proofby mathematical by mathematical induction.pdf

Section 1.6 Proof by Induction 1

Section 1.Section 1.Section 1.Section 1.6666: : : : Proof Proof Proof Proof by Mathematical by Mathematical by Mathematical by Mathematical Induction Induction Induction Induction Purpose of SectionPurpose of SectionPurpose of SectionPurpose of Section: To introduce the Principle of Mathematical Induction, both weak and the strong versions, and show how certain types of theorems can be proven using this technique. IntroductionIntroductionIntroductionIntroduction

The Principle of Principle of Principle of Principle of Mathematical Mathematical Mathematical Mathematical IIIInductionnductionnductionnduction is a method of proof normally used to prove that a proposition is true for all natural numbers 1,2,3, , although there are many variations of the basic method. The method is particularly important in discrete mathematics, and one often sees theorems proven by induction in areas like computer science. The technique is so intuitive and familiar that it sometimes is used without reference to its use. For example, suppose someone tells you they are going to color the natural numbers 1,2,3, with some color and that the number 1 will be colored blue, and that if a given number is colored blue, then the next number will also be colored blue. Is there any doubt in your mind that all the numbers will be colored blue? Of course not. This is the induction axiom. And the good thing is you dont have to proof it. It is an axiom1.

1 In 1889 Italian mathematician Guiseppe Peano (1858-1932) published a list of five

axioms which define the natural numbers. Peanos 5th axiom is called the induction axiom, which states that any property which belongs to 1 and also to the successor of any number which has the property belongs to all numbers.


Margin Note: Margin Note: Margin Note: Margin Note: Dont confuse mathematical induction with inductive reasoning associated with the natural sciences. Inductive reasoning in the sciences is a scientific method whereby one induces general principles from observations. Mathematical induction is not the same thing: it is a deductive form of reasoning used to establish the validity of a proposition for all natural numbers. Mathematical induction provides a convenient way to establish that a statement is true for all natural numbers 1,2, . The following statements are prime candidates for proof by mathematical induction. For all natural numbers n , 21 3 5 (2 1)n n+ + + + = If A is a set contains n elements, then the collection of all subsets of A contain 2n elements. 1 3 5 2 1 1

2 4 6 2 3 1n

n n

+

for all natural numbers n .

Here, then is how the method of mathematical induction works. Mathematical InductionMathematical InductionMathematical InductionMathematical Induction

The Principle of Mathematical Induction is a method of proof for verifying that a proposition ( )P n is true for all natural numbers n . The methodology for proving theorems by induction is as follows. Methodology ofMethodology ofMethodology ofMethodology of Mathematical InductionMathematical InductionMathematical InductionMathematical Induction To verify that a proposition ( )P n holds for all natural numbers n , the Principle of Mathematical InductionPrinciple of Mathematical InductionPrinciple of Mathematical InductionPrinciple of Mathematical Induction consists of successfully carrying out the following two steps.

i Base CaseBase CaseBase CaseBase Case:::: Prove that (1)P is true.

i Induction Step:Induction Step:Induction Step:Induction Step: Assume that ( )P n is true for an arbitrary n , then prove that ( 1)P n + is true.

Note:Note:Note:Note: There are several; modifications of the basic induction proof stated here. For example, there is no reason the base case starts with (1)P . If the base case is replaced by the verification of ( )P a , where " "a is any integer (positive or negative), one would conclude ( )P n true for all n a . Also, if the


induction step is replaced by the implication ( ) ( 2)P n P n + , one would conclude ( )P n true for (1), (3),..., (2 1),...P P P n + Also, sometimes the base case consists in verifying more than one proposition, maybe (1), (2)P P and (3)P are required for the initial step (look at Example 7 in this section). Margin NotMargin NotMargin NotMargin Note:e:e:e: Induction works like dominos. You tip over the first domino; the first domino tips over the second one; the second one tips over the third one; and so on. You get the idea. Eventually, all dominoes are tipped over. TheoremTheoremTheoremTheorem 1 1 1 1 (Famous Identity(Famous Identity(Famous Identity(Famous Identity by I by I by I by Inductionnductionnductionnduction))))

If n is a positive integer, then ( )11 22

n nn

++ + + = .

Proof: Proof: Proof: Proof: We denote ( )P n as the statement to be proved:

( )1( ) : 1 22

n nP n n

++ + + =

Base CaseBase CaseBase CaseBase Case:::: Clearly (1)P is true since2

(1)P says 1 (2)12

=

Induction StepInduction StepInduction StepInduction Step:::: We assume ( )P n true for an arbitrary n :

( )1( ) : 1 22

n nP n n

++ + + = (assume true)

Adding 1n + to each side of this equation, we get:

( )11 2 ( 1) ( 1)2

( 1) 2( 1)2

( 1)( 2)2

n nn n n

n n n

n n

++ + + + + = + +

+ + +=

+ +=

which is the statement ( 1)P n + . Hence, we have proven that ( 1)P n + is true. By induction the result is proven.

2 The reader can verify that P(2) and P(3) are also true, but that isnt relevant to proof by induction.


Note:Note:Note:Note: Do we have to prove that the principle of mathematical induction holds? The answer is no. We accept mathematical induction as a logical axiom in much the same way as we accept the classical rules of Aristotelian logic.

Famous Story in Famous Story in Famous Story in Famous Story in MathematicsMathematicsMathematicsMathematics

The identity

( )11 22

n nn

++ + + =

can also be proven using the idea Gauss had when he was 9 years old and impressed his teacher by summing 1 2 100 5050.+ + + =

Indeed: ( ) ( ) ( )1 2 100 1 100 2 99 50 51 50 101 5050.+ + + = + + + + + + = = Theorem Theorem Theorem Theorem 2 (Induction in Calculus) 2 (Induction in Calculus) 2 (Induction in Calculus) 2 (Induction in Calculus) Prove that for every natural numbers n , we have

( ) ( )( ) : n x xnd xeP n x n edx = + Proof: Proof: Proof: Proof: We show ( )P n is true for all natural numbers n by induction.

Base StepBase StepBase StepBase Step: If 1n = and from the product rule for differentiation, we can write

( ) ( )1x x x xd xe dx e e x edx dx

= + = + .

Induction Step: Induction Step: Induction Step: Induction Step: Assuming

( ) ( )( ) : n x xnd xeP n x n edx = + true for an arbitrary n , we compute


( ) ( )

( )( )

( )

1

1( 1) :

(induction assumption)

(product rule) 1

n x n x

n n

x

x x

x

d xe d xedP ndx dx dx

dx n e

dxx n e e

x n e

+

+

+ =

= +

= + +

= + +

which proves ( 1)P n + . Hence the theorem is proved.

Theorem Theorem Theorem Theorem 3333 (Proving Closed Forms of Series by Induction) (Proving Closed Forms of Series by Induction) (Proving Closed Forms of Series by Induction) (Proving Closed Forms of Series by Induction) For any positive integer n :

1

2

0

111

nnk n

k

xx x x x

x

+

=

+ + + + =

Proof: Proof: Proof: Proof:

Letting ( )P n be the statement

1

2 1( ) : 11

nn xP n x x x

x

+

+ + + + =

we verify

Base CaseBase CaseBase CaseBase Case:::: In this problem the initial step starts at 0n = due to the way ( )P n is defined. It is not necessary, but we evaluate both (0)P and (1)P .

0 1 21 1 1(0) : 1 1 (1) : 1 1

1 1 1x x xP P x xx x x

+

= = = + = = +

Induction Induction Induction Induction Step:Step:Step:Step: Assuming ( )P n is true for any natural number n , we have

1

2 111

nn xx x x

x

+

+ + + + =

Adding 1nx + to each side of this equation, gives


12 1 1

1 1

1 2 1

2

1111 ( 1)

11

11

1

nn n n

n n

n n n

n

xx x x x x

x

x x x

x

x x x

x

x

x

++ +

+ +

+ + +

+

+ + + + + = +

+ =

+ =

=

which says the ( 1)P n + is true. By induction the theorem is proven.

Margin Note: Margin Note: Margin Note: Margin Note: How do we know if the proof of a theorem is correct? It would be nice if we could feed a theorem into a computer and let the machine verify if the proof, but except for very simple cases, this is not feasible. Many theorems are easy to follow, but many are extremely difficult. When Andrew Wiles proved Fermats Last Theorem in 1993, he presented his results to a group of experts. No one could verify on the spot if the proof was correct due to its length (100 pages) and complexity. Only after the proof was analyzed by a committee of six specialists over several months was the theorem validated. Theorem Theorem Theorem Theorem 4444 (Inequality by Induction) (Inequality by Induction) (Inequality by Induction) (Inequality by Induction) The inequality 22n n> holds for 5n . Proof: Proof: Proof: Proof: Defining 2( ) : 2nP n n> we prove: Base Case:Base Case:Base Case:Base Case: 5 2(5) : 2 32 25 5P = > = . Induction Step:Induction Step:Induction Step:Induction Step: ( ) ( )1P n P n + for 5n . We must show

( )22 12 2 1 , 5n nn n n+> = + . We write

1

2

2 2

2

2

2

2

2 2 22 (induction hypothesis)

5 (we assume 5)2 32 1

( 1)

n n

n

n n

n n n

n n n

n n

n

+=

>

= +

+

= + +

> + +

= +

Hence ( 1)P n + is true and so by induction ( )P n true for all 5n .


Theorem Theorem Theorem Theorem 5555 (Tower of (Tower of (Tower of (Tower of Benares Benares Benares Benares Theorem) Theorem) Theorem) Theorem)

The Tower of Benares3 puzzle (or tower of Hanoi) consists moving a collection of disks from one peg onto another, where one is only allowed to move one disk at a time and no larger disk can ever be above a smaller disk.

The number of steps required to move n disks from one peg to another peg (either one) is 2 1n . ProofProofProofProof: Let ( ) the number of moves for disks is 2 -1nP n n= We prove the induction steps:

( )( ) ( )1 is true

1

P

P n P n +

i ( )1P is true since for a single disk it takes 12 1 1 = step to move one disk from one peg to another. i Assume ( )P n is true. That is, it takes 2 1n steps to transfer n disks. We now show it requires 12 1n+ steps to move 1n + disks. Let 1n + disks be stacked on pet A . We move the top n disks to one of the other pegs, say peg C (which takes 2 1n steps by assumption), and then move the largest disk, still sitting on peg A , to peg B (which takes 1 step).

3 According to legend, the Temple at Benares in ancient India marked the center of the world.

Within the temple priests moved golden disks from one diamond needle to another. God placed 64 gold disks on one needle at the time of creation. It was said that when the temple priests completed their task the universe would come to an end. Since it takes 264 -1 moves to complete the task and assuming the priest move one disk per second, it would take roughly 585 billion years to move the disks from one needle to another..


Finally, we move the n disks sitting on peg C to the top of the largest disk resting on peg B (another 2 1 steps)n . Hence, we have the 1n + disks sitting on pole B in the proper arrangement. We are done. Adding up these steps, we find

( ) ( )

1

number of steps required to move 1 disks 2 1 1 2 1

2 2 12 1

n n

n

n

n

+

+ = + +

=

=

Hence, we have proven ( )1P n + is true so by induction the number of steps required to move n disks from one pole to another is 2 1n for any natural number n .

Is Mathematics Based on Logic?Is Mathematics Based on Logic?Is Mathematics Based on Logic?Is Mathematics Based on Logic? In the late 1800s and early 1900s a few mathematicians and logicians like Dedekind, Frege, Hilbert, Russell, and Whitehead tried to construct arithmetic from formal logical principles and axioms. The Italian logician Giuseppe Peano (1858-1932) formulated five axioms from which one could deduce the natural numbers 1, 2, 3, . . This brings up the philosophical question of what exactly should be the starting point of mathematics and arithmetic? To some mathematical intuitionists like Kronecker and Poincare, felt the natural numbers are as intuitive and basic as one can get and should act as the starting point, rather than being formulated on less intuitive logical axioms. Logicians would disagree.

The type of induction discussed thus far is called weak inductionweak inductionweak inductionweak induction. We now introduce the concept of strong induction, although technically the two methods are equivalent. Strong InductionStrong InductionStrong InductionStrong Induction

The Principle of Mathematical Induction stated thus far is sometimes called weak induction in contrast to a variation of it called strong induction. The two types of induction are actually equivalent but sometimes weak induction doesnt fit into the proof in a natural way whereas strong induction does.


Methodology oMethodology oMethodology oMethodology of (Strong) Mathematical Inductionf (Strong) Mathematical Inductionf (Strong) Mathematical Inductionf (Strong) Mathematical Induction To verify a proposition ( )P n holds for all natural numbers n , the Principle of Principle of Principle of Principle of (Strong) Mathematical Induction(Strong) Mathematical Induction(Strong) Mathematical Induction(Strong) Mathematical Induction consists of successfully carrying out the following steps. 1111. . . . Base CaseBase CaseBase CaseBase Case:::: Prove that (1)P is true. 2. Induction Step2. Induction Step2. Induction Step2. Induction Step: Show that for all n (1) (2) ( ) ( 1).P P P n P n +

A fundamental result in number theory is the FundamentFundamentFundamentFundamental Theorem of al Theorem of al Theorem of al Theorem of ArithmeticArithmeticArithmeticArithmetic, which can be proven by strong induction.

NoteNoteNoteNote: To prove that a proposition ( )P n is true for all natural numbers does not mean you have to use induction, but generally induction is the most effective route. Also, if you use some other method in lieu of induction, you are might be using some fact in your proof that does require induction.

Theorem Theorem Theorem Theorem 6 6 6 6 ( ( ( (Fundamental Theorem of ArithmeticFundamental Theorem of ArithmeticFundamental Theorem of ArithmeticFundamental Theorem of Arithmetic)))) Every natural number 2n can be written as the product of prime numbers. For example

2 2350 2 5 7, 1911 3 7 13= = .

Proof: Proof: Proof: Proof: Denoting

( )P n = n can be written as the product of prime numbers

we prove:

Base CaseBase CaseBase CaseBase Case:::: (2)P holds since 2 is a prime number.

Induction Step:Induction Step:Induction Step:Induction Step: For an arbitrary 2,3,...n = we prove

(2) (3) ( ) ( 1)P P P n P n + ....

Assume (2), (3), ( )P P P n is true. That is, each natural number 2,3, n can be written as the product of primes.

We now consider two cases.


Case 1: Suppose 1n + is a prime number. In this case ( 1)P n + is true and thus the induction step is proven since any prime number 1n + can be written as a product of primes, namely 1 1n n+ = + .

Case 2: Suppose 1n + is not prime, which means we can factor 1n qp+ = , where clearly both of the factors ,p q must be less than 1n + and greater and or equal to 2. (For example if 1 15n + = , we would could factor 15 5 3= where both factors 3,5 are less than 15 and greater than or equal to 2). But since

,p q are less than 1n + , the induction hypothesis (all ( )P k are true for 1k n< + ) states they can both be written as the product of primes, say

1 2 1 2m np p p p q q q q= =

But if both and p q can be written as the product of primes so can

( )( )1 1 1 21 m nn pq p p p q q q+ = = . Hence ( 1)P n + is true and so by the principle of strong induction ( )P n is true for all 2n .

History of Mathematical Induction History of Mathematical Induction History of Mathematical Induction History of Mathematical Induction Although some elements of mathematical induction have been hinted at since the time of Euclid, one of the oldest argument using induction goes back to the Italian mathematician Francesco Maurolico, who used induction in 1575 to prove that the sum of the first n odd natural numbers is 2n . The method was later discovered independently by the Swiss mathematician John Bernoulli, and French mathematicians Blaise Pascal (1623-1662) and Pierre de Fermat (1601-1665). Finally, in 1889 the Italian logician Guiseppe Peano (1858-1932) laid out five axioms for deducing the natural numbers, of which his fifth axiom was the Principle of Mathematical Induction. Hence, from that point of view induction is a formal axiom of arithmetic.

Theorem Theorem Theorem Theorem 7777 (Solution of a Recurrence Relation)(Solution of a Recurrence Relation)(Solution of a Recurrence Relation)(Solution of a Recurrence Relation) Suppose a sequence 0 1, ,..., ,...nu u u is defined by the recursion relation with initial conditions:

1 13 2n n nu u u+ = 0 12, 3u u= =

Find the sequence , 1,2,...n

u n = that satisfies these equations. Doing a few computations we find 2 1 03 2 5,u u u= = 3 7,u = 4 9,u = 5 33u = and thus a reasonable guess would be 2 1n

nu = + . . . . To show ( ) : 2 1nnP n u = + satisfies the


recurrence relation for all 0n , we use strong induction starting with initial step 0.n = .

Base CaseBase CaseBase CaseBase Case: : : : 00(0) : 2 1 3P u = + = (check)

Induction Step: Induction Step: Induction Step: Induction Step: Assuming (0), (1) ,..., ( )P P P n true, we can write

112 1, 2 1n nn nu u = + = +

and from the recurrence relation, we have

( ) ( )1 1

1

1

3 2

3 2 1 2 2 1

3 2 2 12 1

n n n

n n

n n

n

u u u+

+

=

= + +

= +

= +

which proves ( 1).P n + Hence, by strong induction we have that ( ) : 2 1nnP n u = + satisfies the recurrence relation for all 0n .

This next example shows a variation of the base step from previous examples. Each problem is different and you must adjust the induction proof accordingly.

Theorem Theorem Theorem Theorem 8888 (Modifying the Base Step) (Modifying the Base Step) (Modifying the Base Step) (Modifying the Base Step) You are given two rulers: one is 3 units long, the other 5. Show that you can measure any unit distance greater than or equal to 8 using only these two rulers.

Proof:Proof:Proof:Proof: Let ( )P n be the proposition

( ) any integer length 8 can be measured with rulers of length 3 and 5P n n= It is useful to see the following pattern that develops.


( )( )( )

(8) 5 3(9) 3 3 3(10) 5 5(11) (8) 3 5 3 3(12) (9) 3 3 3 3 3(13) (10) 3 5 5 3... ... ...

PPPP P

P P

P P

= +

= + +

= +

= + = + +

= + = + + +

= + = + +

This pattern will serve as an aid in deciding the base and induction steps which is often the most difficult part in an induction proof.

Base Step:Base Step:Base Step:Base Step: For the base step, we verify the first three propositions:

( )(8) 5 3, (9) 3 3 3, 10 5 5P P P= + = + + = + Induction Step:Induction Step:Induction Step:Induction Step: We now prove the induction step

( )(8), (9), ..., ( ) 1 , 10P P P n P n n +

To prove this we make the simple observation that if ( )2P n is true (i.e. a length of 2n can be measured with rulers of length 3 and 5), then ( )1P n + is also true since a length of 1n + is 3 units longer than 2n . [ For example

( ) ( ) ( )9 3 3 3 and 12 3 3 3 3P P= + + = + + + . Hence, we know ( )11P is true since ( )8P is true, and ( )12P is true since ( )9P is true, and so on. ] Hence if

(8), (9), ..., ( )P P P n 10n

is true so is ( )1P n + and so by induction ( )P n is true for all natural numbers n .

There are several variations of the basic method of mathematical induction. One such variation is double induction.

Double InductionDouble InductionDouble InductionDouble Induction

Sometimes we would like to prove a proposition ( ),P m n involving two natural numbers m by iterating the inductive process. This is done by performing an induction on one of the variables, say m , and then an induction


on n . This strategy is called double inductiondouble inductiondouble inductiondouble induction and is carried out by the following steps.

( )( ) ( )( ) ( )

1. Prove 1,1 is true2. Prove ,1 1,13. Prove , , 1 for all natural numbers

P

P m P m

P m n P m n m

+

+

You can think of double induction as proving ( ),P m n at all points in the first quadrant of the xy -plane with integer coordinates.


ProblemsProblemsProblemsProblems

1. (Proof by Induction)(Proof by Induction)(Proof by Induction)(Proof by Induction) Prove by weak or strong induction the following propositions.

a) ( ) ( )2 2 2 2 1 2 11 2 36

n n nn

+ ++ + + + =

b) 3n n is divisible by 3 for 1n .

c) ( ) ( ) ( )0

1 1 22

n

ia ib n a nb

=

+ = + +

d) 2 !n n< ( 4)n

f) 1 12, 5n nu u u= + = has the solution 2 3nu n= + for 2n .

g) ( ) ( )1

4 3 2 1n

kk n n

=

=

h) 5 divides 7 2n n for all positive integers.

i) 2n n>

j) ( )cos sin cos sinni n i n + = + (De Moivres Theorem)

k) sin sinnx n x

l) n na b a b< < ( ,a b positive real numbers)

m) ( ) ( )1ln 1 1 !n

n n

n

dx n x

dx+

=

2. (Graphical Induction)(Graphical Induction)(Graphical Induction)(Graphical Induction) Draw some lines in the plane as shown in Figure 1 and color adjoining regions either red or blue. Show that it is possible to draw an arbitrary number of lines in such a way that two adjoining regions are not


colored the same color. Hint: Define ( )P n as the proposition it is possible to draw n lines in the desired manner for arbitrary n .

Figure 1

3. (Clever Mary)(Clever Mary)(Clever Mary)(Clever Mary) To prove the identity

( )0

12

n

k

n nk

=

+=

Mary evaluated the left-hand side of the equation for 0,1, 2n = getting

n 0 1 2 F(n) 0 1 3

. She then fit a polynomial to these three points, getting

( )1( )2

n nf n += . Mary turned this into her professor. Is her proof4 valid?

4 This problem is based on a problem in the book A = B by Marko Petkovsek, Doron Zeilberger and

Herbert Wilf. (This amazing book, incidentally, can be downloaded free on the internet.)


4. ((((ProofProofProofProofs without Wordss without Wordss without Wordss without Words)))) They say a good picture is worth a thousand words, but in mathematics it might be closer to a million. For the figures below, describe why the figure provides a visual proof of the statement.

a) 2 2 3a b c+ = b) ( )11 22

n nn

++ + + =

c) 21 3 5 (2 1)n n+ + + + = d) 2 2 ( )( )x y x y x y = +


e) 2 2 ( )( )x y x y x y = + f) 2

a bab +

g) 2 2

2

2 2a a

x ax x

+ = +

h) ( )1 / /0

1p q q pt t dt+ =

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