Chapter%206%20Mathematical%20Induction

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Chapter 6 Mathematical Induction

• 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

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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 S1, 2, 3 is well ordered, while
  Z, Q, R are not well ordered.

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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
• P(1) is true and
• The implication
• If
  P(k), then P(k1)
• is true for every positive integer k,
• then P(n) is true for every positive integer n.

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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
• P(1) is true and
• ?k?N, P(k) ? P(k1) is true,
• then ?n?N, P(n) is true.

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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(k1).
• 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(k1) for an
  arbitrary positive integer k, the statement P(k)
  is called the inductive hypothesis.

6
Example

• Result. For every positive integer n,
  123nn(n1)/2.
• Proof. Since 1(1)(2)/2, the statement is true
  for n1. Assume that
• 123kk(k1)/2,
• Where k is a positive integer, we want to show
  that
• 123(k1)(k1)(k
  2)/2.
• Thus
• 123(k1) (123k)(k1)
• k(k1)/2(k1)
• (k1)(k2)/2.
• By the Principle of Mathematical Induction,
• 123nn(n1)/2
• For every positive integer n.
  
  
  
• 
  
  
• Note the last sentence in the proof of Result is
  typical of the last sentence of every proof using
  mathematical induction.

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Example

• Result. For every positive integer n,
• 1222n2n(n1)(2n1)/6.
• Proof. Exercise.

8
Example

• Result. For every positive integer n,
• Proof. Since , the
  formula holds for n1. Assume
• That
  for a positive integer k.
• We want to show that

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Proof Cont.

• By the Principle of Mathematical Induction,
• For every positive integer n.
  
  
  
• 
  
  

Observe that
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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
• P(m) is true and
• The implication
• If P(k), then P(k1).
• is true for every integer k ? S,
• then P(n) is true for every integer n ? S.

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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
• P(m) is true and
• ?k?S, P(k) ? P(k1) is true,
• then ?n?S, P(n) is true.

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Example

• Result. For every nonnegative integer n, 2n gt n.
• Proof. We proceed by induction. The inequality
  holds for n0 since 20gt0. Assume that 2kgtk, where
  k is a nonnegative integer. We want to show that
  2k1gtk1. We consider two cases k0 and k?1.
• Case 1 We assume k0, then 2k1 2gt1k1.
• Case 2 We assume that k ?1. Then 2k1
  2(2k)gt2kkk ? k1.
• By Principle of Mathematical Induction, 2n gt
  n for every nonnegative integer n.
• 
  
  

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Example

• Result. For every integer n ? 5, 2n gt n2.
• Proof. Exercise.

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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
• P(1) is true and
• The implication
• If P(i) for every integer I with 1?i ?k,
  then P(k1).
• 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
• P(1) is true and
• (2)?k?N, P(1)?? P(2) ? . . . ?P(k) ? P(k1) is
  true,
• then ?n?N, P(n) is true.

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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
• P(m) is true and
• The implication
• If P(i) for every integer I with m ? i ? k,
  then P(k1).
• is true for every integer k ? S,
• then P(n) is true for every integer n? S.

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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 ak1 is
  expressed in terms of a1,a2,,ak and, more
  generally, for ngtk, an is expressed in terms of
  a1,a2,,an-1.

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Example

• Result
• A sequence an is defined recursively by
• a11, a23, and an2an-1-an-2 for n?3.
• Then an2n-1 for all n ?N.
• Proof. We proceed by induction. Since
  a12(1)-11, the formula holds for n1. Assume
  for an arbitrary positive integer k that ai2i-1
  for all integers i with 1 ? i ? k. We want to
  show that ak12(k1)-12k1. If k1, then
  ak1a22(1)13. Since a23, it follows that
  ak12k1 when k1. hence we may assume that k
  ?2. Since k1 ?3, it follows that
  ak12ak-ak-12(2k-1)-(2k-3)2k1, which is the
  desired result. By the Strong Principle of
  Mathematical Induction, an2n-1 for all n ?N.
• 
  
  

18
Example

• Result. A sequence an is defined recursively by
  
• a11, a24, and an2an-1-an-22 for n?3.
• Then ann2 for all n ?N.
• Proof. Exercise.