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

 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.

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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:

• 4/n has a 3-term representation. (Erdos and Straus)
• 5/n has a 3-term represntation (Sierpinski)
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 Scienceeppstein@ics.uci.edu    http://www.ics.uci.edu/~eppstein/
```

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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.

## 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).

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:
• The first reason is a practical one.
Suppose you have 5 sacks of grain to share between 8 people, so each would receive 5/8 of a sack of grain in terms of present-day fractions. How are you going to do it simply, without using a calculator? You could try pouring the 5 sacks of grain into 8 heaps and, by carefully comparing them, perhaps by weighing them against each other, balance them so they are all the same! But is there a better way?
• The second is that is much easier to compare fractions using Egyptian fractions than it is by using our present-day notation for fractions! For instance:
Which is bigger:
5/8 or 4/7?
but remember - you are not allowed to use your calculator to answer this!
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): 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:

• Every fraction T/B can be written as a sum of unit fractions...
• .. and each can be written in an infinite number of such ways!
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

• T/B<1 and
• if T=1 the problem is solved since T/B is already a unit fraction, so
• we are interested in those fractions where T>1.

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

• There is always a finite list of unit fractions whose sum is any given fraction T/B
• We can find such a sum by taking the largest unit fraction at each stage and repeating on the remainder (the greedy algorithm)
• The unit fractions so chosen get smaller and smaller (and so all are unique)

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 3T\  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 3 212| . . . . . . . . . . 4 - - - 3 - 3 - - - 2 - 4 - - - 4 -13| . . . . . . . . . . . 4 3 3 3 3 3 3 3 2 3 2 2 - 3 3 3 214| . . . . . . . . . . . . 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 3 218| . . . . . . . . . . . . . . . . 4 - - - 4 - 3 - - - 4 -19| . . . . . . . . . . . . . . . . . 3 3 3 4 3 3 4 3 3 4 320| . . . . . . . . . . . . . . . . . . 4 - 4 - - - 4 - 4 -21| . . . . . . . . . . . . . . . . . . . 4 5 - 3 4 - - 4 -22| . . . . . . . . . . . . . . . . . . . . 5 - 4 - 4 - 3 -23| . . . . . . . . . . . . . . . . . . . . . 3 4 4 3 3 4 324| . . . . . . . . . . . . . . . . . . . . . . 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 + 1/20and4/5 = 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 + 1/62  2/5 = 1/3 + 1/153  2/7 = 1/4 + 1/284  2/9 = 1/5 + 1/45 = 1/6 + 1/185  2/11 = 1/6 + 1/666  2/13 = 1/7 + 1/917  2/15 = 1/8 + 1/120 = 1/9 + 1/45 = 1/10 + 1/30 = 1/12 + 1/208  2/17 = 1/9 + 1/1539  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 + 1/62  2/5 = 1/3 + 1/153  2/7 = 1/4 + 1/284  2/9 = 1/5 + 1/45 = 1/6 + 1/185  2/11 = 1/6 + 1/666  2/13 = 1/7 + 1/917  2/15 = 1/8 + 1/120 = 1/9 + 1/45 = 1/10 + 1/30 = 1/12 + 1/208  2/17 = 1/9 + 1/1539  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 20T\                                                                       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.

• Can you find some patterns in the second table above?
• Can you write any of them in mathematical (algebraic) terms?
http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fractions/egyptian.html

-----------------------------------------------------------------------------------------------
http://www.math.buffalo.edu/mad/Ancient-Africa/mad_egyptian-fractions.html

Egyptian Fractons

You 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 fractions

1/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 means

3 + 1/13 + 1/17 + 1/173 = 3.141527 which approximates to 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 calculus
`http://www.math.buffalo.edu/mad/Ancient-Africa/mad_egyptian-fractions.html------------------------------------------------------------------http://mathforum.org/epigone/historia_matematica/stangtroastrimpTo discuss the later time period, I like Howard Eves' explanation of ageneral Egyptian fraction rule. Eves reported that the 400 AD Akhmim P. contained such a general rule.  I first came across this rule in 1962, ina history of math  class, as Eves stated in the simple form:n/pq = 1/pr + 1/qrwhere r = (p + q)/nThe 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 + 1/422/91, r = (20/2) = 10, or 2/91 = 1/70 + 1/130For myself, I doubt that Ahmes thought in this manner. I suspect  twoother 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 tablessince its. conversions of  4/17, 8/17, 3/19, 7/19 and 9/19 areinteresting, using methods that improve upon the RMP optimal seriesKevin Brown, a number theorist, was a major contributorhttp://www.ecst.csuchico.edu/~atman/Misc/horus-eye.htmlThank you all. once again, for considering Egyptian fractions, not intheir usual modern recreational  math context, or as a step child ofGreek math, or limited to hieroglyphic script, but vividly read by  royalhigher math, proto-number theory.  My examples are sources three MiddleKingdom 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 GardnerSacramento, CA http://mathforum.org/epigone/historia_matematica/stangtroastrimp----------------------------------------------------------------------------http://www.math.buffalo.edu/mad/Ancient-Africa/best-egyptian-fraction.html`

The Best Egyptian Fractons

It 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

`----------------------------------------------http://www.mathpages.com/home/rhind.htmThe 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 leastin 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 acumbersome style of writing that persisted for so many centuries justout 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 wefind 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 ofmy land was flooded.  One of the earliest written records fromancient Egypt (transcribed circa 1650 BC from a source believed todate 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 theexpression 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 implicitlytreated as a single family.Of the remaining table entries, the next is 2/5, to which theyassigned the expression 1/3 + 1/15.  All but one of the remainingdenominators in the table that are divisible by 5 are assigned asimple multiple of this expression, i.e., for 2/(5k) they used1/3k + 1/15k.  Similarly they assigned 1/4 + 1/28 to the tableentry 2/7, and then "seived out" all the remaining denominatorsdivisible by 7 using expressions of the form 1/4k + 1/28k.Finally, they assigned 1/6 + 1/66 to the table entry 2/11 andthen used 1/6k + 1/66k for 2/11k with k=5.The prime 11 seems to be where they stoped this procedure, whichis consistent with that fact that the table extended only to denominators up to 101, so all the composites are seived out bythe primes less than 11.  It's remarkable that the Egyptians of 1850 BC (and probably much earlier) had already developed thiscrude version of the "Sieve of Eratosthenes", and seemed tohave 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 theychose to treat it as a multiple of 11.  Also, the compositenumbers 35, 91, and 95 were evidently not treated as composites,but were assigned unique representations.  Nevertheless, theoverall impression is very strong that they consciously seivedout the multiples of the smaller primes up to the square root ofthe 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 to2/23, although it may be coincidental.)  Once these primes, andtheir multiples, have been resolved, the table entries for theremaining prime denominators suggest that the Egyptians determinedthe representations by using the identity               2         1       2a - p              ---   =   ---  +   -------                     (2)               p         a         apwhere "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 is31 = 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   890On this basis, it's possible to summarize the 2/n table in theRhind Papyrus by giving the values of a,b,(c,(d)) for each primep such that 2/p = 1/a + 1/b + (1/c + (1/d)).  These values arepresented 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  890exceptional cases:   35     25       30   42   91     49       70  130   95     25       60  380  570  101   1111      606  101  202  303This 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 threeand 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          7and the representation appearing the the Rhind Papyrus is the onewith a/x = 7.  In all, the Egyptians used the solution with theminimum a/x for the "large" primes      13, 17, 19, 29, 31, 37, 41, 43, 59, 67, 73, 79, 83, 97whereas they missed it for the primes                  47, 53, 61, 71, 89In 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 2/29is 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 thefour exceptional cases.  The first three are the composites 35,91, and 95, that for some reason were not seived out like the restof 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 kindsof "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 tothe Egyptians.  In particular, the Harmonic Mean certainly LOOKSEgyptian, given their affinity for unit fractions.In any case, notice that G(p,q) is not only the geometric mean ofp and q, it's also the geometric mean of A(p,q) and H(p,q).  Inother words, for any p,q we have                     G = sqrt(pq) = sqrt(AH)which follows simply because pq = AH.  In other words, AH givesan alternative decomposition of the composite number pq.  Thisleads 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 fractionbecause p+q is even.  This formula yields the Rhind Papyrusrepresentations        2       1   /  1     1 \         1      1       ---  =  --- (  --- + --- )   =   ---- + ----       5*7      6   \  5     7 /         30     42and        2        1   /  1     1  \         1       1       ----  =  --- (  --- + ---- )   =   ---- +  ---       7*13      10  \  7     13 /         70     130       Thus we can say that every composite entry in the Rhind Papyrus2/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   606This entry can actually be constructed by formula (2) with a=606 andthe partition 1111 = 202 + 303 + 606, but it seems to stand out fromthe other table entries due to the fact that it's a simple multipleof 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 thatit 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 morethan 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 tosuggest a greater number-theoretic sophistication than is generallycredited to the ancient Egyptians.  Whether they originated theseideas or borrowed them, perhaps from the Babylonians, is unclear.(We shouldn't overlook the possibility that the Babylonians borrowedthem 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.