Inside the auditorium the noise was deafening. As the arch-deacon rose slowly from his chair in the middle of a bank of chairs to the left side of the stage and rang the bell, a hush fell over the audience. The arch-deacon, acting as master of ceremonies, introduced the competitors, Nicoli Tartaglia and Antonio Maria Fiore, who sat at tables covered with books and papers in the middle of the stage. Tartaglia said nothing, simply nodding his bushy head of unkempt hair to the introductory remarks. Fiore walked out to the apron of the stage and thanked the organizers and the arch-deacon for his generous words of introduction. Tartaglia, who looked more like a bear than a man, appeared to be upset, nervous and pale. Fiore seemed more self assured...

Patrick Coulton

This is the rest of the story:

Scipione Del Ferro was a profeesor of mathematics at Bologna, Italy. Del Ferro had rediscovered the trick for solving equations of the form

Tartaglia had claimed to be able to solve a cubic equation with linear term missing. Namely, the equation (which I shall call Tartaglia's equation)

Antonio Maria Fiore, a student of Del Ferro, believed that Tartaglia was lying, and so he challenged him to a contest in Venice. In fact, Tartaglia only knew the general solution to the the equation of Del Ferro, but upon being challenged, he was able to discover at the twelfth hour a solution for Tartaglia's equation. Tartaglia easily won the contest since he was able to solve two different kinds of cubic equations.

The solutions are given by

and

But how were these solutions discovered and what do they mean? Consider the following problem:

or

This equation is easily factored as

Then using the principle a(b) = 0 implies a = 0 or b =0 we have

which gives solutions x=-1,-2. On the other hand, the quadratic formula gives solutions

or

and

As we can see these two methods give the same answer. But why? The easiest explanation is a geometric one. If you graph the quadratic function (we will assume that a > 0)

you will obtain a parabola with a minimum point on the x-axis of

The two solutions occur where the graph crosses the x-axis. These intersection points are symmetric and at a distance of

and try to rewrite the left hand side as a square. Note that

so we can add 9/4 to both sides and obtain

We now compute the square root of each side to get

and

Solving for x gives the usual solution. This method is completely equivalent to the quadratic formula. We can now write the function as

This is called completing the square. The minimum value of the function is clearly -1/4 and it occurs when x =- 3/2. The roots of the function clearly occur when (x +3/2)² is equal to 1/4. If you were to use this method and write the solution in terms of the constants a,b,c you would obtain the quadratic formula.

Now consider the following problem extracted from an ancient Babylonian text:

I have added the area and two thirds of the length of the side of my square and the result is 0;35 (ed. in sexagesimal notation). What is the side of my square?

We have that the area of the square is some unknown x squared, or A=x². To this we must add two thirds of the unknown side x The resultant sum is 35/60. The sexagesimal system is a base 60 system and the notation 0;35 indicates 0 + 35/60. We can write this in modern notation as

The text offers the following solution:

You must take two thirds of the number 1 which is 0;40. You must take one half of this which is 0;20. You must multiply this by itself 0;20 to get 0;6,40. Add this result to 0;35 to get 0;41,40 . Look now and see that 0;50 is the side of the square with area 0;41;40. The half 0;20 of 0;40 you must subtract from 0;50 to find that 0;30 is the side of the square.

Below I have supplied the text with modern fractional notation in parentheses.

You must take two thirds of the number 1 which is 0;40 (ed. 2/3). You must take one half of this which is 0;20 (ed. 1/3). You must multiply this by itself 0;20 (ed. 1/3) to get 0;6,40 (ed. [6/60] + [40/3600] = [360 + 40]/3600) = 400/3600 = 1/9). Add this result to 0;35 (ed. 35/60) to get 0;41,40 (ed. [41/60] + [40/3600] = 2500/3600) . Look now and see that 0;50 (ed. 50/60) is the side of the square with area 0;41;40. The half 020 (ed. 1/3) of 0;40 (ed. 2/3) you must subtract from 0;50 to find that 0;30 (ed. 1/2) is the side of the square.

As you can see from this exposition, algebra in the ancient world was quite a windy subject. It is also difficult to see the solution process that the author employs for this problem. In fact, one is lead to believe that the solution is simply guessed, and the result is then presented in this archaic fashion. But this is far from the case as you will presently see. One possible explanation for this appraoch may be propensity that the ancients had for making mathematics seem mystical and inaccessible. This made mathematicians appear to be powerful and wise. The secrecy and mysticism of the Pythagoreans led to the distruction of their colony on the Italian penisula. Even at the height of the Renaissance, Tartaglia and Fiore jealously guarded their new results in the theory of equations. Another explanation for the style of this author is that without symbolic manipulation, algebra is difficult to describe.

Let us return to the text and attempt to solve the equation

using the approach of this ancient author.

• We first observe that the author computes one half of the coefficient of the linear term. In his case, that is one half of 2/3. But in our case, that would be one half of 3, or 3/2.
• Next, the author squares this value. In our case, we obtain 9/4 as the square of 3/2.
• The author then adds this number to the number which is the result of the summation in the given problem. In our case, this number is -2. We add 9/4 to -2 and obtain 1/4.
• Next we observe that 1/4 is the square of 1/2.
• Finally, we must subtract the value 3/2 from 1/2 to obtain -1 which is one of the solutions.

If you review the method of completing the square, you will see that this is exactly the same method except that it is written in prose. We can conclude from this that the ancient Babylonians were familiar with a method of solution equivalent to the quadratic formula. One final note: the Babylonians would not have considered negative numbers to be solutions.

Homework Assigment 1

1. How many different methods are there for solving the quadratic formula?
2. List the methods that you know and explain.
3. Which method is the best to teach students?
4. Use the method of completing the square to solve and verify the solution using the quadratic formula.
   • x² + 6x = -5
   • x² - 2x = -1
   • x² = -1
5. Using your notes on completing the square, write these solutions out in prose as we did in the text.

Essentially algebra includes the study of number systems along with the usual operations of addition and multiplication. Of course, modern algebra has abstracted these notions to include, not just numbers, but sets of all kinds, and not just addition and multiplication, but operations of all kinds. More concretely, algebra involves the solution of equations of unknowns in these systems. Linear equations are the first topic of study in algebra. Next, solutions of quadratic equations are studied. Some special cases of cubic equations are then considered. But the typical high school/ middle school curriculum in algebra will end with these topics. Supplementary topics sometimes covered will included matrices which are helpful in solving systems of linear equations.

The solution of the general cubic equation is quite complicated and is rarely studied outside of advanced course work in mathematics. Modern computer algebra systems can solve most cubics easily and so the theory of cubic equations is not as well studied as it once was. It is perhaps somewhat surprising that the Babylonians studied the cubic equation at all.

We have already seen that the Babylonians did not do algebra in a symbolic way but rather wrote out problems and solutions completely in prose. This was, in fact, the standard in mathematics until about the sixteenth century. Symbolic algebra was slow in arriving and it may very well be that it leads to many learning obstacles for our students. We will return to this subject when we study the transition to symbolic algebra.

The Babylonians used a grouping numeral system somewhat like the system of roman numerals. Since the Babylonians had developed the use of clay tablets for recording permanent records, a triangular reed or stylus was used to write on the soft clay. Then the clay was baked. One advantage of this system is that records could be maintained for a long time and some of those records have come down to us.

The Babylonians used a sexagesimal, or base 60 system. The Babylonians were excellent astronomers. Many of their discoveries were passed down to the Greeks, Persians and Arabs and eventually found their way into modern astronomy. As you might know, there are 360 degrees in a circle. This was an invention of the Babylonians. It is probably based on two important facts:

• There are 365 days in a year,
• There are many divisors of 60.

One such method of computation is called the method of doubling which reduces multiplication to the addition of a column of numbers. For examle we will use the method to compute the product of 21 and 35. To do this we will double 21 and halve 35. The list is as follows

• 21 x 35
• 42 x 17 (but note we lost a sum of 21 since 35 is odd)
• 84 x 8 (we lost a sum of 42 since 17 is not even)
• 168 x 4
• 336 x 2
• 672 x 1

  21
  42
672

which has total 735. Note that this system is very convenient for grouped number systems. A grouped number system means that we have different symbols for each power of the system base and that we group the different powers together. The Roman Numeral system is a good example of this type of numeral system. The values used are typically

• 1 is I
• 10 is X
• 100 is C
• 1000 is M

• 5 is V
• 50 is L
• 500 is D.

• XXI x XXV
• XXXXIIx XVII (but note we lost a sum of XXI since XXV is odd)
• LXXXIIII x VIII (we lost a sum of XXXXII since XVII is not even)
• CXXXXXXVIII x IIII, or CLXVIII
• CCCXXVVIIIIII x II, or CCCXXXVIII
• DCLXXII

  XXI
  XXXXII
DCLXXII

which is DCCXXXV

Multiplication in the Babylonian system would not be any harder if one were used to the sexagesimal system. The Egyptians, Greeks, and Romans also used grouping systems but these were primarily in base 10. The Chinese and Japanese used a place system similar to our modern place system, but since there was no zero a symbol was used to indicate the power of ten for each digit. This system is called a Multiplicative grouping system, but it eventually gave rise to the modern Hindu-Arabic place system.

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