MAT 300, Mathematical Structures
Dr. L. Mantini
Problem Set 7 Solutions
Spring 2010
5.1
: 2, 4, 6, 14;
5.2
: 2, 4, 5, 6, 8, 9, 11, 17, 18
5.1.2 Is each relation
f
a function?
(a) We have
A
=
{
a,b,c,d
}
,
B
=
{
x,y,z
}
, and
f
(
a
) =
y
,
f
(
b
) =
x
, and
f
(
c
) =
y
.
NO
,
this is not a function since no value of
f
(
d
) is given.
(b) Let
W
be the set of all English words and
A
the set of all letters of the alphabet.
Deﬁne
f
:
W
→
A
and
g
:
W
→
A
by
f
=
{
(
w,a
)

letter
a
occurs in word
w
}
and
g
=
{
(
w,a
)

letter
a
is the ﬁrst letter of word
w
}
.
NO
,
f
is not a function since a
word with more than one letter has more than one value.
YES
,
g
is a function since
every English word has a unique ﬁrst letter.
(c) Let
P
=
{
J,M,S,F
}
, and arrange the letters in
P
around a circle. Then deﬁne
R
=
{
(
p,q
)
∈
P
×
P

p
is immediately to the right of
q
}
.
YES
,
R
:
P
→
P
is a
function since each letter in
P
has a unique nearest neighbor to the right.
5.1.4 (3 points) Determine the requested values of the given functions.
(a) Let
N
be the set of all countries and
C
the set of all cities. Let
H
:
N
→
C
be deﬁned
by
H
(
n
) = capital of
n
. What is
H
(Italy)?
Solution
:
H
(Italy) = Rome.
(b) Let
A
=
{
1
,
2
,
3
}
and
B
=
P
(
A
), the set of all subsets of
A
. Let
F
:
B
→
B
be deﬁned
by
F
(
X
) =
A
\
X
. What is
F
(
{
1
,
3
}
)?
Solution
:
F
(
{
1
,
3
}
) =
A
\ {
1
,
3
}
=
{
2
}
.
(c) Let
f
:
R
→
R
×
R
where
f
(
x
) = (
x
+1
,x

1). What is
f
(2)?
Solution
:
f
(2) = (3
,
1).
5.1.6 Let
f
and
g
be the functions from
R
to
R
given by
f
(
x
) = 1
/
(
x
2
+2) and
g
(
x
) = 2
x

1.
Find
f
◦
g
and
g
◦
f
.
Solution
: (
f
◦
g
)(
x
) =
f
(2
x

1) =
1
(2
x

1)
2
+ 2
=
1
4
x
2

4
x
+ 3
, and (
g
◦
f
)(
x
) =
g
(
f
(
x
)) = 2
f
(
x
)

1 =
2
x
2
+ 2

1 =

x
2
x
2
+ 2
.
The next exercise uses the deﬁnition of a
constant function
, which has two equivalent
versions. We may use either version.
Deﬁnition 1:
Let
A
and
B
be sets. A function
f
:
A
→
B
is
constant
if there exists a
b
∈
B
such that
f
(
x
) =
b
for all
x
∈
A
.
Deﬁnition 2:
Let
A
and
B
be sets. A function
f
:
A
→
B
is
constant
if there exists
x
1
∈
A
such that
f
(
x
) =
f
(
x
1
) for all
x
∈
A
.
5.1.14 (6 points) Suppose
A
is a nonempty set and
f
:
A
→
A
.
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View Full Document(a) Suppose there is some
a
∈
A
such that
f
(
x
) =
a
for all
x
∈
A
. Prove that for all
g
:
A
→
A
,
f
◦
g
=
f
.
Proof
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 Spring '07
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 Math, Functions and mappings, Empty function, Dr. L. Mantini

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