Babylonian Reciprocals (Base 60)
Date: 04/08/99 at 13:19:43 From: Sam Bobek Subject: Babylonian division (base 60) Hello, I am a student at Oregon State University, and am currently working on ancient mesopotamian mathematics. I have come across a stumbling point though, in the area of division. In order to divide, they used reciprocals (a/b = a*(1/b)). The problem is that I don't have the tables they used to "cheat," so I have to determine the value of the reciprocal. The fraction that I have to rationalize is 1/(30;4). The actual division problem is (38,9;16)/(30;4). If you could help me out, it would be greatly appreciated.
Date: 04/09/99 at 09:04:50 From: Doctor Rick Subject: Re: Babylonian division (base 60) Hi, Sam, This is an interesting topic! I wish I knew more about it ... I believe your notation is that a semicolon separates the ones from the one-sixtieths, so your number 30;4 is 30 + 4/60, or 451/15. The reciprocal you are looking for is 15/451. However, I suspect the Babylonian reciprocal tables would have had only integers. From what I have read, for instance in the MacTutor History of Mathematics archive Web site at http://www-groups.dcs.st-and.ac.uk/~history/HistoryTopics.html - look for Babylonian and Egyptian mathematics - it looks as if the Babylonians would not even have all the integers in their table; only those whose only prime factors are 2, 3, and 5. This is because only these numbers would have terminating reciprocals. If you know more about how the Babylonians worked, I'd like to hear it! I can only guess how they would divide by 30;4. Perhaps they did the analog of what I tell students to do in base 10: multiply numerator and denominator by 10 until the denominator is an integer. This (with 60 instead of 10) would transform your problem into (38,9,16)/(30,4). The denominator is now 30*60+4 = 1804. Unfortunately, 1804 = 2^2 * 11 * 41. It would not be in the Babylonians' table of reciprocals (if I understand correctly). Looking in my table of factors and primes, I find that the nearest number whose only prime factors are 2, 3, and 5 is 1800. Since 1800 = 3600/2, the reciprocal of 1800 is 2/3600, or (;0,2). But of course this corresponds to approximating 30;4 in your problem by just 30. You were probably hoping to do better than this. The next thing to do would be to try multiples of 1804 and see if any of them lie closer to a number in our table. But my factor table only goes up to 2000, so I'm stuck. So let's try to do better than I think the Babylonians would have done. Let's find the reciprocal of 15/451, as you asked. I won't attempt to figure out how the Babylonians might have done it, since they probably wouldn't have tried. We start by multiplying 15/451 by 60; the integer part of this is the 1/60's place: 60 * 15 / 451 = 1 + 449/451 ==> 1/60 + ... If we multiply the fractional part remaining by 60, the integer part of this is the 1/60^2 place: 60 * 449 / 451 = 59 + 331/451 ==> 59/60^2 + ... Let's just go one more place: 60 * 331 / 451 = 44 + 16/451 ==> 44/60^3 + ... Our answer is: 1/(30;4) = 15/451 = 1/60 + 59/60^2 + 44/60^3 + ... = ;1,59,44,... You can see that this is just a bit less than ;2, which is 1/30. If you wanted, you could work out the exact repeating sexagesimal by continuing until the same remainder appears twice. I hope this helps. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum