CM360X. Notes on the Middle Ages and
Renaissance.
1
Some opinions.
All would-be historians of medieval mathematics must ask themselves where to
look for their subject matter.
[Thakkar (2009), p. 618 — proposing theology textbooks as a promising
source.]
[T]he mechanical and physical science of which the present day is so proud
comes to us through an uninterrupted sequence of almost imperceptible refinements from the doctrines professed in the Schools of the Middle Ages. The
so-called intellectual revolutions consisted,in most cases, of nothing but an evolution developing over long periods of time. The so-called Renaissances were
frequently nothing but unjust and sterile reactions. Finally, respect for tradition
is an essential condition for all scientific progress.
[Duhem (1991), p.9.]
In fact, it had been one of the traditions of Scholasticism from the twelfth
century onward to employ the so-called sic et non method, advocated especially
by Abelard; its principle was that in dealing with a given subject all the opinions
that had ever been pronounced about it and all the arguments that could be
advanced for or against a certain view were enumerated and discussed as fully
as possible...This method, of course, presented great advantages; it bespoke a
striving after objectivity and it helped to prevent an idea, once it had been
pronounced, from falling into oblivion again. It is, however, obvious that if
the method were applied too thoroughly, the disadvantages would be bound to
preponderate.
[Dijksterhuis (1986), p.167.]
The elements of mathematics, that is to say number and measure, termed
arithmetic and geometry, discourse with supreme truth on discontinuous and
continuous quantities. Here no one argues that twice three makes more or less
than six, nor that a triangle has angles smaller than two right angles, but with
eternal silence, every dissension is destroyed, and in tranquility these sciences
are relished by their devotees.
[Leonardo da Vinci notes in ‘Paris Manuscript M’ (around 1500), online at
http://www.universalleonardo.org/trail.php?trail=545&work=323.]
Behold, the art which I present is new, but in truth so old, so spoiled and
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defiled by the barbarians, that I considered it necessary, in order to introduce
an entirely new form into it, to think it out and publish a new vocabulary,
having got rid of all its pseudo-technical terms lest it should retain its filth and
continue to stink in the old way...And yet underneath the Algebra or Almucabala
which they lauded and called ‘the great art’, all Mathematicians recognized that
incomparable gold lay hidden, though they used to find very little.
[Viéte, The Analytic Art, in Klein (1968), pp. 318-9.]
It has become a matter of common usage to call the barbarous age that time
which extends from about 900 or a thousand years up to about 150 years past,
since men were for 700 or 800 years in the condition of imbeciles without the
practice of letters or sciences...but although the afore-mentioned preceding times
could call themselves a wise age in respect to the barbarous age just mentioned,
nevertheless we have not consented to the definition of such a wise age, since
both taken together are nothing but the true barbarous age in comparison to that
unknown time at which we state that it [i.e. the true wise age] was, without
any doubt, in existence.
[Stevin, Géographie, quoted in Klein (1968) p.187.]
2
Some extracts.
A. From Oresme, Quaestiones super Euclidem.
Next we inquire whether an addition to any magnitude could be made by
proportional parts.
First, it is argued that it cannot be, since then it would follow that a magnitude could be capable of being increased to an actual infinity. This consequence
contradicts what Aristotle says in the third book of Physica and also Campanus’
statement [in his commentary on the Common Notions in Euclid I], where he
distinguishes between a number and a magnitude, in that a number can increase
indefinitely and not decrease indefinitely, but the reverse is true of a magnitude.
Proof of the consequence: From the fact that the addition takes place indefinitely it follows that the increase too takes place indefinitely.
Against this it is argued: anything that is taken away from one magnitude
can be added to another. It is possible to take away from a magnitude an infinite
number of proportional parts1 , therefore it is also possible to prove that it can
be increased by an infinite number of parts.
Secondly, it must be noted that if an addition were made to infinity by
proportional parts in a ratio of equality or of greater inequality, the whole would
become infinite; if, however, this addition should be made [by proportional parts]
in a ratio of lesser inequality, the whole would never become infinite, even if
the addition continued into infinity. As will be declared afterward, the reason is
because the whole will bear a certain finite ratio to the first [magnitude] assumed
to which the addition is made... [Here follow some definitions on fractions.]
1 This is the substance of the previous ‘question’, which deals with the successive subtractions which exhaust a magnitude, as in Euclid X.1.
2
The first proposition is that if a one-foot quantity should be assumed and
an addition were made to it into infinity according to a subdouble [that is onehalf] proportion so that one-half of one foot is added to it, then one-fourth,
then one-eighth, and so on into infinity by halving the halves [lit. doubling the
halves], the whole will be exactly twice the first [magnitude] assumed. This is
clear, because if from something one takes away successively these parts, then
[one is left with nothing, and so] from the double quantity one has taken away
the double, as appears by question 1 [which was about subtraction]; and so by
a similar reasoning, if they are added.
The second proposition is this, that if a quantity, such as one foot, were
assumed, then a third were added and then after a third [of that] and so into
infinity, the whole would be precisely one foot and a half, or in the sesquialterate
proportion [this is the medieval terminology for the ratio of 3 to 2]. Furthermore,
this rule should be known: We must see how much the second part falls short of
the first part, and how much the third falls short of the second, and so on with
the others, and denominate this by its denomination,and then the ratio of the
whole aggregate to the quantity [first] assumed will be just as a denominator
to a numerator. [This looks very obscure, but the meaning seems to be that, if
your ratio is a fraction q so the series is a + aq + aq 2 + . . ., then the ‘falling short’
is 1 − q; and you invert this — exchange the denominator and the numerator
1
, which is the ratio of the sum to the first term a. There
— to get the sum 1−q
is no proof.] For example: in the above the second part, which is a third of the
first, differs from the first by two thirds, therefore the ratio of the whole to the
first part or the the assumed quantity is as 3 to two and this is sesquialterate.
The third proposition is this: It is possible that an addition should be made,
though not proportionally, to any quantity by ratios of lesser inequality, and
yet the whole would become infinite. For example, let a one-foot quantity be
assumed to which one-half of a foot is added during the first part of an hour,
then one-third of a foot in another, then one-fourth, then one-fifth, and so on
into infinity following the series of numbers, I say that the whole would become
infinite, which is proved as follows: There exist infinite parts any one of which
will be greater than one-half foot and [therefore] the whole will be infinite. The
antecedent is obvious, since one-quarter and one-third are greater than onehalf; similarly from one-fifth to one-eighth is greater than one-half, and from
one-ninth to one-sixteenth, and so on into infinity...
B. Problems:
(i) from Leonardo of Pisa (‘Fibonacci’), Liber abaci, p.566. A quadratic
equation in the style of Arabic algebra.
I separated 12 into two parts, and I multiplied one by the other, and that
which resulted I divided by the difference between the parts, and 21 4 [i.e. 4 12 ]
resulted; you put the thing for the lesser part, and you multiply one by the
other, namely the 12 minus the thing, yielding 12 things minus the census [this
means the square of the thing], and you divide by the difference between the
portions, namely between the thing and 12 minus the thing, that is 12 minus
3
two things, and because you know that the result of this division is 12 4, you
multiply the 12 4 by 12 minus the two things, yielding 54 minus 9 things that are
equal to 12 things minus the census. Therefore you restore the census and the 9
things to both parts yielding the census plus 54 equal to 21 roots [at this point
we have x2 + 54 = 21x — ‘roots’ is the same as ‘things’]; therefore from the
square of half of the mumber of roots, namely from 14 110, you subtract the 54
leaving 14 56; the root of this, that is 12 7, you subtract from half of the number
of roots, namely 12 10, leaving 3 for the posed thing, namely for the lesser part,
therefore the greater part is 9.
(ii) from later ‘abbacus books’, linear equations.
1. A tree is 1/3 and 1/4 underground and above ground it is 30 braccia [unit
of length]. I want to know how long it is altogether?
2. A man had a denaro and another came to him and he asked, ‘I have one
denaro. How much do you have?’ And he replied as follows, ‘I have so much
that with the same amount and with one half of what I have and with a quarter
and with your denaro it would be 100.’ How much did he have?
[I’ll answer this — a typical example of ‘false position’ — and you can look
at the others. Suppose the amount was 4 — you choose an amount which makes
the fractions easy to work with. Then what he has, plus the same, plus one half,
plus one quarter, is 4 + 4 + 2 + 1 = 11. But we want it to be 99 (to make up
100 with the other’s denaro). So we must multiply by 9, and the answer is
4 × 9 = 36.]
3. A worker can do a job in 4 days and another worker can do the same job
in 5 days. I want to know putting these two masters on the same job, how long
will it take?
4. How much does 87 gold florins 35 s. 6 d. earn in 2 years 7 months and
15 days at 10 per cent simple interest? [Don’t try to answer this! I don’t even
know how many s. or d. there are in a gold florin, and I think the answer
is nasty. Questions involving interest are obviously important, and there are
plenty of these.]
C. Piero della Francesca’s description of a perspective construction.
(Note the practical geometry which the painter is advised to use; and the
amount of detail (the ultimate source is said to be Euclid’s Optics); and compare
Dürer’s picture.)
Now to demonstrate the way that I intend to follow I shall give two or three
examples of plane surfaces in order that by means of these you may more readily
understand the diminution of bodies. So let there be constructed in proper form
a square surface which is BCDE and then mark the point A which will be the
eye and it will be as far back as you wish to stand to see this surface at the
point A. Now fasten a nail or, if you will, a nail with a very fine silken thread,
the hair of a horse’s tail would be good, especially when it is far from the line.
A line FG is then drawn parallel to BC which will be the picture plane between
the eye and the surface . On this surface mark a point M which must be made
on each surface and on each body. It makes no difference where you make it
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because it is a certain limit as you will understand as you go on. Now one will
need a strip of wood that is very thin and straight. Then take one of these
strips and lay it such that it is flush with FG and make sure that it is firmly
positioned. Then take one end of the thread and take it to B of the place and
where it touches the strip of wood make a point B. Then stretch the thread to C
and where it touches the wooden strip draw M. Now mark an A on the wooden
strip called the wooden strip A and this strip is then taken away and laid to the
side. This is the wooden strip that indicates width. Now one needs to see how
much higher DE of this plane BCDE is than BC. One therefore positions A as
high above the line CE as one wants to stand to look at the plane in question,
while neither moving towards or away from FG which marks the picture plane.
After the eye has been fixed, as I described earlier, one takes a strip of paper
and one places it contiguous to FG and draws EC which which meets the strip
of paper at the point A, which will be the (paper) strip A. Then one takes the
thread to E and at the place where it touches the strip of paper one makes the
points C and B at this same spot. One then takes away the strip of paper and
one makes with it another that is identical to it with the same markings. And it
will likewise be marked A as in the other one. Then draw a straight line in the
place where you wish to make the perspectivally diminished plane, namely the
line EG and divide it in two at point M and above M draw a perpendicular to I
which will then become FH and GI. Then take the two strips of paper marked
with A: one is placed contiguous with to GI and (the point) A of both lies on
the line FG. Then one takes the strip of wood for width and one lays it over
the two strips of paper such that one goes through E and D of the two strips
(of paper) and M lies on the line MN and where the D of the the strip of wood
touches the place make a point D and where E touches draw E. One now brings
the wooden strip further down. such that it passes through B and C of the two
strips of paper and M lies on the line MN and where B falls mark the point B
and where C meets the strip of wood make the point C and the plane is drawn.
Take away the strip of wood and draw BC,BD,DE,EC which is the diminished
square plane which we said we would make.
(From On the Perspective of Painting (c.1474), book 3. Online at http://
www.mmi.unimaas.nl/people/Veltman/articles/perspectives/art36.htm)
D. Tartaglia; his ‘rhyme’ for solving the cubic, and a problem solved.
When the cube and the things together
Are equal to some discrete number
[To solve x3 + cx = d,]
Find two other numbers differing in this one.
Then you will keep this as a habit
That their product should always be equal
Exactly to the cube of the third of the things.
[Find u, v such that u − v = d and uv = (c/3)3 .]
The remainder then as a general rule
Of their cube roots subtracted
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Will be equal
√ principal thing.
√ to your
[Then x = 3 u − 3 v.]
Exercise. (i) Use this method to solve the equation ‘cube and three things
equal to four’, or x3 + 3x = 4. (Hint: You are given u − v and uv; find u + v.)
(ii) Why don’t you get the obvious answer 1?
(iii) Try to prove that x as given in Tartaglia’s formulation is a solution
of the general cubic equation ‘cube and things equal to numbers’ by ‘modern’
algebra.
[A ‘hard’ example from Tartaglia.] And if it were 1 cube plus 1 thing equal
to 11, it would be necessary to find two numbers or quantities such that one is
11 more than the other, and that the product of the one by the other should
1
, that is the cube of the third of the things, whence operating as above it
be 27
31
will be found that our thing is R/ u. cube R/ 30 108
plus 5 12 minus R/ u. cube
1
31
R/ 30 108 minus 5 2 and not other...
The ‘u.’ in the above is for ‘universal’; the whole means simply ‘cube root’.
‘R/ ’ is a common sign for ‘root’ at this time. We can recognize Tartaglia’s
solution, in our notation, as
sr
sr
31
31
1
1
3
3
30
30
+5 −
−5
108
2
108
2
E. From Viète, The Analytic Art.
Book II Zetetic XVII. Given the difference between the roots and the difference between their cubes, to find the roots. [Try to read through this text to
see what it means, if possible, before consulting the notes below.]
Let B be the difference between the roots and D solid the difference between
the cubes. The roots are to be found.
Let the sum of the roots be E. Therefore E + B will be twice the greater
root and E − B twice the smaller. [Why?] The difference between the cubes of
these is B in E squared 6 + B cubed 2 which is consequently equal to D solid
8.
( D solid 4 )
Therefore
−B cube
B3
equals E squared
The squares being given, the root is given, and the difference between the
roots and their sum being given, the roots are given.
Accordingly the difference of the cubes quadrupled, minus the cube of the
difference of the sides, being divided by the difference of the sides tripled, there
results the square of the sum of the sides.
If B is 6, D solid is 504, the sum of the sides 1N , 1Q equals 100.
Notes. A ‘Zetetic’ is Viète’s word for a method of finding out. In his
notation the ‘roots’ are lines, so the sum of their cubes is a ‘solid’, which is why
he calls it ‘D solid’; his rule is that (as the Greeks prescribed) you must always
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keep track of the dimensions of quantities and not set lines equal to solids, for
example. B and D solid are denoted by consonants, because they are known;
while E is a vowel, because it is unknown. Numbers come after the letters, so
that ‘E squared 6’ means what we’d call 6E 2 .
F. Stevin discovers recurring decimals — from The Dime.
It sometimes happens that the quotient cannot be expressed by whole numbers, like 0.4 divided by 0.03, [I am not using Stevin’s slightly messy notation,
which it’s hard to typeset] as follows:
1/ 1/ 1/ (1
4/ 0/ 0/ 0/ 0 0 0 (1 3 3 3
3/ 3/ 3/ 3/
Where there it appears that there will come infinitely many threes, there
will always be left 31 . In this case one can approximate as closely as the matter
requires, omitting the remainder. It is true that 13.33 13 , or 13.333 13 would be
the exact answer, but our aim is to operate in this Dime by whole numbers, for
we see that in business one takes no account of the thousandth part of a speck,
or a grain, etc...
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