CS 70
Discrete Mathematics and Probability Theory
Fall 2010
Tse/Wagner
Note 3
Induction
Induction is an extremely powerful tool in mathematics. It is a way of proving propositions that hold for all
natural numbers:
1)
∀
k
∈
N
, 0
+
1
+
2
+
3
+
···
+
k
=
k
(
k
+
1
)
2
2)
∀
k
∈
N
, the sum of the first
k
odd numbers is a perfect square.
3)
Any graph with
k
vertices and
k
edges contains a cycle.
Each of these propositions is of the form
∀
k
∈
N
P
(
k
)
. For example, in the first proposition,
P
(
k
)
is the
statement 0
+
1
+
···
+
k
=
k
(
k
+
1
)
2
,
P
(
0
)
says 0
=
0
(
0
+
1
)
2
,
P
(
1
)
says 0
+
1
=
1
(
1
+
1
)
2
, etc. The
principle of
induction
asserts that you can prove
P
(
k
)
is true
∀
k
∈
N
, by following these three steps:
Base Case:
Prove that
P
(
0
)
is true.
Inductive Hypothesis:
Assume that
P
(
k
)
is true.
Inductive Step:
Prove that
P
(
k
+
1
)
is true.
The principle of induction formally says that if
P
(
0
)
and
∀
n
∈
N
(
P
(
n
) =
⇒
P
(
n
+
1
))
, then
∀
n
∈
N
P
(
n
)
.
Intuitively, the base case says that
P
(
0
)
holds, while the inductive step says that
P
(
0
) =
⇒
P
(
1
)
, and
P
(
1
) =
⇒
P
(
2
)
, and so on. The fact that this “domino effect" eventually shows that
∀
n
∈
N
P
(
n
)
is what the principle
of induction (or the induction axiom) states. In fact, dominoes are a wonderful analogy: we have a domino
for each proposition
P
(
k
)
. The dominoes are lined up so that if the
k
th
domino is knocked over, then it in turn
knocks over the
k
+
1
st
. Knocking over the
k
th
domino corresponds to proving
P
(
k
)
is true. So the induction
step corresponds to the fact that the
k
th
domino knocks over the
k
+
1
st
domino. Now, if we knock over the
first domino (the one numbered 0), then this sets off a chain reaction that knocks down all the dominoes.
Let’s see some examples.
Theorem:
∀
k
∈
N
,
k
∑
i
=
0
i
=
k
(
k
+
1
)
2
.
Proof
(by induction on
k
):
• Base Case:
P
(
0
)
asserts:
0
∑
i
=
0
i
=
0
(
0
+
1
)
2
. This clearly holds, since the left and right hand sides both
equal 0.
CS 70, Fall 2010, Note 3
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
• Inductive Hypothesis: Assume
P
(
k
)
is true. That is,
k
∑
i
=
0
i
=
k
(
k
+
1
)
2
.
• Inductive Step: We must show
P
(
k
+
1
)
. That is,
k
+
1
∑
i
=
0
i
=
(
k
+
1
)(
k
+
2
)
2
:
k
+
1
∑
i
=
0
i
= (
k
∑
i
=
0
i
)+(
k
+
1
)
=
k
(
k
+
1
)
2
+(
k
+
1
)
(by the inductive hypothesis)
= (
k
+
1
)(
k
2
+
1
)
=
(
k
+
1
)(
k
+
2
)
2
.
Hence, by the principle of induction, the theorem holds.
♠
Note the structure of the inductive step. You try to show
P
(
k
+
1
)
under the assumption that P
(
k
)
is true.
The idea is that
P
(
k
+
1
)
by itself is a difficult proposition to prove. Many difficult problems in EECS are
solved by breaking the problem into smaller, easier ones. This is precisely what we did in the inductive step:
P
(
k
+
1
)
is difficult to prove, but we were able to recursively define it in terms of
P
(
k
)
.
We will now look at another proof by induction, but first we will introduce some notation and a definition
for divisibility. We say that integer
a
divides
b
(or
b
is divisible by
a
), written as
a

b
, if and only if for some
integer
q
,
b
=
aq
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '11
 Rau
 Mathematical Induction, Inductive Reasoning, Natural number, Mathematical proof, inductive hypothesis

Click to edit the document details