Math Games

Egyptian Fractions

Ed Pegg Jr., June 21, 2004

<>Find three different whole numbers, all less than 600, so that their reciprocals add up to 2/95. For example, you might choose 48 and 4560<>.  As reciprocals, 1/48 + 1/4560 = 2/95.  That doesn't solve the problem, since 4560 exceeds 600.  How might you solve this problem?  A bit of algebra, perhaps?  Number theory?  A computer search?

Amazingly, this problem was solved 4000 years ago, without any of those techniques. 



  Fraction 
   Ahmes (Rhind) Papyrus, 
1/a + 1/b as a + b
   2/p = 1/A + (2A - p)/Ap, 
with A = (p+1)/2
   Other representations 
of interest
2/3 2 + 6 2 + 6
2/5 3 + 15 3 + 15
2/7 4 + 28 4 + 28
 6 + 14 + 21
2/9 6 + 18 5 + 45
2/11 6 + 66 6 + 66
2/13 8 + 52 + 104 7 + 91
 10 + 26 + 65
2/15 10 + 30 8 + 120
 12 + 20
2/17 12 + 51 + 68 9 + 153
2/19 12 + 76 + 114 10 + 190
2/21 14 + 42 11 + 231
 15 + 35
2/23 12 + 276 12 + 276
2/25 15 + 75 13 + 325
2/27 18 + 54 14 + 378

2/29 24 + 58 + 174 + 232 15 + 435
2/31 20 + 124 + 155 16 + 496
2/33 22 + 66 17 + 561
2/35 30 + 42 18 + 630
 20 + 140
2/37 24 + 111 + 296 19 + 703
2/39 26 + 78 20 + 780
 52 + 60 + 65
2/41 24 + 246 + 328 21 + 861
2/43 42 + 86 + 129 + 301 22 + 946
2/45 30 + 90 23 + 1035
 36 + 60
2/47 30 + 141 + 470 24 + 1128
2/49 28 + 196 25 + 1225
 42 + 98 + 147
2/51 34 + 102 26 + 1326
2/53 30 + 318 + 795 27 + 1431
2/55 30 + 330 28 + 1540
 40 + 88
2/57 38 + 114 29 + 1653
2/59 36 + 236 + 531 30 + 1770
2/61 40 + 244 + 488 + 610 31 + 1891
2/63 42 + 126 32 + 2016
 56 + 72
2/65 39 + 195 33 + 2145
 45 + 117
2/67 40 + 335 + 536 34 + 2278
2/69 46 + 138 35 + 2415
2/71 40 + 568 + 710 36 + 2556
2/73 60 + 219 + 292 + 365 37 + 2701
2/75 50 + 150 38 + 2850
 60 + 100
2/77 44 + 308 39 + 3003
 63 + 99
2/79 60 + 237 + 316 + 790 40 + 3160
2/81 54 + 162 41 + 3321
2/83 60 + 332 + 415 + 498 42 + 3486
2/85 51 + 255 43 + 3655
 55 + 187
2/87 58 + 174 44 + 3828
2/89 60 + 356 + 534 + 890 45 + 4005
2/91 70 + 130 46 + 4186
 7 + 637
2/93 62 + 186 47 + 4371
2/95 60 + 380 + 570 48 + 4560
2/97 56 + 679 + 776 49 + 4753
2/99 66 + 198 50 + 4950
 90 + 110
2/101 101 + 202 + 303 + 606 51 + 5151

Wilbur Knorr, writing in Historia Mathematica in 1982, showed 2/nth table was constructed by two basic methods, one for 2/p series and another for  2/pq series.



2/pq = (2/A) x (A/p)  where A = (p + 1).Akhmim P. Eves suggests  n/pq = 1/pr + 1/qr, where r = (p + q)/n

The rule seems to have been used in the RMP for the unusual 2/35 and 2/91 series, as given by:

2/35, r = (12/2) = 6, or 2/35 = 1/30 + 1/42
2/91, r = (20/2) = 10, or 2/91 = 1/70 + 1/130
2/21 = 2/8 × ((7+1)/21) = 1/4 × (1/3 + 1/21) = 1/12 + 1/84

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http://compgeom.cs.uiuc.edu/~jeffe/wagon/s848.html
Famous unsolved problem (see [Guy, Unsolved Problems in Number Theory, Springer; or Klee & Wagon, Old and New Unsolved Problems in Plane Geometry and Number Theory, MAA]: Does the odd greedy algorithm for getting a representation of m/n (n odd) as a sum of distinct odd unit fractions terminate always? ("Greedy" means: at each step chose the largest 1/m that will fit, with m odd and different from previous choices.) My PoW was motivated by my discovery a year ago that 3/179 gives quite interesting results on the odd greedy algorithm! It yields 19 terms, beginning with:
63, 2731, 5963959, . . .
The last term has almost 500,000 digits! More amazing, the numerators of the remainders (which decrease in the unrestricted greedy algorithm), are
3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 2, 3, 4, 1
Quite a pattern! What is going on?Now, some correspondents last year found better representations: Kevin Brown (Seattle) found: 63, 1611, 3759 and Milo Gardner (Sacramento) found: 63, 1253, 11277. Several of you (Ron Hardin, AT&T Labs, Research; John Guilford, HP, Everett, Wash.; David Eppstein, UC Irvine Dept. of Information & Computer Science; Charles Rees, Univ. of New Orleans) found Brown's answer, so it might well be the case that Brown's solution is the simplest 3-term representation there is. [It is. See below.] There is no 2-term rep'n (exercise). Some of you might be interested in the question of what algorithms the Egyptians used when computing Egyptian fractions. Many of them are immortalized on the Rhind papyrus, and Milo Gardner has made an extensive investigation. Contact him for details at gardnerm@ecs.csus.edu. Eppstein also found that 3/179 =
1/171 + 1/209 + 1/285 + 1/895 + 1/1611 + 1/1969 + 1/2685
which might be the smallest max-denominator. [It is.] Eppstein's home page has misc. Egyptian fraction stuff (http://www.ics.uci.edu/~eppstein/numth/egypt/) and his article in Mathematica in Education and Research [vol 4, no. 2, 1995, pp 5-15] discusses Mathematica implementations of a variety of algorithms. I guess I should throw my own code down. It implements the odd greedy method, together with an option for seeing the numerators of the remainders, which might, if a proof is ever found, play a role in the proof that this halts. The code is from a revision of my book Mathematica in Action (Springer). Now, Neil Sloane and Ron Hardin (AT&T) conjecture that every 3/b (b not divisible by 3) has a 3-term representation as a sum of distinct odd unit fractions. This is clearly true in the unrestricted case (greedy algorithm, numerators of remainders decrease!). Their conjecture nicely complements famous conjectures in this area: Richard Guy (Calgary) observes: 3/(6n+1) = 1/(2n+1) + 1/(2n+1)(4n+1) + 1/(4n+1)(6n+1), and some others along these lines. David Bailey (NASA Ames), using customized high-precision arithmetic on an SGI work station and aware of my desire to "stump" the odd greedy algorithm, found the following monster: 3/2879. The odd greedy rep'n has 21 terms; the last term has 3,018,195 digits; and the sequence of numerators of the remainders is
3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 1
I believe with this computation David gets the record for the largest Egyptian fraction computation ever done. I think my 3/179 was a record when I did it last year. Nice! Eppstein then found: 1/963 + 1/308053 + 1/2772477 and 1/1615 + 1/5355 + 1/14395 + 1/20153 + 1/25911 + 1/43185 + 1/48943 + 1/54701 + 1/60459 for this example, the latter being the unique example with min-max-denominator. It would be nice to have some heuristics to predict where the bad examples for the odd greedy algorithm are. In particular, I would like one that stumps David Bailey's software in the sense that he will be unable to tell whether the algorithm halts.
Late News: Boy, #847 generated a lot of correspondence! David Bailey reports that 5/5809 stumps his program, which is limited to about 15,000,000,000-digit precision. He gets:
1/163 + 1/1125979 + [21 terms omitted] + 1/s + ...
where the integer s is approximately 6.875 x 1012565952. So here we have an example of a conceptually very simple algorithm, and a specific input, and we have no idea, and may never have any idea, whether the algorithm halts. This is very nice indeed. Of course, I have no doubt that nice rep'ns exist. The issue is that odd greedy is very mysterious. The remainder-numerators for 5/5809 look like: 5, 6, 7, 8,.... 20, 21, which is as far as I checked. So I guess a weaker conjecture than the main one here is:
Can one disprove the possibility that the numerators of the remainders in an odd greedy instance look like 5,6,7,8,.........all the way to infinity?
Eppstein is working on a Mathematica package that incorporates a variety of methods. David Bailey (and I) continue to wonder about 5/5809's behavior under odd greedy. His program checks to 60,000,000 digits, but it did not halt. So, if this halts, it leads to numbers having more than 60,000,000 digits. Eppstein's package has no trouble finding reps. of 5/5809 such as: 1/897 + 1/6739 + 1/20217 and 1/2051 + 1/2925 + 1/5567 + 1/5985 + 1/7325. Eppstein adds that his small max-denominator result reported in previous mailing is provably optimal. Readers interested in his general-purpose Egyptian fractions package should contact him: eppstein@euclid.ics.uci.edu. NEW! David Eppstein, using modular reduction (continually reduce the denominator modulo, say, 30!) was able to prove that the sequence does indeed halt, and that the numerators of the remainders look like 5, 6, 7, ..., 29, 30, 1. (David actually used the modulus 37943838567204570000 = 30!/224·310·53·7.) This is very clever and completely avoids the ridiculously large numbers one gets without the reduction. Bailey got up to 60,000,000 digits in the 24th denominator, and then was forced to stop. But now we now that a few more steps does indeed lead to termination in this one case. Nice work, David.
[The following is an email message from David Eppstein to Stan Wagon outlining his results for 3/179:]Jeff Erickson forwarded me your above-titled message on odd Egyptian fraction representations of 3/179. You asked to be sent any solutions.As I believe you know, I have a large collection of algorithms for constructing Egyptian fractions, published in Mathematica in Educ. & Res., and online at http://www.ics.uci.edu/~eppstein/numth/egypt/. Most of these methods produce some even denominators, but a few do not. (Beware, I probably introduced some typos transferring these from my Mac to my Sun.)EgyptOddGreedy[3/179] didn't terminate in a reasonable amount of time. I assume this is why you chose this particular fraction to ask about...The proofs I know of that odd representations exist are I think pretty similar to the method I implemented as EgyptRemainder, which takes as a second argument a "powerful" number (with lots of divisors so that there are lots of ways to form sums of divisors). If the powerful number is odd, you get odd representations, modulo some recombination that happens because of possible repeated terms. EgyptRemainder[3/179,3*3*3*5*5*7] yielded the following:
1/105 + 1/189 + 1/525 + 1/33831 + 1/93795
1/105 + 1/189 + 1/525 + 1/31325 + 1/120825
1/105 + 1/175 + 1/675 + 1/33831 + 1/93795
1/105 + 1/175 + 1/675 + 1/31325 + 1/120825
1/75 + 1/525 + 1/675 + 1/33831 + 1/93795
1/75 + 1/525 + 1/675 + 1/31325 + 1/120825
1/75 + 1/315 + 1/4725 + 1/33831 + 1/93795
1/75 + 1/315 + 1/4725 + 1/31325 + 1/120825
1/63 + 1/1575 + 1/4725 + 1/33831 + 1/93795
1/63 + 1/1575 + 1/4725 + 1/31325 + 1/120825
EgyptShort[3/179,3], a close-to-brute-force search for representations with few terms (the second argument is the number of terms) produced a number of three-term representations, among them the odd
1/61 + 1/2745 + 1/491355
1/63 + 1/1253 + 1/11277
1/63 + 1/1611 + 1/3759
These are therefore the only odd three-term representations. The last of these is particularly succinct. There is only one two-term representation of 3/179, not odd. A small modification of EgyptSmallMult (another brute force method, which combines small multiples of the original fraction in order to get something that differs from the input by a fraction not divisible by the original denominator) shows that
3/179 = 1/895 + 1/1611 + 1/1969 + 1/2685 + 7/495,
3/179 = 1/179 + 1/537 + 1/895 + 1/1253 + 1/2327 + 1/2685 + 3/455,
and that any other set of odd unit fractions having a difference from 3/179 that is not itself a multiple of 1/179 must include a higher denominator. Calling EgyptShort on 7/495 produces the representations
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/77 + 1/1485 + 1/2079
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/77 + 1/1575 + 1/1925
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/77 + 1/1683 + 1/1785
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/81 + 1/891 + 1/1485
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/81 + 1/935 + 1/1377
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/81 + 1/561 + 1/1683
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/81 + 1/693 + 1/1071
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/81 + 1/765 + 1/935
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/87 + 1/495 + 1/1595
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/91 + 1/585 + 1/693
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/95 + 1/495 + 1/627
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/99 + 1/275 + 1/2475
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/99 + 1/285 + 1/1881
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/99 + 1/297 + 1/1485
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/99 + 1/315 + 1/1155
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/99 + 1/385 + 1/693
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/99 + 1/429 + 1/585
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/105 + 1/315 + 1/693
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/105 + 1/385 + 1/495
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/117 + 1/195 + 1/2145
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/117 + 1/209 + 1/1235
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/117 + 1/221 + 1/935
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/135 + 1/165 + 1/1485
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/135 + 1/189 + 1/693
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/135 + 1/209 + 1/513
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/143 + 1/195 + 1/495
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/165 + 1/225 + 1/275
1/895 + 1/1611 + 1/1969 + 1/2685 + 1/171 + 1/209 + 1/285
which have the minimum max denominator (2685) of any odd representation of 3/179. Unless I missed one in going through the results, all other representations with the same max denominator have more terms.--
David Eppstein UC Irvine Dept. of Information & Computer Science
eppstein@ics.uci.edu http://www.ics.uci.edu/~eppstein/
http://compgeom.cs.uiuc.edu/~jeffe/wagon/s848.html
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http://www.ics.uci.edu/~eppstein/numth/egypt/
Nowadays, we usually write non-integer numbers either as fractions (2/7) or decimals (0.285714). The floating point representation used in computers is another representation very similar to decimals. But the ancient Egyptians (as far as we can tell from the documents now surviving) used a number system based on unit fractions: fractions with one in the numerator. This idea let them represent numbers like 1/7 easily enough; other numbers such as 2/7 were represented as sums of unit fractions (e.g. 2/7 = 1/4 +1/28). Further, the same fraction could not be used twice (so 2/7 = 1/7 + 1/7 is not allowed). We call a formula representing a sum of distinct unit fractions an Egyptian fraction.This notation is cumbersome and difficult to compute with, so the Egyptian scribes made large tables so they could look up the answers to arithmetic problems. However there is also some interesting mathematics associated with the problem of converting modern fraction notation into the Egyptian form. A number of famous mathematicians have looked at this problem, and invented different ways of doing this conversion process. Each of these methods has advantages and disadvantages in terms of the complexity of the Egyptian fraction representations it produces and in terms of the amount of time the conversion process takes. There are still some unsolved problems about whether some of these processes finish, or whether they get into an infinite loop.To investigate some of these questions, I wrote a Mathematica notebook, now called "Algorithms for Egyptian Fractions", which as the title implies implements on the computer some of these computation methods. A version of this notebook was published as "Ten Algorithms for Egyptian Fractions" in Mathematica in Education and Research, vol. 4, no. 2, 1995, pp. 5-15, available online through MathSource.Since then I have made several changes including improvements to the binary remainder method and two new sections on reverse greedy methods and brute force searches. I am making this updated version available here.Algorithms for Egyptian fractions: Transcriptions of network conversations: Other sites of interest: http://www.ics.uci.edu/~eppstein/numth/egypt/
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Accurate reckoning: the entrance into knowledge of all existing things and all obscure secrets. -- Ahmes, 1600 BC
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http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fractions/egyptian.html
Contents of this page
An Introduction to Egyptian Mathematics Some of the oldest writing in the world is on a form of paper made from papyrus reeds that grew all along the Nile river in Egypt. [The image is a link to David Joyce's site on the History of Maths at Clarke University.] The reeds were squashed and pressed into long sheets like a roll of wall-paper and left to dry in the sun. When dry, these scrolls could be rolled up and easily carried or stored.
Some of the papyrus scrolls date back to about 2000 BC, around the time of the construction of the larger Egyptian pyramids. Because there are deserts on either side of the Nile, papyrus scrolls have been well preserved in the dry conditions. So what was on them? How to preserve a body as a mummy? Maybe it was how to construct the extensive system of canals used for irrigation across Egypt or on storage of grain in their great storage granaries? Perhaps they tell how about navigation and how to build boats (again out of papyrus reeds which float very well, and pictures of these boats have been found in many Egyptian tombs)?
The surprising answer is that the oldest ones are about mathematics!
Henry Rhind and his Papyrus scroll One of the papyrus scrolls, discovered in a tomb in Thebes, was bought by a 25 year old scotsman, Henry Rhind at a market in Luxor, Egypt, in 1858. After his death at the age of 30, the scroll found its way to the British Museum in London in 1864 and remained there ever since, being referred to as the Rhind Mathematical Papyrus (or RMP for short). So what did it say? The hieroglyphs (picture-writing) on the papyrus were only deciphered in 1842 (and the Babylonian clay-tablet cuneiform writing was deciphered later that century). It starts off by saying that the scribe "Ahmes" is writing it about 1600 BC but that he had copied it from "ancient writings" so it probably goes back to at least 2000BC and probably further. The picture is also a link so click on it to go to the St Andrews MacTutor biography of Ahmes. Since early civilizations would need to predict the start of spring accurately in order to sow seeds, then a large part of such mathematical writing has applications in astronomy. Also, calculations were needed for surveying (geometry) and for building and for accounting. However, quite a lot of the problems in the RMP are arithmetic puzzles - problems posed just for the fun of solving them! On this page we will look at how the Egyptians of 5000 years ago worked with fractions.
Egyptian Fractions The Egyptians of 3000 BC had an interesting way to represent fractions.
Although they had a notation for 1/2 and 1/3 and 1/4 and so on (these are called reciprocals or unit fractions since they are 1/n for some number n), their notation did not allow them to write 2/5 or 3/4 or 4/7. Instead, they were able to write any fraction as a sum of unit fractions where all the unit fractions were different. For example,
3/4 = 1/2 + 1/4 6/7 = 1/2 + 1/3 + 1/42

Writing a fraction as a sum of distinct unit fractions is called writing it as an Egyptian Fraction.Why use Egyptian fractions today? For two very good reasons: On this page we see how both of these work in Egyptian fractions.A practical use of Egyptian Fractions So suppose Fatima has 5 sacks of grain to share among the 8 workers who have helped dig in her field this week and clear the irrigation channels. Pause for a minute and decide how YOU would solve this problem before reading on..... First Fatima sees that they all get at least half a sack of grain, so she gives all 8 of them half a sack each, with one sack left.
Now it is easy to divide one sack into 8, so they get an extra eighth of a sack each and all the sacks are divided equally between the 5 workers. On the picture here they each receive one red part (1/2 a sack) and one green part (1/8 of a sack): 4 sacks split into halves and 1 split into eighths and 5/8 = 1/2 + 1/8
    Things to do
  1. Suppose Fatima had 3 sacks of grain to share between 4 people. How would she do it?
  2. ...and what if it was 2 sacks amongst 5 people?
  3. ...or 4 sacks between 5 people?
  4. What about 13 loaves to share among 12 people? We could give them one sack each and divide the 13th into 13 parts for the final portion to give to everyone.
    Try representing 13/12 as 1/2 + 1/3 + 1/* . What does this mean - that is, how would you divide the sacks using this representation?
    Was this easier?
Comparing Egyptian fractions Which is larger: 3/4 or 4/5? We could use decimals so that 3 /4 =0.75 whereas 4/5 =0.8 so we can see that 4/5 is bigger. Could you do this without converting to decimals?
What common fraction could we convert both 3/4 and 4/5 into?
20 would do so that
3/4 = 15/20 whereas
4/5 = 16/20
so again we can easily see that 4/5 is larger than 3/4. Using Egyptian fractions we write each as a sum of unit fractions:
3/4 = 1/2 + 1/4
4/5 = 1/2 + 3/10 and, expanding 3/10 as 1/4 + 1/20 we have
4/5 = 1/2 + 1/4 + 1/20
We can now see that 4/5 is the larger - by exactly 1/20.
    Things to do
  1. Which is larger: 4/7 or 5/8?
  2. Which is larger: 3/11 or 2/7?
Different representations for the same fraction We have already seen that 3/4 = 1/2 + 1/4
Can you write 3/4 as 1/2 + 1/5 + 1/* ?
What about 3/4 as 1/2 + 1/6 + 1/* ?
How many more can you find? Here are some results that mathematicians have proved: Now let's examine each of these in turn and I'll try to convince you that each is true for all fractions T/B<1. You can skip over these two sub-sections if you like. Each fraction has an infinite number of Egyptian fraction forms To see why the second fact is true, consider this:
1 = 1/2 + 1/3 + 1/6 (*)
So if
3/4 = 1/2 + 1/4
then by diving the equation (*) by 4 we have:
1/4 = 1/8 + 1/12 + 1/24
which we can then feed back into our Egyptian fraction for 3/4:
3/4 = 1/2 + 1/4
3/4 = 1/2 + 1/8 + 1/12 + 1/24
But now we can do the same thing for the final fraction here: 1/24 since
1/24 = 1/48 + 1/72 + 1/144
and so
3/4 = 1/2 + 1/8 + 1/12 + 1/48 + 1/72 + 1/144
Now we can repeat the process by again expanding the last term: 1/144 and so on for ever!
Each time we get a different set of unit fractions which add to 3/4!
This shows conclusively once we have found one way of writing T/B as a sum of unit fractions, then we can derive as many other representations as we wish! Every fraction T/B<1 has an Egyptian Fraction form We now show there is always at least one sum of unit fractions whose sum is any given fraction T/B<1 by actually showing how to find such a sum. The proof was given by Fibonacci in the same book (Liber Abaci) produced in 1202 that introduced the rabbit problem involving the Fibonacci Numbers. Remember that The method is to find the biggest unit fraction we can and take it from T/B. With what is left, we repeat the process. We will show that this series of unit fractions always decreases, never repeats a fraction and eventually will stop. Such processes are now called algorithms and this is an example of a greedy algorithm since we (greedily) take the largest unit fraction we can and then repeat on the remainder. Let's look at an example before we present the proof: 521/1050.
521/1050 is less than one-half (since 521 is less than a half of 1050) but it is bigger than one-third. So the largest unit fraction we can take away from 521/1050 is 1/3:
521/1050 = 1/3 + R
What is the remainder?
521/1050 - 1/3 = 57/350
So we repeat the process on 57/350:
This time the largest unit fraction less than 57/350 is 1/7 and the remainder is 1/50.
So
521/1050 = 1/3 + 1/7 + 1/50
The sequence of remainders is important in the proof that we do not have to keep on doing this for ever for some fractions T/B:
521/1050, 57/350, 1/50
in particular, although the denominators of the remainders are getting bigger, the important fact that is true in all cases is that the numerator of the remainder is getting smaller. If it keeps decreasing then it must eventually reach 1 and the process stops. Now let's see how we can show this is true for all fractions T/B. We want
T/B = 1/u1 + 1/u2 + ... + 1/un
where u1 < u2 < ... < un
Also, we are choosing the largest u1 at each stage.
What does this mean?
It means that
1/u1 < T/B
but that 1/u1 is the largest such fraction. For instance, we found that 1/3 was the largest unit fraction less than 521/1050. This means that 1/2 would be bigger than 521/1050.
In general, if 1/u1 is the largest unit fraction less than T/B then
1/u1-1 > T/B
Since T>1, neither 1/u1 nor 1/u1-1 equal T/B. What is the remainder? It is
T/B - 1/u1 = (T*u1 - B)/(B*u1)
Also, since
1/(u1-1) > T/B
then multiplying both sides by B we have
B/(u1-1) > T
or, multiplying both sides by (u1-1) and expanding the brackets, then adding T and subtracting B to both sides we have:
B > T*(u1 - 1)
B > T*u1 - T
T > T*u1 - B
Now T*u1 - B was the numerator of the remainder and we have just shown that it is smaller than the original numerator T. If the remainder, in its lowest terms, has a 1 on the top, we are finished. Otherwise, we can repeat the process on the remainder, which has a smaller denominator and so the remainder when we take off its largest unit fraction gets smaller still. Since T is a whole (positive) number, this process must inevitably terminate with a numerator of 1 at some stage. That completes the proof that However, the egyptian fraction produced by the greedy method may not be the shortest such fraction. Here is an example:
by the greedy method, 4/17 reduces to
4/17 = 1/5 + 1/29 + 1/1233 + 1/3039345
whereas we can also check that
4/17 = 1/5 + 1/30 + 1/510
    Things to do
  1. What does the greedy method give for 5/21?
    What if you started with 1/6 (what is the remainder)?
  2. Can you improve on the greedy method's solution for 9/20 (that is, use fewer unit fractions)? [Hint: Express 9 as a sum of two numbers which are factors of 20.]
  3. The numbers in the denominators can get quite large using the greedy method: What does the greedy method give for 5/91?
    Can you find a two term Egyptian fraction for 5/91?
    [Hint: Since 91 = 7x13, try unit fractions which are multiples of 7.]
So let's now concentrate on the shortest Egyptian fractions for any given fraction. Shortest Egyptian Fractions Here is the complete list of all the shortest representations of T/B for B up to 11. We use a list notation here to make the unit fractions more readable. For instance, above we saw that:
4/5 = 1/2 + 1/4 + 1/20
which we will write as:
4/5 = [2,4,20]
2/3 = [2,6]
2/5 = [3,15]
2/7 = [4,28]
2/9 = [5,45] = [6,18]
2/11 = [6,66]

3/4 = [2,4]
3/5 = [2,10]
3/7 = [3,11,231] = [3,12,84] = [3,14,42] = [3,15,35] = [4,6,84] = [4,7,28]
3/8 = [3,24] = [4,8]
3/10 = [4,20] = [5,10]
3/11 = [4,44]

4/5 = [2,4,20] = [2,5,10]
4/7 = [2,14]
4/9 = [3,9]
4/11 = [3,33]

5/6 = [2,3]
5/7 = [2,5,70] = [2,6,21] = [2,7,14]
5/8 = [2,8]
5/9 = [2,18]
5/11 = [3,9,99] = [3,11,33] = [4,5,220]

6/7 = [2,3,42]
6/11 = [2,22]

7/8 = [2,3,24] = [2,4,8]
7/9 = [2,4,36] = [2,6,9]
7/10 = [2,5]
7/11 = [2,8,88] = [2,11,22]

8/9 = [2,3,18]
8/11 = [2,5,37,4070] = [2,5,38,1045] = [2,5,40,440] = [2,5,44,220]
= [2,5,45,198] = [2,5,55,110] = [2,5,70,77] = [2,6,17,561] = [2,6,18,198]
= [2,6,21,77] = [2,6,22,66] = [2,7,12,924] = [2,7,14,77] = [2,8,10,440]
= [2,8,11,88]

9/10 = [2,3,15]
9/11 = [2,4,15,660] = [2,4,16,176] = [2,4,20,55] = [2,4,22,44] = [2,5,10,55]

10/11 = [2,3,14,231] = [2,3,15,110] = [2,3,22,33]
8/11 has an unusually large number of different (shortest) reprentations! Here is a table of the lengths of the shortest Egyptian Fractions for all fractions T/B (Top over Bottom) where the denominator B takes all values up to 30: Shortest Egyptian Fractions lengths for fraction T/B
KEY: - means the fraction T/B is not in its lowest form (eg 9/12).
Find its lowest form P/Q (9/12=3/4) and then look up that fraction
. means the fraction T/B is bigger than 1. Try B/T instead!
Otherwise, the number is the *minimum* number of unit fractions
that are needed to sum to T/B.
Find T down the side and B across the top
\B: 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 3
T\ 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0
2| 2 - 2 - 2 - 2 - 2 - 2 - 2 - 2 - 2 - 2 - 2 - 2 - 2 - 2 -
3| . 2 2 - 3 2 - 2 2 - 3 2 - 2 2 - 3 2 - 2 2 - 2 2 - 2 2 -
4| . . 3 - 2 - 2 - 2 - 3 - 2 - 3 - 2 - 2 - 2 - 3 - 2 - 3 -
5| . . . 2 3 2 2 - 3 2 3 2 - 2 3 2 2 - 2 3 3 2 - 2 2 2 2 -
6| . . . . 3 - - - 2 - 3 - - - 2 - 3 - - - 2 - 2 - - - 2 -
7| . . . . . 3 3 2 3 2 2 - 3 3 3 2 3 2 - 3 3 2 3 2 2 - 3 2
8| . . . . . . 3 - 4 - 3 - 2 - 4 - 3 - 2 - 2 - 3 - 3 - 3 -
9| . . . . . . . 3 4 - 3 2 - 2 2 - 4 2 - 3 3 - 3 2 - 2 3 -
10| . . . . . . . . 4 - 3 - - - 3 - 2 - 2 - 3 - - - 2 - 2 -
11| . . . . . . . . . 3 3 3 3 3 3 2 3 2 2 - 3 2 3 3 3 2 3 2
12| . . . . . . . . . . 4 - - - 3 - 3 - - - 2 - 4 - - - 4 -
13| . . . . . . . . . . . 4 3 3 3 3 3 3 3 2 3 2 2 - 3 3 3 2
14| . . . . . . . . . . . . 3 - 4 - 4 - - - 3 - 3 - 2 - 4 -
15| . . . . . . . . . . . . . 4 4 - 4 - - 3 4 - - 2 - 2 2 -
16| . . . . . . . . . . . . . . 5 - 3 - 3 - 4 - 3 - 3 - 3 -
17| . . . . . . . . . . . . . . . 3 4 3 3 3 4 3 3 3 3 3 3 2
18| . . . . . . . . . . . . . . . . 4 - - - 4 - 3 - - - 4 -
19| . . . . . . . . . . . . . . . . . 3 3 3 4 3 3 4 3 3 4 3
20| . . . . . . . . . . . . . . . . . . 4 - 4 - - - 4 - 4 -
21| . . . . . . . . . . . . . . . . . . . 4 5 - 3 4 - - 4 -
22| . . . . . . . . . . . . . . . . . . . . 5 - 4 - 4 - 3 -
23| . . . . . . . . . . . . . . . . . . . . . 3 4 4 3 3 4 3
24| . . . . . . . . . . . . . . . . . . . . . . 4 - - - 4 -
25| . . . . . . . . . . . . . . . . . . . . . . . 4 4 3 4 -
26| . . . . . . . . . . . . . . . . . . . . . . . . 4 - 4 -
27| . . . . . . . . . . . . . . . . . . . . . . . . . 4 5 -
28| . . . . . . . . . . . . . . . . . . . . . . . . . . 5 -
29| . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
So the "smallest" fraction that needs three terms is T=4 B=5 i.e. 4/5
In fact there are two ways to write 4/5 as a sum of threeunit fractions:4/5 = 1/2 + 1/4 + 1/20
and
4/5 = 1/2 + 1/5 + 1/10
Finding patterns for shortest Egyptian Fractions There seem to be lots of patterns to spot in the table above.
The top row, for instance, seems to have the pattern that 2/B can be written as a sum of just 2 unit fractions (providing that B is odd since otherwise, 2/B would not be in its "lowest form"). The odd numbers are those of the form 2i+1 as i goes from 1 upwards. Let's list some of these in full:i 2/(2i+1)=
1 2/3 = 1/2 + 1/6
2 2/5 = 1/3 + 1/15
3 2/7 = 1/4 + 1/28
4 2/9 = 1/5 + 1/45 = 1/6 + 1/18
5 2/11 = 1/6 + 1/66
6 2/13 = 1/7 + 1/91
7 2/15 = 1/8 + 1/120 = 1/9 + 1/45 = 1/10 + 1/30 = 1/12 + 1/20
8 2/17 = 1/9 + 1/153
9 2/19 = 1/10 + 1/190
Let's concentrate on the first sum on each line since some of the fractions above have more than one form as a sum of two unit fractions. It looks as if 2/(2i+1) = 1/(i+1) + 1/X
Can you spot how we can use (2i+1) and i to find the missing X?
Here is the table again with the (2i+1) and i parts in bold and the X part in italics :1 2/3 = 1/2 + 1/6
2 2/5 = 1/3 + 1/15
3 2/7 = 1/4 + 1/28
4 2/9 = 1/5 + 1/45 = 1/6 + 1/18
5 2/11 = 1/6 + 1/66
6 2/13 = 1/7 + 1/91
7 2/15 = 1/8 + 1/120 = 1/9 + 1/45 = 1/10 + 1/30 = 1/12 + 1/20
8 2/17 = 1/9 + 1/153
9 2/19 = 1/10 + 1/190
Yes! Just multiply them together!
[On the first row, 3x2=6 and the final unit fraction is 1/6.] On row i, we have 2/(2i+1) = 1/(i+1) + 1/X and so we have noticed that X is the product of (2i+1) and (i+1).
So it looks like we may have the pattern: 2/(2i+1) = 1/(i+1) + 1/[(i+1)(2i+1)]
We can check it by simplifying the fraction on the right and seeing if it reduces to the one on the left: 1/(i+1) + 1/[(i+1)(2i+1)]
= (2i+1 + 1)/[(i+1)(2i+1)]
= (2i+2)/[(i+1)(2i+1)]
= 2(i+1)/[(i+1)(2i+1)]
= 2/(2i+1) which is the fraction we wanted!!
So algebra has shown us that the formula is always true.How many Egyptian Fractions of shortest length are there for T/B? Here is a table like the one above, but this time the entries are the NUMBER of ways we can write T/B as a sum of the MINIMUM number of unit fractions.
For instance, we have seen that 4/5 can be written with a minimum of 2 unit fractions, so 2 appears in the first table under T/B=4/5.
But we saw that 4/5 has two ways in which it can be so written, so in the following table we have entry 2 under T/B=4/5.
2/15 needs at least 2 unit fractions in its Egyptian form: here are all the variations:2/15 = 1/8 + 1/120
= 1/9 + 1/45
= 1/10 + 1/30
= 1/12 + 1/20
so it has four representations. In the table below, under T/B=2/15 we have the entry 4:NUMBER of Shortest Egyptian Fractions:
\B 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
T\ 2: 1 - 1 - 1 - 2 - 1 - 1 - 4 - 1 - 1 -
3: . 1 1 - 6 2 - 2 1 - 8 3 - 2 1 - 8 4
4: . . 2 - 1 - 1 - 1 - 4 - 3 - 4 - 1 -
5: . . . 1 3 1 1 - 3 2 5 1 - 1 5 2 1 -
6: . . . . 1 - - - 1 - 1 - - - 1 - 2 -
7: . . . . . 2 2 1 2 2 1 - 6 5 3 1 5 2
8: . . . . . . 1 - 15 - 3 - 2 - 21 - 2 -
9: . . . . . . . 1 5 - 1 1 - 1 1 - 14 1
10: . . . . . . . . 3 - 1 - - - 3 - 1 -
11: . . . . . . . . . 2 1 1 2 2 1 1 3 1
12: . . . . . . . . . . 3 - - - 1 - 1 -
13: . . . . . . . . . . . 6 2 1 1 2 1 5
14: . . . . . . . . . . . . 1 - 2 - 5 -
15: . . . . . . . . . . . . . 5 4 - 3 -
16: . . . . . . . . . . . . . . 39 - 1 -
17: . . . . . . . . . . . . . . . 1 3 2
18: . . . . . . . . . . . . . . . . 1 -
19: . . . . . . . . . . . . . . . . . 1
    Things to do
  1. 3/7 needs at least how many unit fraction in its Egyptian Fraction form?
    Use the first table to find the answer (3).
  2. Can you find an example with that number of unit fractions which sum to 3/7?
    Hint: try 3/7 = 1/3 + 1/12 + 1/X. What is X?
  3. How many others can you find?
    Hint: try 3/7 = 1/3 + 1/11 + 1/X. What is X this time?
    The total number of different answers (6) is given in the second table
The above is just a resource for you to experiment with and to check your own results. http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fractions/egyptian.html

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http://www.math.buffalo.edu/mad/Ancient-Africa/mad_egyptian-fractions.html
Egyptian FractonsYou may find it amazing that fractions, as we know them, barely existed, for the european civilization, until the 17'th century. Even in the 19th century, a method called russian peasant fractions, was the same used by the Europeans since they met the African, and the Egyptians at least since 4000BC in Egypt. As the method was found on several papyrus, we now call this technique egyptian fractions. Let us agree to call a number a unit fraction if it has form 1/n , where n is a positive integer. An egyptian fraction is an expression of the sum of unit fractions1/a + 1/b + 1/c + ... , where the denominators a,b,c, ... are increasing.An egyptian number is any number equal which can be expressed as the sum of an integer plus the sum of an Egyptian fraction. Here are some egyptian fractions:1/2 + 1/3 (so 5/6 is an egyptian number), 1/3 + 1/11 + 1/231 (so 3/7 is an egyptian number), 3 + 1/8 + 1/60 + 1/5280 (so 749/5280 is an egyptian number). The egyptians also made note of the fraction 2/3.1/5 + 1/37 + 1/4070 and 1/6 + 1/22 + 1/66 are regarded as different egyptian fractions even though the sum of each is 5/22.The earliest records of egyptian fractions date to nearly 3900 years ago in the papyrus copied by Ahmes (sometimes called Ahmos - ref1, ref2) purportedly from records at least 300 years earlier. It is conjectured that the mysterious, so called, meaningless, egyptian triple 13, 17, 173 actually means3 + 1/13 + 1/17 + 1/173 = 3.141527 which approximatesto 4 places!!!
(considerably better than the usual 3.16 credited to the egyptians.)
However, I have not been able to locate a credible reference for the triple 13, 17, 173 in this context!!!)However, to the victor goes the spoils; i.e., prior to the "discovery" of the Rhind papyrus, egyptian fractions were thought, by european mathematicians, to come from the Greeks. Even the name pi is Greek.The rule of Egyptian fractions requires us to write only unit fractions, integers, and their sums. So if a duke is awarded 3/7'th of the conquered land, the quanity might be represented as (1/4 + 1/7 + 1/28)'th of the conquered land, which is a bit better than
(1/3 + 1/11 + 1/231)'th of the conquered land, but still awkward. Until the 18'th century, when our present method, from India and also thousands of years old, of writing any integer in the numerator became popular in Europe, Egyptian fractions were the primary method of writing non-integer numbers.THEOREM. Every rational number is an egyptian number.The modern proof of the Theorem was discovered in 1880, but European's have known how to compute Egyptian numbers since Fibonacci in the 12'th century. Before exhibiting the rule we make a convention. Given a non-integer r, let [r] denote the smallest integer > r.Suppose p/q < 1 is written in lowest terms then there is an egyptian fraction with at most p terms and whose sum is p/q (see the proof below). Let r/s = p/q -1/[q/p]. So p/q = 1/[q/p] + r/s. If r=1, we are done; otherwise, repeat the process. Here are some examples:1. Consider p/q = 4/23. Since 23/4 = 5.65, [23/4] = 6. Compute 4/23 - 1/6 as a fraction, and get 1/138. Thus, 4/23 = 1/6 + 1/138.2. Consider 5/22. [22/5] = 5. 5/22 - 1/5 = 3/110. Now [110/3] = 37. But
3/110 - 1/37 = 1/4070. So 5/22 = 1/5 + 1/37 + 1/4070.3. With a little ingenuity, you can determine other egyptian fractions whose sum is 5/22. For example, let's start, for no special reason, with 1/6 instead of 1/5. 5/22 - 1/6 = 2/33. [33/2] = 17 and 2/33 - 1/17 = 1/561. Thus, 5/22 = 1/6 + 1/17 + 1/561.4. In the last expression for 5/22, keep 1/6 but exchange 1/17 for 1/22. This means we compute 2/33 - 1/22 = 1/66. Thus, 5/22 = 1/6 + 1/22 + 1/66. This expression is more satisfactory since the denominators are not as large as in the proceeding two cases..Note: There is significant interest in determining which expression is "best" or what the egyptians would have used. We discuss this on the page "The Best Egyptian Fraction."About the proof of the theorem:Notice that when 0 < p < q are integers with only 1 as a common divisor (i.e., p/q < 1 is written in lowest terms) the construction in the 2nd paragraph after the theorem gives, via mathematical induction, the proof of the theorem, since for p[q/p]-q < p.In spite of the Theorem, there is very little interest in egyptian fractions (or even their modernized version - continued fractions) today, and only infinite series, a topic of elementary calculus could indicate their passage (recall from calculushttp://www.math.buffalo.edu/mad/Ancient-Africa/mad_egyptian-fractions.html

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http://mathforum.org/epigone/historia_matematica/stangtroastrimp

To discuss the later time period, I like Howard Eves' explanation of a
general Egyptian fraction rule. Eves reported that the 400 AD Akhmim P.
contained such a general rule. I first came across this rule in 1962, in
a history of math class, as Eves stated in the simple form:

n/pq = 1/pr + 1/qr

where r = (p + q)/n

The rule seems to have been used in the RMP for the unusual 2/35 and
2/91 series, as given by:

2/35, r = (12/2) = 6, or 2/35 = 1/30 + 1/42
2/91, r = (20/2) = 10, or 2/91 = 1/70 + 1/130

For myself, I doubt that Ahmes thought in this manner. I suspect two
other algebraic identity ideas as being more likely to have computed
2/35 and 2/91. One is the product of the arithmetic mean = (p + q)/2
and the geometric mean = pq^1/2 , as Howard Eves discusses (in the
same paragraph that he reviews the Moscow Mathematical Papyrus
and the apparent use of the general formula V = h(a^2 + ab+ b^2)/3).
The second algebraic identity may have been a trail and error form of
thinking (as Ahmes knew by his use of the method of false position).

Finally the 400 AD Akhmim P. offers other clues to reading Egyptian
fraction in historical ways. The AP n/3 - n/32 tables were published by
Knorr in 1982. I have condensed the AP to its n/17 and n/19 tables
since its. conversions of 4/17, 8/17, 3/19, 7/19 and 9/19 are
interesting, using methods that improve upon the RMP optimal series
Kevin Brown, a number theorist, was a major contributor

http://www.ecst.csuchico.edu/~atman/Misc/horus-eye.html

Thank you all. once again, for considering Egyptian fractions, not in
their usual modern recreational math context, or as a step child of
Greek math, or limited to hieroglyphic script, but vividly read by royal
higher math, proto-number theory. My examples are sources three Middle
Kingdom mathematical texts and one 400 AD Coptic/Greek mathematical text,
restated as should have been compared and jointly analyzed, beginning
130- 140 years ago. Comments are welcomed.

Best Regards to all,

Milo Gardner
Sacramento, CA

http://mathforum.org/epigone/historia_matematica/stangtroastrimp
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http://www.math.buffalo.edu/mad/Ancient-Africa/best-egyptian-fraction.html
The Best Egyptian FractonsIt is our intention to use some of the recent work of Milo Gardner and others to understand Egyptian Fractions. Below is a list, found on the Rhind (Ahmes) Papyrus, used for 2/n where n is an odd n umber from 3 to 101.  

2/3 = 1/2 + 1/6

 
2/5 = 1/3 + 1/15

 
2/7 = 1/4 + 1/28

 

2/9 = 1/6 + 1/18

 
2/11 = 1/6 + 1/66

 
2/13 = 1/8 + 1/52 + 1/104

 

2/15 = 1/10 + 1/30

 
2/17 = 1/12 + 1/51 + 1/68

 
2/19 = 1/12 + 1/76 + 1/114

 

2/21= 1/14 + 1/42

 
2/23 = 1/12 + 1/276

 
2/25 = 1/15 + 1/75

 

2/27 = 1/18 + 1/54

 
2/29 = 1/24 + 1/58 + 1/174 + 1/232

 

2/31 = 1/20 + 1/124 + 1/155

 
2/33 = 1/22 + 1/66

 
2/35 = 1/25 + 1/30 + 1/42

 

2/37 = 1/24 + 1/111 + 1/296

 
2/39 = 1/26 + 1/78

 
2/41 = 1/24 + 1/246 + 1/328

 

2/43 = 1/42 + 1/86 + 1/129 + 1/301

 
2/45 = 1/30 + 1/90

 

2/47 = 1/30 + 1/141 + 1/470

 
2/49 = 1/28 + 1/196

 
2/51 = 1/34 + 1/102

 

2/53 = 1/30 + 1/318 + 1/795

 
2/55 = 1/30 + 1/330

 
2/57 = 1/38 + 1/114

 

2/59 = 1/36 + 1/236 + 1/531

 
2/61 = 1/40 + 1/244 + 1/488 + 1/610

 

2/63 = 1/42 + 1/126

 
2/65 = 1/39 + 1/195

 
2/67 = 1/40 + 1/335 + 1/536

 

2/69 = 1/46 + 1/138

 
2/71 = 1/40 + 1/568 + 1/710

 
2/73 = 1/60 + 1/219 + 1/292 + 1/365

 

2/75 = 1/50 + 1/150

 
2/77 = 1/44 + 1/308

 
2/79 = 1/60 + 1/237 + 1/316 + 1/790

 

2/81 = 1/54 + 1/162

 
2/83 = 1/60 + 1/332 + 1/415 + 1/498

 

2/85 = 1/39 + 1/195

 
2/87 = 1/58 + 1/174

 
2/89 = 1/60 + 1/356 + 1/534 + 1/890

 

2/91 = 1/70 + 1/130

 
2/93 = 1/62 + 1/186

 
2/95 = 1/60 + 1/380 + 1/570

 

2/97 = 1/56 + 1/679 + 1/776

 
2/99 = 1/66 + 1/198

 
2/101 = 1/101 + 1/202 + 1/303 + 1/606
from Milo Gardner:I would like to add a simpler 'Occam's Razor' view of 2/pq series found in the Rhind Mathematical Papyrus [RMP]. Rather than 2/pq = (1/q + 1/pq)2/(p + 1) as used for all but four series, I suggest that: 2/pq = 2/A x A/pq where A = (p + 1) and (p + q) covers all but 2/95. As discussed previously, 2/95 is simple a mod 5 version of 2/p, where 2/p = 1/A + (2A -p)/Ap with
p = 19, A = 12, with aliquot parts of A ,3 + 2, used as follows:
2/19 = 1/12 + (3 + 2)/(12*19) = 1/12 + 1/76 + 1/114
such that: 2/95 = 1/5 x (1/12 + 1/76 + 1/114) = 1/60 + 1/380 + 1/570 as Ahmes may have mentally computed.In conclusion, I would like to add that Fibonacci used a parametric version of the 2/p rule, extending Ahmes' range for A, from p/2 < A , p, with A being a highly composite number, to p/n < A < 2p for solving n/p series.
Regards,
Milo Gardner 12/16/99
http://www.math.buffalo.edu/mad/Ancient-Africa/best-egyptian-fraction.html
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http://www.mathpages.com/home/rhind.htm
The Rhind Papyrus 2/N TableOne of the most puzzling episodes in the history of human thought is
the 2000-year reign of Egyptian unit fractions. We can, at least
in part, reconstruct the arithmetical manipulations involved, but
the underlying reason or motive for expressing fractional quantities
as sums of unit fractions remains mysterious. Was it simply a
cumbersome style of writing that persisted for so many centuries just
out of deference to traditional forms, or did it express an actual way
of thinking that has since been forgotten?

At the beginning of almost every general history of mathematics we
find a description of how the ancient Egyptians operated with fractions
almost exclusively in terms of UNIT fractions. For example, instead
of saying 2/5 of my land was flooded, they would say 1/3 + 1/15 of
my land was flooded. One of the earliest written records from
ancient Egypt (transcribed circa 1650 BC from a source believed to
date from around 1850 BC or earlier) is known as the Rhind Mathematical
Papyrus, and contains a table expressing fractions of the form 2/n as
sums of two, three, or four unit fractions with distinct denominators.
The table covers 2/n for n up to 101, although the fractions with
"even" denominators, e.g., 2/4, 2/6, etc, are omitted, showing that
they clearly perceived the obvious equivalence of these with the
reduced forms 1/2, 1/3, etc.

The first entry in the table is 2/3, to which they assigned the
expression 1/2 + 1/6. Every other table entry of the form 2/(3k)
is assigned the expression 1/2k + 1/6k, which suggests they
consciously treated all denominators divisible by 3 as a single
family, just as all denominators divisible by 2 were implicitly
treated as a single family.

Of the remaining table entries, the next is 2/5, to which they
assigned the expression 1/3 + 1/15. All but one of the remaining
denominators in the table that are divisible by 5 are assigned a
simple multiple of this expression, i.e., for 2/(5k) they used
1/3k + 1/15k. Similarly they assigned 1/4 + 1/28 to the table
entry 2/7, and then "seived out" all the remaining denominators
divisible by 7 using expressions of the form 1/4k + 1/28k.
Finally, they assigned 1/6 + 1/66 to the table entry 2/11 and
then used 1/6k + 1/66k for 2/11k with k=5.

The prime 11 seems to be where they stoped this procedure, which
is consistent with that fact that the table extended only to
denominators up to 101, so all the composites are seived out by
the primes less than 11. It's remarkable that the Egyptians of
1850 BC (and probably much earlier) had already developed this
crude version of the "Sieve of Eratosthenes", and seemed to
have a grasp of the difference between prime and composite
numbers.

Admittedly the seive is not perfect, at least not according to our
present understanding. For one thing, the number 55 should have
been seived out as a multiuple of 5, but for some reason they
chose to treat it as a multiple of 11. Also, the composite
numbers 35, 91, and 95 were evidently not treated as composites,
but were assigned unique representations. Nevertheless, the
overall impression is very strong that they consciously seived
out the multiples of the smaller primes up to the square root of
the largest denominator in the table, and then treated the
remaining primes with unique representations.

As we've seen, for each of the small primes 3,5,7,11 the Egyptians
expressed 2/p as a sum of two unit fractions using the simple formula

2 1 1
--- = -------- + ---------- (1)
p (p+1)/2 p(p+1)/2

(The same formula also applies to the expression they assigned to
2/23, although it may be coincidental.) Once these primes, and
their multiples, have been resolved, the table entries for the
remaining prime denominators suggest that the Egyptians determined
the representations by using the identity

2 1 2a - p
--- = --- + ------- (2)
p a ap

where "a" is just some nice round number a > p/2. To find the
remaining terms, you partition the quantity 2a - p into one, two,
or three distinct parts such that each part is a divisor of a. (That's
why it's good to choose a nice round number for a, so it has lots of
divisors.) For example, with n=89 we chose a=60, which gives the
difference 31. Thus, we need to express 31 as a sum of three or fewer
distinct integers each of which divides 60. One such partition is
31 = 15 + 10 + 6, which leads to the representation that appears in
the Rhind Papyrus for 2/89:

2 1 1 1 1
--- = --- + --- + --- + ---
89 60 356 534 890

On this basis, it's possible to summarize the 2/n table in the
Rhind Papyrus by giving the values of a,b,(c,(d)) for each prime
p such that 2/p = 1/a + 1/b + (1/c + (1/d)). These values are
presented in the table below.

TABLE 1: Summary of Rhind Papyrus 2/n Representations

p 2a-p a b c d Also covers these
--- ------ --- --- --- --- -------------------
3 1 2 6 all multiples of 3
5 1 3 15 25, 65, 85
7 1 4 28 49, 77
11 1 6 66 55

23 1 12 276

13 3 8 52 104
17 7 12 51 68
19 5 12 76 114
31 9 20 124 155
37 11 24 111 296
41 7 24 246 328
47 13 30 141 470
53 7 30 318 795
59 13 36 236 531
67 13 40 335 536
71 9 40 568 710
97 15 56 679 776

29 19 24 58 174 232
43 41 42 86 129 301
61 19 40 244 488 610
73 47 60 219 292 365
79 41 60 237 316 790
83 37 60 332 415 498
89 31 60 356 534 890

exceptional cases:

35 25 30 42
91 49 70 130
95 25 60 380 570

101 1111 606 101 202 303


This table raises two obvious questions. First, assuming the
Egyptians used something like formula (2) to determine their general
unit fraction representations for 2/p where p is a "large" prime, how
did they select the value of "a" and the partition of 2a - p from
the available possibilities? Remarkably, if you examine all the
possibilities using a computer, and limit yourself to just the three
and four-term representations where the smallest number x in the
partition of 2a-p is greater than 1, then in most cases the expression
appearing in the Rhind Papyrus is the one for which a/x is minimized.
For example, the only possible solutions for p=43 are

partition of 2n-p
p a 2a-p x y z a/x
--- --- ----- ---- ---- ---- -----
43 24 5 2 3 12
43 28 13 2 4 7 14
43 30 17 2 15 15
43 30 17 2 5 10 15
43 36 29 2 9 18 18
43 42 41 6 14 21 7

and the representation appearing the the Rhind Papyrus is the one
with a/x = 7. In all, the Egyptians used the solution with the
minimum a/x for the "large" primes

13, 17, 19, 29, 31, 37, 41, 43, 59, 67, 73, 79, 83, 97

whereas they missed it for the primes

47, 53, 61, 71, 89

In these "missed" cases they missed the minimums by 2, 6, 1, 3, and 1
respectively.

Another interesting fact that appears from a review of all the
possible representations for each prime is that p=29 is the first
prime for which there is no three-term representation of 2/p (with
the restrictions noted above). Thus, it's not surprising that 2/29
is the first entry in the Rhind Papyrus where a four-term
representation is used.

The second major question raised by Table 1 is how to explain the
four exceptional cases. The first three are the composites 35,
91, and 95, that for some reason were not seived out like the rest
of the composites. From out point of view the case 2/95 = 2/(5*19)
should have been seived out by the small prime p=5, giving it a
representation of 1/3k + 1/15k with k=19. Instead, we find that
its representation was evidently based on the "large" prime p=19,
i.e., it is of the form 1/12k + 1/76k + 1/114k with k=5.

The cases 2/35 and 2/91 are even more unusual, and in a sense
these are the most intriguing entries in the table. These are the
only two composites whose representations are not simple multiples
of the representations of one of their prime factors. Remarkably,
in these two cases it appears the Egyptians reverted from the normal
multiplicative decomposition to what might be called a "harmonic-
airthmetic" decomposition.

Recall that the ancient Greeks had definitions for various kinds
of "means", including the

Arithmetic Mean: A(p,q) = (p+q)/2

Geometric Mean: G(p,q) = sqrt(pq)

Harmonic Mean: H(p,q) = 2/(1/p + 1/q)

It's believed the Greeks inherited these definitions from the
Babylonians, but it's certainly possible they were also known to
the Egyptians. In particular, the Harmonic Mean certainly LOOKS
Egyptian, given their affinity for unit fractions.

In any case, notice that G(p,q) is not only the geometric mean of
p and q, it's also the geometric mean of A(p,q) and H(p,q). In
other words, for any p,q we have

G = sqrt(pq) = sqrt(AH)

which follows simply because pq = AH. In other words, AH gives
an alternative decomposition of the composite number pq. This
leads to the formula

2 2 2 / 1 1 \
--- = -------------- = ----- ( --- + --- ) (3)
pq A(p,q) H(p,q) p + q \ p q /

where of course the leading factor on the right is a unit fraction
because p+q is even. This formula yields the Rhind Papyrus
representations

2 1 / 1 1 \ 1 1
--- = --- ( --- + --- ) = ---- + ----
5*7 6 \ 5 7 / 30 42

and

2 1 / 1 1 \ 1 1
---- = --- ( --- + ---- ) = ---- + ---
7*13 10 \ 7 13 / 70 130

Thus we can say that every composite entry in the Rhind Papyrus
2/n table is based on a decomposition of n into its prime factors.
In most cases the simple geometric decomposition pq was used, but
in two cases they used the arithmetic-harmonic decomposition A*H.

(As to why the numbers 35, 91, (and 95) might have been singled out
for special treatment, see Appendix I )

This leaves only the final entry in the 2/n table, which gives

2 1 1 1 1
--- = --- + --- + --- + ---
101 101 202 303 606

This entry can actually be constructed by formula (2) with a=606 and
the partition 1111 = 202 + 303 + 606, but it seems to stand out from
the other table entries due to the fact that it's a simple multiple
of 1/n. This entry may have been just a formality, suggesting that for
any n not covered in the table (i.e., larger than 100), we can use the
four-term expansion

2/n = 1/n + 1/2n + 1/3n + 1/6n (4)

so this effectively "completes" the table, allowing us to say that
it provides a unit fraction representation of 2/n for ALL integers n.
Interestingly, formula (4) can be seen as an illustration of the
"perfectness" of the number 6, in the sense that the sum of the
divisors equals double the number, i.e., 1+2+3+6 = 12 = 2*6.

In summary, the 2/n table of the Rhind Papyrus, which dates from more
than a thousand years before Pythagoras, seems to show an awareness
of prime and composite numbers, a crude version of the "Sieve of
Eratosthenes", a knowledge of the arithmetic, geometric, and harmonic
means, and of the "perfectness" of the number 6. This all seems to
suggest a greater number-theoretic sophistication than is generally
credited to the ancient Egyptians. Whether they originated these
ideas or borrowed them, perhaps from the Babylonians, is unclear.
(We shouldn't overlook the possibility that the Babylonians borrowed
them from the Egyptians.)

http://www.mathpages.com/home/rhind.htm
-------------------------------------------------------------


Bruce Friedman, mathorigins.com, http://www.mathorigins.com/image grid/BRUCE OLD_005.htm.

Milo Gardner, blah, http://www.math.buffalo.edu/mad/Ancient-Africa/gardner.EMLRL.160202.doc.

Mathematica Code:

Many of the pages at mathworld.wolfram.com have extensive notebooks.  Code used for this column can be found at the Turmites entry. Code for the Busy Beaver Turing machine is available from NKS Online Downloadable Programs.

References:

Leonid Bunimovich, "Many Dimensional Lorentz Cellular Automata and Turing Machines," Int Jour of Bifurcation and Chaos, 6 (1996), 1127-1135.

Heiko Hamann, "Definition and Behavior of Langton's Ant in Three Dimensions," Complex Systems, 3 (2003), 263-268.

Christopher Langton, "Studying Artificial Life with Cellular Automata," Physica D, 22 (1986), 120-149.

MacTutor History of Mathematics. "Alan Turing." http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Turing.html.

James Propp, "Further Ant-ics," Mathematical Intelligencer, 16 (1994), 37-42.

Paul Rendell, "A Turing Machine implemented in Conway's Game of Life," http://rendell.server.org.uk/gol/tm.htm.

Softology, Visions of Chaos, http://www.softology.com.au/voc.htm.

Eric W. Weisstein et al. "Turmite." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/Turmite.html.

Stephen Wolfram, History of Turing machines, A New Kind of Science, http://www.wolframscience.com/nksonline/page-930c-text.

Zillions of Games, Turing Machine, http://www.zillions-of-games.com/cgi-bin/zilligames/submissions.cgi/21399?do=show;id=10.

Math Games archives.

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Ed Pegg Jr. is the webmaster for mathpuzzle.com. He works at Wolfram Research, Inc. as the administrator of the Mathematica Information Center.