CSE
1400
Applied Discrete Mathematics
Mathematical Induction
Department of Computer Sciences
College of Engineering
Florida Tech
Fall
2011
Induction
1
Summations
1
The Tower of Hanoi and Mersenne Numbers
7
Rabbit Population Growth and Fibonacci Numbers
8
Induction on Arithmetic and Geometric Sequences
10
Problems on Induction
13
Abstract
The word induction has many meanings. For us it is a formal proof
process that a predicate
p
(
n
)
is
True
for all natural numbers
n
be
longing to some set, most often the set of natural numbers
N
=
{
0, 1, 2, . . .
}
.
The principle of induction is: If
set
X
contains
zero
and if
x
∈
X
implies its successor
x
+
1
∈
X
, then every natural number is in
X
, that
is
X
=
N
.
Induction
Induction often involves sums:
If the sum of
n
term equals some formula, then the sum of
n
+
1 terms
equals the same formula with appropriate substitutions.
Induction often involves recurrsion:
A function
f
(
n
)
satisfies a recurrence relation.
An inductive proof of the truth of
predicate
p
(
n
)
for all natural numbers
n
has two steps.
1
.
A proof that
p
(
0
)
is
True
.
2
.
A proof of the truth of the condi
tional statement
If
p
(
n
)
is
True
,
then
p
(
n
+
1
)
is
True
.
A short notation for this is
p
(
0
)
p
(
n
)
→
p
(
n
+
1
)
∴
(
∀
n
∈
N
)
p
(
n
)
Which is read "
p
(
0
)
is
True
; and if
p
(
n
)
is
True
, then
p
(
n
+
1
)
is
True
. Therefore,
p
(
n
)
is
True
for all natural numbers
n
."
Summations
Consider the predicate:
The sum of the first
n
natural numbers is
n
(
n

1
)
/2.
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cse 1400 applied discrete mathematics
mathematical induction
2
You could use the notation
p
(
n
)
to
denote the truth value of the statement
“The sum of the first
n
natu
ral numbers is
n
(
n

1
)
/2.”
For instance, the sum of the first 5 natural numbers is
0
+
1
+
2
+
3
+
4
=
10
=
5
·
(
5

1
)
/2
=
5
·
4/2
Since the sum of the first 5 natural numbers is 10, the sum of the first
6 natural numbers is
10
+
5
=
15
=
6
×
5
2
The heart of mathematical induction relies on (
1
) noticing a pattern
for small initial cases, and (
2
) establishing a link allowing a proof of
of an instance from the truth of previous instances.
In this case, the pattern is
The sum of
n
terms in an arithmetic
progression is
n
times the average of
the first and last term.
0
+
1
+
2
+
3
+
4
+
· · ·
+ (
n

1
) =
n
×
n

1
2
which can be phrases as “the sum of the first
n
natural numbers is
n
times
n

1 divided by 2.” or
p
(
n
)
for short.
To establish the link between the truth of
p
(
n
)
and
p
(
n
+
1
)
re
quires a little algebra. We must generalize the instance above where
the step was from 5 to 6.
If the sum of the first
n
natural numbers
The sum of the first
n
+
1 natural
numbers is the sum of the first
n
natural numbers plus
n
, the
(
n
+
1
)
st
natural number.
is
n
(
n

1
)
/2, then the sum of the first
n
+
1 natural numbers is
n
(
n

1
)
/2 plus
n
. That is, if
0
+
1
+
2
+
3
+
4
+
· · ·
+ (
n

1
) =
n
(
n

1
)
2
then
[
0
+
1
+
2
+
· · ·
+ (
n

1
)] +
n
=
n
(
n

1
)
2
+
n
=
n
n

1
2
+
1
= (
n
+
1
)
n
2
This shows that if the function
n
(
n

1
)
/2 correctly computes the
value in the
n
th
case, then it correctly computes the value in the
(
n
+
1
)
st
case.
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 Fall '11
 Shoaff
 Math, Mathematical Induction, Recursion, Natural number, Fibonacci number, Applied Discrete Mathematics Mathematical Induction

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