McCombs Math 381
Mathematical Induction
Main Idea:
We will develop a technique for proving statements about the set
of positive integers.
Proof Template:
To prove a statement of the form,
“Every positive integer
n
satisfies the propositional function
P n
( )
.”
Basis Step:
Show that
P
1
( )
is true.
Induction Hypothesis:
Assume that
P k
( )
is true.
Inductive Step:
Show that
P k
( )
!
P k
+
1
(
)
is true.
Examples
Use mathematical induction to prove each of the following.
1.
For every integer
n
>
0
,
10
2
n
!
1
is divisible by 11.
Let
P n
( )
:10
2
n
!
1
is divisible by 11.
Basis Step:
P
1
( )
:10
2
!
1
=
100
!
1
=
99
Clearly, 11 divides 99.
Induction Hypothesis:
Assume
P k
( )
is true.
That is, assume
10
2
k
!
1
is divisible by 11.
This means that
10
2
k
!
1
=
11
a
, for some integer
a
.
Inductive Step:
Show
P k
( )
!
P k
+
1
(
)
.
10
2
k
!
1
=
11
a
"
100 10
2
k
!
1
(
)
=
100 11
a
(
)
10
2
10
2
k
!
1
(
)
=
100 11
a
(
)
10
2
k
+
2
!
10
2
=
100 11
a
(
)
10
2
k
+
2
!
100
+
99
=
100 11
a
(
)
+
99
10
2
k
+
2
!
1
=
100 11
a
(
)
+
99
10
2
k
+
1
(
)
!
1
=
11 100
a
+
9
(
)
Therefore,
10
2
k
+
1
(
)
!
1
is divisible by 11.
Thus,
P k
( )
!
P k
+
1
(
)
.

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