HYPATIA

(370-415 A.D.)

by Alice Cotton

According to
historians, Theon, a noted Greek mathematician and astronomer had an intense
desire to create a "perfect human being" which manifested itself in
the upraising of his daughter, Hypatia. During a time when women were
considered less than human, such an idea was indeed revolutionary.

Historians seem
to agree that Theon guided the entire scope of her education and immersed her
in an atmosphere of learning, questioning and exploration. Hypatia received formal
training in the arts, literature, science and philosophy as well as lessons in
speech, rhetoric, the power of words, the power of hypnotic suggestion, the
proper use of voice, and the gentle tones considered pleasing. She was also
involved in vigorous physical training in order to perfect her body as well as
her mind.

As a result of
her training, her remarkable intelligence and her talent Hypatia turned out to
be a brilliant, charismatic and beautiful woman. Like her father she became
involved in an academic life at the University of Alexandria, the greatest seat
of learning in the world at the time. The university was staffed by top
scholars from all over the world and Hypatia made her mark as a distinguished
and highly sought after figure in mathematics, medicine and philosophy. It
seems she was a very popular lecturer and teacher in the areas of mathematics,
philosophy, astronomy and simple mechanics. Students came from all over to hear
her speak.

As no copies of
her written work exist today, (no doubt burned during the Christian
extermination of all "pagan" works), it is believed her mathematical
focus was inspired by Diophantus, who was an important mathematician who lived
and worked in Alexandria shortly before her time.
Much of what is known about her work comes from the letters of one of her
pupils, Synesius of Cyrene.

She wrote a
commentary on the first six books of Diophantus' Arithmetic . She also wrote a
treatise on the Conics of Apollonius (here she discusses the work of a
prominent Alexandrian who lived 500 years before her time). Hypatia seemed to
be aware of the importance of conic sections and curves. It is noted that an
interest in these shapes didn't reemerge again until the 17th Century.

Theon also taught
Hypatia to rebel against *religious dogmatism and *to think independently. She became a
leader of the Neo-Platonic school of philosophy (scientific rationalism), This
philosophy ran counter to the doctrinaire beliefs of the dominant Christian
religion of the time and she was seen as a threat during a time when the
struggle for political power in Alexandria became

heated. It seems
she was a close friend of Orestes, Roman governor of Alexandria. Orestes and
Cyril, Bishop of Alexandria, were politically opposed and since Hypatia was not
popular with the Christian religious leadership she became a easy mark for
Cyril.

Apparently
Hypatia did not disguise her contempt for the Christian religion and was
outspoken on a number of unpopular issues. For example, she is quoted as saying
such things as:

On Christianity:

“Why should
learning, authority, antiquity, birth, rank, the system of empire which has
been growing up, fed by the accumulated wisdom of ages - why, I say, should any
of these protect your life a moment from the fury of any beggar who believes
the Son of God died for him as much as for you, and that he is your equal, if
not your superior, in the sight of his low-born and illiterate deity?”

She was an
elitist:

“Me
struggle is simply one between the aristocracy and the mob - between wealth,
refinement, art, learning, all that makes a nation great, and the savage herd
of child-breeders below, the many ignoble, who were meant for labor for the
noble few."

She was a
feminist:

“Would the
empire of the world restore my lost self-respect - my just pride? Would it save
my cheek from blushes every time I recollected that I bore the hateful and
degrading name of wife? The property, the puppet of a man submitting to his
pleasure -bearing his children-wearing myself out with all the nauseous cares
of wifehood-no longer able to glory in myself, pure and self-sustained, but
forced every day and night to recollect that my very beauty is no longer the
sacrament of Athene's love for me, but the plaything of a man .......”

As with anyone
who is too different (women were not supposed to be independent and powerful),
too self-assured and too outspoken they often become an easy target for zealots
and fanatical dogmatists. It is said that Cyril spread rumors about Hypatia
that fueled the self-righteous indignation of Christian extremists in
Alexandria. The focus of their anger manifested in her murder where it is said
an angry mob attacked and killed her. There are several stories regarding how
she was murdered, but all stories seem to agree on one important point - that
she was, in fact, murdered.

Since there are
no copies of her work to draw from, I decided to look into the work of
Diophantus to get some idea of the mathematics she might have been involved in.
Historians believe Diophantus was probably an Egyptian who received a Greek
education in Alexandria. His ideas were very new for the times. He started the
evolution of our algebraic notation and he was the first to pose and solve
problems that called for solutions in integers or rational numbers.

Typical linear
Diophantine equations (which are the simplest Diophantine equations) involve
such problems as:

ax + by = c where
a, b and c are integers. We want to find the solution for x and y.

Diophantine
solutions are restricted to positive, rational integers. He called impossible
(negative or irrational) solutions "absurd" because there was no
concept of these at that time. Though there may be many solutions to a problem,
Diophantus was satisfied to find only one solution. There was not a systematic
theory involved in solving his equations; each equation was dealt with
individually and required it's ownspecial technique.

Since the time of
Diophantus many mathematicians have pursued the study of his equations. In
India, Brahmagupta was the first to get all possible solutions of integral
value (an integer or whole number). Nowadays, in honor of Diophantus, any
equation in one or more unknowns that is to be solved for integral values of
the unknowns is called a Diophantine equation.

Here is a sample
problem:

In a corral there
are cowboys and an odd number of horses. There are 20 legs in all; how many
belong to horses?

The Diophantine
equation lay out would look like this:

4*h* + 2*c* = 20 because each horse has 4 legs and each cow has 2
legs.

One way to solve
this problem would be by guessing and checking or, more systematically, by
setting up a table:

Here I
substituted 1... 9 in place of *c* (not going over 9 because h would no longer be positive - see
below):

*c* 1 2
3 4 5 6 7 8
9 10

*h* 9/2
4
7/2 3 5/2
2 3/2 1 1/2
0

Or a table could
be set up substituting 1….4 for h (not going over 4 because *c* would no longer be positive - see below):

h
1 2
3 4
5

c
8
6 4 2
0

This Diophantine
equation has four solutions that are positive integers (though Diophantus may
have been satisfied with just one solution): (c, h) = (2,4), (4,3), (6,2) and
(8, 1). However, the problem states that the corral contained an ODD number of
horses and that there were cowboys meaning more than one cowboy, then the
answer has to be 3 horses (12 legs) and 4 cowboys (8 legs), hence (3 x 4) + (4
x 2) = 20

12 + 8 = 20

For
middle-schoolers, this problem could be presented and the algebra brought in
afterwards OR in the context of
history, this Diophantine equation could be introduced and students could try to figure out
how to apply it on their own (or in groups, of course) and then later discuss
explorations and ideas, after
which other problems could be introduced.

Another aspect of
solving an ax + by = c equation involves creating a way to know when an
equation has a solution and when it does not.

This involves
finding the greatest common divisor (GCD) of the numbers involved so that if the GCD of (a,b) will
not divide c evenly, then there are no solutions and if the GCD of (a, b) will divide c evenly then
there is a solution.

For example in
the 1st problem 4h + 2c = 20. So the question is whether (4,2) will divide 20. The GCD for (4,2) is 2,
and 20 divided by 2 = 10 which means there is a solution for the problem.

Here is another
equation with a less obvious solution: 5x + 6y = 17. The GCD for (5,6) is 1 so
there is a solution.

How about 3n + 6m
= 11 - The GCD for (3,6) is 3 and 3 into11 doesn't have a positive number
solution so there are no solutions to this equation.

Other equations
to try Do these have solutions?

14x + 34y = 90

14x + 35y = 91

14x + 36y = 93

The question now
is why does this work? The proof provided by the book was very abstract and
would be difficult for middle-schoolers to understand, so I referred back to
some previous problems to see if I could understand it in a visual manner. I
tried comparing one equation that worked and one that had no solutions:

A) 5x+6y= 17 and
B) 3h+6c= 11 and C) 4h+2c= 20

Observations and
further explorations: I can see that I cannot get a combination to work for 3h
+ 6c = 11, 1 got one solution for 5x + 6y = 17 and 4 solutions for 4h + 2c =
20.

How does this tie
in with the greatest common divisor idea? I can see that in equation B) the two
numbers 3 and 6 can fit into each other (2 threes can fit into 6) with 3 as the
common divisor. All combinations of 3 and 6 under 11 are

3 + 3 + 3 = (too small)

6 + 3 = 9 (too small)

3 + 3 + 3 + 3 =
12 (which is over)

I can see that in
equation A) the common divisor is 1 and this seems to work ... once only:
6+6+5= 17

6 + 5 + 5 = 16 is
too small

6 + 6 + 6 = 18 is
too large

I wonder if there
is only one solution for any (a, b) that divides into c evenly that equals 1?

I'll try another:
2x + 5y = 21

Combinations
include: 5+5+5+2+2+2=21 5 + (2 x 8) = 21 all others are too large or too small

So there are two
solutions for this equation which means an equation where (a, b) divided
into c = 1 is not restricted to
only one solution.

In equation Q it
is easy to see there are several possible combinations with 2 as the GCD
because 2's fit into 4s and combinations of both can fit into the even number
20.

I can see where
these might be the kinds of explorations middle-schoolers can do from guess and
check to making systematic tables, to drawing pictures or diagrams, to asking
questions and finding correlations, to applying the algebra.

Diophantine
equations can be used to help solve certain kinds of story problems such as:

On a
machine-graded test a student gets x 5 point questions right and y 10point
questions right with a total grade of 97. She thinks something is wrong. Why?

In the stock
market John trades (buys or sells) x shares of Canadian off stock at 6 cents a
share and trades y shares of mining stock at 14 cents a share, for a net loss
of 4 cents. Find a solution for (x, y). Find a solution in case the total
number of shares that changed hands was 286.

A baseball
pitcher strikes out s players on 3 pitches each and walks w players of 4
pitches each, for a total of 17 pitches. Find s and w. How many possible
solutions does this problem have?

Diophantine
equations do not end with this one simple format but continue on dealing with
squares and powers of 4 as well as solutions for 3 variables (where one
variable is eliminated algebraically so you end up with a linear Diophantine
equation).

Note: I thought
it was interesting that one of the "Sampler of Problems" homework
assignments I chose happened to be a Diophantine problem with 3 variables that
is discussed in the Burton text! I had spent a considerable amount of time on
this problem, sound one solution and had wondered ever since how it might be
solved algebraically. Amazingly enough, here it is! According to the text there
are three solutions. The problem I am referring to is: If a cock is worth 5
coins, a hen 3 coins and three chickens together 1 coin, how many cocks, hens
and chickens, totaling 100, can be bought for 100 coins?

The text shows
you set up the equation this way:

5x + 3y + 1/3z =
100 and then x + y + z = 100 which can be rewritten as z=100-x-y

This set up
eliminates one of the unknowns by turning the z in the first

equation into 100
- y - x as shown in the second equation. So then you have:

5x + 3y + 1/3 (100 - x - y) = 100

or 7x + 4y = 100

It took me awhile
to figure out how you got from 5x + 3y + 1/3 (100 - x - y)

100 to the
equation 7x + 4y = 100 but with help of some wonderful friends, I

found out how to
do it:

You multiply the first equation 5x + 3y + 1/3 z = 100 times three. This

effectively gets
rid of the fraction and makes it possible to get rid of the z easily:

3 x Ox + 3Y + 1/3z = 100) = 15x + 9Y + z = 300

Then I bring in the 100 - x - y in place of z:

15x + 9y + 1(100 - x - y) = 300

(15x - x) + (9y - y) + 100 = 300

14x + 8y + 100 = 300

14x + 8y = 300 - 100

14x + 8y = 200

If you divide the entire equation by two you get

7x+4y= 100

Once you have the
Diophantive linear sequence you can proceed as before to find the solutions.
For example for solution x = 4. (7 x 4) + 4y = 100 4y = 100 - 28 4y = 72 y= 18
Since x+y+z= 100

andz= 100-x- y
then z 100 - 4 - 18 z 78

Then I plugged in
all the numbers to check the solution (4, 18, 78)

(5 x 4) + (3 x 18) + (1/3 x 78) = 100

20 + 54 + 26 = 100

This solution
does work.

This is the same
solution I got on my homework assignment as well. There are

two others (8,
11, 8 1) and (12, 4, 84) which I did not get.

Since this
process is very abstract and doesn't provide much of a visual I included the
homework assignment (see attached- it starts on the back of the first page) to
show a solution (though laborious) that made sense to me.

I found another
such problem found in the Burton text Problems section and decided to try it
out using the algebraic method and using my method:

A hundred bushels
of grain are distributed among 100 persons in such a way that each man receives
3 bushels, each woman 2 bushels, and each child half a bushel. How many men,
women, and children are there?

It seems you
would set this up the same way: 3m + 2b + 1/2c= 100 and m+b+c = 100 so to
eliminate one of the unknowns you would do this: c = 100-m-b

so 3m + 2b + 1/2
(100 - m - b) = 100

I would follow
through the problem this way in order to get to the 2 variable diophantine
equation:

2x (3m + 2b+ 1/2z
= 100) = 6m + 4b + z=200

6m + 4b + M00 - m
- b) = 200

(6m - m) + (4b -
b) + 100 = 200

5m + 3b = 200 -
100

5m + 3b = 100

From here I can
find the solutions:

First of all I
know there can be a solution because GCD (5,3) is 1 which will divide 100. I
started with a table to find the solutions:

M 1
2 3
4

b 95/3 30 85/3 80/3

I stopped at four
because I could see that the number had to be a multiple of 5 (8m) subtracted
from 100 divided by 3 (3b) in order to work so I simply listed the multiples of
5 subtracted from 100 until I found all the ones that would divide by 3:

100 - 5 = 95 100
- 10 = 90 100 - 15 = 85 100 - 20 = 80 100 - 25 = 75

100 - 30 = 70 100
- 35 = 65 100 - 40 = 60 100 - 45 = 55 100 - 50 = 50

100 - 55 = 45 100
- 60 = 40 100 - 65 = 35 100 - 70 = 30 100 - 75 = 25

100 - 80 = 20 100
- 85 = 15 100 - 90 = 10 100 - 95 = 5

From here I find
which of the above 3 will divide into and I get (90, 75, 60, 45, 30, 15) Since
90 represents 100 - (5 x 2) and so on the table will now look like this:

m
2
5
8
11
14 17

b
30
25
20
15 10
5

So there are 6
solutions here. To find the z for each I used 100 - m - b and got

z
68
70
72
74 76
78

Observations:

- m numbers go up in thirds starting at 2 with a range of (2, 17)

- b numbers go down by fifths starting at 30 with a range of (30, 5)

- z numbers go up by twos starting a 68 with a range of (68, 78)

I decided to try
my method for a solution and see how it compares. I played with numbers until I
discovered the following:

These
combinations of ratios that equal 100 bushels for 100 people show the
following:

men =
5 women =
25
children = 70

This matches the
second solution in the table above. My method works to find one solution in a
nice visual way, but does not make it easy to find the other solutions as
readily as the algebra. However, with enough time, the other solutions can be
found. For example:

These
combinations lead to the solution men = 17 women = 5 children = 78 which is the
sixth solution in the table above. I am sure I could find all six solutions
using my method.

This is great fun
discovering that these algebraic solutions match the solutions I get with a method
I created. I feel tremendous satisfaction in constructing my own mathematics
and even more when I can make connections with other mathematical models. This
is what I want to share with my students. I believe these kinds of experiences
provide the kind of inspiration and motivation that will keep students (myself
included) involved in math for a lot longer and in a much more meaningful way.

CONICS: Hypatia
was also interested in conics but I couldn't find much about it save a few
models and diagrams in Math Equals. I included the shapes of the four cuts made
into cones that form a circle, an ellipse, parabola and a hyperbola. These are
formed by visualizing a knife cutting perfectly straight through the cone
(forming a plane intersection with the cone). It is said that the Greeks
were fascinated with relationships among conic sections, perhaps a
foreshadow of events to come, for later, at the beginning of the 17th C, the
mathematics of conic sections became very important.

This concludes my
overview of Hypatia's life and the sampling of mathematics she was involved in.

BIBLIOGRAPHY

__Women in
Mathematics, Lynn M. Osen, MIT Press, c. 1974__

__Hypatia's
Heritage, Margaret Alic, Beacon Press, c 1986__

__Hypatia,
Charles Kingsley, Hurst and Company, c.1910__

__Math Equal,
Teri Perl, Addison -Wesley, c. 1978__

__ __

__Discovering
Number Theory, John and Margaret Maxfield, W.B. Saunders, c. 1972__

__Elementary
Number Theory, second Edition, Underwood Dudley, W.H. Freeman and Company c.
1978__

__The History of
Mathematics, second edition, David M. Burton, WBC Publishers c. 1988__