Math 350
Advanced Calculus I
Fall 2009
Homework Solutions
Section 7
FUNCTIONS
7.7
Classify each function as injective, surjective, bijective, or none of these.
(e)
f
:
N
→
Z
defined by
f
(
n
) =
n
2
−
n
.
Injectivity.
This function is injective but not surjective. To show that
f
is injective, we need
to show
(
∀
n
1
∈
N
) (
∀
n
2
∈
N
) [
f
(
n
1
) =
f
(
n
2
)
→
n
1
=
n
2
]
.
Proof:
Suppose
n
1
, n
2
∈
N
are arbitrary. Suppose
f
(
n
1
) =
f
(
n
2
)
.
We need to show
n
1
=
n
2
.
Since
f
(
n
1
) =
n
2
1
−
n
1
and
f
(
n
2
) =
n
2
2
−
n
2
, then
f
(
n
1
) =
f
(
n
2
) implies
n
2
1
−
n
1
=
n
2
2
−
n
2
(
n
2
1
−
n
2
2
)
−
(
n
1
−
n
2
) = 0
(
n
1
−
n
2
)(
n
1
+
n
2
)
−
(
n
1
−
n
2
) = 0
(
n
1
−
n
2
)(
n
1
+
n
2
−
1) = 0
This implies
n
1
−
n
2
= 0
or
n
1
+
n
2
−
1 = 0
.
Since
n
1
and
n
2
are natural numbers, then both are
≥
1 so
n
1
+
n
2
≥
2, which means
n
1
+
n
2
−
1
6
= 0. Therefore
n
1
−
n
2
= 0 so
n
1
=
n
2
, as required. Therefore
f
is injective.
Surjectivity.
To show that
f
is not surjective, we need to show
(
∃
k
∈
Z
) (
∀
n
∈
N
) [
f
(
n
)
6
=
k
]
.
That is, we need to show there exists
k
∈
Z
such that
f
(
n
)
6
=
k
for all
n
∈
N
.
Proof:
Find a
k
.
For all
n
∈
N
,
f
(
n
) =
n
(
n
−
1)
≥
0. This tells
us we should choose
k <
0.
For instance,
choose
k
=
−
1.
Verify
k
satisfies:
(i) Domain:
k
∈
Z
(ii) Prop. Function: (
∀
n
∈
N
) [
f
(
n
)
6
=
k
]
Clearly
k
=
−
1
∈
Z
. To prove (ii), suppose
n
is an arbitrary natural number. Then
f
(
n
) =
n
2
−
n
=
n
(
n
−
1)
≥
0
,
so
f
(
n
)
6
=
−
1 =
k
, as required.
Thus
f
is not surjective.
1
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Section 7
•
Functions
Math 350 Homework Solutions page 2
7.7
(continued)
(f)
f
: [3
,
∞
)
→
[5
,
∞
) defined by
f
(
x
) = (
x
−
3)
2
+ 5
Injectivity.
This function is injective. To show that
f
is injective, we need to show
°
∀
x
1
∈
[3
,
∞
)
¢ °
∀
x
2
∈
[3
,
∞
)
¢
[
f
(
x
1
) =
f
(
x
2
)
→
x
1
=
x
2
]
.
Proof:
Suppose
x
1
, x
2
∈
[3
,
∞
) are arbitrary. Suppose
f
(
x
1
) =
f
(
x
2
)
.
We need to show
x
1
=
x
2
.
Since
f
(
x
1
) =
f
(
x
2
), then
(
x
1
−
3)
2
+ 5 = (
x
2
−
3)
2
+ 5
(
x
1
−
3)
2
= (
x
2
−
3)
2
Taking square roots of both sides of this equation, we obtain

x
1
−
3

=

x
2
−
3

.
Since
x
1
, x
2
≥
3, then
x
1
−
3
≥
0 and
x
2
−
3
≥
0, so

x
1
−
3

=
x
1
−
3 and

x
2
−
3

=
x
2
−
3.
The above equality then simplifies to
x
1
−
3 =
x
2
−
3
x
1
=
x
2
Therefore
f
is injective.
Surjectivity.
To show that
f
is surjective, we need to show
°
∀
y
∈
[5
,
∞
)
¢
(
∃
x
∈
[3
,
∞
)
¢
[
f
(
x
) =
y
]
.
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