HOME

Egypt
Origins

# Eleanor Robson and Plimpton 322

### Part VI - Pythagoreans Through Reciprocal Pairs

Robson borrowed Neugebauer's igum and igibum as always representing reciprocals of one another and thus always yielding sexagesimal 1 when multiplied together. This starts her off, and she proceeds into her arguments from there. She illustrates from YBC 6967 because it fits with her requirements of an area involving the OB "1," or "60."  She could have just as well started with any problem involving an area of 1 (60). I provided some of those solutions in the preceding paper. She states that

if we revert to the normal definition of reciprocal pairs, whereby their product is 1 (or some even power of 60), then our method will serve to generate "Pythagorean" triples nicely.

In order to arrive at this conclusion she goes through a mental gyration. She takes the 5 X 12 igibum-igum figure, which had an area of sexagesimal 1, (60), and turns it into a square with the same area. Now it is no longer 5 X 12; it is an area with √60 as one side. Then she arbitrarily takes the intermediate results of the solution of YBC 6967, 3;30 and 8;30 and arrives at a set of numbers which are Pythagorean: 3;30, 8;30 and √60. However, this is not a Pythagorean as we understand it, since the maneuver does not generate an integer.

If you examine the YBC 6967 solution you will see that the area of the large square is 8;30 X 8;30, and the area of the small square is 3;30 X 3;30. The difference between the square of these two numbers is 60. [8;30 (8.5) and 3;30 (3.5), square the numbers, subtract them, and arrive at 60: 72.25 - 12.25 = 60.] In fact, Robson is telling us that the geometric solution to YBC 6967 gives us a Pythagorean relationship with the large square representing the diagonal, and the small square representing the short side. Again, this is not a Pythagorean as we understand it.

This manipulation has nothing to do with the YBC 6967, nor does it have anything to do with Pythagorean triples. It is merely an arrangement that Robson invented, from the numbers suggested by YBC 6967. She could have used a solution for a 10 X 6 igibum-igum (x - y = 4). Then the area of the large square would be 1 15 or 64, and the area of the small square would be 4: 64 - 4 = 60. Hence she would have 8 X 8 and 2 X 2, and she could say that 8, 2 and √60 make a Pythagorean.

Without prior warning of what she is doing she now takes the number 12, and its reciprocal 5, (5/60 = 1/12), and goes through the process of creating another Pythagorean triple. These numbers again were suggested by YBC 6967 but do not have any relationship to that problem. The two sides of YBC 6967, 12 and 5, are confused by Robson as the number 12 and its reciprocal, 5. But the reciprocal is 5/60, not 5, although one would represent this number as sexagesimal 5. She subtracts the two numbers, (alternating between sexagesimal and decimal): 12 - 0;05 = 12- 1/12 = 11;55 = 11.091666 . . .

She now mysteriously divides this by 2, to obtain 5;57 30 = 5.958333 . . . and states that this is the smaller square side. I assume she is mathematically duplicating what was done with the 8;30 and 3;30 of the previous example. By smaller square side Robson means the smaller side of a Pythagorean. By the larger square side she means the larger side of a Pythagorean. Her discussion is cloudy at this point.

I spent some time trying to understand where she obtained the 2. I then realized she was following the discussion by Neugebauer on his pg 41. If one takes the ratio of D/L for the diagonal and long sides of a Pythagorean triple, with his "p" and "q" solutions, one arrives at D/L = 1/2 (p/q + q/p). She set up 1/2 (12 - 0;05). So her 12 of this example is equal to  Neugebauer's "p/q" and her 0;05 is Neugebauer's "q/p"

She then squares 5;57 30 = 35;30 06 15 = 35.50173511. She adds 1 as in the YBC 6967 example to obtain 35;30 06 15 + 1 = 36.50173511. She then finds the "square side" (actually square root) of this number to obtain 6;02 30 = 6.0416666 . . .

She then states that she has thusly generated a nice mathematical triple: 6;02 30, 5;57 30, and 1, or 6.0416666, 5.958333 and 1.

She refrains from calling this set of numbers a Pythagorean triple. In order to get a Pythagorean triple from this string she first multiplies by 2 to get 12;05 - 11;55 - 2. (Note how this duplicates her previous numbers: 12 - 0;05 and 11;55.) She then multiplies by 12 to get 2 25;0 - 2 23;0 - 24. Indeed, this is a Pythagorean triple, 145 - 143 - 24.

She follows with a similar routine for the numbers 5 and 0;12 or 5 and 0.2. Lastly she multiplies by 5 to obtain the well known Pythagorean triple, 13 - 12 - 5.

Let's examine what she did. YBC 6967 has nothing to do with this process, except it was the path by which she got herself into this routine. This is shown by her choice of 12 and 5, the igibum and igum of that illustration. Without regard for any smaller or larger squares we take her steps:

First she takes a number and its reciprocal: x and 1/x.

Second, she takes the difference of these two numbers, x - 1/x.

Third, she divides the difference of these two numbers (x - 1/x) by 2 = (x - 1/x)/2.

Fourth, she squares this number [(x - 1/x)/2]2.

Fifth, she adds 1, [(x - 1/x)/2]2 + 1.

Sixth, she takes the square root of this number √{[(x - 1/x)/2]2 + 1}.

Seventh, she then states that from this process she has generated a mathematical triple.

Eighth, she then multiplies by appropriate numbers to obtain a Pythagorean triple.

Let's try it in decimal, arbitrarily with the number 7, within the limits of my desk calculator:

1. Take the number 7.

2. 1/7 = 0.142857142

3. 7 - 1/7 = 6.857142857

4. ÷ 2 = 3.428571429

5. Square this number = 11.75510204

6. Add 1 = 12.75510204

7. Take the Square Root of this number = 3.571428571

If you multiply the appropriate lines by 7 you will obtain:

from line 7, = 25,

from line 4, = 24, and

from line 1, = 7.

Indeed, this is a Pythagorean triple, 25 - 24 - 7.

Well! I for one did not know that one could generate Pythagorean triples from a number and its reciprocal. That was a major discovery. (Neugebauer already knew this. See discussion on his pg 41.)

I decided to try my hand at it. Using her method I generated Pythagorean triples for the first 22 whole numbers. (22 was strictly arbitrary to fit onto one sheet of paper.) I shall shortly expand on the meaning of "p" and "q."

### First 22 Whole Numbers

(p = x, q = 1)

 x 1/x (x - 1/x)/2 a/b (x + 1/x)/2 c/b [(x + 1/x)/2]2 x Mult Prod a b c p/q q/p (p2 - q2)/2pq (p2 + q2) /2pq [(p2 +q2) /2pq]2 Short Side (p2 - q2) Long Side 2pq Diag. (p2 + q2) 2 0.50000000 0.75000000 1.25000000 1.56250000 2 4 4 3 5 3 0.33333333 1.33333333 1.66666667 2.77777778 1 3 3 4 5 4 0.25000000 1.87500000 2.12500000 4.51562500 2 8 8 15 17 5 0.20000000 2.40000000 2.60000000 6.76000000 1 5 5 12 13 6 0.16666667 2.91666667 3.08333333 9.50694444 2 12 12 35 37 7 0.14285714 3.42857143 3.57142857 12.75510204 1 7 7 24 25 8 0.12500000 3.93750000 4.06250000 16.50390625 2 16 16 63 65 9 0.11111111 4.44444444 4.55555556 20.75308642 1 9 9 40 41 10 0.10000000 4.95000000 5.05000000 25.50250000 2 20 20 99 101 11 0.09090909 5.45454545 5.54545455 30.75206612 1 11 11 60 61 12 0.08333333 5.95833333 6.04166667 36.50173611 2 24 24 143 145 13 0.07692308 6.46153846 6.53846154 42.75147929 1 13 13 84 85 14 0.07142857 6.96428571 7.03571429 49.50127551 2 28 28 195 197 15 0.06666667 7.46666667 7.53333333 56.75111111 1 15 15 112 113 16 0.06250000 7.96875000 8.03125000 64.50097656 2 32 32 255 257 17 0.05882353 8.47058824 8.52941176 72.75086505 1 17 17 144 145 18 0.05555556 8.97222222 9.02777778 81.50077160 2 36 36 323 325 19 0.05263158 9.47368421 9.52631579 90.75069252 1 19 19 180 181 20 0.05000000 9.97500000 10.02500000 100.50062500 2 40 40 399 401 21 0.04761905 10.47619048 10.52380952 110.75056689 1 21 21 220 221 22 0.04545455 10.97727273 11.02272727 121.50051653 2 44 44 483 485

Note that she presents (x - 1/x)/2 and (x + 1/x)/2, (semi-difference and semi-sum) as well as the square of the latter, [(x + 1/x)/2]2 as proposed columns in the restoration of Plimpton 322.  I show them in the Table above.

If you take Robson's numbers you will quickly discover that her (x - 1/x)/2 is the ratio of S/L, where S is the short side of the Pythagorean triangle and L is the long. Her first example is 143/24 = 5.958333 . . . The ratio of (x + 1/x)/2 is the ratio of D/L where D is the diagonal: 145/24 = 6.0416666  . . .

Line numbers 5 and 12 show Robson's examples; you can see how they fit with the other Pythagorean triplets.

Here I have labeled the results with a, b, and c to avoid confusion, and generalize the process.

If  you examine the Table you will see that those numbers that are odd always have themselves as a multiplier. I illustrated with 7. You will also see that the even numbers always have a multiplier of 2 X themselves; 44 for 22. This generates the series of numbers I labeled "Product." You will also note that the examples used by Robson had a multiplier of 24 (2 X 12) for 12, and 5 for 5.

Robson summarized the meaning of the columns in her Table 6, and her proposed restoration of Plimpton 322.

1. The first two columns contain the reciprocal pairs.

2. The third and fourth their semi-differences and semi-sums.

3. The fifth contains the areas of the large squares, found by adding 1 to the semi-difference (if the reciprocals are unknown), or by squaring the semi-sum (if the reciprocals are known.) (I show the square of the semi-sum, as does Robson.)

4. The next two columns show the two square-sides (small and large) multiplied by successive factors to get the shortest possible strings.

5. The penultimate shows the products of those factors, namely the corresponding long sides.

These last two statements show the short, diagonal, and long sides respectively of the Pythagorean triplets. (Not to be confused with my Table above.)

Now let us back calculate. Plimpton 322 contained the short side and the diagonal of a set of Pythagorean triplets. From the Pythagorean rule Robson could calculate the long side. (As have many others.)

To calculate the "x" values she took the fourth column because she knew the D/L ratio, and the value of D and L. (It made no sense to take the fifth column since that is merely the square of the fourth.) I shall illustrate with the number 7 again. D/L = 3.57142857 = (x + 1/x)/2. To solve this I must set up a quadratic:

(x2 + 1)/2x = 3.57142857

x2 + 1 = 7.14285714x

x2 - 7.14285714x + 1 = 0

x = 7, confirming that the calculation is correct.

Thus Robson was able to calculate the "x" values for all 15 of the Plimpton 322 lines (after she appropriately corrected the errors.)

However, she could have determined the "x" values from Neugebauer's "p" and "q" values. Perhaps she did, but did not tell us.

If we substitute "p" and "q" for Robson's reciprocals we find that "p/q" = x. and "q/p" = 1/x. This turns the third column with (x - 1/x)/2 into (p/q - q/p)/2. Multiplying out these substitutions we get (p2 - q2)/2pq. The fourth column gives (p2 + q2)/2pq.

But we know the meaning of these substitutions. They are Neugebauer's values in his mathematical treatment of Old Babylonian Pythagoreans. (p2 - q2) = the short side; (p2 + q2) = the diagonal; 2pq = the long side.

Robson's reciprocal numbers are nothing more than the numbers generated by Neugebauer through his "p" and "q" values. In her examples above she unknowingly chose p/q with q = 1, as I have shown in the Table.

Robson's reciprocals duplicated Neugebauer's solutions.

The method of reciprocals is nothing more than another way of symbolizing the generation of Pythagorean triples, while the logic and the results are the same: (x - 1/x)/2 = (p2 - q2)/2pq.

This is an attractive way to define Pythagorean triples. Except that we should be aware of what we are doing. If I take Neugebauer's equation, (p2 - q2)/2pq, and divide by "pq" I revert to (p/q - q/p)/2. If I then define "x"  as p/q and "1/x" as q/p I have duplicated Robson's mathematical expressions.

We could credit Robson for offering another mathematical approach for generating Pythagorean triples but Neugebauer had already done it. Her organizing Plimpton 322 into an orderly array was merely a repeat of his work. (Which he did not consciously discuss.) Therefore, she has done nothing that was not already known. Perhaps she has made us more aware of it.

Prior to this work I knew only a mathematical method using the Incircle of a triangle for generating Pythagorean triples. See:

http://www.egyptorigins.org/InCircle.pdf

However, huge theoretical implications follow from the generation of Pythagorean triples. As we can see, the theoretical generation of the relationships, (p2 - q2)/2pq, (p2 + q2)/2pq, and 2pq, follow directly from the incircle geometric analysis. The question before us is just how much did the ancient scribes know about these theoretical relationships.

The interested reader can generate a huge array of Pythagorean triplets from arbitrary "p" and "q" values. And know what they mean from the incircle geometry, or the simple mathematical statements. This recognition is not possible from the use of reciprocals.

An example with "q" = 2 is illustrated in:

http://www.egyptorigins.org/pythq2.htm

Robson goes on to justify her use of reciprocals.