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Egyptian Fractions
The ancient Egyptians only used fractions of the form 1 /n so any other fraction
had to be represented as a sum of such unit fractions and, furthermore, all theunit fractions were different!Why? Is this a better system than our present day one? In fact, it is for some
tasks.This page explores some of the history and gives you a summary of computersearches for such representations. There's lots of investigations to do in thisarea of maths suitable for 8-10 year olds as well as older students and it isalso designed as a resource for teachers and educators.
Contents of this page
The icon means there is a Things to do investigation at the end of thesection.
indicates an on-line interactive calculator is provided for the section.
An Introduction to Egyptian Mathematics
Henry Rhind and his Papyrus scroll
Egyptian Fractions
Why use Egyptian fractions today?
A practical use of Egyptian Fractions
Comparing Egyptian fractionsA Calculator to convert a Fraction to an Egyptian Fraction
Different representations for the same fraction
Each fraction has an infinite number of Egyptian fraction forms
Every ordinary fraction has an Egyptian Fraction form
Fibonacci's Method a.k.a. the Greedy Algorithm
A Proof optional
Shortest Egyptian Fractions
The greedy method and the shortest Egyptian fractionA Calculator for Shortest Egyptian Fractions
The number of Shortest Length Egyptian FractionsShortest Egyptian Fractions lengths for fraction T/B
Are there fractions whose shortest EF length is 3 (4, 5, ..) ?
Finding patterns for shortest Egyptian Fractions
How many Egyptian Fractions of shortest length are there for T/B?
Fixed Length Egyptian Fractions
Egyptian fractions for 4/n and the Erdös-Straus Conjecture
5/n = 1/x + 1/y + 1/z?Smallest Denominators
The 2/n Table of the Rhind Papyrus
Links and References
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An Introduction to EgyptianMathematics
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 linkto David Joyce's site on the History of Maths at ClarkeUniversity.] The reeds were squashed and pressed intolong sheets like a roll of wall-paper and left to dry inthe sun. When dry, these scrolls could be rolled up andeasily carried or stored.
Some of the papyrus scrolls date back to about 2000BC, around the time of the construction of the largerEgyptian pyramids. Because there are deserts on
either side of the Nile, papyrus scrolls have been wellpreserved in the dry conditions.
So what was on them do you think? How to preserve abody as a mummy? Maybe it was how to construct theextensive system of canals used for irrigation acrossEgypt or on storage of grain in their great storagegranaries? Perhaps they tell how to build boats out of papyrus reeds which float very well because 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 a25 year old Scotsman, Henry Rhind at a market in Luxor, Egypt, in 1858. Afterhis death at the age of 30, the scroll found its way to the British Museum inLondon in 1864 and remained there ever since, being referred to as the RhindMathematical Papyrus (or RMP for short).
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So what did it say?
The hieroglyphs (picture-writing) on the papyrus were only deciphered in 1842
(and the Babylonian clay-tablet cuneiform writing was deciphered later thatcentury).
It starts off by saying that the scribe "Ahmes" is writing it about 1600 BC butthat he had copied it from "ancient writings" so it probably goes back to atleast 2000BC and probably further. The picture is also a link so click on it to goto the St Andrews MacTutor biography of Ahmes.
Since early civilisations would need to predict the start of spring accurately inorder to sow seeds, then a large part of such mathematical writing hasapplications in astronomy. Also, calculations were needed for surveying
(geometry) and for building and for accounting. However, quite a lot of theproblems in the RMP are arithmetic puzzles - problems posed just for the funof solving them!
On this page we will look at how the Egyptians of 4000 years ago worked withfractions.
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 as we would today.
Instead, they were able to write any fraction as a sum of unit fractions whereall the unit fractions were different.
For example,3 /4 =
1 /2 +1 /4
6 /7 =1 /2 +
1 /3 +1 /42
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A fraction written as a sum of distinct unit fractions is called an EgyptianFraction.
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 eachwould 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 trypouring the 5 sacks of grain into 8 heaps and, by carefully comparingthem, perhaps by weighing them against each other, balance them sothey are all the same! But is there a better way? We will see that usingunit fractions makes this easier.The second reason is that it is much easier to compare fractions using
Egyptian fractions than it is by using our present-day notation forfractions! For instance:Which is bigger: 5 /8 or 4 /7?
but remember - you are not allowed to use your calculator to answer this!Again unit fractions can make this much simpler.
On this page we see how both of these work in Egyptian fractions.
A practical use of Egyptian Fractions
So suppose Fatima has 5 loaves of bread to share among the 8 workers whohave helped dig her fields this week and clear the irrigation channels. Pausefor a minute and decide how YOU would solve this problem before readingon.....
First Fatima sees that they all get at least half a loaf, so she gives all 8 of them half a loaf each, with one whole loaf left.Now it is easy to divide one loaf into 8, so they get an extra eighth of a loaf each and all the loaves are divided equally between the 5 workers. On thepicture here they each receive one red part (1 /
2
a loaf) and one green part (1 /8of a loaf):
and 5 /8 =1 /2 +
1 /8
Things to do
1. Suppose Fatima had 3 loaves to share between 4 people. How would she do it?
2. ...and what if it was 2 loaves amongst 5 people?
3. ...or 4 loaves between 5 people?
4. What about 13 loaves to share among 12 people? We could give them one loaf each
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and divide the 13th
into 13 parts for the final portion to give to everyone.
Try representing13
/12
as1/2 +
1/3 +
1/* . What does this mean - that is, how would you
divide the loaves 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 =75 /100 whereas 4 /5 =0.8 = 0.80 =80 /100 so we can see that 80 (hundredths) is bigger than 75 and we can now
see that 4/5 is bigger than 3/4.
Could you do this without converting to decimals?We could try using ordinary fractions as follows:What common fraction could we convert both 3 /4 and 4 /5 into? 20ths would do:3 /4 =
15 /20 whereas4 /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 have4 /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
or2/7?
A Calculator to convert a Fraction to an EgyptianFraction
An Egyptian Fraction forT
/B is a sum of unit fractions, all different , whose sumis T /B. Enter your fraction in the boxes below and the click on the Convert to
an Egyptian fraction button and an equivalent Egyptian fraction will beprinted in the RESULTS window.
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Further down this page is another calculator which will find all the shortestEgyptian Fractions but this calculator is quicker if you just want one. Themethod used in this calculator is the Greedy Algorithm which we will examinein more detail below but the disadvantage of this method is that sometimes itwill fail if a denominator gets too large.
C A L C U L A T O R
Convert to an Egyptian fraction
R E S U L T S Clear
Different representations for thesame 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 eachis true for all fractions T /B less than one (so that T, the number on top, is
smaller than B, the bottom number).
You can skip over these two sub-sections if you like.
Each fraction has an infinite number of Egyptian
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fraction forms
To see why the second fact is true, consider this:
1 = 1 /2 +1 /3 +
1 /6 (*)
So if3
/4 =1
/2 +1
/4 Let's use (*) to expand the final fraction 1 /4:
So let's divide equation (*) by 4:1 /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, dividing equation(*) by 24 this time. Since we are choosing the largest denominator to expand,it will be replaced by even larger ones so we won't repeat any denominatorsthat we have used already:
1 /24 =1 /48 +
1 /72 +1 /144
and so3 /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!If T=1 already (so we have 1 /B) then using (*) we can always start off the
process by dividing (*) by B to get an initial 3 unit fractions that sum to 1 /B.
Every ordinary fraction has anEgyptian Fraction form
We now show there is always at least one sum of unit fractions whose sum isany given fraction T /B<1 by actually showing how to find such a sum.
Fibonacci's Method a.k.a. the Greedy Algorithm
This method and a proof are given by Fibonacci in his book Liber Abaci produced in 1202, the book in which he mentions the rabbit problem involvingthe Fibonacci Numbers.
Remember that
T/B<1 and
if T=1 the problem is solved since T /B is already a unit fraction, so
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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
and hence its other name - the greedy algorithm.With what is left, we repeat the process. We will show that this series of unitfractions 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 thenrepeat 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 from521 /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 is1 /7 and the remainder is
1 /50.
How do we know it is 7? Divide the bottom (larger) number, 350, bythe top one, 57, and we get 6.14... . So we need a number largerthan 6 (since we have 6 + 0.14) and the next one above 6 is 7.)
So 521 /1050 =1 /3 +
1 /7 +1 /50
The sequence of remainders is important in the proof that we do not have tokeep 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 theremainder is getting smaller . If it keeps decreasing then it must eventuallyreach 1 and the process stops.
Practice with these examples and then we'll have a look at finding short Egyptian fractions.
Things to do
1. What does the greedy method give for5/21
?
What if you started with
1
/6 (what is the remainder)?
2. Can you improve on the greedy method's solution for9/20
(that is, use fewer unit
fractions)? [Hint: Express 9 as a sum of two numbers which are factors of 20.]
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3. The numbers in the denominators can get quite large using the greedy method: What
does the greedy method give for5/91
?
Can you find a two term Egyptian fraction for5/91
?
[Hint: Since 91 = 7x13, try unit fractions which are multiples of 7.]
A Proof
This section is optional: click on the button see the proof. Show the proof The next section explores the shortest Egyptian fractions for any givenfraction.
Shortest Egyptian Fractions
The greedy method and the shortest Egyptianfraction
However, the Egyptian fraction produced by the greedy method may not be theshortest 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 that4 /17 =1 /5 +
1 /30 +1 /510
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. Forinstance, 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]
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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] =
[3,4,7,924]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) representations!
A Calculator for Shortest Egyptian Fractions
The calculator below will find all the Egyptian fractions of shortest length foran ordinary fraction.
C A L C U L A T O R
Find the shortest Egyptian fractions for
R E S U L T S Clear
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The number of Shortest Length EgyptianFractions
Here is a table of the lengths of the shortest Egyptian Fractions for allfractions 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 e.g. 9 /12 so 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!
numberis the minimum number of unit fractions that are needed to sum toT /B.
Find T (top or numerator) down the side and B (bottom or denominator) 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 -
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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
Are there fractions whose shortest EF length is 3 (4, 5, ..) ?
From the table above, we see 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 three unit fractions:4 /5 =
1 /2 +1 /4 +
1 /20 and 4 /5 =1 /2 +
1 /5 +1 /10
There are many other fractions whose shortest EF has 3 unit fractions. Thosewith a denominator 10 or less are:
4 /5 3 /7
5 /7 6 /7
7 /8 7 /9
8 /9 9 /10
Is there a fraction that needs 4 unit fractions?Yes! 8 /11 canot be written as a sum of less than 4 unit fractions, for instance
8 /11 =1 /2 +
1 /6 +1 /22 +
1 /66
and there are 15 other EFs of length 4 for this fraction.Other fractions with a denominator 20 or less that need at least 4 unitfractions are:
8 /11 9 /11
10 /11 12 /13
13 /14 15 /16
8 /17 14 /17
15 /17 9 /19
14 /19 15 /19
17 /19 18 /19
This leads us naturally to ask:
Is there a fraction that needs 5 unit fractions?Yes! The smallest numerator and denominator are for the fraction 16 /17
16 /17 =1 /2 +
1 /3 +1 /17 +
1 /34 +1 /51
and there are 38 other EFs of length 5 for this fraction.Other fractions with a denominator 40 or less that need 5 unit fractions are:
16 /17 21 /23
22 /23 27 /29
28 /29 30 /31
32 /34 33 /34
36 /37 37 /38
38 /39
Continuing:
The smallest fraction needing 6 unit fractions is
77
/79
77 /79 =1 /2 +
1 /3 +1 /8 +
1 /79 +1 /474 +
1 /632
and there are 159 other EFs of length 6 for this fraction.Other fractions with a denominator up to 130 that need 6 unit fractions are:
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77 /79 101 /107
102 /103 104 /107
106 /107 108 /109
112 /113 115 /118
117 /118 119 /127
123 /127
The smallest fraction needing 7 unit fractions is 732/733
732 /737 =1 /2 +
1 /3 +1 /8 +
1 /45 +1 /6597 +
1 /25655 +1 /30786
and many other EFs of length 7 for this fraction.The smallest fraction needing 8 unit fractions is 27538/27539.
Mr. Huang Zhibin ( ) of China in April 2014 has verified that this fractionneeds 8 unit fractions and gives this example:
27538 /27539 =1 /2+
1 /3+1 /7+
1 /43+1 /1933+
1 /14893663+1 /1927145066572824+
1 /212829231672162931784
Beyond 8 unit fractions is unknown territory!
A097049 has the numerators and A097048 the denominators of these"smallest" fractions which need at least 2,3,4,5,... terms in any EgyptianFraction.
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 form2i+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
42 /9 =
1 /5 +1 /45
2 /9 =1 /6 +
1 /18
52
/11 =1
/6 +1
/66
6 2 /13 =1 /7 +
1 /91
7
2 /15 =1 /8 +
1 /1202 /15 =
1 /9 +1 /45
2 /15 =1 /10 +
1 /302 /15 =
1 /12 +1 /20
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 /?
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Can you spot how we can use (2i+1) and i to find the missing number?Here is the table again with the (2i+1) and i+1 parts in red and the ? numberis in green :
i2/2i+1
=1/i+1 +
1/?
12/3
=1/2 +
1/6
22/5
=1/3 +
1/15
3
2
/7 =
1
/4 +
1
/28
42/9
=1/5 +
1/45
=1/6 +
1/18
52/11
=1/6 +
1/66
62/13
=1/7 +
1/91
72/15
=1/8 +
1/120
=1/9 +
1/45=
1/10 +
1/30=
1/12 +
1/20
82/17
=1/9 +
1/153
92/19
=1/10 +
1/190
Yes! Just multiply the red numbers i+1 and 2i+1 to get the green ones!
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 reducesto 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
So algebra has shown us that the formula is always true.
How many Egyptian Fractions of shortest lengthare there for T/B?
Here is a table like the one above, but this time each entry is a count of allthe 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 theentry 4:
NUMBER of Shortest Egyptian Fractions:
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Things to do
1. 1 =1/2 +
1/3 +
1/6
shows 3 different unit fractions with a sum of 1 whereas
1 =1/2 +
1/4 +
1/10
+1/12
+1/15
is a set of 5 unit fractions.a. In how many ways can you make write 1 as a sum of 4 different unit fractions?
b. How many other ways can you find to write 1 as a sum of 5 unit fractions? (There
are more than 10 but less than 100.)
Check your answers at A006585 .
2. Suppose we now allowed unit fractions to be repeated in the above question e.g.
1 =1/2 +
1/4 +
1/4 =
1/3 +
1/3 +
1/3
There are a total now of 14 ways to write 1 as a sum of 4 unit fractions which includes all
those solutions you found in the first question. What are they?
3. Is it always possible to find n different unit fractions that sum to 1 no matter what n is?
Can you give a formula for the n unit fractions or a method of constructing them for
certain values of n?
4. Difficult!
Here are the EFs for 1 with the smallest numbers (that is, the largest denominator is
smallest) of various lengths:
Length Denominators
Numbers whose reciprocals sum to 1
3 2,3,6
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4 2,4,6,12
5 2,4,10,12,15
6 3,4,6,10,12,15
7 3,4,9,10,12,15,18
8
4,5,6,9,10,15,18,20
3,5,9,10,12,15,18,20
94,6,8,9,10,12,15,18,24
4,5,8,9,10,15,18,20,24
10 5,6,8,9,10,12,15,18,20,24
116,7,8,9,10,12,14,15,18,24,28
5,7,8,9,10,14,15,18,20,24,28
5,6,8,9,10,15,18,20,21,24,28
12
6,7,8,9,10,14,15,18,20,24,28,30
4,8,9,10,12,15,18,20,21,24,28,30
a. Are there any patterns here that you can use to extend this table?
b. The list of the largest numbers in each of these cases is
6,12,15,15,18,20,24,24,28,30,... .
How does it continue? Check your answer with A030659
5. Fibonacci's Greedy algorithm to find Egyptian fractions with a sum of 1 is as follows:
Choose the largest unit fraction we can, write it down and subtract it
Repeat this on the remainder until we find the remainder is itself a unit
fraction not equal to one already written down.
At this point we could stop or else continue splitting the unit fraction into
smaller fractions.
To use this method to find a set of unit fractions that sum to 1:
So we would start with1/2 as the largest unit fraction less than 1:
1 =1/2 + (
1/2 remaining)
so we repeat the process on the remainder: the largest fraction less than1/2 is
1/3:
1 =1/2 +
1/3 + (
1/6 remaining).
We could stop now or else continue with1/7 as the largest unit fraction less than
1/6 ...
1 =1/2 +
1/3 +
1/7 + ...
Find a few more terms, choosing the largest unit fraction at each point rather than
stopping.
The infinite sequence of denominators is called Sylvester's Sequence.
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Check your answers at A000058 in Sloane's Online Encyclopedia of Integer
Sequences.
6. Investigate shortest Egyptian fractions for3/n:
a. Find a fraction of the form3/n that is not a sum of two unit fractions.
b. Is it always possible to write3/n as a sum of three unit fractions ?
Give a formula for the different cases to verify your answer.
7. Find a value for n where4/n cannot be expressed as a sum of two unit fractions.
Egyptian fractions for 4/n and the Erdös-StrausConjecture
Although many fractions of the form 4/n can be written as a sum of just twounit fractions, others, such as 4/5 and 4/13 need three or more.
In 1948, the famous mathematician Paul Erdös (1913-1996) together with E.G. Straus suggested the following:
The Erdös-Straus Conjecture:Every fraction 4 /n can be written as a sum of three unit fractions.
It has been verified that 3 unit fractions can found for all values of n up to1014 but as yet no one has proved it true for all values of n nor has anyonefound a number n for which it is not true.The Calculator above shows that for any given n there are many ways tochoose the whole numbers, x, y and z for the three unit fraction denominators.Using the calculator above, can you find patterns for some values of n, x, yand z?
For instance: among all the result of threefractions summing to 4 /n when n is even, we
have:
n x y z
6 3 4 12
8 4 5 20
10 5 6 30
12 6 7 42
...
How would youwrite this patternmathematically?
Here is a list of all the 3-term Egyptian fractions for 4 /n for n from 5 to 15.
4/5=1/2 + 1/4 + 1/201/2 + 1/5 + 1/10
4/6=
1/2 + 1/7 + 1/42
1/2 + 1/8 + 1/241/2 + 1/9 + 1/181/2 + 1/10 +1/151/3 + 1/4 + 1/12
1/2 + 1/15 +1/210
4/10=
1/3 + 1/16 +1/2401/3 + 1/18 + 1/901/3 + 1/20 + 1/60
1/3 + 1/24 + 1/401/4 + 1/7 + 1/1401/4 + 1/8 + 1/401/4 + 1/10 + 1/201/4 + 1/12 + 1/151/5 + 1/6 + 1/30
1/3 + 1/34 +
4/13=
1/4 + 1/18 +1/4681/4 + 1/20 +1/130
1/4 + 1/26 +1/521/5 + 1/10 +1/130
1/4 + 1/29 +1/8121/4 + 1/30 +
1/4 + 1/61 +1/36601/4 + 1/62 +1/1860
1/4 + 1/63 +1/12601/4 + 1/64 +1/9601/4 + 1/65 +1/7801/4 + 1/66 +1/660
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4/7=
1/2 + 1/16 +1/1121/2 + 1/18 +1/631/2 + 1/21 +1/421/3 + 1/6 + 1/14
4/8=
1/3 + 1/7 + 1/421/3 + 1/8 + 1/241/3 + 1/9 + 1/181/3 + 1/10 +
1/151/4 + 1/5 + 1/201/4 + 1/6 + 1/12
4/9=
1/3 + 1/10 +1/901/3 + 1/12 +1/361/4 + 1/6 + 1/361/4 + 1/9 + 1/12
4/11=
1/11221/3 + 1/36 +1/3961/3 + 1/42 +1/1541/3 + 1/44 +1/1321/4 + 1/9 + 1/3961/4 + 1/11 + 1/441/4 + 1/12 + 1/33
4/12=
1/4 + 1/13 +
1/1561/4 + 1/14 + 1/841/4 + 1/15 + 1/601/4 + 1/16 + 1/481/4 + 1/18 + 1/361/4 + 1/20 + 1/301/4 + 1/21 + 1/281/5 + 1/8 + 1/1201/5 + 1/9 + 1/451/5 + 1/10 + 1/301/5 + 1/12 + 1/201/6 + 1/7 + 1/421/6 + 1/8 + 1/241/6 + 1/9 + 1/181/6 + 1/10 + 1/15
4/14=
1/4201/4 + 1/32 +1/2241/4 + 1/35 +1/1401/4 + 1/36 +1/1261/4 + 1/42 +1/841/4 + 1/44 +1/771/5 + 1/12 +
1/4201/5 + 1/14 +1/701/5 + 1/20 +1/281/6 + 1/9 +1/1261/6 + 1/12 +1/281/6 + 1/14 +1/211/7 + 1/8 + 1/56
4/15=
1/4 + 1/68 +1/5101/4 + 1/69 +1/4601/4 + 1/70 +1/4201/4 + 1/72 +1/3601/4 + 1/75 +1/3001/4 + 1/76 +1/285
1/4 + 1/78 +1/2601/4 + 1/80 +1/2401/4 + 1/84 +1/2101/4 + 1/85 +1/2041/4 + 1/90 +1/1801/4 + 1/96 +1/1601/4 + 1/100 +1/1501/4 + 1/105 +
1/1401/4 + 1/108 +1/1351/4 + 1/110 +1/1321/5 + 1/16 +1/2401/5 + 1/18 + 1/901/5 + 1/20 + 1/601/5 + 1/24 + 1/401/6 + 1/11 +1/1101/6 + 1/12 + 1/601/6 + 1/14 + 1/351/6 + 1/15 + 1/301/7 + 1/10 + 1/42
1/8 + 1/10 + 1/241/9 + 1/10 + 1/18
Can you spot any further patterns here?Use the Calculator above to help with your investigations.
If you do find any more, let me know (see contact details at the foot of thispage) and I will put your results here.If we can find a set of cases that cover all values of n, then we have a proof of the Erdös-Straus conjecture.Show more on this Conjecture
Things to do
1. The number of solutions to4/n as a sum of 3 unit fractions is:
The first value is4/3:
1 solution for
4
/3 =
1
/1+
1
/4+
1
/12
1 solution for4/4 =
1/2+
1/3+
1/6
2 ways for4/5
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5 ways for4/6 =
2/3
...
The series of counts is (0,0), 1, 1, 2, 5, ...
How does it continue?
Check your answers with A073101 in Neil Sloane's Online Encyclopedia of Integer
Sequences.If the Erdös-Straus Conjecture is true then the only zeroes in the whole infinite seriesare for n=1 and 2.
With thanks to Robert David Acker, Jr. for suggesting this topic.
5/n = 1/x + 1/y + 1/z?
Another famous mathematician, Sierpinski suggested in 1956 that the sameapplied to all fractions of the form 5 /n, that is that each of these also can be
expressed as a sum of 3 unit fractions.
There are:0 solutions for 5 /2
1 solution for 5 /3:5 /3=
1 /1+1 /2+
1 /6
2 for 5 /4:5 /4=
1 /1+1 /5+
1 /20 and 1 /1+1 /6+
1 /12;
1 for 5 /5; what is it?
The number of solutions this time is the series 0,1,2,1,1,3,5,9,6,3,12,... whichis A075248 in Neil Sloane's Online Encyclopedia of Integer Sequences. If theconjecture is true, then there are no zeroes in this series apart from thestarting value.
Things to do
1.
a. Find the single set of 3 unit fractions with a sum of 5/6.
b. Find the three sets of 3 for5/7.
c. What formulae can you find for special cases of5/n as a sum of 3 unit fractions?
2. From the table of lengths of the shortest Egyptian fractions above, find a fraction that
needs 5 unit fractions for its Egyptian fraction.
3. Can you find a fraction that cannot be written using less than 6 unit fractions for its
Egyptian fraction?
4. Investigate Egyptian fractions which have only odd denominators.
Is it possible to find a sum of odd Egyptian fractions for every fractiona/b?
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The above will give you some ideas for your own experiments and theReferences below point to more information and ideas.Happy calculating!
Smallest Denominators
Apart from the shortest Egyptian fractions (those with the fewest unitfractions), we can also look for the smallest numbers in the denominators.As we saw at the start of the Fixed Length Egyptian Fractions section above,the smallest denominators do not always appear in the shortest Egyptianfractions.
The shortest for 8/11 is8/11 = 1/2 + 1/6 + 1/22 + 1/66and 15 others, but this one has the fewest numbers with just 4 unit
fractions but it includes a denominator of 66;The EF for 8/11 with smallest numbers has no denominator larger than 44and there are two such EFs both containing 5 unit fractions (out of the667 of length 5):8/11 = 1/2 + 1/11 + 1/12 + 1/33 + 1/44 and8/11 = 1/3 + 1/4 + 1/11 + 1/33 + 1/44
Here is a list of the EF's for 1 of various lengths with smallest denominators:Length Denominators with sum of 1
3 2 3 6
4 2 4 6 12
5 2 4 10 12 156 3 4 6 10 12 15
7 3 4 9 10 12 15 18
83 5 9 10 12 15 18 204 5 6 9 10 15 18 20
94 5 8 9 10 15 18 20 244 6 8 9 10 12 15 18 24
10 5 6 8 9 10 12 15 18 20 24
115 6 8 9 10 15 18 20 21 24 285 7 8 9 10 14 15 18 20 24 286 7 8 9 10 12 14 15 18 24 28
124 8 9 10 12 15 18 20 21 24 28 306 7 8 9 10 14 15 18 20 24 28 30
4 8 9 11 12 18 20 21 22 24 28 30 334 8 10 11 12 15 20 21 22 24 28 30 33
4 9 10 11 12 15 18 20 21 22 28 30 33
So of all the EFs for 1 with 3 fractions, the smallest has all denominators nobigger than 6.Of those EFs for 1 with 4 fractions, the smallest has no denominator biggerthan 12. and for 5 fractions, the smallest has no denominator bigger than 15.The series of these smallest maximum denominators (the minimax solution) inthe EFs for 1 of various lengths is given by:6, 12, 15, 15, 18, 20, 24, 24, 28, 30, 33, 33, 35, 36, 40, 42, ... A030659.
The 2/n table of the Rhind Papyrus
Here is the Table at the start of the Rhind mathematical papyrus. It is a table
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of unit fractions for 2 /n for the odd values of n from 3 to 101.Sometimes the shortest Egyptian fraction is ignored in the table in favour of alonger decomposition. Only one sum of unit fractions is given when several arepossible. The scribe tends to favour unit fractions with even denominators,since this makes their use in multiplication and division easier. The Egyptianmultiplication method was based on doubling and adding, in exactly the sameway that a binary computer uses today, so it is easy to double when the unit
fractions are even.Also, he prefers to use smaller numbers. Their method of writing numerals wasdecimal more like the Roman numerals than our decimal place system though.He seems to reject any form that would need a numeral bigger than 999. Allthe shortest forms and alternative shortest forms are given here in an extracolumn.
2 =
1 +
1 +
1 +
1n a b c d
n a b c d shortest?
5 3 15 √
7 4 28 √
9 6 18 √ 5 45
11 6 66 √
13 8 52 104 × 7 91
15 10 30 √ 8 1209 4512 20
17 12 51 68 × 9 153
19 12 76 114 × 10 190
21 14 42 √ 11 23112 8415 35
23 12 276 √
25 15 75 √ 13 325
27 18 54 √ 15 13524 378
29 24 58 174 232 × 15 435
31 20 124 155 × 16 496
33 22 66 √
21 77
18 19817 561
35 30 42 √ 21 10520 14018 630
37 24 111 296 × 19 703
39 26 78 √ 24 10421 27320 780
41 24 246 328 × 21 861
43 42 86 129 301 × 22 946
45 30 90 √
36 6035 6327 13525 22524 360231035
n a b c d shortest?
47 30 141 470 √ 241128
49 28 196 √ 251225
51 34 102 √
30 17027 459261326
53 30 318 795 × 271431
55 30 330 √
40 8833 165
281540
57 38 114 √
33 20930 570291653
59 36 236 531 × 301770
61 40 244 488 610 × 311891
63 42 126 √
56 7245 105
36 25235 31533 693322016
65 39 195 √
45 11735 455332145
67 40 335 536 × 342278
69 46 138 √
39 29936 828352415
71 40 568 710 ×362556
73 60 219 292 365 ×37
n a b c d shortest?
77 44 308 √ 63 9942 46239 3003
79 60 237 316 790 × 40 3160
81 54 162 √ 45 40542 113441 3321
83 60 332 415 498 ×
42 3486also166 249498
85 51 255 √
55 187
45 76543 3655
87 58 174 √ 48 46445 130544 3828
89 60 356 534 890 ×45 4005184 of length 3
91 70 130 √ 52 36449 63746 4186
93 62 186 √
51 527
48 148847 4371
95 60 380 570 ×
60 22857 28550 95048 4560
97 56 679 776 × 49 4753
99 66 198 √
90 11063 23155 49554 59451 168350 4950
101 202 303 606 × 51 5151
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2701
75 50 150 √
60 10045 22542 35040 60039 975382850
All those with n a multiple of 3 follow the same pattern:
2 /3n =1 /2n +
1 /6n
But there are still some mysteries here.For instance why choose
2 /95 =1 /60 +
1 /380 +1 /570
instead of the much simpler
2
/95 =1
/60 +1
/228?
Why stop at 103? There is a sum for 2 /103 with two unit fractions but it
contains a four digit number:
2 /103 =1 /52 +
1 /5356
and all of the other 65 of length 3 contain a denominator of at least 1236.The one with this least maximum denominator is: 2 /103 =
1 /60 +1 /515 +
1 /1236
There are only two of length 4 that don't use four digit numbers:
2 /103 =1 /103 +
1 /206 +1 /309 +
1 /6182 /103 =
1 /72 +1 /309 +
1 /824 +1 /927.
The Calculator above may help with your investigations.
Things to do
1. Is there a pattern common to all the2
/5n forms in the papyrus table?
2. Is there a pattern common to all the2/7n
forms in the papyrus table?
3. Which fractions in the table could be found by the Fibonacci method?
Links and References
David Eppstein of University of California, Irvine has a host of links on allsorts of information on Egyptian Fractions and a comprehensive guide tothe different algorithms that can be used to write your own Egyptian
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Fraction computer programs, although his are described using manydifferent techniques available in the Mathematica package and there arereferences to C and C++ sources too.Dr Scott William's page on The Rhind 2/n Table has a list of the fractions2 /n written as Egyptian fractions in the Rhind papyrus that we mentioned
at the start of this page and that is given in full earlier on this page. Healso includes a discussion and analysis of the fractions chosen and
suggestions of the methods the Egyptians might have used. He has someinteresting pages on African mathematics and mathematicians fromancient times to today.Eric Weisstein's Mathworld article on Egyptian Fractions has manyreferences too.Fibonacci on Egyptian Fractions M. Dunton and R. E. Grimm, Fibonacci Quarterly vol 4 (1966), pages 339-353. Here Grimm and Dunton give anEnglish translation and explanation using modern notation of the sectionin chapter 7 of Fibonacci's Liber Abaci which gives methods of expressinga fraction as a sum of unit fractions. Fibonacci deals with several special
cases called distinctions before giving the "greedy" algorithm above asthe seventh and general method. Download this paper in PDFThe Rhind Mathematical Papyrus G Robins, C Shute, British Museum Press,1987, (88 pages, paperback) is highly recommended for its explanationsof the arithmetic methods that may have been used in the 2/n table andthe other tables and problems in the papyrus. It has excellent colourphotographs of the papyrus and many illustrations. Buy it from theAmazon.co.uk site [use the link above] as it is much cheaper than theAmazon.com site!
The following two books are recommended if you want to read more about theextraordinary Hungarian mathematician Paul Erdös
The Man Who Loved Only Numbers The Story of Paul Erdos and the Searchfor Mathematical Truth by P Hoffmann, Fourth Estate (1999) paperbackMy Brain Is Open: The Mathematical Journeys of Paul Erdos B Schechter,Simon & Schuster (2000) paperback
or try this highly acclaimed DVD:N is a Number: A Portrait of Paul Erdös (2007) Region 1, USA and Canadaonly, for NSTC (non-EU) TVs.
On the History of Egyptian mathematics, I recommend:
Mathematics in the Time of the Pharaohs by Richard J Gillings, Dover,1972 is an inexpensive and readable account of the mathematics in theRhind Papyrus, it contents and methods. Recommended!The Exact Sciences in Antiquity by Otto Neugebauer, Dover, secondedition 1969, is another great book covering not only Egyptian arithmeticbut also the Babylonian, Sumerian and Greek contributions to bothnumber notation and arithmetic as well as astronomy. It is about thehistory of the mathematics more than the maths itself and is now, rightly,a classic on this subject.
© 1996 2014 Dr Ron Knott last update: 21 April 2014