CS 70
Discrete Mathematics for CS
Spring 2004
Papadimitriou/Vazirani
Lecture 2
This lecture covers induction, which is the most common technique for proving universally quantified sen
tences over the natural numbers and other discrete sets of objects (lists, trees, tilings, graphs, programs, etc.).
We need induction, or something like it, because any attempt at proof by enumeration of possible worlds
is doomed to failure—with infinitely many objects, there are infinitely many possible worlds that can be
defined.
The principle of induction
The principle of induction is an axiom of the natural numbers.
Recall from Lecture 1 that the natural
numbers satisfy Peano’s axioms (where we have written
n
+
1 instead of
s
(
n
)
for clarity):
0 is a natural number
If
n
is a natural number,
n
+
1 is a natural number
These axioms don’t quite
define
the natural numbers, because they include no statement of the kind, “
...
and by the way, there are no other natural numbers.” This makes it impossible to use these axioms alone to
prove that any property holds for all natural numbers
n
.
The principle of induction essentially fills this gap. Informally, it says the following:
If you can prove that some property holds for 0, and you can prove that the property holds for
n
+
1 if it holds for
n
, then you have proved that the property holds for all the natural numbers.
To state this principle formally, let
P
(
n
)
be an arbitrary proposition about the natural number
n
. (For exam
ple,
P
(
n
)
might be “2
n
>
n
”.) The principle of induction is as follows:
Axiom 2.1 (Induction)
: For any property
P
, if
P
(
0
)
and
∀
n
∈
N
(
P
(
n
) =
⇒
P
(
n
+
1
))
, then
∀
n
∈
N
P
(
n
)
.
This says that if
P
(
0
)
holds, and
P
(
0
) =
⇒
P
(
1
)
, and
P
(
1
) =
⇒
P
(
2
)
, and so on, then
P
(
n
)
must be true
for all
n
. This seems pretty reasonable. (Later we will see that the axiom has an alternative form that seems
even more reasonable.)
Inductive proofs
The standard first example is to verify the wellknown formula for the sum of the first
n
positive integers:
Theorem 2.1
:
∀
n
∈
N
n
∑
i
=
1
i
=
n
(
n
+
1
)
/
2
[Note:
∑
n
i
=
1
i
means 1
+
2
+
···
+
n
but is more precise. When
n
=
0 the summation has no terms so is 0;
when
n
=
1 it has just the term
i
=
1 so the sum is 1.]
In many, but not all, cases of proof by induction, the property
P
used in the induction is exactly the one we
want to prove. That is the case here:
P
(
n
)
is the proposition that
∑
n
i
=
1
i
=
n
(
n
+
1
)
/
2.
CS 70, Spring 2004, Lecture 2
1
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To construct an inductive proof, first we need to establish the
base case
P
(
0
)
, then we need to prove that
BASE CASE
P
(
n
+
1
)
follows from
P
(
n
)
. The latter proof is called the
inductive step
, and usually involves all the work.
INDUCTIVE STEP
The key idea of induction, though, is that this step is easier than proving the whole universal proposition
from scratch because you get to
assume
that
P
(
n
)
is true when proving
P
(
n
+
1
)
. This assumption is called
the
inductive hypothesis
.
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 Spring '05
 HUANG
 Mathematical Induction, Natural number, Peano axioms, inductive hypothesis, inductive step

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