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Advanced Calculus I
Spring 2008
Homework Solutions 1
Textbook Chapter 7
7.7 Classify each function as injective, surjective, bijective, or none of these.
(a)
f
: N
→
Z
deﬁned by
f
(
n
) =
n
2

n
Solution
This function is injective but not surjective, hence is not bijective.
Injectivity.
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.
We show that there exists
k
∈
Z such that
f
(
n
)
6
=
k
for all
n
∈
N.
Choose
k
=

1. Clearly
k
=

1
∈
Z. 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.
(d)
f
: [3
,
∞
)
→
[5
,
∞
) deﬁned by
f
(
x
) = (
x

3)
2
+ 5.
Solution
This function is both injective and surjective, hence is bijective.
Injectivity.
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

x
1

3

=

x
2

3

upon taking square roots of both sides of the equality. 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 and so
x
1
=
x
2
. Therefore
f
is injective.
Surjectivity.
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 Spring '08
 QIAN
 Calculus

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