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Chapter 6 Mathematical Induction

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. Section 6.1 The Principle of mathematical induction.

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Chapter 6 Mathematical Induction

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  1. 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

  2. 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.

  3. 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(k+1) is true for every positive integer k, then P(n) is true for every positive integer n.

  4. 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(k+1) is true, then nN, P(n) is true.

  5. 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.

  6. 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. # Note: the last sentence in the proof of Result is typical of the last sentence of every proof using mathematical induction.

  7. Example Result. For every positive integer n, 12+22+…+n2=n(n+1)(2n+1)/6. Proof. Exercise.

  8. Example Result. For every positive integer n, Proof. Since , the formula holds for n=1. Assume That for a positive integer k. We want to show that

  9. Proof Cont. By the Principle of Mathematical Induction, For every positive integer n. # Observe that

  10. 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(k+1). is true for every integer k  S, then P(n) is true for every integer n  S.

  11. 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(k+1) is true, then nS, P(n) is true.

  12. 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. #

  13. Example Result. For every integer n  5, 2n > n2. Proof. Exercise.

  14. 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(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 • 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.

  15. 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(k+1). is true for every integer k  S, then P(n) is true for every integer n S.

  16. 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.

  17. 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. #

  18. 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.

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