Transcript Mathematical Proofs

Chapter 6 Mathematical Induction
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6.1 The Principle of mathematical induction
6.2 A More General Principle of Mathematical Induction
6.3* Proof by Minimum Counterexample
6.4 The Strong Principle of Mathematical Induction
Section 6.1 The Principle of mathematical induction
Let A be a nonempty set of real numbers. A number m  A is called a
least element of A if xm for every x A. Some nonempty sets of
real numbers have a least element; others do not. For example: N
has a smallest element 1, while Z has no least element.
Theorem. If a set A of real numbers has a least element, then A has a
unique least element.
A nonempty set S of real numbers is said to be well-ordered if every
nonempty subset of S has a least element.
For example: S={1, 2, 3} is well ordered, while Z, Q, R are not well
ordered.
The principle of Mathematical Induction
The Well-Ordering Principle
The set N of positive integers is well-ordered.
The principle of Mathematical Induction
For each positive integer n, let P(n) be a statement. If
(1) P(1) is true and
(2) The implication
If P(k), then P(k+1)
is true for every positive integer k,
then P(n) is true for every positive integer n.
The principle of Mathematical Induction
More symbolically, the principle of Mathematical Induction can be
stated as
For each positive integer n, let P(n) be a statement. If
(1) P(1) is true and
(2) kN, P(k)  P(k+1) is true,
then nN, P(n) is true.
Induction Proof
That is, nN, P(n) can be proved to be true if
• We can show that P(1) is true and
• We can establish the truth of the implication
If P(k), then P(k+1).
for every positive integer k.
A proof using the principle of Mathematical Induction is called an
induction proof or a proof by induction. The verification of the truth
of P(1) in an induction proof is called the base step, basis step, or
the anchor of the induction. The implication “If P(k), then P(k+1)” for
an arbitrary positive integer k, the statement P(k) is called the
inductive hypothesis.
Example
Result. For every positive integer n, 1+2+3+…+n=n(n+1)/2.
Proof. Since 1=(1)(2)/2, the statement is true for n=1. Assume that
1+2+3+…+k=k(k+1)/2,
Where k is a positive integer, we want to show that
1+2+3+…+(k+1)=(k+1)(k+2)/2.
Thus
1+2+3+…+(k+1)= (1+2+3+…+k)+(k+1)
= k(k+1)/2+(k+1)
=(k+1)(k+2)/2.
By the Principle of Mathematical Induction,
1+2+3+…+n=n(n+1)/2
For every positive integer n.
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Note: the last sentence in the proof of Result is typical of the last
sentence of every proof using mathematical induction.
Example
Result. For every positive integer n,
12+22+…+n2=n(n+1)(2n+1)/6.
Proof. Exercise.
Example
Result. For every positive integer n,
1
1
1
n

  

.
2 3 3 4
(n  1)  (n  2) 2n  4
1
1
1


Proof. Since 2  3 2 1  4 6 . , the formula holds for n=1. Assume
1
1
1
k





That 2  3 3  4
for a positive integer k.
(k  1)  (k  2) 2k  4
1
1
1
k 1
k 1






.
We want to show that 2  3 3  4
(k  2)  (k  3) 2(k  1)  4 2k  6
Proof Cont.
Observe that
1
1
1
k 1
k 1

  


.
2 3 3 4
(k  2)  (k  3) 2(k  1)  4 2k  6
1
1
1
1
[

  
]
2 3 3 4
(k  2)  (k  3) ( k  2)  ( k  3)
k
1


2(k  2) (k  2)  (k  3)
k 2  3k  2

2(k  2)(k  3)
k 1

2k  6
By the Principle of Mathematical Induction,
1
1
1
n

  

.
2 3 3 4
(n  1)  (n  2) 2n  4
For every positive integer n.
#
Section 6.2 A More General Principle
of Mathematical Induction.
We now describe an analogous technique to verify the truth of a
statement of the following type where m denotes some fixed integer:
For every integer nm, P(n).
Theorem. For each integer m, the set S={ I  Z : I  m } is well ordered.
Theorem. The Principle of Mathematical Induction.
For a fixed integer m, Let S={ I  Z : I  m }. For each integer n  S, let
P(n) be a statement. If
(1) P(m) is true and
(2) The implication
If P(k), then P(k+1).
is true for every integer k  S,
then P(n) is true for every integer n  S.
The Principle of Mathematical Induction
For a fixed integer m, Let S={ I  Z : I  m }. For each integer n  S, let
P(n) be a statement. If
(1) P(m) is true and
(2) kS, P(k)  P(k+1) is true,
then nS, P(n) is true.
Example
Result. For every nonnegative integer n, 2n > n.
Proof. We proceed by induction. The inequality holds for n=0 since
20>0. Assume that 2k>k, where k is a nonnegative integer. We want
to show that 2k+1>k+1. We consider two cases: k=0 and k1.
Case 1: We assume k=0, then 2k+1 =2>1=k+1.
Case 2: We assume that k 1. Then 2k+1 =2(2k)>2k=k+k  k+1.
By Principle of Mathematical Induction, 2n > n for every nonnegative
integer n.
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Example
Result. For every integer n  5, 2n > n2.
Proof. Exercise.
Section 6.4 The Strong Principle of Mathematical Induction
Theorem The Strong Principle of Mathematical Induction.
For each positive integer n, let P(n) be a statement. If
(1) P(1) is true and
(2) The implication
If P(i) for every integer I with 1i k, then P(k+1).
is true for every positive integer k,
then P(n) is true for every positive integer n.
The Strong Principle of Mathematical Induction.
For each positive integer n, let P(n) be a statement. If
(1) P(1) is true and
(2)kN, P(1) P(2)  . . . P(k)  P(k+1) is true,
then nN, P(n) is true.
A More General Strong Principle of Mathematical Induction
Theorem The Strong Principle of Mathematical Induction.
For a fixed integer m, Let S={ I  Z : I  m }. For each integer n  S, let
P(n) be a statement. If
(1) P(m) is true and
(2) The implication
If P(i) for every integer I with m  i  k, then P(k+1).
is true for every integer k  S,
then P(n) is true for every integer n S.
Sequence
Suppose we are considering a sequence a1,a2,…,an of numbers. One
way of defining a sequence {an} is to specify explicitly the nth term
an.
A sequence can also be defied recursively. In a recursively defined
sequence {an}, only the first term or first few terms are defined
specifically, say a1,a2,…,ak for some fixed k N. Then ak+1 is
expressed in terms of a1,a2,…,ak and, more generally, for n>k, an is
expressed in terms of a1,a2,…,an-1.
Example
Result
A sequence {an} is defined recursively by
a1=1, a2=3, and an=2an-1-an-2 for n3.
Then an=2n-1 for all n N.
Proof. We proceed by induction. Since a1=2(1)-1=1, the formula holds
for n=1. Assume for an arbitrary positive integer k that ai=2i-1 for all
integers i with 1  i  k. We want to show that ak+1=2(k+1)-1=2k+1. If
k=1, then ak+1=a2=2(1)+1=3. Since a2=3, it follows that ak+1=2k+1
when k=1. hence we may assume that k 2. Since k+1 3, it follows
that ak+1=2ak-ak-1=2(2k-1)-(2k-3)=2k+1, which is the desired result.
By the Strong Principle of Mathematical Induction, an=2n-1 for all n
N.
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Example
Result. A sequence {an} is defined recursively by
a1=1, a2=4, and an=2an-1-an-2+2 for n3.
Then an=n2 for all n N.
Proof. Exercise.