Download The Root of the Problem: A Brief History of Equation Solving

The Root of the Problem:
A Brief History of Equation Solving
Alison Ramage
Department of Mathematics and Statistics
University of Strathclyde
[email protected]
http://www.mathstats.strath.ac.uk/
Mathematical Association 2009 – p.1/30
Background and References
• Great Moments in Mathematics
• Mathematics and Mathematicians
H. Eves
• The Mathematics of Great Amateurs
• A History of Mathematics
P.Dedron & J. Hard
J.L. Coolidge
C. Boyer & U.C. Merzbach
• MacTutor History of Mathematics Archive
http://www-groups.dcs.st-and.ac.uk:80/∼history/
Mathematical Association 2009 – p.2/30
Background and References
• Great Moments in Mathematics
• Mathematics and Mathematicians
H. Eves
P.Dedron & J. Hard
• The Mathematics of Great Amateurs
• A History of Mathematics
J.L. Coolidge
C. Boyer & U.C. Merzbach
• MacTutor History of Mathematics Archive
http://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
• Mathematics and Mathematicians
H. Eves
P.Dedron & J. Hard
• The Mathematics of Great Amateurs
• A History of Mathematics
J.L. Coolidge
C. Boyer & U.C. Merzbach
• MacTutor History of Mathematics Archive
http://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 other
histories.”
Asimov (1920-1992)
Mathematical Association 2009 – p.2/30
Notation
• RHETORICAL
Rhind Papyrus c. 1650 BC
Demochares has lived a fourth of his life as a boy,
a fifth as a youth, a third as a man and has spent
thirteen years in his dotage. How old is he?
Mathematical Association 2009 – p.3/30
Notation
• RHETORICAL
Rhind Papyrus c. 1650 BC
Demochares has lived a fourth of his life as a boy,
a fifth as a youth, a third as a man and has spent
thirteen years in his dotage. How old is he?
• SYNCOPATED
Diophantus c. 250 AD
1 2 3 4 5 6 7 8 9 10
α β γ δ ǫ ζ ξ η θ ι
ς: unknown
∆γ : unknown squared κγ : unknown cubed
κγ α∆γ ιγςη
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
knowns=consonants
Viète
1570
unknowns=late letters
knowns=early letters
Descartes
1630
>,<
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 certain
simple 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 certain
simple rules with meaningless marks on paper.”
Hilbert (1862-1943)
“A formal manipulator in mathematics often
experiences the discomforting feeling that his pencil
surpasses him in intelligence.”
Eves (1911-2004)
Mathematical Association 2009 – p.5/30
Archimedes (287 BC - 212 BC)
• Greek mathematician and
astronomer
• lived and worked in Syracuse
• invented war machines
• On Floating Bodies
• slain by a Roman soldier
On the Measurement of the Circle
Geometric methods for calculating square roots using
circles with circumscribed hexagons (similar to the
Babylonians)
Mathematical Association 2009 – p.6/30
Archimedes (287 BC - 212 BC)
• Greek mathematician and
astronomer
• lived and worked in Syracuse
• invented war machines
• On Floating Bodies
• slain by a Roman soldier
On the Measurement of the Circle
Geometric methods for calculating square roots using
circles with circumscribed hexagons (similar to the
Babylonians)
“ǫνρηκα!”
Mathematical Association 2009 – p.6/30
Diophantus (c. 200 BC - 284 BC)
• Greek mathematician
• lived and worked in Alexandria
• allowed positive rationals as
solutions and coefficients
Arithmetica
Collection of 130 problems in solving equations (although
only 6 of the original 13 books survive). Introduced
algebraic 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 sage
After attaining half the measure of his father’s life
chill fate took him.
After consoling his fate by the science of numbers
for 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 sage
After attaining half the measure of his father’s life
chill fate took him.
After consoling his fate by the science of numbers
for four years, he ended his life.
1
1
1
1
3
L + L + L + 5 + L + 4 = L ⇔ L = 9 ⇔ L = 84
6
12
7
2
28
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 proof
of 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 proof
of this, but the margin is too narrow to contain it.”
Pierre de Fermat (1601-1665)
“ Thy soul, Diophantus, be with Satan
because 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 and
longitudes for 2402 localities
as a basis for a world map
• wrote about sundials and the
Jewish 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 and
longitudes for 2402 localities
as a basis for a world map
• wrote about sundials and the
Jewish 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
squares equal to numbers
roots equal to numbers
squares and roots equal to numbers
squares and numbers equal to roots
roots and numbers equal to squares
x2 = 5x
x2 = 4
5x = 15
x2 + 10x = 39
x2 + 21 = 10x
3x + 4 = x2
Mathematical Association 2009 – p.11/30
Science of Reunion and Opposition
squares equal to roots
squares equal to numbers
roots equal to numbers
squares and roots equal to numbers
squares and numbers equal to roots
roots and numbers equal to squares
x2 = 5x
x2 = 4
5x = 15
x2 + 10x = 39
x2 + 21 = 10x
3x + 4 = x2
“ With a name like this under your belt, you can bluff
your way past even a bona fide mathematician.”
Bluffer’s Guide to Maths
Mathematical Association 2009 – p.11/30
Completing the Square
x2 + 10x = 39
e
h
s
g
x
f
5/2
Mathematical Association 2009 – p.12/30
Completing the Square
1. x2 term: centre square s
x2 + 10x = 39
e
h
s
g
x
f
5/2
Mathematical Association 2009 – p.12/30
Completing the Square
1. x2 term: centre square s
x2
2. 10x: add e, f , g, h, each 5/2 units wide
(so area of each is 10x/4 units)
+ 10x = 39
e
h
s
g
x
f
5/2
Mathematical Association 2009 – p.12/30
Completing the Square
1. x2 term: centre square s
x2
2. 10x: add e, f , g, h, each 5/2 units wide
(so area of each is 10x/4 units)
+ 10x = 39
3. total area of solid figure is 39 units
e
h
s
g
x
f
5/2
Mathematical Association 2009 – p.12/30
Completing the Square
1. x2 term: centre square s
x2
2. 10x: add e, f , g, h, each 5/2 units wide
(so area of each is 10x/4 units)
+ 10x = 39
3. total area of solid figure is 39 units
e
4. complete the square: add four
squares, each of area 25/4 units
h
s
g
x
f
5/2
Mathematical Association 2009 – p.12/30
Completing the Square
1. x2 term: centre square s
x2
2. 10x: add e, f , g, h, each 5/2 units wide
(so area of each is 10x/4 units)
+ 10x = 39
3. total area of solid figure is 39 units
e
4. complete the square: add four
squares, each of area 25/4 units
h
s
g
x
f
5. area of large square is
39 + 25 = 64 units
5/2
Mathematical Association 2009 – p.12/30
Completing the Square
1. x2 term: centre square s
x2
2. 10x: add e, f , g, h, each 5/2 units wide
(so area of each is 10x/4 units)
+ 10x = 39
3. total area of solid figure is 39 units
e
4. complete the square: add four
squares, each of area 25/4 units
h
s
g
x
f
5. area of large square is
39 + 25 = 64 units
5/2
6. side of large square has length
x + 5 = 8 units
Mathematical Association 2009 – p.12/30
Completing the Square
1. x2 term: centre square s
x2
2. 10x: add e, f , g, h, each 5/2 units wide
(so area of each is 10x/4 units)
+ 10x = 39
3. total area of solid figure is 39 units
e
4. complete the square: add four
squares, each of area 25/4 units
h
s
g
x
f
5. area of large square is
39 + 25 = 64 units
5/2
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 Abaci
Proved no root to x3 + 2x2 + 10x = 20 can be constructed by
ruler 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:
φn − (1 − φn )
√
F (n) =
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 chess
illustrated by Leonardo da Vinci
Summa de arithmetica, geometrica,
proportioni et proportionalita
Summary of current work in arithmetic, algebra, geometry
and trigonometry.
Mathematical Association 2009 – p.15/30
Pacioli’s notation
6.p.R.10
18.m.R.90
108.m.R.3240.p.R.3240.m.R.90
hoc 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 form
x3 + mx = n algebraically
• no knowledge of negative numbers
• solution by radicals: find the roots by adding,
subtracting, multiplying, dividing and taking roots of the
coefficients.
• did not publish any of his work
• passed his greatest discovery on to his student Fior
from his deathbed
Mathematical Association 2009 – p.17/30
Scipione dal Ferro (1465-1526)
• Professor in Bologna, Italy
• solved depressed cubic equations of the form
x3 + mx = n algebraically
• no knowledge of negative numbers
• solution by radicals: find the roots by adding,
subtracting, multiplying, dividing and taking roots of the
coefficients.
• did not publish any of his work
• passed his greatest discovery on to his student Fior
from his deathbed
“ It is often more convenient to possess the ashes
of great men than to possess the men themselves
during 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-solving
contest with Fior by solving other
types 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 magna
Methods 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 magna
Methods for cubics and quartics (done by Ferrari).
“I swear to you, by God’s holy Gospels, and as a true
man of honour, not only never to publish your
discoveries, if you teach me them, but also I promise
you, and I pledge my faith as a true Christian, to note
them down in code, so that after my death no one will
be able to understand them.”
Mathematical Association 2009 – p.19/30
Cardano-Tartaglia Method for Cubics
“Quando che’l cubo con le cose appresso
Se agguaglia a qualche numero discreto
Trovati dui alte differenti in esso
Dapoi terrai, questo per consueto, . . . ”
Mathematical Association 2009 – p.20/30
Cardano-Tartaglia Method for Cubics
“Quando che’l cubo con le cose appresso
Se agguaglia a qualche numero discreto
Trovati dui alte differenti in esso
Dapoi terrai, questo per consueto, . . . ”
When the cube and its things near
Add to a new number, discrete,
Determine two new numbers different
By that one; this feat
Will be kept as a rule
Their product always equal, the same,
To the cube of a third
Of the number of things named.
Then generally speaking,
The remaining amount
Of the cube roots subtracted
Will be your desired count.
x3 + mx = n
a3 − b3 = n
a3 b3
=
m 3
3
x=a−b
Mathematical Association 2009 – p.20/30
Solving cubics
• AIM:
solve
x3 + mx = n
(1)
Mathematical Association 2009 – p.21/30
Solving cubics
• AIM:
• IDEA:
solve
x3 + mx = n
(1)
use identity (a − b)3 + 3ab(a − b) = a3 − b3
(2)
Mathematical Association 2009 – p.21/30
Solving cubics
• AIM:
x3 + mx = n
solve
• IDEA:
(1)
use identity (a − b)3 + 3ab(a − b) = a3 − b3
• METHOD: choose a and b so
3ab = m,
(2)
a3 − b3 = n
3
3
m
m
m
6
3
b=
⇒ a3 −
=
n
⇒
a
−
na
−
= 0.
3
3a
27a
27
solve this quadratic in a3 :
a=
n
+
2
r n 2
2
+
m 3
3
!1
3
,
b=
!1
r n
n 2 m 3 3
+
− +
2
2
3
Mathematical Association 2009 – p.21/30
Solving cubics
• AIM:
x3 + mx = n
solve
• IDEA:
(1)
use identity (a − b)3 + 3ab(a − b) = a3 − b3
3ab = m,
• METHOD: choose a and b so
(2)
a3 − b3 = n
3
3
m
m
m
6
3
b=
⇒ a3 −
=
n
⇒
a
−
na
−
= 0.
3
3a
27a
27
solve this quadratic in a3 :
a=
n
+
2
r n 2
2
+
m 3
3
!1
3
,
b=
• from (2), cubic (1) has solution
!1
r n
n 2 m 3 3
+
− +
2
2
3
x=a−b
Mathematical Association 2009 – p.21/30
Solving cubics cont.
x=
n
+
2
r n 2
• Case 1
2
+
m 3
3
! 13
−
n
− +
2
r n 2
2
+
m 3
3
! 13
x3 + mx = n
x3 + 6x = 20 ⇒ m = 6, n = 20
√ 13 √ 13
x = 10 + 108 − −10 + 108 = 2
Mathematical Association 2009 – p.22/30
Solving cubics cont.
x=
! 13
r n
n 2 m 3
+
+
+
2
2
3
• Case 2
! 13
r n
n 2 m 3
− +
+
2
2
3
x3 = mx + n
(use x = a + b)
x3 = 15x + 4 ⇒ m = 15, n = 4
√
√
13
13
x = 2 + −121 − 2 − −121
Mathematical Association 2009 – p.23/30
Solving cubics cont.
x=
! 13
r n
n 2 m 3
+
+
+
2
2
3
• Case 2
! 13
r n
n 2 m 3
− +
+
2
2
3
x3 = mx + n
(use x = a + b)
x3 = 15x + 4 ⇒ m = 15, n = 4
√
√
13
13
x = 2 + −121 − 2 − −121
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) + y 2
= (px2 − qx + p2 − r) + 2yx2 + 2py + y 2
= (p + 2y)x2 − qx + (p2 − r + 2py + y 2 )
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) + y 2
= (px2 − qx + p2 − r) + 2yx2 + 2py + y 2
= (p + 2y)x2 − qx + (p2 − r + 2py + y 2 )
for any
choose y so that RHS is a perfect square: set
q 2 − 4(p + 2y)(p2 − r + 2py + y 2 ) = 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) + y 2
= (px2 − qx + p2 − r) + 2yx2 + 2py + y 2
= (p + 2y)x2 − qx + (p2 − r + 2py + y 2 )
for any
choose y so that RHS is a perfect square: set
q 2 − 4(p + 2y)(p2 − r + 2py + y 2 ) = 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 ever
mathematical journal
• results rejected by the French
Academy as illegible
• died of tuberculosis
• only honoured after his death
On algebraic equations in which the impossibility of solving
the 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 ever
mathematical journal
• results rejected by the French
Academy as illegible
• died of tuberculosis
• only honoured after his death
On algebraic equations in which the impossibility of solving
the general equation of the fifth degree is demonstrated
Quintics are insoluble via method of radicals.
“Abel has left mathematicians enough to keep them
busy for 500 years.”
Hermite (1822-1901)
Mathematical Association 2009 – p.25/30
Evariste Galois (1811-1832)
• French student and soldier
• Cauchy lost his French Academy
paper, Fourier died before receiving it
• imprisoned for treason
• persecution complex
• killed in a duel
Researches on the algebraic solution of equations
Gave a method of determining when a general equation
could 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 Academy
paper, Fourier died before receiving it
• imprisoned for treason
• persecution complex
• killed in a duel
Researches on the algebraic solution of equations
Gave a method of determining when a general equation
could be solved by radicals: Galois theory of groups.
“. . . had the brilliant idea of just making up fictitious
numbers to supply answers to problems which
didn’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 in
1665
• Professor at Cambridge
• Warden of the Royal Mint
De analysi per aequationes numero terminorum infinitas
Iterative method for finding approximate roots of algebraic
equations.
Mathematical Association 2009 – p.27/30
Sir Isaac Newton (1642-1727)
• born in Grantham
• studied at Cambridge University
• sent home due to the Plague in
1665
• Professor at Cambridge
• Warden of the Royal Mint
De analysi per aequationes numero terminorum infinitas
Iterative method for finding approximate roots of algebraic
equations.
“Newton was of the most fearful, cautious and
suspicious temper that I ever knew.”
Whiston (1667-1752)
Mathematical Association 2009 – p.27/30
Johann Carl Friedrich Gauss (1777-1855)
• Numerical methods for solving
equations
• Gaussian Elimination
• Iterative methods:
Jacobi, Gauss-Seidel, . . .
Mathematical Association 2009 – p.28/30
Johann Carl Friedrich Gauss (1777-1855)
• Numerical methods for solving
equations
• Gaussian Elimination
• Iterative methods:
Jacobi, Gauss-Seidel, . . .
“I have had my results for a long time: but I do not
yet know how to arrive at them.”
Gauss
Mathematical Association 2009 – p.28/30
Some final thoughts . . .
“Life is good for only two things, discovering
mathematics and teaching mathematics.”
Siméon-Denis Poisson (1781-1840)
Mathematical Association 2009 – p.29/30
Some final thoughts . . .
“Life is good for only two things, discovering
mathematics 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, discovering
mathematics 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 the
worst mistakes that can be made in his subject, and
how 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 room
looking 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 room
looking for a black hat which isn’t there.”
Charles Darwin (1809-1882)
“A mathematician is a machine for turning coffee
into theorems.”
Paul Erdös (1913-1996)
Mathematical Association 2009 – p.30/30
Some more final thoughts . . .
“A mathematician is a blind man in a dark room
looking for a black hat which isn’t there.”
Charles Darwin (1809-1882)
“A mathematician is a machine for turning coffee
into 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