CS 70
Discrete Mathematics and Probability Theory
Spring 2012
Alistair Sinclair
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, Spring 2012, Note 3
1
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• 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 computer
science 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
.
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 Spring '08
 PAPADIMITROU
 Mathematical Induction, Inductive Reasoning, Natural number, Mathematical proof, inductive hypothesis

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