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PIBONACCI 8VUMBERS by N. N. VOrob'ev - - CONTENTS - FOREWORD 5 INTRODUCTION 7 CHAPTER I 12 The Simplest Properties of Fibonacci Numbers CHAPTER II 29 Number-theoretic Properties of Fibonacci Numbers CHAPTER III 39 Fibonacci Numbers and Continued Fractions CHAPTER IV 56 Fibonacci Numbers and Geometry CHAPTER V 66 Conclusion

Fibonacci Numbers

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N. N. VOrob'ev




CHAPTER I 12The Simplest Properties of Fibonacci Numbers

CHAPTER II 29Number-theoretic Properties of Fibonacci Numbers

CHAPTER III 39Fibonacci Numbers and Continued Fractions

CHAPTER IV 56Fibonacci Numbers and Geometry

CHAPTER V 66Conclusion

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A translation of "Chisla fibonachchi"(Moscow-Leningrad, Gostekhteoretizdat, 1951)

Translated from the Russian by Halina MossTranslation edited by Ian N. Sneddon

Additional editing by Robert R. Prechter, Jr.Math typeset by Barbara Anderson

Text typeset by Paula RobersonCover and graphics by Roberta Machcinski

Original translation copyright ® 1961Pergamon Press, Ltd.

ISBN: 0-932750-03-6Library of Congress Catalog Card Number: 83-60612

Editor's Note:The notation for binomial coefficients is thatused in the original text. For instance, on p. 18,

the expression C~ is used, whereas common English

usage would show bca .


In elementary mathematics, there are manydifficult and interesting problems not connectedwith the name of an individual, but ratherpossessing the character of "mathematicalfolklore". Such problems are scattered throughoutthe wide li terature of popular (or, simply,entertaining!) mathematics, and often it is verydifficult to establish the source of a particularproblem.

These problems often circulate in severalversions. Sometimes several such problems combineinto a single more complex one; sometimes theoppos i te happens and one problem spl i ts up intoseveral simple ones. Thus it is often difficultto distinguish between the end of one problem andthe beginning of another. We should consider thatin each of these problems we are dealing withlittle mathematical theories, each with its ownhi story, its own complex of problems and its owncharacteristic methods, all, however, closelyconnected with the history and methods of "greatmathematics".

The theory of Fibonacci numbers is just such atheory. Derived from the famous "rabbi t problem",going back nearly 750 years, Fibonacci numbers,even now, provide one of the most fascinatingchapters of elementary mathematics. Problemsconnected with Fibonacci numbers occur in manypopular books on mathematics, are discussed atmeetings of school mathematical societies, andfeature in mathematical competitions.

The present book contains a set of problemswhich were the themes of several meetings of theschoolchildren's mathematical club of LeningradState University in the academic year 1949-50. Inaccordance with the wishes of those taking part,the questions discussed at these meetings weremostly number-theoretical, a theme which isdeveloped in greater detail here.

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This book is designed basically for pupils of16 or 17 years of age in a high school. Theconcept of a limit is encountered only in examples7 and 8 in chapter III. The reader who is notacquainted with this concept can omit thesewithout prejudice to his understanding of whatfollows. That applies also to binomialcoefficients (I, example 8) and to trigonometry(IV, examples 2 & 3). The elements which arepresented of the theories of divisibility and ofcontinued fractions do not presuppose anyknowledge beyond the limits of a school course.

Those readers who develop an interest in theprinciple of constructing recurrent series arerecommended to read the small but comprehensivebooklet by A.I. Markushevich, "RecurrentSequences" (Vozvratnyye posledovatel' nosti)(Gostekhizdat, 1950). Those who become interestedin facts relating to the theory of numbers arereferred to textbooks in this subject.


!. The ancient world was rich in outstandingmathematicians. Many achievements of ancientmathematics are admired to this day for theacuteness of mind of their authors, and the namesof Euclid, Archimedes and Hero are known to everyeducated person.

Things are different as far as the mathematicsof the Middle Ages is concerned. Apart fromVieta, who lived as late as the sixteenth century,and mathematicians closer in time to us, a schoolcourse of mathematics does not mention a singlename connected with the Middle Ages. This is, ofcourse, no accident. In that epoch, the sciencedeveloped extremely slowly and mathematicians ofreal stature were few.

Greater, then, is the interest of the workLiber Abacci (lOa book about the abacus"), writtenby the remarkable Italian mathematician, Leonardoof Pisa, who is better known by his nicknameFibonacci (an abbreviation of filius Bonacci).This book, written in 1202, has survived in itssecond version, published in 1228.

Liber Abacci is a voluminous work, containingnearly all the arithmetical and algebraicknowledge of those times. It played a notablepart in the development of mathematics in WesternEurope in subsequent centuries. In particular, itwas from this book that Europeans becameacquainted with the Hindu (Arabic) numerals.

The theory contained in Liber Abacci isillustrated by a great many examples, which makeup a significant part of the book.

Let us cons ider onewhich can be found onmanuscript of 1228:

of these examples,pages 123-124 of


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How many pairs of rabbits are born of one pairin a year?


A pair1


This problem is stated in the form:

Someone placed a pair of rabbits in a certainplace, enclosed on all sides by a wall, tofind out how many pairs of rabbits will beborn there in the course of one year, it beingassumed that every month a pair of rabbitsproduces another pair, and that rabbits beginto bear young two months after their own birth.

As the first pair produces issue in the firstmonth, in this month there will be 2 pairs.Of these, one pair, namely the first one,gives birth in the following month, so that inthe second month there will be 3 pairs. Ofthese, 2 pairs will produce issue in thefollowing month, so that in the third month 2more pairs of rabbits will be born, and thenumber of pairs of rabbits in that month willreach 5; of which 3 pairs will produce issuein the fourth month, so that the number ofpairs of rabbits will then reach 8. Of these,5 pairs will produce a further 5 pairs, which,added to the 8 pairs, will give 13 pairs inthe fifth month. Of these, 5 pairs do notproduce issue in that month but the other 8do, so that in the sixth month 21 pairsresult. Adding the 13 pairs that will be bornin the seventh month, 34 pairs are obtained;added to the 21 pairs born in the eighth monthit becomes 55 pairs in that month; this, addedto the 34 pairs born in the ninth month,becomes 89 pairs; and increased again by 55pairs which are born in the tenth month, makes144 pairs in that month. Adding the 89further pairs which are born in the eleventhmonth, we get 233 pairs, to which we add,lastly, the 144 pairs born in the finalmonth. We thus obtain 377 pairs. This is thenumber of pairs procreated from the first pairby the end of one year.

First (Month)2












Figure 1

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From Figure 1 we see how we arrive at it: weadd to the first number the second one i.e .• 1and 2; the second one to the third; theth i rd to the fourth; the fourth to thefifth; and in this way. one after another.until we add together the tenth and theeleventh numbers (144 and 233) and obtain thetotal number of rabbits (377); and it ispossible to do this in this order for aninfinite number of months.

£. We now pass from rabbits to numbers andexamine the following numerical sequence


sequence (1) • the condition (2) is obviouslyinadequate. and we must establish certainsupplementary conditions. For example. we can fixthe first few terms of sequence (1). How many ofthe first terms of sequence (1) must we fix sothat it is possible to calculate all its followingterms. using only condition (2)?

We begin by pointing out that not every termof sequence (1) can be obtained by (2). if onlybecause not all terms of (1) have two precedingones. For instance. the first term of thesequence has no terms preceding it, and the secondterm is preceded by only one. This means that inaddition to condition (2) we must know the firsttwo terms of the sequence in order to define it.

This means that for the unique construction of

We note that we cannot calculate the terms ofsequence (1) by condition (2) above.

in which each term equals the sum of two precedingterms. i.e .• for any n>2.

2. 5. 7. 12. 19. 31. 50 •...•1. 3. 4. 7. 11. 18. 29 •...•-1. -5. -6. -11. -17. '" and so on.

~. Let us now turn to the important particularcase of sequence (1). where ul = 1 and u2 = 1. Aswas pointed out above. condition (2) enables us tocalculate successively the terms of this series.It is easy to verify that in this case the first13 terms are the numbers 1. 1. 2. 3. 5, 8. 13. 21.34. 55. 89. 144. 233. 377. which we already met inthe rabbi t problem. To honor the author of theproblem. sequence (1) when ul = u2 = 1 is calledthe Fibonacci seguence. and its terms are know asFibonacci numbers.

Passing thus from two neighboring terms to theone immediately following them. we can reach theterm with any required suffix and calculate it.

Fibonacci numbers possess many interesting andimportant properties. which are the subject ofthis book.

This is obviously sufficient to enable us tocalculate any term of sequence (1). Indeed. u3can be calculated as the sum of the prescribedul and u2' u4 as the sum of u2 and thepreviously calculated u3. u5 as the sum of thepreviously calculated u3 and u4' and so on "inthis order up to an infinite number of terms".


possible to make up any number ofnumerical sequences satisfying thisFor example.

un = un-l + un-2'

It isdifferentcondi tion.

Such sequences. in which each term is defined assome function of the previous ones. areencountered often in mathematics. and are calledrecurrent seguences. The process of successivedefini tion of the elements of such sequences isitself called the recurrence process. and equation(2) is called a recurrence relation. The readercan find the elements of the general theory ofrecurrent sequences in the book by Karkushevichmentioned in the Foreword.

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THE SIMPLEST PROPERTIES OF FIBONACCI NUMBERSAdding these equations term by term. we obtain therequired result.

1. To begin with. we shall calculate the sum ofthe first n Fibonacci numbers. We shall show that


Further. subtracting (S) from (4) term by term.we get

as was required.

u2 + u4 + ... + u2n = u2n+2 - 1 - u2n = u2n+l - 1.

u1 + u2 + u3 + ... + u2n = u2n+2 - 1;

subtracting equation (4) from the above equation.we obtain

From section 1 we have

u2 + u4 + ... + u2n = u2n+1 - 1.

~. The sum of Fibonacci numbers with even suffixes

(3)ul + u2 + ... + un = un+2 - 1.

Indeed. we have:

Adding up these equations term by term. we obtain

ul + u2 + ... + un = un+2 - u2.

and all that remains is to remember that u2 = 1.

u1 - u2 + u3 - u4 + ... + u2n-1 - u2n =

= -u2n-1 + 1.

Now. let us add u2n+l to both sides of (6):

ul - u2 + u3 - u4 + ... - u2n + u2n+1 =


l. The sum of Fibonacci numbers with odd suffixes = u2n + 1.(7)

u1 + u3 + Us + ... + u2n-1 = u2n· (4) Combining (6) and (7). we get for the sum ofFibonacci numbers with alternating signs:

To establish this equation. we shall writeul - u2 + u3 - u4 + ... + (-l)n+lun =

= (-1)n+1un_1 + 1. (8)

~. The formulas (3) and (4) were deduced by meansof the term-by-term addition of a whole series ofobvious equations. A further example of theapplication of this procedure is the proof of theformula for the sum of squares of the first n

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Fibonacci numbers:

2 2 2u1 + u2 + ... + un = unun+1'

We note that

ukuk+1 uk-1 uk = uk(uk+1 - uk-I)2- = uk'


In the second part, the truth of theproposition is assumed for a certain arbitrary(but fixed) number n, and from this assumption,often called the inductive assumption, thededuction is made that the proposition is alsotrue for the number n + 1. The second part of theproof is called the inductive transition.


Adding up the equations2

u1 = ulu2,

2u2 = u2 u3

2u3 = u3 u4

2un = unun+1 - un-lun

term by term, we obtain (9).

i. Many relationships between Fibonacci numbersare conveniently proved with the aid of the methodof induction.

The essence of the method of induction is asfollows. In order to prove that a certainproposition is correct for any natural number, itis sufficient to establish:

(a) that it holds for the number 1;

(b) that from the truth of the propositionfor an arbitrary natural number n follows itstruth for the number n + 1.

Any inductive proof of a proposition true forany natural number consists, therefore, of twoparts.

In the first part of the proof, the truth ofthe proposition is established for n = 1. Thetruth of the proposition for n = 1 is sometimescalled the basis of induction.

The detailed presentation of the method ofinduction and numerous examples of the applicationof different forms of this method can be found inI. S. Sominski i, "The Method of MathematicalInduction". Thus, in particular, the vers ion ofthe method of induction with the inductivetransition "from nand n + 1 to n + 2" employed byus below is given in Sominskii' s book on page 9and is illustrated there on page 16 by problems 18and 19.

We prove by induction the following importantformula:


We shall carry out the proof of this formulaby induction on m. For m = 1, this formula takesthe form

which is obviously true. For m = 2, formula (10)is also true, because

= un-l + un + un = un+l + un-

Thus the basis of the induction is proved.The inductive transition can be proved in thisform: supposing formula (10) to be true for m = kand for m = k + 1, we shall prove that it alsoholds when m = k + 2.

Thus, let

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§.. The following formula will be found useful inwhat follows:


Adding the last two equations term by term, weobtain

2 nun+1 = unun+2 + (-1) .

Let us prove it by induction over n.(12) takes the form


For n=l,

and this was the required result.

2u2 = ulu3 - I,

which is obvious.


Putting m = n in formula (10), we obtain We nowcertain n.it, we obtain



toproved for

both sidesa


or 2 nUn+l + un+lun+2 = Unun+2 + un+lun+2 + (-1)

(11) or

From this last equation, it is obvious that u2nis divisible by un. In the next chapter weshall prove a much more general result.



2 nun+lun+3 = un+2 + (-1) .

un = un+l - un-I,

formula (11) can be rewritten thus:


2 n+lun+2 = un+lun+3 + (-1)

Thus, the inductive transition is established andformula (12) is proved for any n.


2 2u2n = un+1 - un-I,

1. In a similar way, it is possible to establishthe following properties of Fibonacci numbers:

2u1u2 + u2u3 + u3u4 + ... + u2nu2n+l = u2n+1 - I,

nUl + (n-1)u2 + (n-2)u3 + ... + 2un-l + un =

= un+4 - (n+3).3

un-I·3 3

un+l + un

Similarly (taking m = 2n), it can be shown that


i. e. , the difference of the squares of twoFibonacci numbers whose positions in the sequencediffer by two is again a Fibonacci number.

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The proofs are left to the reader.

~. It turns out that there is a connectionbetween the Fibonacci numbers and another set ofremarkable numbers, the binomial coefficients.Let us set out the binomial coefficients in thefollowing triangle, called Pascal's triangle:


C~ ci

0 1 2C2 C2 C2

C~ C~2



To prove the general proposition, it issufficient to show that the sum of all numbersmaking up the (n-2)th and the (n-l)th diagonal ofPascal's triangle is equal to the sum of thenumbers making up the nth diagonal.

On the (n-2)th diagonal we have the numbers

o Cl C2Cn-3' n-4' n-5' ... ,

and on the (n-l)th diagonal the numbers

cg-2' Ca-3' C~-4' ...

The sum of all these numbers can be written thus:

Cg-2 + (Cg-3 + Ca-3) + (Ca-4 + C~_4) + ... (13)

But for binomial coefficients

We shall show that the sum of numbers lyingalong a certain rising diagonal is a Fibonaccinumber.

The straight lines drawn through the numbersof this triangle at an angle of 45 degrees to therows we shall call the "rising diagonals" ofPascal's triangle. For instance, the straightlines passing through numbers I, 4, 3, or I, 5, 6,1, are rising diagonals.

lying on the nth



i+l= Ck+l '


(1 + k-i) =


k(k-l) (k-i+l)1.2 i

(k+l)k(k-l) (k-i+l)= 1.2 ..... i.(i+l)

i . e., the sum of the numbersdiagonal of the triangle.

k(k-l) ... (k-i+l)= 1.2..... i

k(k-l) ... (k-i+l) (k-i)+ =1. 2. . .. . i. (i+1)

Expression (13) therefore equals

cg_l + Ca-2 + C~_3 + ... ,

k(k-l) ... (k-i+l)=---=-=-_---.:--:--1.2..... i

Cg-2 = Cg_l = 1



topmost rising diagonalmerely I, the first

second diagonal also



11 11211 3/3/11 4 6 41/ 5~0 10r···....... 6 15 20

Indeed, the first andof Pascal's triangle isFibonacci number. Theconsists of 1.

i.e. ,

Page 10: Fibonacci Numbers

To see this, we investigate various sequencessatisfying the relationship (2). We shall callall such sequences solutions of equation (2).

Making use of formulas (4), (5), (6) andsimilar ones, the reader can easily obtain furtheridentities connecting Fibonacci numbers withbinomial coefficients.

From this proof and formula (3) we immediatelyget: The sum of all binomial coefficients lyingabove the nth rising diagonal of Pascal's triangle(inclusive of that diagonal) equals un+2 - 1.

~. So far, we have defined Fibonacci numbers by arecurrence procedure, i. e., induct i vely, by theirsuffixes. It turns out, however, that anyFibonacci number can also be defined directly, asa function of its suffix.


v" = v" + v"n n-1 n-2

v' + v" = (v' + v" ) + (v' + v" ).n n n-1 n-l n-2 n-2

Adding these two equations term by term, we get

v' = v' + v'n n-1 n-2



Proof: From the conditions stated in the lemma,we have

as was required.

Lemma 2. If the sequences V' and V" are solutionsof (2), then their sum V' + V" (i. e., the sequencev' + v" v' + v" v' + v" ... )is also a solution

1 I' 2 2' 3 3'of (2).

cVn = cVn_2 + cVn-l'





Thus, the lemma is proved.

where cl and c2 are constants. It istherefore usual to speak of (14) as the generalsolution of the equation (2).

Now, let V' and V" be two solutions ofequation (2) which are not proportional. We shallshow that any sequence V which is a solution ofequation (2) can be written in the form

, I

In future we shall denote the sequences

vI' v2 ' v3 ' " .. ,

v' , v'I' v2 ' 3' . " . ,

v" v" v" ... ,I' 2' 3'

by V, V' and V" respectively.

To begin with, we shall prove two simplelemmas.

c V' + c V"1 1 2 2'(14)

The proof of (15) is carried out by assuming theopposite.

First of all, we shall prove that if solutionsof (2) V' and V" are not proportional, thenLemma 1. If V is the solution of equation (2) and

c is an arbitrary number, then the sequence cV(i. e. , the sequence cVl' cV2' cV3' ... ) isalso a solution of equation (2).

Proof: Multiplying the relationship

vn = vn-2 + vn-l

v' v'1 2vnF;;;1 2

(15 )

term by term by c, we get

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On writing down the derived proportion, we get

For solutions V' and V.. of (2) which are notproportional, let


v'v" - v"v'1 2 1 2


v v..2 1

v v..1 2

cl =------v'v" v"v'

1 2 1 2

In view of condition (15). the simultaneousequations (17) are soluble with respect to c1and c2' no matter what the numbers VI and v2are:


v· v·1 2

v" = v" (16)1 2


[By the condi tion (15). the denominator does notequal zero].

v· + v· v·122

v" + v" = V"122

or, taking into accountsolutions of equation (2),

that V' and v.. are Substituting the values of c1calculated in (14) • we obtainrepresentation of the sequence V.


c2 thusrequired

to describe allis sufficient to

which are not

v' v'3 2

v" = v"3 2

Similarly, we convince ourselves (by induction)thatv· v' v'

3 4 nv" = v" = = V" =

3 4 n

Thus, it follows from (16) that the sequences V'and V" are proportional, which contradicts theassumption. This means that (15) is true.

Now, let us take a certain sequence V, whichis a solution of the equation (2). This sequence,as was pointed out in section 2 of theIntroduction, is fully defined if its two firstterms, VI and v2' are given.

Let us find such c1 and c2' that

This means that in ordersolutions of equation (2), itfind any two solutions of itproportional.

Let us look for these solutions ~ong

geometric progressions. In accordance with lenuna1, it is sufficient to limit ourselves to theconsideration of only those progressions whosefirst term is equal to unity. Thus, let us takethe progression

1. q, q2, ...

In order that this progression should be asolution of (2), it is necessary that for any n,the equali ty

qn-2 + qn-1 = qn

should be fulfilled. Or, dividing by qn-2,

(17)c v· + c v" = v1 2 2 2 2'

1 + q = q2.

The roots of this quadratic equation, i.e.,

Then. on the bas i s of lenunas 1 and 2, c1V· +c2V" gives us the sequence V.

1 + 152

and 1 - .f52

Page 12: Fibonacci Numbers

are solutions of (2). As the progressions foundby us have different conunon ratios and aretherefore not proportional, formula (18) gives usall solutions of equation (2).

In particular, for certain values of cl andc2. formula (18) should also give us theFibonacci series. For this, as was pointed outabove, it is necessary to find cl and c2 fromthe equations



For instance, we can find the sum

u3 + u6 + Ug + ... + u3n·

10. with the help of Binet's formula, it is easyto find the sums of many series connected withFibonacci numbers.

Formula (19) is called Binet's formula inhonor of the mathematician who first proved it.Obviously, similar formulas can be derived forother solutions of (2). The reader should do itfor the sequences introduced in section 2 of theIntroduction.


(Ii lS)n_{1 2 1StUn = 15

i.e. ,


geometricof (2).


required conunon ratios of theWe shall denote them by a and B

Note that aB = -1.

We have thus obtained twoprogressions which are solutionsTherefore, all sequences of the form

will be theprogressions.respectively.


We have


u3 + u6 + ... + u3n


3n••• - B ) ,

B3n+3 _ B3 ).B3 - 1

+ a3n _ B3n =15

the geometric

+ ...

sununedor. havinginvolved,

136=./5(a +ac1 + c2 = 1,

c1 1 + 15+ c2 1 - 15 = l.2 2

Having solved them, we get

i.e. from the simultaneous equations

= 1 + /5(1 + l5)n-1 _ 1 - 15(1 15)n-1,2./5 2 2/5 2


c1 = 1 + 152 15 •


1 - 152 15' But

a3 _ 1 = a + a2 - 1 = a + a + 1 - 1 = 2a.

and similarly. B3 - 1 = 2B. Therefore,

= .1..la3n+3 _ a3 B3n+3 - B3).u3 + u6 + ... + u3n /5\ 2a - 2B

Page 13: Fibonacci Numbers





Using this theorem, itcalculate Fibonacci numberslogarithmic tables.

It is not hard to prove the following theorem.

The reader who is acquainted with the theoryof limits will be able to show, by slightlyaltering the proof of this theorem, that

lim lun - ani = o.n_ CID

Theorem. The Fibonacci number un is the nearestwhole number to the nth term an of the geometricprogression whose first term is a/fS and whosecommon ratio equals a.

Proof: Obviously it is sufficient to establishthat the absolute value of the difference betweenun and an is always less than 1/2. But


- an! = a~Bn Ii- = lan_~_Bnl = I~n.

As B -0.618 ... , therefore IBI < I, and thatmeans that for any n, IBln<l and even more so(since J5> 2) l§.1 <1.. The theorem is proved.

5 2

12. It is relevant to ask the question: howquickly do Fibonacci numbers grow with increasingsuffix? Binet's formula gives us a sufficientlyfull answer even to this question.

We note that

or after cancellations,


+ u6

+ ... + u3n

= k(a3n


- a2

; B3n+2 + B2)=

_ l/a3n+2 B3n+2 a2 _ B2)- 2\ 15 - 15 =

11. As another example of the application ofBinet's formula, we shall calculate the sum of thecubes of the first n Fibonacci numbers.


For instance, let us calculate u14 (u14the answer to the Fibonacci problem aboutrabbits):

log j5 = 0.34949;


or, using formula (8) and the results of thepreceding section,

3 3 3 .!(U3n+2 -1 1)u1 + u2 +... + un = 5 2 + 3[1 + (_l)n+ un_I] =

u3n+2 + (-1)n+16un_1 + 5= 10

J5 = 2.2361,

a = 1 + 15 = 1.6180,2

log a = 0.20898;


Page 14: Fibonacci Numbers


a14log 15 = 14 x 0.20898 - 0.34949 = 2.5762,



The nearest whole number to 376.9 is 377; thisis u14'

When calculating Fibonacci numbers of verylarge suffixes, we can no longer calculate all thefigures of the number by means of available tablesof logarithms; we can only indicate the first fewfigures of it, so that the calculation turns outto be approximate.

Before we continue the study of Fibonaccinumbers, we shall remind the reader of some of thesimplest facts from the theory of numbers.

1. First, we shall indicate the process offinding the greatest common divisor of numbers aand b.

As an exercise, the reader should prove thatin the decimal system, un for n~ 17 has no morethan n/4 and no fewer than nl5 figures. And ofhow many figures will uIOOO consist?

Suppose we divide a byto qo and a remainderbqo + rl and 0 < rl < b.qo = O. -

b with a quotient equalrl' Obviously, a =

Note that if a < b,

Let us further divide b by rl and let usdenote the quotient by ql and the remainder by r2'Obviously b = rlql + r2, and 0<r2<rl' Sincerl<b, therefore ql +O. Then, dividing rl by r2,we shall find q2 ~ 0 and r3 such that rl = q2r2 +r3 and 0< r3< r2' We proceed in this manner for aslong as it is possible to continue the process.

. Sooner or later our process must terminate,sInce all the positive whole numbers rl' r2'r3, ... are different, and everyone of them issmaller than b. That means that their number doesnot exceed b, and the process should terminate nolate~ than at the bth step. But it can onlytermInate when a certain division proves to becarried out perfectly, i. e., the remainder turnsout equal to zero and it wi 11 be impos sible todivide anything by it.

The process thus described bears the name ofEuclidean Algorithm. As a result of itsapplication we obtain the following sequence ofequations:

Page 15: Fibonacci Numbers


1, itcan bealways




We now establish several simple properties ofthe greatest common divisor of two numbers.

Subsequently, however, an important differenceof the geometrical process from Euclid's algorithmfor natural numbers is revealed. The sequence ofremainders obtained from the subtraction ofsegments might not terminate, as the process ofsuch subtraction can turn out to be capable ofbeing continued indefinitely. This will happen ifthe chosen segments are incommensurable.

From the considerations infOllows that two segments whoseexpressed by whole numberscommensurable.

Indeed, let us examine two segments, one oflength a, the other of length b. Let us subtractthe second segment from the first as many times asit is possible (if b> a, obviously we cannot do iteven once) and denote the length of the remainderby rl' Obviously r1 < b. Now, let us subtractfrom the segment of length b the segment of lengthrl as many times as possible, and let us denotethe newly obtained remainder by r2' Carrying onin this manner, we obtain a sequence of remainderswhose lengths, evidently, decrease. Up to thispoint, the resemblance to Euclid's algorithm iscomplete.

55 = 5 x 11.

Thus, (u20, ulS) = 5 = uS, The fact thatthe greatest common divisor of two Fibonaccinumbers turned out to be again a Fibonacci numberis not accidental. It will be shown later thatthat is always the case.

l. There is an analogy between Euclid's algorithmand a process in geometry whereby the commonmeasure of two commensurable segments is found.



We have thus proved that the Euclideanalgori thm when applied to the natural numbers aand b really does lead to their greatest commondivisor. This greatest common divisor of thenumbers a and b is denoted by (a, b).

Let d be a certain common divisor of a and b.From the first equation of (20), we notice thatrl should be divisible by d. But in that case,on the basis of the second equation of (20), r2is divisible by d. Similarly (induction) we provethat d "goes into" r3' ... , r n-1 and, finally,r n ·

Let us examine the last non-zero remainderr n . Obviously r n-1 is divisible by r n . Letus now take the last but one equation in (20). Onits right-hand side both terms are divisible byr n ; therefore r n-2 is divisible by r n .Similarly, we show step by step (induction) thatr n-3' r n-4' and finally a and baredivisible by r n . Thus, r n is a common divisorof a and b. Let us show that r n is the greatestcommon divisor of a and b. In order to do this,it is sufficient to show that any common divisorof a and b will also divide r n .


610 = 55 x 11 + 5,

6765 = 610 x 11 + 55,

As an example, let us find (u20' u15) =(6765, 610):

Page 16: Fibonacci Numbers

--......,,"! !


subsequent steps of this algorithm wi 11consecutively the second, third. etc.,

of the set (20). The last non-zerois still r n , and this means that (a.

c, b).

Thegive usequationsremainderb) = (a +

I. (a, bc) is divisible by (a, b). Indeed, b,and therefore bc, is divisible by (a, b); a isdivisible by (a, b) for obvious reasons. Thismeans, according to the proofs in section I, that(a, bc) is divisible by (a, b) also.

!. (ac, bc) = (a, b)c

Proof: Let the equations (20) describe theprocess of finding (a, b). Kultiplying each ofthese equations by c throughout, we shall, as iseasily verified, obtain a set of equationscorresponding to the Euclidean algorithm asapplied to the numbers ac and bc. The lastnon-zero remainder here will be equal to rnc,Le., (a, b)c.

A useful exercise for the reader would be toprove this theorem on the sole basis of theresults of sections 3-6, i.e., without a repeatedreference to the idea of the Euclidean algorithmand to the set (20).

We now consider certain properties ofFibonacci numbers concerning their divisibility.

!. Theorem. If n is divisible by m, then un isalso divisible by um'

Proof: Let n be divisible by m, i.e., let n =mml' We shall carry out the proof by inductionover mI' For m1 = I, n = m, so that in thiscase it is obvious that un is divisible byUrn. We now suppose that umm1 is divisibleby Urn and consider Um(m1+1). ButUrn(m1+1) = umm1 + m and, according to (10),

nature ofof their

side of thisby 11m' Thefactor, i . e. ,the inductiveUm(m1+1) , is

the theorem is

We prove that for a compound n other than 4.un is a compound number.

!. The topic of the arithmeticalFibonacci numbers (i.e., the naturedivisors) is of great interest.

The first term of the right-handequation is obviously divisiblesecond term contains umm1 as ais divisible by Urn according toassumption. Hence, their sum,divisible by Urn as well. Thusproved.

1. If c is divisible by b, then (a, b) = (a + c,b) •

~. a is divisible by b only if (a, b) b. Thisis obvious.

Proof: Suppose that the application of theEuclidean algorithm to the numbers a and b leadsto the set of equations (20). Let us apply thealgorithm to the numbers a + c and b. Since c isdivisible by b, as given, we can put c = c1b .The first step of the algorithm gives us theequation

i. If (a, c) = I, then (a, bc) = (a, b). Indeed,(a, bc) divides (ab, bc), according to section 3.But (ab, bc) = (a, c)b = 1 x b = b, in view ofsection 4. Thus, b is divisible by (a, bc). Onthe other hand, (a, bc) divides a. By section I,this means that (a, bc) divides (a, b) also. Andsince according to section 3, (a, b) divides (a,bc) as well, then (a, b) = (a, be).

Page 17: Fibonacci Numbers


Indeed, for such an n we can write n =n1n2' where 1 < n1 < n, 1 < n2 < n and eithern1>2 or n2>2. To be definite, let n1>2.Then, according to the theorem just proved, unis divisible by unI' while 1 < unl < un'and this means that un is a compound number.

10. Theorem. Neighboring Fibonacci numbers areprime to each other.

or, by equation (10),

(Urn' un) = (UnqO-1Ur1 + UnqOUr1+1, un)'

or by sections 7 and 8,

or by sections 10 and 5,

Proof: Let un and un+1 have a certain commondivisor d> I, in contradiction of what the theoremstates. Then their difference un+1 unshould be divisible by d. And since un+1 - un

Un-I' then un-1 should be divisible by d.Similarly, we prove (induction) that un-2,un-3' etc., and finally u1' will be divisibleby d. But u1 = 1. Therefore it cannot bedivided by d > 1. The incompatibility thusobtained proves the theorem.

Similarly, we prove that


11. Theorem. For any m, n (Urn, un) = u(m,n)·

Proof: To be defini te, we suppose m> n, and applythe Euclidean algorithm to the numbers m and n:

Combining all these equations, we get

(Urn, un) = (urt ' urt_1),



But we have proved that

In particular, from the above proof we have aconverse of the theorem in section 8: if un isdivisible by Urn' then n is divisible by m. Forif un is in fact divisible by um' thenaccording to section 6,

and since rt-1 is divisible by rt, urt_1is also divisible by Urt. Therefore (Urt,Urt_1) = Urt. Noting, finally, that rt= (m, n), we obtain the required result.

m = nqo + r1' where 0< r1< n ,

As we already know, rt is the greatestcommon divisor of m and n.

Thus, m = nqo + r1; this means that

Page 18: Fibonacci Numbers


Combining (21) and (22) we get


At the same time, the reader should prove thatno Fibonacci number exists that would give aremainder of 4 when divided by 8; also, thatthere are no odd Fibonacci numbers divisible by 17.

to discoverleast onenumber m.

It would therefore be interestingwhether it is possible to find atFibonacci number divisible by a givenIt turns out that this is possible.

~. Let us now take a certain whole number m. Ifthere exists even one Fibonacci number undivisible by m, it is possible to find as manysuch Fibonacc i numbers as des ired. For example,such will be the numbers u2n' u3n' u4n'

In view of this, the divisibility of Fibonaccinumbers can be studied by examining thedivisibility of their suffixes.

i.e., m = (n,m), which means that n is divisibleby m.

12. Combining the theorem in section 8 and thecorollary to the theorem in section 11, we have:un is divisible by Um if, and only if, n isdivisible by m.

Let us find, for instance, some "signs ofdivisibility" of Fibonacci numbers. By "sign ofdivisibility", we mean a sign to show whether anyparticular Fibonacci number is divisible by acertain given number.

Let U be the remainder of the division of u bym, and let us write down a sequence of pairs ofsuch remainders:


A Fibonacci number is even if, and only if,its suffix is divisible by 3.

A Fibonacci number is divisible by 3 if, andonly if, its suffix is divisible by 4.

A Fibonacci number is divisible by 4 if, andonly if, its suffix is divisible by 6.

A Fibonacci number is divisible by 5 if, andonly if, its suffix is divisible by 5.

A Fibonacci number is divisible by 7 if, andonly if, its suffix is divisible by 8.

The proofs of all these signs of divisibilityand all similar ones can be carried out easily bythe reader, with the help of the proposition putforward at the beginning of the section, and byconsidering the third, fourth, sixth, fifth,eighth, etc. Fibonacci numbers respectively.

If we regard pairs [a1' b1] and [a2' b2] asequal when a1 = a2 and bl = b2' the number ofdifferent pairs of remainders of division by me~uals m2 . If, therefore, we take the firstm + 1 terms of the sequence (23), there must beequal ones among them.

Let [Uk' Uk+l] be the first pair thatrepeats itself in the sequence (23). We shallshow that this pair is [I, 1]. Indeed, let ussuppose the opposite, Le., that the firstrepeated pair is the pair [Uk' Uk+l] , where k> 1.Let us find in (23) a pair [Uj, Uj+l] (j> k) equalto the pair [Uk' Uk+l]. Since Uj_1 = Uj+l - Uj anduk-l = uk+1 - Uk, and Uj+1 = Uk+1 and Uj = Uk' theremainders of division of Uj_l and uk-l by mareequal, i.e., Uj_l = Uk-l. However, it also followsthat [Uk-I' Uk] = [Uj-l' Uj] , but the pair[Uk-l, Uk] is situated in the sequence (23) earlierthan [Uk' Uk+l] and therefore [Uk' Uk+l] is not thefirst pair that repeated itself, which contradicts

Page 19: Fibonacci Numbers



Let us see how we can find the partialdenominators of such an expansion of the ordinaryfraction alb.



ql + 1

q2 + 1

q3 +, 1, , ,




We consider the expression

The process of transformation of a certainnumber into a continued fraction is called thedevelopment of this number into a continuedfraction.

Sometimes continued fractions are also knownas chain fractions. They are of use in a wideassortment of mathematical problems. The readerwho wants to study them in greater detail isreferred to A.Ya. Khinchin. "Chain Fractions".

The expression (24) is called a continuedfraction and the numbers qO' q1' ...• qn arecalled the partial denominators of this fraction.

where q1' q2' . . . • qn are whole pos i ti venumbers and qo is a whole non-negative number.Thus. in contrast to the numbers q1. q2 •...•qn' the number qo can equal zero. We shallk.eep this somewhat special position of the numberqo in mind. and not mention it specially on eachoccasion.

our premise. This means that the supposition k> 1is wrong. and therefore k = 1.

We have thus proved the following theorem:

Note that this theorem does not state anythingabout exactly which Fibonacci number will bedivisible by m. It only tells us that the firstFibonacci number divisible by m should not beparticularly large.

Thus [1. 1] is the first pair that repeatsitself in (23). Let the repeated pair be in thetth place (in accordance with what was establishedearlier we can regard 1 < t < m2 + 1). i.e .•rUt. Ut+l] = [1. 1]. This means that bothUt and Ut+l. when divided by m. give 1 as aremainder. It follows that their difference isexactly divisible by m. But Ut+1 Ut =Ut-l. so that the (t - l)th Fibonacci number isdivisible by m.

Theorem: Whatever the whole number m. at leastone number divisible by m can be found among thefirst m2 Fibonacci numbers.

Page 20: Fibonacci Numbers





The Euclidean algorithm as applied to a givenpair of natural numbers a and b is realized in acompletely definite and unique way. The partialdenominators of the development of alb into acontinuous fraction are also defined in a uniqueway by the system of equations describing thisalgori thm. Any rational fraction alb, therefore,can be expanded into a continued fraction in oneand only one way.

,l. Let

...... 1+­qn

be a certain continued fraction, and let usconsider the following numbers:

+_1_q2 +,,

, 1'+­qn

By the very sense of the Euclidean algorithm,qn > 1. (If qn were equal to unity, then r n-1would equal r n , and r n-2 would have beendivisible by rn-l exactly, i.e., the wholealgorithm would have terminated one stepearlier.) This means that in place of qn' wecan consider the expression (qn 1) + 1/1,i.e., consider (qn-1) the last but one partialdenominator, and 1 the last. Such a conventionturns out to be convenient for what follows.

Continuing this process to the end (induction)we arrive, as is seen easily, at the equation



The first of these equations gives us

We consider the Euclidean algorithm as appliedto the numbers a and b.

and therefore,

a 1b"= qO + q1 + 1

q2 +Lr2r3

so that

a 1b"= qo + 1 .

ql +­r1r2

From the third equation of (25), we deduce

r1 r3 1-= q2 +-= q2 +­r2 r2 r2


But it follows from the second equation of set(25) that

a rl 1b = qo + b = qo + b •


Page 21: Fibonacci Numbers



(30)qO(q1q2 + 1) + q2

qIq2 + 1

is easily verified.

The basis of the induction is thus proved.


Let us now suppose that the equations (27).(28) and (29) are true and let us cons ider theconvergent fraction

The equation

P2Q1 - P1Q2 = (_1)1

The fraction Pl/QI is in its lowest termsaccording to the definition. and equal fractionsin their lowest terms have equal numerators andequal denominators. This means that PI =

qOq1 + 1 and Q1 = q1·

The greatest common di vi sor of the numbersqO(qlq2 + 1) + q2 and qlq2 + 1 equals(Q2. Q1Q2 + 1) on the basis of section 7 ofII. and on the bas i s of the same propos i tion. italso equals (Q2. 1). i.e .• 1. This means thatthe fraction on the right-hand side in (30) is inits lowest terms. and therefore


Since the numbers qOql + 1 and q1prime to each other. the fraction qOq1 +is reduced to its lowest terms. q1





the transition from Pk to Pk+lQk Qk+1of the last of

took part in thefraction. i.e .•

Note that

Qk+1 = Qkqk+l + Qk-1.

Pk+1Qk - PkQk+1 = (_I)k.

We prove all these equations simultaneously byinduction over k.

Lemma. For every continued fraction (26) thefollowing relationships obtain:

We shall prove them first for k = 1.

is realized by the replacementthose partial denominators whichconstruction of this convergentqk. by qk + __1__


l. The following lemma plays an important role inthe theory of continued fractions.

are called convergent fractions of the continuedfraction w.

Pn_ = w.Qn


These numbers. written down in the form ofordinary simple fractionsI I

, I

Page 22: Fibonacci Numbers





continuedit follows


follows directly from the inductive(29). In this way. the inductive

is established and the whole lemma is

But in view of what was proved above.

This simple yet important observation will bemade more exact later in the book.

Since partial denominators offractions are positive whole numbers.from the above lemma that:

and (32)assumptiontransitionproved.

Theorem. If a continued fraction has n partialdenominators and each of these partial denominatorsequals unity. the fraction equals un+1.


!. We now apply the lemma of section 3 todescribe all continued fractions with partialdenominators equal to unity. For such fractionswe have the following interesting theorem.

To complete the proof of the inductivetransition. it remains to show that

Pk+l Pk (_l)k

Qk+l - Qk = QkQk+l·


expressionstheintonot comeQk-1' then


by qk+1 + _1_;qk+2

since qk+l doesfor Pk. Qk. Pk-1.

The transition from Pk+1 to Pk+2 according toQk+1 Qk+2

the observation made above is brought about by thereplacement of qk+1 in the expression for Pk+1


Pk+2 Pk(qk+1 + ~) + Pk-l--=Qk+2 Qk(qk+1 + qk:2) + Qk-l

or. remembering the inductive assumptions in (27)and (28).

and that

Thus the right-hand side of (31) is in itslowest terms. and (31) is therefore an equationbetween two fractions reduced to their lowestterms. This means that

Pk+2 Pk+1qk+2 + PkQk+2 = Qk+1qk+2 + Qk·

Let us suppose that the numbers Pk+1qk+2 +Pk and Qk+1qk+2 + Qk have a certain commondivisor d> 1. The expression

We now prove that the right-hand fraction in(31) is in its lowest terms. For this it issufficient to prove that its numerator anddenominator are mutually prime.

should then be divisible by d. But by theinductive assumption (29). this expression equals(_l)k+l and cannot be divided by d.

, I

Page 23: Fibonacci Numbers





From the results of the lemma of section 3, it iseasy to detect that in view of (36)

Po~Po' Pi~P1' P2~P2'


Qo> QO' Qi> Q1' Q2~ Q2' ...

Obviously, the smallest value of any partialdenominator is unity. This means that if all thepartial denominators of a certain continuedfraction are unity, the numerators anddenominators of its convergent fractions increasemore slowly than those of the convergent fractionsof any other continued fraction.

Lemma. If the continued fraction w has as itspartial denominators the numbers qo, q1' q2'... , qn' while

qo = q1 = q2 = = qi-1 = qi+1 =

Let us estimate to what extent this increaseis slowed down. Obviously, discounting thecontinued fractions whose partial denominators areuni ty, the slowest to increase are the numeratorsand denominators of the convergent fractions ofthat continued fraction one of whose partialdenominators is 2 and the remaining ones unity.Such continued fractions are also connected withFibonacci numbers as shown by the following lemma:

Let us denote the convergent fractions of w by

Po PI P2QO' Q1' Q2'

and the convergent fractions of w' byP' P' P'

0 1 2-, -, -,Q' Q' Q'

0 1 2


fraction withObviously,convergent

Qn+1 = QnQn+1 +This means that


w= qo + 1

q1 + 1q2 +

w'= q' + 10 q' 1+1 q' +2

~. Suppose we are given two continued fractions wand w':



Similarly, Q1 = I, Q2 = 1 andQn-1 = Qn + Qn-1' so that Qn = un'

_ un+1an --u;-'

The reader should compare this result withformulas (12) and (29).


a =1+1.=1.,211


therefore, PI = I, P2 = 2. Further, Pn+l = PnQn+1+ Pn-1 = Pn + Pn-1' Therefore (compare I, section8), Pn = un+1'

Proof: Let us denote the continuedn unit partial denominators by an'aI, a2' ... , an are consecutivefractions of an'

Page 24: Fibonacci Numbers




11 + ......

...... ...... + 1 + _1__2 + 1



i+1 partialdenominators

ui+2un-i+2 + ui+lun-i=

ui+lun-i+2 + uiun-i

(ui + ui+l)un-i+2 + (ui-l + ui)un-i= ui+lun-i+2 + ~iun-i

Thus, the inductive transition has been provedand so has the whole 1enuna.

ui+lun-i+2 + uiun-iui un-i+2 + ui-lun-i

and similarly,

Corollary. If not.all the partial denominators ofthe continued fraction ware unity, qo "I: 0, andthere are no less than n of these partialdenominators, then, on writing w in the form of anordinary fraction P/Q, we have

1 + 1

i partial . ·1· . ·1· +.-:~.~ ~ ..denominators + 1 +~


The continued fraction below the dotted linein (38) is, by (37), equal to

It can obviously be considered thus:

ui+lun-i+2 + uiun-iuiun-i+2 + ui-lun-i

The whole fraction (38) therefore equals

11 + -----=-----




Let us take the continued fraction

= qn = I, qi = 2 (i F 0)

Thus, the basis of induction has been proved..Let us now suppose that for any n,

i partial{ 1 +


denominators 1 + ..... , ,'+ 1 + 1

2 + _1_an_i

1 1 1w = 1 + = 1 + un-1 = 1 + =1 2 + 2un + un-1

2 --+ an-l un un


1un un+2+un

= +-- =un+2 un+2

or, putting Uo = 0,

u2un+2 + ulunw =

ulun+2 + uOun

or, in view of what was proved at the beginning ofthis section,

= ui+lun-i+3 + ui un-i+1ui un-i+3 + ui-lun-i+1

w = 1 + 12 +--

n-l partial{l +"'+ 1denominator Jl


Proof of this lenuna is carried out by inductionover i. If i = I, then for any n,


Page 25: Fibonacci Numbers






=P2n+2 P2n+1 P2n+lQ2n+2 - Q2n+1 + Q2n+1


... ,


1· We shall call the expression


an infinite continued fraction.

The definitions and results of the precedingsections can be extended quite naturally toinfinite continued fractions.

We shall show that this sequence has a limit.

be a sequence (obviously an infinite one) of theconvergent fractions of the fraction (39).

Wi th thl' sal'm l' . d .n mIn , we examIne separatelythe sequences

Po P2 P2nQo

,Q2' ••• t



PI P3 P2n+1Ql' Q3'

... ,Q2n+l'

From (33) and (34),

= -1 1Q Q + > 02n+2 2n+1 Q2n+lQ2n

This means that the sequence (41) is an increasingone. In the same way, it follows from


fraction an has n partialthe number of steps of theas applied to the numbers a


The first part of the theorem can bequite simply. It is sufficient to take asFibonacc i number followi ng b, i. e., un+l'

This theorem means that thealgorithm as applied to neighboringnumbers is, in a sense, the "longest".

un+l _-u-- - an'


The continueddenominators, i.e.,Euclidean algorithmand b equals n - 1.


To prove the second part of the theorem, wesuppose the contrary, i . e., that the number ofsteps of this particular algorithm is not lessthan n - 1. Let us expand the ratio alb into thecontinued fraction w. Obviously w will have noless than n partial denominators (in fact one morethan the length of the Euclidean algorithm). As bis not a Fibonacci number, not all the partialdenominators of w will be unity, and therefore,according to the corollary of the lemma in section5, b > un' which contradicts the conditions ofthe theorem.

~' Theorem. For a certain a, the number of stepsin the Euclidean algorithm applied to the numbersa and b equals n - 1 if b = un' and for any a itis less than n - 1 if b <un'

Q> un+l'

Proof:proveda theThen,

A $ubstantial role is played here, of course,by the lemma of section 3, on the basis of whichwe obtain only fractions in their lowes~ terms inthe process of "contracting" a continued fractioninto a vulgar one. Therefore no diminution ofnumerators and denominators of the fractionsobtained due to "canceling" will take place.

Page 26: Fibonacci Numbers


Since a rational number can always be expandedinto a finite continued fraction. it follows fromthe foregoing proof that it cannot be expandedinto an infini te continued fraction. It followsthat the value of an infinite continued fractionmust necessarily be an irrational number.

The theory of expansions of irrational numbersinto continuous fractions represents a branch ofnumber theory which is rich in content andinteresting in its results. We shall not delvedeeply into this theory. but we shall consideronly one example connected with Fibonacci numbers.

the same limit. which is also. obviously. thelimit of (40). This limit is called the value ofthe infinite continued fraction (39).

Let qO. ql. q2' ... and qa. ql. q2 •... betheir corresponding partial denominators. Weshall show that it follows from the equati on w =w· that qo = qa. ql = ql' q2 = q2 •... and so on.

Let us prove now that any number can be thevalue of no more than one continued fraction. Letus take for this purpose two continued fractions wand w· (it does not matter whether they are finfteor infinite).

Indeed. qo is the integral part of the numberwand qa is the integral part w·. so that qO = qb.Further. the continued fractions wand w· can berepresented respectively in the form

where wI and wl are again continued fractions. Itfollows from w = w· and qo = qO that wI = wl also.,This means that the integral parts of the numberswI and wI are also equal. i. e. • ql = q'l'Continuing these arguments (induction) we see thatq2 = q2' q3 = q3' etc.


1 1=----- - <0

Q2n+3Q2n+2 Q2n+2Q2n+1

1 1---<-.Qn+1Qn n2=


From the above considerations. it is possibleto conclude that the sequences (41) and (42) have

From (33) and (34).

and therefore. as n increases. the absolute valueof the difference of (n+1) th and nth convergentfractions tends to zero.

P2n P2m+1--<--Q2n Q2m+1'

Comparing (43). (44) and (45). we obtain


and from the fact that (41) increases and (42)decreases. it follows that

and let us take the odd number k to be greaterthan 2n and 2m+1. It follows from (33) that

Any term of the sequence (42) is greater thanany term of the sequence (41). Indeed. let usexamine the numbers

that the sequence (42) is a decreasing one.


Page 27: Fibonacci Numbers

Therefore, in view of the results of section 4,

8. Let us find the value of the infinitecontinued fraction


55= 34 = 1.6176,

and we obtain

and a = 1.6180. As we see, the error is less than0.1'1...

lim an = a.n-.ao

Of the errors involved in the approximatecalculation of irrational numbers by convergentfractions, it turns out that the number arepresents the worst case. Any other number isdescribable by means of its convergent fractionsin some sense more exactly than a. However, weshall not stop to consider this circumstance,interesting though it be.

The theorem that has been proved means thatthe ratio of neighboring Fibonacci numbersapproaches a as their suffixes increase. Thisresult can be used for the approximate calculationof the number a. (Compare the calculation of unin I section 12.) This calculation produces a, . .very small error, even when small Flbonacclnumbers are taken. For example (correct to thefifth decimal place),

( its absolutecontinues to

to infinity


I1 + 11 + 1 + ...

As we have proved above, this value is lim an'n_aoLet us calculate this limit.

an+llim un+1 = -- + 9n+1

lim an lim 15 =n_ao un n_ao~/5+ en

a + 6n+1/5lim (a + en+1/5)

an n_ao an= lim

6n/5 =(1 +

9n/5 )"n_ao1 + lim

an n_ao --;n


anun = 15 + 6n ,

where 16nl<1/2, whatever n is.

As has already been established in I, section12, un is the nearest whole number to an/IS; thismeans that


O.=n_ao an

Also, for the same reasons,


0,----n_ao an

But 9n+1 /5 is a bounded quantityvalue is less than 2} and anincrease indefinitely as n tends(because a> 1). This means that

Page 28: Fibonacci Numbers


From well known


Figure 3

2 . 3600R sIn 2.10 '

We now calculate sin 180 ,

formulas of trigonometry we have

sin 360 = 2 sin 180 cos 180 ,

cos 360 = 1 - 2 sin2l8° ,

i.e., it is 2R sin 180 ,

then it follows from (*) that

sin 720 = cos 180 f 0,

so that


The side alO of the regular decagon (figure3) inscribed in a circle of radius R is .qual to

£. The Golden section appears quite frequently ingeometry.




1 + .(52 = a

Figure 2



2 2(1 + J5)x = -1 + 15 = (-1 + /5) (l + /5) =




x2 = 1 - x.

The positive root of (47) is -1 + 15 so that2

the ratios in proportion (46) are equal to

each. Such a division (at point Cl) is calledthe median section. It is also called the Goldensection.

If the negative value of the root of theequation (47) is taken, the point of section C2lies outside the segment AB (this kind of divisionis called external section in geometry), as shownin figure. 2. It is easily shown that here, too,we are dealing with the Golden Section:



For this purpose, we denote the length of thegreater part of the segment by x. Obviously, thelength of the smaller part will be equal to 1 - xand the conditions of our problem give us theproportion,

1 x

1. Let us divide the unit segment AB into twoparts in such a way (figure 2) that the greaterpart is the mean proportional of the smaller partand the whole segment.



Page 29: Fibonacci Numbers



Figure 4



But from the definition of the Golden Section,

It is left to the reader to prove that the

The angle AFD equals 1080 , and the angle ADFequals 36°. Therefore, according to the sine rule,

AD sin 10So sin 720 1 + IS-= = = 2 cos 36° = 2 = a.AF sin 360 sin 36° 4

Since it is obvious that AF = AC, then

AD AD-- = a,AF AC

and the segment AD is divided at C according tothe Golden Section.

ACCD = a.

Noting that AB = CD, we obtain

AC ABAB = BC = a.

Thus, of the segments BC, AB, AC, AD, each is atimes greater than the preceding one.



-1 - f54

/5, X3 =


-1 +

4x2 =

other words, a10 equals the largerradius of the circle, which has

by means of the Golden section.


1Xl =-,



and therefore sin 18° is one of the roots of theequation

Factorizing the left-hand side of the latterequation, we obtain


1=4x(l-2x2 ),

15-1 15-1 RalO = 2R 4 = R 2 = a'

Inof thedivided

~. Let us examine a regular pentagon.diagonals form a regular pentagonal star.


1 = 4 sin 180 (l

In practice, in calculating alO we can usethe ratio of neighboring Fibonacci numbers (I,section 12 or III, section 8) instead of a, andreckon approximately that alO is 8/13 R or even5/8 R.

As sin 180 is a positive number, other than112, therefore sin 18° = 15 - 1 .


8x3 - 4x + 1 = O.

(2x - 1)(4x2 + 2x - 1) = 0,

Page 30: Fibonacci Numbers


Figure 6

Figure 5





Figure 7




This means that

ABAD = a.


also, AD = AE = EF since AEFD is a square.

Now. let the ratio of the sides of a rectanglebe equal to a. (We shall call such rectangles"Golden Section rectangles" for short.) We nowprove that after inscribing the largest possiblesquare into a Golden Section rectangle (figure 7),we again obtain a Golden Section rectangle.

EF AB - EB a2EB = = - l.EB

But a2 - 1 = a, so that

EFEB = a.

It is shown in figure 8 how a Golden Sectionrectangle can be "nearly completely" exhausted bymeans of squares I. II, III. ... Each successivetime a square is inscribed. the remaining figureis a Golden Section rectangle.

also holds.


AD-= aAE




If a rectangle whose s ides are to each otheras neighboring Fibonacci numbers is divided intosquares (figure 6). then on the basis of III.section 4. all squares except the two smallestones are different.


The arguments in II. section 2 show that sucha process in the case of whole a and b correspondsto the Euclidean algorithm as applied to thesenumbers. The numbers of squares of equal size isin this case (III. section 1) equal to thecorresponding partial denominators of theexpansion of alb into a continued fraction.

4. Let us take a rectangle with sides a and bandproceed to inscribe in it the largest possiblesquares. as shown in figure 5.



Page 31: Fibonacci Numbers

Figure 9

Various idealist philosophers of ancient andmedieval times raissd the outward beauty of GoldenSection rectangles and other figures which conform

i. Golden section rectangles seem "proportional"and are pleasant to look at. Things of this shapeare convenient in use. Therefore, many"rectangular" objects of everday use (books,matchboxes, suitcases and similar things) aregiven this particular form.


Figure 10

to the rules of median section into an aestheticand e.ven a philosophic principle. They tried toexplaln natural and social phenomena in terms ofthe Golden Section and certain other numberrelationships, and they carried out all kinds ofmystic "operations" on the number a and itsconvergent fractions. It is clear that such"theories" have nothing in common with science.

Figure 11


To do this, we take a square of side 8 and cutit up into 4 parts as shown in figure 10. We putthe parts together to form a rectangle (figure 11)of sides 13 and 5, i.e., of area equal to 65.

Q.. We shall round off our presentation wi th ali ttle geometrical joke. We shall demonstrate a"proof" that 64 = 65.

The explanation of this phenomenon, puzzlingat first sight, is easily found. The point isthat the points A, B, C and D in figure 11 do notreally lie on the same straight line, but are thevertices of a parallelogram, whose area is exactlyequal to the "extra" unit of area.

Figure 8






The reader should compare these arguments withsections 4 and 8 of the preceding chapter. Wenote that if a Golden section rectangle I andsquares II and III are inscribed in a square, asshown in figure 9, the remaining rectangle turnsout to be a Golden Section rectangle also. Theproof of this is left to the reader.





Page 32: Fibonacci Numbers


This plausible but misleading "proof" of astatement which is known beforehand to beincorrect (such "proofs" are called sophisms) canbe carried out even more "convincingly" if we takea square of s ide equal to some Fi bonacc i numberwith a sufficiently large even suffix. u2n.instead of a square with side 8. Let us cut upthis square into parts (figure 12):

Figure 12

and let us put these parts together to form arectangle (figure 13):

Figure 13

The "empty space" in the form of a parallelogramstretched along the diagonal of the rectangle isequal in area to unity. according to I. section 6.It is easily calculated that the greatest width ofthis slit. i.e .• the height of the parallelogram.is equal to


Ju 2 + u22n 2n-2

If. therefore. we take a square with side 21cm and "convert.. it into a rectangle with sides 34cm and 13 em. the greatest width of the slit isfound to be

i.e .• about 0.4 nun. which is difficult to detectby eye.

Page 33: Fibonacci Numbers



Not all the problems connected with Fibonaccinumbers can be solved as easily as the ones wehave considered. We shall indicate severalproblems the answers to which are either not knownat all or can only be obtained by quitecomplicated means with the application of muchmore powerful methods of investigation.

1. Let un be divisible by a certain primenumber p, while none of the Fibonacci numberssmaller than un is divisible by p. In this casewe shall call the number p "the proper divisor ofun" . For example, 11 is the proper divi sor ofu10' 17 is the proper divisor of u9' and so on.

It turns out that any Fibonacci number exceptul' u2' u6 and u12 possesses at least oneproper divisor.

I. The natural question arises:suffix n of the Fibonacci numberdivisor is the given prime number p?


is theproper

From II, section 13 we know that n < p2 . Itis pos sible to prove that n< p + 1. Furthermore,it is possible to establish that if p is of theform St ± 1, then up_1 is divisible by p, and ifp is of the form st ± 2, then Up+l is divisibleby p. However, we have no formula to indicate thesuffix of the term with the given proper divisor p.

I. We have proved in II, section 9 that allFibonacci numbers with composite suffixes, exceptu4' are composite themselves. The converse isnot true, since, for example, u19 = 4181 = 37 x113. The question arises: is the number of allprime Fibonacci numbers finite or infinite? Inother words, is there among all the primeFibonacci numbers a greatest one? At this momentthis question is still far from being solved.