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1 Section 3.3 Mathematical Induction

2 Technique used extensively to prove results about large variety of discrete objects Can only be used to prove results obtained some other way - NOT a tool for developing theorems

3 Application of Mathematical Induction Many theorems state that propositional function P(n) is true for all positive integers n Mathematical induction is a technique used for proving theorems of this kind Such propositions take the form  nP(n), where n is a positive integer (n  N)

4 Steps in Inductive Proof Basis step: show that P(1) is true Inductive step: show that the implication P(n)  P(n+1) is true for every positive integer n Expressed as a rule of inference, the inductive proof technique can be stated as: [P(1)   n(P(n)  P(n+1))]   nP(n)

5 Applying Inductive Proof Basis step is simple: substitute 1 for n, and show that P(n) is true for n=1 To prove inductive step: –Assume P(n) is true (inductive hypothesis) –Show that, under this condition, P(n+1) must also be true

6 Analysis of Mathematical Induction At first glance, mathematical induction looks like circular reasoning, but it isn’t: –We don’t assume P(n) is ALWAYS true –Just need to show that P(n+1) must be true IF P(n) is true

7 Analysis of Mathematical Induction When using mathematical induction to prove a theorem, we show that P(1) is true; –Since P(1) implies P(2), P(2) is true; –Since P(2) implies P(3), P(3) is true, etc. –So P(k) is true for any positive integer k

8 Validity of Mathematical Induction Validity is based on the well-ordering property: –Every non-empty set of non-negative integers has a least element If we know that P(1) is true and P(n)  P(n+1) is true for all positive integers, we can show that P(n) must be true for all positive integers, as shown on the next several slides

9 Validity of Mathematical Induction Suppose there is at least one positive integer for which P(n) is false (assume  p) This means S, the set of positive integers for which P(n) is false, is non-empty By the well-ordering property, S has a least element, denoted k We know k can’t be 1, because we know P(1) is true

10 Validity of Mathematical Induction Since k is positive and k > 1, k-1 must be positive Since k-1 < k, then k-1  S, so P(k-1) must be true Since the implication P(k-1)  P(k) is also true, P(k) must be true This is a contradiction; hence, P(n) is true for all positive integers n

11 Example 1: Prove that the sum of the first n odd positive integers is n 2 Thus P(n) = sum of 1st n odd positive integers = n 2 Basis step: –P(1) = sum of 1st 1 odd positive integers = 1: TRUE Inductive step must show that  n(P(n)  P(n+1)) is true

12 Example 1: Inductive Step Suppose P(n) is true for some value of n; in other words: 1 + 3 + 5 + … + (2n - 1) = n 2 where (2n-1) is the nth odd positive integer; the number obtained by adding 2 a total of n-1 times to 1

13 Example 1: Inductive Step We must show that, if P(n) is true, P(n+1) is true: P(n+1) = 1+3+5+ … (2n-1) + (2n+1) = (n+1) 2 –Assuming P(n) is true, it follows that: 1+3+5+…+(2n-1)+(2n+1) = [1+3+…+(2n-1)] + (2n+1) = n 2 + (2n+1) = n 2 + 2n + 1 = (n+1) 2 –This shows that P(n+1) follows from P(n); since P(1) and  n(P(n)  P(n+1) )are true, P(n) is true

14 Example 2: Prove that the sum of the first n even positive integers is n(n+1) Basis Step: Even though 1 is not an even integer, we still prove P(1) - –We’re not dealing with n as a value, but n as a position –2 is the nth positive even integer when n=1 –Since P(1) = 1(1+1) = 2, the proposition is true for P(1) (the sum of 2 is 2)

15 Example 2: Inductive Step We must prove  n(P(n)  P(n+1)) Suppose P(n) is true - this means: 2 + 4 + … + 2n = n (n + 1) So we need to demonstrate that P(n+1) is true if P(n) is true

16 Example 2: Inductive Step 2+4+ … +2n+2(n+1) = (n+1)((n+1)+1) = (n+1)(n+2) = (2+4+ … +2n) + (2n+2) = n(n+1) + (2n+2) = n 2 + 3n +2 = (n+1)(n+2) So P(n+1) follows from P(n): Since P(1) and  n(P(n)  P(n+1)) are true, P(n) is true

17 Example 3: Proving Inequalities Prove P(n) = n < 2 n for all positive integers n Basis step: P(1) = 1 < 2 1 = 1 < 2 TRUE Inductive step: assuming P(n) is true, show that (n+1) < 2 n+1, given n < 2 n –Can add 1 to both sides of P(n), that is: n+1 < 2 n + 1 –Since 1 <= 2 n, we get: n+1 < 2 n +1 <= 2 n +2 n = 2 n+1

18 Example 4: Show that 2 n > n 2 whenever integer n > 4 Since we’re only interested in n>=5, basis step is to prove P(5): 2 5 > 5 2 32 > 25 TRUE Inductive step: we need to prove that: 2 n+1 > (n + 1) 2 assuming 2 n > n 2 is true

19 Example 4: inductive step 2 n+1 = 2 * 2 n > (n +1) 2 = n 2 + 2n + 1 working from the right: n 2 +2n+1 1) n 2 +2n+1 < n 2 +3n n 2 +3n 3)

20 Example 4: inductive step So (n+1) 2 < 2n 2 < 2*2 n since n 2 < 2 n by the inductive hypothesis, and multiplying both sides by 2 doesn’t change the inequality which means (n+1) 2 < 2 n+1 if n 2 < 2 n reversing the inequalities: 2 n+1 > (n+1) 2 if 2 n > n 2 so P(n) is true

21 Example 5: Prove n 3 -n is divisible by 3 where n is a positive integer Basis: P(1) = 1 3 - 1 = 0 is divisible by 3 - true, since 0 is divisible by anything Inductive step: Assuming n 3 -n is divisible by 3, prove P(n+1): (n+1) 3 - (n+1) is divisible by 3

22 Example 5 inductive step Expanding the polynomial: (n 3 +3n 2 +3n+1) - (n+1) = n 3 +3n 2 +2n = n 3 +3n 2 +3n-n = (n 3 - n) + 3(n 2 + n) The first term in this sum is divisible by 3 according to the inductive hypothesis The second term is divisible by 3 by definition So P(n) is true if (P(1)   n(P(n)  P(n+1)) is true

23 Adjusting Basis Step We have already seen that the basis step can be adjusted if P(1) is not part of the universe of discourse for P(n) For example, we have already seen P(5) used as basis In the next example, we’ll use P(0): this tactic is useful to prove P(n) for the series n=k, k+1, k+2...

24 Example 6: Show that 1+2+2 2 + … +2 n = 2 n+1 -1 for all non-negative integers n Basis step: P(0) = 2 0 = 1 = 2 1 - 1 Inductive step: assuming P(n) is true, need to prove: P(n+1) = 1+2+2 2 + … +2 n +2 n+1 = 2 (n+1)+1 -1 = 2 n+2 - 1

25 Example 6: inductive step Can use inductive hypothesis to substitute for first several terms: 1+2+2 2 + … +2 n +2 n+1 = (1+2+2 2 + … + 2 n )+2 n+1 = (2 n+1 - 1) + 2 n+1 = 2 * 2 n+1 - 1 = 2 n+2 - 1 So the theorem is proven

26 Geometric Progression A geometric progression is a sequence of the form a, ar, ar 2, …, ar n …, where a and r are real numbers The previous example was a geometric progression with a=1 and r=2

27 Example 7: proving formula for sum of a geometric progression The formula for the sum of a geometric progression is: n  ar j = ar 0 + ar 1 + ar 2 + … + ar n = (ar n+1 - a) / (r-1) when r  1 j=0 To prove formula using induction, let P(n) be the proposition that the sum of the first n+1 terms in this formula is correct

28 Example 7: Basis Step Prove P(0): a = (ar 1 - a)/(r - 1) = a(r - 1)/(r - 1) = a

29 Example 7: Inductive Step Assume P(n) is true; thus ar 0 + ar 1 + ar 2 + … + ar n = (ar n+1 - a) / (r-1) To show that, given this, P(n+1) is true, add ar n+1 to both sides, obtaining: ar 0 + ar 1 + ar 2 + … + ar n + ar n+1 = (ar n+1 - a) / (r-1) + ar n+1

30 Example 7: Inductive Step Rewriting right-hand side: (ar n+1 -a) / (r-1)+ar n+1 = ((ar n+1 -a) / (r-1))+ar n+1 ((r-1)/(r-1)) = ((ar n+1 -a) / (r-1))+((ar n+2 -ar n+1 )/(r-1) = (ar n+2 -a) / (r-1) Combining left and right we have: a + ar + ar 2 + … + ar n +ar n+1 = (ar n+2 -a) / (r-1) So if P(n) is true, P(n+1) must also be true & formula for sum of geometric series is correct

31 Harmonic Numbers Harmonic numbers are the values H k, where k=1, 2, 3 … defined by: H k = 1 + 1/2 + 1/3 + … + 1/k Examples: H 2 = 1 + 1/2 = 3/2 H 4 = 1 + 1/2 + 1/3 + 1/4 = 25/12 H 6 = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 = 147/60

32 Example 8: show that H 2 n >= 1 + n/2 whenever n is a positive integer Basis: P(0) = H 2 0 = H 1 >= 1 + 0/2 TRUE Inductive: Assuming P(n) is true, prove H 2 n+1 >= 1 + (n+1)/2 H 2 n+1 =1+1/2+1/3+ … +1/2 n +1/2 n +1+ … 1/2 n+1 = H 2 n + 1/2 n + 1 + … + 1/2 n+1 >=(1 + n/2) + 2 n (1/2 n+1 ) >=(1 + n/2) + 1/2 = 1 + (n+1)/2 so the inequality is valid for all non-negative integers n

33 Example 9: number of subsets in a finite set Show that if S is a finite set with n elements, S has 2 n subsets Basis: P(0) = set with 0 elements has 2 0 (1) subset - true, it has itself Inductive: Assuming P(n), prove P(n+1) - set with n+1 elements has 2 n+1 subsets

34 Example 9: inductive step Let T be a set with n+1 elements We can write T=S  {a} where a  T and S=T-{a} For each subset of S there are exactly 2 subsets of T - one containing {a} and one not containing {a} Since there are 2n subsets of S (inductive hypothesis), there are 2*2 n or 2 n+1 subsets of T - so theorem is proven

35 Second principle of mathematical induction Different form of mathematical induction Assume P(k) true for k = 1 … n, then show P(k+1) must also be true based on this assumption –Basis step: show P(1) is true –Inductive step: show [P(1)  P(2)  …  P(n)]  P(n+1)

36 Example 10 Prove that every amount of postage >= 12 cents can be formed using 4-cent and 5-cent stamps Basis: Postage of 12 cents can be formed using 3 4-cent stamps

37 Example 10: inductive step Assume P(n) true If at least one 4-cent stamp was used, can get n+1 by replacing the 4-cent stamp with a 5-cent stamp: –P(13) is 5, 4, 4 –P(14) is 5, 5, 4 –P(15) is 5, 5, 5

38 Example 10: inductive step Since n>=12, at least 3 5-cent stamps were used; replace these with 4 4-cent stamps to get n+1: P(16) = 4, 4, 4, 4 Applying second principle of mathematical induction we see that P(12) through P(15) can be formed using 4 and 5-cent stamps - this is basis step, where n=15

39 Example 10: inductive step Let n>=15 Assume we can form postage of k cents, where 12<=k<=n To form postage of n+1 cents, use stamps that form n-3 cents + 1 4-cent stamp This works for any value of n

40 Section 3.3 Mathematical Induction - ends -

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