Homework1-Solution Key
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Course
2000
Subject
Mathematics
Date
Apr 3, 2024
Type
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15
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Homework 1 - Solution Key
Due Date: September 6, 2023
•
Please submit your answers as a single consolidated PDF file.
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•
Questions 1 through 23 are valued at 4 points each.
•
Question 24 is valued at 8 points.
1.
Which of the following sets are equal?
A
=
{
a, b, c, d
}
B
=
{
d, e, a, c
}
C
=
{
d, b, a, c
}
D
=
{
a, a, d, e, c, e
}
Solution:
Two sets are equal if and only if they contain exactly the same elements,
regardless of their order or repetitions.
(a) Comparing sets
A
and
B
:
Set
B
contains an element
e
which is not present in set
A
. Therefore,
A
and
B
are not equal.
(b) Comparing sets
A
and
C
:
Both sets contain the elements
a, b, c,
and
d
(though in different or-
ders). Hence,
A
and
C
are equal.
(c) Comparing sets
A
and
D
:
Set
D
contains the element
e
, and even though there are repetitions
of the elements
a
and
e
, repetitions do not make any difference in set
theory. However, the presence of the element
e
in
D
and its absence
in
A
means that
A
and
D
are not equal.
(d) Comparing sets
B
and
C
:
Set
B
contains an element
e
which is not present in set
C
. Therefore,
B
and
C
are not equal.
1
(e) Comparing sets
B
and
D
:
Both sets contain the elements
a, c, d,
and
e
(though
B
does not have
repetitions). Hence,
B
and
D
are equal.
(f) Comparing sets
C
and
D
:
Set
D
contains the element
e
which is not present in set
C
. Therefore,
C
and
D
are not equal.
In summary A and C are equal, and B and D are equal.
2. Write in words how to read each of the following out loud:
(a)
{
x
∈
R
+
|
0
< x <
1
}
(b)
{
x
∈
R
|
x
≤
0 or
x
≥
1
}
(c)
{
n
∈
Z
|
n
is a factor of 6
}
(d)
{
n
∈
Z
+
|
n
is a factor of 6
}
Solution:
(a) The set of all positive real numbers
x
such that
x
is greater than 0
and less than 1.
(b) The set of all real numbers
x
such that
x
is less than or equal to 0
or
x
is greater than or equal to 1.
(c) The set of all integers
n
such that
n
is a factor of 6.
(d) The set of all positive integers
n
such that
n
is a factor of 6.
3.
(a) Is 2
∈ {
2
}
?
(b) How many elements are in the set
{
2
,
2
,
2
,
2
}
?
(c) How many elements are in the set
{
0
,
{
0
}}
?
(d) Is
{
0
} ∈ {{
0
}
,
{
1
}}
?
(e) Is 0
∈ {{
0
}
,
{
1
}}
?
Solution:
(a) Yes, 2 is an element of the set
{
2
}
.
(b) In set theory, duplicate elements in a set are not counted multiple
times. The set
{
2
,
2
,
2
,
2
}
is equivalent to the set
{
2
}
. Thus, there is
only 1 element in the set.
(c) The set
{
0
,
{
0
}}
contains two distinct elements: the number 0 and
the set containing the number 0. Thus, there are 2 elements in the
set.
(d) Yes,
{
0
}
is an element of the set
{{
0
}
,
{
1
}}
.
(e) No, 0 is not an element of the set
{{
0
}
,
{
1
}}
.
Instead,
{
0
}
is an
element of the set, which means that the set contains the set that
has the number 0, not the number 0 itself.
2
4. Which of the following sets are equal?
A
=
{
0
,
1
,
2
}
B
=
{
x
∈
R
| −
1
≤
x <
3
}
C
=
{
x
∈
R
| −
1
< x <
3
}
D
=
{
x
∈
Z
| −
1
< x <
3
}
E
=
{
x
∈
Z
+
| −
1
< x <
3
}
Solution:
A
and
D
are equal as both contain the integers 0, 1, and 2.
E
is a subset of
A
and
D
, but not equal to them since it only contains pos-
itive integers.
B
and
C
are not equal to any of the sets above because they
contain all real numbers between their specified ranges, not just integers.
5. For each integer
n
, let
T
n
=
{
n, n
2
}
. How many elements are in each of
T
2
, T
−
3
, T
1
and
T
0
? Justify your answers.
Solution:
For each set
T
n
:
T
2
=
{
2
,
2
2
}
=
{
2
,
4
}
contains 2 distinct elements: 2 and 4.
T
−
3
=
{−
3
,
(
−
3)
2
}
=
{−
3
,
9
}
contains 2 distinct elements: -3 and 9.
T
1
=
{
1
,
1
2
}
=
{
1
,
1
}
contains only 1 distinct element: 1.
T
0
=
{
0
,
0
2
}
=
{
0
,
0
}
contains only 1 distinct element: 0.
6. Use the set-roster notation to indicate the elements in each of the following
sets.
(a)
S
=
{
n
∈
Z
|
n
= (
−
1)
k
,
for some integer
k
}
.
(b)
T
=
{
m
∈
Z
|
m
= 1 + (
−
1)
i
,
for some integer
i
}
.
(c)
U
=
{
r
∈
Z
|
2
≤
r
≤ −
2
}
.
(d)
V
=
{
s
∈
Z
|
s >
2 or
s <
3
}
.
(e)
W
=
{
t
∈
Z
|
1
< t <
−
3
}
.
(f)
X
=
{
u
∈
Z
|
u
≤
4 or
u
≥
1
}
.
Solution:
(a)
S
represents numbers that are powers of
−
1. For even powers, the
value is 1, and for odd powers, the value is -1. So,
S
=
{−
1
,
1
}
.
(b) Adding 1 to powers of
−
1 gives us 2 for even powers and 0 for odd
powers. So,
T
=
{
0
,
2
}
.
(c) There are no integers between 2 and -2 inclusive. So,
U
=
∅
.
(d) This set includes all integers. So,
V
=
Z
.
(e) There are no integers between 1 and -3 exclusive. So,
W
=
∅
.
(f) This set includes all integers.
3
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7. Let
A
=
{
c, d, f, g
}
,
B
=
{
f, j
}
, and
C
=
{
d, g
}
.
Answer each of the
following questions. Give reasons for your answers.
(a)
Is
B
⊆
A
?
(b)
Is
C
⊆
A
?
(c)
Is
C
⊆
C
?
(d)
Is
C
a proper subset of
A
?
Solution:
(a) No, because the element
j
is in
B
but not in
A
.
(b) Yes, because every element in
C
(that is,
d
and
g
) is also in
A
.
(c) Yes, because every set is a subset of itself.
(d) Yes, because every element in
C
is also in
A
, and
A
has elements not
in
C
.
8.
(a) Is 3
∈ {
1
,
2
,
3
}
?
(b) Is 1
⊆ {
1
}
?
(c) Is
{
2
} ∈ {
1
,
2
}
?
(d) Is
{
3
} ∈ {
1
,
{
2
}
,
{
3
}}
?
(e) Is 1
∈ {
1
}
?
(f) Is
{
2
} ⊆ {
1
,
{
2
}
,
{
3
}}
?
(g) Is
{
1
} ⊆ {
1
,
2
}
?
(h) Is 1
∈ {{
1
}
,
2
}
?
(i) Is
{
1
} ⊆ {
1
,
{
2
}}
?
(j) Is
{
1
} ⊆ {
1
}
?
Solution:
(a) Yes, because 3 is an element of the set
{
1
,
2
,
3
}
.
(b) No, because 1 is an element and not a subset.
(c) No, because
{
2
}
is a set and there are no sets as elements in
{
1
,
2
}
.
(d) Yes, because the set
{
3
}
is an element of the set
{
1
,
{
2
}
,
{
3
}}
.
(e) Yes, because 1 is an element of the set
{
1
}
.
(f) No, because
{
2
}
is a set and is not a subset of
{
1
,
{
2
}
,
{
3
}}
. It’s an
element, not a subset.
(g) Yes,
{
1
}
is a subset of
{
1
,
2
}
.
(h) No, because while the set
{
1
}
is an element, 1 is not.
(i) Yes. To determine if
{
1
} ⊆ {
1
,
{
2
}}
, we check if every element in
{
1
}
is also an element of
{
1
,
{
2
}}
. Since 1 is an element of
{
1
,
{
2
}}
, the
statement is true.
4
(j) Yes. A set is always a subset of itself. Therefore,
{
1
} ⊆ {
1
}
is true.
9.
(a) Is ((
−
2)
2
,
−
2
2
) = (
−
2
2
,
(
−
2)
2
)?
(b) Is (5
,
−
5) = (
−
5
,
5)?
(c) Is
(
3
√
8
−
9
,
3
√
−
1
)
= (
−
1
,
−
1)?
(d) Is
−
2
−
4
,
(
−
2)
3
=
(
3
6
,
−
8
)
?
Solution:
(a) No, because (4
,
−
4)
̸
= (
−
4
,
4).
(b) No, because (5
,
−
5)
̸
= (
−
5
,
5).
(c) Yes, both ordered pairs are (
−
1
,
−
1).
(d) Yes, both ordered pairs are (0
.
5
,
−
8).
10. Let
A
=
{
w, x, y, z
}
and
B
=
{
a, b
}
.
Use the set-roster notation to write each of the following sets, and indicate
the number of elements that are in each set:
(a)
A
×
B
(b)
B
×
A
(c)
A
×
A
(d)
B
×
B
Solution:
(a)
A
×
B
=
{
(
w, a
)
,
(
w, b
)
,
(
x, a
)
,
(
x, b
)
,
(
y, a
)
,
(
y, b
)
,
(
z, a
)
,
(
z, b
)
}
Number of elements: 8.
(b)
B
×
A
=
{
(
a, w
)
,
(
a, x
)
,
(
a, y
)
,
(
a, z
)
,
(
b, w
)
,
(
b, x
)
,
(
b, y
)
,
(
b, z
)
}
Number of elements: 8.
(c)
A
×
A
=
{
(
w, w
)
,
(
w, x
)
,
(
w, y
)
,
(
w, z
)
,
(
x, w
)
,
(
x, x
)
,
(
x, y
)
,
(
x, z
)
,
(
y, w
)
,
(
y, x
)
,
(
y, y
)
,
(
y, z
)
,
(
z, w
)
,
(
z, x
)
,
(
z, y
)
,
(
z, z
)
}
Number of elements: 16.
(d)
B
×
B
=
{
(
a, a
)
,
(
a, b
)
,
(
b, a
)
,
(
b, b
)
}
Number of elements: 4.
11. Determine if the following sets are finite or infinite:
(a)
A
=
{
x
∈
R
|
x
2
= 4
}
(b)
B
=
{
x
∈
Z
|
x
2
≤
25
}
(c)
C
=
{
x
∈
R
|
x
2
=
x
}
Solution:
(a) The solutions to the equation
x
2
= 4 are
x
= 2 and
x
=
−
2. Hence,
A
=
{
2
,
−
2
}
which is finite.
5
(b) The integers that satisfy this inequality are
−
5
,
−
4
,
−
3
,
−
2
,
−
1
,
0
,
1
,
2
,
3
,
4
,
5.
Thus,
B
is finite with 11 elements.
(c) The solutions to the equation
x
2
=
x
are
x
= 0 and
x
= 1.
So,
C
=
{
0
,
1
}
which is finite.
12. Identify if the following sets are subsets of
N
,
Z
,
Q
, or
R
:
(a)
A
=
{
1
,
2
,
3
,
4
}
(b)
B
=
{−
3
,
−
2
,
−
1
}
(c)
C
=
{
1
2
,
1
,
3
2
}
(d)
D
=
{
π, e
}
Solution:
(a)
A
is a subset of
N
,
Z
,
Q
, and
R
.
(b)
B
is a subset of
Z
,
Q
, and
R
but not
N
.
(c)
C
is a subset of
Q
and
R
but not
N
or
Z
.
(d)
D
is a subset of
R
but not
N
,
Z
, or
Q
. Both
π
and
e
are irrational
numbers, meaning they cannot be expressed as fractions of integers.
13. For each integer
k
, let
M
k
=
{
k
−
1
, k
+ 1
}
. List the elements of:
(a)
M
5
(b)
M
−
2
(c)
M
0
Solution:
(a)
M
5
=
{
4
,
6
}
(b)
M
−
2
=
{−
3
,
−
1
}
(c)
M
0
=
{−
1
,
1
}
14. Define the set using set-builder notation:
(a)
S
=
{
2
,
4
,
6
,
8
,
10
}
(b)
T
=
{
1
,
4
,
9
,
16
,
25
}
Solution:
(a)
S
=
{
x
|
x
= 2
n,
1
≤
n
≤
5
, n
∈
Z
}
(b)
T
=
{
x
|
x
=
n
2
,
1
≤
n
≤
5
, n
∈
Z
}
15. If
A
=
{
a, b, c
}
, list all possible two-element subsets of
A
.
Solution:
(a)
{
a, b
}
6
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(b)
{
a, c
}
(c)
{
b, c
}
16. Let
A
=
{
2
,
3
,
4
}
and
B
=
{
6
,
8
,
10
}
. Define a relation
R
from
A
to
B
as
follows:
For all (
x, y
)
∈
A
×
B,
(
x, y
)
∈
R
if and only if
y
x
is an integer.
(a) Determine the validity of the following:
Is 4
R
6?
Is 4
R
8?
Is (3
,
8)
∈
R
?
Is (2
,
10)
∈
R
?
(b) Write
R
as a set of ordered pairs.
(c) Identify the domain and co-domain of
R
.
(d) Draw an arrow diagram for
R
.
Solution:
(a)
4
R
6 is false because
6
4
= 1
.
5 which is not an integer.
4
R
8 is true because
8
4
= 2 which is an integer.
(3
,
8)
/
∈
R
because
8
3
is not an integer.
(2
,
10)
∈
R
because
10
2
= 5 which is an integer.
(b) The relation
R
as a set of ordered pairs is:
R
=
{
(2
,
6)
,
(2
,
8)
,
(2
,
10)
,
(3
,
6)
,
(4
,
8)
}
(c) The domain of
R
is:
{
2, 3, 4
}
and the co-domain is:
{
6, 8, 10
}
.
(d) Diagram:
17. Let
C
=
D
=
{−
3
,
−
2
,
−
1
,
1
,
2
,
3
}
. Define a relation
S
from
C
to
D
such
that for all (
x, y
)
∈
C
×
D
, (
x, y
)
∈
S
if and only if
1
x
−
1
y
is an integer.
7
(a) Determine the validity of the following:
Is 2
S
2?
Is
−
1
S
−
1?
Is (3
,
3)
∈
S
?
Is (3
,
−
3)
∈
S
?
(b) Write
S
as a set of ordered pairs.
(c) Identify the domain and co-domain of
S
.
(d) Sketch an arrow diagram for
S
.
Solution:
(a)
2
S
2 is true since
1
2
−
1
2
= 0.
−
1
S
−
1 is true since
1
−
1
−
1
−
1
= 0.
(3
,
3)
∈
S
is true since
1
3
−
1
3
= 0.
(3
,
−
3)
/
∈
S
since
1
3
−
1
−
3
=
2
3
which is not an integer.
(b)
S
as a set of ordered pairs is:
S
=
{
(
−
3
,
−
3)
,
(
−
2
,
−
2)
,
(
−
2
,
2)
,
(
−
1
,
−
1)
,
(
−
1
,
1)
,
(1
,
−
1)
,
(1
,
1)
,
(2
,
−
2)
,
(2
,
2)
,
(3
,
3)
}
(c)
Domain of
S
:
{−
3
,
−
2
,
−
1
,
1
,
2
,
3
}
.
Co-Domain of
S
:
{−
3
,
−
2
,
−
1
,
1
,
2
,
3
}
.
(d) Diagram
18. Define a relation
R
from
R
to
R
as follows:
For all (
x, y
)
∈
R
×
R
, (
x, y
)
∈
R
means that
y
=
x
2
.
(a) Is (2
,
4)
∈
R
? Is (4
,
2)
∈
R
? Is (
−
3)
R
9? Is 9
R
(
−
3)?
(b) Draw the graph of
R
in the Cartesian plane.
Solution
:
8
(a) For the relation
R
from
R
to
R
defined as
y
=
x
2
:
Is (2
,
4)
∈
R
? Yes, because 4 = 2
2
.
Is (4
,
2)
∈
R
? No, because 2
̸
= 4
2
.
Is (
−
3)
R
9? Yes, because 9 = (
−
3)
2
.
Is 9
R
(
−
3)? No, because
−
3
̸
= 9
2
.
(b) Graphical representation of the relation
R
:
19. Let
A
=
{
4
,
5
,
6
}
and
B
=
{
5
,
6
,
7
}
and define relations
R
,
S
, and
T
from
A
to
B
as follows:
For all (
x, y
)
∈
A
×
B
, (
x, y
)
∈
R
means that
x
≥
y
.
(
x, y
)
∈
S
means that
x
−
y
2
is an integer.
T
=
{
(4
,
7)
,
(6
,
5)
,
(6
,
7)
}
.
(a) Draw arrow diagrams for
R
,
S
, and
T
.
(b) Indicate whether any of the relations
R
,
S
, and
T
are functions.
Solution
:
9
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(a) Use these sets for the arrow diagrams:
•
Relation
R
:
{
(5,5), (6,5), (6,6)
}
•
Relation
S
:
{
(4,6), (5,5), (5,7), (6,6)
}
•
Relation
T
:
{
(4,7), (6,5), (6,7)
}
(b) Function check:
•
Relation
R
is not a function.
•
Relation
S
is not a function.
•
Relation
T
is not a function.
Detailed explanation
: For a relation to be considered a function:
i.
Every element from the domain must be associated with
at least one element from the codomain.
•
This means that for every element
x
in set
A
, there must exist
an element
y
in set
B
such that the pair (
x, y
) is part of the
relation.
This ensures that every input
x
has some output
y
. It doesn’t allow for “orphan” elements in the domain that
don’t relate to any element in the codomain.
ii.
No element from the domain can be associated with
more than one element from the codomain.
•
This implies that for each element
x
in set
A
, there can be
only one unique element
y
in set
B
such that the pair (
x, y
)
is part of the relation. In other words, each input
x
can have
only one specific output
y
. This is the defining feature of a
function; you cannot have a single input giving two different
outputs.
For the relations
R
,
S
, and
T
given:
•
Relation
R
:
–
R
=
{
(5
,
5)
,
(6
,
5)
,
(6
,
6)
}
–
The element 4 from set
A
does not relate to any element in
B
. Hence, condition 1 is violated.
–
The element 6 from set
A
is related to both 5 and 6 from set
B
. Hence, condition 2 is violated.
–
Therefore,
R
is not a function.
•
Relation
S
:
–
S
=
{
(4
,
6)
,
(6
,
6)
,
(6
,
4)
}
–
The element 5 from set
A
does not have any relationship in
S
. Condition 1 is not satisfied.
–
The element 6 from set
A
is related to both 6 and 4 from set
B
. Hence, condition 2 is violated.
–
Therefore,
S
is not a function.
•
Relation
T
:
–
T
=
{
(4
,
7)
,
(6
,
5)
,
(6
,
7)
}
10
–
The element 5 from set
A
does not have any relationship in
T
. Thus, condition 1 is not satisfied.
–
The element 6 from set
A
is related to both 5 and 7 from set
B
. Hence, condition 2 is violated.
–
Therefore,
T
is not a function.
20. Find four relations from
{
a, b
}
to
{
x, y
}
that are not functions from
{
a, b
}
to
{
x, y
}
.
Solution
:
The Cartesian product of
{
a, b
}
and
{
x, y
}
is:
{
a, b
} × {
x, y
}
=
{
(
a, x
)
,
(
a, y
)
,
(
b, x
)
,
(
b, y
)
}
From this product, the relations that are not functions include:
R
1
=
{
(
a, x
)
,
(
a, y
)
,
(
b, x
)
,
(
b, y
)
}
R
2
=
{
(
a, x
)
,
(
a, y
)
}
R
3
=
{
(
b, x
)
,
(
b, y
)
}
R
4
=
∅
In addition to the relations that are not functions, the relations from
{
a, b
}
to
{
x, y
}
that are functions include:
F
1
=
{
(
a, x
)
,
(
b, x
)
}
F
2
=
{
(
a, x
)
,
(
b, y
)
}
F
3
=
{
(
a, y
)
,
(
b, x
)
}
F
4
=
{
(
a, y
)
,
(
b, y
)
}
21. Define a relation
T
from
R
to
R
as follows: For all real numbers
x
and
y
,
(
x, y
)
∈
T
means that
y
2
−
x
2
= 1
.
Is
T
a function? Explain.
Solution
:
Given the relation
T
=
{
(
x, y
)
|
x, y
∈
R
and
y
2
−
x
2
= 1
}
we need to determine if
T
is a function.
Plugging in
x
= 1 into the equation
y
2
−
x
2
= 1, we get:
y
2
= 2
which has two solutions:
y
=
√
2
11
and
y
=
−
√
2
This indicates that the ordered pairs (1
,
√
2) and (1
,
−
√
2) both belong to
the relation
T
. Hence, for the single
x
value of 1, there exist two distinct
y
values.
This violates the definition of a function, which states that every element
in the domain must map to exactly one element in the codomain. In this
case, for
x
= 1, there are two different
y
values, so
T
is not a function.
22. Let
A
=
{−
1
,
0
,
1
}
and
B
=
{
t, u, v, w
}
.
Define a function
F
:
A
→
B
by the following arrow diagram:
(a) Write the domain and co-domain of
F
.
(b) Find
F
(
−
1),
F
(0), and
F
(1).
Solution
:
(a)
Domain of F is
A
=
{−
1
,
0
,
1
}
Co-domain of F is
B
=
{
t, u, v, w
}
(b) Using the arrow diagram, we can determine the following mappings:
F
(
−
1) =
u
F
(0) =
w
F
(1) =
u
23. Let
C
=
{
1
,
2
,
3
,
4
}
and
D
=
{
a, b, c, d
}
.
Define a function
G
:
C
→
D
by the following arrow diagram:
12
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(a) Write the domain and co-domain of
G
.
(b) Find
G
(1),
G
(2),
G
(3), and
G
(4).
Solution:
(a)
Domain of G is
C
=
{
1
,
2
,
3
,
4
}
Co-domain of G is
D
=
{
a, b, c, d
}
(b) Using the arrow diagram, we can determine the following mappings:
G
(1) =
c
G
(2) =
c
G
(3) =
c
G
(4) =
c
24. Let
X
=
{
2
,
4
,
5
}
and
Y
=
{
1
,
2
,
4
,
6
}
.
Which of the following arrow diagrams determine functions from
X
to
Y
?
(a)
(b)
13
(c)
(d)
(e)
14
Solution
: Only (d)
15
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