cis the set of
all possible differences between elements of set Aand set B.
Question 14
Question
Let A={x∈Z| −3≤x≤3}and B={x∈Z|2≤x≤7}. Determine the
set A∪Band A∩B.
Solution
Step 1: To find the union of sets Aand B, we need to combine all elements that
are in either set.
A∪B={x∈Z| −3≤x≤7}
Step 2: To find the intersection of sets Aand B, we need to find the common
elements in both sets.
A∩B={x∈Z|2≤x≤3}
Therefore, we have A∪B={−3,−2,−1,0,1,2,3,4,5,6,7}and A∩B=
{2,3}.
Question 15
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|5≤x≤15}. Find the set
(A∩B)∪(A∩Bc).
Solution
Step 1: First, we find A∩B, which is the intersection of sets Aand B.
A∩B={x∈Z|5≤x≤10}={5,6,7,8,9,10}
8
Step 2: Next, we find Bc, which is the complement of set B.
Bc={x∈Z|x < 5 or x > 15}={... −3,−2,−1,0,1,2,3,4,16,17,18, ...}
Step 3: Then, we find A∩Bc, which is the intersection of set Aand the
complement of set B.
A∩Bc={x∈Z|2≤x < 5}={2,3,4}
Step 4: Finally, we find (A∩B)∪(A∩Bc), which is the union of A∩Band
A∩Bc.
(A∩B)∪(A∩Bc) = {5,6,7,8,9,10}∪{2,3,4}={2,3,4,5,6,7,8,9,10}
Therefore, (A∩B)∪(A∩Bc) = {2,3,4,5,6,7,8,9,10}.
Question 16
Question
Let A={x∈R| −2≤x≤2}and B={x∈R|0< x ≤4}. Determine the
complement of A∪B.
Solution
To find the complement of A∪B, we first need to find A∪B.
Step 1: Find A∪B.
A∪B={x∈R| −2≤x≤2 or (0 < x ≤4)}
Therefore, A∪B={x∈R| −2≤x≤4}.
Step 2: Find the complement of A∪B. The complement of A∪Bis the
set of all elements that are in the universal set Rbut not in A∪B.
Therefore, the complement of A∪Bis {x∈R|x < −2 or x > 4}.
Question 17
Question
Let A={1,2,3,4}and B={3,4,5,6}. Find (A∪B)∩(A−B).
Solution
Step 1: Find A∪B.
A∪B={1,2,3,4}∪{3,4,5,6}
={1,2,3,4,5,6}
9
Step 2: Find A−B.
A−B={1,2,3,4}−{3,4,5,6}
={1,2}
Step 3: Find (A∪B)∩(A−B).
(A∪B)∩(A−B) = {1,2,3,4,5,6}∩{1,2}
={1,2}
Therefore, (A∪B)∩(A−B) = {1,2}.
Question 18
Question
Let A={x∈Z|1≤x≤10},B={x∈Z|6≤x≤15}, and C={x∈Z|
8≤x≤12}. Find the set (A∩B)∪C.
Solution
Step 1: First, find the intersection of sets Aand B:
A∩B={x∈Z|6≤x≤10}
Step 2: Next, find the union of the intersection with set C:
(A∩B)∪C={x∈Z|6≤x≤10}∪{x∈Z|8≤x≤12}
={x∈Z|6≤x≤12}
Step 3: Therefore, the set (A∩B)∪Cis {x∈Z|6≤x≤12}.
Question 19
Question
Let A={x∈R|x2−4x+ 3 >0}and B={x∈R|x2−5x+ 6 <0}. Find
the intersection of sets Aand B.
Solution
To find the intersection of sets Aand B, we need to solve for xin the inequalities
defining each set and then find the values of xthat satisfy both inequalities.
Step 1: Find the values of xthat satisfy the inequality x2−4x+3 >0
in set A.
10
To solve x2−4x+ 3 >0, we first factor the quadratic expression if possible.
In this case, the expression factors as: (x−3)(x−1) >0. Next, we determine
the critical points by setting each factor equal to zero: x−3=0⇒x= 3 and
x−1=0⇒x= 1.
We can now test each interval created by these critical points to determine
when the inequality holds true. The solution is A= (1,3).
Step 2: Find the values of xthat satisfy the inequality x2−5x+6 <0
in set B.
To solve x2−5x+6 <0, we again factor the quadratic expression if possible.
In this case, the expression factors as: (x−2)(x−3) <0. Next, we determine
the critical points by setting each factor equal to zero: x−2=0⇒x= 2 and
x−3=0⇒x= 3.
We can now test each interval created by these critical points to determine
when the inequality holds true. The solution is B= (2,3).
Step 3: Find the intersection of sets Aand B.
To find the values of xthat satisfy both Aand B, we need to find the
intersection of the intervals (1,3) and (2,3). The only intersection point between
the two sets is the interval (2,3).
Therefore, the intersection of sets Aand Bis (2,3).
Question 20
Question
Let A={x|x∈Z,2≤x≤7}and B={x|x∈Z,5≤x≤10}. Determine
A∩Band A∪B.
Solution
Step 1: Determine A∩B. Since A={2,3,4,5,6,7}and B={5,6,7,8,9,10},
we have:
A∩B={x|x∈Aand x∈B}
Therefore, A∩B={5,6,7}.
Step 2: Determine A∪B. The union of sets Aand Bis denoted by A∪B
and is defined as:
A∪B={x|x∈Aor x∈B}
Thus, A∪B={2,3,4,5,6,7,8,9,10}.
Question 21
Question
Let A={x|xis a prime number less than 10}and B={3,5,7,9}. Find the
Cartesian product A×B.
11
Solution
Step 1: A={2,3,5,7}since these are the prime numbers less than 10.
Step 2: A×B={(2,3),(2,5),(2,7),(2,9),(3,3),(3,5),(3,7),(3,9),
(5,3),(5,5),(5,7),(5,9),(7,3),(7,5),(7,7),(7,9)}.
Step 3: Therefore, A×B={(2,3),(2,5),(2,7),(2,9),(3,3),(3,5),(3,7),(3,9),
(5,3),(5,5),(5,7),(5,9),(7,3),(7,5),(7,7),(7,9)}.
Question 22
Question
Let A={x∈Z|4≤x≤10}and B={x∈Z|7≤x≤15}. Find the set
A∩B.
Solution
Step 1: First, let’s list the elements of sets Aand B.
A={4,5,6,7,8,9,10}
B={7,8,9,10,11,12,13,14,15}
Step 2: Now, we will find the intersection of sets Aand B, denoted by A∩B,
which contains elements that are in both Aand B.
A∩B={7,8,9,10}
Therefore, the set A∩Bis {7,8,9,10}.
Question 23
Question
Let A={x|x2−5x+ 6 = 0}and B={x|x2−3x−4 = 0}. Find the set
A∩B.
Solution
Step 1: Find the solutions for the equations in set Aand set B.
For set A: The equation x2−5x+6 = 0 can be factored as (x−2)(x−3) = 0.
So, the solutions are x= 2 and x= 3.
For set B: The equation x2−3x−4 = 0 can be factored as (x+1)(x−4) = 0.
So, the solutions are x=−1 and x= 4.
Step 2: Determine the intersection of sets Aand B. The intersection of two
sets Aand Bis the set of elements that are common to both sets. In this case,
A∩B={2,3} ∩ {−1,4}={}.
Therefore, the set A∩Bis the empty set ∅.
12
Question 24
Question
Let A={n∈Z|3≤n≤10}and B={n∈Z|6≤n≤15}. Determine A∪B
and A∩B.
Solution
Step 1: To find A∪B, we need to determine all elements that are in either
set Aor set B(or both). Step 2: Set A={3,4,5,6,7,8,9,10}and set B=
{6,7,8,9,10,11,12,13,14,15}. Step 3: The union A∪Bincludes elements that
are in set Aor set B. Step 4: Therefore, A∪B={3,4,5,6,7,8,9,10,11,12,13,14,15}.
Step 5: To find A∩B, we need to determine all elements that are in both set
Aand set B.Step6 : T heintersectionA∩Bincludes elements that are in both
set Aand set B.Step7 : T heref ore,A∩B={6,7,8,9,10}.
Question 25
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Define the set C={x∈Z|1≤
x≤10}. Find (A∪B)∩C.
Solution
Step 1: First, let’s find A∪B, which is the union of sets Aand B.
A∪B={1,2,3,4,5}∪{3,4,5,6,7}
A∪B={1,2,3,4,5,6,7}
Step 2: Next, let’s find (A∪B)∩C, which is the intersection of the set we
found in Step 1 with set C.
(A∪B)∩C={1,2,3,4,5,6,7}∩{1,2,3,4,5,6,7,8,9,10}
(A∪B)∩C={1,2,3,4,5,6,7}
Therefore, (A∪B)∩C={1,2,3,4,5,6,7}.
Question 26
Question
Let A={x∈R|x2<4}and B={x∈R|x > 0}. Determine the set A∩B.
13
Solution
Step 1: We first determine the elements of set A. Since x2<4, we have
−2<x<2. Thus, A= (−2,2).
Step 2: Next, we determine the elements of set B. Since x > 0, the elements
of set Bare B= (0,∞).
Step 3: Now, we find the intersection of sets Aand B.
A∩B= (−2,2) ∩(0,∞)
Step 4: To find the intersection, we take the elements that are common in
both sets. Since 0 <2 and −2<0, the intersection is all the real numbers
greater than 0 but less than 2.
A∩B= (0,2)
Question 27
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|4≤x≤12}. Find the
intersection, union, and set difference A\B.
Solution
Step 1: Find the intersection A∩B.
A={1,2,3,4,5,6,7,8,9,10}
B={4,5,6,7,8,9,10,11,12}
A∩B={4,5,6,7,8,9,10}
Step 2: Find the union A∪B.
A∪B={1,2,3,4,5,6,7,8,9,10,11,12}
Step 3: Find the set difference A\B.
A\B={1,2,3}
Therefore, the intersection A∩B={4,5,6,7,8,9,10}, the union A∪B=
{1,2,3,4,5,6,7,8,9,10,11,12}, and the set difference A\B={1,2,3}.
Question 28
Question
Let A={1,2,3,4,5},B={3,4,5,6,7}, and C={2,4,6,8}. Compute the
following set: (A∩B)∪(A∩C).
14
Solution
Step 1: Find A∩B.
A∩B={3,4,5}
Step 2: Find A∩C.
A∩C={2,4}
Step 3: Compute (A∩B)∪(A∩C).
(A∩B)∪(A∩C) = {3,4,5}∪{2,4}
Step 4: Simplify the union of the two sets.
(A∩B)∪(A∩C) = {2,3,4,5}
Therefore, (A∩B)∪(A∩C) = {2,3,4,5}.
Question 29
Question
Let A,B, and Cbe sets such that A⊆B. Prove or disprove the following
statement: (A∩C)⊆(B∩C).
Solution
To prove or disprove the statement (A∩C)⊆(B∩C), we will consider both
cases.
Case 1: Proving the Statement
We will prove that if A⊆B, then (A∩C)⊆(B∩C).
Step 1: Take an arbitrary element xfrom (A∩C). Since x∈(A∩C), we
have x∈Aand x∈C.
Step 2: Since A⊆B, the fact that x∈Aimplies x∈B.
Step 3: Now, we have shown that x∈Band x∈C, which implies x∈
(B∩C).
Step 4: Since our choice of xwas arbitrary, we have shown that if A⊆B,
then (A∩C)⊆(B∩C).
Therefore, the statement (A∩C)⊆(B∩C) is true if A⊆B.
Case 2: Disproving the Statement
We will now show a counterexample to disprove the statement for the general
case where A⊈B.
Step 1: Let A={1,2},B={1,2,3},C={2,3}.
Step 2: Then, (A∩C) = {2}and (B∩C) = {2,3}.
Step 3: It is clear that {2}⊈{2,3}, so the statement (A∩C)⊆(B∩C) is
not true when A⊈B.
Therefore, the statement (A∩C)⊆(B∩C) is true if A⊆B, and false in
general.
15
Solution
Step 1: We first find the solutions to the inequalities x2−4x+ 3 >0 and
x2−5x+ 6 >0.
For x2−4x+ 3 >0, we can factor this as (x−1)(x−3) >0. The solutions
are 1 <x<3.
For x2−5x+ 6 >0, we can factor this as (x−2)(x−3) >0. The solutions
are x < 2 or x > 3.
Step 2: Now, we find the intersection of Aand B.
The solutions to A∩Boccur when both inequalities are simultaneously
satisfied, so we look for the values of xwhich are in both 1 <x<3 and x < 2
or x > 3.
Taking the intersection of the intervals, we have 1 < x < 2 or 3 < x < 3,
which simplifies to 1 <x<2.
Therefore, A∩B={x∈R|1<x<2}.
Question 3
Question
Let A={x|1≤x≤5}and B={x|3≤x≤7}. Find the intersection of the
two sets A∩B.
Solution
Step 1: Write out the elements of sets Aand B.
A={1,2,3,4,5}, B ={3,4,5,6,7}
Step 2: Identify the common elements in sets Aand B.
A∩B={3,4,5}
Therefore, A∩B={3,4,5}.
Question 4
Question
Let A={x∈Z|0≤x≤10}and B={x∈Z|5≤x≤15}. Find the
intersection A∩Band the complement of A∪B.
Solution
Step 1: Find the intersection A∩B.
A∩B={x∈Z|0≤x≤10}∩{x∈Z|5≤x≤15}
={x∈Z|5≤x≤10}
2
Step 2: Find the complement of A∪B. To find the complement of A∪B,
we need to consider all the elements that are not in the set A∪B.
A∪B={x∈Z|0≤x≤10 or 5 ≤x≤15}
={0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
So, the complement of A∪Bis all integers outside the set {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}:
Z− {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
Question 5
Question
Let A={x∈R| −3≤x < 5}and B={x∈R| −1< x ≤4}. Find the sets
(A∩B)∪(A∩Bc) and (A∪B)∩(Ac∪B).
Solution
Let’s first find A∩Band A∩Bc:
A∩B: Elements that are in both Aand B.
A∩Bc: Elements that are in Abut not in B.
Step 1: Find A∩B.
A∩B={x∈R| −3≤x < 5 and −1< x ≤4}
A∩B={x∈R| −1<x<4}
Step 2: Find A∩Bc.
A∩Bc={x∈R| −3≤x < 5 and (x≤ −1 or x > 4)}
A∩Bc={x∈R| −3≤x < −1 or 4 <x<5}
Now, let’s find (A∩B)∪(A∩Bc):
(A∩B)∪(A∩Bc) = {x∈R| −1<x<4 or −3≤x < −1 or 4 <x<5}
(A∩B)∪(A∩Bc) = {x∈R| −3≤x < 5}=A
Next, let’s find (A∪B)∩(Ac∪B):
(A∪B)∩(Ac∪B) = {x∈R|(−3≤x < 5 or −1< x ≤4) and (x < −3 or x≥4)}
(A∪B)∩(Ac∪B) = {x∈R| −3≤x < −1 or 4 ≤x < 5}
Therefore,
(A∩B)∪(A∩Bc) = A
and
(A∪B)∩(Ac∪B) = {x∈R|(−3≤x < −1) or (4 ≤x < 5)}
.
3
Question 6
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Find the
intersection A∩Band the complement of A∩B.
Solution
Step 1: Find A∩B.
A={1,2,3,4,5,6,7,8,9,10}
B={5,6,7,8,9,10,11,12,13,14,15}
A∩B={5,6,7,8,9,10}
Step 2: Find the complement of A∩B. The universal set in this case is
defined as {x∈Z|1≤x≤15}. The complement of A∩Bis given by:
({1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}−{5,6,7,8,9,10})
which simplifies to
{1,2,3,4,11,12,13,14,15}
Question 7
Question
Let A={x∈N|1≤x≤10}and B={x2|x∈N, x < 6}. Find the set A∩B.
Solution
Step 1: Find the elements of set B.
B={x2|x∈N, x < 6}
={12,22,32,42,52}
={1,4,9,16,25}
Step 2: Find the intersection of sets Aand B.
A∩B={x|x∈Aand x∈B}
={1,4,9}
Therefore, A∩B={1,4,9}.
Question 8
Question
Let A={x∈R|1≤x≤5}and B={x∈R|3≤x≤7}. Find the set A∩B.
4
Solution
To find the intersection of sets Aand B, we need to determine the elements
that are common to both sets.
A={x∈R|1≤x≤5}={1,2,3,4,5}
B={x∈R|3≤x≤7}={3,4,5,6,7}
Step 1: List the elements of set Aand set B.
A={1,2,3,4,5}
B={3,4,5,6,7}
Step 2: Identify the common elements in Aand B.
The intersection of sets Aand B, denoted A∩B, is the set containing
elements that are in both Aand B.
A∩B={3,4,5}
Therefore, the set A∩Bconsists of the elements {3,4,5}.
Question 9
Question
Let A={x∈Z:xis a prime number less than 10}and B={2,3,5,7}. Find
the following sets: 1. A∪B2. A∩B3. A\B4. B\A
Solution
1. To find A∪B, we need to find all the elements that are in either set Aor set
B(or both).
A∪B={2,3,5,7}∪{2,3,5,7}={2,3,5,7}
2. To find A∩B, we need to find all the elements that are in both set A
and set B.
A∩B={2,3,5,7}∩{2,3,5,7}={2,3,5,7}
3. To find A\B, we need to find all the elements that are in set Abut not
in set B.
A\B={2,3,5,7}\{2,3,5,7}={}
4. To find B\A, we need to find all the elements that are in set Bbut not
in set A.
B\A={2,3,5,7}\{2,3,5,7}={}
5
Question 10
Question
Let A={x∈Z| −3≤x≤3}and B={x∈Z| −2≤x≤2}. Determine the
set (A∪B)∩(A∩B).
Solution
Step 1: First, find A∪B.
A∪B={−3,−2,−1,0,1,2,3} ∪ {−2,−1,0,1,2}
A∪B={−3,−2,−1,0,1,2,3}
Step 2: Next, find A∩B.
A∩B={−3,−2,−1,0,1,2,3} ∩ {−2,−1,0,1,2}
A∩B={−2,−1,0,1,2}
Step 3: Finally, find (A∪B)∩(A∩B).
(A∪B)∩(A∩B) = {−3,−2,−1,0,1,2,3} ∩ {−2,−1,0,1,2}
(A∪B)∩(A∩B) = {−2,−1,0,1,2}
Therefore, (A∪B)∩(A∩B) = {−2,−1,0,1,2}.
Question 11
Question
Let A={x∈R|1≤x≤5}and B={x∈R|3≤x≤7}. Find the set A∩B.
Solution
Step 1: Find the intersection of sets Aand Busing the given intervals.
A∩B={x∈R|1≤x≤5}∩{x∈R|3≤x≤7}
Step 2: Determine the overlapping interval between the two sets. The in-
tersection of the two intervals will be the set where they overlap, which is [3,5]
since 3 ≤x≤5.
Therefore, the set A∩Bis:
A∩B={x∈R|3≤x≤5}
6
Question 12
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Find each of the
following sets:
a) A∪B
b) A∩B
c) A\B
d) B\A
Solution
a) To find A∪B, we need to take the union of sets Aand B, which includes all
elements that are in Aor in Bor in both.
A∪B={x∈Z|1≤x≤15}
A∪B={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
b) To find A∩B, we need to find the intersection of sets Aand B, which
includes all elements that are in both Aand B.
A∩B={x∈Z|5≤x≤10}
A∩B={5,6,7,8,9,10}
c) To find A\B, we need to find the elements that are in Abut not in B.
A\B={x∈Z|1≤x < 5}
A\B={1,2,3,4}
d) To find B\A, we need to find the elements that are in Bbut not in A.
B\A={x∈Z|11 ≤x≤15}
B\A={11,12,13,14,15}
Question 13
Question
Let A={1,2,3,4}and B={3,4,5,6}. Define the set C={x∈Z|x=
a−b, for some a∈Aand b∈B}. Find C.
7
Solution
Step 1: To find the set C, we need to consider all possible differences between
elements of set Aand set B.
Step 2: Calculate the differences for all pairs of elements (a, b) where a∈A
and b∈B.
Step 3: Differences for (1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6),
(3,3), (3,4), (3,5), (3,6), (4,3), (4,4), (4,5), (4,6) are −2, −3, −4, −5, −1, −2,
−3, −4, 0, −1, −2, −3, 1, 0, −1, −2 respectively.
Step 4: Therefore, set C={−5,−4,−3,−2,−1,0,1}, where Cis the set of
all possible differences between elements of set Aand set B.
Question 14
Question
Let A={x∈Z| −3≤x≤3}and B={x∈Z|2≤x≤7}. Determine the
set A∪Band A∩B.
Solution
Step 1: To find the union of sets Aand B, we need to combine all elements that
are in either set.
A∪B={x∈Z| −3≤x≤7}
Step 2: To find the intersection of sets Aand B, we need to find the common
elements in both sets.
A∩B={x∈Z|2≤x≤3}
Therefore, we have A∪B={−3,−2,−1,0,1,2,3,4,5,6,7}and A∩B=
{2,3}.
Question 15
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|5≤x≤15}. Find the set
(A∩B)∪(A∩Bc).
Solution
Step 1: First, we find A∩B, which is the intersection of sets Aand B.
A∩B={x∈Z|5≤x≤10}={5,6,7,8,9,10}
8
Step 2: Next, we find Bc, which is the complement of set B.
Bc={x∈Z|x < 5 or x > 15}={... −3,−2,−1,0,1,2,3,4,16,17,18, ...}
Step 3: Then, we find A∩Bc, which is the intersection of set Aand the
complement of set B.
A∩Bc={x∈Z|2≤x < 5}={2,3,4}
Step 4: Finally, we find (A∩B)∪(A∩Bc), which is the union of A∩Band
A∩Bc.
(A∩B)∪(A∩Bc) = {5,6,7,8,9,10}∪{2,3,4}={2,3,4,5,6,7,8,9,10}
Therefore, (A∩B)∪(A∩Bc) = {2,3,4,5,6,7,8,9,10}.
Question 16
Question
Let A={x∈R| −2≤x≤2}and B={x∈R|0< x ≤4}. Determine the
complement of A∪B.
Solution
To find the complement of A∪B, we first need to find A∪B.
Step 1: Find A∪B.
A∪B={x∈R| −2≤x≤2 or (0 < x ≤4)}
Therefore, A∪B={x∈R| −2≤x≤4}.
Step 2: Find the complement of A∪B. The complement of A∪Bis the
set of all elements that are in the universal set Rbut not in A∪B.
Therefore, the complement of A∪Bis {x∈R|x < −2 or x > 4}.
Question 17
Question
Let A={1,2,3,4}and B={3,4,5,6}. Find (A∪B)∩(A−B).
Solution
Step 1: Find A∪B.
A∪B={1,2,3,4}∪{3,4,5,6}
={1,2,3,4,5,6}
9
Step 2: Find A−B.
A−B={1,2,3,4}−{3,4,5,6}
={1,2}
Step 3: Find (A∪B)∩(A−B).
(A∪B)∩(A−B) = {1,2,3,4,5,6}∩{1,2}
={1,2}
Therefore, (A∪B)∩(A−B) = {1,2}.
Question 18
Question
Let A={x∈Z|1≤x≤10},B={x∈Z|6≤x≤15}, and C={x∈Z|
8≤x≤12}. Find the set (A∩B)∪C.
Solution
Step 1: First, find the intersection of sets Aand B:
A∩B={x∈Z|6≤x≤10}
Step 2: Next, find the union of the intersection with set C:
(A∩B)∪C={x∈Z|6≤x≤10}∪{x∈Z|8≤x≤12}
={x∈Z|6≤x≤12}
Step 3: Therefore, the set (A∩B)∪Cis {x∈Z|6≤x≤12}.
Question 19
Question
Let A={x∈R|x2−4x+ 3 >0}and B={x∈R|x2−5x+ 6 <0}. Find
the intersection of sets Aand B.
Solution
To find the intersection of sets Aand B, we need to solve for xin the inequalities
defining each set and then find the values of xthat satisfy both inequalities.
Step 1: Find the values of xthat satisfy the inequality x2−4x+3 >0
in set A.
10
To solve x2−4x+ 3 >0, we first factor the quadratic expression if possible.
In this case, the expression factors as: (x−3)(x−1) >0. Next, we determine
the critical points by setting each factor equal to zero: x−3=0⇒x= 3 and
x−1=0⇒x= 1.
We can now test each interval created by these critical points to determine
when the inequality holds true. The solution is A= (1,3).
Step 2: Find the values of xthat satisfy the inequality x2−5x+6 <0
in set B.
To solve x2−5x+6 <0, we again factor the quadratic expression if possible.
In this case, the expression factors as: (x−2)(x−3) <0. Next, we determine
the critical points by setting each factor equal to zero: x−2=0⇒x= 2 and
x−3=0⇒x= 3.
We can now test each interval created by these critical points to determine
when the inequality holds true. The solution is B= (2,3).
Step 3: Find the intersection of sets Aand B.
To find the values of xthat satisfy both Aand B, we need to find the
intersection of the intervals (1,3) and (2,3). The only intersection point between
the two sets is the interval (2,3).
Therefore, the intersection of sets Aand Bis (2,3).
Question 20
Question
Let A={x|x∈Z,2≤x≤7}and B={x|x∈Z,5≤x≤10}. Determine
A∩Band A∪B.
Solution
Step 1: Determine A∩B. Since A={2,3,4,5,6,7}and B={5,6,7,8,9,10},
we have:
A∩B={x|x∈Aand x∈B}
Therefore, A∩B={5,6,7}.
Step 2: Determine A∪B. The union of sets Aand Bis denoted by A∪B
and is defined as:
A∪B={x|x∈Aor x∈B}
Thus, A∪B={2,3,4,5,6,7,8,9,10}.
Question 21
Question
Let A={x|xis a prime number less than 10}and B={3,5,7,9}. Find the
Cartesian product A×B.
11
Solution
Step 1: A={2,3,5,7}since these are the prime numbers less than 10.
Step 2: A×B={(2,3),(2,5),(2,7),(2,9),(3,3),(3,5),(3,7),(3,9),
(5,3),(5,5),(5,7),(5,9),(7,3),(7,5),(7,7),(7,9)}.
Step 3: Therefore, A×B={(2,3),(2,5),(2,7),(2,9),(3,3),(3,5),(3,7),(3,9),
(5,3),(5,5),(5,7),(5,9),(7,3),(7,5),(7,7),(7,9)}.
Question 22
Question
Let A={x∈Z|4≤x≤10}and B={x∈Z|7≤x≤15}. Find the set
A∩B.
Solution
Step 1: First, let’s list the elements of sets Aand B.
A={4,5,6,7,8,9,10}
B={7,8,9,10,11,12,13,14,15}
Step 2: Now, we will find the intersection of sets Aand B, denoted by A∩B,
which contains elements that are in both Aand B.
A∩B={7,8,9,10}
Therefore, the set A∩Bis {7,8,9,10}.
Question 23
Question
Let A={x|x2−5x+ 6 = 0}and B={x|x2−3x−4 = 0}. Find the set
A∩B.
Solution
Step 1: Find the solutions for the equations in set Aand set B.
For set A: The equation x2−5x+6 = 0 can be factored as (x−2)(x−3) = 0.
So, the solutions are x= 2 and x= 3.
For set B: The equation x2−3x−4 = 0 can be factored as (x+1)(x−4) = 0.
So, the solutions are x=−1 and x= 4.
Step 2: Determine the intersection of sets Aand B. The intersection of two
sets Aand Bis the set of elements that are common to both sets. In this case,
A∩B={2,3} ∩ {−1,4}={}.
Therefore, the set A∩Bis the empty set ∅.
12
Question 24
Question
Let A={n∈Z|3≤n≤10}and B={n∈Z|6≤n≤15}. Determine A∪B
and A∩B.
Solution
Step 1: To find A∪B, we need to determine all elements that are in either
set Aor set B(or both). Step 2: Set A={3,4,5,6,7,8,9,10}and set B=
{6,7,8,9,10,11,12,13,14,15}. Step 3: The union A∪Bincludes elements that
are in set Aor set B. Step 4: Therefore, A∪B={3,4,5,6,7,8,9,10,11,12,13,14,15}.
Step 5: To find A∩B, we need to determine all elements that are in both set
Aand set B.Step6 : T heintersectionA∩Bincludes elements that are in both
set Aand set B.Step7 : T heref ore,A∩B={6,7,8,9,10}.
Question 25
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Define the set C={x∈Z|1≤
x≤10}. Find (A∪B)∩C.
Solution
Step 1: First, let’s find A∪B, which is the union of sets Aand B.
A∪B={1,2,3,4,5}∪{3,4,5,6,7}
A∪B={1,2,3,4,5,6,7}
Step 2: Next, let’s find (A∪B)∩C, which is the intersection of the set we
found in Step 1 with set C.
(A∪B)∩C={1,2,3,4,5,6,7}∩{1,2,3,4,5,6,7,8,9,10}
(A∪B)∩C={1,2,3,4,5,6,7}
Therefore, (A∪B)∩C={1,2,3,4,5,6,7}.
Question 26
Question
Let A={x∈R|x2<4}and B={x∈R|x > 0}. Determine the set A∩B.
13
Solution
Step 1: We first determine the elements of set A. Since x2<4, we have
−2<x<2. Thus, A= (−2,2).
Step 2: Next, we determine the elements of set B. Since x > 0, the elements
of set Bare B= (0,∞).
Step 3: Now, we find the intersection of sets Aand B.
A∩B= (−2,2) ∩(0,∞)
Step 4: To find the intersection, we take the elements that are common in
both sets. Since 0 <2 and −2<0, the intersection is all the real numbers
greater than 0 but less than 2.
A∩B= (0,2)
Question 27
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|4≤x≤12}. Find the
intersection, union, and set difference A\B.
Solution
Step 1: Find the intersection A∩B.
A={1,2,3,4,5,6,7,8,9,10}
B={4,5,6,7,8,9,10,11,12}
A∩B={4,5,6,7,8,9,10}
Step 2: Find the union A∪B.
A∪B={1,2,3,4,5,6,7,8,9,10,11,12}
Step 3: Find the set difference A\B.
A\B={1,2,3}
Therefore, the intersection A∩B={4,5,6,7,8,9,10}, the union A∪B=
{1,2,3,4,5,6,7,8,9,10,11,12}, and the set difference A\B={1,2,3}.
Question 28
Question
Let A={1,2,3,4,5},B={3,4,5,6,7}, and C={2,4,6,8}. Compute the
following set: (A∩B)∪(A∩C).
14
Solution
Step 1: Find A∩B.
A∩B={3,4,5}
Step 2: Find A∩C.
A∩C={2,4}
Step 3: Compute (A∩B)∪(A∩C).
(A∩B)∪(A∩C) = {3,4,5}∪{2,4}
Step 4: Simplify the union of the two sets.
(A∩B)∪(A∩C) = {2,3,4,5}
Therefore, (A∩B)∪(A∩C) = {2,3,4,5}.
Question 29
Question
Let A,B, and Cbe sets such that A⊆B. Prove or disprove the following
statement: (A∩C)⊆(B∩C).
Solution
To prove or disprove the statement (A∩C)⊆(B∩C), we will consider both
cases.
Case 1: Proving the Statement
We will prove that if A⊆B, then (A∩C)⊆(B∩C).
Step 1: Take an arbitrary element xfrom (A∩C). Since x∈(A∩C), we
have x∈Aand x∈C.
Step 2: Since A⊆B, the fact that x∈Aimplies x∈B.
Step 3: Now, we have shown that x∈Band x∈C, which implies x∈
(B∩C).
Step 4: Since our choice of xwas arbitrary, we have shown that if A⊆B,
then (A∩C)⊆(B∩C).
Therefore, the statement (A∩C)⊆(B∩C) is true if A⊆B.
Case 2: Disproving the Statement
We will now show a counterexample to disprove the statement for the general
case where A⊈B.
Step 1: Let A={1,2},B={1,2,3},C={2,3}.
Step 2: Then, (A∩C) = {2}and (B∩C) = {2,3}.
Step 3: It is clear that {2}⊈{2,3}, so the statement (A∩C)⊆(B∩C) is
not true when A⊈B.
Therefore, the statement (A∩C)⊆(B∩C) is true if A⊆B, and false in
general.
15
Solution
Step 1: We first find the solutions to the inequalities x2−4x+ 3 >0 and
x2−5x+ 6 >0.
For x2−4x+ 3 >0, we can factor this as (x−1)(x−3) >0. The solutions
are 1 <x<3.
For x2−5x+ 6 >0, we can factor this as (x−2)(x−3) >0. The solutions
are x < 2 or x > 3.
Step 2: Now, we find the intersection of Aand B.
The solutions to A∩Boccur when both inequalities are simultaneously
satisfied, so we look for the values of xwhich are in both 1 <x<3 and x < 2
or x > 3.
Taking the intersection of the intervals, we have 1 < x < 2 or 3 < x < 3,
which simplifies to 1 <x<2.
Therefore, A∩B={x∈R|1<x<2}.
Question 3
Question
Let A={x|1≤x≤5}and B={x|3≤x≤7}. Find the intersection of the
two sets A∩B.
Solution
Step 1: Write out the elements of sets Aand B.
A={1,2,3,4,5}, B ={3,4,5,6,7}
Step 2: Identify the common elements in sets Aand B.
A∩B={3,4,5}
Therefore, A∩B={3,4,5}.
Question 4
Question
Let A={x∈Z|0≤x≤10}and B={x∈Z|5≤x≤15}. Find the
intersection A∩Band the complement of A∪B.
Solution
Step 1: Find the intersection A∩B.
A∩B={x∈Z|0≤x≤10}∩{x∈Z|5≤x≤15}
={x∈Z|5≤x≤10}
2
Step 2: Find the complement of A∪B. To find the complement of A∪B,
we need to consider all the elements that are not in the set A∪B.
A∪B={x∈Z|0≤x≤10 or 5 ≤x≤15}
={0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
So, the complement of A∪Bis all integers outside the set {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}:
Z− {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
Question 5
Question
Let A={x∈R| −3≤x < 5}and B={x∈R| −1< x ≤4}. Find the sets
(A∩B)∪(A∩Bc) and (A∪B)∩(Ac∪B).
Solution
Let’s first find A∩Band A∩Bc:
A∩B: Elements that are in both Aand B.
A∩Bc: Elements that are in Abut not in B.
Step 1: Find A∩B.
A∩B={x∈R| −3≤x < 5 and −1< x ≤4}
A∩B={x∈R| −1<x<4}
Step 2: Find A∩Bc.
A∩Bc={x∈R| −3≤x < 5 and (x≤ −1 or x > 4)}
A∩Bc={x∈R| −3≤x < −1 or 4 <x<5}
Now, let’s find (A∩B)∪(A∩Bc):
(A∩B)∪(A∩Bc) = {x∈R| −1<x<4 or −3≤x < −1 or 4 <x<5}
(A∩B)∪(A∩Bc) = {x∈R| −3≤x < 5}=A
Next, let’s find (A∪B)∩(Ac∪B):
(A∪B)∩(Ac∪B) = {x∈R|(−3≤x < 5 or −1< x ≤4) and (x < −3 or x≥4)}
(A∪B)∩(Ac∪B) = {x∈R| −3≤x < −1 or 4 ≤x < 5}
Therefore,
(A∩B)∪(A∩Bc) = A
and
(A∪B)∩(Ac∪B) = {x∈R|(−3≤x < −1) or (4 ≤x < 5)}
.
3
Question 6
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Find the
intersection A∩Band the complement of A∩B.
Solution
Step 1: Find A∩B.
A={1,2,3,4,5,6,7,8,9,10}
B={5,6,7,8,9,10,11,12,13,14,15}
A∩B={5,6,7,8,9,10}
Step 2: Find the complement of A∩B. The universal set in this case is
defined as {x∈Z|1≤x≤15}. The complement of A∩Bis given by:
({1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}−{5,6,7,8,9,10})
which simplifies to
{1,2,3,4,11,12,13,14,15}
Question 7
Question
Let A={x∈N|1≤x≤10}and B={x2|x∈N, x < 6}. Find the set A∩B.
Solution
Step 1: Find the elements of set B.
B={x2|x∈N, x < 6}
={12,22,32,42,52}
={1,4,9,16,25}
Step 2: Find the intersection of sets Aand B.
A∩B={x|x∈Aand x∈B}
={1,4,9}
Therefore, A∩B={1,4,9}.
Question 8
Question
Let A={x∈R|1≤x≤5}and B={x∈R|3≤x≤7}. Find the set A∩B.
4
Solution
To find the intersection of sets Aand B, we need to determine the elements
that are common to both sets.
A={x∈R|1≤x≤5}={1,2,3,4,5}
B={x∈R|3≤x≤7}={3,4,5,6,7}
Step 1: List the elements of set Aand set B.
A={1,2,3,4,5}
B={3,4,5,6,7}
Step 2: Identify the common elements in Aand B.
The intersection of sets Aand B, denoted A∩B, is the set containing
elements that are in both Aand B.
A∩B={3,4,5}
Therefore, the set A∩Bconsists of the elements {3,4,5}.
Question 9
Question
Let A={x∈Z:xis a prime number less than 10}and B={2,3,5,7}. Find
the following sets: 1. A∪B2. A∩B3. A\B4. B\A
Solution
1. To find A∪B, we need to find all the elements that are in either set Aor set
B(or both).
A∪B={2,3,5,7}∪{2,3,5,7}={2,3,5,7}
2. To find A∩B, we need to find all the elements that are in both set A
and set B.
A∩B={2,3,5,7}∩{2,3,5,7}={2,3,5,7}
3. To find A\B, we need to find all the elements that are in set Abut not
in set B.
A\B={2,3,5,7}\{2,3,5,7}={}
4. To find B\A, we need to find all the elements that are in set Bbut not
in set A.
B\A={2,3,5,7}\{2,3,5,7}={}
5
Question 10
Question
Let A={x∈Z| −3≤x≤3}and B={x∈Z| −2≤x≤2}. Determine the
set (A∪B)∩(A∩B).
Solution
Step 1: First, find A∪B.
A∪B={−3,−2,−1,0,1,2,3} ∪ {−2,−1,0,1,2}
A∪B={−3,−2,−1,0,1,2,3}
Step 2: Next, find A∩B.
A∩B={−3,−2,−1,0,1,2,3} ∩ {−2,−1,0,1,2}
A∩B={−2,−1,0,1,2}
Step 3: Finally, find (A∪B)∩(A∩B).
(A∪B)∩(A∩B) = {−3,−2,−1,0,1,2,3} ∩ {−2,−1,0,1,2}
(A∪B)∩(A∩B) = {−2,−1,0,1,2}
Therefore, (A∪B)∩(A∩B) = {−2,−1,0,1,2}.
Question 11
Question
Let A={x∈R|1≤x≤5}and B={x∈R|3≤x≤7}. Find the set A∩B.
Solution
Step 1: Find the intersection of sets Aand Busing the given intervals.
A∩B={x∈R|1≤x≤5}∩{x∈R|3≤x≤7}
Step 2: Determine the overlapping interval between the two sets. The in-
tersection of the two intervals will be the set where they overlap, which is [3,5]
since 3 ≤x≤5.
Therefore, the set A∩Bis:
A∩B={x∈R|3≤x≤5}
6
Question 12
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|5≤x≤15}. Find each of the
following sets:
a) A∪B
b) A∩B
c) A\B
d) B\A
Solution
a) To find A∪B, we need to take the union of sets Aand B, which includes all
elements that are in Aor in Bor in both.
A∪B={x∈Z|1≤x≤15}
A∪B={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
b) To find A∩B, we need to find the intersection of sets Aand B, which
includes all elements that are in both Aand B.
A∩B={x∈Z|5≤x≤10}
A∩B={5,6,7,8,9,10}
c) To find A\B, we need to find the elements that are in Abut not in B.
A\B={x∈Z|1≤x < 5}
A\B={1,2,3,4}
d) To find B\A, we need to find the elements that are in Bbut not in A.
B\A={x∈Z|11 ≤x≤15}
B\A={11,12,13,14,15}
Question 13
Question
Let A={1,2,3,4}and B={3,4,5,6}. Define the set C={x∈Z|x=
a−b, for some a∈Aand b∈B}. Find C.
7
Solution
Step 1: To find the set C, we need to consider all possible differences between
elements of set Aand set B.
Step 2: Calculate the differences for all pairs of elements (a, b) where a∈A
and b∈B.
Step 3: Differences for (1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6),
(3,3), (3,4), (3,5), (3,6), (4,3), (4,4), (4,5), (4,6) are −2, −3, −4, −5, −1, −2,
−3, −4, 0, −1, −2, −3, 1, 0, −1, −2 respectively.
Step 4: Therefore, set C={−5,−4,−3,−2,−1,0,1}, where Cis the set of
all possible differences between elements of set Aand set B.
Question 14
Question
Let A={x∈Z| −3≤x≤3}and B={x∈Z|2≤x≤7}. Determine the
set A∪Band A∩B.
Solution
Step 1: To find the union of sets Aand B, we need to combine all elements that
are in either set.
A∪B={x∈Z| −3≤x≤7}
Step 2: To find the intersection of sets Aand B, we need to find the common
elements in both sets.
A∩B={x∈Z|2≤x≤3}
Therefore, we have A∪B={−3,−2,−1,0,1,2,3,4,5,6,7}and A∩B=
{2,3}.
Question 15
Question
Let A={x∈Z|2≤x≤10}and B={x∈Z|5≤x≤15}. Find the set
(A∩B)∪(A∩Bc).
Solution
Step 1: First, we find A∩B, which is the intersection of sets Aand B.
A∩B={x∈Z|5≤x≤10}={5,6,7,8,9,10}
8
Step 2: Next, we find Bc, which is the complement of set B.
Bc={x∈Z|x < 5 or x > 15}={... −3,−2,−1,0,1,2,3,4,16,17,18, ...}
Step 3: Then, we find A∩Bc, which is the intersection of set Aand the
complement of set B.
A∩Bc={x∈Z|2≤x < 5}={2,3,4}
Step 4: Finally, we find (A∩B)∪(A∩Bc), which is the union of A∩Band
A∩Bc.
(A∩B)∪(A∩Bc) = {5,6,7,8,9,10}∪{2,3,4}={2,3,4,5,6,7,8,9,10}
Therefore, (A∩B)∪(A∩Bc) = {2,3,4,5,6,7,8,9,10}.
Question 16
Question
Let A={x∈R| −2≤x≤2}and B={x∈R|0< x ≤4}. Determine the
complement of A∪B.
Solution
To find the complement of A∪B, we first need to find A∪B.
Step 1: Find A∪B.
A∪B={x∈R| −2≤x≤2 or (0 < x ≤4)}
Therefore, A∪B={x∈R| −2≤x≤4}.
Step 2: Find the complement of A∪B. The complement of A∪Bis the
set of all elements that are in the universal set Rbut not in A∪B.
Therefore, the complement of A∪Bis {x∈R|x < −2 or x > 4}.
Question 17
Question
Let A={1,2,3,4}and B={3,4,5,6}. Find (A∪B)∩(A−B).
Solution
Step 1: Find A∪B.
A∪B={1,2,3,4}∪{3,4,5,6}
={1,2,3,4,5,6}
9
Step 2: Find A−B.
A−B={1,2,3,4}−{3,4,5,6}
={1,2}
Step 3: Find (A∪B)∩(A−B).
(A∪B)∩(A−B) = {1,2,3,4,5,6}∩{1,2}
={1,2}
Therefore, (A∪B)∩(A−B) = {1,2}.
Question 18
Question
Let A={x∈Z|1≤x≤10},B={x∈Z|6≤x≤15}, and C={x∈Z|
8≤x≤12}. Find the set (A∩B)∪C.
Solution
Step 1: First, find the intersection of sets Aand B:
A∩B={x∈Z|6≤x≤10}
Step 2: Next, find the union of the intersection with set C:
(A∩B)∪C={x∈Z|6≤x≤10}∪{x∈Z|8≤x≤12}
={x∈Z|6≤x≤12}
Step 3: Therefore, the set (A∩B)∪Cis {x∈Z|6≤x≤12}.
Question 19
Question
Let A={x∈R|x2−4x+ 3 >0}and B={x∈R|x2−5x+ 6 <0}. Find
the intersection of sets Aand B.
Solution
To find the intersection of sets Aand B, we need to solve for xin the inequalities
defining each set and then find the values of xthat satisfy both inequalities.
Step 1: Find the values of xthat satisfy the inequality x2−4x+3 >0
in set A.
10
To solve x2−4x+ 3 >0, we first factor the quadratic expression if possible.
In this case, the expression factors as: (x−3)(x−1) >0. Next, we determine
the critical points by setting each factor equal to zero: x−3=0⇒x= 3 and
x−1=0⇒x= 1.
We can now test each interval created by these critical points to determine
when the inequality holds true. The solution is A= (1,3).
Step 2: Find the values of xthat satisfy the inequality x2−5x+6 <0
in set B.
To solve x2−5x+6 <0, we again factor the quadratic expression if possible.
In this case, the expression factors as: (x−2)(x−3) <0. Next, we determine
the critical points by setting each factor equal to zero: x−2=0⇒x= 2 and
x−3=0⇒x= 3.
We can now test each interval created by these critical points to determine
when the inequality holds true. The solution is B= (2,3).
Step 3: Find the intersection of sets Aand B.
To find the values of xthat satisfy both Aand B, we need to find the
intersection of the intervals (1,3) and (2,3). The only intersection point between
the two sets is the interval (2,3).
Therefore, the intersection of sets Aand Bis (2,3).
Question 20
Question
Let A={x|x∈Z,2≤x≤7}and B={x|x∈Z,5≤x≤10}. Determine
A∩Band A∪B.
Solution
Step 1: Determine A∩B. Since A={2,3,4,5,6,7}and B={5,6,7,8,9,10},
we have:
A∩B={x|x∈Aand x∈B}
Therefore, A∩B={5,6,7}.
Step 2: Determine A∪B. The union of sets Aand Bis denoted by A∪B
and is defined as:
A∪B={x|x∈Aor x∈B}
Thus, A∪B={2,3,4,5,6,7,8,9,10}.
Question 21
Question
Let A={x|xis a prime number less than 10}and B={3,5,7,9}. Find the
Cartesian product A×B.
11
Solution
Step 1: A={2,3,5,7}since these are the prime numbers less than 10.
Step 2: A×B={(2,3),(2,5),(2,7),(2,9),(3,3),(3,5),(3,7),(3,9),
(5,3),(5,5),(5,7),(5,9),(7,3),(7,5),(7,7),(7,9)}.
Step 3: Therefore, A×B={(2,3),(2,5),(2,7),(2,9),(3,3),(3,5),(3,7),(3,9),
(5,3),(5,5),(5,7),(5,9),(7,3),(7,5),(7,7),(7,9)}.
Question 22
Question
Let A={x∈Z|4≤x≤10}and B={x∈Z|7≤x≤15}. Find the set
A∩B.
Solution
Step 1: First, let’s list the elements of sets Aand B.
A={4,5,6,7,8,9,10}
B={7,8,9,10,11,12,13,14,15}
Step 2: Now, we will find the intersection of sets Aand B, denoted by A∩B,
which contains elements that are in both Aand B.
A∩B={7,8,9,10}
Therefore, the set A∩Bis {7,8,9,10}.
Question 23
Question
Let A={x|x2−5x+ 6 = 0}and B={x|x2−3x−4 = 0}. Find the set
A∩B.
Solution
Step 1: Find the solutions for the equations in set Aand set B.
For set A: The equation x2−5x+6 = 0 can be factored as (x−2)(x−3) = 0.
So, the solutions are x= 2 and x= 3.
For set B: The equation x2−3x−4 = 0 can be factored as (x+1)(x−4) = 0.
So, the solutions are x=−1 and x= 4.
Step 2: Determine the intersection of sets Aand B. The intersection of two
sets Aand Bis the set of elements that are common to both sets. In this case,
A∩B={2,3} ∩ {−1,4}={}.
Therefore, the set A∩Bis the empty set ∅.
12
Question 24
Question
Let A={n∈Z|3≤n≤10}and B={n∈Z|6≤n≤15}. Determine A∪B
and A∩B.
Solution
Step 1: To find A∪B, we need to determine all elements that are in either
set Aor set B(or both). Step 2: Set A={3,4,5,6,7,8,9,10}and set B=
{6,7,8,9,10,11,12,13,14,15}. Step 3: The union A∪Bincludes elements that
are in set Aor set B. Step 4: Therefore, A∪B={3,4,5,6,7,8,9,10,11,12,13,14,15}.
Step 5: To find A∩B, we need to determine all elements that are in both set
Aand set B.Step6 : T heintersectionA∩Bincludes elements that are in both
set Aand set B.Step7 : T heref ore,A∩B={6,7,8,9,10}.
Question 25
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Define the set C={x∈Z|1≤
x≤10}. Find (A∪B)∩C.
Solution
Step 1: First, let’s find A∪B, which is the union of sets Aand B.
A∪B={1,2,3,4,5}∪{3,4,5,6,7}
A∪B={1,2,3,4,5,6,7}
Step 2: Next, let’s find (A∪B)∩C, which is the intersection of the set we
found in Step 1 with set C.
(A∪B)∩C={1,2,3,4,5,6,7}∩{1,2,3,4,5,6,7,8,9,10}
(A∪B)∩C={1,2,3,4,5,6,7}
Therefore, (A∪B)∩C={1,2,3,4,5,6,7}.
Question 26
Question
Let A={x∈R|x2<4}and B={x∈R|x > 0}. Determine the set A∩B.
13
Solution
Step 1: We first determine the elements of set A. Since x2<4, we have
−2<x<2. Thus, A= (−2,2).
Step 2: Next, we determine the elements of set B. Since x > 0, the elements
of set Bare B= (0,∞).
Step 3: Now, we find the intersection of sets Aand B.
A∩B= (−2,2) ∩(0,∞)
Step 4: To find the intersection, we take the elements that are common in
both sets. Since 0 <2 and −2<0, the intersection is all the real numbers
greater than 0 but less than 2.
A∩B= (0,2)
Question 27
Question
Let A={x∈Z|1≤x≤10}and B={x∈Z|4≤x≤12}. Find the
intersection, union, and set difference A\B.
Solution
Step 1: Find the intersection A∩B.
A={1,2,3,4,5,6,7,8,9,10}
B={4,5,6,7,8,9,10,11,12}
A∩B={4,5,6,7,8,9,10}
Step 2: Find the union A∪B.
A∪B={1,2,3,4,5,6,7,8,9,10,11,12}
Step 3: Find the set difference A\B.
A\B={1,2,3}
Therefore, the intersection A∩B={4,5,6,7,8,9,10}, the union A∪B=
{1,2,3,4,5,6,7,8,9,10,11,12}, and the set difference A\B={1,2,3}.
Question 28
Question
Let A={1,2,3,4,5},B={3,4,5,6,7}, and C={2,4,6,8}. Compute the
following set: (A∩B)∪(A∩C).
14
Solution
Step 1: Find A∩B.
A∩B={3,4,5}
Step 2: Find A∩C.
A∩C={2,4}
Step 3: Compute (A∩B)∪(A∩C).
(A∩B)∪(A∩C) = {3,4,5}∪{2,4}
Step 4: Simplify the union of the two sets.
(A∩B)∪(A∩C) = {2,3,4,5}
Therefore, (A∩B)∪(A∩C) = {2,3,4,5}.
Question 29
Question
Let A,B, and Cbe sets such that A⊆B. Prove or disprove the following
statement: (A∩C)⊆(B∩C).
Solution
To prove or disprove the statement (A∩C)⊆(B∩C), we will consider both
cases.
Case 1: Proving the Statement
We will prove that if A⊆B, then (A∩C)⊆(B∩C).
Step 1: Take an arbitrary element xfrom (A∩C). Since x∈(A∩C), we
have x∈Aand x∈C.
Step 2: Since A⊆B, the fact that x∈Aimplies x∈B.
Step 3: Now, we have shown that x∈Band x∈C, which implies x∈
(B∩C).
Step 4: Since our choice of xwas arbitrary, we have shown that if A⊆B,
then (A∩C)⊆(B∩C).
Therefore, the statement (A∩C)⊆(B∩C) is true if A⊆B.
Case 2: Disproving the Statement
We will now show a counterexample to disprove the statement for the general
case where A⊈B.
Step 1: Let A={1,2},B={1,2,3},C={2,3}.
Step 2: Then, (A∩C) = {2}and (B∩C) = {2,3}.
Step 3: It is clear that {2}⊈{2,3}, so the statement (A∩C)⊆(B∩C) is
not true when A⊈B.
Therefore, the statement (A∩C)⊆(B∩C) is true if A⊆B, and false in
general.
15
Question 30
Question
Let A={1,2,3,4,5}and B={3,4,5,6,7}. Define the set C=A∪Band
D=A∩B. Find the set Esuch that E= (A∪B)−(A∩B).
Solution
Step 1: Find C=A∪B
C=A∪B={1,2,3,4,5}∪{3,4,5,6,7}={1,2,3,4,5,6,7}
Step 2: Find D=A∩B
D=A∩B={1,2,3,4,5}∩{3,4,5,6,7}={3,4,5}
Step 3: Find E= (A∪B)−(A∩B)
E= (A∪B)−(A∩B) = {1,2,3,4,5,6,7}−{3,4,5}={1,2,6,7}
Therefore, the set Eis {1,2,6,7}.
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