=
k
+ 2
2(
k
+ 1)
k
+ 1
2
k
·
k
2
+ 2
k
(
k
+ 1)
2
?
=
k
+ 2
2(
k
+ 1)
k
+ 2
2(
k
+ 1)
=
k
+ 2
2(
k
+ 1)
Therefore,
k
+ 1
∈
S
and
S
=
{
n
∈
N
:
n
≥
2
}
.
20. Let
S
be the set of positive integers for which the equation holds.
Step 1.
1
∈
S
:
(cos
x
+
i
sin
x
)
1
= cos
x
+
i
sin
x
.
Therefore,
1
∈
S
.
2
Step 2.
Assume that the positive integer
k
∈
S
. That is, assume
(cos
x
+
i
sin
x
)
k
= cos
kx
+
i
sin
kx
Step 3.
Prove that
k
+ 1
∈
S
.
(cos
x
+
i
sin
x
)
k
+1
=
(cos
x
+
i
sin
x
)(cos
x
+
i
sin
x
)
k
= (cos
x
+
i
sin
x
)(cos
kx
+
i
sin
kx
)
=
cos
x
cos
kx

sin
x
sin
kx
+
i
(sin
x
cos
kx
+ cos
x
sin
kx
)
=
cos(
k
+ 1)
x
+
i
sin(
k
+ 1)
x
Therefore,
k
+ 1
∈
S
and
S
=
N
.
21(b). The inequality does not hold for
n
= 1
,
2
,
3.
Prove that
n
2
≤
2
n
for all
n
≥
4.
Let
S
be the set of positive integers for which the inequality holds.
Step 1.
4
∈
S
:
4
2
= 16
≤
2
4
. Therefore,
4
∈
S
.
Step 2.
Assume that the positive integer
k
∈
S
. That is, assume
k
2
≤
2
k
,
(
k
≥
4).
Step 3.
Prove that
k
+ 1
∈
S
.
(
k
+ 1)
2
=
k
2
+ 2
k
+ 1
≤
2
k
+ 2
k
+ 1
≤
2
k
+ 2
k
= 2
·
2
k
= 2
k
+1
(here we used
2
k
+ 1
≤
2
k
)
Therefore,
k
+ 1
∈
S
and
S
=
{
n
∈
N
:
n
≥
4
}
.
Section 11
3(c).
x
negationslash
= 0.
Suppose
(1
/x
) = 0.
Then,
1 =
x
parenleftbigg
1
x
parenrightbigg
= 1
·
0 = 0
which is false.
Therefore,
(1
/x
)
negationslash
= 0.
x
=
x
·
1 =
x
parenleftbigg
1
x
·
1
1
/x
parenrightbigg
=
parenleftbigg
x
·
1
x
parenrightbigg
1
1
/x
= 1
·
1
1
/x
=
1
1
/x
4. Given
x
≥
0.
x
≤
epsilon1
for every
epsilon1 >
0
You've reached the end of your free preview.
Want to read all 4 pages?
 Fall '08
 Staff
 Integers, Natural number, Prime number