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

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 3term representation there is. [It is. See below.]
There is no 2term 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 maxdenominator. [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 515] 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 3term 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 3term representation. (Erdos and Straus)
 5/n has a 3term 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 highprecision 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
minmaxdenominator.
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 15,000,000,000digit 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 remaindernumerators 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 maxdenominator result reported
in previous mailing is provably optimal. Readers interested in his
generalpurpose 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
abovetitled 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 closetobruteforce search for representations
with few terms (the second argument is the number of terms) produced a
number of threeterm 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 threeterm representations. The last
of these is particularly succinct. There is only one twoterm
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

http://www.ics.uci.edu/~eppstein/numth/egypt/
Nowadays, we usually write noninteger 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.
515, 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:
 Znám's
Problem, an open problem in number theory related to Egyptian
fractions.
 Web
Mathematica applet for the greedy Egyptian fraction algorithm.
 This week's
finds in Egyptian fractions, John Baez.
 Madison
Capps' science fair project.
 Binary
Egyptian Fractions, paper by Croot et al.
 Egyptian
Fractions page by Ron Knott.
 Lecture
notes from a section on Egyptian fractions in a history of math
course by Carl Eberhart, U. Kentucky, including a Maple implementation
of the greedy algorithm.
 Best of
Mathnerds: the magnificent seven asks for seventerm Egyptian
fraction decompositions of 1, and describes a method for finding
decompositions of any fraction using a method based on Farey sequences
(essentially equivalent to the
continued fraction method).
 The
1998 British Informatics Olympiad used a sample programming
question on Egyptian fractions, with sample solutions including a Java
applet for calculating 4term representations of a value m/n by
choosing a value d (sequentially testing d=1, d=2, d=3 etc), finding
the divisors of nd (using a simple brute force loop rather than any
kind of factorization technique), and looping through all permutations
of the divisors testing which ones have a prefix that sums to md
(rather than using any kind of dynamic program).
 Binary
Egyptian Fractions, Ernest S. Croot III, David E. Dobbs, John B.
Friedlander, Andrew J. Hetzel, and Francesco Pappalardi.

The Distribution of Prime Primitive Roots and Dense Egyptian Fractions,
dissertation by Greg Martin.
 Jim Loy's
Egyptian fraction page.

Julian Steprans uses Egyptian fractions for a homework assignment.
 The ErdosStrauss
conjecture. Allan Swett confirms that 4/n=1/x+1/y+1/z is solvable
for n up to 1012.
 Kris'
Egyptian fraction applet. Seems to be using binary remainders?
 A
loving look at the unitary partition problem. Which numbers can be
partitioned into distinct positive integers, the reciprocals of which
add to one?
 An odd
Egyptian puzzle with many
solutions. Stan Wagon's problem of the week #848, on odd
representations of 3/179.
 K. S. Brown's extensive work on Egyptian fractions includes pages
speculating on how
the Egyptians did it as well as several on more numbertheoretic
concerns: these pages on unit fractions and
Fibonacci, the
two Ohm problem, counting unit
fraction partitions of unity, minimizing
denominators in unit fraction expansions, minimizing the max
denominator in a tterm expansion of 1, odd greedy
stubbornness, irrationality
of quadratic sums, and wagon trains and
sticky wickets
 The CECM Inverse
Symbolic Calculator also mentions including Egyptian fraction
routines.
 Milo Gardner has done extensive research on the historical
methods used by the Egyptians to construct their tables of fractions..
 Terrance
Nevin uses greedy Egyptian fraction methods as a basis for
investigating the dimensions of the Egyptian pyramids.
 The
Magma symbolic algebra system uses the
splitting method for Egyptian fractions as an example of its
sequence manipulation features.
 Is there a
harmonic integer? Stan Wagon asks if any integer k has an Egyptian
fraction representation

1 

1 

1 

1 

1 
k = 

+ 

+ 

+ 

+ ··· + 


2 

3 

4 

5 

n 
(Hint: consider the largest power of two less than or equal to n.)
 Unit
Fractions from Eric Weisstein's treasure trove of Mathematics.

Undergraduate projects. This list of possible projects includes a
very brief mention of Egyptian fractions.

Problem: sum of unit fractions. Which numbers n/(n+1) have a
twoterm Egyptian fraction representation? Problem from a U. Georgia
summer course in mathematical problem solving.
http://www.ics.uci.edu/~eppstein/numth/egypt/

Accurate reckoning: the entrance into knowledge of all existing things
and all obscure secrets.  Ahmes, 1600 BC
<>
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
wallpaper 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 (picturewriting) on the papyrus
were only deciphered in 1842 (and the Babylonian claytablet
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 presentday 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 presentday 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  Suppose Fatima had
3 sacks of grain to share between 4 people. How would she do it?
 ...and what if it was 2 sacks amongst 5 people?
 ...or 4 sacks between 5 people?
 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  Which is larger:
4/7 or 5/8?
 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 subsections 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 onehalf (since 521 is less than a half of 1050)
but it is bigger than onethird. 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/u11 > T/B
Since T>1, neither 1/u1 nor 1/u11 equal T/B. What is the remainder?
It is
T/B  1/u1 = (T*u1  B)/(B*u1)
Also, since
1/(u11) > T/B
then multiplying both sides by B we have
B/(u11) > T
or, multiplying both sides by (u11) 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 What does the
greedy method give for 5/21?
What if you started with 1/6 (what is the remainder)?
 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.]
 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  3/7 needs at least
how many unit fraction in its Egyptian Fraction form?
Use the first table to find the answer (3).
 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?
 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/AncientAfrica/mad_egyptianfractions.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 