Harriot and binary numbers

The paper J W Shirley, Binary numeration before Leibniz, Amer. J. Physics 19 (8) (1951), 452-454 contains an interesting look at some mathematics which appears in the hand written papers of Thomas Harriot. Using the photographs of the two original Harriot manuscript pages reproduced in Shirley's paper, we explain how Harriot was doing arithmetic with binary numbers.

Leibniz is credited with the invention of binary arithmetic, that is arithmetic using base 2. Laplace wrote:-

Leibniz saw in his binary arithmetic the image of Creation. ... He imagined the Unity represented God, and Zero the void; that the Supreme Being drew all beings from the void, just as unity and zero express all numbers in his system of numeration. This conception was so pleasing to Leibniz that he communicated it to the Jesuit, Grimaldi, president of the Chinese tribunal for mathematics, in the hope that this emblem of creation would convert the Emperor of China, who was very fond of the sciences ...

However, Leibniz was certainly not the first person to think of doing arithmetic using numbers to base 2. Many years earlier Harriot had experimented with the idea of different number bases. In the papers which he left unpublished at his death there are calculations which show that he:-

... had experimented with number systems and for his own satisfaction had evolved and considered working with not only binary systems, but ternary, quaternary, quinternary, and higher systems as well. ... Harriot did nothing beyond solve the problem of how various number systems could be used. He saw no practical applications for his theory in his day, and he dropped it.

What did he examine? On one sheet of paper Harriot has written

Reductio

  1101101

    64
    32
     8
     4
     1
   109

Here is he reducing the binary number 1101101 to a decimal. Underneath this calculation Harriot illustrates the inverse process of converting the decimal 109 into the binary number 1101101.

Conversio

  109
   64   |  7              1101101
   45
   32   |  6
   13
    8   |  4
    5
    4   |  3
    1   |  1

Notice that Harriot has kept a record of the positions of the 1s in the binary expression.

On another page Harriot has six different examples of calculating with binary numbers. Under the title Subductionis exempla (subtraction examples) he has two calculations

Subductionis exempla 

  10110010
    111011
   1110111


  10101001
    110111
   1110010

The first is 178 - 59 = 119 in binary, the second is 169 - 55 = 114 in binary.

Next Harriot gives two examples of addition under the heading Additionis exempla.

Additionis exempla 

    111011
   1110111
  10110010

    110111
   1110010
  10101001

The first is 59 +119 = 178 in binary, the second is 55 + 114 = 169 in binary.

The final two calculations that Harriot carries out on this page involve multiplications. The first appears under the heading Multiplicatio.

Multiplicatio 

        1101101
        1101101
        1101101
      1101101
     1101101
   1101101
  1101101      
 10111001101001

This is precisely how one would expect to multiply binary numbers. Notice of course that to multiply binary numbers we only need to be able to multiply by 0 and 1, so effectively multiplication is reduced to addition. Napier would have approved!

The final sum is headed Aliter, cum additione successiva, meaning "another way, by successive addition".

Aliter, cum additione successiva 

         1101101
         1101101
         1101101
      10001000
     10110001
   10011001
  10111001      
  10111001101001

Here Harriot has successively added 1101101 moved the appropriate number of places to the left. For example the second row is 1101101 + 11011 and the remaining 01 will be the rightmost two digits of the answer, the third row is 1101101 + 1000100 with the remaining 0 as the next digit in the answer, the fourth row is 1101101 + 101100 with the remaining 01 as the next two digits in the answer, and the final row is 1101101 + 1001100 with the remaining 1 as the next digit in the answer. The answer to the multiplication is now the final row together with the remaining digits mentioned above 10111001(1)(01)(0)(01) = 10111001101001.

Try multiplying the binary numbers 11000011000 and 11001010101 using this last method of Harriot. What do these numbers have to do with Harriot?

JOC/EFR August 2006

The URL of this page is:
http://www-history.mcs.st-andrews.ac.uk/Extras/Harriot_binary_numbers.html