=
(
k
+ 1)[(
k
+ 1) + 1]
2
,
i.e., to show
1 + 2 + 3 +
...
+
k
+
k
+ 1
?
=
(
k
+ 1)(
k
+ 2)
2
,
i.e., by using the induction assumption (1), to show
k
(
k
+ 1)
2
+
k
+ 1
?
=
(
k
+ 1)(
k
+ 2)
2
,
or
k
(
k
+ 1) + 2(
k
+ 1)
2
?
=
(
k
+ 1)(
k
+ 2)
2
,
or
k
2
+ 3
k
+ 2
2
?
=
(
k
+ 1)(
k
+ 2)
2
,
or
k
2
+ 3
k
+ 2
2
?
=
(
k
+ 1)(
k
+ 2)
2
,
or, to show
(
k
+ 1)(
k
+ 2)
2
?
=
(
k
+ 1)(
k
+ 2)
2
,
Y es
!
=
⇒
(
k
+ 1)
∈
S.
∴
S
=
N
.
[Example 2]
Prove that
1 + 2 + 2
2
+ 2
3
+
...
+ 2
n

1
= 2
n

1
,
∀
n
∈
N
.
Proof:
Let
S
be the set of positive integers for which the equation holds.
Step 1. Show that
1
∈
S
1
?
= 2
1

1
Y es
!
=
⇒
1
∈
S.
4
Step 2. Assume that
k
∈
S
Namely, we assume that
1 + 2 + 2
2
+ 2
3
+
...
+ 2
k

1
= 2
k

1
.
(2)
Step 3. Prove that
k
+ 1
∈
S
for the
k
in Step 2
To show:
1 + 2 + 2
2
+ 2
3
+
...
+ 2
k

1
+ 2
k
?
= 2
k
+1

1
,
i.e., to show
1 + 2 + 2
2
+ 2
3
+
...
+ 2
k

1
+ 2
k
?
= 2
k
+1

1
i.e., by using the induction assumption (2), to show
(2
k

1) + 2
k
?
= 2
k
+1

1
,
or
2
·
2
k

1
?
= 2
k
+1

1
,
or, to show
2
k
+1

1
?
= 2
k
+1

1
Y es
!
=
⇒
(
k
+ 1)
∈
S.
∴
S
=
N
.
Some generalization
Theorem 1.2
Let
S
be a subset of the set
N
=
{
1
,
2
,
3
, ...
}
. If
1.
m
∈
S
2.
k
∈
S
with
k
≥
m
implies
k
+ 1
∈
S
.
Then
S
=
{
n
∈
N

n
≥
m
}
.
[Example 3]
Prove that
n
2
≤
2
n
,
for any integer n
≥
4
.
Proof:
Let
S
:=
{
n
∈
N

n
2
≤
2
n
}
. When
n
= 3, 3
2
= 9
≤
2
3
= 8 does not hold. Then the
smallest number in
S
must be greater than 3.
Step 1. Show that
4
∈
S
4
2
?
≤
2
4
Y es
!
=
⇒
4
∈
S.
5
Step 2. Assume that
k
∈
S
with
k
≥
4 Namely, we assume that for any integer
k
≥
4,
k
2
≤
2
k
.
(3)
Step 3. Prove that
k
+ 1
∈
S
for the
k
in Step 2
To show:
(
k
+ 1)
2
≤
?
2
k
+1
,
i.e., to show
k
2
+ 2
k
+ 1
≤
?
2
k
+1
,
i.e., by using the induction assumption (3), it suffices to show
2
2
k
+ 2
k
+ 1
?
≤
2
·
2
k
,
or, by subtracting 2
k
in both sides, to show
2
k
+ 1
?
≤
2
k
.
(4)
We use mathematical induction again to prove (4). Let
e
S
=
{
n
∈
N

2
n
+1
≤
2
n
, n
≥
4
}
.
Step 3a. Show
4
∈
e
S
2
·
4 + 1 = 9
≤
2
4
= 16
Y es
!
=
⇒
4
∈
e
S.
Step 3b. Assume that
k
∈
e
S
with
k
≥
4
2
k
+ 1
≤
2
k
.
(5)
Step 3c. Show
k
+ 1
∈
e
S
for the
k
in Step 3b
To show:
2(
k
+ 1) + 1
?
≤
2
k
+1
,
i.e.,
2
k
+ 3
?
≤
2
·
2
k
,
2
Here we have used a logical fact: if
A
≤
B
and
B
≤
C
, then
A
≤
C
.
6
i.e., by the induction assumption (5), it suffices to show
2
k
+ 2
?
≤
2
·
2
k
,
or, by subtracting 2
k
in both sides, to show
2
?
≤
2
k
,
Y es
!
=
⇒
k
+ 1
∈
e
S.
Then
e
S
=
{
n
∈
N

n
≥
4
}
, and hence (4) is proved. We conclude
∴
S
=
{
n
∈
N

n
≥
4
}
Appendix
Some problems did in the classroom
3.1.1. True or false:
(a) If
S
is a nonempty subset of
N
, then there exists an element
m
∈
S
such
that
m
≥
k
for all
k
∈
S
.
You've reached the end of your free preview.
Want to read all 9 pages?
 Fall '08
 Staff
 Natural number, Prime number