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Chapter : Mixing Up Ancient and
Modern
1 Introduction: Pythagoras’ and Euclid’s Theorems.
2 Numbers and Geometry,
3 Algebraic Equations and Geometry,
4 Babylonians and Second Degree Algebraic Equations,
5 Numbers, Basis and Polynomials,
Exercises.
1.
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Chapter 1
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Mixing Up Ancient and
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Modern
1.1 Introduction: Pythagoras’ and Euclid’s
Theorems
The Indian mathematical doctrines, as well as the Assyrian-Baby-
lonian, have an undoubted cultural charm, because they do not seem
a self-contained body, built on Axioms and Theorems (as in the Greek
tradition), but arise as one set of insights to be introduced ad hoc
in the context of specific needs. The Pythagorean Theorem, as pre-
sented in the Elements of Euclid, requires a long elaboration and is
introduced after the study of the properties of triangles, of Euclid’s
Theorems . . . The Indians conceived it in a hybrid form (horrifying
for the Greeks) but effective from a practical point of view1 .
With reference to Fig. 1.1, we note that the surface of the in-
ner square (Qin ) is linked to that of the outer square (Qout ) by the
obvious identity
Qin = Qout − 4 T (1.1.1)
where T represents the area of the triangles with edges a, b. The
1
There is an astonishing number of “independent” proofs of the Theorem.
A. Bogomolny reported 122 different demonstrations in http://www.cut-the
knot.org/pythagoras/index.shtml.
1
2.
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Figure 1.1: An algebraic proof of the Pythagorean Theorem.
surface of the outer square is
Qout = (a + b)2 (1.1.2)
and being
ab
Qin = c2 , T = , (1.1.3)
2
it follows
2 2 ab
c = (a + b) − 4 (1.1.4)
2
thus eventually yielding the identity
c2 = a2 + b2 (1.1.5)
which once a and b are known, specifies the edge c. It represents the
Pythagorean Theorem.
An Euclidean Bourbakist2 would have good reasons to object,
because the rules of the game were not respected. We have indeed
2
Bourbakist is an adjective indicating a mathematician belonging to a school
of extreme mathematical rigor. The name Nicolas Bourbaki is the collective
pseudonym chosen by an authoritative group of French mathematicians, who in-
3.
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Mixing Up Ancient and Modern 3
forgotten to prove that the four triangles of sides a, b are equal and
that are right angled, we have not respected the rule according to
which the use of algebra in a geometric demonstration is contrary to
the spirit of the geometrical proof itself... Despite these shortcuts,
the method is undoubtedly effective. Perhaps it was an intuition of
this type that brought Pythagoras (Assyrians and Indians as well,
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before him) to formulate the Theorem in 6th century BC.
We could take advantage of the same procedure to deduce the
well known Euclid’s Theorems.
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Figure 1.2: Euclid Theorems.
In Fig. 1.2 we keep the height relative to the c side of one of the
triangles inside the square of the Fig. 1.1 and, denoting by p and
q = c − p, the projections of the catheti a and b (see the figure) on
the hypotenuse c we obtain
h2 + p 2 = a 2 , h2 + (c − p)2 = b2 , (1.1.6)
tended to reformulate the teaching of Mathematics on new grounds which accepts
only a well organized system of Axioms, namely the opposite of what we have
discussed so far.
4.
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which hold as a consequence of the fact that the triangles of edges
(h, a, p), (h, b, q) are right angled. Subtracting the respective sides of
the two previous equalities, we arrive at the following result
2cp − c2 = a2 − b2 (1.1.7)
which once embedded with the Pythagorean Theorem yields
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cp = a2 , cq = b2 . (1.1.8)
Furthermore, once combining the previous identities, we end up with
h2 = a2 − p2 = (c − p)p, h2 = pq. (1.1.9)
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The last two identities are the Euclid’s first and second Theorems,
whose meaning is illustrated in Fig. 1.2.
We have therefore reversed the point of view expressed by Greek
mathematics according to which the Pythagorean Theorem is de-
duced from those of Euclid.
The ancient Greeks suffered from many idiosyncrasies. With re-
gard to geometry, they fixed precise rules, which determined the
genesis of the great questions that lasted for millennia. According to
Greek mathematicians, the tools allowed for geometric constructions
were the ruler and the compass. This prescription prevented the pos-
sibility of squaring the circle, namely of constructing a square with
the area equivalent to that of a circle. The proof that π is a tran-
scendental number was stimulated by the impossibility of squaring
the circle according to the ancient Greeks prescription. The solution
of the problem required more than two thousand years. Entire new
fields of Math had to be explored, astronomical distances in math
knowledge had to be covered, before stating that circle cannot be
squared employing Euclidean tools only. The answer was contained
in the transcendental nature of π, namely that it is not the solution
of any algebraic equation with rational coefficients. A statement so
apparently conceptually far from the formulation of the problem it
solved. This is a funny aspect of Math, it happened also for the solu-
tion of the last Fermat Theorem, whose solution required the creation
of new chapters in mathematical thought, as we will comment in the
forthcoming parts of this book.
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1.2 Numbers and Geometry
The ancient Egyptians too knew Pythagoras’ Theorem before
Pythagoras himself. They transformed it into a technological tool.
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Figure 1.3: Egyptian or Pythagorean triad for a = 16, b = 63, c = 65
and a2 = 33, b2 = 56, c2 = c = 65.
According to Fig. 1.3 a cord can be divided by a number of
equidistant knots, say 12, so as to form three consecutive groups of
3, 4 and 5. This triad constitutes the so-called Egyptian or Pythago-
rean triad and, since
32 + 42 = 52 , (1.2.1)
the nodes mark the sides of a right-angled triangle and therefore can
be used to “square” blocks of stone to be used in constructions.
The Assyrian-Babylonians had gone further: a concrete proof of
their knowledge is codified in their texts on clay tablets (see the Fig.
1.4 below).
Documents, dating between 2000 and 1500 BC, testify that they
knew the properties of the cords in a circle, the volume of the pyramid
and . . . the Pythagorean Theorem. Regarding the latter, they listed
“tables”, on clay tablets, which described Pythagorean triads, or
three integers that satisfy the Pythagorean Theorem. The titanic
computational effort (for the times) had been largely justified by the
practical outcome. Today we would not need it, but we could also
ask ourselves what links the Egyptian triad to others, such as those
6.
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Figure 1.4: Clay tablet.
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listed below
(3, 4, 5) (5, 12, 13) (7, 8, 15)
(1.2.2)
(7, 24, 25) (12, 35, 37) (15, 36, 39).
To understand this link, we write our triad (3, 4, 5) as a column vector
3
ε= 4 (1.2.3)
5
and ask ourselves the problem of seeking a suitable transformation
which by acting on the vector “returns” another Pythagorean term,
for example
5
L · ε = 12 . (1.2.4)
13
By skipping the various, not entirely trivial, aspects of the search for
the solution, it is possible to prove that this transformation exists
and that can be expressed in terms of a matrix, given by
1 −2 2
L = 2 −1 2 . (1.2.5)
2 −2 3
In more general terms it can be proved that the solution is not unique
and other two (independent) matrices accomplish the same task,
7.
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Mixing Up Ancient and Modern 7
namely
1 2 2 −1 2 2
U = 2 1 2 , R = −2 1 2 (1.2.6)
2 2 3 −2 2 3
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The previous matrices are called Hall matrices3 , they were in-
troduced in the early 1970s of the last century as the result of a non-
trivial effort, which demonstrates how a problem, several thousand
years old, has continued to arouse interest (and not only historical)
until recent times.
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1.3 Algebraic Equations and Geometry
The nuisance of the Greeks for everything that could not be
constructed with ruler and compass has already been underscored.
√
While they recognized that an irrational number, like 2, can be
constructed with Euclidean tools (as shown in the following figure),
they badly accepted its irrationality: a stain in the Pythagorean uni-
verse. It was kept as a secret, in the circle of initiates, and those
spreading it out were punished with death!
The construction of this irrational number involves a fairly simple
procedure (see Fig. 1.5):
1. Define an oriented axis, that we will say the line of the numbers,
fix on this a point A that stands out 1 from the origin O;
2. Construct a segment BA of length 1, perpendicular to the line;
3. Determine
√ the length of the OB segment, which we know to
be 2 as a consequence of the Pythagorean Theorem;
3
See e.g. A. Hall, Genealogy of Pythagorean Triads, Classroom Notes 232,
The Mathematical Gazette, vol. 54, 390, 1970, pp. 377–379 (and reprinted in
Biscuits of Number Theory, editors Arthur T. Benjamin and Ezra Brown) or B.
Berggren, Pytagoreiska triangular, Tidskrift for Elementar Matematik, Fysik och
Kemi, 17, 1934, pp. 129–139. For a more recent study see J. Miki, A Note on
the Generation of Pythagorean Triples, MAT-KOL (Banja Luka), XXIV, 1, 2018,
pp. 41–51 www.imvibl.org/dmbl/dmbl.htm, doi: 10.7251/MK1801041M.
8.
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Figure 1.5: Geometrical constructions of some irrational numbers.
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4. The point B can be projected on the axis OA by carrying a
rotation pointing
√ the compass at O, we have determined the
position of 2 on the line of numbers;
5. This procedure has allowed the construction of a “quadratic”
irrational number using the means of the Euclidean paradigms.
We can push the argument even further, a quadratic irrational
can be viewed as the root of a second degree algebraic equation. The
just foreseen simple procedure can be extended to display the geo-
metrical nature an equation of second degree.
Figure 1.6: Geometrical point of view of a second degree algebraic
equation.
9.
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Mixing Up Ancient and Modern 9
To accomplish this task we refer to Fig. 1.5 and follow the in-
structions given below.
1. Build a segment, bounded by the ends A, B;
2. Construct a segment of length < AB
2 , perpendicular to AB at
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one of the extremes, say B, and indicate the upper end with
C;
3. Construct a semicircle of diameter equal to AB;
4. Move BC, parallel to itself, inside the circle, indicating with
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P the point of contact with the circumference and with H the
point lying on the diameter AB, so that BC = P H;
5. From elementary geometry we know that the AP B triangle is
right angled;
6. Apply the second Euclid Theorem and get
AH · HB = P H 2 (1.3.1)
7. Use the following definitions
AH = x HB = AB − x AB = s P H = p (1.3.2)
8. Then get p
p= xs − x2 (1.3.3)
which is a second degree algebraic equation, with a transparent
geometrical meaning in the spirit of the ancient Greek mathe-
matics.
Quadratic irrationals are accordingly nicely fitted within the geomet-
rical cage.
Cubic irrationals are even
√ more scandalous than their quadratic
3
counterparts. The
√ number 2 cannot be constructed with ruler and
compass. That 3 2 is irrational it is easy to prove it, in light of what
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the Greeks did not know. If it were rational, we would be authorized
to write √3 m
2 = , ∀m, n ∈ Z (1.3.4)
n
according to which, after taking the cube of both sides, we find
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m3 = 2n3 = n3 + n3 (1.3.5)
and we might accordingly have stated the existence of a triple (m, n, p)
satisfying the identity
m3 = n3 + p3 (1.3.6)
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In clear contrast with the Fermat Theorem (not important if p = n).
Therefore, through
√ the “reductio ad absurdum” procedure we have
shown that 3 2 is an irrational number (a nice exercise for the reader
is
√ to check whether the same method could be used to prove that
n
2 is irrational . . . ).
Let’s ask ourselves if there is a “super-Pythagorean” triangle
whose sides satisfy for example the identity (see also4 Fig. 1.7)
a3 + b3 = c3 . (1.3.7)
The use of the algebraic rule of the decomposition of the sum of two
cubes
a3 + b3 = (a + b) a2 − ab + b2 ,
(1.3.8)
eventually yields
c3 = (a + b) a2 − ab + b2 .
(1.3.9)
Albeit “Super-Pythagorean” triangle, it satisfies the Theorem of co-
sine5
c2 = a2 − 2 ab cos(φ) + b2 (1.3.10)
4
It should be noted that numbers representing the sides of a triangle are pos-
itive and satisfy the triangular inequalities, namely a + b > c. Under these
assumptions the Fermat Theorem is straightforwardly proved using elementary
means.
5
Sometimes defined as Carnot Theorem, it was however well known to Euclid.
11.
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Figure 1.7: Super-Pythagorean Theorem.
which, combined with the fundamental identity for Fermat triangles,
yields
c a2 − ab + b2
= 2 . (1.3.11)
a+b a − 2 ab cos(φ) + b2
By assuming that the triangle is isosceles with a = b = l, we reduce
the above identity to
c l2 − l2 + l2 1 c 1
= 2 = ⇒ = .
2l l − 2l2 cos(φ) + l2 2 − 2 cos(φ) l 1 − cos(φ)
(1.3.12)
On the other side, it is also true that
c 3
l3 + l3 = c3 ⇒ =2 (1.3.13)
l
which yields
√
3 1
2= . (1.3.14)
1 − cos(φ)
Therefore, if we construct an isosceles triangle of side 1 (with an
angle at the vertex of about φ = 1.364 rad) we have automatically
12.
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obtained a super-Pythagorean triangle√ and we have geometrically
3
constructed the irrational number 2 . . . Problem solved? Obvi-
ously not!!! At least according to the prescriptions of the ancient
Greeks. In the construction we did not use either row or compass
and we made everything depending on the solution of an equation of
third degree, unknown to the Greeks and not-soluble with Euclidean
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instruments. The ignorance of the algebraic equations higher than
the second caused the Greeks not a little trouble. During a plague
that raged in Athens, the Delo oracle ruled that the scourge would be
removed if the volume of the temple dedicated to Apollo had dou-
bled. It was a cube with a side of length l. In modern terms the
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problem of the oracle is reduced to the solution of the equation
x3 = 2 l 3 (1.3.15)
where x is the side of the cube, doubling the original volume. The
problem, fatally (it must be said), is that of defining the cube root of
2, which, as we have seen, was not easy for the Greeks. The solution
frustrated the best minds at the dawn of Western thought and the
plague further raged. A mathematician by the name of Menecmo
(about 320 AC) proposed a solution of geometric nature, through
the use of conics. Let’s try to follow an argument that makes us
understand how the problem can be solved in these terms.
We rewrite our equation in the form
x2 2l2
= (1.3.16)
l x
then we set
x2 2l2
y= ⇒ y= . (1.3.17)
l x
The intersection between the parabolas and the hyperbola will pro-
vide the value of the unknown x. Even combining the previous rela-
tionships we can get another parabola
y 2 = 2 lx (1.3.18)
which intersected with one of the previous conics will provide the
same solution, as it has been shown in Fig. 1.8.
13.
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35
30 y=x 2 / l
25 y=2l2 / x
20 y= 2 lx
y(x,l)
15
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10
5 P
0
0 2 4 6 8 10
x
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Figure 1.8: Intersection between the parabolas (1.3.17) and the hy-
perbola (1.3.18) for l = 2. P ≡ (2.5198, 3.1748) is the intersection
point.
The problem was solved, other proposed solutions (even one by
Plato) will not be reported here. The chronicles do not tell if the
plague had diminished or not. We could, however, venture a (mod-
ern) hypothesis: infectious diseases can be described in terms of the
predator-prey mechanism, the predator in this case is the bacillus of
the plague and prey is man. From a mathematical point of view,
after a long enough time, the system reaches a state of balance and
the disease recedes. If the time necessary to find the solution was
comparable to that characterizing the dynamics of the predator-prey
interaction the Greeks might have convinced themselves that it was
just the strength of the conics, to defeat the raging morbus. In
addition we have not too much arguments to say that they were
not right. It may be argued that Menecno did not use Euclidean
tools, but conics. The rebuttal is straightforward, conics too can be
constructed with ruler and compass. In the 18th century, the Italian
mathematician, Lorenzo Mascheroni showed in his book “La Geome-
tria del Compasso” that the compass was sufficient to accomplish the
Euclidean geometric construction. One of the problem treated in the
book was even that of duplicating the cube by the use of compass.
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1.4 Babylonians and Second Degree
Algebraic Equations
Since we have quoted the equations of second degree and the at-
titude of the people of pre-Hellenic culture to have a more flexible
attitude towards Mathematics itself, to which they looked for prac-
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tical reasons, we believe it is appropriate to say something about the
algebraic abilities of the Babylonians. They would have certainly
solved the problem of Menecno without resorting to conics. We con-
sider it absolutely noteworthy that the solution of the second degree
equations was known to the Babylonians, who had made it a working
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tool6 . Probably the Babylonians had come to understand that the
solution of a second degree equation is reducible to the extraction
of a square root and to the solution of a first degree equation. We
will discuss the so-called Babylonian algorithm for calculating square
roots in the next chapter.
In order to better appreciate the Babylonian method, let us re-
member that the logical steps that lead to the solution of one second
degree equation are the following:
1. Factorize the second degree polynomial as
" 2 2 #
b c b b −4ac
(ax2 + bx + c) = a x2 + x+ = a x+ − ;
a a 2a 4a2
(1.4.1)
2. Keep the square root and determine the two distinct roots of
the equation
√
−b+ ∆
2 2
b −4ac
x+ = 2
x+
b
= ⇒ 2a√ , ∆ = b −4ac .
2a 4a2 −b− ∆ 4a2
x− =
2a
(1.4.2)
The example which we report below, and found on a tablet dating
back to 4000 BC, is presumably an exercise proposed for educational
6
This aspect will be commented further in this book.
15.
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Mixing Up Ancient and Modern 15
purposes. It represents an extremely important example to under-
stand the path of evolution of Mathematics.
The problem is formulated as follows:
I added 7 times the side of my square to 11 times its surface and
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I got 6.25, how much is the side?
Translated in a modern mathematical language the solution of
the exercise is provided by the roots of the second degree algebraic
equation
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11x2 + 7x = 6.25. (1.4.3)
The tablet does not contain a general formula for the solution of
second degree equations but a series of instructions reported below.
For an easy comparison with the “modern rule”, bear in mind that
a = 11, b = 7, c = −6.25. (1.4.4)
The solution recipe follows the steps listed below.
1. Multiply 11 and 6.25: (−a · c = 68.75);
b
2. Divide 7 by 2: = 3.5 ;
2
2
b
3. Square it: = 12.25 ;
4
4. Add it to the result in step 1:
b2 − 4ac
12.25+68.75 = 81 ⇒ = 81 ;
4
r !
√ b2 − 4ac
5. Take the square root: 81 = 9 = ;
4
6. Subtract the result in 2 from 9:
r !
b2 −4ac b
9−3.5 = 5.5 ⇒ − = 5.5 ;
4 2
The cooking recipe ends up with the crucial point:
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16 Vedic Mathematics
7. Find the number which multiplied by 11 returns 5.5, it will
provide the length of the square side (this instruction amounts
to
r ! √
1 b2 − 4ac b −b + ∆
11 · 0.5 = 5.5 ⇐= − ⇐= x+ =
a 4 2 2a
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(1.4.5)
which is the positive root of the second degree equation).
That’s the tool kit!
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No explanation why it goes that way, no attempt to frame it into a
wider context. We have noticed that only one root has been reported.
Although natural to discard the negative root for the solution of a
geometric problem, it seems however that it was systematic, even
when two positive solutions were admitted. This is supposedly due
to the fact that those ancient mathematicians did not know that,
in the process of extracting the square root, one can have positive
and negative numbers as well. Apart from these details, it is truly
amazing that a problem, posed (and solved in its essential lines) 5000
years ahead of the era of the great algebraic masters of the modern
era, has been “forgotten” for a few millennia.
Babylonians were also able to calculate cubic roots. They might
have been able to solve quite straightforwardly the problem of cube
duplication. They had “tables” of exact cubic roots, if the number
from which to extract the root had not been listed they used an
algorithm, which from our (modern) perspective is reduced to the
identity
√ a √
r
3 3
a= 3 · b (1.4.6)
b
where b is chosen to be an exact cube root.
The example we give is taken from a tablet in which it was re-
quired to evaluate the cubic root of 729000 that is written as
√ √ √
r
3 3 729000 3 3
729000 = · 27000 = 27 · 30 = 90 (1.4.7)
27000
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Mixing Up Ancient and Modern 17
The scribe, presenting the exercise, took care to choose known roots,
if they were not, the procedure to be followed is a sort of interpola-
tion, as we will see in the next chapters.
Let us now summarize what has been discussed so far. We gave
a bird’s eye view, in perfect Assyrian-Babylonian style, over four
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millennia of Mathematics. In our “tour” we mixed ancient and mod-
ern, without taking care of the logical connection and resorting to
concepts (eventually true) used, for our purposes, in “extemporane-
ous” form, that is, not inserted in a coherent “corpus”, such as the
canonized one in the Elements of Euclid.
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1.5 Numbers, Basis and Polynomials
One of the highest accomplishments of Mathematics has been the
discovery of the decimal numeral system which is the way suggested
by Arabs and Indus to express numbers in the current notation. Any
number can be ordered in powers of 10, accordingly the number 231
is written as
231 = 2 · 102 + 3 · 10 + 1 · 100 = {2, 3, 1}10 . (1.5.1)
The notation we have foreseen consists of a series of numbers (the
composing digits) in brace brackets and of a further number (the
base), appended as lower index to the right bracket. The further
rule is that the inner bracket numbers are non negative integers less
than 10 (all digits 0, 1, . . . , 9). According to these prescriptions the
base is not unique (10 was perhaps chosen in correspondence of the
number of human fingers) but any other positive integer may be
employed as well. The use of the base 7 yields
{2, 3, 1}7 = 2 · 72 + 3 · 7 + 1 · 70 = 120. (1.5.2)
The use of a more generic p and an inner bracket triad a, b, c arranged
as
{a, b, c}p = ap2 + bp + c (1.5.3)
represents another way of expressing a second degree polynomial.
The use of this notation suggests that the numbers (a2 , 2a, 1)p are
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