(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:


4h + 2c = 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



            - 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.




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