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Section 3.3: Mathematical Induction

Section 3.3: Mathematical Induction. Mathematical induction is a proof technique that can be used to prove theorems of the form: n Z + ,P(n)

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Section 3.3: Mathematical Induction

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  1. Section 3.3: Mathematical Induction Mathematical induction is a proof technique that can be used to prove theorems of the form: nZ+,P(n) We have already proved many such theorems using universal generalization. For example, we proved nZ+, [there is a prime greater than n]. When we proved such a theorem, we would start our proof with “Let n be a positive integer”. Then we would proceed to show that P(n) is true using only the fact that n is a positive integer. That is, we proved P(n) for an arbitrary n  Z+. Sometimes it may be difficult to prove P(n) for an arbitrary n. In such cases we may be able to apply mathematical induction. We use mathematical induction when it is easier to see why P(n) is true if we know P(n – 1) is true. That is, mathematical induction is incremental.

  2. A proof by mathematical induction of nZ+,P(n) consists of 2 steps: Basis Step: Show that P(1) is true. Inductive Step: Show kZ+, [P(k)  P(k + 1)] . Once the two steps have been shown, we can conclude nZ+,P(n). That is, the rule of inference we are exploiting is: [P(1)  kZ+(P(k)  P(k + 1))]  nZ+,P(n) Remark: Note that in the inductive step of a proof by mathematical induction, we do not assume that kZ+,P(k) is true. This is what we are trying to prove. What we do is to show that if P(k) is true then P(k + 1) is also true for any positive integer k. Once the two steps have been established, we can see that P(n) is true for all n  Z+. First, P(1) is true because we showed that in the basis step. Then the inductive step tells us P(1)  P(2). Since P(1) is true, so is P(2). Then the inductive step says P(2)  P(3) …

  3. Ex: The sum of the first n odd positive integers is n2. Basis Step: [Show that P(1) is true] P(1) is the proposition that the sum of the first 1 odd positive integer is 12. This is true. Inductive Step: [Show kZ+, [P(k)  P(k + 1)]] Let k be a positive integer. Assume that P(k) is true. That is, assume that the sum of the first k odd positive integers is k2. Now we must show that P(k + 1) is true. That is, we wish to show that the sum of the first k + 1 odd positive integers is (k + 1)2. So we assumed that 1 + 3 + … + (2k – 1) = k2. We wish to show that 1 + 3 + … + (2k – 1) + (2k + 1) = (k + 1)2. Well 1 + 3 + … + (2k – 1) + (2k + 1) = k2 + (2k + 1) = (k + 1)(k + 1). So the sum of the first k + 1 odd positive integers is (k + 1)2. Hence by the principle of mathematical induction, the sum of the first n odd positive integers is n2.

  4. Ex: n < 2n for all positive integers n. Basis Step: [Show that P(1) is true] P(1) is the proposition 1 < 21. This is true since 1 < 2. Inductive Step: [Show kZ+, [P(k)  P(k + 1)]] Let k be a positive integer. Assume that P(k) is true. That is, assume that k < 2k. Now we must show that P(k + 1) is true. That is, we wish to show that k + 1 < 2k + 1. Since we assumed that k < 2k then k + 1 < 2k + 1. Now since k is a positive integer, 1 < 2k so k + 1 < 2k + 1 < 2k + 2k = 2* 2k = 2k + 1. So k + 1 < 2k + 1. This completes the inductive step. Hence by the principle of mathematical induction, n < 2n for all positive integers n. Sometimes we wish to show that P(n) is true for all natural numbers n. To do this, we simply shift the basis step to show P(0) is true.

  5. Ex: nk = 0 2k = 2n+1 – 1 for all natural numbers n. Basis Step: [Show that P(0) is true] P(0) is the proposition 0k = 0 2k = 20+1 – 1 = 2 – 1 = 1. This is true since 20 = 1. Inductive Step: [Show jN, [P(j)  P(j + 1)]] Let j be a natural number. Assume that P(j) is true. That is, assume that jk = 0 2k = 2j+1 – 1. Now we must show that P(j + 1) is true. That is, we wish to show that j+1k = 0 2k = 2(j+1)+1 – 1. Since jk = 0 2k = 2j+1 – 1 then j+1k = 0 2k = 2j+1 – 1 + 2j+1 = 2*2j+1 – 1 = 2j+2 – 1. So j+1k = 0 2k = 2(j+1)+1 – 1. This completes the inductive step. Hence by the principle of mathematical induction, nk = 0 2k = 2n+1 – 1 for all natural numbers n. We can shift the basis step in other ways as well. To prove that P(n) is true for all n  {b, b + 1, b + 2, etc.}, use P(b) for the basis step.

  6. Ex: 2n < n! for all positive integers n greater than 3. Basis Step: [Show that P(4) is true] P(4) is the proposition 24 < 4!. This is true since 16 < 24. Inductive Step: [Show j{4, 5, 6, …}, [P(j)  P(j + 1)]] Let j be a positive integer greater than 3. Assume that P(j) is true. That is, assume that 2j < j!. Now we must show that P(j + 1) is true. That is, we wish to show that 2j+1 < (j + 1)!. Since 2j < j! then 2j+1 < 2*j!. And since j > 3, then j + 1 > 4 so j + 1 > 2. Hence 2j+1 < 2*j! < (j + 1)*j! = (j + 1)!. So 2j+1 < (j + 1)!. This completes the inductive step. Hence by the principle of mathematical induction, 2n < n! for all positive integers n greater than 3.

  7. Strong Induction Note that after we have proved a result by mathematical induction, we can see that P(n) is true for all positive integers n because: P(1) was shown true in the basis step. P(2) is true since the inductive step says P(1)  P(2) and P(1) is true. P(3) is true since the inductive step says P(2)  P(3) and P(2) is true. … We can see from this process, that when P(k) is in question, we already know P(1), P(2), …, P(k – 1) are true. But we only need to know that P(k – 1) is true to apply the inductive step. We could however use the stronger inductive step that would require that P(1), P(2), …, P(k – 1) are all true in order to conclude that P(k) is true.

  8. A proof by strong induction of nZ+,P(n) consists of 2 steps: Basis Step: Show that P(1) is true. Inductive Step: Show kZ+, [(P(1)  P(2)  …  P(k))  P(k + 1)] . Once the two steps have been shown, we can conclude nZ+,P(n). That is, the rule of inference we are exploiting is: [P(1)  kZ+((P(1)  P(2)  …  P(k))  P(k + 1))]  nZ+,P(n) Remark: Note that with strong induction, in the inductive step we assume a stronger hypothesis. Instead of just assuming that P(k) is true then showing that P(k + 1) follows, we assume that P(x) is true for all x from 1 to k inclusive. This is why it is called strong induction. Once the two steps have been established, we can see that P(n) is true for all n  Z+. First, P(1) is true because we showed that in the basis step. Then the inductive step tells us P(1)  P(2). Since P(1) is true, so is P(2). Then the inductive step says [P(1)  P(2)]  P(3) …

  9. Ex: Any integer greater than 1 can be written as a product of primes. Basis Step: [Show that P(2) is true] P(2) is the proposition “2 can be written as a product of primes”. This is true; 2 is a product of 1 prime. Inductive Step: [Show j{2, 3, 4, …}, [(P(2)  P(3)  …  P(j))  P(j + 1)]] Let j be an integer greater than 1. Assume that P(x) is true for all x from 2 to j inclusive. That is, assume all integers from 2 to j inclusive can be written as a product of primes. Now we must show P(j + 1) is true. We must show that j + 1 can be written as a product of primes. There are two cases to consider. If j + 1 is prime then we are done. If j + 1 is not prime then j + 1 = a*b for some integers a and b where 1 < a < j + 1 and 1 < b < j + 1. So both a and b are between 2 and j inclusive. So P(a) and P(b) are true. Hence both a and b can be written as a product of primes. So j + 1 can be written as a product of primes. Hence by the principle of mathematical induction, any integer greater than 1 can be written as a product of primes.

  10. The Well-Ordering Property The Well-Ordering Property: A set S satisfies the well-ordering property if every non-empty subset of S has a “least” element. We will discuss the well-ordering property more after we cover relations. For now, we note that the well-ordering property is possessed by the set of natural numbers. That is, any non-empty subset of the set of natural numbers has a “least” element. It turns out that mathematical induction can be applied to any well-ordered set. The set of positive integers is well-ordered. Any set of the form {b, b + 1, b + 2, … } where b is an integer satisfies the well-ordering property. The set of integers does not satisfy the well-ordering property (not with the usual definition of “least”).

  11. We will now prove the principle of mathematical induction. Proof: Let P(n) be a proposition and assume that P(1) is true and also that kZ+[P(k)  P(k + 1)]. We will show that nZ+P(n) is true. Assume, to the contrary, that nZ+P(n) is false. That is, there is some positive integer c such that P(c) is false. Now consider the set F = {xZ+ | P(x) is false}. This is a non-empty subset of the set of positive integers (non-empty since c is in it). Hence by the well-ordering property of the set of positive integers, the set F has a least element. Call this least element d. Clearly d  1 since we know P(1) is true. So d > 1. But then d – 1 Z+. Also d – 1  F since d was the least element in F. So P(d – 1) is true. But since d – 1 Z+ then P(d – 1)  P(d). And since P(d – 1) is true, then P(d) is true. But this contradicts the fact that d is a member of f. So it must be the case that nZ+P(n) is true.

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