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The Root of the Problem:A Brief History of Equation Solving
Alison RamageDepartment of Mathematics and Statistics
University of Strathclyde
[email protected]://www.mathstats.strath.ac.uk/
Mathematical Association 2009 – p.1/30
Background and References
• Great Moments in Mathematics H. Eves
• Mathematics and Mathematicians P.Dedron & J. Hard
• The Mathematics of Great Amateurs J.L. Coolidge
• A History of Mathematics C. Boyer & U.C. Merzbach
• MacTutor History of Mathematics Archivehttp://www-groups.dcs.st-and.ac.uk:80/∼history/
Mathematical Association 2009 – p.2/30
Background and References
• Great Moments in Mathematics H. Eves
• Mathematics and Mathematicians P.Dedron & J. Hard
• The Mathematics of Great Amateurs J.L. Coolidge
• A History of Mathematics C. Boyer & U.C. Merzbach
• MacTutor History of Mathematics Archivehttp://www-groups.dcs.st-and.ac.uk:80/∼history/
• Bluff Your Way in Maths R. Ainsley
Mathematical Association 2009 – p.2/30
Background and References
• Great Moments in Mathematics H. Eves
• Mathematics and Mathematicians P.Dedron & J. Hard
• The Mathematics of Great Amateurs J.L. Coolidge
• A History of Mathematics C. Boyer & U.C. Merzbach
• MacTutor History of Mathematics Archivehttp://www-groups.dcs.st-and.ac.uk:80/∼history/
• Bluff Your Way in Maths R. Ainsley
“Mathematics is a unique aspect of human thought,and its history differs in essence from all otherhistories.”Asimov (1920-1992)
Mathematical Association 2009 – p.2/30
Notation
• RHETORICAL Rhind Papyrus c. 1650 BCDemochares has lived a fourth of his life as a boy,a fifth as a youth, a third as a man and has spentthirteen years in his dotage. How old is he?
Mathematical Association 2009 – p.3/30
Notation
• RHETORICAL Rhind Papyrus c. 1650 BCDemochares has lived a fourth of his life as a boy,a fifth as a youth, a third as a man and has spentthirteen years in his dotage. How old is he?
• SYNCOPATED Diophantus c. 250 AD
1 2 3 4 5 6 7 8 9 10α β γ δ ǫ ζ ξ η θ ι
ς: unknown ∆γ: unknown squared κγ : unknown cubed
κγα∆γιγςη x3 + 13x2 + 8x
unknown cubed 1, unknown squared 13, unknown 8
Mathematical Association 2009 – p.3/30
Notation cont.
• SYMBOLIC
+, − Widman 1489
√ Rudolf 1525
= Recorde 1557
unknowns=vowels Viète 1570knowns=consonants
unknowns=late letters Descartes 1630knowns=early letters
>,< Harriot 1631
Mathematical Association 2009 – p.4/30
Notation cont.
×, ∼, π Oughtred 1640
∞ Wallis 1650
f(x) Euler 1750
n! Kramp 1800
Mathematical Association 2009 – p.5/30
Notation cont.
×, ∼, π Oughtred 1640
∞ Wallis 1650
f(x) Euler 1750
n! Kramp 1800
“Mathematics is a game played according to certainsimple rules with meaningless marks on paper.”Hilbert (1862-1943)
Mathematical Association 2009 – p.5/30
Notation cont.
×, ∼, π Oughtred 1640
∞ Wallis 1650
f(x) Euler 1750
n! Kramp 1800
“Mathematics is a game played according to certainsimple rules with meaningless marks on paper.”Hilbert (1862-1943)
“A formal manipulator in mathematics oftenexperiences the discomforting feeling that his pencilsurpasses him in intelligence.”Eves (1911-2004)
Mathematical Association 2009 – p.5/30
Archimedes (287 BC - 212 BC)
• Greek mathematician andastronomer
• lived and worked in Syracuse
• invented war machines
• On Floating Bodies
• slain by a Roman soldier
On the Measurement of the CircleGeometric methods for calculating square roots usingcircles with circumscribed hexagons (similar to theBabylonians)
Mathematical Association 2009 – p.6/30
Archimedes (287 BC - 212 BC)
• Greek mathematician andastronomer
• lived and worked in Syracuse
• invented war machines
• On Floating Bodies
• slain by a Roman soldier
On the Measurement of the CircleGeometric methods for calculating square roots usingcircles with circumscribed hexagons (similar to theBabylonians)
“ǫνρηκα!”
Mathematical Association 2009 – p.6/30
Diophantus (c. 200 BC - 284 BC)
• Greek mathematician
• lived and worked in Alexandria
• allowed positive rationals assolutions and coefficients
ArithmeticaCollection of 130 problems in solving equations (althoughonly 6 of the original 13 books survive). Introducedalgebraic symbolism and Diophantine equations.
Mathematical Association 2009 – p.7/30
How old was Diophantus?
Here lies Diophantus, the wonder behold.Through art algebraic, the stone tells how old:’God gave him his boyhood one-sixth of his life,One twelfth more as youth while whiskers grew rife;And then yet one-seventh ere marriage begun;In five years there came a bouncing new son.Alas, the dear child of master and sageAfter attaining half the measure of his father’s lifechill fate took him.After consoling his fate by the science of numbersfor four years, he ended his life.
Mathematical Association 2009 – p.8/30
How old was Diophantus?
Here lies Diophantus, the wonder behold.Through art algebraic, the stone tells how old:’God gave him his boyhood one-sixth of his life,One twelfth more as youth while whiskers grew rife;And then yet one-seventh ere marriage begun;In five years there came a bouncing new son.Alas, the dear child of master and sageAfter attaining half the measure of his father’s lifechill fate took him.After consoling his fate by the science of numbersfor four years, he ended his life.
1
6L +
1
12L +
1
7L + 5 +
1
2L + 4 = L ⇔
3
28L = 9 ⇔ L = 84
Mathematical Association 2009 – p.8/30
Marignal notes in Arithmetica
Fermat’s Last Theorem
If an integer n is greater than 2, then
an + bn = cn
has no solutions in non-zero integers a, b, and c.
Mathematical Association 2009 – p.9/30
Marignal notes in Arithmetica
Fermat’s Last Theorem
If an integer n is greater than 2, then
an + bn = cn
has no solutions in non-zero integers a, b, and c.
“ I have assuredly found an admirable proofof this, but the margin is too narrow to contain it.”Pierre de Fermat (1601-1665)
Mathematical Association 2009 – p.9/30
Marignal notes in Arithmetica
Fermat’s Last Theorem
If an integer n is greater than 2, then
an + bn = cn
has no solutions in non-zero integers a, b, and c.
“ I have assuredly found an admirable proofof this, but the margin is too narrow to contain it.”Pierre de Fermat (1601-1665)
“ Thy soul, Diophantus, be with Satanbecause of the difficulty of your theorem.”Maximus Planudes (1260-1330)
Mathematical Association 2009 – p.9/30
Mohamed ibn-Muso al-Khwarizmi (c.790-840)
• Arabian librarian
• lived and worked in Howarizmi
• calculated latitudes andlongitudes for 2402 localitiesas a basis for a world map
• wrote about sundials and theJewish calendar
Hisâb al-jabr w’almuqâbalah
Mathematical Association 2009 – p.10/30
Mohamed ibn-Muso al-Khwarizmi (c.790-840)
• Arabian librarian
• lived and worked in Howarizmi
• calculated latitudes andlongitudes for 2402 localitiesas a basis for a world map
• wrote about sundials and theJewish calendar
Hisâb al-jabr w’almuqâbalah
al-Khwarizmi ≡ algorithm, al-jabr ≡ algebra
Mathematical Association 2009 – p.10/30
Science of Reunion and Opposition
squares equal to roots x2 = 5xsquares equal to numbers x2 = 4roots equal to numbers 5x = 15squares and roots equal to numbers x2 + 10x = 39squares and numbers equal to roots x2 + 21 = 10xroots and numbers equal to squares 3x + 4 = x2
Mathematical Association 2009 – p.11/30
Science of Reunion and Opposition
squares equal to roots x2 = 5xsquares equal to numbers x2 = 4roots equal to numbers 5x = 15squares and roots equal to numbers x2 + 10x = 39squares and numbers equal to roots x2 + 21 = 10xroots and numbers equal to squares 3x + 4 = x2
“ With a name like this under your belt, you can bluffyour way past even a bona fide mathematician.”Bluffer’s Guide to Maths
Mathematical Association 2009 – p.11/30
Completing the Square
x2 + 10x = 39
x
5/2
s
e
fh
g
1. x2 term: centre square s
Mathematical Association 2009 – p.12/30
Completing the Square
x2 + 10x = 39
x
5/2
s
e
fh
g
1. x2 term: centre square s
2. 10x: add e, f , g, h, each 5/2 units wide
(so area of each is 10x/4 units)
Mathematical Association 2009 – p.12/30
Completing the Square
x2 + 10x = 39
x
5/2
s
e
fh
g
1. x2 term: centre square s
2. 10x: add e, f , g, h, each 5/2 units wide
(so area of each is 10x/4 units)
3. total area of solid figure is 39 units
Mathematical Association 2009 – p.12/30
Completing the Square
x2 + 10x = 39
x
5/2
s
e
fh
g
1. x2 term: centre square s
2. 10x: add e, f , g, h, each 5/2 units wide
(so area of each is 10x/4 units)
3. total area of solid figure is 39 units
4. complete the square: add four
squares, each of area 25/4 units
Mathematical Association 2009 – p.12/30
Completing the Square
x2 + 10x = 39
x
5/2
s
e
fh
g
1. x2 term: centre square s
2. 10x: add e, f , g, h, each 5/2 units wide
(so area of each is 10x/4 units)
3. total area of solid figure is 39 units
4. complete the square: add four
squares, each of area 25/4 units
5. area of large square is
39 + 25 = 64 units
Mathematical Association 2009 – p.12/30
Completing the Square
x2 + 10x = 39
x
5/2
s
e
fh
g
1. x2 term: centre square s
2. 10x: add e, f , g, h, each 5/2 units wide
(so area of each is 10x/4 units)
3. total area of solid figure is 39 units
4. complete the square: add four
squares, each of area 25/4 units
5. area of large square is
39 + 25 = 64 units
6. side of large square has length
x + 5 = 8 units
Mathematical Association 2009 – p.12/30
Completing the Square
x2 + 10x = 39
x
5/2
s
e
fh
g
1. x2 term: centre square s
2. 10x: add e, f , g, h, each 5/2 units wide
(so area of each is 10x/4 units)
3. total area of solid figure is 39 units
4. complete the square: add four
squares, each of area 25/4 units
5. area of large square is
39 + 25 = 64 units
6. side of large square has length
x + 5 = 8 units
7. solution is x = 3
Mathematical Association 2009 – p.12/30
Leonardo Pisano Fibonacci (1170-1250)
• Italian traveller
• originally from Pisa
• brought Arabic maths to Europe
Liber AbaciProved no root to x3 + 2x2 + 10x = 20 can be constructed byruler and compass: gave solution x ≃ 1.3688081075.
Mathematical Association 2009 – p.13/30
Fibonacci numbers
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .
recurrence relation:
F (n + 2) = F (n + 1) + F (n)
closed form solution:
F (n) =φn − (1 − φn)
√5
golden ratio φ = 1.6180339887 . . .
rabbits, honey bees, seed heads, pine cones,petal arrangements, vegetables, shell spirals. . .
Mathematical Association 2009 – p.14/30
Luca Pacioli (1445-1517)
• Franciscan friar in Sansepolcro,Italy
• the father of accounting
• unpublished treatise on chessillustrated by Leonardo da Vinci
Summa de arithmetica, geometrica,proportioni et proportionalita
Summary of current work in arithmetic, algebra, geometryand trigonometry.
Mathematical Association 2009 – p.15/30
Pacioli’s notation
6.p.R.1018.m.R.90
108.m.R.3240.p.R.3240.m.R.90hoc est 78.
(6 +√
10)(18 −√
90) = (108 −√
3240 +√
3240 −√
900) = 78
“. . . , one cannot give general rules except that, sometimes,by trial, . . . in some particular cases.”
Mathematical Association 2009 – p.16/30
Scipione dal Ferro (1465-1526)
• Professor in Bologna, Italy
• solved depressed cubic equations of the formx3 + mx = n algebraically
• no knowledge of negative numbers
• solution by radicals: find the roots by adding,subtracting, multiplying, dividing and taking roots of thecoefficients.
• did not publish any of his work
• passed his greatest discovery on to his student Fiorfrom his deathbed
Mathematical Association 2009 – p.17/30
Scipione dal Ferro (1465-1526)
• Professor in Bologna, Italy
• solved depressed cubic equations of the formx3 + mx = n algebraically
• no knowledge of negative numbers
• solution by radicals: find the roots by adding,subtracting, multiplying, dividing and taking roots of thecoefficients.
• did not publish any of his work
• passed his greatest discovery on to his student Fiorfrom his deathbed
“ It is often more convenient to possess the ashesof great men than to possess the men themselvesduring their lifetime.”Jacobi (1804-1851)
Mathematical Association 2009 – p.17/30
Niccolo Fontana Tartaglia (1500-1557)
• The Stammerer
• teacher in Venice and Verona
• self-taught mathematician
• won a public equation-solvingcontest with Fior by solving othertypes of cubic
• swore Cardan to secrecy
New Problems and Inventions
Personal insults about Cardan.
Mathematical Association 2009 – p.18/30
Girolamo Cardano (1501-1576)
• doctor and lawyer in Milan
• detailed autobiography
• family problems
• cast horoscopes
Ars magnaMethods for cubics and quartics (done by Ferrari).
Mathematical Association 2009 – p.19/30
Girolamo Cardano (1501-1576)
• doctor and lawyer in Milan
• detailed autobiography
• family problems
• cast horoscopes
Ars magnaMethods for cubics and quartics (done by Ferrari).
“I swear to you, by God’s holy Gospels, and as a trueman of honour, not only never to publish yourdiscoveries, if you teach me them, but also I promiseyou, and I pledge my faith as a true Christian, to notethem down in code, so that after my death no one willbe able to understand them.”
Mathematical Association 2009 – p.19/30
Cardano-Tartaglia Method for Cubics
“Quando che’l cubo con le cose appressoSe agguaglia a qualche numero discretoTrovati dui alte differenti in essoDapoi terrai, questo per consueto, . . . ”
Mathematical Association 2009 – p.20/30
Cardano-Tartaglia Method for Cubics
“Quando che’l cubo con le cose appressoSe agguaglia a qualche numero discretoTrovati dui alte differenti in essoDapoi terrai, questo per consueto, . . . ”
When the cube and its things near x3 + mx = n
Add to a new number, discrete,Determine two new numbers different a3 − b3 = n
By that one; this featWill be kept as a ruleTheir product always equal, the same, a3b3 =
(
m
3
)3
To the cube of a thirdOf the number of things named.Then generally speaking,The remaining amountOf the cube roots subtracted x = a − b
Will be your desired count.Mathematical Association 2009 – p.20/30
Solving cubics
• AIM: solve x3 + mx = n (1)
• IDEA: use identity (a− b)3 +3ab(a− b) = a3 − b3 (2)
Mathematical Association 2009 – p.21/30
Solving cubics
• AIM: solve x3 + mx = n (1)
• IDEA: use identity (a − b)3 + 3ab(a − b) = a3 − b3 (2)
• METHOD: choose a and b so 3ab = m, a3 − b3 = n
b =m
3a⇒ a3 −
m3
27a3= n ⇒ a6 − na3 −
m3
27= 0.
solve this quadratic in a3:
a =
(
n
2+
√
(n
2
)2
+
(m
3
)3
)1
3
, b =
(
−n
2+
√
(n
2
)2
+
(m
3
)3
)1
3
Mathematical Association 2009 – p.21/30
Solving cubics
• AIM: solve x3 + mx = n (1)
• IDEA: use identity (a − b)3 + 3ab(a − b) = a3 − b3 (2)
• METHOD: choose a and b so 3ab = m, a3 − b3 = n
b =m
3a⇒ a3 −
m3
27a3= n ⇒ a6 − na3 −
m3
27= 0.
solve this quadratic in a3:
a =
(
n
2+
√
(n
2
)2
+
(m
3
)3
)1
3
, b =
(
−n
2+
√
(n
2
)2
+
(m
3
)3
)1
3
• from (2), cubic (1) has solution x = a − b
Mathematical Association 2009 – p.21/30
Solving cubics cont.
x =
(
n
2+
√
(n
2
)2
+(m
3
)3
)1
3
−
(
−n
2+
√
(n
2
)2
+(m
3
)3
)1
3
• Case 1x3 + mx = n
x3 + 6x = 20 ⇒ m = 6, n = 20
x =(
10 +√
108)
1
3
−(
−10 +√
108)
1
3
= 2
Mathematical Association 2009 – p.22/30
Solving cubics cont.
x =
(
n
2+
√
(n
2
)2
+(m
3
)3
)1
3
+
(
−n
2+
√
(n
2
)2
+(m
3
)3
)1
3
• Case 2x3 = mx + n
(use x = a + b)
x3 = 15x + 4 ⇒ m = 15, n = 4
x =(
2 +√−121
)
1
3 −(
2 −√−121
)
1
3
Mathematical Association 2009 – p.23/30
Solving cubics cont.
x =
(
n
2+
√
(n
2
)2
+(m
3
)3
)1
3
+
(
−n
2+
√
(n
2
)2
+(m
3
)3
)1
3
• Case 2x3 = mx + n
(use x = a + b)
x3 = 15x + 4 ⇒ m = 15, n = 4
x =(
2 +√−121
)
1
3 −(
2 −√−121
)
1
3
PROBLEM: x = 4 is a solution, equation not insoluble!
Mathematical Association 2009 – p.23/30
Ferrari’s Method for Quartics
depressed quartic: x4 + px2 + qx + r = 0
Mathematical Association 2009 – p.24/30
Ferrari’s Method for Quartics
depressed quartic: x4 + px2 + qx + r = 0
two useful results:
(x2 + p)2 = px2 − qx + p2 − r complete the square
(x2 + p + y)2 = (x2 + p)2 + 2y(x2 + p) + y2
= (px2 − qx + p2 − r) + 2yx2 + 2py + y2
= (p + 2y)x2 − qx + (p2 − r + 2py + y2) for any
Mathematical Association 2009 – p.24/30
Ferrari’s Method for Quartics
depressed quartic: x4 + px2 + qx + r = 0
two useful results:
(x2 + p)2 = px2 − qx + p2 − r complete the square
(x2 + p + y)2 = (x2 + p)2 + 2y(x2 + p) + y2
= (px2 − qx + p2 − r) + 2yx2 + 2py + y2
= (p + 2y)x2 − qx + (p2 − r + 2py + y2) for any
choose y so that RHS is a perfect square: set
q2 − 4(p + 2y)(p2 − r + 2py + y2) = 0
cubic equation in y, solve by C-T formula
Mathematical Association 2009 – p.24/30
Ferrari’s Method for Quartics
depressed quartic: x4 + px2 + qx + r = 0
two useful results:
(x2 + p)2 = px2 − qx + p2 − r complete the square
(x2 + p + y)2 = (x2 + p)2 + 2y(x2 + p) + y2
= (px2 − qx + p2 − r) + 2yx2 + 2py + y2
= (p + 2y)x2 − qx + (p2 − r + 2py + y2) for any
choose y so that RHS is a perfect square: set
q2 − 4(p + 2y)(p2 − r + 2py + y2) = 0
cubic equation in y, solve by C-T formula
take square root of both sides ⇒ quadratic equation for x!Mathematical Association 2009 – p.24/30
Niels Henrik Abel(1802-1829)
• Norwegian pauper and invalid
• published papers in the first evermathematical journal
• results rejected by the FrenchAcademy as illegible
• died of tuberculosis
• only honoured after his death
On algebraic equations in which the impossibility of solvingthe general equation of the fifth degree is demonstrated
Quintics are insoluble via method of radicals.
Mathematical Association 2009 – p.25/30
Niels Henrik Abel(1802-1829)
• Norwegian pauper and invalid
• published papers in the first evermathematical journal
• results rejected by the FrenchAcademy as illegible
• died of tuberculosis
• only honoured after his death
On algebraic equations in which the impossibility of solvingthe general equation of the fifth degree is demonstrated
Quintics are insoluble via method of radicals.
“Abel has left mathematicians enough to keep thembusy for 500 years.”Hermite (1822-1901)
Mathematical Association 2009 – p.25/30
Evariste Galois (1811-1832)
• French student and soldier
• Cauchy lost his French Academypaper, Fourier died before re-ceiving it
• imprisoned for treason
• persecution complex
• killed in a duel
Researches on the algebraic solution of equationsGave a method of determining when a general equationcould be solved by radicals: Galois theory of groups.
Mathematical Association 2009 – p.26/30
Evariste Galois (1811-1832)
• French student and soldier
• Cauchy lost his French Academypaper, Fourier died before re-ceiving it
• imprisoned for treason
• persecution complex
• killed in a duel
Researches on the algebraic solution of equationsGave a method of determining when a general equationcould be solved by radicals: Galois theory of groups.
“. . . had the brilliant idea of just making up fictitiousnumbers to supply answers to problems whichdidn’t otherwise have a solution.” Bluffers’ Guide
Mathematical Association 2009 – p.26/30
Sir Isaac Newton (1642-1727)
• born in Grantham
• studied at Cambridge University
• sent home due to the Plague in1665
• Professor at Cambridge
• Warden of the Royal Mint
De analysi per aequationes numero terminorum infinitasIterative method for finding approximate roots of algebraicequations.
Mathematical Association 2009 – p.27/30
Sir Isaac Newton (1642-1727)
• born in Grantham
• studied at Cambridge University
• sent home due to the Plague in1665
• Professor at Cambridge
• Warden of the Royal Mint
De analysi per aequationes numero terminorum infinitasIterative method for finding approximate roots of algebraicequations.
“Newton was of the most fearful, cautious andsuspicious temper that I ever knew.”Whiston (1667-1752)
Mathematical Association 2009 – p.27/30
Johann Carl Friedrich Gauss (1777-1855)
• Numerical methods for solvingequations
• Gaussian Elimination
• Iterative methods:Jacobi, Gauss-Seidel, . . .
Mathematical Association 2009 – p.28/30
Johann Carl Friedrich Gauss (1777-1855)
• Numerical methods for solvingequations
• Gaussian Elimination
• Iterative methods:Jacobi, Gauss-Seidel, . . .
“I have had my results for a long time: but I do notyet know how to arrive at them.”Gauss
Mathematical Association 2009 – p.28/30
Some final thoughts . . .
“Life is good for only two things, discoveringmathematics and teaching mathematics.”Siméon-Denis Poisson (1781-1840)
Mathematical Association 2009 – p.29/30
Some final thoughts . . .
“Life is good for only two things, discoveringmathematics and teaching mathematics.”Siméon-Denis Poisson (1781-1840)
“Now I will have less distraction.”Leonhard Euler (1707-1783)
Mathematical Association 2009 – p.29/30
Some final thoughts . . .
“Life is good for only two things, discoveringmathematics and teaching mathematics.”Siméon-Denis Poisson (1781-1840)
“Now I will have less distraction.”Leonhard Euler (1707-1783)
“An expert is someone who knows some of theworst mistakes that can be made in his subject, andhow to avoid them.”Werner Karl Heisenberg (1901-1976)
Mathematical Association 2009 – p.29/30
Some more final thoughts . . .
“A mathematician is a blind man in a dark roomlooking for a black hat which isn’t there.”Charles Darwin (1809-1882)
Mathematical Association 2009 – p.30/30
Some more final thoughts . . .
“A mathematician is a blind man in a dark roomlooking for a black hat which isn’t there.”Charles Darwin (1809-1882)
“A mathematician is a machine for turning coffeeinto theorems.”Paul Erdös (1913-1996)
Mathematical Association 2009 – p.30/30
Some more final thoughts . . .
“A mathematician is a blind man in a dark roomlooking for a black hat which isn’t there.”Charles Darwin (1809-1882)
“A mathematician is a machine for turning coffeeinto theorems.”Paul Erdös (1913-1996)
“A talk in mathematics should be one of four things:beautiful, deep, surprising . . . or short.”Michel Mendés-France (1935-)
Mathematical Association 2009 – p.30/30