First Awakenings and Fibonacci: the Dark Ages Refers to the Period Ca

First Awakenings and Fibonacci:

The Dark Ages refers to the period ca. 450-1050 in Europe. Originally it was applied to the period after the fall of Rome and before the Italian Renaissance (ca. 1400) but present usage refers to the period ca 1050-1400 as the Middle Ages to acknowledge a somewhat more enlightened period. After the fall of Rome, the only unifying force was the Christian Church. The Church did support education of the clergy and there were monastic schools where the trivium (rhetoric, grammar and logic) was taught and “logic” was little emphasized. Later the quadrivium (arithmetic, astrology, harmonia and geometry) was also taught and lay people might hire teachers. The books available were few: It was the fall of Toledo in 1085 and of Sicily in 1091 to the Christians that gave Europe access to the learning and the libraries of the Arab world and marked the end of the Dark ages. Cordoba’s library had 600,000 volumes The Dark Ages were marked by invasions: the first ones helped undermine the Roman empire; later invasions were by Magyars, Saracens and especially the Vikings. There was great insecurity and little political unity. One exception was the Carolingian dynasty in France. The grandfather of Charlemagne (Charles Martel) had helped to stop the advance of the Moors at the Battle of Tours in 732. On Christmas Day 800 AD, Pope Leo III crowned Charlemagne the first Holy Roman Emperor. There was a brief period, the “Carolingian Enlightenment” when education and learning were encouraged. Alcuin of York was recruited to oversee the education and he decided every abbey was to have its own school and the curriculum was to include the quadrivium and trivium both. At first it was the church educating clergy but slowly interest in education grew. Read about Gerard of Cremona (1114-1187) and Adelard of Bath who translated some of the great works into Latin and made them available to Europeans. Some of the translations of Greek works were of Arabic translations and some were of the original Greek. Gradually the a great heritage of learning trickled into Europe. Leonardo of Pisa (Fibonacci) (1175-1250) wrote Liber Abaci and Liber Quadratorum. In the former he was roughly the first to introduce “arabic” numbers into Europe. This work clearly demonstrated the superiority of the new number system over Roman nu- merals although Gerard of Cremona (1114-1187) introduced them to Toledo (Spain) in his many translations. In this work Fibonacci also introduced the practise of expressing fractions as one number witten over another. He however wrote in the reversed order of the Arab authors: 1 7 7 1 5=5+ + 3 10 10 30 Although Fibonacci did transmit Arabic mathematics to Europe he also was an original mathematician in his own right. Read how Fibonacci solved a problem proposed by John of Palermo who was a member the retinue of Emperor Frederick II (Holy Roman Emperor 1194-1250) x2 +5= u2 and x2 5 = v2 (x = 41/12). This lead him to write Liber Quadratorum where he solved systems− like

x2 + x = u2 x2 x = v2 − 2

(x = 25/24). Read Fibonacci’s solution on page 281. He was happy just to ﬁnd one solution. Fibonacci also solved the cubic x3 +2x2 + 10x = 20. How many solutions? One because the derivative 3x2+4x+10 > 0 for all x and so f(x)= x3+2x2+10x 20 is always 3 − increasing. We don’t need to use calculus: f(x)=(x +2/3) + (10 4/9)x 20 8/27 by “completing the cube.” Fibonacci proved that there was no rational− solution− − and then found the solution 1.3688081075 correct to 9 decimals. Fibonacci did not reveal his method but of course f(1.3688081074) < 0 and f(1.3688081076) > 0. Example: Let’s try the cubic equation x3 +3x2 +5x = 8 by Fibonacci’s method. First check for a rational solution a/b where gcd(a,b) = 1. Substitute to ﬁnd that a3 +3a2b +5ab2 =8 b3 From this we see that b3 divides a3 +3a2b +5ab2. Therefore b divides a3 +3a2b +5ab2 and so b a3 and so b a. But gcd(a,b)=1 and so b = 1. That says that there is an integer solution| of our equation.| However f(x) = x3 +3x±2 +5x 8=(x + 1)3 +2x 9 is an increasing function and so there is one and only one root− and it must be positive− but f(1) = 1 > 0 and so there is no integer root and hence no rational root. We can guess a root like x =0.9 and compute f(.9) = 0.343 so that the solution is greater than 0.9 but less than 1. Fibonacci went on to show− that his root is not of the form

a √b, √a √b, a √b ± ± q ± where a and b are rational numbers. These numbers are all constructible and so Fibonacci made the initial steps to showing that the root is not constructible. In fact a solutions of a cubic equation with integer coefficients is constructible if and only if it is rational. It is not difficult to see that our solution cannot be of the form √y where y is rational because if it were then ± √y[y2 +5]+3y =8 ± and this would say that √y is rational but we already showed that the equation has no rational solution. Besides Fibonacci there was Jordanus Nemorarius or Jordanus de Nemore (ca 1225) who wrote De Numeris Datis. This book marks the earliest European use of letters to represent unknowns. Jordanus also wrote a treatise on Euclidean geometry De trian- gulus which did not include new results but new proofs. European mathematics was reawakening but not flourishing yet. 6.3 The Fibonacci Sequence: This sequence was first mentioned by Fibonacci in the following context A man put one pair of rabbits in a certain place surrounded by a wall. How many pairs of rabbits can be produced from that pair in a year, if the nature of these rabbits is such that every month each pair bears a new pair which from the second month becomes productive? After two months there is the original pair and a newborn pair: that is 2 pairs. At the end of the third month there is a newborn pair from the original pair and the two pairs of the second month: 3 pairs. At the end of the fourth month there will be two 3 newborn pairs from the adult pairs alive two months ago plus the three of the month before: 5 pairs. At any stage the number of pairs at the end of the nth month is the number alive at the end of the (n-1)st month plus one new pair for every adult pair that was around in the (n-2)nd month: 1, 2, 3, 5, 8, 13, 21, .... At the end of 12 months there are 377 pairs ( see the test page 267). We write: F1 = 1, F2 = 1, F3 = 2, F4 = 3, F5 =5, ,F6 =8, .... This is an example of a sequence that can be defined inductively by the formula

Fn = Fn−1 + Fn−2 for all n 3. Combined with the initial definitions F1 = 1, F2 = 1 this completely specifies the≥ sequence. Despite the name Fibonacci sequence the sequence was really first brought to attention by Edouard Lucas (1842-1891) who named it after Fibonacci because of the above cited example but he Fibonacci is not responsible for even the fundamental results. Theorem: Two consecutive Fibonacci numbers are relatively prime: gcd(Fn+1,Fn)=1.) Proof: Let d be any common divisor of Fn+1 and Fn Since Fn−1 = Fn+1 Fn, we − have d Fn−1. Now apply the same reasoning again to conclude that d Fn−2. And so on. | | Therefore d F2 =1 and so d = 1 and gcd(Fn+1,Fn)=1. | ± Read in the book a straightforward argument (which adds Fk = Fk+2 Fk+1, 1 k n) − ≤ ≤ n F1 + F2 + F3 + ... + Fn−1 + Fn = Fk = Fn+2 1 (1) − Xk=1 which generalizes easily to: if n¿m

Fk = Fn+2 Fm+1 − kX=m Consider Problem 8 page 272.

n 2 2 2 2 2 2 F1 + F2 + F3 + ... + Fn−1 + Fn = Fk = Fn+1Fn (2) Xk=1

2 We write Fk = Fk(Fk+1 Fk−1) for k 2. Add these up − ≥ 2 F2 = F3F2 F2F1 2 − F3 = F4F3 F3F2 2 − F4 = F5F4 F4F3 . − . 2 Fn = Fn+1Fn FnFn−1 n − 2 2 Fk = Fn+1Fn F2F1 + F = Fn+1Fn − 1 Xk=1 4

A further property is

2 n−1 Fn = Fn+1Fn−1 +( 1) for n 2 (3) − ≥ We will try a proof by induction on n. To initialize the argument consider the case n = 2: 2 1 F2 = 1 and we compare that to F3F1 +( 1) =2 1 = 1 The formula is therefore true for n = 2. Assume it is true for n and consider− the− formula in the case n + 1 replaces n.

2 n Fn = Fn+2Fn +( 1) +1 − Expand the left side.

2 2 Fn+1 =(Fn+2 Fn)(Fn + Fn−1) = Fn+2Fn Fn +(Fn+2 Fn)Fn−1 − − − n−1 = Fn+2Fn [Fn+1Fn−1 +( 1) ]+(Fn+2 Fn)Fn−1 − n − − = Fn+2Fn +( 1) +(Fn+2 Fn Fn+1)Fn−1 − − − and, since the expression in parentheses is 0, this veriﬁes the induction step. Fn+1 The Golden Ratio: = α where α = (1+ √5)/2 1.619 is the golden ratio. Fn ≈ Veriﬁcation: We will consider the reciprocal.

2 n−1 n−1 Fn Fn Fn+1Fn−1 +( 1) Fn−1 ( 1) = = − = + − Fn+1 Fn+1Fn Fn+1Fn Fn Fn+1Fn Now apply this iteratively with

n+m−1 Fn+m Fn+m−1 ( 1) = + − Fn+m+1 Fn+m Fn+m+1Fn+m n+m−2 n+m−1 Fn+m−2 ( 1) ( 1) = + − + − Fn+m−1 Fn+mFn+m−1 Fn+m+1Fn+m n−1 n+m−2 n+m−1 Fn−1 ( 1) ( 1) ( 1) = + − + ... + − + − Fn Fn+1Fn Fn+mFn+m−1 Fn+m+1Fn+m

The sum consists of m terms. Clearly Fn n 1 for every n and so that sum is at most ≥ − k=m 1 k(k 1) Xk=n − and this partial sum goes to 0 as n by comparison with the series k−2. This shows → ∞ that Fn/Fn+1 converges. Therefore the reciprocal also converges to α say. What does it converge to?

Fn+1 Fn + Fn−1 lim = lim n→∞ Fn n→∞ Fn 1 α =1+ α 5 so that α2 α 1 = 0. Since α> 0 we have the result. 2 − − In Problem 5, p272 the following theorem is stated If gcd(m,n)= d then gcd(Fn,Fm) = Fd. For example gcd(F9,F6) = F3 or gcd(34,8) =2. Possibly of interest is the formula

Fn = Fn+1−kFk + Fn−kFk−1 for 1 k n 1 ≤ ≤ − Fibonacci and the Pythagorean Problem: Recall that at triple of positive integers (x,y,z)isa Pythagorean triple if x2 + y2 = z2 and a Pythagorean triple is primitive if gcd(x,y) =1. Euclid’s Elements and Diophantus’ Arithmetica both stated the following method for generating Pythagorean triples x =2st, y = s2 t2 z = s2 + t2 − What Fibonacci showed was that all primitive Pythagorean triples were of this form. His proof may be from Arab mathematicians who apparently knew this result. We assume that the last listed member of the triple is the largest. Lemma Let (x,y,z) be a primitive Pythagorean triple. Then x or y is even and the other is odd. Proof: Clearly at most one of x and y can be even since gcd(x,y) =1. Suppose that neither was even: x =2h +1 and y =2k +1. Then z2 =4h2 +4h +1+4k2 +4k +1=4(h2 + h + k2 + k)+2 This says that 2 z2 but 4 ∤ z2 which is impossible because z2 is a perfect square. 2 We shall choose| x to be even. Theorem The triple (x,y,z) is a primitive Pythagorean triple if and only if there exist two relative prime integers s and t so that s>t> 0 and x =2st y = s2 t2 and z = s2 + t2 − where one of s and t is even and the other is odd. Proof: Clearly if s and t exist then (x,y,z) is a Pythagorean triple. (See the text, page 296). We need only check that it is primitive. If d is a common divisor of x and y then d2 z2, and so d z. Therefore d (s2 + t2) and this says d is not even (because s and t are even| and odd).| Also d (s2 t|2) so d 2s2 and d 2t2 and so d s2 and d t2. If d 2 then this says s2 and t2 have| a common− prime| factor.| That is impossible| (gcd(| s,t) =1)≥ d = 1. Conversely± suppose that (x,y,z) is a primitive Pythagorean. We would like to set s2 =(z + y)/2 and t2 =(z y)/2 but of course that is problematic since we don’t know either of these is a perfect− square. We therefore introduce z + y z y u = v = − 2 2 Because z and y are odd, u and v are integers. Observe that gcd(u,v) =1 because if u and v have a common factor d then d (u+v) that is d z and similarly d y and so d = 1. Moreover | | | ± z + y z y z2 y2 x 2 uv = − = − = 2 2 4 2 6

That is uv is a perfect square. This and the fact gcd(u,v) =1 implies u and v are perfect squares: u = s2 and v = t2. To check this observe that every prime p in the prime decomposition of uv must appear to an even power: p2ν uv. Of course p must divide either u or v but because gcd(u,v) =1 it can’t divide both| and so if p2ν divides u or v. Therefore every prime in the prime decomposition of u appears to an even power and similarly for v. We have therefore verified the relationships between x, y, z and s and t. We further know that gcd(s2,t2) =1 and so gcd(s,t)=1. We also know s>t> 0 because u>v> 0 by their definitions. Finally it remains to check that s or t is even but not both. Certainly gcd(s,t) =1 implies they are not both even. On the other hand if both were odd then z2 = s2 + t2 would be even which we know is not possible for a primitive Pythagorean triple. So exactly one of s and t is even. 2 Theorem: If (x,y,z) is a Pythagorean triple than either x or y is divisible by 3. Proof: We may suppose that (x,y,z) is primitive. By the preceding Theorem x =2st and y = s2 t2 = (s t)(s + t). Thus we need only check that s or t or s t or s + t − − − must be divisible by 3. Every integer is of the form 3m or 3m +1 or 3m 1 for some integer m because the set of all multiples of 3 forms a lattice of integers from− which all integers differ by at most one. We may suppose that neither s nor t is divisible by 3. If s =3m +1 and t =3n +1 then s t is divisible by 3. If s =3m +1 and t =3n 1 then s + t is divisible by 3. Similarly if−s =3m 1 and so either x or y is divisible by− 3. 2 Read about Fibonacci’s solution to the− tournament problem on page 297-298.