The Quiet Milleniumc© Ken W. Smith, 2012

Last modified on March 22, 2012

Contents

3 The Quiet Millenium 23.1 Mathematics in China . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

3.1.1 Chinese History . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.1.2 Nine Chapters and Nine Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33.1.3 Chinese mathematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43.1.4 The Chinese Remainder Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

3.2 Mathematics in India . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93.2.1 A brief history of India . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93.2.2 Brahmagupta . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93.2.3 Bhaskara . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3.3 Mathematics in Arabia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.3.1 History of Arabia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.3.2 Al-Khwarizmi and others in the School of Wisdom . . . . . . . . . . . . . . . . . . 123.3.3 Omar Khayyam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.3.4 The Method of False Position . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3.4 Math in Europe during the Middle Ages . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.4.1 A summary mathematical work prior to the Middle Ages . . . . . . . . . . . . . . 173.4.2 Fibonacci and the reawakening of Europe . . . . . . . . . . . . . . . . . . . . . . . 173.4.3 The rise of universities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.4.4 The rise of algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.4.5 The Renaissance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.5 The cubic and quartic equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.5.1 The Italians of the Sixteenth Century . . . . . . . . . . . . . . . . . . . . . . . . . 203.5.2 The quadratic equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.5.3 The general solution to the cubic equation . . . . . . . . . . . . . . . . . . . . . . . 213.5.4 The necessity of complex numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.5.5 Rational Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.5.6 Quartic Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.6 The French and the early Seventeenth Century (under construction) . . . . . . . . . . . . 26

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3 The Quiet Millenium

The development of mathematics paused and stagnated in Europe between 400 AD and 1400 AD. Whilewe consider this “Quiet Millenium” in Eurpoe, we also take a moment to fill in what is known about thedevelopment of mathematics outside of Europe prior to the European Renaissance.

Three civilizations developed mathematics along strands that were originally independent and even-tually began to intertwine.

China had its own mathematics which, in the time of the Babylonians and Egyptians, may have beensomewhat similar to the ancient mathematics of those two empires. The mathematics of China developedindependently of the western cultures (with some possible interaction with India) until the Middle Ageswhen there was some commerce with Europe.

There was probably some commerce between India and Greece prior to Alexander’s invasion in 327BC but certainly after that invasion, there was considerable Greek influence in India.

In the seventh century AD we see the rise of Islam in Arabia and the quick conquest of north Africanregions including Alexandria. The arabic culture soon recognized the learning in Alexandria and beganto translate and then build on the Greek mathematics.

We will look at all three cultures, China, India and Arabia, before moving back to Europe and themathematics of the late Middle Ages.

3.1 Mathematics in China

We briefly summarize the mathematics in China. (My sources are chapter 7 in the textbook by Eves,section 5.5 of Burton, pages 318-349 of Swetz and Wikipedia’s articles on Chinese history and ChineseMathematics.)

3.1.1 Chinese History

Ancient China, until 600 BCThe culture of ancient China, beginning around 1600 BC, rivaled that of Egypt and Babylon. Un-

fortunately, most of the writings from the time have disappeared because they were written on bamboo,which easily decayed and because there was a deliberate attempt in the Qin dynasty period to destroyall ancient texts.

Ancient China include three major dynasties. In ancient times there was apparently the Xia dynasty(2100-1600 BC?). Details on it are somewhat mythical; it is apparently mentioned in writings from theShang dynasty (1600 – 1027 BC) which followed. the Shang dynasty is the earliest dynasty for which wehas strong evidence. It was followed by the The Zhou (Chou) dynasty (1027 BC – 256 BC).

The Zhou dynasty, 1027 BC – 256 BCDuring the Zhou dynasty we see the rise of Confucianism. This time brings in a golden age of

philosophy similar to that occurring in Greece about the same time. (Confucius lived from 551-479 BC,roughly a century after Thales of Miletus in Greece and roughly the same time as Siddharta GautamaBuddha in India.)

During the Zhou dynasty we see the Hundred Schools of Thought, during which numerous schoolsand philosophies arose. This is a time in which, as in Greece, there is a leisure class with time to askabout the meaning of life and the make up of the universe.

Four schools of thought (or philosophy) rose during this time: Confucianism, Mohism, Taoism andLegalism. This is sometimes viewed as “Classical China” (Eves says that period begins around 600 BCand continues to 221 AD.)

The beginning of Imperial ChinaThe Qin (Chin) dynasty was short – fifteen years – but influential (221 – 206 BC). It began with the

conquest of all of China and the unification of China. It ended with the death of the first emperor andquickly morphed into the Han Dynasty.

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During this dynasty we have the beginning of the Great Wall. We also have the ”book burning”program which eliminated many ancient works (and eliminated many scholars of the time.)

Surviving this program is the book “Nine Chapters of the Mathematical Art”.)The Han dynasty continued the unification of China (206 BC – 221 AD). During this time Con-

fucianism became the state religion. This dynasty traded with Rome (in the second or third centuryAD.)

Imperial China and the Tang DynastyThe Han dynasty collapsed in 221 AD and was followed by an Imperial China lasted about two

thousand years. It included the powerful Tang Dynasty (618 – 907), and eventually three final dynastiesduring which the Europeans traded with China and exchanged information:

1. Yuan Dynasty (1271 – 1368)

2. Ming Dynasty (1368 – 1644)

3. Qing Dynasty (1644 – 1911)

In the twelfth century AD, Genghis Khan rode out of Mongolia with a great army and began the conquestof China. His grandson Kublai Khan (1216-1294) conquered all of China and much of the nearby territory,setting up the Yuan dynasty.

Notable events in Chinese history (from a western viewpoint) include the visits of Marco Polo (1275-1292) to the courts of Kubali Khan.

3.1.2 Nine Chapters and Nine Sections

“Nine Chapters of the Mathematical Art” was an ancient Chinese text on mathematics that predates theQin dynasty. Ideas in the text may be as old as 500 BC. to 1000 BC.

The contents of the Nine Chapters are discussed in some depth in Swetz, pages 330-337. The NineChapters include surveying area (for agriculture), proportions and percentages (for business), arithmeticand geometric progressions and the “rule of 3”, square root and cube root computations (mainly formeasuring land?), computations of volumes with engineering applications, the method of false positionto solve linear equations (this was called “excess and deficiency” since they would employ two guesses,one high and one low.) There were problems on right angles, applying the Pythagorean theorem.

This work includes

1. Early decimal positional system. This includes the abacus where the digits were visible as beadson strings.

2. Magic squares and number play (see Eves, 238, exercise 7.3 on magic squares.)

3. The rule of 3 and rule of 9

4. The rule of false position (also used by the Greeks)

5. Linear equations

6. Pythagorean Theorem (this was chapter nine of the “Nine Chapters”)

7. Estimates for π.

The Sea Island Mathematical Manual was a commentary on the Nine Chapters and is named for afirst problem that involves measuring the size of an island.

In this manual we have some approximations for π.

During the Imperial period (221 AD - 1911 AD) we see a significant mathematical work called Math-ematical Treatise in Nine Sections, not be confused with the much earlier Nine Chapters!

This later work includes

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1. Solutions to some cubic equations

2. a symbol for zero

3. Horner’s method for evaluating a polynomial

4. decimal fractions

5. some linear algebra

6. negative numbers

7. more on π.

8. Pascal’s triangle (see p. 218 in Eves.)

The Chinese Remainder Theorem appears here. Also in the Nine Sections is an attempt at a solution toa cubic equation (says Eves, p. 215.)

The Nine Chapters has considerable rhetorical and syncopated algebra?

3.1.3 Chinese mathematics

We examine some of the Chinese contributions to mathematics.One ancient idea is the magic square, dating back to prehistoric times. Indeed, the first magic square

was supposedly given to an earlier emperor by a Lo river tortoise.A magic square is an array of integers in which the row sums, column sums and diagonal sums are all

the same.Here is the Lo shu magic square technique. (From Swetz, pages 320-1.)Consider the first nine natural numbers:

9 8 7 6 5 4 3 2 1

Write the first nine natural numbers in order in a square:

7 4 18 5 29 6 3

Notice that we use the Chinese habit of going down the page and then moving from right to left!Now rotate this 90 degrees counterclockwise into a diamond:

14 2

7 5 38 6

9

Exchange opposite corners of the diamond:

94 2

3 5 78 6

1

Compress (don’t rotate!) into a square (so that the 3 on the left slides in under the 4 and 8, etc.):

4

4 9 23 5 78 1 6

We can make even bigger squares – let’s do a 9 × 9 square the Lo shu way. Write out the first 81natural numbers in a square.

73 64 55 46 37 28 19 10 174 65 56 47 38 29 20 11 275 66 57 48 39 30 21 12 376 67 58 49 40 31 22 13 477 68 59 50 41 32 23 14 578 69 60 51 42 33 24 15 679 70 61 52 43 34 25 16 780 71 62 53 44 35 26 17 881 72 63 54 45 36 27 18 9

Then take each row and turn it into a square as before. For example, the second row from the topbecomes

56 29 265 38 1174 47 20

=⇒

229 11

56 38 2065 46

74

=⇒

7429 11

20 38 5665 47

2

=⇒29 74 1120 38 5665 2 47

Now we have 9 3 × 3 magic squares, each with exactly one of the first nine integers in it. For example,the square above, has the number 2 in it, so we will think of it as the 3× 3 subsquare with label 2.

Go back to our original 3× 3 magic square and view it this way:

4 9 2

3 5 7

8 6 1

Use this guide to place the nine 3× 3 subsquares into position. For example, since we found a magicsquare with a two in it:

29 74 1120 38 5665 2 47

we then place these numbers around the 2 in the big square:

29 74 1120 38 56

4 9 65 2 47

3 5 7

8 6 1

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The abacusA major tool the Chinese used for their accounting systems was the abacus. This is a system of beads

on strings which replicates our base ten positional arithmetic.

(This image comes from the 12th edition of the Encyclopdia Britannica or earlier and was found on Wikipedia.)

The linear equation, Pythagorean theorem and Pascal’s triangleThe ancient Chinese used the method of false position to solve linear equations in one unknown. The

ancient Chinese could apparently solve a system of linear equations using a counting board and operationssimilar to our row reduction.

Ancient Chinese could take square roots and seemed to have the Pythagorean theorem.

The ancient Chinese seem to have had a version of the binomial theorem (“Pascal’s triangle”.)

3.1.4 The Chinese Remainder Theorem

The ancient Chinese had a method for counting large numbers according to the remainders under divisionby various small numbers. For example, a large number x might have the property that it was odd, was1 more than a multiple of 3, was 2 more than a multiple of five and 3 more than a multiple of 7 and 4more than a multiple of 11. What is x?

In modern notation we would writex ≡ 1 mod 2

x ≡ 1 mod 3

x ≡ 2 mod 5

x ≡ 3 mod 7

x ≡ 4 mod 11

We can solve these congruences a pair at a time. For example, the first pair of congruences is equivalentto x ≡ 1 mod 6 while the third and fourth congruences are equivalent to x ≡ 17 mod 35. (Details onthe method of solving these pairs will come later.) So our congruences reduce to

x ≡ 1 mod 6

x ≡ 17 mod 35

x ≡ 4 mod 11

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Again we take pairs of congruences and replace the first two by

x ≡ 157 mod 210

and finally solvex ≡ 157 mod 210

x ≡ 4 mod 11

to getx ≡ 367 mod 2310.

In other words, the set of all solutions to the original five congruences is 367+2310k where k is an integer.

How is this done?The main engine in this mathematical theory is the Euclidean algorithm.

The CRT and the Euclidean algorithmWe solve the pair of congruences

x ≡ a mod m (1)

x ≡ b mod n

in the following manner. Let d represent (m,n), the greatest common divisor of m and n and let `represent [m,n], the least common multiple of m and n.

Since d divides both m and n then x ≡ a mod d, x ≡ b mod d are necessary for the existence ofsolutions to equation (1) and so a, and b must be congruent modulo d. We suppose then that a ≡ bmod d, or equivalently, that d|(b− a).

Lemma 1. Let s, t be integers such that d = ms+ nt. Then

x0 =m

dbs+

n

dat (2)

is a solution to the congruences and it is unique modulo ` = [m,n].

Proof. First note that since d|m and d|n then x0 is an integer. Note also that, since 1 =m

ds+

n

dt,

x− a =m

dbs+

n

dat− a =

m

dbs+ a(

nt

d− 1).

We replace 1 bym

ds+

n

dt to obtain

x− a =m

dbs+ a(

n

dt− m

ds− n

dt) =

m

dbs− m

das =

m

ds(b− a) =

b− ad

ms.

Since d divides b− a, thenb− ad

is an integer and so x− a is an integer times ms. Therefore m|(x− a)

and so x ≡ a mod m.Similarly, x ≡ b mod n.Two solutions x0 and x1 (to a pair of congruences) will differ by a multiple of the LCM of m and n.

Indeed, given one solution x0, the set of all solutions to the congruences is {x0 + k` : k ∈ Z} where `represents LCM(m,n). 2

We can generalize the Chinese Remainder Theorem to more than two congruences with a slightly morecomplicated formulae. But it is my experience that the best way to solve a collection of simultaneouscongruences is to solve them a pair at a time.

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Let’s do the example above; we wish to solve

x ≡ 1 mod 2 (3)

x ≡ 1 mod 3

x ≡ 2 mod 5

x ≡ 3 mod 7

x ≡ 4 mod 11

We probably don’t need the Euclidean algorithm to tell us that 1 = 3 − 2 is the GCD of 2 and 3 andso we can replace the first two congruences by x ≡ 1 mod 6. But the second pair, x ≡ 2 mod 5, x ≡ 3mod 7 take a moment. Using the Euclidean algorithm we have that 1 = (−2) · 7 + 3 · 5 is the GCD of 5and 7 and so x = (2)(−2)(7) + (3)(3)(5) = 45 − 28 = 17 is a solution mod 35. So our five congruenceshas become the triplet

x ≡ 1 mod 6

x ≡ 17 mod 35

x ≡ 4 mod 11

We tackle the first two congruence here by using the Euclidean algorithm on the moduli 6 and 35 anddiscovering that 1 = (−1)(35) + (6)(6) is the GCD of 35 and 6 and so x = (1)(−1)(35) + (17)(6)(6) = 577mod 210 is a solution to the first pair. But 577 ≡ 157 mod 210 and so we will use x = 157 instead. Thuswe have, finally, just a pair of congruences

x ≡ 157 mod 210

x ≡ 4 mod 11

Again, we use the Euclidean algorithm on the moduli 210 and 11 and write 1 = (1)(210)+(−19)(11) and soset x equal to x = (4)(1)(210) + (157)(−19)(11) = −31973. We reduce this answer mod LCM(210, 11) =2310 and obtain our final answer x ≡ 367 mod 2310.

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3.2 Mathematics in India

3.2.1 A brief history of India

At the time of Alexander the Great (about 330 BC) India was a collection of small kingdoms or maha-janapadas. Alexander, after conquering Persia attempted to conquer India. Although Alexander’s Indiacampaign was initially successful, he eventually retreated from India, apparently because India was sofar from Greece and it was difficult to maintain his armies at that distance. The defeat of some of thesmall mahajanapadas by Alexander, and his ensuing retreat left a political vacuum filled by the Mauryaempire which united India for several centuries afterwards.

During the Maurya empire and afterwards, the Greek influence on India included Greek mathematics.

Gupta EmpireAround 320 AD, six centuries later, India was once again united in the Gupta empire. This empire

led to the “Golden Age” of India which included a variety of educational and artistic advances andachievements.

Our modern symbols first appear in the Indian positional notation. As in Babylon, there are earlyappearances of the “Pythagorean” theorem. Early Indian mathematics, like that of Babylon, focused onreligious requirements.

An ancient Hindu religious ritual, the Srauta, involved geometry in building an altar. The Sulvasutraswere mathematical texts intended to aid in the construction of these altars. It included some earlyexamples of the Pythagorean theorem and some square roots.

Later mathematics focused on astronomy and developed as astronomy developed. Varahamihira (505?-587? AD) lived during the Gupta period (or “Golden Age”) of India. During this time, astronomy was themain motivation for mathematical development in India. There was an observatory at Ujjain in centralIndia (today in the state of Madhya Pradesh.) It was a leading mathematical center there and several leadastronomers at this observatory made major contributions to Indian mathematics. Varahamihira was oneof the early astronomers there. Around this time, an early work, “Knowledge of the Sun”, was written.As the mathematics of astronomy was introduced, so too were the basic concepts of trigonometry.

Aryabhata studied arithmetic and geometric series. He examined the equation

ax+ by = c (4)

where a, b, c are given integers and we seek integer solutions for x and y. This equation can be solved bythe Euclidean algorithm.

There was some early syncopated algebra and a larger emphasis on algebra than in the Greek math-ematics. Because of the algebra, Indian mathematicians were more willing to accept negative numbersand irrational numbers.

There was some early trigonometry.

3.2.2 Brahmagupta

A century later, the astronomer Brahmagupta (598?-688?) wrote a number of mathematical works. Inhis most famous work, he does algebra and diophantine equations, looks at the Pythagorean Theorem,etc. His works include the number zero.

Brahmagupta gave a full solution to the Diophantine equation 4. (A full solution can be obtainedthrough the Euclidean algorithm.)

They extended Heron’s formula to area of a cyclic quadrilateral.

K2 = (s− a)(s− b)(s− c)(s− d)

where K is the area and s the semiperimeter.

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3.2.3 Bhaskara

Five hundred years later, around 1150 AD, the mathematician and astronomer Bhaskara wrote a seriesof works on astronomy and mathematics. One work, called Lilavati, is addressed to his daughter and isintended to teach her mathematics. He begins some of his problems with the phrase, “Tell me, beautifulmaiden with the beaming eyes....”

He has the “Behold!” dissection proof of the Pythagorean Theorem. He uses syncopated algebra. Hedevelops the rule of 9 and the method of false position (guess an answer and then correct at the end. (Inthis way the ”false positions” works in place of a variable.) He uses the method of inversion (start withfinal result and reverse the steps.)

These Indian works

1. preserved the mathematics of Greece,

2. received some mathematical ideas from China,

3. included syncopated algebra along with the method of false position,

4. introduced our modern numeral system, including zero.

The Indian mathematicians also worked with square roots of rationals (these are called surds if thesquare root is irrational.)

Our modern positional number system, along with the symbols, comes from India. Some of theancient mathematical texts (such as a commentary on Bhaskara’s Lilavati) include instructions on howto multiply numbers in the positional system.

Bhaskara taught the method of inversion, that is, working backwards from an answer. If one wereasked, “What number when multiplied by 3, then added to 5 and divided by 2 is equal to 19?” then onereverses the process, first multiplying 19 by 2, etc.

Bhaskara did a number of computations with surds. For example, he developed the identity√a±√b =

√(a+

√a2 − b)/2±

√(a−

√a2 − b)/2. (5)

Pell’s equationBrahmagupta, then Bhaskara, developed solutions to what is now called Pell’s equation. We fix an

integer n (such as n = 2) and compute solutions to

x2 − ny2 = 1 (6)

The method developed by these Indian mathematicians is call the cakravala. It uses the identity (knownto Diophantus) that

(a2 + b2)(c2 + d2) = (ac± bd)2 + (ad∓ bc)2. (7)

Brahmagupta extended this identity to

(a2 − nb2)(c2 − nd2) = (ac+ nbd)2 + (ad− nbc)2. (8)

If (a, b) and (c, d) are solutions to equation (6) then equation (8) demonstrates that

(a, b) ? (c, d) := (ac+ nbd, ad+ bc) (9)

are also solutions. This process creates an operation on solutions to equation (6) that is closed under thisoperation. For example, if n = 2 then choose solution (3, 2) to the equation x2 − 2y2 = 1. Using the pairof solutions (3, 2) and (3, 2) (yes, we can use the same solution!) we obtain the solution

(3, 2) ? (3, 2) := (9 + 8, 6 + 6) = (17, 12) (10)

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With that solution we can obtain another

(3, 2) ? (17, 12) := (51 + 48, 34 + 36) = (99, 70) (11)

and so on, creating an infinite family of solutions to the equation x2 − 2y2 = 1.

Since we can always rewrite equation (6) in the form n =x2 − 1

y2, then the solution to Pell’s equation

give an infinite number of approximations,x

y, to√n. For example, with n = 2 we see that

17

12is close to

√2 and

99

70is closer still!

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3.3 Mathematics in Arabia

3.3.1 History of Arabia

After the introduction of Islam and consolidation of tribes in the Arabian peninsula, a newly arisingArab empire conquered Alexandria around 641 AD. The Arab empire assimilated much of the learningof Alexandria and eventually created a School of Wisdom and an observatory in Baghdad. Baghdadreplaced Alexandria as the center of knowledge an learning.

The Arab culture retained the Greek mathematical knowledge and commented on it. Around 766AD, the work of the Indian astronomer Brahmagupta was translated into Arabic. An observatory wasbuilt in Baghdad at that time.

3.3.2 Al-Khwarizmi and others in the School of Wisdom

Mohammed al-Khowarizmi (780-850) was a member of Baghdad’s House of Wisdom and authored anumber of mathematical tracts and books. He solved linear and quadratic equations and used the Indiannumerals which, with little change, are the ones we use today. He translated Greek works and occasionallyimproved on them.

One of his books discussed “restoration” and “balancing”; by ”restoration”, Al-Khwarizmi meantmoving a negative term to the other side of an equal sign and making it positive there. The work he usedfor ”restoration” was the Arabic word “al-jabr” and that word was used in the title of his book. LaterEuropeans reading his works would speak of “algebra” and it is from the title of that book that we getour modern word. It is also from Khowarizmi’s name that we get the word “algorithm.”

Despite all his work with algebra, Khowarizmi’s algebra was still rhetorical, written out in words.Khowarizmi also wrote works on arithmetic, trigonometry, astronomy and geography.

Others from the School of WisdomOther Arab mathematicians living in Baghdad and teaching at the School of Wisdom include

1. ibn Qurra (826-901) who was an astronomer and algebraist who translated many Greek works.

2. Abul Wafa (940-998) translated works of Diophantus and extended Ptolemy’s computations of sinevalues, creating a table in which the sines of angles (at increments of 1/4 degree) were computedto nine decimal places (Eves, page 231.) Wafa introduced the tangent function and eventually theArabs worked with a trigonometry involving all six trig functions (Eves, p. 235.)

3. al-Karkhi (953-1029) was the successor to al-Khowarizmi (according to an article by Pazwash andMavrigian in Swetz’s book, p. 293.) He wrote on algebra and knew of Pascal’s triangle and thebinomial theorem. He demonstrated a “lattice” or “sieve” method for multiplication of Indiannumerals in base ten positional notation.

4. ibn Sina (Avicenna) (980-1037) wrote on mathematics, philosophy, medicine, geography, Islam andpoetry. He was later known in Europe as Avicenna, a latin version of the Arab ibn Sina. Hewrote commentaries on Appollonius’s Conics and demonstrated how one might draw parts of conicsections using the Euclidean tools.

3.3.3 Omar Khayyam

Omar Khayyam (about 1100 AD) may be the most famous Arabian mathematician. He is rememberedfor his poem, The Rubaiyat but he also wrote a number of works on algebra. He apparently knewof Appollonius’s work on conic sections and commentaries of Arabic mathematicians (such as those ofAvicenna) on conic sections and Khayyam used these ideas in solving cubic equations.

Khayyam had a number of methods for solving cubic equations. For example, according to theWikipedia article on Khayyam and attached picture, Khayyam would solve a cubic of the form x3+ax2 = bby drawing a parabola satisfying x2 = ay (in blue in the drawing below)

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(This drawing released into the public domain by Peter Kuiper)

and a circle (in purple) of diameterb

a2touching the origin. The line segment (in red) on the x-axis would

be the solution to the cubic equation.

Here is a beautiful solution of Khayyam’s to all cubic equations (from Eves, problem 7.15, page 247).This solution uses the Euclidean tools and one additional piece, the ability to draw a particular conicsection, a hyperbola.

Given the cubic x3 + b2x+ a3 = cx2, we construct the following geometric figure.

1. Construct a line segment from A to B of length a3/b2.

2. Continue the line to the point C so that BC = c.

3. On the line segment AC construct a semicircle. (It’s radius is then half of a3/b2 + c.)

4. Construct a perpendicular to AC at the point B. This line touches the circle at D, so that thesegment BD has length

√c(a3/b2).

5. Mark a point E on BD so that BE has length b.

6. Draw a line parallel to AC through E.

This gives us the figure below.

A a3/b2

b

cB C

D

E

7. Consider the rectangle with base AB and altitude BE. Its area is (a3/b2)b = a3/b. Find a point Gon BC so that the rectangle with base of length BG and height of length DE has the same area.(In the picture below, we want the two rectangles DELH and ABEM in blue, to have the samearea.)

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H

A a3/b2

b

B C

D

EM L

G

8. Finally, draw the unique hyperbola through H which has lines EL and ED as asymptotes. This hy-perbola intersects the circle at J . (In the figure below, I’ve drawn the hyperbola and its asymptotesin red.)

9. Drop a perpendicular from J to the line AC. Let N represent the point of intersection of thatperpendicular with AC.

J

H

A a3/b2

b

B C

D

EM L

N

K

G

The claim is that the line segment BN is the solution to the cubic equation! In other words, using BNas a length, we have

(BN)3 + b(BN) + a3 = c(BN)2.

The hyperbola plays an important place here. By definition of the hyperbola, the product of thelengths EK and KJ will be equal to the product DE and BG (the area of the upper right blue triangle.)(In general, a hyperbola with x and y axis as asymptotes will have an equation of form xy = k; that is theproducts of the x and y coordinates is a constant. Here we might imagine (EK,KJ) as the “coordinate”of the point J and (BG,ED) as coordinate of the point H.) The rectangle in green, below, has the samearea as the rectangle in blue.

14

J

H

A a3/b2

b

B C

D

EM L

N

K

G

Using line segments as lengths, we might write these rectangle areas as

(EK)(KJ) = (BG)(DE) = (AB)(BE). (12)

.Add the rectangle BNKE to both of these rectangles to create a new pair of rectangles with equal

area: the rectangle with base BN and altitude NJ has the same area as the rectangle ANKM. Thus

(AN)(BE) = (BN)(NJ). (13)

and so(BE)/(BN) = (NJ)/(AN). (14)

But we know from our Euclidean tools that the square of the length of NJ is equal to the productsof the lengths AN and NC, that is

(AN)(NC) = (NJ)2. (15)

Divide this equation by (AN)2. This forces (NC)/(AN) = (NJ)2/(AN)2. Using equation (14) wesee that

(NC)/(AN) = (NJ)2/(AN)2 = (BE)2/(BN)2. (16)

Since (BE)2/(BN)2 = (NC)/(AN) then

(BE)2(AN) = (BN)2(NC). (17)

We constructed BE so that it had length b and we constructed BC so that it had length c. Substituting,we have

b2(AN) = (BN)2(c−BN). (18)

Since AB has length a3/b2 then AN has length a3/b2 +BN . Substitute a3/b2 +BN for AN and simplifythe right-hand side to get

b2(a3/b2 +BN) = c(BN)2 − (BN)3. (19)

Multiply through the b2 term in the left-hand side and we have

a3 + b2(BN) = c(BN)2 − (BN)3 (20)

which is what we wished to show.

15

3.3.4 The Method of False Position

Guessing a value for the answer, computing with that value and then modifying the guess dependingon the computation, is sometimes called the method of false position. In this method, the “guess”or false number takes the place (“position”) of a variable. Diophantus used this method in some of hisproblems and the Chinese, Indians and Arabs also used this method.

For example, suppose we are to solve the following problem: “One-third of a number plus one-half ofthat number is fifty.”

The modern student writes 13x+ 1

2x = 50 and solves for x. In the method of false position, one noticesthat the fractions 1

3 and 12 can be written with common denominator 6 and so one might start by saying,

“Let my number be six.” Then “one-third of my number plus one-half of that number is 3+2 = 5.” Sincewe really want the final answer to be 50, not five, we multiply our false guess by ten and so check outthat our solution should really be 60 instead of six.

Eves, p 246, has an exercise (7.14) on “double false position.” In this method, used by the Arabs, oneestimates the root of a function by interpolation, guessing two inputs x1, x2 which give positive and nega-tive outputs, say f(x1) > 0 > f(x2). Then the slope of the line joining points P (x1, f(x1)) and Q(x2, f(x2))is

f(x1)− f(x2)

x1 − x2and so the equation for the line joining the two points is

f(x1)− f(x2)

x1 − x2=y − f(x2)

x− x2.

Setting y = 0, we find the x-intercept of that line to be

x3 :=x2f(x1)− x1f(x2)

f(x1)− f(x2).

This is a good estimate for the zero of our equation. If we want more accuracy than this number givesus, we evaluate our function at x3, and then continue the process through another step, using either x1or x2.

For example, if we want to find a zero of f(x) = x3 − 36x + 72 we might notice that f(2) = 8 whilef(3) = −9. So using x1 = 2 and x2 = −9 we have

x3 =3 · 8 + 2 · 9

8 + 9=

42

17≈ 2.47.

We could then evaluate our cubic polynomial f(x) at x3 and discover that f(42

17) = −9144

173. Since this

new output is negative, we can use this new value in place of x2 and go again.This work was all done, of course, without modern synthetic algebra, but used a natural understanding

of the line segment joining the two points P (x1, f(x1)) and Q(x2, f(x2)).

Additional online information about Arab mathematics is available at Wikipedia and the MacTutorarticle on Arab mathematics. There is also (here at www.jphogendijk.nl) a nice bibliography and set oflinks on Arab mathematics.

16

3.4 Math in Europe during the Middle Ages

3.4.1 A summary mathematical work prior to the Middle Ages

We review math history to this point. Before the European Middle Ages, one can briefly summarizemathematics as developing in a series of stages. Primitive societies developed simple numeral systems(including our decimal positional notation) in order to count sheep, count people, collect taxes. Large em-pires developed more detailed business mathematics, including compound interest, the quadratic formula,the Pythagorean theorem, formulae for area and volume. Two major early societies (where mathematicalrecords have been found) were the Egyptian and Babylonian empires.

The Greeks took mathematics to a higher level. Beginning around 800 BC, the Greeks asked philo-sophical questions. Why? Is it always this way? Is there a pattern? Is there an underlying principle?The philosopher Thales (b. before 600 BC in Miletus?) is the first recorded philospher. His emphasison reasoning and concepts made mathematics part of his philosophical thinking. The Pythagoreans, acult or commune started by Pythagoras, worshiped number and emphasized deduction. For the firsttime (as far as we know) mathematical concepts were deduced (proven.) Concepts, not procedures, wereemphasized. Understanding was valued over rote-learning. To the Greeks, mathematics was real, it wasvisual. A number was the same as length. A number squared was a square of those sides. Lets look atsome of the Greek mathematics. (Why would there be no 4th power?)

The Roman empire adopted many of the Greek ideas (including Greek engineering) but mathematicalexploration died.

The Roman Empire replaced Alexander’s Greek empire and for five hundred years (beginning in 44BC with Julius Caesar’s rise to power and ending in 467 BC with the overthrow of Rome by Germanictribes), the Roman empire dominated the Mediterranean world and had influence as far away as India andChina. During the growth of the Roman empire, Greek was slowly replaced by Latin and the writings ofthe Greek scholars disappeared in Europe. For a time, Greek education continued in Alexandria, Egypt,across the Mediterranean from Europe but Alexandria was overrun several times, eventually sacked byArabian forces in 641 AD.

The Eastern Roman empire (the Byzantine Empire) lasted another thousand years after the fall ofRome.

After the fall of Rome the center of mathematical learning moved to the Arabian peninsula where aSchool of Wisdom was created in Baghdad. There is Baghdad, as in Alexandria, scholars migrated tolearn the newest ideas and librarians collected works from all over the world. In Baghdad many ancientGreek works, included Euclid’s Elements, were translated in Arabic and eventually scholars added to thiswork with their own ideas and commentaries.

Meanwhile mathematics and science languished in Europe. In Rome there were translations (intoLatin) of Aristotle’s writings on science, but very little material survived. Beginning around 1000 AD,Europe began a slow revival now called the High Middle Ages. Population in Europe increased, Muslimforces in Spain and Turkey began to retreat, Viking raids from the north ceased and the feudal systemwas replaced by a variety of kingdoms. Eventually schools of scholars grew up in major urban areas suchas Bologna, Italy and Paris, France. Commerce extended into Arabia and eventually India.

3.4.2 Fibonacci and the reawakening of Europe

Until the Renaissance period (beginning about 1400 AD), mathematical concepts were preserved (andoccasionally extended) by the Hindu and Arab cultures. Around 1150, these ideas, including translationsof the earlier Greek writers, began to filter back into Europe. One of the first European travelers to impactEurope’s view of mathematics was Fibonacci (1175-1250.) He was probably the best mathematician ofthe middle ages.

He was born Leonardo son of Bonnacci, of Pisa, Italy (“son of Bonnacci” = fi Bonnacci.) He wasborn in Pisa where his father was a merchant and brought up in Boughie on the north coast of Africa.

17

He had an early interest in algebra and toured the Mediterranean, learning about Eastern and Arabicmathematics. He wrote works on algebra (using ideas from the Arabs.) He wrote works on geometry.His algebra was rhetorical.

In 1202 he published Liber abaci which promoted the Hindu-Arabic methods of calculation. Hestrongly advocated the Hindu-Arabic numerals. In these works he included a number of problems whichwere popular in Europe. One involved a sequence now named after him. He also wrote (1220) Practicageometriae, and (1225) Liber quadratorum on analysis.

He solved a problem on roots of a cubic where he found nine decimal place approximation of theanswer and attempted to prove the root was not constructible! (Some of these problems are on pages263, 284 of Eves.)

Fibonacci’s writings stirred up an interest in the Hindu-Arabic numeral system. Slowly over the nextcenturies, the numeral system would filter into Europe, slowly replacing the Roman numerals, especiallyamong the accountants and bookkeepers in commercial centers.

After Fibonacci, there is not a great deal of new mathematical development in Europe but the knowl-edge of the Greeks begins to make its way back to Europe via Arab translations. The next few centuriessee many Arab writing translated into Latin; many of the Arab writing are translations of earlier Greekwork such as Euclid’s Elements. As interest in scholarship arose, Europe saw the first small communitiesdevoted to learning, reading and analyzing the ancient ideas. These communities would eventually becalled “universities”.

3.4.3 The rise of universities

Prior to 1200 AD, much of the ancient learning of Europe was preserved in Christian monasteries.Around 1200 AD, teachers and their students would meet to study ancient texts and expound on them.As was common in that day, the scholars would form a guild, or union, to promote and protect theirinterests and profession. This union was called (in Latin) “universitas”, meaning a group of people with acommon goal and collective legal rights. Scholars could charge fees; scholars were free to explore learningwithout hindrance; students and scholars were free to move about the country and occasionally move toa new location. The collection of scholars could certify students as having mastered a certain body ofknowledge and so students were granted degrees and some became “masters” and went on to becomescholars themselves.

Most schools were relatively small and focused on a particular topic (law, medicine, art, theology andphilosophy.) The schools congregated together and eventually the separate colleges (universities) in onelocation were identified as a single university.

The major universities during the high middle ages were Bologna, beginning around 1088 and Paris,beginning around 1150. When English scholars were expelled from Paris in 1167, they moved to Oxford,England (a small town on the Ox River) and created a school there, now Oxford University. As thatuniversity grew in size and reputation, its eventual growth in size led to a number of buildings calledcolleges (including one association called “New College”, so named because it was so brand-new andcutting edge – created in 1379.) In 1209, a “town-and-gown” conflict between the members of OxfordUniversity and the town of Oxford led to an exit of scholars for the community of Cambridge (a smalltown on the River Cam) where Cambridge University was created. (The Foundations of Modern Sciencein the Middle Ages, by Edward Grant describes the seeds of modern science during this time.)

3.4.4 The rise of algebra

Nicholas Oresme (1323-1382) was an example of the intellectual developments by the fourteenth century.He studied at the College of Navarre, an institution for poor students, part of the University of Paris. Hewrote a number of mathematical works and translated Aristotle. He analyzed carefully the possibility ofthe earth rotating, instead of the sun and stars moving in the heavens, and concluded that there was nostrong evidence for either case. He discusses infinite sums, apparently arguing that the harmonic series

18

grows beyond bound but finding a sum for the series

∞∑k=1

k

2k.

In the fifteenth century Regiomantus (1436-1476) wrote numerous tracts on mathematics and as-tronomy, including trigonometry and spherical trig. In a work on algebra he introduced a syncopatedalgebra.

Europeans learned from the Arabs to work through algebraic processes and, over time, developed asyncopated algebra, while still relying heavily on geometric reasoning. Pascioli (1445-1509), Bombelliand later Vieti and Stifel suggested notation that converged towards our modern notation.

Pascioli used latin abbreviations for words, writing p for piu, m for meno; these would eventuallyevolve into our plus and minus signs. He wrote co for cosa, meaning an unknown quantity and thenwrote ce, cu and cece for the square, cube and fourth power of the unknown. (Eves, p. 267.)

3.4.5 The Renaissance

The fifteenth century became a time of dramatic change in Europe, the flowering of the Renaissanceperiod. The old, permanent, unchangeable universe suddenly seemed very changeable, very different.Aristotle’s view of the universe as eternal and unchanging gave way to an interest in exploration andchange.

A number of events contributed to this. During the fourteenth century, the black plague devastatedEurope, destroying villages and feudal estates and forcing some reorganization and renewal. Exploration(for the purpose of commerce or missionary enterprise) led to discoveries of new kingdoms (China) andeventually a new continent (in 1492.)

The Gutenberg printing press (in 1440) led to rapid dissemination of new ideas. At the same time,Italy and the rest of Europe were flooded with translations of Arab manuscripts, many coming to Italyafter the fall of Constantinople in 1453. At the same time, scholars disputed the authority of bothAristotle (in science) and the Catholic Church (in religion), leading to new ideas and eventually social,political and religious revolutions. Out of these disputes, the Protestant reformation began in 1517.

The old Julian calendar had served Europe for a fifteen hundred years but was clearly flawed andneeded to be replaced.

Eventually the telescope was invented and Galileo (1564-1642) discovered that the heavens were muchmore intricate than anyone had believed.

All of these events created change. The exploration of the new world and a variety of scientificdiscoveries led to the expansion of universities and an emphasis on intellectual pursuits. ?

19

3.5 The cubic and quartic equations

3.5.1 The Italians of the Sixteenth Century

One of the first truly new concepts in mathematics occurred in Italy during the 1500s when Italianscholars, using the Arab algebra and ideas with quadratic and cubic equations first solved completelythe cubic equation and then shortly after that found the full solution to the quartic (“fourth degree”)equation.

The main players in this story are Scipione del Ferro (1465-1526), Girolamo Cardano (1501-1576),Tartaglia (1500-1557), and Ludovico Ferrari (1522-1565, a student of Cardano.)

The story begin with del Ferro, a teacher at the University of Bologna, who apparently completed thesolution of the cubic equation, most likely building on ideas of Khayyam and other Arab mathematicians.Del Ferro’s student, Antonio Fior, challenged Tartaglia to a public math solving contest (very common inRenaissance Italy) involving problems all required solving cubic equations. Tartaglia, under this pressure,found a solution to the cubic “depressed” equation and won the contest.

After this, Cardano “stole” this solution from Tartaglia and generalized it, publishing the results in1545. Ferrari, Cardano’s student, then went on to solve the quartic equation.

In this section we work through the details of the solutions to the cubic equation and briefly outlinethe solution to the quartic.

3.5.2 The quadratic equation

There are four steps to finding the zeroes of a quadratic polynomial.

1. First divide by the leading term, making the polynomial monic.

2. Then, given x2 + a1x + a0, substitute x = y − a12

to obtain an equation without the linear term.

(This is the “depressed” equation.)

3. Solve then for y as a square root. (Remember to use both signs of the square root.)

4. Once this is done, recover x using the fact that x = y − a12

.

For example, let’s solve2x2 + 7x− 15 = 0.

First, we divide both sides by 2 to create an equation with leading term equal to one:

x2 +7

2x− 15

2= 0.

Then replace x by x = y − a12

= y − 7

4to obtain:

y2 =169

16

Solve for y:

y =13

4or − 13

4

Then, solving back for x, we have

x =3

2or − 5.

This method is equivalent to “completing the square” and is the steps taken in developing the much-memorized quadratic formula. For example, if the original equation is our “high school quadratic”

ax2 + bx+ c = 0

20

then the first step creates the equation

x2 +b

ax+

c

a= 0.

We then write x = y − b

2aand obtain, after simplifying,

y2 − b2 − 4ac

4a2= 0

so that

y = ±√b2 − 4ac

2a

and so

x = − b

2a±√b2 − 4ac

2a.

The solutions to this quadratic depend heavily on the value of b2 − 4ac. We give this a name (thediscriminant) and a symbol (∆) and so we discuss the discriminant

∆ = b2 − 4ac. (21)

If the coefficients of the polynomial are integers and ∆ is a perfect square integer, we have rational roots.If the discriminant is positive, we have real roots. If the discriminant is zero, we have a single root. Ifthe discriminant is negative, we have imaginary roots.

We note for later that if the discriminant ∆ = b2− 4ac is equal to zero then we have a single root andso our polynomial is a perfect square.

3.5.3 The general solution to the cubic equation

Every polynomial equation involves two steps to turn the polynomial into a slightly simpler polynomial.

1. First divide by the leading term, creating a monic polynomial (in which the highest power of xhas coefficient one.) This does not change the roots.

2. Then, given xn+an−1xn−1 +an−2x

n−2 + ...a1x+a0, substitute x = y− an−1n

to obtain an equation

without the term of degree n− 1. (This is the depressed polynomial.) Since this step is reversible,solutions to the “depressed equation” give us solutions to the original equation.

Here are the first steps in del Ferro’s method of solving the cubic. First, we do the two automatic steps:

1. Divide by the leading term, creating a cubic polynomial x3+a2x2+a1x+a0 with leading coefficient

one.

2. Then substitute x = y− a23

to obtain an equation without the term of degree two. This creates an

equation of the form x3 + Px−Q = 0.

The Renaissance mathematicians del Ferro, Tartaglia and Cardano would rewrite this equation in theform x3 +Px = Q or x3 = Px+Q depending on the sign of P . (They assumed all numbers were positiveand so put Px on one side or the other to keep that convention.) They then noticed (!) the followingalgebra identity:

(a± b)3 = 3ab(a± b) + a3 ± b3. (22)

We have some choices here so I will use the plus sign. I will also move 3ab(a + b) over to the left side.Here is the version I want:

(a+ b)3 − 3ab(a+ b) = a3 + b3. (23)

21

Given x3 + Px = Q, setx := a+ b

and notice that if

ab =−P3

(24)

anda3 + b3 = Q (25)

then a solution a, b to equations (24) and (25) gives a solution x = a+b to the cubic equation x3+Px = Q.

We can solve equations (24) and (25) by substitution. For example, we might replace a by−P3b

(using

equation 24) and then substitute into equation 25 so that we obtain

(−P3b

)3 + b3 = Q.

Multiplying by b3, we have

(−P3

)3 + (b3)2 = Qb3

which is a quadratic equation in b3. Rewrite this equation as

(b3)2 −Qb3 − (P

3)3 = 0 (26)

so that

b3 =Q

2±√

∆ (27)

where the discriminant ∆ is

∆ := (P

3)3 + (

Q

2)2.

Now the quadratic equation (26) has two solutions. Which do we use? It turns out that it doesn’tmatter which of the two solutions we use since if

b3 =Q

2+√

∆

then (since a3 + b3 = Q)

a3 =Q

2−√

∆.

Thus we can choose b3 to be one solution to the quadratic equation (26) and choose a3 to be the other.However we still need to solve for a and b and there are three possible solutions for b in equation (27)!

There are several places where the Italian mathematicians struggled. Sometimes the discriminant inthe quadratic equation (26) was negative so the solutions involved “imaginary” numbers and they weren’tsure what to do with these. Furthermore, even if the solutions to the quadratic equation were real, they

then only found one solution to b3 =Q

2+√

∆ and they eventually needed three solutions to this equation.

3.5.4 The necessity of complex numbers

We will press on and finish this work by jumping ahead to Euler’s day and using his understanding ofcomplex numbers.

Using Taylor series, Euler developed the following wonderful formula:

eiθ = cos θ + i sin θ (28)

22

This formula describes much phenomena in mathematics, intertwining complex numbers, exponentialfunctions and trigonometric functions. In particular, this gives every complex number a polar formz = reiθ = r cos θ + i(r sin θ).

In order to solve the cubic equation, we will create the element

ω := e2πi/3 = −1

2+

√3

2i. (29)

(Note that ω has polar from r = 1 and θ = 2π/3 so that ω lies on the unit circle in the second quadrant.)The importance of ω is that ω3 = 1 and (ω2)3 = 1 so that 1, ω and ω2 are all cube roots of 1!

Using ω, we can see that any real number has three cube roots. Given one cube root, β, the otherswill be ωβ and ω2β. For example, 3

√8 really has three meanings (!); the three cube roots of eight are

2, 2ω and 2ω2.We use ω to finish equation (27) by picking one cube root of −Q/2±

√∆. We call that one particular

solution β. The other solutions to the equation (27) are b = ωβ and b = ω2β.

Each solution to (27) gives a value for a so that ab = P/3. Let α :=P

3β. Then

α− β, αω2 − βω, and αω − βω2

are all solutions to the cubic equation x3 + Px = Q.

Euler’s example.We work through an example due to Euler: We find all solutions to

x3 − 6x = 4. (30)

Here P = −6 and Q = 4 and so the discriminant is

∆ = −8 + 4 = −4 so√

∆ = 2i.

Therefore b3 = −2± 2i.We choose a sign and solve the cubic equation b3 = −2+2i. Write −2+2i in polar form as

√8 (e

3π4 i).

Then b3 =√

8 eπ4 i has a particular solution

β =√

2 eπ4 i = 1 + i.

Set

α =−2

β=−2√

2e−

π4 i = −

√2 e−

π4 i =

√2 e

3π4 i.

(Or, in cartesian form, α = −1 + i.) Then

x = α− β = (−1 + i)− (1 + i) = −2

is one solution to the cubic equation (30).But there are other solutions to equation (30)! If x1 = α− β is a solution then so are

x2 = αω2 − βω =√

2(e3π4 ie

4π3 i − eπ

4 ie2π3 i) =

√2(e

25π12 i − e 11π

12 i) =√

2(eπ12 i + e

−π12 i)

andx3 = αω − βω2 =

√2(e

3π4 ie

2π3 i − eπ

4 ie4π3 i) =

√2(e

17π12 i − e 19π

12 i) =√

2(e−7π12 i + e

7π12 i)

By Euler’s formula, eθi + e−θi = 2 cos θ. So our answers are equivalent to

x2 = 2√

2 cos(π

12) and x3 = 2

√2 cos(

7π

12).

23

The value of cos( π12 ) is not immediate, but we can find it from a trig identity. Since

cos(2θ) = 2 cos2 θ − 1

then

cos2 θ =1 + cos 2θ

2.

Thus

cosπ

12=

√1

2+

√3

4

and

x2 =√

8

√1

2+

√3

4=

√4 + 2

√3.

In a similar manner we can find x3. Our final answers are

x2 =

√4 + 2

√3 and x3 = −

√4− 2

√3.

Another example from Euler.We solve Euler’s cubic:

x3 − 6x = 9. (31)

Since (a− b)3 + 3ab(a− b) = a3 − b3, we set

3ab = −6 and a3 − b3 = 9.

(Let P := −6;Q := 9.) Thena = −2/b; (−2/b)3 − b3 = 9

so−8− b6 = 9b3 or b6 + 9b3 + 8 = 0.

View this as a quadratic in b3 so that

(b3)2 + 9b3 + 8 = 0 =⇒ (b3 + 8)(b3 + 1) = 0. (32)

Therefore either b3 = −8 or b3 = −1.Suppose b3 = −8 and presumably b = −2. Then a = 1 and x = a− b = 3.If b = −1 then a = −2 and x = a− b = 3, so choosing the other factor does not give new information.

We have found one solution, x = 3.But what about other solutions? Set β = −2 as a solution to the cubic equation (32), so that β3 = −8

and let α = 1 be the corresponding choice for a. Then b = βω = −2ω is also a solution to that cubic andin this case a = αω2 = ω2 is the corresponding choice for a. Then a second solution is

x = a− b = ω2 + 2ω = (−1

2−√

3

2i) + (−1 +

√3i) =

1

2(−3 +

√3 i).

If instead we have a = αω and b = βω2 then the final solution is

x = a− b = ω + 2ω2 = (−1

2+

√3

2i) + (−1−

√3i) =

1

2(−3−

√3 i).

So we have found all three solutions:

3,1

2(−3 +

√3 i). and

1

2(−3−

√3 i).

24

3.5.5 Rational Solutions

We can avoid the lengthy computations, above, if we are lucky enough to find a rational solution to ourpolynomial. For example, in the last problem, if we had merely stumbled on the root x = 3, we couldhave divided the cubic polynomial x3 − 6x− 9 by x− 3 and rewritten it as

x3 − 6x− 9 = (x− 3)(x2 + 3x+ 3).

The quadratic equation, applied to x2 + 3x + 3 would have given us the final two solutions without theextra work.

So how do we find these rational solutions when they occur?

If x = pq is a rational solution to the polynomial equation f(x) = 0 then qx − p is a factor of the

polynomial f(x) and so we can use long division to write f(x) = (qx− p)g(x) where g(x) is a polynomialof smaller degree.

We teach a version of this method in high school when students learn to solve quadratic equations byfactoring. For example, one might solve the equation 3x2 − 2x − 8 = 0 by factoring the left-hand sideinto (3x+ 4)(x− 2), obtain solutions x = − 4

3 and x = 2 and so avoid the quadratic formula.

We can try this method on polynomials of higher degree with integer coefficients. Descartes wouldpoint out, in the 1600s, that if f(x) = anx

n + an−1xn−1 + ...a1x + a0 has a root x = p/q then q must

divide an and p must divide a0 (Notice how this works on the quadratic 3x2−2x−8.) With then a shortlist of possible rational solutions, Descartes was willing to try these solutions and exhaust the possibilitiesfor a rational root.

The modern version of this is to pull out a graphing calculator, graph the polynomial equation y = f(x)and hope that the calculator identifies a nice rational (or even integer!) root. For example, with Euler’scubic x3− 6x− 9 , we discover that x = 3 is a root. When then divide the polynomial by x− 3 to obtaina quadratic polynomial and now we can go ahead and use the quadratic formula.

This method is much faster than the general method, but it requires that we be “lucky” and stumbleupon a root.

3.5.6 Quartic Polynomials

After del Ferro and Tartaglia solved the general cubic equation (and the result was released to the publicby Cardano), mathematicians then concentrated on the quartic equation. Cardano’s student, Ferrari,around 1540, suggested a general method for solving the depressed quartic.

Given the depressed quarticx4 + px2 + qx+ r = 0

rewrite the problem so that it appears to involve a square:

(x2 + p)2 = px2 − qx+ (p2 − r).

Now insert a new variable y and expand (x2 + p+ y)2 with the hopes that the right side, (p+ 2y)x2−qx+ (p2 − r + 2py + y2) is a perfect square.

By the quadratic formula, the right-hand side will be a perfect square when the discriminant 4(p +2y)(p2 − r + 2py + y2)− q2 is zero.

So we first need to solve4(p+ 2y)(p2 − r + 2py + y2)− q2 = 0 (33)

for y. Since equation (33) is a cubic equation, we know how to do that!

In other words, given a quartic, we make it monic, and then turn it into a depressed equation, inserty, and solve the associated cubic for y. Once this is done, we solve the quadratic equation (x2 +p+y)2 =(p+ 2y)x2 − qx+ (p2 − r + 2py + y2) for x.

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This is lengthy and tedious, but it works in general! (Well, it works if one is willing to believe incomplex numbers.)

By 1545, after Cardano published Ferrari’s work (with his permission, apparently!) mathematiciansbegan to search for solutions to the general fifth-degree or quintic equation. It would take the work ofAbel, Galois and others to show that in fact a general solution to the quintic is impossible.

We will save that story for another time.

3.6 The French and the early Seventeenth Century (under construction)

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