ANALYSIS
EXERCISE 5–SOLUTIONS
1.
(a) Base case :
n
= 10; 2
10
= 1024
>
1000 = 10
3
. The statement is
true.
Induction step: Suppose that 2
k
> k
3
(
k
≥
10).
We have to
prove that 2
k
+1
>
(
k
+ 1)
3
. By the assumption
2
k
+1
= 2
·
2
k
>
2
k
3
.
So it is sufficient to prove that 2
k
3
>
(
k
+1)
3
which is equivalent
to
k
3

3
k
2

3
k

1
>
0. Since
k
≥
10 we have that
k

3
≥
7
and
k
2
≥
100. Therefore we have
k
3

3
k
2

3
k

1 = (
k

3)(
k
2

3)

10
which is evidently positive. This proves the statement for
n
=
k
+1. By the principle of mathematical induction the statement
is true for all natural
n
≥
10.
(b) Base case :
n
= 9; 2
9
= 512
>
4
·
9
2
+ 1. The statement is true.
Induction step: Suppose that 2
k
>
4
k
2
+ 1 (
k
≥
9). We have
to prove that 2
k
+1
>
4(
k
+ 1)
2
+ 1. By the assumption 2
k
+1
=
2
·
2
k
>
2(4
k
2
+ 1), so it is sufficient to prove that 2(4
k
2
+ 1)
>
4(
k
+ 1)
2
+ 1 which is easy to see taking the difference between
the lefthand side and the righthand side.
This proves the
statement for
n
=
k
+ 1.
By the principle of mathematical
induction the statement is true for all natural
n
≥
9.
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 Winter '10
 Spyros
 Natural number, Prime number, Rational number, ANALYSIS EXERCISE 5–SOLUTIONS

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