Fibonacci

70
Fibonacci Numbers and the Golden Ratio Emily Cookson April 19, 2008

Transcript of Fibonacci

Page 1: Fibonacci

Fibonacci Numbers and the Golden Ratio

Emily Cookson

April 19, 2008

Page 2: Fibonacci

Abstract

The Fibonacci numbers have some interesting properties and applications.This project begins by looking at some basic properties and moves on tolook at divisibility properties. The golden ratio and its relationship withcontinued fractions and geometry is then discussed and some claims of

Fibonacci numbers in Architecture are investigated.

Page 3: Fibonacci

Contents

1 Introduction 51.1 Leonardo Pisano Fibonacci . . . . . . . . . . . . . . . . . . . 51.2 The Fibonacci and Lucas Sequences . . . . . . . . . . . . . . 61.3 Binet’s Formula . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Basic Properties of Fibonacci Numbers 102.1 Negation Formula . . . . . . . . . . . . . . . . . . . . . . . . . 102.2 Cassini’s Identity . . . . . . . . . . . . . . . . . . . . . . . . . 112.3 Catalan’s Identity . . . . . . . . . . . . . . . . . . . . . . . . 122.4 Fibonacci and Lucas Relationships . . . . . . . . . . . . . . . 142.5 Matrix Theory . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3 Divisibility Properties 163.1 Divisibility Lemmas . . . . . . . . . . . . . . . . . . . . . . . 163.2 Greatest Common Divisor Theorem . . . . . . . . . . . . . . 183.3 Divisibility Corollaries . . . . . . . . . . . . . . . . . . . . . . 193.4 Primes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

4 The Golden Ratio and Continued Fractions 214.1 Limit of Consecutive Fibonacci Numbers . . . . . . . . . . . . 214.2 The Golden Ratio . . . . . . . . . . . . . . . . . . . . . . . . 234.3 Continued Fractions . . . . . . . . . . . . . . . . . . . . . . . 244.4 Q-polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . 274.5 Lame’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 304.6 Pell’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . 32

5 Golden Geometry 355.1 The Golden Rectangle . . . . . . . . . . . . . . . . . . . . . . 355.2 The Golden Triangle . . . . . . . . . . . . . . . . . . . . . . . 375.3 Logarithmic Spirals . . . . . . . . . . . . . . . . . . . . . . . . 405.4 The Pentagon . . . . . . . . . . . . . . . . . . . . . . . . . . . 435.5 The Pentagram . . . . . . . . . . . . . . . . . . . . . . . . . . 44

1

Page 4: Fibonacci

6 Fibonacci Numbers in Architecture 466.1 Ancient Egypt . . . . . . . . . . . . . . . . . . . . . . . . . . 466.2 Ancient Greece . . . . . . . . . . . . . . . . . . . . . . . . . . 526.3 Medieval Islamic Architecture . . . . . . . . . . . . . . . . . . 556.4 Le Corbusier . . . . . . . . . . . . . . . . . . . . . . . . . . . 596.5 Inspired by Nature . . . . . . . . . . . . . . . . . . . . . . . . 62

7 Conclusion 65

A The First 50 Fibonacci and Lucas Numbers 66

Bibliography 67

2

Page 5: Fibonacci

List of Figures

1.1 Leonardo of Pisa . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.1 Paradox gap . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.2 Paradox overlap . . . . . . . . . . . . . . . . . . . . . . . . . . 12

4.1 Graph of limit τ . . . . . . . . . . . . . . . . . . . . . . . . . 224.2 Twist of Cassini’s identity . . . . . . . . . . . . . . . . . . . . 224.3 The golden section . . . . . . . . . . . . . . . . . . . . . . . . 23

5.1 Construction of a golden rectangle . . . . . . . . . . . . . . . 365.2 Inflation of the golden rectangle . . . . . . . . . . . . . . . . . 375.3 Inflation of the golden triangle . . . . . . . . . . . . . . . . . 385.4 Spiral with an expansion of τ in 90◦ turn . . . . . . . . . . . 405.5 Logarithmic spiral proper to the golden triangle . . . . . . . . 415.6 Spiral with an expansion of τ in 108◦ turn . . . . . . . . . . . 415.7 Spiral with an expansion of τ in 360◦ turn . . . . . . . . . . . 415.8 Spiral constructed from a sequence of golden rectangles . . . 425.9 Fibonacci Spiral . . . . . . . . . . . . . . . . . . . . . . . . . 425.10 Construction of a pentagon . . . . . . . . . . . . . . . . . . . 435.11 Pentagon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445.12 Incommensurability . . . . . . . . . . . . . . . . . . . . . . . 45

6.1 Great Pyramid of Giza . . . . . . . . . . . . . . . . . . . . . . 476.2 Pyramid structure . . . . . . . . . . . . . . . . . . . . . . . . 476.3 Constructing the golden section . . . . . . . . . . . . . . . . . 486.4 The Parthenon . . . . . . . . . . . . . . . . . . . . . . . . . . 526.5 Parthenon claims of τ . . . . . . . . . . . . . . . . . . . . . . 526.6 Construction of Proposition 11 . . . . . . . . . . . . . . . . . 546.7 Complete set of girih tiles . . . . . . . . . . . . . . . . . . . . 556.8 Rhombus tiling . . . . . . . . . . . . . . . . . . . . . . . . . . 566.9 Kite and Dart tiling . . . . . . . . . . . . . . . . . . . . . . . 566.10 Spandrel from the Darb-i-Imam shrine . . . . . . . . . . . . . 586.11 Tile mappings . . . . . . . . . . . . . . . . . . . . . . . . . . . 586.12 Defect and correction . . . . . . . . . . . . . . . . . . . . . . . 596.13 Model male . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

3

Page 6: Fibonacci

6.14 Le Modulor . . . . . . . . . . . . . . . . . . . . . . . . . . . . 616.15 Design for South-west facade of Governor’s Palace . . . . . . 616.16 Spiral Cafe . . . . . . . . . . . . . . . . . . . . . . . . . . . . 626.17 Radiograph of Nautilus pompilius shell . . . . . . . . . . . . . 636.18 Eden Project . . . . . . . . . . . . . . . . . . . . . . . . . . . 646.19 Core model . . . . . . . . . . . . . . . . . . . . . . . . . . . . 646.20 Double spiral pattern . . . . . . . . . . . . . . . . . . . . . . . 646.21 Coneflower . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

4

Page 7: Fibonacci

Chapter 1

Introduction

1.1 Leonardo Pisano Fibonacci

Leonardo of Pisa, or Leonardo Pisano in Italian, is better known by hisnickname Fibonacci. Fibonacci is a contraction of the Latin “filius Bonacci”,which means “the son of Bonaccio”. Fibonacci himself occasionally used thename Bigollo, which means a traveller.

Figure 1.1: Leonardo of Pisa[CR]

Leonardo of Pisa (1170-1250) was born in Pisa, Italy, but was educatedin North Africa. His father, Guilielmo Bonaccio, was a customs inspector inAlgeria. Fibonacci was taught mathematics in Bugia, a town on the BarbaryCoast which is now the Algerian port city of Bejaia, and travelled with hisfather around the Mediterranean coast. He would have encountered severaldifferent mathematical systems and recognised the advantages of the Hindu-Arabic system over all the others. Around 1200, he ceased his travels, andreturned to Pisa.

Fibonacci wrote several mathematical texts after returning to Italy. Inhis Liber Abaci (1202) he introduced the Hindu-Arabic numerical system

5

Page 8: Fibonacci

and Arabic mathematics to Europe. The Fibonacci numbers are namedafter Fibonacci, whose Liber Abaci posed a rabbit problem which is thebasis of the Fibonacci sequence. [RK, CR]

1.2 The Fibonacci and Lucas Sequences

The Fibonacci numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, ... where, given thetwo starting values, each number is the sum of the two preceding numbers.The first 50 Fibonacci numbers are given in Appendix A. This sequence ofnumbers is defined by the recurrence relation

Fn+1 = Fn + Fn−1 (1.1)

where F0 = 0, F1 = 1, and n ≥ 1.The original problem that Fibonacci investigated in his Liber Abaci was

about breeding rabbits. According to Burton [DB], it read as follows:

A man put one pair of rabbits in a certain place entirely sur-rounded by a wall. How many pairs of rabbits can be producedfrom that pair in a year, if the nature of these rabbits is such thatevery month each pair bears a new pair which from the secondmonth on becomes productive?

Assuming the original pair are an adult pair and defining adult and youngpairs to be rabbits older than and younger than one month old respectively,we obtain the following table:

Table 1.1: Growth of Rabbit Colony [DB]Months Adult Pairs Young Pairs Total

0 1 0 11 1 1 22 2 1 33 3 2 54 5 3 85 8 5 136 13 8 217 21 13 348 34 21 559 55 34 89

10 89 55 14411 144 89 23312 233 144 377

By solving this rabbit problem, Fibonacci calculated 13 consecutiveterms of the Fibonacci sequence:

6

Page 9: Fibonacci

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377.

The recurrence relation was also deduced from the rabbit problem. If wedenote the total number of pairs at the end of month n by Tn and thenumber of adult pairs at the end of month n by An, then clearly

Tn = Tn−1 +An−1, (n ≥ 1).

As young pairs become adult pairs after a month, it is obvious that An =Tn−1. Hence

Tn = Tn−1 + Tn−2

where T0 = 1, T1 = 2, and n ≥ 2, which is comparable to the Fibonaccisequence where Fn+2 = Tn.

The French mathematician, Edouard Lucas (1842-1891), discovered thatthe sequence 2, 1, 3, 4, 7, 11, ... has lots of similar properties to, andconnections with, the Fibonacci numbers. These numbers are called Lucasnumbers, defined by

Ln+1 = Ln + Ln−1 (1.2)

where L0 = 2, L1 = 1, and n ≥ 1. The first 50 Lucas numbers are given inAppendix A.

A generalised Fibonacci sequence is defined by

Gn+1 = Gn +Gn−1 (1.3)

where G0 = a and G1 = b, for any a, b ∈ Z, and n ≥ 1. The first two values,a and b, are known as the seed. The Fibonacci sequence is a generalised Fi-bonacci sequence where the seed values are given as 0 and 1. The generalisedFibonacci sequence with seed values 2 and 1 is the Lucas sequence.

1.3 Binet’s Formula

We can work out any Fibonacci number using the recurrence relation (1.1)but it takes a long time to work out Fn for large n as one would have to workout all numbers in the sequence from the seed values up to Fn−1. Fortu-nately, there is a formula for Fn, called Binet’s formula, which involves onlyn and does not need any other Fibonacci values. According to Knott [RK],Binet’s formula was known to Euler and De Moivre and was rediscovered in1843 by Jacques Binet (1786-1856). It is defined as follows:

Fn =1√5

((1 +√

52

)n−

(1−√

52

)n). (1.4)

7

Page 10: Fibonacci

Proof. Let Fk = xk, where k ≥ 2. Using rule (1.1),

Fk+1 = Fk + Fk−1

⇒ xk+1 = xk + xk−1

Since xk−1 6= 0, dividing through by xk−1 gives us

x2 = x+ 1

⇒ x2 − x− 1 = 0

⇒ x =1±√

52

Take 1+√

52 = τ , 1−

√5

2 = σ, say. Therefore the Fibonacci sequence must beof the form Fn = ατn + βσn. Using the Fibonacci seed, we obtain

F0 = α+ β = 0; F1 = ατ + βσ = 1.

Substitution givesατ − ασ = α(τ − σ) = 1.

Henceα =

1τ − σ

=1√5.

Therefore

Fn = ατn + βσn = ατn − ασn =1√5

(τn − σn).

That is,

Fn =1√5

((1 +√

52

)n−

(1−√

52

)n)or, in terms of τ , noting σ = −1

τ

Fn =1√5

(τn −

(−1τ

)n).

We can note from this proof that since all generalised Fibonacci se-quences follow the same recurrence relation then any generalised Fibonaccisequence must be of the form

Gn = ατn + βσn. (1.5)

Here are a few useful facts about τ and σ:

τ + σ = 1, τ − σ =√

5, τσ = −1, τ2 = τ + 1, σ2 = σ + 1. (1.6)

A similar formula to Binet’s Fibonacci Formula can be found for Lucasnumbers:

Ln = τn + σn (1.7)

8

Page 11: Fibonacci

Proof. By (1.5), Ln = ατn + βσn where α, β are constants to be found andτ = 1+

√5

2 , σ = 1−√

52 . Using the Lucas seed, we obtain

L0 = α+ β = 2; L1 = ατ + βσ = 1.

Substitution gives

ατ + (2− α)σ = α(τ − σ) + 2σ = 1.

Hence

α =1− 2στ − σ

=√

5√5

= 1.

Therefore

Ln = ατn + βσn = ατn + (2− α)σn = τn + σn.

9

Page 12: Fibonacci

Chapter 2

Basic Properties of FibonacciNumbers

2.1 Negation Formula

We previously only defined the Fibonacci sequence for n ≥ 1, given the seedvalues, but there are in fact Fibonacci numbers for negative n too. Therecurrence relation (1.1) can be rearranged to Fn−1 = Fn+1−Fn and can beused to extend the sequence backwards, giving F−n as 1,−1, 2,−3, 5,−8, . . .for n ≥ 1. The general formula is

F−n = (−1)n+1Fn, n ≥ 1. (2.1)

Proof. F−1 = F1 −F0 = 1 = (−1)2F1 and F−2 = F0 −F−1 = −1 = (−1)3F2

are both true therefore the equation is true for n = 1 and n = 2. Assume itis true for n ≤ k, then

F−k = (−1)k+1Fk

andF−(k−1) = (−1)kFk−1.

Subtracting one from the other gives

F−k+1 − F−k = (−1)k(Fk−1 + Fk).

Using rule (1.1),F−k−1 = (−1)kFk+1

and henceF−(k+1) = (−1)k+2Fk+1.

Therefore, by induction, property (2.1) is true for n ≥ 1.

10

Page 13: Fibonacci

This property (2.1) can also be proved using Binet’s formula. Similarly,Binet’s formula can be used to prove that Lucas numbers for negative n canbe found using the following negation formula:

L−n = (−1)nLn, n ≥ 1. (2.2)

Proof. Using Binet’s formula,

L−n = τ−n + σ−n =1τn

+ (−τ)n = (−1)n((−1τ

)n+ τn

)= (−1)n(σn + τn) = (−1)nLn.

2.2 Cassini’s Identity

One may observe, when looking at the first few Fibonacci numbers, that thesquare of any Fibonacci number appears to be one integer away from theproduct of the preceding and succeeding numbers. For example, looking atn = 4, 32 = 9 and 2×5 = 10. This actually applies to all Fibonacci numbersand is called Cassini’s identity:

Fn+1Fn−1 − F 2n = (−1)n. (2.3)

Proof. F2F0 − F 21 = 0 − 1 = −1 = (−1)1 is true therefore the equation is

true for n = 1. Assume it is true for n = k, then

Fk+1Fk−1 − F 2k = (−1)k.

Using (1.1),

Fk+1(Fk+1 − Fk)− Fk(Fk+2 − Fk+1) = (−1)k

⇒ F 2k+1 − FkFk+2 = (−1)k

⇒ Fk+2Fk − F 2k+1 = −(−1)k = (−1)k+1.

Similarly,

(Fk + Fk−1)Fk−1 − Fk(Fk−1 + Fk−2) = (−1)k

⇒ F 2k−1 − FkFk−2 = (−1)k

⇒ FkFk−2 − F 2k−1 = −(−1)k = (−1)k−1.

Therefore, by induction, property (2.3) is true for all n, where Fn is a Fi-bonacci number.

Paradox

It follows from Cassini’s identity that if you divide an Fn × Fn square intopieces as in Figure 2.1 and rearrange the pieces into a rectangle of sides Fn−1

and Fn+1 then the areas of the square and of the rectangle will differ by 1,whatever the side of the original square.

11

Page 14: Fibonacci

Figure 2.1: Paradox gap[TK]

When n is even we get a gap, as in Figure 2.1, whereas when n is oddwe get an overlap, as in Figure 2.2.

Figure 2.2: Paradox overlap[TK]

Because the gap, or overlap, always has an area of 1 it will be less visibleas n increases. The paradox is that a diagram can appear to prove thatFn−1Fn+1 = F 2

n for some n as, for large n, the gap, or overlap, is so narrowthat it is imperceptible to the eye. F6 = 8 is often used as an example wherethe gap is hardly noticeable.

2.3 Catalan’s Identity

From closer study of the first few numbers of the Fibonacci sequence, itappears that there is a relationship between the square of a Fibonacci num-ber and the product of any two numbers equidistant from that Fibonaccinumber. One may observe that the difference is always a square number.

12

Page 15: Fibonacci

Example 2.3.1. Consider n = 3. Cassini’s identity tells us that the squareof F3 is ±1 from the product of the preceding and succeeding numbers. Letus now consider the product of the numbers that are 0 ≤ r ≤ 7 either sideof F3 in the Fibonacci sequence and the difference from the square of F3.

Table 2.1: Difference between the product F3−rF3+r and the square of F3

r F3−rF3+r Fr F 23 − F3−rF3+r

0 2× 2 = 4 0 01 1× 3 = 3 1 -12 1× 5 = 5 1 13 0× 8 = 0 2 -44 1× 13 = 13 3 95 −1× 21 = −21 5 -256 2× 34 = 68 8 647 −3× 55 = −165 13 -169

Observe that the difference between the product F3−rF3+r and the squareof F3 for 0 ≤ r ≤ 7 is ±F 2

r .

The result from this example can actually be extended to all r and n.Letting r = 1 gives Cassini’s identity. The general case is called Catalan’sidentity:

F 2n − Fn+rFn−r = (−1)n−rF 2

r . (2.4)

Proof. Using Binet’s formula,

F 2n − Fn+rFn−r =

(1√5

(τn − σn))2

− 1√5

(τn+r − σn+r)1√5

(τn−r − σn−r)

=15((τn − σn)2 − (τn+r − σn+r)(τn−r − σn−r)

)=

15((τ2n − 2τnσn + σ2n)− (τ2n − τn+rσn−r − τn−rσn+r + σ2n)

)=

15

(τn+rσn−r − 2τnσn + τn−rσn+r)

=15

(τn−rσn−r(τ r − σr)2)

= (τσ)n−r(τ r − σr)2

5= (−1)n−rF 2

r .

13

Page 16: Fibonacci

2.4 Fibonacci and Lucas Relationships

We have seen that the Fibonacci and Lucas sequences are both generalisedFibonacci sequences. The Lucas numbers have lots of properties that aresimilar to those of Fibonacci numbers. The two sequences are, in fact, evenmore intimately connected. Two formulae relating Fibonacci and Lucasnumbers are

Ln = Fn−1 + Fn+1 (2.5)

and5Fn = Ln−1 + Ln+1 (2.6)

for all integers n.We can use Binet’s formula to prove property (2.5) as follows:

Proof.

Fn+1 + Fn−1 =1√5

((τn+1 − σn+1) + (τn−1 − σn−1)

)=

1√5

(τn−1(τ2 + 1)− σn−1(σ2 + 1)

)=

1√5

(τn−1(τ(τ − σ))− σn−1(σ(τ − σ))

)=

1√5

(τn−1(

√5τ)− σn−1(−

√5σ))

= τn + σn

= Ln.

Property (2.6) can be proved by induction as follows:

Proof. 5F0 = 0 = −1 + 1 = L−1 + L1 and 5F1 = 5 = 2 + 3 = L0 + L2 areboth true therefore the equation is true for n = 0 and n = 1. Assume it istrue for n ≤ k, then

5Fk = Lk−1 + Lk+1

and5Fk−1 = Lk−2 + Lk.

Adding the two together gives

5Fk + 5Fk−1 = Lk−1 + Lk+1 + Lk−2 + Lk.

Using rules (1.1) and (1.2),

5Fk+1 = (Lk−1 + Lk−2) + (Lk+1 + Lk) = Lk + Lk+2.

14

Page 17: Fibonacci

Therefore rule (2.6) is true for n ≥ 0. Using rules (2.1) and (2.2), we canprove (2.6) for n = −k, k ≥ 1 as follows:

5F−k = (−1)k+1(Lk−1 + Lk+1)

= (−1)k+1((−1)−(k−1)L−(k−1) + (−1)−(k+1)L−(k+1))

= L−(k+1) + L−(k−1).

Therefore, by induction, 5Fn = Ln−1 + Ln+1 for all integers n.

2.5 Matrix Theory

We can express the recurrence relation (1.1) in terms of 2×2 matrices usingthe Fibonacci Q-matrix. The following property is taken from [RJ]. Rule(1.1) becomes

(Fn+1

Fn

)= Q

(FnFn−1

)where Q ≡

(1 11 0

).

Recall Fn = 0, 1, 1 for n = 0, 1, 2. Then, from the recurrence relation, wehave (

Fn+1

Fn

)= Qn−1

(11

)and (

FnFn−1

)= Qn−1

(10

)which gives us

Qn =(Fn+1 FnFn Fn−1

), n ≥ 1. (2.7)

Many properties can be derived using matrix properties.

Example 2.5.1. Using the fact that AnAm = Am+n for any square matrixA, one can derive

F 2n + F 2

n−1 = F2n−1. (2.8)

Proof. Q2n = QnQn

⇒(F2n+1 F2n

F2n F2n−1

)=(Fn+1 FnFn Fn−1

)(Fn+1 FnFn Fn−1

)=(

F 2n+1 + F 2

n Fn+1Fn + FnFn−1

FnFn+1 + Fn−1Fn F 2n + F 2

n−1

).

15

Page 18: Fibonacci

Chapter 3

Divisibility Properties

Considering that the Fibonacci numbers are based on a recurrence relation,one does not expect the numbers to have any particular divisibility proper-ties. However, there are actually several associated divisibility properties.In this chapter, I will prove the greatest common divisor theorem using [FS]and Chapter 6 of [SV] as guides. I will be using the notation (a, b) to denotethe greatest common divisor of a and b.

3.1 Divisibility Lemmas

The proof of the greatest common divisor theorem requires knowledge of thefollowing three Lemmas.

Lemma 3.1.1.Gn+m = Fm−1Gn + FmGn+1 (3.1)

where G1, G2, ... is any generalised Fibonacci sequence.

Proof. Gn+1 = F0Gn+F1Gn+1 and Gn+2 = F1Gn+F2Gn+1 are true becauseF0 = 0 and F1 = F2 = 1, and therefore the equation is true for m = 1 andm = 2. Assume it is true for m ≤ k, then

Gn+k = Fk−1Gn + FkGn+1

andGn+(k−1) = Fk−2Gn + Fk−1Gn+1.

By addition,

Gn+(k−1)+Gn+k = (Fk−2+Fk−1)Gn+(Fk−1+Fk)Gn+1 = FkGn+Fk+1Gn+1 = Gn+(k+1).

We know this is true from the definition of a generalised Fibonacci sequence(1.3) so the equation holds for m = k + 1. Therefore, by induction, rule(3.1) is true for all integers m,n ≥ 1.

16

Page 19: Fibonacci

Taking Gi = Fi, we obtain

Fn+m = Fm−1Fn + FmFn+1 (3.2)

andF2n = Fn−1Fn + FnFn+1 = Fn(Fn−1 + Fn+1) = FnLn. (3.3)

Lemma 3.1.2. If m is divisible by n, then Fm is divisible by Fn:

n | m⇒ Fn | Fm. (3.4)

Proof. Let m = rn, r ∈ Z. It follows from rule (3.3) that F2n is divisible byFn so the statement holds for r = 2. Assume it is true for r = k. It followsfrom rule (3.2) that if Fkn is divisible by Fn, then so is F(k+1)n since

F(k+1)n = Fkn+n = Fn−1Fkn + FnFkn+1.

Therefore, by induction, Frn is divisible by Fn for all integers r ≥ 2.

Lemma 3.1.3. Fn and Fn+1 are relatively prime:

(Fn, Fn+1) = 1, ∀n. (3.5)

A simple proof for Lemma 3.1.3 uses the Euclidean algorithm. Thefollowing algorithm is taken from [DK].

Definition 3.1.1 (The Modern Euclidean Algorithm). Given nonnegativeintegers u and v, this algorithm finds their greatestcommon divisor.

A1: [v = 0?] If v = 0, the algorithm terminates with u as the answer.A2: [Take u mod v.] Set r ← u mod v, u← v, v ← r, and return to A1.

(3.6)

The following proof of Lemma 3.1.3 is based on a proof in Chapter 9 of[TK].

Proof. Consider u = Fn+1, v = Fn. Using the Euclidean Algorithm (3.6) tofind (u, v), we have:

Fn+1 = 1 · Fn + Fn−1

Fn = 1 · Fn−1 + Fn−2

...F4 = 1 · F3 + F2 = 1 · 2 + 1 = 3F3 = 1 · F2 + F1 = 1 · 1 + 1 = 2F2 = 1 · F1 + F0 = 1 · 1 + 0 = 1.

The algorithm terminates here as v = 0. So we take u = F1 = 1 as theanswer.

17

Page 20: Fibonacci

3.2 Greatest Common Divisor Theorem

Theorem 3.2.1.(Fm, Fn) = F(m,n) (3.7)

Proof. Let m > n. Applying the Euclidean algorithm (3.6) to m and n givesus:

m = np0 + r1

n = r1p1 + r2

r1 = r2p2 + r3...

rt−2 = rt−1pt−1 + rt

rt−1 = rtpt

(3.8)

where rt = (m,n).We can see from the first line that m = np0 + r1, so

(Fm, Fn) = (Fnp0+r1 , Fn)

which, by (3.2), gives us

(Fm, Fn) = (Fnp0−1Fr1 + Fnp0Fr1+1, Fn).

Because Fn | Fnp0 by (3.4), this becomes

(Fm, Fn) = (Fnp0−1Fr1 , Fn).

Using (3.5) we can see that (Fnp0−1, Fnp0) = 1. By (3.4) we know that Fnis a factor of Fnp0 . Therefore, since Fnp0−1 and Fnp0 are relatively prime,Fnp0−1 and Fn are also relatively prime. Hence we obtain

(Fm, Fn) = (Fr1 , Fn).

Now, by returning to the equations generated in (3.8) and repeatedly sub-stituting these equations into the last expression, we can similarly provethat

(Fm, Fn) = (Frt−1 , Frt).

From (3.8) we know that rt−1 = rtpt, so

(Frt−1 , Frt) = Frt

and hence(Fm, Fn) = Frt .

Therefore, as rt = (m,n), we find that

(Fm, Fn) = F(m,n).

18

Page 21: Fibonacci

3.3 Divisibility Corollaries

The following corollary is an interesting corollary of Theorem 3.2.1.

Corollary 3.3.1. If (m,n) = 1, 2 or 5, then Fmn is divisible by FmFn.

Proof. If (m,n) = 1 or (m,n) = 2 then (Fm, Fn) = 1 by Theorem 3.2.1.Fmn is divisible by Fm and Fn by Lemma 3.1.2. It follows that Fmn is alsodivisible by the product FmFn as Fm and Fn are relatively prime. Consider(m,n) = 5 with m = 5a, n = 5kb, and (5, a) = (5, b) = (a, b) = 1. ByTheorem 3.2.1,

(Fm, Fn) = F(m,n) = 5.

By Lemma 3.1.2, Fmn is divisible by Fm, Fn, and 5k+1. Hence Fmn is alsodivisible by the product FmFn.

Conversely, it can be proved that

Corollary 3.3.2. If Fmn is divisible by FmFn, then (m,n) = 1, 2 or 5.

A proof can be found in [SV].

3.4 Primes

Dunlap [RD] comments that “the problem of divisibility of a set of numbersalso raises the question of prime factors.” Previously we found that if m isdivisible by n then Fm is divisible by Fn. It appears to imply that if m is aprime then Fm is a prime.

Corollary 3.4.1. If m is not a prime then Fm is not a prime unless m = 4.

Proof. Let m be divisible by v, and let v be divisible by w. Then, by Lemma3.1.2, Fm is divisible by Fv and Fw. Therefore, if m is not a prime, then Fmis a prime if and only if Fv = Fw = 1. As this only occurs when both v andw can only be 1 or 2, then m can only be 1, 2 or 4. Hence F4 = 3 is the onlycase where m is not prime and Fm is prime.

However, the converse is not always true.

Example 3.4.1. 19 is a prime but F19 is not a prime because F19 = 4181 =113× 37.

19

Page 22: Fibonacci

An interesting theorem also related to primes, which appears in [RK], isCarmichael’s theorem.

Theorem 3.4.1 (Carmichael’s Theorem). Fn has at least one prime factorthat is not a factor of any earlier Fibonacci number for all Fn, with theexception of the following special cases:

1. F1 = 1, which has no prime factors;

2. F2 = 1, which has no prime factors;

3. F6 = 8, which only has a prime factor of 2 (which is F3); and

4. F12 = 144, which only has 2 and 3 as prime factors (which are F3 andF4 respectively).

20

Page 23: Fibonacci

Chapter 4

The Golden Ratio andContinued Fractions

4.1 Limit of Consecutive Fibonacci Numbers

The sequence of ratios of consecutive Fibonacci numbers converges to a limitτ :

τ = limn→∞

Fn+1

Fn

= limn→∞

Fn + Fn−1

Fn

= limn→∞

1 + limn→∞

Fn−1

Fn

= 1 +1

limn→∞

FnFn−1

= 1 +1τ.

It follows that the limit τ satisfies the equation

τ2 − τ − 1 = 0 (4.1)

which has roots 1+√

52 and 1−

√5

2 . One may note we have seen these beforein the proof of Binet’s formula (1.4). Taking the positive root gives us

τ =1 +√

52

= 1.618 . . . .

We will later prove that this is the golden ratio.

21

Page 24: Fibonacci

Figure 4.1: Graph of limit τ

One may observe that if the ratio of consecutive Fibonacci numbersconverges to a limit then one must also find that

limn→∞

(Fn+1

Fn− FnFn−1

)= 0.

This observation is related to Cassini’s identity (2.3). It follows from the

Figure 4.2: Twist of Cassini’s identity

identity that, for n > 1,

Fn+1

Fn− FnFn−1

=(−1)n

FnFn−1

which converges quickly to zero, sign oscillating, as n tends to infinity.

22

Page 25: Fibonacci

4.2 The Golden Ratio

Two quantities are in the golden ratio if the ratio between the sum of thosequantities and the larger quantity is the same as the ratio between the largerquantity and the smaller quantity.

Definition 4.2.1. The golden section is a line divided into two segmentsaccording to the golden ratio.

Figure 4.3: The golden section

This can be expressed as follows:

a+ b

a=a

b= τ, where a > b.

Rearranging gives us a = bτ , so substitution gives us:

bτ + b

bτ= τ ⇒ τ + 1

τ= τ.

Hence one can see thatτ2 = τ + 1 (4.2)

andτ = 1 +

1τ. (4.3)

As equation (4.2) and equation (4.1) are the same, this proves the remarkableresult that the golden ratio τ is the same as the limit τ , where τ = 1+

√5

2 =1.618 . . . .

Definition 4.2.2. The golden ratio is the ratio τ : 1.

The golden ratio is often denoted by the Greek letter φ, named after theGreek sculptor Phidias. It is said that Phidias widely used the golden ratioin his works of sculpture, although there is little evidence to show this.

23

Page 26: Fibonacci

4.3 Continued Fractions

Definition 4.3.1. A finite simple continued fraction is a fraction of theform:

x = a0 +1

a1 +1

a2 +1

...

an−1 +1an

. (4.4)

where a0, a1, . . . , an are integers, all of which are positive, except possibly a0.

This definition is taken from [DB]. It is convention to denote a simplecontinued fraction by a symbol which displays its partial quotients:

[a0; a1, a2, a3, . . . , an].

Theorem 4.3.1. Any rational number can be written as a finite simplecontinued fraction.

Proof. Let ab , where b > 0 be any rational number. The Euclidean algorithm

(3.6) gives us the equations:

a = ba0 + r1

b = r1a1 + r2

r1 = r2a2 + r3...

an−2 = rn−1an−1 + rn

rn−1 = rnan

(4.5)

where 0 < rn < rn−1 < · · · < r3 < r2 < r1 < b. Note that since each rk isa positive integer then the coefficients a1, . . . , an are positive. Rearrangingthe equations (4.5) gives us:

a

b= a0 +

r1b

= a0 +1br1

b

r1= a1 +

r2r1

= a1 +1r1r2

r1r2

= a2 +r3r2

= a2 +1r2r3

...rn−2

rn−1= an−1 +

rnrn−1

= an−1 +1

rn−1

rnrn−1

rn= an.

(4.6)

24

Page 27: Fibonacci

If we substitute the second equation in (4.6) into the first, then

a

b= a0 +

1a1 + 1

r1r2

.

Continuing in this way, by repeatedly substituting, we find that

a

b= a0 +

1

a1 +1

a2 +1

a3 +1...

an−1 +1an

.

It follows from this proof, based on [DB], that the partial denominatorsa0, . . . , an are the same as the quotients that occur when the Euclideanalgorithm (3.6) is applied to a and b.

This explains how one can discover how a rational number ab , where

a < b, can be expressed as a simple continued fraction by forming a rectanglewith height a and width b and dividing it firstly into a1 squares of widtha, then dividing the remainder into a2 squares, and repeatedly dividingremainders into ak squares until there is no remainder. This method isshown in the next two examples and can be used as a simple pictorial wayto find the partial quotients a1, . . . , an of a rational number if n is small.

Example 4.3.1.713

= [0; 1, 1, 6]

7

13

6

1

Example 4.3.2.949

= [0; 5, 2, 4]

25

Page 28: Fibonacci

9

494

1

Any number x ∈ R can, in fact, be expressed as a simple continuedfraction as irrational numbers can be expressed as infinite simple continuedfractions of the form:

x = a0 +1

a1 +1

a2 +1. . .

. (4.7)

Every irrational number has a unique representation as an infinite con-tinued fraction. Burton proves this in [DB]. However, if x is rational, therepresentation [a0; a1, a2, ..., an] is unique if and only if an 6= 1, as in thiscase x = [a0; a1, a2, ..., an−1 + 1] as well.

Example 4.3.3. Representations of 310 :

[0; 3, 2, 1] =1

3 +1

2 +11

=1

3 +13

= [0; 3, 3]

The golden ratio, τ , is an irrational number and so can be expressed asan infinite simple continued fraction. By repeatedly substituting τ into thedenominator in equation (4.3) we find

τ = 1 +1τ

= 1 +1

1 +1τ

= · · · = 1 +1

1 +1

1 +1

1 +1

1 +1

1 +1. . .

which shows us that the continued fraction representation of τ is

τ = [1; 1, 1, 1, 1, 1, . . . ]. (4.8)

It appears to follow that there does not exist another number which is furtherfrom a simple rational approximation than the golden ratio. It is for thisreason that some consider τ to be the ‘most irrational’ number. However,

26

Page 29: Fibonacci

by definition, a number is either rational or irrational and cannot be moreor less so.

In Chapter 3, we found that the Euclidean algorithm for Fn+1 and Fnproduces the equations:

Fn+1 = 1 · Fn + Fn−1

Fn+ = 1 · Fn−1 + Fn−2

...F4 = 1 · F3 + F2 = 1 · 2 + 1 = 3F3 = 1 · F2 + F1 = 1 · 1 + 1 = 2F2 = 1 · F1 + F0 = 1 · 1 + 0 = 1.

So, we may writeFn+1

Fn= [1; 1, . . . , 1]

where there are n partial denominators all equal to 1. Hence, a continuedfraction representation of the ratio of consecutive Fibonacci numbers is

Fn+1

Fn= [1; 1, . . . , 1]︸ ︷︷ ︸

n digits

. (4.9)

Note that this is not a unique representation and we may also write

Fn+1

Fn= [1; 1, . . . , 1, 2]︸ ︷︷ ︸

n−1 digits

. (4.10)

4.4 Q-polynomials

The Q-polynomials are intimately related to continued fractions. We need tointroduce Q-polynomials and develop some of their properties to further ourinvestigation into continued fractions and their connections with Fibonaccinumbers. This section is based on section 4.5.3 of [DK].

Definition 4.4.1. The polynomials Qn(x1, . . . , xn) of n variables, for n ≥ 0,are called ‘continuants’, and are defined by the rule

Qn(x1, . . . , xn) =

1, if n = 0x1, if n = 1x1Qn−1(x2, . . . , xn) +Qn−2(x3, . . . , xn), if n > 1.

(4.11)

27

Page 30: Fibonacci

Consider ab where a < b. The following lemma shows that Q-polynomials

are connected to such fractions. As a0 = 0, for simplicity let us define

[x1, . . . , xn] =1

x1 +1

x2 +1...

xn−1 +1xn

. (4.12)

Note that if n = 0 then [x1, . . . , xn] is taken to be 0.

Lemma 4.4.1. The basic property of the Q-polynomials is that

[x1, . . . , xn] =Qn−1(x2, . . . , xn)Qn(x1, . . . , xn)

, n ≥ 1. (4.13)

Proof. For n = 1,

[x1, . . . , xn] =1x1

=Qn−1(x2, . . . , xn)Qn(x1, . . . , xn)

.

Assuming that the statement is true for n = k,

x0 + [x1, . . . , xk] = x0 +Qk−1(x2, . . . , xk)Qk(x1, . . . , xk)

by (4.11)

=Qk−1(x2, . . . , xk) + x0Qk(x1, . . . , xk)

Qk(x1, . . . , xk)

=Qk+1(x0, . . . , xk)Qk(x1, . . . , xk)

.

Thus

[x0, . . . , xk] =1

x0 + [x1, . . . , xk]

=1

Qk+1(x0, . . . , xk)Qk(x1, . . . , xk)

=Qk(x1, . . . , xk)Qk+1(x0, . . . , xk)

which, by renumbering the indices of the x values, gives us

[x1, . . . , xk+1] =Qk(x2, . . . , xk+1)Qk+1(x1, . . . , xk+1)

.

Therefore, by induction, the initial statement (4.13) is true.

28

Page 31: Fibonacci

According to Knuth [DK], Euler made the following surprising observa-tion which curiously connects Q-polynomials with Fibonacci numbers:

In general, Qn(x1, . . . , xn) is the sum of all terms obtainable bystarting with x1x2 . . . xn and deleting zero or more non-overlappingpairs of consecutive variables xjxj+1; there are Fn+1 such terms.

The following lemma follows from this observation made by Euler:

Lemma 4.4.2. The Q-polynomials are symmetrical:

Qn(x1, . . . , xn) = Qn(xn, . . . , x1). (4.14)

By combining Definition 4.4.1 and Lemma 4.4.2 we obtain the followinglemma:

Lemma 4.4.3.

Qn(x1, . . . , xn) = xnQn−1(x1, . . . , xn−1) +Qn−2(x1, . . . , xn−2), n > 1.(4.15)

Let qn = Qn(x1, . . . , xn) and pn = Qn−1(x2, . . . , xn). By combiningDefinition 4.4.1 and Lemma 4.4.3 we can express qn and pn as follows:

Definition 4.4.2.

qn =

1, if n = 0x1, if n = 1xnqn−1 + qn−2, if n > 1.

Definition 4.4.3.

pn =

0, if n = 01, if n = 1xnpn−1 + pn−2, if n > 1.

The Q-polynomials also satisfy the following important lemma:

Lemma 4.4.4.

Qn(x1, . . . , xn)Qn(x2, . . . , xn+1)−Qn+1(x1, . . . , xn+1)Qn−1(x2, . . . , xn) = (−1)n, n ≥ 1.(4.16)

Proof. Using Definitions 4.4.2 and 4.4.3, we can express the statement as

qnpn+1 − qn+1pn = (−1)n, n ≥ 1.

The statement is true for n=1 as

q1p2 − q2p1 = x1x2 − (x2x1 + 1) = −1 = (−1)1.

29

Page 32: Fibonacci

Assuming that the statement is true for n = k, we consider the case wheren = k + 1:

qk+1p(k+1)+1 − q(k+1)+1pk+1

= qk+1pk+2 − qk+2pk+1

= qk+1(xk+2pk+1 + pk)− (xk+2qk+1 + qk)pk+1

= qk+1pk − qkpk+1

= −(qkpk+1 − qk+1pk)

= −(−1)k

= (−1)k+1.

Therefore, by induction, the initial statement (4.16) is true.

4.5 Lame’s Theorem

An interesting theorem relating to Fibonacci numbers and the Euclideanalgorithm was proved by Lame around 1845. The execution time of theEuclidean algorithm (3.6) depends on the number of times A2 is performed.When applying the Euclidean algorithm to two integers one may observethat the worst case for the number of steps required is when the each quotientak is 1. From (4.9) we can see that this is when the inputs are consecutiveFibonacci numbers. The following theorem and proof are taken from section4.5.3 of [DK]. Knuth [DK] states that:

This theorem has the historical claim of being the first practicalapplication of the Fibonacci sequence; since then many otherapplications of Fibonacci numbers to algorithms and to the studyof algorithms have been discovered.

Theorem 4.5.1. For n ≥ 1, let u and v be integers with u > v > 0 suchthat Euclid’s algorithm applied to u and v requires exactly n division steps,and such that u is as small as possible satisfying these conditions. Thenu = Fn+2 and v = Fn+1.

Proof. We found previously that the positive partial denominators a0, . . . , anof a simple continued fraction are the same as the quotients that occur whenthe Euclidean algorithm (3.6) is applied to u and v. We also found thatconsidering only cases with an ≥ 2 and n ≥ 1 is the same as consideringall rational numbers as the case where an = 1 is not unique. Note that asu > v > 0 then for v

u we find a0 = 0. If the partial quotients obtained whenusing the Euclidean algorithm are a1, a2, . . . , an, then we have, by (4.13),

v

u= [a1, a2, . . . , an] =

Qn−1(a2, . . . , an)Qn(a1, a2, . . . , an)

. (4.17)

30

Page 33: Fibonacci

By (4.16), Qn−1(a2, . . . , an) and Qn(a1, a2, . . . , an) are relatively prime. Thistells us that and the fraction on the right-hand side of (4.17) is in lowestterms. Therefore

u = Qn(a1, a2, . . . , an)d and v = Qn−1(a2, . . . , an)d, (4.18)

where d = (u, v). By (4.18), we must have u = Qn(a1, . . . , an)d, wherea1, a2, . . . , an, and d are positive integers and an ≥ 2. Since, by Defini-tion 4.4.1, Qn(a1, . . . , an) is a polynomial with positive coefficients, involv-ing all of the variables, the minimum value is achieved only when a1 =1, . . . , an−1 = 1, an = 2, d = 1. Hence, u is as small as possible whenu = Qn(1, . . . , 1, 2︸ ︷︷ ︸

n digits

). By Definition 4.4.1,

Qn(1, . . . , 1, 2︸ ︷︷ ︸n digits

) =

2, if n = 13, if n = 2Qn−1(1, . . . , 1, 2︸ ︷︷ ︸

n−1 digits

) +Qn−2(1, . . . , 1, 2︸ ︷︷ ︸n−2 digits

), if n > 2.

So, by letting u = un, we find

u = un =

2, if n = 13, if n = 2un−1 + un−2, if n > 2.

By comparing this result to the Fibonacci recurrence relation (1.1) one cansee that u = Fn+2. When u is as small as possible,

v = Qn−1(1, . . . , 1, 2︸ ︷︷ ︸n−1 digits

) = un−1 = F(n−1)+2.

Therefore, for n ≥ 1, u = Fn+2 and v = Fn+1.

The following corollary, takem from [DK], follows from Theorem 4.5.1.

Corollary 4.5.1. If 0 ≤ u, v < N , the number of division steps requiredwhen the modern euclidean algorithm is applied to u and v is at most(logτ (

√5N)

)− 2.

Proof. By Lame’s Theorem (Theorem 4.5.1), the maximum number of steps,n, occurs when u = Fn and v = Fn+1, where n is as large as possible withFn+1 < N . Recall Fn = 1√

5(τn − σn) but σn√

5is always small enough, since

n ≥ 0, so that we have Fn = τn√5

rounded to the nearest integer. Since

Fn+1 < N , we have σn+1√

5< N , and hence n + 1 < logτ (

√5N). Thus

n < logτ (√

5N)− 1 or n ≤ logτ (√

5N)− 2.

31

Page 34: Fibonacci

The following corollary, taken from [TK], is a corollary of Theorem 4.5.1which is also sometimes referred to as Lame’s Theorem.

Corollary 4.5.2. The number of divisions needed to compute (u, v) by theEuclidean algorithm is no more than five times the number of decimal digitsin v, where u ≥ v ≥ 2.

One can see that this is true as follows:By Theorem 4.5.1, Fn+1 ≤ v < N . Using log rules observe that:

logτ (√

5v) =log10(

√5v)

log10(τ)

=log10

√5 + log10 v

log10(τ)

≈ 4.785 ·(

log10(√

5) + log10(v))

≈ 4.785 · log10 v + 1.672.

Now consider Corollary 4.5.1 where Fn+1 ≤ v < N . Then n ≤ logτ (√

5v)−1 < 5 · log10 v + 1. Define d as the number of decimal digits in v. Observed− 1 ≤ log10 v < d and thus n < 5d+ 1 or n ≤ 5d.

A full proof of the bound using the Euclidean algorithm can be found in[TK].

4.6 Pell’s Equation

Pell’s equation, named after John Pell (1611-1685), is a Diophantine equa-tion of the form x2 −Dy2 = N , where x, y, and N are integers and D > 0is a non-square natural number. The most common case is when N = ±1.

Recall Definition 4.3.1 and note that

[x0;x1, . . . , xn] = x0 + [x1, . . . , xn].

We previously observed that pnqn

= [x1, . . . , xn]. So, by letting

rn = x0qn + pn,

we obtainrnqn

= x0 +pnqn

= [x0;x1, . . . , xn].

Clearly, by substitution and induction, we can define rn as follows:

Definition 4.6.1.

rn =

x0, if n = 0x0x1 + 1, if n = 1xnrn−1 + rn−2, if n > 1.

32

Page 35: Fibonacci

By combining Lemma 4.4.4 and Definition 4.6.1 we obtain the followingLemma.

Lemma 4.6.1.rnqn−1 − rn−1qn = (−1)n−1. (4.19)

Proof. Recall rn = x0qn + pn. By (4.16),

qn−1pn − qnpn−1 = (−1)n−1.

Hence

(−1)n−1 = x0qnqn−1 + qn−1pn − x0qnqn−1 − qnpn−1

= (x0qn + pn)qn−1 − (x0qn−1 + pn−1)qn= rnqn−1 − rn−1qn.

The following theorems and lemma have been taken from Burton [DB].The proofs have been omitted in the interest of brevity and may be foundin Chapter 13 of [DB].

Theorem 4.6.1. The kth convergent of the simple continued fraction[x0;x1, . . . , xn] has the value rk

qk(0 ≤ k ≤ n).

Theorem 4.6.2. If rq is a convergent of the continued fraction expansion

of√D, then x = r, y = q is a solution of one of the equations

x2 −Dy2 = c,

where |k| < 1 + 2√D.

Lemma 4.6.2. Let the convergents of the continued fraction expansion of√D be rk

qk. If m is length of the period of expansion of

√D, then

r2km−1 −Dq2km−1 = (−1)km, (k = 1, 2, 3, . . . ).

Lemma 4.6.2 relies on the knowledge that the continued fraction of aquadratic surd always becomes periodic at some term xk+1, where xk+1 =2x0, so √

D = [x0;x1, . . . , xk, 2x0].

It follows from these theorems and this lemma that one can find solutionsto equations of the form

x2 −Dy2 = ±1

using the convergent rkqk

and the continued fraction expansion of√D. There-

fore one can solve Pell’s equation by finding an appropriate continued frac-tion.

33

Page 36: Fibonacci

Example 4.6.1. √5 = [2; 4]

so m = 1 and, by Lemma 4.6.2, x = rk, y = qk, k = 1, 2, 3, . . . , are solutionsof

x2 − 5y2 = ±1.

We can see this is true for the first few convergents:

9/4 : 92 − 5(42) = 1,

38/17 : 382 − 5(172) = −1,

161/72 : 1612 − 5(722) = 1.

Special case

There is a relationship between Fibonacci numbers and a special case ofPell’s equation where D = 5 and N = ±4. The following theorems aretaken from Chapter 27 of [TK].

Theorem 4.6.3. The Pell’s equation x2 − 5y2 = 4 is solvable in positiveintegers if and only if x = L2n and y = F2n, where n ≥ 1.

Theorem 4.6.4. The Pell’s equation x2 − 5y2 = −4 is solvable in positiveintegers if and only if x = L2n−1 and y = F2n−1, where n ≥ 1.

Using substitution and Binet’s formula one can see that x = Ln, y = Fnis a solution of x2 − 5y2 = ±4 for all n ≥ 1:

x2 − 5y2 = (τn + σn)2 − 5(

1√5

(τn − σn))2

= τ2n + 2τnσn + σ2n − (τ2n − 2τnσn + σ2n)= 4(τσ)n

= 4(−1)n

=

{4 if n = even,−4 if n = odd.

It can be proved using number theory that the Fibonacci and Lucasnumbers are the only integer solutions to these equation.

34

Page 37: Fibonacci

Chapter 5

Golden Geometry

In this chapter we explore some of the two-dimensional geometry related tothe golden ratio. The chapter is based on Chapter 13 of [SV] and Chapter3 of [RD].

5.1 The Golden Rectangle

Definition 5.1.1. A golden rectangle is a rectangle where the lengths of theadjacent sides are in the golden ratio.

Constructing a Golden Rectangle

One can construct a golden rectangle as follows:

1. Construct a square ABCD;

2. Bisect AB and label the midpoint E;

3. Use a compass to draw an arc radius EC with centre E from C tobaseline level;

4. Extend the baseline to meet the arc and label the intersection F ;

5. Extend the top line of the square, draw a line perpendicular to thebaseline at F , and label the intersection G.

35

Page 38: Fibonacci

Figure 5.1: Construction of a golden rectangle[RD]

The rectangle AFGD is a golden rectangle as AF : AD = τ : 1.

Proof. From the first three steps we know AB = BC = CD = AD; AE =EB = 1

2AB; and EF = EC. So, using Pythagoras theorem, we can see that

EC2 = EB2 +BC2

=(

12AB

)2

+AB2

=14AB2 +AB2

=54AB2

Therefore, EC =√

52 AB. Hence, EF =

√5

2 AD, AF = 12(1+

√5)AD = τAD,

and BF = −12(1−

√5)AD = −σAD = 1

τAD.

Inflation

Consider the golden rectangle ABCD in Figure 5.2. Divide the rectangleinto a square whose sides have length AD and a smaller golden rectangle.Repeatedly dividing the smaller rectangle into another square and rectanglefinds smaller golden rectangles. The successively smaller golden rectanglesconverge to O. Note that, by Definition 4.2.1,

AD

EB=AE

EB= τ

36

Page 39: Fibonacci

and

(AB)(AD) : (BC)(EB) = (AB)(AD) : (AD)(EB)= AB : EB

=AB

AD:EB

AD

= τ :1τ

= τ2 : 1.

Clearly, the inflation of one golden rectangle to the smaller golden rectanglemeans a reduction in the linear dimensions of the rectangle by a factor of τand a reduction in the area by a factor of τ2.

Figure 5.2: Inflation of the golden rectangle

The golden rectangle is the only rectangle with the property that takinga square from it leaves a similar rectangle. The diagonals of alternatinggolden rectangles in the inflation process are perpendicular, always either inline with the diagonal of the original rectangle or the diagonal of the smallerrectangle. Dunlap [RD] states that the point O where the diagonals of thegolden rectangles meet was named the Eye of God by the mathematicianClifford A Pickover.

5.2 The Golden Triangle

Definition 5.2.1. The golden triangle is an isosceles triangle such that thelengths of the equal sides and the base are in the golden ratio.

Inflation

Consider the golden triangle ACD in Figure 5.3. By Definition 5.2.1, ADCD =

τ . Divide AD such that Ab and bD are in the golden ratio, i.e. ADAb = Ab

bD = τ.So,

AD

bD= τ2.

37

Page 40: Fibonacci

HenceCD

bD=(CD

AD

)(AD

bD

)=(

)(τ2) = τ.

Therefore, CbD is clearly similar to ACD and therefore is also a goldentriangle.

Figure 5.3: Inflation of the golden triangle[SV]

The procedure can be repeated to find smaller golden triangles, i.e. Dba,bap, and so on. The successively smaller, similar triangles converge to O.Note that

AC : CD = AD : CD = τ

and12

(AC)(CD) :12

(CD)(bD) = AC : bD = AD : bD = τ2.

Clearly, the inflation of a golden triangle to a smaller golden triangle meansa reduction in the linear dimensions of the golden triangle by a factor of τand a reduction in the area of the golden triangle by a factor of τ2.

In Figure 5.3, Vajda [SV] shows that the median from D in the triangleADC, the median from b in triangle CbD, and the median from a in thetriangle Dab, and so on, all meet at O.

Note that the shape AbC that joins onto the smaller triangle to completethe larger triangle is called the golden gnomon and has the same inflationrelationship as the golden triangle.

38

Page 41: Fibonacci

Trigonometric Formula for Fn

In Figure 5.3, ∠DAC is 36◦, which is π5 radians. This angle is closely related

to τ as the angles involved in a regular pentagon and in the golden triangleand gnomon are all multiples of π

5 . In addition, we find that cos π5 = τ2 . The

following proof is based on Chapter 25 of [TK].

Proof. Let θ = π10 . Then 2θ + 3θ = π

2 , so

sin(2θ) = sin(π

2− 3θ

)= − cos

((π2− 3θ

)+π

2

)= cos(3θ).

Note that cos(θ) 6= 0. Hence

sin(2θ) = cos(3θ)

⇒ 2 sin(θ) cos(θ) = 4 cos3(θ)− 3 cos(θ)

⇒ 4 cos2(θ)− 2 sin(θ)− 3 = 0

⇒ 4 sin2(θ) + 2 sin(θ)− 1 = 0

⇒ sin(θ) =−1±

√5

4.

Since sin(θ) > 0, it follows that

sin(θ) = −σ2

=1

2τ.

Therefore

cos(π

5

)= 1−2 sin2

( π10

)= 1−2

(σ2

4

)=

2− σ2

2=

2− (1 + σ)2

=1− σ

2=τ

2.

Using the formulae (1.6) in chapter 1, we find

cos(

3π5

)= 4 cos3

(π5

)− 3 cos

(π5

)= 4

(τ2

)3− 3

(τ2

)=τ(τ2 − 3)

2=τ(τ − 2)

2=

1− τ2

2.

Using these results, a trigonometric formula for Fn can be found:

Fn =1√5

(τn − σn) =2n(cosn

(π5

)− cosn

(3π5

))√

5. (5.1)

39

Page 42: Fibonacci

5.3 Logarithmic Spirals

Definition 5.3.1. Logarithmic spirals are given by the polar equation:

r = aebθ, −∞ < θ <∞, (5.2)

where a and b are arbitrary positive constants.

Logarithmic spirals are also known as equiangular spirals because eachray from the origin cuts the spiral in a constant angle α, where b = cotα.

Loeb [AL] notes that a logarithmic spiral can be drawn through suc-cessive dynamically related points. By connecting the opposite vertices ofthe squares in a progression of golden rectangles (DEFGHIJ . . . in Figure5.2) one can indeed construct a logarithmic spiral. This spiral is sometimescalled the golden spiral.

From any point on the golden spiral to another point on the spiral,lengths are affected by a scaling factor of τ for every rotation of 90◦, whichis π

2 radians. As Knott [RK] notes, it follows that the polar coordinateschange from (r, θ) to (rτ, θ + π

2 ). Therefore, by setting one point as (1, 0),the points on the spiral are summarised by:

r = τn and θ = nτ

2.

Hence we obtain the single equation

r = τ2θπ

which is a logarithmic spiral where a = 1 and b = 2π log τ .

By converting into Cartesian coordinates (x = r cos θ, y = r sin θ), Ihave adapted Knott’s excel spreadsheet [RK] in the section called ‘Two-dimensional Geometry and the Golden section’ to produce a plot of thisspiral where −14.5 ≤ θ ≤ 6.25 (Figure 5.4).

Figure 5.4: Spiral with an expansion of τ in 90◦ turn

40

Page 43: Fibonacci

Similarly, by connecting the acute vertices of the golden triangles in aprogression of inflated triangles (ACDbap . . . in Figure 5.3) one can con-struct a logarithmic spiral as in Figure 5.5. Loeb [AL] states that for thegolden triangle, every rotation of 108◦, which is 3π

5 radians, is accompaniedby a scaling factor of τ . So, we obtain the equation

r = τ5θ3π

which is a logarithmic spiral where a = 1 and b = 53π log τ (Figure 5.6).

Figure 5.5: Logarithmic spiralproper to the golden triangle

[RD]

Figure 5.6: Spiral with an expan-sion of τ in 108◦ turn

One may observe in a similar way that a spiral where every full turn(rotation of 360◦ or 2π radians) is accompanied by a scaling factor of τ hasthe equation

r = τθ2π

which is a logarithmic spiral where a = 1 and b = 12π log τ (Figure 5.7).

Knott [RK] calls this spiral the Phi Spiral.

Figure 5.7: Spiral with an expansion of τ in 360◦ turn

41

Page 44: Fibonacci

Approximations of the golden spiral

By connecting the opposite vertices of the squares in a progression of goldenrectangles (DEFGHIJ . . . in Figure 5.2) with circular arcs one can con-struct a spiral as in Figure 5.8 which is a very close approximation to thegolden spiral (Figure 5.4). This spiral is not a logarithmic spiral as the angleα at which a ray cuts the spiral is not constant.

Figure 5.8: Spiral constructed from a sequence of golden rectangles[RD]

A Fibonacci spiral is created by connecting the opposite vertices of Fi-bonacci squares as in Figure 5.9. The spiral is not a logarithmic spiral as, forevery 90◦ turn, a Fibonacci spiral does not expand by a constant factor butby a changing factor related to the ratios of consecutive Fibonacci terms.However, since the limit of ratios of consecutive Fibonacci terms is τ , theFibonacci spiral is a good approximation to the golden spiral.

Figure 5.9: Fibonacci Spiral

42

Page 45: Fibonacci

5.4 The Pentagon

Constructing a Pentagon

The construction of a regular pentagon relies upon the construction of twoline segments in the golden ratio. The following method is based on Ap-pendix 1 of [RD]:

1. Construct a square ABCD;

2. Bisect AB and label the midpoint E;

3. Use a compass to draw arc 1 as in Figure 5.10 with radius EC andcentre E;

4. Extend the baseline to meet arc 1 and label the intersection F ;

5. Use a compass to draw an arc 2 with radius AB and centre A;

6. Use a compass to draw an arc 3 with radius AB and centre F andlabel the intersection of arcs 2 and 3 as G;

7. Use a compass to draw an arc 4 with radius AF and centre G, labelthe intersection of arcs 2 and 4 as H, and label the intersection of arcs3 and 4 as I;

8. Connect points AGFIH to form a regular pentagon.

Figure 5.10: Construction of a pentagon[RD]

43

Page 46: Fibonacci

5.5 The Pentagram

The diagonals of the pentagon form a pentagram. Vajda [SV] states thatthe diagonals and the sides of the pentagon are in the golden ratio.

Proof. Consider regular pentagon ABCDE in Figure 5.11. Note that allinterior angles in a regular pentagon are 3π

5 radians and all interior anglesin a triangle sum to π radians.

Figure 5.11: Pentagon[SV]

Clearly, ADE ∼= CAB and Ad = dB. Therefore

∠AED =3π5⇒ ∠EAD =

π

5⇒ ∠AdB =

3π5.

Hence

∠cAd = ∠EAB − ∠EAc− ∠dAB

= ∠EAB − ∠EAD − ∠CAB

=3π5− 2

(π5

)=π

5.

DAC is an isosceles triangle, so

∠ADC = ∠ACD =2π5.

By five-fold symmetry, bCD ∼= eDC and ∠bCd = ∠cAd. So, ∠bCD = π5

and bCD and eDC are isosceles triangles similar to ADC. Noting similartriangles,

AC : DC = DC : eC.

44

Page 47: Fibonacci

Letting ABCDE be a unit pentagon

AC : 1 = 1 : (AC − 1)

so AC = τ and eC = τ−1 = 1τ . Therefore, in general, AC : DC = τ : 1.

However, further analysis shows even more golden ratios in Figure 5.11.Note that the isosceles triangles mentioned are golden triangles. Therefore,by five-fold symmetry, it is clear that each diagonal is crossed by two otherdiagonals and that these crossing points mark the golden section of the line.In fact, ADC is a golden triangle with CbD and CeD as its smaller goldentriangles and aDb and aCe as their smaller golden triangles. It is thenobvious that all of the segments in the pentagram are related by the goldenratio.

Figure 5.12: Incommensurability

By repeating this process for the inner pentagon, we can see that eachcoloured line in Figure 5.12 is in the golden ratio with the ones sharing avertex with it. By repeating this process ad infinitum we have a geometricproof that the golden ratio is incommensurable.

45

Page 48: Fibonacci

Chapter 6

Fibonacci Numbers inArchitecture

There are appears to be two kinds of so-called ‘Golden’ architecture to con-sider when looking at Fibonacci numbers and the golden ratio in architec-ture. Firstly, there is architecture with an alleged link to Fibonacci numbersand the golden ratio. Secondly, there is modern architecture that was de-signed using Fibonacci numbers and the golden ratio. In this chapter, severalclaims of ‘Golden’ architecture will be investigated in chronological order ofbuild, looking at the mathematics behind the claims.

6.1 Ancient Egypt

The Great Pyramid

The golden ratio allegedly appears in the ratio of dimensions of the GreatPyramid of Giza, which was built around 2500BC. According to Livio [ML],the lengths of sides of the base of the Great Pyramid vary from 755.43 feetto 756.08 feet. Livio states that the average of the lengths is 755.79 feet andthe height of the pyramid is 481.4 feet. In Figure 6.2, h represents height,b is half the base, and a is the slant height of the Great Pyramid. UsingPythagoras’ theorem, we find that a = 612.05 feet. This gives a ratio of

a

b=

612.05377.90

= 1.62,

which differs from the golden ratio by approximately 0.1 percent. Note thatan approximation to τ is the best that an architect can achieve since τ is anirrational number. Therefore the golden ratio does seem to appear in theratio of dimensions of the Great Pyramid. The question that comes to mindis whether this is just a coincidence.

There is no original documentary evidence to support any claims thatthe golden ratio was used in any of the Egyptian’s designs. Several texts

46

Page 49: Fibonacci

Figure 6.1: Great Pyramid of Giza[NT]

Figure 6.2: Pyramid structure[TK]

use Herodotus’ statement as supporting historical documentation of the in-tentions of the designers to claim that the golden ratio is present in theGreat Pyramid by design even though Herodotus lived around 485-425 BC.According to Markowsky [GM], David Burton, author of one such book,writes:

Herodotus related in one passage that the Egyptian priests toldhim that the dimensions of the Great Pyramid were so chosenthat the area of a square whose side was the height of the greatpyramid equaled the area of a face triangle.

If the passage is true then, assuming Herodotus had reliable primarysources, Herodotus’ statement may be used to support the claim as thepassage implies that the ratio of the slant height of a face to half the lengthof the base is the golden ratio.

Proof. In Figure 6.2, if the area of a square whose side is equal to h is equalto the area of a triangular face, then h2 = ab. Using Pythagoras’ theorem,we have a2 = h2 + b2. So, by substitution, we obtain a2 = ab+ b2. Dividingthrough by b2 we get (a

b

)2=(ab

)+ 1.

Let x =(ab

). We obtain the quadratic equation

x2 − x− 1 = 0,

which we previously discovered has the golden ratio as its only positive root.Therefore a : b = τ : 1.

Livio and Markowsky [ML, GM] both note this proof and challenge theclaim by looking at translations of Herodotus’s original text. Livio [ML]states that the translation reads as follows:

It is a square, eight hundred feet each way, and the height thesame.

47

Page 50: Fibonacci

This passage does not imply that the ratio of the slant height of a face to halfthe length of the base is the golden ratio. Furthermore, the Great Pyramid isnowhere near 800 feet high and the side 2b is significantly less than 800 feet.Therefore, Herodotus’ statement cannot be used as supporting historicaldocumentation of the intentions of the designers. Hence, I conclude thatthere is no evidence to suggest that the golden ratio is present in the GreatPyramid by design.

There is no original documentary evidence that the Egyptians knewabout the golden ratio. However, one may still ask whether the Egyptianscould have known about the golden ratio and constructed the Pyramid usingit. The Egyptians could have constructed the golden ratio using the follow-ing basic method using the basic tools they had but there is no evidencethat they knew about this method.

Constructing the Golden section

Figure 6.3: Constructing the golden section[RK]

One may construct the golden section as in Figure 6.3 in the followingmanner using only a marker and a compass or some string:

1. Draw a line of length 12AB at right angles to AB vertically from B to

a new point T;

2. Draw a line from A to T;

3. Draw an arc with centre T from B to AT and mark the intersectionV;

4. Draw an arc with centre A from V to AB and mark the intersectionG.

The point G divides the original line AB into the golden section, where theratio of AB to AG is the golden ratio.

In the case of the pyramid, it is the ratio of the slant height and halfthe base that is in the golden ratio so they would have had to combine themethod above with Pythagoras’ theorem to work out how high to build the

48

Page 51: Fibonacci

pyramid or with trigonometry to work out the angle at which to slant togive a right-angled triangle. However, the Egyptians did not know aboutPythagoras’ theorem or trigonometry. The Egyptians may have been ableto construct the golden ratio with this method but it does not necessarilyfollow that they were therefore able to build the Great Pyramid with thismethod.

With this method the Egyptians may have been able to construct thegolden section but this does not mean that they knew the value of τ . Todetermine whether they could have known the value of τ , one has to lookat Egyptian mathematics.

Egyptian Fractions

Most of what we know about Egyptian mathematics is taken from the RhindPapyrus. The papyrus tells us that the Egyptians used integers and unitfractions and were familiar with addition, subtraction, and geometric andarithmetic progression.

The Egyptians are likely to only have discovered the value of τ if theyhad discovered the limit of the sequence of ratios of consecutive Fibonaccinumbers, in which case they would have to know about the Fibonacci num-bers. Rossi and Tout [RT] note that there is no evidence that the Egyptiansknew about the Fibonacci numbers but conclude that, considering how fa-miliar the ancient Egyptian scribes were with numerical series, it is notinconceivable for the Egyptians to have discovered the Fibonacci series. As-suming they knew about the Fibonacci numbers, one then wonders whetherthey could have, firstly, found the ratios of consecutive numbers and, sec-ondly, discovered the limit τ . To attempt to answer this, one has to look atEgyptian fractions.

The Egyptians appear to have used a number system based on unitfractions (fractions with one in the numerator). The only exception was 2

3 .An Egyptian fraction is a fraction written as a sum of distinct unit fractions.

Example 6.1.1. Example from the Rhind Papyrus [FG]:

261

=140

+1

244+

1488

+1

610.

49

Page 52: Fibonacci

The first few ratios of consecutive Fibonacci numbers can be expressedusing Egyptian fractions as follows:

12

=12

23

=12

+16

35

=12

+110

58

=12

+110

+140

813

=12

+110

+165

1321

=12

+110

+165

+1

273

Rossi and Tout [RT] noted the following pattern. Starting with 12 , by

adding a unit fraction whose denominator is a product of discontinuous pairsof terms we obtain 1

2 ,35 ,

813 , . . . , and the intermediate values can be calculated

by adding to the previous ratio a unit fraction whose denominator is givenby the multiplication of the two terms of the ratio. This can be written inmathematical terms as follows:

Theorem 6.1.1. The ratios of consecutive Fibonacci numbers can be ex-pressed as:

FnFn+1

=

n2∑i=1

1F2i−1F2i+1

if n is even

n−12∑i=1

1F2i−1F2i+1

+1

FnFn+1if n is odd

where n > 1.

Proof. Firstly, consider when n is even. Let n = 2k; k ∈ Z+. UsingCassini’s identity,

F2k+1F2k−1 − F 22k = (−1)2k

⇒ F2k−1

F2k− F2k

F2k+1=

(−1)2k

F2kF2k+1

⇒ F2k

F2k+1=F2k−1

F2k− (−1)2k

F2kF2k+1.

HenceF2k

F2k+1=F2k−1

F2k− 1F2kF2k+1

. (6.1)

50

Page 53: Fibonacci

Now consider when n is odd. Let n = 2k + 1, k ∈ Z+. Similarly, byusing Catalan’s identity, we find

F2k+1

F2k+2=

F2k

F2k+1+

1F2k+1F2k+2

. (6.2)

Assuming Theorem 6.1.1 is true for n = 2k,

F2k+1

F2k+2=

k∑i=1

1F2i−1F2i+1

+1

F2k+1F2k+2. (6.3)

Therefore if the statement is true for n is even then it is true for n is odd.Consider the case where n = 2k + 2. Using (6.1) and (6.2) We find:

F2k+2

F2k+3=F2k+1

F2k+2− 1F2k+2F2k+3

=F2k

F2k+1+

1F2k+1F2k+2

− 1F2k+2F2k+3

=F2k

F2k+1+

F2k+3 − F2k+1

F2k+1F2k+2F2k+3

=F2k

F2k+1+

1F2k+1F2k+3

by (1.1).

Assuming Theorem 6.1.1 is true for n = 2k,

F2k+2

F2k+3=

k∑i=1

1F2i−1F2i+1

+1

F2k+1F2k+3.

HenceF2(k+1)

F2(k+1)+1=

k+1∑i=1

1F2i−1F2i+1

.

The statement is true for n = 2 asF2

F3=

12

=1

F1F3.

So, by induction, the statement is true for n is even. Therefore, by (6.3),the statement is true for all integers n > 1.

By Theorem 6.1.1, all the ratios of consecutive Fibonacci numbers canbe expressed in Egyptian fractions. However, even if the Egyptians diddiscover the pattern, it is not obvious from this pattern that the sequenceof ratios will converge. There is also no evidence that the Egyptians knewabout limits or irrational numbers. So, it is unlikely that they would havefound the limit τ . Therefore, I conclude it is unlikely that the Egyptiansknew about the value of τ .

I do not believe that the Egyptians knew about the golden ratio or usedit intentionally in their designs. I agree with G. S. Toomer’s comment [FG]that:

51

Page 54: Fibonacci

The truth is that Egyptian mathematics remained at much toolow a level to be able to contribute anything of value. The sheerdifficulties of calculation with such a crude numerical system andprimitive methods effectively prevented any advance or interestin developing science for its own sake.

6.2 Ancient Greece

The Parthenon

It is said that the golden ratio was used in the building of the Parthenon,built at Athens in the 5th century BC. Several texts claim that the originaldimensions of the Parthenon fitted almost precisely into a golden rectangle.To support this claim a figure similar to Figure 6.5 is often included. Huntley[HH] writes:

While its triangular pediment was still intact, its dimensionscould be fitted almost exactly into a Golden Rectangle,. . . . Itstands therefore as another example of the aesthetic value of thisparticular shape.

Markowsky [GM] points out that parts of the Parthenon actually fall outsidethe golden rectangle.

Figure 6.4: The Parthenon Figure 6.5: Parthenon claims of τ[RK]

Two main weaknesses, as noted by Livio [ML], of claims about the pres-ence of the golden ratio in architecture on the basis of dimensions aloneare:

1. They may involve number juggling;

2. They may overlook inaccuracies in measurements.

52

Page 55: Fibonacci

These weaknesses are apparent when investigating the claim that manydimensions of the Parthenon are related to the golden ratio as in Figure 6.5.Markowsky [GM] comments:

The dimensions of the Parthenon vary from source to sourceprobably because different authors are measuring between dif-ferent points. With so many numbers available a golden ratioenthusiast could choose whatever numbers gave the best result.

He goes on to give an example of quoted measurements which give ratios farfrom the golden ratio. One also has to note that the triangular pediment isnot intact and the base is slightly curved so the measures are all approximate,which leaves them open to inaccuracies and number juggling.

Huntley [HH] comments that “the proportions of the well-known Parthenonbear witness to the influence exerted by the golden rectangle on Greek ar-chitecture”. However, there is no original documentary evidence that theratio was used by the Greeks in Art or Architecture and we have noted thatdimensions alone are not sufficient evidence. Therefore, I do not agree withHuntley.

There is not enough evidence in the dimensions to prove or disprove theclaim previously mentioned. However, one may still look at the likeliness oftruth in the claim. One may question if the Greeks knew about the goldenratio and if they could, or would, have used it in their architecture.

Euclid

We know that the Greeks knew about the Golden ratio around 300 BC be-cause it first appeared in Euclid’s Elements around that time. The followingdefinitions and propositions are taken from [FG]. In Book V I, Euclid givesthe following definition.

Definition 6.2.1. A straight line is said to have been cut in extreme andmean ratio when, as the whole line is to the greater segment, so is the greaterto the less.

Euclid calls the golden ratio the extreme and mean ratio. In Book V I,Euclid also proposes a method of constructing the golden ratio.

Proposition 6.2.1 (Proposition 30). To cut a given finite straight line inextreme and mean ratio

Euclid begins the proposition by constructing a figure which he also usedin Proposition 11, Book II.

Proposition 6.2.2 (Proposition 11). To cut a given straight line so thatthe rectangle contained by the whole and one of the segments is equal to thesquare on the remaining segment.

53

Page 56: Fibonacci

Figure 6.6: Construction of Proposition 11[FG]

Euclid’s proof of Proposition 11 can be found in [FG].In Proposition 30, Euclid proves Proposition 11 slightly differently to his

original proof. He uses the result that the square AG is equal in area to therectangle BK in Figure 6.6 to prove that this construction cuts the line ABin the golden ratio. Euclid reasons that since AG and BK are equal in areaand equiangular, then the sides about the equal angles in these shapes arereciprocally proportional. Hence KH

HG = AHHB . As KH = AB and HG = AH

thenAB

AH=AH

HB, where AB > AH > HB.

Therefore, the straight line AB has been cut in the golden ratio at H.We have seen that Euclid knew about the golden ratio and knew a

method to construct it. The Greeks may have had the knowledge to useeither Euclid’s method or the method that the Egyptians may have usedto construct the golden ratio, so one may conclude that it is possible thatthe Greeks could have used the golden ratio in their architectural designs.However, Euclid’s coverage of the topic is brief and does not in any wayimply either that the golden ratio is of any significant importance or that itshould have an application in architecture.

The Parthenon was built before Euclid’s time and although Euclid’sElements collected existing mathematical knowledge there is no evidencethat this particular knowledge was known about or considered anythingmore than a mathematical definition and proposition when the Parthenonwas built. However, as they had the mathematical knowledge to deduce themethod of dividing a line using a compass and a string, even if they did

54

Page 57: Fibonacci

not about Euclid’s method in the 5th century BC it is still possible thatthe Greeks may have had the knowledge to construct the Parthenon withdimensions in the golden ratio. It is difficult to conclude whether or not theGreeks would have constructed the Parthenon using the ratio.

6.3 Medieval Islamic Architecture

Girih Tiling

In February 2007, the journal Science published an article [PL] claimingthat by AD 1200 Islamic girih line patterns were reconceived as decoratedtessellations using a set of five equilateral polygons, which the authors ofthe article call Girih tiles.

Figure 6.7: Complete set of girih tiles[PL]

The tiles are a decagon, a pentagon, a hexagon, a bowtie, and a rhombus(see Figure 6.7). Every edge of each polygon has the same length and twogirih lines intersect the midpoint of every edge at 72◦ and 108◦ angles. Girihlines are lines which decorate the tiles. The decoration may differ slightlyfrom pattern to pattern and not all the tiles are necessarily used in eachpattern. In most patterns, the girih and decoration are visible but theboundaries of the tiles are not usually shown.

Most girih tilings in Islamic architecture were periodic. There are noknown examples of girth tilings which when extended would be non-periodicover the entire plane. However, the article claims that a non-periodic girihtiling could be possible. It is this non-periodic girih tiling that is of interestto us. The article [PL] states that

. . . [an] innovation arising from the application of girih tiles wasthe use of self-similarity transformation (the subdivision of largegirih tiles into smaller ones) to create overlapping patterns at

55

Page 58: Fibonacci

two different length scales, in which each pattern is generated bythe same girih tile shapes.

It goes on to say that

a subdivision rule, combined with decagonal symmetry, is suffi-cient to construct perfect quasi-crystalline tilings – patterns withinfinite perfect quasi-periodic translational order and crystallo-graphically forbidden rotational symmetries, such as pentagonalor decagonal. . . .

Quasi-periodic order means that tile shapes repeat with incommensuratefrequencies. Quasi-periodic tiling may exhibit local periodicity under sometransformations but the entire tiling is non-periodic.

Penrose Tiling

Penrose tiling is an example of quasi-crystalline tiling named after RogerPenrose, who investigated the topic in the 1970s. Penrose tiles are an ape-riodic set of tiles which give only non-periodic tilings. One type of Penrosetiling is the Rhombus tiling made up of oblate and prolate rhombus tiles(Figure 6.8).

Figure 6.8: Rhombus tiling[RD]

A second type of Penrose tiling is the Kite and Dart tiling (Figure 6.9).

Figure 6.9: Kite and Dart tiling[RD]

56

Page 59: Fibonacci

Both the oblate rhombus and the dart can be divided into two goldengnomons (obtuse isosceles triangles with base to side ratio of τ : 1). Boththe prolate rhombus and the kite can be divided into two golden triangles(isosceles triangles with side to base ratio of τ : 1). The ratio of the numbersof the two different tile shapes in each type of tiling mentioned tends to thegolden ratio as N →∞, where N is the total number of tiles. Hence, bothtilings mentioned are intimately connected to the golden ratio.

Penrose tilings can be constructed by matching rules, where only certainedges can join together, or self-similar subdivisions. Penrose tilings can berescaled while still maintaining the correct ratio of tile shapes to producea Penrose tiling. The tilings can be repeatedly rescaled indefinitely. Thisform of subdivision follows from the fact that the tile shapes are composed ofgolden triangles and gnomons. Recall from Chapter 5 that golden trianglesand gnomons can be dissected into smaller triangles that are golden gnomonsand golden triangles.

The authors of the article [PL] discovered a possible link between Kiteand Dart Penrose tiling and some girih tiling in medieval Islamic architec-ture. This link interests us because Penrose tiling is closely related to thegolden ratio. Note that the potential relationship between girih tiling andthe golden ratio is due to the properties of the shapes chosen for tiling andthat the Islamic architects are unlikely to have known about this indirectrelationship.

Darb-i-Imam Shrine

The most relevant architectural example in the article [PL] is the Darb-iImam shrine in Isfahan in Iran, built in 1453. The tiling uses the decagon,hexagon, and bowtie tile shapes. Self-similar subdivision constructs an ar-bitrarily large Darb-i Imam pattern. The authors claim that both the largeand small girih tile patterns on the Darb-i Imam can be mapped completelyinto Penrose tiling (figure 6.11). Since Penrose tilings are closely related tothe golden ratio this implies that the Drab-i Imam patterns are related tothe golden ratio. Also, the ratio of hexagons to bowties tends to the goldenratio as N → ∞, where N is the number of tiles. Hence, the pattern isintimately connected to the golden ratio.

57

Page 60: Fibonacci

Figure 6.10: Spandrel from the Darb-i-Imam shrine[PL]

Figure 6.11: Tile mappings[PL]

58

Page 61: Fibonacci

The Darb-i Imam pattern shows us that the Islamic architects had all theelements needed to construct perfect quasi-crystalline patterns. However,there are problems with the Darb-i Imam pattern which shows that theydid not have a complete understanding of the elements when they made thepattern. The problems are:

1. There are 11 mismatches out of 3700 Penrose tiles;

2. There is no evidence that they developed the matching rule method;

3. A small arrangement of large tiles that does not appear in the subdi-vided pattern was used to start construction, instead of a single girihtile, and so the tiling is not strictly self-similar.

The mismatches are all of the form shown in Figure 6.12 which can beeasily corrected by rearranging a few tiles. This mistake would not have beenmade if they had combined the matching rule and subdivision methods.

Figure 6.12: Defect and correction[PL]

Further research needs to be done to see whether any perfect quasi-crystalline tilings exist, whether there are more patterns with a strong linkwith Penrose tilings and the golden ratio, and whether Penrose tiling was,in fact, discovered in the 13th century.

6.4 Le Corbusier

Charles-Edouard Jeanneret-Gris (1887-1965) was a Swiss-born architect. Headopted the name Le Corbusier in the early 1920s. He introduced a newsystem of proportioning called the Modulor which was based on humanmeasurements, Fibonacci numbers, and the golden ratio.

The perfect human body is considered in some texts to be associated tothe golden ratio. In Figure 6.13,

AE

CE≈ τ ≈ CE

AC

59

Page 62: Fibonacci

so the navel divides the height into the golden section. According to Koshy,Figure 6.13 is based on model male proportions.

Figure 6.13: Model male[TK]

Le Modulor

Le Corbusier used this supposed relationship between human measurementsand the golden ratio to develop two series of measurements based on a six-foot (1.83m) man. The series rouge

4, 6, 10, 16, 27, 43, 70, 113, 183, 296,. . .

is based on the height of the navel (1.13m), and the series bleue

13, 20, 33, 53, 86, 140, 226, 366, 592,. . .

is based on the height of the tip of the upraised fingers (2.26m). The redand blue series were created from 113 and 226 respectively by repeatedlydividing and multiplying by the golden ratio and rounding to integers in anattempt to make the series satisfy the generalised Fibonacci equation.

Example 6.4.1. Taking the height of the navel as 113cm and repeatedlydividing by the golden ratio gives

113τ

= 69.8378407 . . . ≈ 70,

113τ

τ= 43.1621592 . . . ≈ 43,

113ττ

τ= 26.6756814 . . . ≈ 27.

60

Page 63: Fibonacci

In Figure 6.14, Le Corbusier shows how consecutive series rouge numbers27, 43, 70, 113, and 183, and consecutive series bleue numbers 86, 140, and226, are supposedly connected to the human stature.

Figure 6.14: Le Modulor[KF]

The Modulor system was based on human proportions and the goldenratio because Le Corbusier wanted to create a proportion system that wasrelated to natural creation. According to Livio [ML], Le Corbusier proposedthat the Modulor could be used to give harmonious proportions to every-thing and provide the model for standardisation. The Modulor system was

Figure 6.15: Design for South-west facade of Governor’s Palace[KF]

put into practice in many of Le Corbusiers architectural designs and projects.For example, Le Corbusier designed the urban layout of Chandigarh in Indiausing the Modulor system. According to Livio [ML], Le Corbusier wrote:

In the general section of the building which involves providingshelter from the sun for the offices and courts, the Modulor will

61

Page 64: Fibonacci

bring textural unity in all places. In the design of the frontages,the Modulor (texturique) will apply its red and blue series withinthe spaces already furnished by the frames.

Le Corbusier designed The Governor’s Palace (Figure 6.15) in 1953 for theChandigarh project. According to Frampton [KF], Le Corbusier designedthe building using the Modulor and then “drastically reduced [it] in size bythe application of different Modulor dimensions in the following year whenit was found to be grossly overblown”.

6.5 Inspired by Nature

Logarithmic spirals, which we met previously in Chapter 5, are common innature. These natural spirals have inspired many artists and architects. Thefollowing two designs have been designed using logarithmic spirals, Fibonaccinumbers and the golden ratio.

Spiral Cafe

The Spiral Cafe, built in 2004 at the Bullring in Birmingham, was designedby Marks Barfield Architects. On their website [MB], Marks Barfield Ar-

Figure 6.16: Spiral Cafe[MB]

chitects say:

The building form is inspired by the 13th century mathemati-cian Leonardo Fibonacci who identified the natural patterns ofgrowth found throughout the universe, from the shapes of seashells and pine cones to fractal patterns within galaxies. Illus-trated graphically, the sequence forms a graceful spiral. Theshape of the Cafe was derived from extruding this golden spiral

62

Page 65: Fibonacci

along a tilting axis to form a simple curved enclosure. Like ashell, the exterior of the cafe is rough, rugged and durable pati-nated copper, whereas the inside is smooth and precious and litby pearl-like glass spheres.

This passage implies that building was designed using the Fibonaccispiral as a close approximation to the golden spiral proper to the goldenrectangle, which is a logarithmic spiral. The building was made to look likea shell as many creatures create shells which are logarithmic spirals. Oneof particular interest to us is the chambered nautilus sea shell. Accordingto Knott [RK], the shell of the Nautilus pompilius (Figure 6.17) grows by afactor of the golden ratio in one turn.

Figure 6.17: Radiograph of Nautilus pompilius shell[HH]

The Core

The Core, built in 2005 at Eden Project in Cornwall, was designed byGrimshaw Architects. On their website [EP], Eden Project Ltd say:

An exhibit in its own right, the Core takes its inspiration from thetree, incorporating a central trunk and canopy roof that shadesthe ground and harvests the sun. The design is based on the Fi-bonacci code, Nature’s fundamental growth blueprint, in whichopposing spirals follow the sequence 0, 1, 1, 2, 3, 5, 8, 13, 21,34... where every number is the sum of the previous two.

The passage and pictures imply that the building was designed usinga pattern similar to the seed arrangement in a sunflower. The seeds in

63

Page 66: Fibonacci

Figure 6.18: Eden Project[EP]

Figure 6.19: Core model[FT]

the head of a sunflower form two sets of oppositely directed logarithmicspirals as in Figure 6.20. The number of spirals are, astonishingly, consec-utive Fibonacci numbers: generally 55 clockwise and 34 counterclockwise.Comparable arrangements occur in pineapples, pinecones, cauliflowers, andconeflowers (Figure 6.21).

Figure 6.20: Double spiral pattern[HH]

Figure 6.21: Coneflower[RK]

We saw in Chapter 4 that the ratios of consecutive Fibonacci numbers arethe convergents of the golden ratio. Hence, many believe this phenomenonis related to the golden ratio. As discussed previously, the golden ratiois an irrational number which may be considered the most irrational as itis the furthest away from a rational number approximation. It has beensuggested that this allows for an optimal packing of seeds as seeds separatedin this way fall into spirals and not regular lines. Knott [RK] explains thesearrangements in more detail and looks at the relationship between the goldenratio and the leaf arrangement in plants, called phyllotaxis.

64

Page 67: Fibonacci

Chapter 7

Conclusion

The Fibonacci numbers form a fascinating sequence which is intimatelylinked to a wide range of material. As we have seen, there are many prop-erties associated with the Fibonacci sequence, and related sequences, andmany theorems and results involving these properties, which surprisinglycover topics such as divisibility, the golden ratio, and continued fractions.There are still many more interesting properties, theorems and results whichI have not covered here. For example, one may consider the sequence undervarious moduli.

We have looked at two-dimensional geometry related to the golden ratioand how some of this geometry is potentially linked to architecture. Thegolden ratio may have appeared in architecture before the 20th century butthis is unlikely to have been intentional and cannot be proved by dimen-sions alone. In the 20th century, architects have been inspired by, and havedesigned using, the Fibonacci numbers and the golden ratio.

There are several other topics to explore, which I have not covered here,that are linked to the golden ratio. Three-dimensional geometry and quasi-crystallography are possible extensions of the concepts discussed in Chapters5 and 6. I have only briefly touched upon the links to nature and this topicis open to further exploration. Claims have also been made suggesting thatthe golden ratio has been used in art and music and that it is the mostaesthetically pleasing ratio. However, there is little evidence to suggest thatthis is the case.

Fibonacci numbers and the golden ratio have interested mathematiciansfor centuries. In doing this project, I have discovered for myself the greatmathematical significance and interest of the sequence and have found thisbroad topic very intriguing.

65

Page 68: Fibonacci

Appendix A

The First 50 Fibonacci andLucas Numbers

n Fn Ln n Fn Ln0 0 2 25 75025 1677611 1 1 26 121393 2714432 1 3 27 196418 4392043 2 4 28 317811 7106474 3 7 29 514229 11498515 5 11 30 832040 18604986 8 18 31 1346269 30103497 13 29 32 2178309 48708478 21 47 33 3524578 78811969 34 76 34 5702887 12752043

10 55 123 35 9227465 2063323911 89 199 36 14930352 3338528212 144 322 37 24157817 5401852113 233 521 38 39088169 8740380314 377 843 39 63245986 14142232415 610 1364 40 102334155 22882612716 987 2207 41 165580141 37024845117 1597 3571 42 267914296 59907457818 2584 5778 43 433494437 96932302919 4181 9349 44 701408733 156839760720 6765 15127 45 1134903170 253772063621 10946 24476 46 1836311903 410611824322 17711 39603 47 2971215073 664383887923 28657 64079 48 4807526976 1074995712224 46368 103682 49 7778742049 17393796001

66

Page 69: Fibonacci

Bibliography

[AL] A L Loeb, Concepts and Images: Visual Mathematics, Birkhauser,1993

(ISBN 081763620X).

[CR] J J O’Connor and E F Robertson, Leonardo Pisano Fibonacci,c©1998

(http://www-history.mcs.st-andrews.ac.uk/Biographies/Fibonacci.html).

[DB] D M Burton, Elementary Number Theory, Wm. C. Brown, c©1988(ISBN 069706896X).

[DK] D E Knuth, The Art of Computer Programming, Volume 2: Seminu-merical Algorithms (2nd Edition), Addison-Wesley, c©1981

(ISBN 0201038226).

[EP] Eden Project Ltd, The Core, c©2007(http://www.edenproject.com/horticulture/1118.html).

[EW] E W Weisstein, Pell Equation, MathWorld–A Wolfram Web Resource(http://mathworld.wolfram.com/PellEquation.html).

[FG] J Fauvel and J Gray, The History of Mathematics: A Reader,Macmillan Press Ltd, 1988

(ISBN 0333427912).

[FS] F E Su, et al., Fibonacci GCD’s, please., Mudd Math Fun Facts(http://www.math.hmc.edu/funfacts).

[FT] M Freiberger and R Thomas,Bridges: mathematical connections in art and music, Millennium

Mathematics Project, University of Cambridge, c©1997-2004(http://plus.maths.org/latestnews/may-aug06/bridges/).

[GM] G Markowsky, Misconceptions about the Golden Ratio, The CollegeMathematics Journal, Vol. 23, No. 1. (Jan., 1992), pp. 2-19

(http://www.jstor.org/).

67

Page 70: Fibonacci

[HH] H E Huntley, The Divine Proportion: a Study in MathematicalBeauty, Dover Publications, Inc., c©1970

(ISBN 0486222543).

[KF] K Frampton, Le Corbusier, Thames & Hudson, c©2001(ISBN 0500203415).

[MB] Marks Barfield Architects, Spiral Cafe, c©2005(http://www.marksbarfield.com/project.php?projectid=11).

[ML] M Livio,The Golden Ratio : The Story of Phi, the Extraordinary Number ofNature, Art and Beauty, Review, c©2002

(ISBN 0747249881).

[NT] Nina Aldin Thune, Kheops pyramid, c©2005(http://en.wikipedia.org/wiki/Image:Kheops-Pyramid.jpg).

[PL] P J Lu, et al., Decagonal and quasi-crystalline tilings in medievalIslamic architecture, Science 315, 1106 (2007)

(http://www.sciencemag.org/).

[RD] R A Dunlap, The Golden Ratio and Fibonacci Numbers,World Scientific, c©1997

(ISBN 9810232640).

[RJ] R C Johnson, Matrix methods for Fibonacci and related sequences,2007

(http://www.dur.ac.uk/bob.johnson/fibonacci/).

[RK] R Knott, Fibonacci Numbers and the Golden Section, c©1996-2008(http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fib.html).

[RT] C Rossi and C A Tout,Were the Fibonacci Series and the Golden Section Known in AncientEgypt?, Historia Mathematica, Volume 29, Issue 2, 2002

(http://www.sciencedirect.com/).

[SV] S Vajda,Fibonacci and Lucas Numbers, and the Golden Section: Theory

and Applications, Ellis Horwood Ltd, 1989(ISBN 0745807151).

[TK] T Koshy, Fibonacci and Lucas Numbers with Applications,John Wiley & Sons, Inc., c©2001

(ISBN 0471399698).

Prepared in LATEX 2ε by EWSC

68