discrete mathematics-mathematical induction

Applied Discrete MathematicsMathematical Induction

William Shoaff

Spring 2008

mailto:[email protected]

http://www.fit.edu/registrar/calendar/

Outline

1 Introduction to Induction

2 Induction on Sequences

3 Induction on Sums

Quotation

The concept of “mathematical induction”should be distinguished from what is usually called

“inductive reasoning” in science.– Don Knuth “The Art of Computer Programming: Fundamental

Algorithms”American computer scientist (1938 – )

Mathematical Induction

Inductive reasoning is a logic used for discovery in science

Repeated observation of consistent results leads to a conjecture

The principle of mathematical induction is an axiom in thetheory of natural numbers

Mathematical induction is a template procedure used toestablish facts about enumerable setsThere is a basis for induction, an observation of the fact insome instanceAnd there is a material implication that “if the fact is true insome instance, then it will be true in the next instance”


Inductive reasoning is a logic used for discovery in science

Repeated observation of consistent results leads to a conjectureThe principle of mathematical induction is an axiom in thetheory of natural numbers



Inductive reasoning is a logic used for discovery in scienceRepeated observation of consistent results leads to a conjecture






Mathematical induction is a template procedure used toestablish facts about enumerable sets

There is a basis for induction, an observation of the fact insome instanceAnd there is a material implication that “if the fact is true insome instance, then it will be true in the next instance”




Mathematical induction is a template procedure used toestablish facts about enumerable setsThere is a basis for induction, an observation of the fact insome instance

And there is a material implication that “if the fact is true insome instance, then it will be true in the next instance”

Observations: Full Binary Trees

The full binary tree of height h = 0 has 1 node

The full binary tree of height h = 1 has 3 nodes

1

2 3

Observations: Full Binary Trees

The full binary tree of height h = 2 has 7 nodes

1

2

4 5

3

6 7

Question

What is the functional relationship between the height h of afull binary tree and the number of nodes n in the tree?

ObservationsHeight Nodesh = 0 n = 1h = 1 n = 3h = 2 n = 7

Hypothesis

The full binary tree of height h has n = 2h+1 − 1 nodesThis statement is true for h = 0, 1 and 2

Height Nodes0 1 = 2(0+1) − 11 3 = 2(1+1) − 12 7 = 2(2+1) − 1

To prove the statement for every integer

show that if it is true for some height h ≥ 0, then it will betrue for height h + 1

Hypothesis

The full binary tree of height h has n = 2h+1 − 1 nodes

This statement is true for h = 0, 1 and 2

Height Nodes0 1 = 2(0+1) − 11 3 = 2(1+1) − 12 7 = 2(2+1) − 1



Hypothesis


Height Nodes0 1 = 2(0+1) − 11 3 = 2(1+1) − 12 7 = 2(2+1) − 1



Hypothesis


Height Nodes0 1 = 2(0+1) − 11 3 = 2(1+1) − 12 7 = 2(2+1) − 1

To prove the statement for every integershow that if it is true for some height h ≥ 0, then it will betrue for height h + 1

Height h, Full Binary Tree Has 2h+1 − 1 Nodes

A full binary tree of height h + 1 is constructed from

1 root nodeA full left subtree of of height hA full right subtree of of height h

root

leftsubtree

rightsubtree

height h

height 1


A full binary tree of height h + 1 is constructed from1 root node

A full left subtree of of height hA full right subtree of of height h

root

leftsubtree

rightsubtree

height h

height 1


A full binary tree of height h + 1 is constructed from1 root nodeA full left subtree of of height h

A full right subtree of of height h

root

leftsubtree

rightsubtree

height h

height 1


A full binary tree of height h + 1 is constructed from1 root nodeA full left subtree of of height hA full right subtree of of height h

root

leftsubtree

rightsubtree

height h

height 1


root

leftsubtree

rightsubtree

height h

height 1

If the left and right subtree both have 2h+1 − 1 nodes, thenthe nodes in the higher tree are

(1 root node) + (nodes in left) + (nodes in right)

1 + (2h − 1) + (2h+1 − 1) = 1 + 2(2h+1 − 1)

= 1 + (2h+2 − 2)

= 2h+2 − 1


root

leftsubtree

rightsubtree

height h

height 1

If the left and right subtree both have 2h+1 − 1 nodes,

thenthe nodes in the higher tree are


1 + (2h − 1) + (2h+1 − 1) = 1 + 2(2h+1 − 1)

= 1 + (2h+2 − 2)

= 2h+2 − 1


root

leftsubtree

rightsubtree

height h

height 1



1 + (2h − 1) + (2h+1 − 1) = 1 + 2(2h+1 − 1)

= 1 + (2h+2 − 2)

= 2h+2 − 1


root

leftsubtree

rightsubtree

height h

height 1



1 + (2h − 1) + (2h+1 − 1)

= 1 + 2(2h+1 − 1)

= 1 + (2h+2 − 2)

= 2h+2 − 1


root

leftsubtree

rightsubtree

height h

height 1



1 + (2h − 1) + (2h+1 − 1) = 1 + 2(2h+1 − 1)

= 1 + (2h+2 − 2)

= 2h+2 − 1

Gauss Fools the Third Grade Teacher

There is an often told story, with many variations, of howyoung Gauss irritated his third grade teacherThe class was asked to compute the value of the sum

1 + 2 + 3 + · · ·+ 99 + 100

Almost immediately Gauss answered 5050Perhaps Gauss knew the total would be the number of terms(100) times the average of the first and last term

1+2+3+ · · ·+99+100 = 100(1 + 100

2

)= 50(101) = 5050


There is an often told story, with many variations, of howyoung Gauss irritated his third grade teacher

The class was asked to compute the value of the sum

1 + 2 + 3 + · · ·+ 99 + 100


1+2+3+ · · ·+99+100 = 100(1 + 100

2

)= 50(101) = 5050



1 + 2 + 3 + · · ·+ 99 + 100


1+2+3+ · · ·+99+100 = 100(1 + 100

2

)= 50(101) = 5050



1 + 2 + 3 + · · ·+ 99 + 100

Almost immediately Gauss answered 5050

Perhaps Gauss knew the total would be the number of terms(100) times the average of the first and last term

1+2+3+ · · ·+99+100 = 100(1 + 100

2

)= 50(101) = 5050



1 + 2 + 3 + · · ·+ 99 + 100


1+2+3+ · · ·+99+100 = 100(1 + 100

2

)= 50(101) = 5050

Generalizing the Problem

Consider the more general sum

0 + 1 + 2 + 3 + · · ·+ (n − 2) + (n − 1)

The sum has n termsThe first term is 0, the last term is (n − 1), and their averageis (n − 1)/2Is it true that

0 + 1 + 2 + 3 + · · ·+ (n − 2) + (n − 1) = n(n − 12

)?



0 + 1 + 2 + 3 + · · ·+ (n − 2) + (n − 1)

The sum has n terms

The first term is 0, the last term is (n − 1), and their averageis (n − 1)/2Is it true that

0 + 1 + 2 + 3 + · · ·+ (n − 2) + (n − 1) = n(n − 12

)?



0 + 1 + 2 + 3 + · · ·+ (n − 2) + (n − 1)

The sum has n termsThe first term is 0, the last term is (n − 1), and their averageis (n − 1)/2

Is it true that

0 + 1 + 2 + 3 + · · ·+ (n − 2) + (n − 1) = n(n − 12

)?



0 + 1 + 2 + 3 + · · ·+ (n − 2) + (n − 1)

The sum has n termsThe first term is 0, the last term is (n − 1), and their averageis (n − 1)/2Is it true that

0 + 1 + 2 + 3 + · · ·+ (n − 2) + (n − 1) = n(n − 12

)?

Prove∑

k = n(n−1)2

To prove

0 + 1 + 2 + 3 + · · ·+ (n − 2) + (n − 1) = n(n − 12

)is true for all natural numbers n, show

Basis: For n = 0

The sum on the left is empty, and so equal to 0

0+ 1+ · · ·+(0− 2)+ (0− 1) = 0 (empty sum)

The right hand side is also equal to 0 for n = 0

0(0− 12

)= 0

Prove∑

k = n(n−1)2

To prove

0 + 1 + 2 + 3 + · · ·+ (n − 2) + (n − 1) = n(n − 12

)is true for all natural numbers n, show

Basis: For n = 0The sum on the left is empty, and so equal to 0

0+ 1+ · · ·+(0− 2)+ (0− 1) = 0 (empty sum)

The right hand side is also equal to 0 for n = 0

0(0− 12

)= 0

Prove∑

k = n(n−1)2

Induction: Pretend

0 + 1 + 2 + · · ·+ (n − 2) + (n − 1) =n(n − 1)

2

for some n ≥ 0Then the next sum is:

[0 + 1 + · · ·+ (n − 1)] + n =n(n − 1)

2+ n

=n(n − 1)

2+

2n2

=n(n + 1)

2

Which shows the next sum equals the functionat the next natural number

Prove∑

k = n(n−1)2

Induction: Pretend

0 + 1 + 2 + · · ·+ (n − 2) + (n − 1) =n(n − 1)

2

for some n ≥ 0

Then the next sum is:

[0 + 1 + · · ·+ (n − 1)] + n =n(n − 1)

2+ n

=n(n − 1)

2+

2n2

=n(n + 1)

2


Prove∑

k = n(n−1)2

Induction: Pretend

0 + 1 + 2 + · · ·+ (n − 2) + (n − 1) =n(n − 1)

2


[0 + 1 + · · ·+ (n − 1)] + n =n(n − 1)

2+ n

=n(n − 1)

2+

2n2

=n(n + 1)

2


Prove∑

k = n(n−1)2

Induction: Pretend

0 + 1 + 2 + · · ·+ (n − 2) + (n − 1) =n(n − 1)

2


[0 + 1 + · · ·+ (n − 1)] + n

=n(n − 1)

2+ n

=n(n − 1)

2+

2n2

=n(n + 1)

2


Prove∑

k = n(n−1)2

Induction: Pretend

0 + 1 + 2 + · · ·+ (n − 2) + (n − 1) =n(n − 1)

2


[0 + 1 + · · ·+ (n − 1)] + n =n(n − 1)

2+ n

=n(n − 1)

2+

2n2

=n(n + 1)

2


Inductive Template For Functional Equality

To prove “f (n) = g(n)” is true for all natural numbers n

Basis: Show: both functions map 0 to the same value, thatis,

f (0) = g(0)

Induction: Show: if f (n) = g(n) for some n ≥ 0, thenf (n + 1) = g(n + 1)

Quotation

The so-called law of induction cannot possibly be a lawof logic, since it is obviously a proposition with a sense.Nor, therefore, can it be an a priori law.

Ludwig Wittenstein, "Tractatus Logico-Philosophicus"Austrian-British philosoper (1889 - 1951)

Checking Solutions to Algebraic Equations

Pretend you want to check that x = 4 is a solution to thelinear equation

3x − 5 = 7

You can do this by substitiuting 4 for x on the left-hand sideand show the result is the right-hand side

3(4)− 5 = 12− 5 = 7


Pretend you want to check that x = (1 +√5)/2 is a solution

to the quadratic equation

x2 = x + 1

You can do this by substitiuting (1 +√5)/2 for x in both the

left-hand and right-hand sides and show the results are equal

The left-hand side is(1 +√5

2

)2

=1 + 2

√5 + 5

4=

3 +√5

2

The right-hand side is

1 +√5

2+ 1 =

1 +√5

2+

22

=3 +√5

2


Pretend you want to check that x = (1 +√5)/2 is a solution

to the quadratic equation

x2 = x + 1

You can do this by substitiuting (1 +√5)/2 for x in both the

left-hand and right-hand sides and show the results are equalThe left-hand side is(

1 +√5

2

)2

=1 + 2

√5 + 5

4=

3 +√5

2


1 +√5

2+ 1 =

1 +√5

2+

22

=3 +√5

2

Checking Solutions to Recurrence Equations

The same basic idea can be used to check a proposed solutionto a recurrence equationSuppose you want to show that mn = 2n − 1 is a solution tothe recurrence

mn = 2mn−1 + 1

You can do this by substitiuting 2n − 1 for mn in the left-handside and 2n−1 − 1 for mn−1 in the right-hand sides and showthe results are equal

The left-hand side is2n − 1


2(2n−1 − 1

)+ 1 = (2n − 2) + 1 = 2n − 1


The same basic idea can be used to check a proposed solutionto a recurrence equation

Suppose you want to show that mn = 2n − 1 is a solution tothe recurrence

mn = 2mn−1 + 1




2(2n−1 − 1

)+ 1 = (2n − 2) + 1 = 2n − 1


The same basic idea can be used to check a proposed solutionto a recurrence equationSuppose you want to show that mn = 2n − 1 is a solution tothe recurrence

mn = 2mn−1 + 1




2(2n−1 − 1

)+ 1 = (2n − 2) + 1 = 2n − 1


There’s is one problem with thisFor any constant c , you can show mn = c2n − 1 is a solutionto the recurrence

mn = 2mn−1 + 1

The left-hand side would be

c2n − 1

And the right-hand side would be

2(c2n−1 − 1

)+ 1 = c2n − 1

An initial value nails down the solution: If m0 = 0, thenc20 − 1 = c − 1 = 0 forces c to be 1


There’s is one problem with this

For any constant c , you can show mn = c2n − 1 is a solutionto the recurrence

mn = 2mn−1 + 1


c2n − 1


2(c2n−1 − 1

)+ 1 = c2n − 1



There’s is one problem with thisFor any constant c , you can show mn = c2n − 1 is a solutionto the recurrence

mn = 2mn−1 + 1


c2n − 1


2(c2n−1 − 1

)+ 1 = c2n − 1


Induction on The Triangular Sequence

Consider the recurrence for the triangular numbers

tn = tn−1 + (n − 1)

To show that tn = n(n − 1)/2 solves the recurrence substituteinto the left-hand and right hand sides

The left-hand side isn(n − 1)/2


(n − 1)(n − 2)2

+ (n − 1) =(n − 1)(n − 2)

2+

2(n − 1)2

=(n − 1)(n − 2) + 2(n − 1)

2

=n(n − 1)

2

To complete the induction, establish the basis:t0 = 0(0− 1)/2 = 0

Induction on The Mersenne Sequence

Consider the recurrence for the Mersenne numbers

mn = 2mn−1 + 1

or, relabeling subscripts

mn+1 = 2mn + 1

If mn = 2n − 1, then

mn+1 = 2mn + 1 = 2(2n − 1) + 1

Which can be rewritten as

mn+1 = 2n+1 − 1

The function 2n − 1 maps 0 to 0 which matches the initialMersenne number m0 = 0Since both the basis and the inductive conditional are bothtrue, the Mersenne numbers mn can be computed by thefunction m(n) = 2n − 1 for all natural numbers n



mn = 2mn−1 + 1


mn+1 = 2mn + 1


mn+1 = 2mn + 1 = 2(2n − 1) + 1


mn+1 = 2n+1 − 1

The function 2n − 1 maps 0 to 0 which matches the initialMersenne number m0 = 0

Since both the basis and the inductive conditional are bothtrue, the Mersenne numbers mn can be computed by thefunction m(n) = 2n − 1 for all natural numbers n



mn = 2mn−1 + 1


mn+1 = 2mn + 1


mn+1 = 2mn + 1 = 2(2n − 1) + 1


mn+1 = 2n+1 − 1

The function 2n − 1 maps 0 to 0 which matches the initialMersenne number m0 = 0Since both the basis and the inductive conditional are bothtrue, the Mersenne numbers mn can be computed by thefunction m(n) = 2n − 1 for all natural numbers n

Induction on The Fermat Sequence

Consider the recurrence for the Fermat numbers

rn = (rn−1 − 1)2 + 1


rn+1 = (rn − 1)2 + 1

If rn = 22n+ 1, then

rn+1 = ((22n+ 1)− 1)2 + 1


rn+1 = 22n+1+ 1

The function 22n+ 1 maps 0 to 3 which matches the initial

Fermat number r0 = 3Since both the basis and the inductive conditional are bothtrue, the Fermat numbers rn can be computed by the functionr(n) = 22n

+ 1 for all natural numbers n



rn = (rn−1 − 1)2 + 1


rn+1 = (rn − 1)2 + 1


rn+1 = ((22n+ 1)− 1)2 + 1


rn+1 = 22n+1+ 1


Fermat number r0 = 3

Since both the basis and the inductive conditional are bothtrue, the Fermat numbers rn can be computed by the functionr(n) = 22n




rn = (rn−1 − 1)2 + 1


rn+1 = (rn − 1)2 + 1


rn+1 = ((22n+ 1)− 1)2 + 1


rn+1 = 22n+1+ 1


Fermat number r0 = 3Since both the basis and the inductive conditional are bothtrue, the Fermat numbers rn can be computed by the functionr(n) = 22n


Quotation

MacPherson told me that my theorem can be viewedas blah blah blah Grothendick blah blah blah, whichmakes it much more respectable. I think someintuition leaks out in every step of an induction proof.

Jim Propp, American mathematician

Induction on Sums

To prove a summation∑

ak can be expressed as a functiong(·) on the natural numbers,

Establish a basis for induction, that is, show0∑

k=0

ak = a0 = g(0)

Prove the inductive conditional: If

Ifn−1∑k=0

ak = g(n),

then

n∑k=0

ak =

[n−1∑k=0

ak

]+ an

= g(n) + an

= g(n + 1)

Induction on Sums




k=0

ak = a0 = g(0)


Ifn−1∑k=0

ak = g(n),

thenn∑

k=0

ak

=

[n−1∑k=0

ak

]+ an

= g(n) + an

= g(n + 1)

Induction on Sums




k=0

ak = a0 = g(0)


Ifn−1∑k=0

ak = g(n),

thenn∑

k=0

ak =

[n−1∑k=0

ak

]+ an

= g(n) + an

= g(n + 1)

Alice Sum Math Induction Proof

The Alice sum formula

n−1∑k=0

1 = n

can be established by mathematical induction

Basis: For n = 0

The sumn−1∑k=0

1 =0−1∑k=0

1

is empty and equal to 0The value on the right hand, n, is 0 too



n−1∑k=0

1 = n


Basis: For n = 0The sum

n−1∑k=0

1 =0−1∑k=0

1

is empty and equal to 0

The value on the right hand, n, is 0 too



n−1∑k=0

1 = n


Basis: For n = 0The sum

n−1∑k=0

1 =0−1∑k=0

1

is empty and equal to 0The value on the right hand, n, is 0 too


Induction: Show that

Ifn−1∑k=0

1 = n,

thenn∑

k=0

1 = n + 1

By computing

n∑k=0

1 =

[n−1∑k=0

1

]+ 1

= n + 1


Induction: Show thatIf

n−1∑k=0

1 = n,

thenn∑

k=0

1 = n + 1

By computing

n∑k=0

1 =

[n−1∑k=0

1

]+ 1

= n + 1



n−1∑k=0

1 = n,

thenn∑

k=0

1 = n + 1

By computing

n∑k=0

1

=

[n−1∑k=0

1

]+ 1

= n + 1



n−1∑k=0

1 = n,

thenn∑

k=0

1 = n + 1

By computing

n∑k=0

1 =

[n−1∑k=0

1

]+ 1

= n + 1

Gauss Sum Math Induction Proof

The Gauss sum formula

n−1∑k=0

k = 0 + 1 + 2 + 3 + · · ·+ (n − 1) =n(n − 1)

2


Basis: For n = 0 the sum on the left is empty, and so equalto 0. And the right hand side is 0(0− 1)/2 = 0 too

Induction: If∑n−1

k=0 k = n(n − 1)/2, then

n∑k=0

k =

[n−1∑k=0

k

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

Zeno Sum Math Induction Proof

The Zeno sum formula

n−1∑k=0

2k = 1 + 2 + 4 + · · ·+ 2n−1 = 2n − 1


Basis: For n = 0 the sum on the left is empty, and so equalto 0. And the right hand side n is 0 too


k=0 2k = 2n − 1, then

n∑k=0

2k =

[n−1∑k=0

2k

]+ 2n = 2n+1 − 1

Sum of the Odd Integers is a Square

The ancient formulan∑

k=1

(2k − 1) = 1 + 3 + 5 + · · ·+ (2n − 1) = n2


Basis: For n = 0 the sum on the left is empty, and so equalto 0. And the right hand side n is 0 too.


n∑k=1

(2k − 1) = 1 + 3 + 5 + · · ·+ (2n − 1) = n2

Induction: If∑n

k=1(2k − 1) = n2, then

n+1∑k=1

(2k − 1) =

[n∑

k=1

(2k − 1)

]+ 2(n + 1)− 1

= n2 + 2(n + 1)− 1

= n2 + 2n + 1

= (n + 1)2


n∑k=1

(2k − 1) = 1 + 3 + 5 + · · ·+ (2n − 1) = n2

Induction: If∑n

k=1(2k − 1) = n2, then

n+1∑k=1

(2k − 1)

=

[n∑

k=1

(2k − 1)

]+ 2(n + 1)− 1

= n2 + 2(n + 1)− 1

= n2 + 2n + 1

= (n + 1)2


n∑k=1

(2k − 1) = 1 + 3 + 5 + · · ·+ (2n − 1) = n2

Induction: If∑n

k=1(2k − 1) = n2, then

n+1∑k=1

(2k − 1) =

[n∑

k=1

(2k − 1)

]+ 2(n + 1)− 1

= n2 + 2(n + 1)− 1

= n2 + 2n + 1

= (n + 1)2

Sum of Cubes is a Squared Triangular Number

The formulan−1∑k=1

k3 =

(12n(n − 1)

)2


Basis: For n = 0 the sum on the left is empty, and so equalto 0. And the right hand side n is 0 too.


n−1∑k=1

k3 =

(12n(n − 1)

)2


k=1 k3 = n2(n − 1)2/4, then

n∑k=1

k3 =

[n−1∑k=1

k3

]+ n3

=

(12n(n − 1)

)2

+ n3

= n2(

(n − 1)2

4+ n)

= n2(

(n + 2n + 1)4

)=

(12n(n + 1)

)2


n−1∑k=1

k3 =

(12n(n − 1)

)2


k=1 k3 = n2(n − 1)2/4, then

n∑k=1

k3

=

[n−1∑k=1

k3

]+ n3

=

(12n(n − 1)

)2

+ n3

= n2(

(n − 1)2

4+ n)

= n2(

(n + 2n + 1)4

)=

(12n(n + 1)

)2


n−1∑k=1

k3 =

(12n(n − 1)

)2


k=1 k3 = n2(n − 1)2/4, then

n∑k=1

k3 =

[n−1∑k=1

k3

]+ n3

=

(12n(n − 1)

)2

+ n3

= n2(

(n − 1)2

4+ n)

= n2(

(n + 2n + 1)4

)=

(12n(n + 1)

)2


n−1∑k=1

k3 =

(12n(n − 1)

)2


k=1 k3 = n2(n − 1)2/4, then

n∑k=1

k3 =

[n−1∑k=1

k3

]+ n3

=

(12n(n − 1)

)2

+ n3

= n2(

(n − 1)2

4+ n)

= n2(

(n + 2n + 1)4

)

=

(12n(n + 1)

)2


n−1∑k=1

k3 =

(12n(n − 1)

)2


k=1 k3 = n2(n − 1)2/4, then

n∑k=1

k3 =

[n−1∑k=1

k3

]+ n3

=

(12n(n − 1)

)2

+ n3

= n2(

(n − 1)2

4+ n)

= n2(

(n + 2n + 1)4

)=

(12n(n + 1)

)2

Sum of Fibonacci Numbers

Recall the Fibonacci sequence

~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉

Also recall the recurrence for the Fibonacci numbers

fn = fn−1 + fn−2, n ∈ {2, 3, 4, . . .}, f0 = 0, f1 = 1

Let’s write the sequence using the names of the numbersrather than their values

~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉


~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉

Consider the sequence of partial sums

S0 = f0 = 0S1 = f0 + f1 = 1S2 = f0 + f1 + f2 = 2S3 = f0 + f1 + f2 + f3 = 4S4 = f0 + f1 + f2 + f3 + f4 = 7S4 = f0 + f1 + f2 + f3 + f4 + f5 = 12

Can you make a conjecture about these sums?


~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉


S0

= f0 = 0S1 = f0 + f1 = 1S2 = f0 + f1 + f2 = 2S3 = f0 + f1 + f2 + f3 = 4S4 = f0 + f1 + f2 + f3 + f4 = 7S4 = f0 + f1 + f2 + f3 + f4 + f5 = 12



~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉


S0 = f0 = 0

S1 = f0 + f1 = 1S2 = f0 + f1 + f2 = 2S3 = f0 + f1 + f2 + f3 = 4S4 = f0 + f1 + f2 + f3 + f4 = 7S4 = f0 + f1 + f2 + f3 + f4 + f5 = 12



~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉


S0 = f0 = 0S1

= f0 + f1 = 1S2 = f0 + f1 + f2 = 2S3 = f0 + f1 + f2 + f3 = 4S4 = f0 + f1 + f2 + f3 + f4 = 7S4 = f0 + f1 + f2 + f3 + f4 + f5 = 12



~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉


S0 = f0 = 0S1 = f0 + f1 = 1

S2 = f0 + f1 + f2 = 2S3 = f0 + f1 + f2 + f3 = 4S4 = f0 + f1 + f2 + f3 + f4 = 7S4 = f0 + f1 + f2 + f3 + f4 + f5 = 12



~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉


S0 = f0 = 0S1 = f0 + f1 = 1S2

= f0 + f1 + f2 = 2S3 = f0 + f1 + f2 + f3 = 4S4 = f0 + f1 + f2 + f3 + f4 = 7S4 = f0 + f1 + f2 + f3 + f4 + f5 = 12



~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉


S0 = f0 = 0S1 = f0 + f1 = 1S2 = f0 + f1 + f2 = 2

S3 = f0 + f1 + f2 + f3 = 4S4 = f0 + f1 + f2 + f3 + f4 = 7S4 = f0 + f1 + f2 + f3 + f4 + f5 = 12



~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉


S0 = f0 = 0S1 = f0 + f1 = 1S2 = f0 + f1 + f2 = 2S3

= f0 + f1 + f2 + f3 = 4S4 = f0 + f1 + f2 + f3 + f4 = 7S4 = f0 + f1 + f2 + f3 + f4 + f5 = 12



~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉


S0 = f0 = 0S1 = f0 + f1 = 1S2 = f0 + f1 + f2 = 2S3 = f0 + f1 + f2 + f3 = 4

S4 = f0 + f1 + f2 + f3 + f4 = 7S4 = f0 + f1 + f2 + f3 + f4 + f5 = 12



~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉


S0 = f0 = 0S1 = f0 + f1 = 1S2 = f0 + f1 + f2 = 2S3 = f0 + f1 + f2 + f3 = 4S4

= f0 + f1 + f2 + f3 + f4 = 7S4 = f0 + f1 + f2 + f3 + f4 + f5 = 12



~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉


S0 = f0 = 0S1 = f0 + f1 = 1S2 = f0 + f1 + f2 = 2S3 = f0 + f1 + f2 + f3 = 4S4 = f0 + f1 + f2 + f3 + f4 = 7

S4 = f0 + f1 + f2 + f3 + f4 + f5 = 12



~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉


S0 = f0 = 0S1 = f0 + f1 = 1S2 = f0 + f1 + f2 = 2S3 = f0 + f1 + f2 + f3 = 4S4 = f0 + f1 + f2 + f3 + f4 = 7S4

= f0 + f1 + f2 + f3 + f4 + f5 = 12



~F = 〈0, 1, 1, 2, 3, 5, 8, 13, 21, . . .〉~F = 〈f0, f1, f2, f3, f4, f5, f6, f7, f8, . . .〉


S0 = f0 = 0S1 = f0 + f1 = 1S2 = f0 + f1 + f2 = 2S3 = f0 + f1 + f2 + f3 = 4S4 = f0 + f1 + f2 + f3 + f4 = 7S4 = f0 + f1 + f2 + f3 + f4 + f5 = 12



n−1∑k=0

fk = fn+1 − 1

Basis: For n = 0

The sum on the left-hand side is empty and soequal to 0The formula on the right-hand side is also equalto 0

f0+1 − 1 = f1 − 1 = 1− 1 = 0

This establishes the basis for induction


n−1∑k=0

fk = fn+1 − 1

Basis: For n = 0The sum on the left-hand side is empty and soequal to 0

The formula on the right-hand side is also equalto 0

f0+1 − 1 = f1 − 1 = 1− 1 = 0



n−1∑k=0

fk = fn+1 − 1

Basis: For n = 0The sum on the left-hand side is empty and soequal to 0The formula on the right-hand side is also equalto 0

f0+1 − 1 = f1 − 1 = 1− 1 = 0



n−1∑k=0

fk = fn+1 − 1

Induction: Pretend thatn−1∑k=0

fk = fn+1 − 1

is true for some n ≥ 0

Then

n∑k=0

fk =

[n−1∑k=1

fk

]+ fn

= [fn+1 − 1] + fn= [fn+1 + fn]− 1= fn+2 − 1


n−1∑k=0

fk = fn+1 − 1


fk = fn+1 − 1

is true for some n ≥ 0Then

n∑k=0

fk =

[n−1∑k=1

fk

]+ fn

= [fn+1 − 1] + fn= [fn+1 + fn]− 1= fn+2 − 1


n−1∑k=0

fk = fn+1 − 1


fk = fn+1 − 1


n∑k=0

fk

=

[n−1∑k=1

fk

]+ fn

= [fn+1 − 1] + fn= [fn+1 + fn]− 1= fn+2 − 1


n−1∑k=0

fk = fn+1 − 1


fk = fn+1 − 1


n∑k=0

fk =

[n−1∑k=1

fk

]+ fn

= [fn+1 − 1] + fn= [fn+1 + fn]− 1= fn+2 − 1


n−1∑k=0

fk = fn+1 − 1


fk = fn+1 − 1


n∑k=0

fk =

[n−1∑k=1

fk

]+ fn

= [fn+1 − 1] + fn

= [fn+1 + fn]− 1= fn+2 − 1


n−1∑k=0

fk = fn+1 − 1


fk = fn+1 − 1


n∑k=0

fk =

[n−1∑k=1

fk

]+ fn

= [fn+1 − 1] + fn= [fn+1 + fn]− 1

= fn+2 − 1


n−1∑k=0

fk = fn+1 − 1


fk = fn+1 − 1


n∑k=0

fk =

[n−1∑k=1

fk

]+ fn

= [fn+1 − 1] + fn= [fn+1 + fn]− 1= fn+2 − 1

Pascal’s Triangle

11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 11 7 21 35 35 21 7 11 8 28 56 70 56 28 8 1...

......

......

......

......

. . .

Pascal’s Triangle: Sum of Rows

There are many identities that can be found in Pascal’s trianglePerhaps the most famous one is found by summing values in arow

1 = 11 +1 = 21 +2 +1 = 41 +3 +3 +1 = 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64


There are many identities that can be found in Pascal’s triangle

Perhaps the most famous one is found by summing values in arow

1 = 11 +1 = 21 +2 +1 = 41 +3 +3 +1 = 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64



1 = 11 +1 = 21 +2 +1 = 41 +3 +3 +1 = 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64



1

= 11 +1 = 21 +2 +1 = 41 +3 +3 +1 = 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64



1 = 1

1 +1 = 21 +2 +1 = 41 +3 +3 +1 = 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64



1 = 11 +1

= 21 +2 +1 = 41 +3 +3 +1 = 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64



1 = 11 +1 = 2

1 +2 +1 = 41 +3 +3 +1 = 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64



1 = 11 +1 = 21 +2 +1

= 41 +3 +3 +1 = 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64



1 = 11 +1 = 21 +2 +1 = 4

1 +3 +3 +1 = 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64



1 = 11 +1 = 21 +2 +1 = 41 +3 +3 +1

= 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64



1 = 11 +1 = 21 +2 +1 = 41 +3 +3 +1 = 8

1 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64



1 = 11 +1 = 21 +2 +1 = 41 +3 +3 +1 = 81 +4 +6 +4 +1

= 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64



1 = 11 +1 = 21 +2 +1 = 41 +3 +3 +1 = 81 +4 +6 +4 +1 = 16

1 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64



1 = 11 +1 = 21 +2 +1 = 41 +3 +3 +1 = 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1

= 321 +6 +15 +20 +15 +6 +1 = 64



1 = 11 +1 = 21 +2 +1 = 41 +3 +3 +1 = 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 32

1 +6 +15 +20 +15 +6 +1 = 64



1 = 11 +1 = 21 +2 +1 = 41 +3 +3 +1 = 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1

= 64



1 = 11 +1 = 21 +2 +1 = 41 +3 +3 +1 = 81 +4 +6 +4 +1 = 161 +5 +10 +10 +5 +1 = 321 +6 +15 +20 +15 +6 +1 = 64


We can make the conjecture

The sum of values in row n of Pascal’s triangle is 2n forn = 0, 1, 2, 3, . . .

When you notice how a row is computed from the previousrow, it becomes clear why the row sums doubleEvery term, except the first and last 1, is added twice togenerate the next row


We can make the conjectureThe sum of values in row n of Pascal’s triangle is 2n forn = 0, 1, 2, 3, . . .




When you notice how a row is computed from the previousrow, it becomes clear why the row sums double

Every term, except the first and last 1, is added twice togenerate the next row


Consider row 6 of Pascal’s Triangle

1 6 15 20 15 6 1

It can be used to generate row 7

1 7 21 35 35 21 7 1

Sum adjacent numbers in row 6 yields

1 1 + 6 6 + 15 15 + 20 20 + 15 15 + 6 6 + 1 11 7 21 35 35 21 7 1

Adding in the initial and terminal 1 in row 7 shows each termin row 6 is summed twice when summing row 7



1 6 15 20 15 6 1


1 7 21 35 35 21 7 1


1

1 + 6

6 + 15 15 + 20 20 + 15 15 + 6 6 + 1 11 7 21 35 35 21 7 1




1 6 15 20 15 6 1


1 7 21 35 35 21 7 1


1

1 + 6

6 + 15 15 + 20 20 + 15 15 + 6 6 + 1 11

7

21 35 35 21 7 1




1 6 15 20 15 6 1


1 7 21 35 35 21 7 1


1

1 + 6 6 + 15

15 + 20 20 + 15 15 + 6 6 + 1 11

7

21 35 35 21 7 1




1 6 15 20 15 6 1


1 7 21 35 35 21 7 1


1

1 + 6 6 + 15

15 + 20 20 + 15 15 + 6 6 + 1 11

7 21

35 35 21 7 1




1 6 15 20 15 6 1


1 7 21 35 35 21 7 1


1

1 + 6 6 + 15 15 + 20

20 + 15 15 + 6 6 + 1 11

7 21

35 35 21 7 1




1 6 15 20 15 6 1


1 7 21 35 35 21 7 1


1

1 + 6 6 + 15 15 + 20

20 + 15 15 + 6 6 + 1 11

7 21 35

35 21 7 1




1 6 15 20 15 6 1


1 7 21 35 35 21 7 1


1

1 + 6 6 + 15 15 + 20 20 + 15

15 + 6 6 + 1 11

7 21 35

35 21 7 1




1 6 15 20 15 6 1


1 7 21 35 35 21 7 1


1

1 + 6 6 + 15 15 + 20 20 + 15

15 + 6 6 + 1 11

7 21 35 35

21 7 1




1 6 15 20 15 6 1


1 7 21 35 35 21 7 1


1

1 + 6 6 + 15 15 + 20 20 + 15 15 + 6

6 + 1 11

7 21 35 35

21 7 1




1 6 15 20 15 6 1


1 7 21 35 35 21 7 1


1

1 + 6 6 + 15 15 + 20 20 + 15 15 + 6

6 + 1 11

7 21 35 35 21

7 1




1 6 15 20 15 6 1


1 7 21 35 35 21 7 1


1

1 + 6 6 + 15 15 + 20 20 + 15 15 + 6 6 + 1

11

7 21 35 35 21

7 1




1 6 15 20 15 6 1


1 7 21 35 35 21 7 1


1

1 + 6 6 + 15 15 + 20 20 + 15 15 + 6 6 + 1

11

7 21 35 35 21 7

1




1 6 15 20 15 6 1


1 7 21 35 35 21 7 1


1 1 + 6 6 + 15 15 + 20 20 + 15 15 + 6 6 + 1 11 7 21 35 35 21 7 1



To make the preceding argument precise we must name thingsWe’ll name the rows and columns by the natural numbers

0 1 2 3 4 5 6 7 8 · · ·0 11 1 12 1 2 13 1 3 3 14 1 4 6 4 15 1 5 10 10 5 16 1 6 15 20 15 6 17 1 7 21 35 35 21 7 18 1 8 28 56 70 56 28 8 1...

......

......

......

......

. . .


To make the preceding argument precise we must name things

We’ll name the rows and columns by the natural numbers

0 1 2 3 4 5 6 7 8 · · ·0 11 1 12 1 2 13 1 3 3 14 1 4 6 4 15 1 5 10 10 5 16 1 6 15 20 15 6 17 1 7 21 35 35 21 7 18 1 8 28 56 70 56 28 8 1...

......

......

......

......

. . .


To make the preceding argument precise we must name thingsWe’ll name the rows and columns by the natural numbers

0 1 2 3 4 5 6 7 8 · · ·0 11 1 12 1 2 13 1 3 3 14 1 4 6 4 15 1 5 10 10 5 16 1 6 15 20 15 6 17 1 7 21 35 35 21 7 18 1 8 28 56 70 56 28 8 1...

......

......

......

......

. . .

Pascal’s Triangle: Binomial Coefficients

We’ll name terms in row n, column m by the symbol(nm

)The symbol

(nm

)is called a binomial coefficient

The symbol(nm

)is read “n choose m”

The binomial coefficient “n choose m” can be expressed interms of factorials (

nm

)=

n!m!(n −m)!



)

The symbol(nm


The symbol(nm



nm

)=

n!m!(n −m)!



)The symbol

(nm


The symbol(nm



nm

)=

n!m!(n −m)!

Pascal’s Triangle

0 1 2 3 4 5 6 7 8 · · ·0(00

)1(10

) (11

)2(20

) (21

) (22

)3(30

) (31

) (32

) (33

)4(40

) (41

) (42

) (43

) (44

)5(50

) (51

) (52

) (53

) (54

) (55

)6(60

) (61

) (62

) (63

) (64

) (65

) (66

)7(70

) (71

) (72

) (73

) (74

) (75

) (76

) (77

)8(80

) (81

) (82

) (83

) (84

) (85

) (86

) (87

) (88

)...

......

......

......

......

. . .

Pascal’s Identity

The relationship between terms in row n − 1 and row n iscalled Pascal’s identityPascal’s identity states that(

n − 1m − 1

)+

(n − 1m

)=

(nm

)That is, the sum of terms in columns m − 1 and m of rown − 1 equals the term in column m row n

Pascal’s Identity

The relationship between terms in row n − 1 and row n iscalled Pascal’s identity

Pascal’s identity states that(n − 1m − 1

)+

(n − 1m

)=

(nm


Pascal’s Identity


n − 1m − 1

)+

(n − 1m

)=

(nm

)

That is, the sum of terms in columns m − 1 and m of rown − 1 equals the term in column m row n

Pascal’s Identity


n − 1m − 1

)+

(n − 1m

)=

(nm


Pascal’s Identity

An example of Pascal’s identity is(74

)=

7!4!3!

=7× 6× 53× 2× 1

= 35

and (75

)=

7!5!2!

=7× 62× 1

= 21

Therefore (85

)=

(74

)+

(75

)= 35 + 21

Pascal’s Identity

Another example of Pascal’s identity is(128

)=

12!8!4!

= 495

and (129

)=

12!9!3!

= 660

Therefore (139

)= 495 + 660 = 1155

Pascal’s Triangle Row Sum

Armed with Pascal’s identity, we can use induction to establishthe sum of values in row n is 2n

To sum the elements in row n we writen∑

k=0

(nk

)=

(n0

)+

(n1

)+

(n2

)+ · · ·+

(n

n − 1

)+

(nn

)Leave the first and last terms alone, but use Pascal’s identityon the middle terms(

n0

)+

[(n − 10

)+

(n − 11

)]+

[(n − 11

)+

(n − 12

)]+ · · ·+

[(n − 1n − 2

)+

(n − 1n − 1

)]+

(nn

)




k=0

(nk

)=

(n0

)+

(n1

)+

(n2

)+ · · ·+

(n

n − 1

)+

(nn

)

Leave the first and last terms alone, but use Pascal’s identityon the middle terms(

n0

)+

[(n − 10

)+

(n − 11

)]+

[(n − 11

)+

(n − 12

)]+ · · ·+

[(n − 1n − 2

)+

(n − 1n − 1

)]+

(nn

)




k=0

(nk

)=

(n0

)+

(n1

)+

(n2

)+ · · ·+

(n

n − 1

)+

(nn

)Leave the first and last terms alone, but use Pascal’s identityon the middle terms(

n0

)+

[(n − 10

)+

(n − 11

)]+

[(n − 11

)+

(n − 12

)]+ · · ·+

[(n − 1n − 2

)+

(n − 1n − 1

)]+

(nn

)


Collect the terms that appear twice(n0

)+

(n − 10

)+2(n − 11

)+· · ·+2

(n − 1n − 2

)+

(n − 1n − 1

)+

(nn

)Recognize that the first two and last two terms are equal to 1and so can be replaced as

2(n − 10

)+ 2(n − 11

)+ · · ·+ 2

(n − 1n − 2

)+ 2(n − 1n − 1

)Therefore, if the sum of terms in row n− 1 is 2n−1, the sum ofterms in row n is 2n



)+

(n − 10

)+2(n − 11

)+· · ·+2

(n − 1n − 2

)+

(n − 1n − 1

)+

(nn

)

Recognize that the first two and last two terms are equal to 1and so can be replaced as

2(n − 10

)+ 2(n − 11

)+ · · ·+ 2

(n − 1n − 2

)+ 2(n − 1n − 1




)+

(n − 10

)+2(n − 11

)+· · ·+2

(n − 1n − 2

)+

(n − 1n − 1

)+

(nn


2(n − 10

)+ 2(n − 11

)+ · · ·+ 2

(n − 1n − 2

)+ 2(n − 1n − 1

)

Therefore, if the sum of terms in row n− 1 is 2n−1, the sum ofterms in row n is 2n



)+

(n − 10

)+2(n − 11

)+· · ·+2

(n − 1n − 2

)+

(n − 1n − 1

)+

(nn


2(n − 10

)+ 2(n − 11

)+ · · ·+ 2

(n − 1n − 2

)+ 2(n − 1n − 1


Epigraph

φ is an H of a lot cooler than π.

Stettner, in Dan Brown’s “The Da Vinci Code”

Cassini’s Identity

Theorem (Cassini’s Identity)

Fn+1Fn−1 − F 2n = (−1)n, for n > 0

Interpret Cassini’s identity as stating the area of a rectanglewith sides Fn−1 and Fn+1 is plus or minus one the area of asquare with sides Fn

Arithmetically, multiply every other Fibonacci number andsubtract the square of the number in the middleThe result alternates between −1 and +1

http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Cassini.html



Fn+1Fn−1 − F 2n = (−1)n, for n > 0


Arithmetically, multiply every other Fibonacci number andsubtract the square of the number in the middle

The result alternates between −1 and +1




Fn+1Fn−1 − F 2n = (−1)n, for n > 0


Arithmetically, multiply every other Fibonacci number andsubtract the square of the number in the middleThe result alternates between −1 and +1



n 0 1 2 3 4 5 6 7 8 9 . . .

Fn 0 1 1 2 3 5 8 13 21 34 . . .

0 · 1− 12 = −12 · 1− 12 = 1

3 · 1− 22 = −15 · 2− 32 = 1

8 · 3− 52 = −1


n 0 1 2 3 4 5 6 7 8 9 . . .

Fn 0 1 1 2 3 5 8 13 21 34 . . .

0 · 1− 12

= −12 · 1− 12 = 1

3 · 1− 22 = −15 · 2− 32 = 1

8 · 3− 52 = −1


n 0 1 2 3 4 5 6 7 8 9 . . .

Fn 0 1 1 2 3 5 8 13 21 34 . . .

0 · 1− 12 = −1

2 · 1− 12 = 1

3 · 1− 22 = −15 · 2− 32 = 1

8 · 3− 52 = −1


n 0 1 2 3 4 5 6 7 8 9 . . .

Fn 0 1 1 2 3 5 8 13 21 34 . . .

0 · 1− 12 = −12 · 1− 12

= 1

3 · 1− 22 = −15 · 2− 32 = 1

8 · 3− 52 = −1


n 0 1 2 3 4 5 6 7 8 9 . . .

Fn 0 1 1 2 3 5 8 13 21 34 . . .

0 · 1− 12 = −12 · 1− 12 = 1

3 · 1− 22 = −15 · 2− 32 = 1

8 · 3− 52 = −1


n 0 1 2 3 4 5 6 7 8 9 . . .

Fn 0 1 1 2 3 5 8 13 21 34 . . .

0 · 1− 12 = −12 · 1− 12 = 1

3 · 1− 22

= −15 · 2− 32 = 1

8 · 3− 52 = −1


n 0 1 2 3 4 5 6 7 8 9 . . .

Fn 0 1 1 2 3 5 8 13 21 34 . . .

0 · 1− 12 = −12 · 1− 12 = 1

3 · 1− 22 = −1

5 · 2− 32 = 1

8 · 3− 52 = −1


n 0 1 2 3 4 5 6 7 8 9 . . .

Fn 0 1 1 2 3 5 8 13 21 34 . . .

0 · 1− 12 = −12 · 1− 12 = 1

3 · 1− 22 = −15 · 2− 32

= 1

8 · 3− 52 = −1


n 0 1 2 3 4 5 6 7 8 9 . . .

Fn 0 1 1 2 3 5 8 13 21 34 . . .

0 · 1− 12 = −12 · 1− 12 = 1

3 · 1− 22 = −15 · 2− 32 = 1

8 · 3− 52 = −1


n 0 1 2 3 4 5 6 7 8 9 . . .

Fn 0 1 1 2 3 5 8 13 21 34 . . .

0 · 1− 12 = −12 · 1− 12 = 1

3 · 1− 22 = −15 · 2− 32 = 1

8 · 3− 52

= −1


n 0 1 2 3 4 5 6 7 8 9 . . .

Fn 0 1 1 2 3 5 8 13 21 34 . . .

0 · 1− 12 = −12 · 1− 12 = 1

3 · 1− 22 = −15 · 2− 32 = 1

8 · 3− 52 = −1

Lewis Carroll’s Favorite Trick

Cassini’s identity is the basis of an absurdity attributed LewisCarroll

Cut the 8× 8 square along the lines indicated below andarrange the pieces into a 5× 13 rectangle to conclude that64 = 65.

Lewis Carroll’s Favorite Trick

Cassini’s identity is the basis of an absurdity attributed LewisCarrollCut the 8× 8 square along the lines indicated below andarrange the pieces into a 5× 13 rectangle to conclude that64 = 65.

Inductive Proof of Cassini’s Identity

Fn+1Fn−1 − F 2n = (−1)n, for n > 0

Replace Fn−1 by Fn+1 − Fn in the left-hand side of Cassini’sproposed identity to obtain

Fn+1Fn−1 − F 2n = Fn+1(Fn+1 − Fn)− F 2

n

= F 2n+1 − Fn+1Fn − F 2

n

= F 2n+1 − Fn(Fn+1 + Fn)

= F 2n+1 − FnFn+2

= −(FnFn+2 − F 2n+1)

= −(Fn+2Fn − F 2n+1)


Fn+1Fn−1 − F 2n = (−1)n, for n > 0


Fn+1Fn−1 − F 2n

= Fn+1(Fn+1 − Fn)− F 2n

= F 2n+1 − Fn+1Fn − F 2

n

= F 2n+1 − Fn(Fn+1 + Fn)

= F 2n+1 − FnFn+2

= −(FnFn+2 − F 2n+1)

= −(Fn+2Fn − F 2n+1)


Fn+1Fn−1 − F 2n = (−1)n, for n > 0


Fn+1Fn−1 − F 2n = Fn+1(Fn+1 − Fn)− F 2

n

= F 2n+1 − Fn+1Fn − F 2

n

= F 2n+1 − Fn(Fn+1 + Fn)

= F 2n+1 − FnFn+2

= −(FnFn+2 − F 2n+1)

= −(Fn+2Fn − F 2n+1)


Fn+1Fn−1 − F 2n = (−1)n, for n > 0

Thus ifFn+1Fn−1 − F 2

n = (−1)n

ThenFn+2Fn − F 2

n+1 = (−1)n+1

Since F2F0 − F1 = 1 · 0− 1 = (−1)1, the basis for induction istrue, so Cassini’s identity holds for all n.

Fundamental Theorem of Arithmetic

The fundamental theorem of arithmetic states that everypositive integer greater than 1 can be written as a (unique)product of primes

As a basis 2 is prime, hence a product of primesPretend that for some n ≥ 2, every integer k = 2, 3, . . . , ncan be written as the product of primesConsider n + 1

If n + 1 is prime, then it is the product of primesIf n + 1 is composite, then n + 1 = ab where 1 < a < n + 1and 1 < b < n + 1Therefore, by the inductive assumption, a and b are theproduct of primes, and so n + 1 is the product of primes


The fundamental theorem of arithmetic states that everypositive integer greater than 1 can be written as a (unique)product of primesAs a basis 2 is prime, hence a product of primes

Pretend that for some n ≥ 2, every integer k = 2, 3, . . . , ncan be written as the product of primesConsider n + 1



The fundamental theorem of arithmetic states that everypositive integer greater than 1 can be written as a (unique)product of primesAs a basis 2 is prime, hence a product of primesPretend that for some n ≥ 2, every integer k = 2, 3, . . . , ncan be written as the product of primes

Consider n + 1



The fundamental theorem of arithmetic states that everypositive integer greater than 1 can be written as a (unique)product of primesAs a basis 2 is prime, hence a product of primesPretend that for some n ≥ 2, every integer k = 2, 3, . . . , ncan be written as the product of primesConsider n + 1




If n + 1 is prime, then it is the product of primes

If n + 1 is composite, then n + 1 = ab where 1 < a < n + 1and 1 < b < n + 1Therefore, by the inductive assumption, a and b are theproduct of primes, and so n + 1 is the product of primes



If n + 1 is prime, then it is the product of primesIf n + 1 is composite, then n + 1 = ab where 1 < a < n + 1and 1 < b < n + 1

Therefore, by the inductive assumption, a and b are theproduct of primes, and so n + 1 is the product of primes

The Fundamental Theorem of the Sum and DifferenceCalculus

Define the falling factorial power nm by

nm = n(n − 1) · · · (n −m + 1)

The following identity is true

mn−1∑k=0

km−1m =n−1∑k=0

k(k − 1) · · · (k −m + 2) = nm

As a basis for induction, when n = 0 the left-hand side equalsthe right-hand side


Pretend

mn−1∑k=0

km−1 = nm

is true for some n ≥ 0

Then

mn∑

k=0

km−1 = mn−1∑k=0

km−1 + mnm−1

= nm + mnm−1

= [n(n − 1) · · · (n −m + 1)] + [m(n(n − 1) · · · (n −m + 2))]= n(n − 1) · · · (n −m + 2)[(n −m + 1) + m]

= (n + 1)n(n − 1) · · · (n −m + 2)= (n + 1)m


Pretend

mn−1∑k=0

km−1 = nm


mn∑

k=0

km−1 = mn−1∑k=0

km−1 + mnm−1

= nm + mnm−1


= (n + 1)n(n − 1) · · · (n −m + 2)= (n + 1)m


Pretend

mn−1∑k=0

km−1 = nm


mn∑

k=0

km−1

= mn−1∑k=0

km−1 + mnm−1

= nm + mnm−1


= (n + 1)n(n − 1) · · · (n −m + 2)= (n + 1)m


Pretend

mn−1∑k=0

km−1 = nm


mn∑

k=0

km−1 = mn−1∑k=0

km−1 + mnm−1

= nm + mnm−1


= (n + 1)n(n − 1) · · · (n −m + 2)= (n + 1)m


Pretend

mn−1∑k=0

km−1 = nm


mn∑

k=0

km−1 = mn−1∑k=0

km−1 + mnm−1

= nm + mnm−1

= [n(n − 1) · · · (n −m + 1)] + [m(n(n − 1) · · · (n −m + 2))]

= n(n − 1) · · · (n −m + 2)[(n −m + 1) + m]

= (n + 1)n(n − 1) · · · (n −m + 2)= (n + 1)m


Pretend

mn−1∑k=0

km−1 = nm


mn∑

k=0

km−1 = mn−1∑k=0

km−1 + mnm−1

= nm + mnm−1


= (n + 1)n(n − 1) · · · (n −m + 2)= (n + 1)m


Pretend

mn−1∑k=0

km−1 = nm


mn∑

k=0

km−1 = mn−1∑k=0

km−1 + mnm−1

= nm + mnm−1


= (n + 1)n(n − 1) · · · (n −m + 2)

= (n + 1)m


Pretend

mn−1∑k=0

km−1 = nm


mn∑

k=0

km−1 = mn−1∑k=0

km−1 + mnm−1

= nm + mnm−1


= (n + 1)n(n − 1) · · · (n −m + 2)= (n + 1)m


The fundamental theorem of the sum and difference calculus is

mn−1∑k=0

km−1 = nm

Notice the analogy with the fundamental theorem of calculus

m∫

xm−1dx = xm

Exponential Beat Powers

The exponent 2n grows faster than the polynomial n2

Prove n2 ≤ 2n for all n > 3

For n = 4 we have 42 = 1624

Pretend n2 ≤ 2n for some n ≥ 4Notice that

(n + 1)2 ≤ 2n2, when n > 3

since

0 ≤ 2n2 − (n + 1)2 = n2 − 2n − 1 = (n − 1)2 − 2

Therefore

(n + 1)2 = 2n2

≤ 2 · 2n

= 2n+1



Prove n2 ≤ 2n for all n > 3For n = 4 we have 42 = 1624


(n + 1)2 ≤ 2n2, when n > 3

since

0 ≤ 2n2 − (n + 1)2 = n2 − 2n − 1 = (n − 1)2 − 2

Therefore

(n + 1)2 = 2n2

≤ 2 · 2n

= 2n+1




Pretend n2 ≤ 2n for some n ≥ 4

Notice that(n + 1)2 ≤ 2n2, when n > 3

since

0 ≤ 2n2 − (n + 1)2 = n2 − 2n − 1 = (n − 1)2 − 2

Therefore

(n + 1)2 = 2n2

≤ 2 · 2n

= 2n+1





(n + 1)2 ≤ 2n2, when n > 3

since

0 ≤ 2n2 − (n + 1)2 = n2 − 2n − 1 = (n − 1)2 − 2

Therefore

(n + 1)2 = 2n2

≤ 2 · 2n

= 2n+1





(n + 1)2 ≤ 2n2, when n > 3

since

0 ≤ 2n2 − (n + 1)2 = n2 − 2n − 1 = (n − 1)2 − 2

Therefore

(n + 1)2

= 2n2

≤ 2 · 2n

= 2n+1





(n + 1)2 ≤ 2n2, when n > 3

since

0 ≤ 2n2 − (n + 1)2 = n2 − 2n − 1 = (n − 1)2 − 2

Therefore

(n + 1)2 = 2n2

≤ 2 · 2n

= 2n+1

Induction Over Products

Define the product notation∏

ak by

n−1∏k=0

ak = a0a1 · · · an−1

Prove by induction that

n∏k=2

(1− 1

k2

)=

n + 12n

for n ≥ 2

For a basis, let n = 2, and find

2∏k=2

(1− 1

k2

)= 1− 1

22 =2 + 12 · 2


Pretend thatn∏

k=2

(1− 1

k2

)=

n + 12n

for n ≥ 2

for some n ≥ 2

Then notice that

n+1∏k=2

(1− 1

k2

)=

[n∏

k=2

(1− 1

k2

)](1− 1

(n + 1)2

)=

[n + 12n

](1− 1

(n + 1)2

)=

12n

((n + 1)2 − 1

(n + 1)

)=

12n

(n2 + 2n(n + 1)

)=

n + 22(n + 1)


Pretend thatn∏

k=2

(1− 1

k2

)=

n + 12n

for n ≥ 2

for some n ≥ 2Then notice that

n+1∏k=2

(1− 1

k2

)=

[n∏

k=2

(1− 1

k2

)](1− 1

(n + 1)2

)=

[n + 12n

](1− 1

(n + 1)2

)=

12n

((n + 1)2 − 1

(n + 1)

)=

12n

(n2 + 2n(n + 1)

)=

n + 22(n + 1)


Pretend thatn∏

k=2

(1− 1

k2

)=

n + 12n

for n ≥ 2


n+1∏k=2

(1− 1

k2

)

=

[n∏

k=2

(1− 1

k2

)](1− 1

(n + 1)2

)=

[n + 12n

](1− 1

(n + 1)2

)=

12n

((n + 1)2 − 1

(n + 1)

)=

12n

(n2 + 2n(n + 1)

)=

n + 22(n + 1)


Pretend thatn∏

k=2

(1− 1

k2

)=

n + 12n

for n ≥ 2


n+1∏k=2

(1− 1

k2

)=

[n∏

k=2

(1− 1

k2

)](1− 1

(n + 1)2

)

=

[n + 12n

](1− 1

(n + 1)2

)=

12n

((n + 1)2 − 1

(n + 1)

)=

12n

(n2 + 2n(n + 1)

)=

n + 22(n + 1)


Pretend thatn∏

k=2

(1− 1

k2

)=

n + 12n

for n ≥ 2


n+1∏k=2

(1− 1

k2

)=

[n∏

k=2

(1− 1

k2

)](1− 1

(n + 1)2

)=

[n + 12n

](1− 1

(n + 1)2

)

=12n

((n + 1)2 − 1

(n + 1)

)=

12n

(n2 + 2n(n + 1)

)=

n + 22(n + 1)


Pretend thatn∏

k=2

(1− 1

k2

)=

n + 12n

for n ≥ 2


n+1∏k=2

(1− 1

k2

)=

[n∏

k=2

(1− 1

k2

)](1− 1

(n + 1)2

)=

[n + 12n

](1− 1

(n + 1)2

)=

12n

((n + 1)2 − 1

(n + 1)

)

=12n

(n2 + 2n(n + 1)

)=

n + 22(n + 1)


Pretend thatn∏

k=2

(1− 1

k2

)=

n + 12n

for n ≥ 2


n+1∏k=2

(1− 1

k2

)=

[n∏

k=2

(1− 1

k2

)](1− 1

(n + 1)2

)=

[n + 12n

](1− 1

(n + 1)2

)=

12n

((n + 1)2 − 1

(n + 1)

)=

12n

(n2 + 2n(n + 1)

)

=n + 2

2(n + 1)


Pretend thatn∏

k=2

(1− 1

k2

)=

n + 12n

for n ≥ 2


n+1∏k=2

(1− 1

k2

)=

[n∏

k=2

(1− 1

k2

)](1− 1

(n + 1)2

)=

[n + 12n

](1− 1

(n + 1)2

)=

12n

((n + 1)2 − 1

(n + 1)

)=

12n

(n2 + 2n(n + 1)

)=

n + 22(n + 1)

discrete mathematics-mathematical induction

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