Strong Mathematical
Induction
Lecture 23
Section 4.4
Tue, Feb 27, 2007
Announcements

Thursday at 4:00 (cookies at 3:30).
The Principle of Strong
Mathematical Induction

Let P(n) be a predicate defined for integers
n. Let a be an integer. If the following two
statements are true
P(a), P(a + 1), …, P(b).
 For all integers k  b, if P(a), P(a + 1), …,
P(k), then P(k + 1).

then the statement

For all integers n  a, P(n) is true.
The Principle of Strong
Mathematical Induction
The range a, a + 1, a + 2, …, b represents
the number of previous cases that the
inductive step depends on.
 Usually this is 1.
 We will see one example where it is 2.

Example: Factoring Integers
Theorem: Every integer n  2 factors into
primes.
 Proof:
 Base case:


Let n = 2. Then n is already a prime, so the
statement is true.
Example: Factoring Integers

Inductive case:
Suppose the statement is true for all
integers from 2 to k, for some k  2.
 Consider k + 1.
 Either k + 1 is prime or it is not prime.
 If it is prime, then we are done.
 If it is not prime, then it factors as a  b, for
some integers a, b, with a, b  2.

Example: Factoring Integers
By the induction hypothesis, a and b
themselves factor into the product of
primes.
 Therefore, k + 1 factors into the product of
primes.

Example: The Fibonacci
Sequence

The Fibonacci sequence {fn} is defined by
 f0 = 0,
 f1 = 1,
 fn = fn – 1 + fn – 2, for all n  2.

The first few terms are
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …
Example: The Fibonacci
Sequence

Theorem: Let
1 5
1 5

and  
.
2
2
Then
for all n  0.
n n
fn 
 
Example: The Fibonacci
Sequence
Proof:
 Base cases:

Let n = 0. Then (0 – 0)/( – ) = 0.
 Let n = 1. Then (1 – 1)/( – ) = 1.


Inductive case:
Suppose the equation holds for all n from 0
to k, for some k  1.
 Consider fk + 1.

A Lemma
Lemma: n + 1 = n + n – 1 and n + 1 = n +
n – 1 for all n  1.
 Proof (for ):
 Base case:

Let n = 1.
 Then a direct calculation shows that 2 = 
+ 1 = 1 + 0.

A Lemma

Inductive step:
Now suppose it is true for some integer k 
1.
 That is, k + 1 = k + k – 1
 Multiply the equation through by .
 The result is k + 2 = k + 1 + k.
 So the statement is true when n = k + 1.


Therefore, the statement is true for all n 
1.
Example: The Fibonacci
Sequence

Back to our theorem…
f k 1  f k  f k 1
 k   k  k 1   k 1


 
 



k
  k 1    k   k 1 
 
 k 1   k 1

.
 
Example: The Fibonacci
Sequence

Therefore, the statement is true for all n 
1.
Example: Trees

A tree is a connected graph that contains
no cycles.
Example: Trees
Notice that if you add an edge to a tree,
you necessarily create a cycle.
 Also notice that if you delete an edge from
a tree, the graph becomes disconnected.
 This suggests that a tree contains exactly
the “right” number of edges.

Example: Trees
Theorem: For all n  1, a tree with n nodes
contains exactly n – 1 edges.
 Proof:
 Base case:

Let n = 1.
 Clearly, there are 0 edges.

Example: Trees

Inductive case:
Suppose that the statement is true for all n
 k, for some k  1.
 Let T be a tree with k + 1 nodes.
 Let v be an arbitrary node in T and let s be
the index of v.
 Let u1, u2, …, us be the adjacent nodes.
 Delete the edges from v to u1, …, us.

Example: Trees
Then each ui is a node of a separate tree
Ti.
 Let ni be the number of nodes in tree Ti.
 Note that n1 + … + ns = k.
 By induction, each tree Ti contains ni – 1
edges.
 Thus, the total number of edges in tree T is
(n1 – 1) + … + (ns – 1) + s,
which equals k.

Example: Trees
Example: Trees
u2
v
u3
u1
Example: Trees
u2
v
u3
u1
Example: Trees
u2
v
u3
u1
4 nodes,
3 edges
Example: Trees
1 node,
0 edges
u2
v
u3
u1
4 nodes,
3 edges
Example: Trees
1 node,
0 edges
u2
v
u3
u1
4 nodes,
3 edges
3 nodes,
2 edges
Example: Trees
1 node,
0 edges
u2
v
u3
u1
4 nodes,
3 edges
3 edges
3 nodes,
2 edges
Example: Trees
u2
v
u3
u1
Total = 3 + 0 + 2 + 3 = 8 edges
Example: Trees
Example: Trees
u2
u1
v
u3
Example: Trees
u2
u1
v
u3
Example: Trees
u2
u1
6 nodes,
5 edges
v
u3
Example: Trees
1 node,
0 edges
u2
u1
6 nodes,
5 edges
v
u3
Example: Trees
1 node,
0 edges
u2
u1
6 nodes,
5 edges
v
u3
1 nodes,
0 edges
Example: Trees
1 node,
0 edges
u2
u1
6 nodes,
5 edges
v
u3
3 edges
1 nodes,
0 edges
Example: Trees
u2
u1
v
u3
Total = 5 + 0 + 0 + 3 = 8 edges
Example: Binary Trees

A binary tree structure consists of a set of
nodes, each with two pointers.
The left pointer points to the left child.
 The right pointer points to the right child.


If there is no child, then the pointer is null.
Example: Binary Trees

Corollary: In a binary tree structure, if there
are n nodes, then there are exactly n + 1
null pointers.
Balanced Binary Strings
Let S be the set of all finite binary strings
that contain an equal number of 0’s and
1’s.
 Define a set T as follows:

The empty string is in T.
 If s  T, then 0s1  T and 1s0  T.
 If s  T and t  T, then st  T.

Balanced Binary Strings
Theorem: T  S.
 Theorem: T = S.

The Paradox of the Pop
Quiz
A professor announces that one day
during the semester he will give a pop
quiz.
 For the quiz to be a “pop” quiz, it must be
unexpected.

The Paradox of the Pop
Quiz

Theorem: For all n  0, the professor
cannot give the pop quiz n days before the
last day of the semester.
The Paradox of the Pop
Quiz
Proof:
 Base case: n = 0.


He cannot give the pop quiz on the last day
of the semester because everyone will be
expecting it then.
The Paradox of the Pop
Quiz

Inductive step:
Suppose he cannot give it on any of the
days 0, 1, 2, …, k days before the end of
the semester, for some k  0.
 Knowing this, the students would be
expecting it on the day k + 1 days before
the end of the semester.
 So he can’t give it on that day.

The Paradox of the Pop
Quiz

By strong induction, for all n  0, he cannot
give the pop quiz n days before the end of
the semester.