Mathematical Induction - UC Denverwcherowi/courses/m3000/lec… ·  · 2004-10-05I often like to think of an analogy to the PMI concerning a string ... not find answers by induction,

Mathematical Induction

The Induction PrincipleIn the axiomatic construction of the natural numbers ℕ = {1, 2, 3, ...} (Peano, 1899), one of the axioms needed is the Principle of Mathematical Induction (PMI):

If S is a subset of ℕ such that; 1) 1 ∈ S, and 2) for all n ∈ ℕ, if n ∈ S then n+1 ∈ S, then S = ℕ.

This principle expresses a fundamental property of the natural numbers which is often used in proving statements about sets of natural numbers.

Definition: A set S of natural numbers satisfying 2) above is called an inductive set.

The Induction PrincipleI often like to think of an analogy to the PMI concerning a string of dominoes standing on end. If you want to know that all the dominoes will fall, you need to know two things: 1) the dominoes are arranged so that if one falls, it knocks over the next one for all the dominoes (the dominoes form an inductive set); and 2) the first dominoe is knocked over.

Regions of a PlaneIn a plane, how many regions are formed by n lines, no three of which meet in a point, and no two of which are parallel?

2 4 7 11

Notice: n # regions 1 1 + 1 2 1 + 1 + 2 3 1 + 1 + 2 + 3 4 1 + 1 + 2 + 3 + 4

Regions of a PlaneIf f(n) = # of regions determined by n lines, the table suggests that

f n = 112⋯n = 1∑i=1

n

i = 1 n n12

= n2n22

.

But how do we prove that this is always true?

Let S = {n ∈ ℕ | f(n) = ½(n2 + n + 2)}. 1. (Base case) Observe that 1 ∈ S, since f(1) = 2 and ½(12 + 1 + 2) = 2. 2. S is an inductive set. We need to show that if m ∈ S (inductive hypothesis) then m+1 ∈ S.

Regions of a PlaneWe assume that the formula is correct for m lines and examine the case of m+1 lines. When we add a new line, it intersects all of the m old lines in a distinct point. This divides the line into m+1 segments. Each of these segments breaks an old region into two regions. So, the new line adds m+1 regions to the previous total. And under the induction hypothesis, the number of regions for m+1 lines is:

f m1=m2m22

m1=m23 m42

=m22 m1m122

=m12m122

.

Thus, S is an inductive set and by the PMI, S = ℕ. That is, the formula is valid for all natural numbers.

The sum of 3 consecutive cubes is divisible by 9.

Let S = {n ∈ ℕ: 9| n3 + (n+1)3 + (n+2)3}.

Observe that 1 ∈ S since 13 + 23 + 33 = 1 + 8 + 27 = 36 and 9|36.

Now assume that m ∈ S, i.e., 9| m3 + (m+1)3 + (m+2)3.Consider m13m23m33=m13m23m39 m227 m27

=m3m13m239m23 m3.

Since 9 divides the sum of the first three terms (induction hypothesis) and 9 divides the last term, 9 divides this final sum.Thus, m+1 ∈ S, and S is an inductive set. By the PMI, S = ℕ and the statement is true for all natural numbers.

∑i=1

n

i2=n n12 n1

6

∑i=1

n

i 2= n n12 n16

}Let S = {n ∈ ℕ:

Observe that 1 ∈ S since 12 = 1 and (1)(2)(3)/6 = 1. Now suppose that m ∈ S and consider:

1222⋯m12=1222⋯m2m12

=mm12 m16

m12=mm12 m16m12

6

=m1[2 m2m6 m6]6

=m1[m22 m3]6

=m1m112m116

.

Thus, m+1 ∈ S, so S is an inductive set. By the PMI, S = ℕ and the formula is valid for all natural numbers.

StampsShow that any postage > 7¢ can be made using only 3¢ and 5¢ stamps.

3¢3¢3¢3¢

3¢3¢3¢

3¢5¢

5¢

3¢3¢5¢

5¢5¢

3¢5¢

5¢

3¢3¢ 3¢

5¢5¢

5¢

8 9 10 11

12 13 14 15

StampsLet S = {postages > 7 which can be made from 3 and 5}

We have already seen that 8 ∈ S.

Suppose that m ∈ S. If a 5¢ is used to make up m¢, then remove the 5¢ stamp and replace it with 2 3¢ stamps. This will give m+1 ∈ S. If no 5¢ is used, then m¢ is made up entirely by 3¢ stamps. Since m > 7, there must be at least 3 3¢ stamps used. Remove 3 3¢ stamps and replace them with 2 5¢ stamps. So in this case also m+1 ∈ S.

Thus, S is an inductive set and by the PMI all natural numbers >7 are in S. ÿ

Size of the Power SetTheorem: If a set A has n elements then P (A) has 2n elements.

Pf: Let S = {n ∈ ℕ: the statement above is true} 1 ∈ S If A has 1 element, i.e., A = {x}, then P (A) = {∅, A} which has 2 = 21 elements.

Suppose m ∈ S. Consider a set A with m+1 elements. Pick one element, say x, in A. There are just two types of subsets of A, those that contain x and those that don't. The subsets that do not contain x are subsets of A' = A – {x} and since A' has m elements, there are 2m of these.

Size of the Power SetThe subsets that contain x are of the form {x} ∪ {a subset of A'}.Thus, there are also 2m of these.In total there are 2m + 2m = 2m+1 subsets of A.So, m+1 ∈ S, and S is an inductive set. By the PMI, S = ℕ and the statement is true for all natural numbers.

All horses have the same colorLet S = {n ∈ ℕ: Any collection of n horses have the same color}

Clearly 1 ∈ S.

Now suppose that m ∈ S. Consider a set of m+1 horses. Cut out one horse from this herd, and you are left with a set of m horses. By the induction hypothesis, they all have the same color. Now, return the horse that was cut out, and cut out a different horse. By the induction hypothesis again, the remaining horses all have the same color, so the horse that was cut out the first time has the same color as all the other horses. Thus, m+1 ∈ S, S is an inductive set and by the PMI all horses have the same color. ??????????

Caveats

As this last example shows, small cases can make a big difference ... making sure that an argument works in the small cases makes a lot of sense.

Also, it should be emphasized that mathematical induction is a proof technique ... it is not a problem solving strategy. You can not find answers by induction, you can only prove that answers you suspect are valid are in fact valid.

Principle of Complete Induction (PCI)Theorem: If S is a set of natural numbers satisfying the condition: for each natural number m, {i ∈ ℕ| i < m} ⊆ S implies m ∈ S;then S = ℕ.

Pf: (By induction) Let K = {n ∈ ℕ| {1,...,n} ⊆ S}. 1 ∈ K : the property that S satisfies holds for all natural numbers, so it holds in particular for m = 1. That is, if {i ∈ ℕ| i < 1} ⊆ S then 1 ∈ S. However, since {i ∈ ℕ| i < 1} = ∅ and ∅ ⊆ S for any set S, we have 1∈ S, and thus, 1 ∈ K since {1} ⊆ S. Now suppose that k ∈ K. That means that {1, ..., k} ⊆ S. The property that S satisfies, with m = k+1, says that k+1 ∈ S. Therefore, {1, ..., k+1} ⊆ S, so k+1 ∈ K. K is thus an inductive set and by the PMI K = ℕ. Thus, ℕ ⊆ S, but we already know that S ⊆ ℕ, so we have S = ℕ.

PCIThe PCI is sometimes called strong induction (and the PMI weak induction) because its hypothesis is a stronger statement ... but some would say that it is in fact a weaker result, because it requires so much more to be assumed in order to prove the same consequence.

We can however ignore this semantical quibble because these statements are logically equivalent. We have just proved the PCI assuming the PMI. We will now prove the PMI assuming the PCI (that is, taking the PCI as an axiom, we will show that the PMI is a theorem). This then shows that these two forms of induction are equally "powerful", and either one may be taken as an axiom in the development of the natural numbers.

PMI – the theoremTheorem: Assume the validity of the PCI. If S is an inductive set of natural numbers that contains 1, then S = ℕ.

Pf: We must show that the set S satisfies the PCI condition: (∀m ∈ℕ)({1,...,m-1}⊆ S ⇒ m ∈ S) Case 1: m = 1 1 ∈ S by assumption. Case 2: m > 1 {1, ..., m-1} is non-empty and contains the integer m-1. {1, ... , m-1}⊆ S ⇒ m-1 ∈ S. Since S is an inductive set, m-1 ∈ S ⇒ m ∈ S. Thus, the condition is satisfied and by the PCI, S = ℕ.

Strong InductionThe PCI can be used in many cases where the PMI would be awkward to use.

Prop: Every natural number greater than 1 has a prime divisor.

Pf: Let S = {n ∈ ℕ| n > 1 and n has a prime divisor}. Note that while 1 ∉ S, 2 ∈ S. For any integer m > 1, assume that {2, ... , m} ⊆ S. Consider m+1. If m+1 is prime, then it has itself as a prime divisor, so m+1 ∈ S. If m+1 is not prime, then m+1 = ab for integers a, b with 1 < a < m+1 (and the same for b, but we don't need that). Thus, a ∈ {2, ..., m} and so a ∈ S. Since a has a prime divisor, m+1 has a prime divisor, i.e., m+1 ∈ S. Thus, by the PCI, S = {n ∈ ℕ| n > 1}.

Inductive DefinitionsIn some situations an infinite set of objects indexed by the natural numbers (i.e., there is a 1st object, a 2nd object, etc.), can be defined in an inductive manner (also called a recursive definition).

Example: Factorials n! = n(n-1)(n-2) ... (3)(2)(1)Can be defined as follows: 1! := 1; n! := n(n-1!)

Thus, 4! = 4(3!) and 3! = 3(2!) and 2! = 2(1!) = 2(1) so 4! = 4(3!) = 4(3(2!)) = 4(3(2(1))) = 4(3)(2)(1)

To prove that a recursive definition is valid for all n, use PMI.

Example Let x

1 = 1, x

2 = 1 and for n ≥ 2, x

n+1 := x

n + 2x

n-1. Prove that x

n is

divisible by 3 iff n is divisible by 3.

We observe that xn+3

= xn+2

+ 2xn+1

= (xn+1

+ 2xn) + 2x

n+1 = 3x

n+1+ 2x

n.

We first prove the necessity (⇐): Assume that 3 | n, i.e., n = 3k. Let S = {k : 3 | x

3k}.

Since x3 = 3, we have 1 ∈ S.

Suppose m ∈ S. That is, 3 | x3m

. By the observation, x3m+3

= 3x3m+1

+ 2x

3m. So, 3 | x

3m+3 = x

3(m+1). Thus, S is inductive and by the PMI, S

= ℕ, i.e., if 3 | n then 3 | xn.

Example (cont.) Let x

1 = 1, x

2 = 1 and for n ≥ 2, x

n+1 := x

n + 2x

n-1. Prove that x

n is

divisible by 3 iff n is divisible by 3.

We observe that xn+3

= xn+2

+ 2xn+1

= (xn+1

+ 2xn) + 2x

n+1 = 3x

n+1+ 2x

n.

Now we show the sufficiency (⇒): Let S = {k : 3 ∤ x

3k-1 and 3 ∤ x

3k-2}.

Since x1 = 1 and x

2 = 1, we have 1 ∈ S.

Suppose m ∈ S. That is, 3 ∤ x3m-1

and 3 ∤ x3m-2

. By the observation, x

3m+2 = 3x

3m + 2x

3m-1, and x

3m+1 = 3x

3m-1 + 2x

3m-2. So,

3 ∤ x3m+2

= x3(m+1)-1

and 3 ∤ x3m+1

= x3(m+1)-2

. Thus, S is inductive and by the PMI, S = ℕ, i.e., if 3 ∤ n then 3 ∤ x

n.

Fibonacci NumbersIn 1202, Leonardo Fibonacci (1170-1250) posed the following famous problem: Suppose a particular breed of rabbit breeds one new pair of rabbits each month, except that a 1-month old pair is too young to breed. Suppose further that no rabbit breeds with any other except its paired mate and that rabbits live forever. Starting with 1 pair of newborn rabbits, how many pairs of rabbits exist after n months? month # rabbit pairs 1 1 2 1 3 2 (originals + newborns) 4 3 (originals + 1 mo. + new) 5 5 ( ... + 2 new) 6 8 ( ... + 3 new)

Fibonacci NumbersLet f

n be the number of pairs of rabbits alive after n months.

Then, f

1 = f

2 = 1 and

fn = f

n-1 + f

n-2 for n > 2.

The numbers fn are known as the Fibonacci numbers.

Problems involving Fibonacci numbers usually are proved by induction due to the recursive definitions, and the PCI is often more useful than the PMI for them.

A RecursionLet F

1= F

2 = 1, and F

n+2 = F

n+1 + F

n + F

n+1F

n.

Prove that Fn = 2f(n) - 1 where f(n) is the nth Fibonacci number.

Pf: Let S = {n ∈ ℕ| Fn = 2f(n) – 1}. Note that 1∈ S.

Suppose that {1,...,m} ⊆ S for m > 1. Consider Fm+1

. F

m+1 = F

m + F

m-1 + F

mF

m-1

= (2f(m)-1) + (2f(m-1)-1) + (2f(m)-1)(2f(m-1)-1) = 2f(m) + 2f(m-1) -2 + 2f(m)+f(m-1) – 2f(m) -2f(m-1) + 1 = 2f(m)+f(m-1) -1 = 2f(m+1) -1.Thus m+1 ∈ S. So, by the PCI S = ℕ.

Well-Ordering PrincipleThis simply stated property of the natural numbers turns out to be quite powerful and equivalent to the PMI.

Theorem (WOP): Every non-empty subset of ℕ has a smallest element.

Pf: Let T be a subset of ℕ. Suppose that T has no smallest element. Let S = ℕ – T.Assume that {1,...,m} ⊆ S. If m+1 ∉ S, then m+1 ∈ T and would be the smallest element in T since all smaller integers are not in T.→← Thus, m+1 ∈ S and by the PCI, S = ℕ. But then, T = ∅. Thus, if T ≠ ∅ then T has a smallest element.

Using the WOPProp: Every natural number greater than 1 has a prime divisor.

Pf: If n is prime, then it is a prime divisor of itself. If n is composite, then n has a factor other than 1 or n. Let s be the smallest of these factors (s exists by WOP). Suppose s is composite. Then s has a divisor d, with 1 < d < s. Since d divides s, d divides n and this contradicts the definition of s. Thus, s is prime.

Division Algorithm for ℕTheorem: Let a and b be natural numbers with b ≤ a. Then there exist q ∈ ℕ and r ∈ ℕ ∪ {0} such that: a = qb + r with 0 ≤ r < b.

Pf: Let T = {s ∈ ℕ | a < sb}. T is non-empty (for instance, a+1 is in T). By the WOP, T contains a smallest element, w.Let q = w-1 and r = a – qb. Since a ≥ b, w > 1, so q ∈ ℕ.Since q < w, a ≥ qb, so r ≥ 0.By the definition of r, a = qb + r.BWOC, suppose r ≥ b.Then r = a – (w-1)b ≥ b, so a ≥ wb →←Therefore, r < b.

Odds and EvensTheorem: Every natural number is either odd or even.

Pf: 1 is odd ( = 2(0) + 1). Let n be a natural number > 1. By the division algorithm for natural numbers, there exist integers q and r so that n = 2q + r with 0 ≤ r < 2.Thus, r = 0 or r = 1. When r = 0, n = 2q is even. When r = 1, n = 2q +1 is odd.