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Section 3.3

Mathematical Induction

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Mathematical Induction

• 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

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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)

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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)

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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

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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

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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

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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

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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

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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

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Example 1: Prove that the sum of the first n odd positive integers is n2

• Thus P(n) = sum of 1st n odd positive integers = n2

• 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

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Example 1: Inductive Step

• Suppose P(n) is true for some value of n; in other words:

1 + 3 + 5 + … + (2n - 1) = n2

where (2n-1) is the nth odd positive integer; the number obtained by adding 2 a total of n-1 times to 1

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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)

= n2 + (2n+1) = n2 + 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

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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)

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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

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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)

= n2 + 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

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Example 3: Proving Inequalities

• Prove P(n) = n < 2n for all positive integers n

• Basis step: P(1) = 1 < 21 = 1 < 2 TRUE

• Inductive step: assuming P(n) is true, show that (n+1) < 2n+1, given n < 2n

– Can add 1 to both sides of P(n), that is:

n+1 < 2n + 1– Since 1 <= 2n, we get:

n+1 < 2n+1 <= 2n+2n = 2n+1

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Example 4: Show that 2n > n2 whenever integer n > 4

• Since we’re only interested in n>=5, basis step is to prove P(5):

25 > 52

32 > 25 TRUE

• Inductive step: we need to prove that:2n+1 > (n + 1)2

assuming 2n > n2 is true

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Example 4: inductive step

2n+1 = 2 * 2n > (n +1)2 = n2 + 2n + 1

working from the right:

n2+2n+1 < n2+2n+n (we know this because n>1)

n2+2n+1 < n2+3n

n2+3n < n2+n2 (because n>3)

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Example 4: inductive step

So (n+1)2 < 2n2 < 2*2n

since n2 < 2n by the inductive hypothesis, and multiplying both sides by 2 doesn’t change the inequality

which means (n+1)2 < 2n+1 if n2 < 2n

reversing the inequalities:2n+1 > (n+1)2 if 2n > n2

so P(n) is true

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Example 5: Prove n3-n is divisible by 3 where n is a positive integer

• Basis: P(1) = 13 - 1 = 0 is divisible by 3 - true, since 0 is divisible by anything

• Inductive step:Assuming n3-n is divisible by 3, prove

P(n+1): (n+1)3 - (n+1) is divisible by 3

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Example 5 inductive step

• Expanding the polynomial:(n3+3n2+3n+1) - (n+1) = n3+3n2+2n

= n3+3n2+3n-n =

(n3 - n) + 3(n2 + 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

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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 ...

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Example 6: Show that 1+2+22+ … +2n = 2n+1-1 for all non-negative integers n

• Basis step: P(0) = 20 = 1 = 21 - 1

• Inductive step: assuming P(n) is true, need to prove:P(n+1) = 1+2+22+ … +2n+2n+1 = 2(n+1)+1-1 = 2n+2 - 1

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Example 6: inductive step

• Can use inductive hypothesis to substitute for first several terms:1+2+22+ … +2n+2n+1 = (1+2+22+ … + 2n)+2n+1

= (2n+1 - 1) + 2n+1

= 2 * 2n+1 - 1

= 2n+2 - 1

• So the theorem is proven

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Geometric Progression

• A geometric progression is a sequence of the form a, ar, ar2, … , arn … , where a and r are real numbers

• The previous example was a geometric progression with a=1 and r=2

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Example 7: proving formula for sum of a geometric progression

• The formula for the sum of a geometric progression is:

n

arj = ar0 + ar1 + ar2 + … + arn = (arn+1 - a) / (r-1) when r 1j=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

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Example 7: Basis Step

• Prove P(0):

a = (ar1 - a)/(r - 1) = a(r - 1)/(r - 1) = a

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Example 7: Inductive Step

• Assume P(n) is true; thus

ar0 + ar1 + ar2 + … + arn = (arn+1 - a) / (r-1)

• To show that, given this, P(n+1) is true, add arn+1 to both sides, obtaining:

ar0 + ar1 + ar2 + … + arn + arn+1 = (arn+1 - a) / (r-1) + arn+1

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Example 7: Inductive Step

• Rewriting right-hand side:(arn+1-a) / (r-1)+arn+1 = ((arn+1-a) / (r-1))+arn+1 ((r-1)/(r-1))

= ((arn+1-a) / (r-1))+((arn+2-arn+1)/(r-1)

= (arn+2-a) / (r-1)

• Combining left and right we have:a + ar + ar2 + … + arn +arn+1 = (arn+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

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Harmonic Numbers

• Harmonic numbers are the values Hk, where k=1, 2, 3 … defined by:

• Hk = 1 + 1/2 + 1/3 + … + 1/k

• Examples:H2 = 1 + 1/2 = 3/2

H4 = 1 + 1/2 + 1/3 + 1/4 = 25/12

H6 = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 = 147/60

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Example 8: show that H2n >= 1 + n/2 whenever n is a positive integer

• Basis: P(0) = H20 = H1 >= 1 + 0/2 TRUE

• Inductive: Assuming P(n) is true, prove H2n+1 >= 1 + (n+1)/2

H2n+1=1+1/2+1/3+ … +1/2n+1/2n+1+ … 1/2n+1

= H2n + 1/2n + 1 + … + 1/2n+1

>=(1 + n/2) + 2n(1/2n+1)

>=(1 + n/2) + 1/2 = 1 + (n+1)/2 so the inequality is valid for all non-negative integers n

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Example 9: number of subsets in a finite set

• Show that if S is a finite set with n elements, S has 2n subsets

• Basis: P(0) = set with 0 elements has 20 (1) subset - true, it has itself

• Inductive: Assuming P(n), prove P(n+1) - set with n+1 elements has 2n+1 subsets

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Example 9: inductive step

• Let T be a set with n+1 elements• We can write T=S{a} where aT 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*2n or 2n+1 subsets of T - so theorem is proven

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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)

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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

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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

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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

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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

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Section 3.3

Mathematical Induction

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