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 the unit 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 computer searches for such representations. There's lots of investigations to do in this area of maths suitable for 8-10 year olds as well as older students and it is also designed as a resource for teachers and educators.

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Contents of this page

The Things To Do icon means there is a Things to do investigation at the end of the section.
Calculator indicates an on-line interactive calculator is provided for the section.
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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. map of upper Nile [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 do you think? 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 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 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).

Rhind papyrus So what did it say?

The hieroglyphs (picture-writing) on the papyrus were only deciphered in 1842 (and the Babylonian clay-tablet cuneiform writing was deciphered later that century).

It starts off by saying that the scribe "Ahmes" is writing it about 1600 BC but that he had copied it from "ancient writings" so it probably goes back to at least 2000BC and probably further. The picture is also a link so click on it to go to the St Andrews MacTutor biography of Ahmes.

Since early civilisations 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 4000 years ago worked with fractions.

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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 where all the unit fractions were different.

For example,

3/4 = 1/2 + 1/4

6/7 = 1/2 + 1/3 + 1/42


A fraction written as a sum of distinct unit fractions is called an Egyptian Fraction.

Why use Egyptian fractions today?

For two very good reasons: On this page we see how both of these work in Egyptian fractions.

A practical use of Egyptian Fractions

So suppose Fatima has 5 loaves of bread to share among the 8 workers who have helped dig her fields 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 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 the picture here they each receive one red part (1/2 a loaf) and one green part (1/8 of a loaf): 4 loaves split into halves and 1 split into eighths

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 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 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 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?

A Calculator to convert a Fraction to an Egyptian Fraction

An Egyptian Fraction for T/B is a sum of unit fractions, all different, whose sum is 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 be printed in the RESULTS window.
Further down this page is another calculator which will find all the shortest Egyptian Fractions but this calculator is quicker if you just want one. The method used in this calculator is the Greedy Algorithm which we will examine in more detail below but the disadvantage of this method is that sometimes it will fail if a denominator gets too large.

C A L C U L A T O R
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R E S U L T S

Different representations for the same fraction

We have already seen that 3/4 = 1/2 + 1/4
Can you write 3/4 as 1/2 + 1/5 + 1/* ?
What about 3/4 as 1/2 + 1/6 + 1/* ?
How many more can you find?

Here are some results that mathematicians have proved:

Now let's examine each of these in turn and I'll try to convince you that each is true for all fractions T/B 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 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
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 denominators that we have used already:
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! 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 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.

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 involving the Fibonacci Numbers.

Remember that

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 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.
How do we know it is 7? Divide the bottom (larger) number, 350, by the top one, 57, and we get 6.14... . So we need a number larger than 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 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.

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 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.]

A Proof

This section is optional: click on the button see the proof.
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


The next section explores the shortest Egyptian fractions for any given fraction.

Shortest Egyptian Fractions

The greedy method and the shortest Egyptian fraction

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

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] = [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 for an ordinary fraction.
C A L C U L A T O R
  the shortest Egyptian fractions for
  
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R E S U L T S

The number of Shortest Length Egyptian Fractions

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 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!
number is the minimum number of unit fractions that are needed to sum to T/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 -
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. Those with 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 unit fractions 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:
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

Beyond 8 unit fractions is unknown territory!

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:
i2/(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
2/9 = 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
2/15 = 1/9 + 1/45
2/15 = 1/10 + 1/30
2/15 = 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/?
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 ? number is in green :
i2/2i+1 = 1/i+1 + 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
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 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
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 each entry is a count of all the 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  -  7  5  3  1  5  2
 8|   .  .  .  .  .  .  1  - 16  -  3  -  2  - 23  -  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

Fixed Length Egyptian Fractions

The shortest Egyptian fractions do not always give the smallest numbers.
For example, the smallest number of unit fractions for 8/11 is 4; there are 16 of them and the one with the smallest numbers (i.e. the one whose largest denominator is the smallest) is 8/11 = 1/2 + 1/6 + 1/22 + 1/66
However, if we look for a larger collection, of 5 unit fractions, we find smaller numbers still:
8/11 = 1/2 + 1/11 + 1/12 + 1/33 + 1/44 and
8/11 = 1/3 + 1/4 + 1/11 + 1/33 + 1/44

Here we investigate Egyptian Fractions with more than the smallest number of reciprocals.
We have already seen that every fraction T/B has an Egyptian Fraction and, what is more, an infinite number of longer and longer Egyptian fraction forms.
So let's see what we can find about the number of Egyptian fractions for t/b of a given length L.

C A L C U L A T O R
Options: leave blank if not needed


Egyptian fractions of length for
--
all denominators ≤          
find no more than
solutions
R E S U L T S

/ 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.
    1. In how many ways can you make write 1 as a sum of 4 different unit fractions?
    2. 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:
    LengthDenominators
    Numbers whose reciprocals sum to 1
    32,3,6
    42,4,6,12
    52,4,10,12,15
    63,4,6,10,12,15
    73,4,9,10,12,15,18
    84,5,6,9,10,15,18,20
    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
    105,6,8,9,10,12,15,18,20,24
    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
    126,7,8,9,10,14,15,18,20,24,28,30
    4,8,9,10,12,15,18,20,21,24,28,30
    1. Are there any patterns here that you can use to extend this table?
    2. 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 with 1/2 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 than 1/2 is 1/3:
    1 = 1/2 + 1/3 + ( 1/6 remaining).
    We could stop now or else continue with 1/7 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.
    Check your answers at A000058 in Sloane's Online Encyclopedia of Integer Sequences.
  6. Investigate shortest Egyptian fractions for 3/n:
    1. Find a fraction of the form 3/n that is not a sum of two unit fractions.
    2. Is it always possible to write 3/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 where 4/n cannot be expressed as a sum of two unit fractions.

Egyptian fractions for 4/n and the Erdös-Straus Conjecture

Although many fractions of the form 4/n can be written as a sum of just two unit 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 to 1014 but as yet no one has proved it true for all values of n nor has anyone found a number n for which it is not true.
The Calculator above shows that for any given n there are many ways to choose 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, y and z?
For instance: among all the result of three fractions summing to 4/n when n is even, we have:
nxyz
63412
84520
105630
126742
...
How would you write this pattern mathematically?
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 + 1/20
1/2 + 1/5 + 1/10
4/6=1/2 + 1/7 + 1/42
1/2 + 1/8 + 1/24
1/2 + 1/9 + 1/18
1/2 + 1/10 + 1/15
1/3 + 1/4 + 1/12
4/7=1/2 + 1/15 + 1/210
1/2 + 1/16 + 1/112
1/2 + 1/18 + 1/63
1/2 + 1/21 + 1/42
1/3 + 1/6 + 1/14
4/8=1/3 + 1/7 + 1/42
1/3 + 1/8 + 1/24
1/3 + 1/9 + 1/18
1/3 + 1/10 + 1/15
1/4 + 1/5 + 1/20
1/4 + 1/6 + 1/12
4/9=1/3 + 1/10 + 1/90
1/3 + 1/12 + 1/36
1/4 + 1/6 + 1/36
1/4 + 1/9 + 1/12
4/10=1/3 + 1/16 + 1/240
1/3 + 1/18 + 1/90
1/3 + 1/20 + 1/60
1/3 + 1/24 + 1/40
1/4 + 1/7 + 1/140
1/4 + 1/8 + 1/40
1/4 + 1/10 + 1/20
1/4 + 1/12 + 1/15
1/5 + 1/6 + 1/30
4/11=1/3 + 1/34 + 1/1122
1/3 + 1/36 + 1/396
1/3 + 1/42 + 1/154
1/3 + 1/44 + 1/132
1/4 + 1/9 + 1/396
1/4 + 1/11 + 1/44
1/4 + 1/12 + 1/33
4/12=1/4 + 1/13 + 1/156
1/4 + 1/14 + 1/84
1/4 + 1/15 + 1/60
1/4 + 1/16 + 1/48
1/4 + 1/18 + 1/36
1/4 + 1/20 + 1/30
1/4 + 1/21 + 1/28
1/5 + 1/8 + 1/120
1/5 + 1/9 + 1/45
1/5 + 1/10 + 1/30
1/5 + 1/12 + 1/20
1/6 + 1/7 + 1/42
1/6 + 1/8 + 1/24
1/6 + 1/9 + 1/18
1/6 + 1/10 + 1/15
4/13=1/4 + 1/18 + 1/468
1/4 + 1/20 + 1/130
1/4 + 1/26 + 1/52
1/5 + 1/10 + 1/130
4/14=1/4 + 1/29 + 1/812
1/4 + 1/30 + 1/420
1/4 + 1/32 + 1/224
1/4 + 1/35 + 1/140
1/4 + 1/36 + 1/126
1/4 + 1/42 + 1/84
1/4 + 1/44 + 1/77
1/5 + 1/12 + 1/420
1/5 + 1/14 + 1/70
1/5 + 1/20 + 1/28
1/6 + 1/9 + 1/126
1/6 + 1/12 + 1/28
1/6 + 1/14 + 1/21
1/7 + 1/8 + 1/56
4/15=1/4 + 1/61 + 1/3660
1/4 + 1/62 + 1/1860
1/4 + 1/63 + 1/1260
1/4 + 1/64 + 1/960
1/4 + 1/65 + 1/780
1/4 + 1/66 + 1/660
1/4 + 1/68 + 1/510
1/4 + 1/69 + 1/460
1/4 + 1/70 + 1/420
1/4 + 1/72 + 1/360
1/4 + 1/75 + 1/300
1/4 + 1/76 + 1/285
1/4 + 1/78 + 1/260
1/4 + 1/80 + 1/240
1/4 + 1/84 + 1/210
1/4 + 1/85 + 1/204
1/4 + 1/90 + 1/180
1/4 + 1/96 + 1/160
1/4 + 1/100 + 1/150
1/4 + 1/105 + 1/140
1/4 + 1/108 + 1/135
1/4 + 1/110 + 1/132
1/5 + 1/16 + 1/240
1/5 + 1/18 + 1/90
1/5 + 1/20 + 1/60
1/5 + 1/24 + 1/40
1/6 + 1/11 + 1/110
1/6 + 1/12 + 1/60
1/6 + 1/14 + 1/35
1/6 + 1/15 + 1/30
1/7 + 1/10 + 1/42
1/8 + 1/10 + 1/24
1/9 + 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 this page) 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.

Here are some simple cases for 4/n that we can easily see have 3 unit fractions since the 3 fractions need not have different denominators:
Continuing in this way, we can eliminate many forms of denominator from the list of those needing verification - but no one has managed to find a proof for all n yet!

In Mordell's Diophantine Equations published by Academic Press in 1969, he showed that we can reduce the unknown (unproven) cases to just those prime denominators n which have a remainder of 1, 112, 132, 172, 192 or 232 when divided by 840 (see A139665).

/ Things to do /

  1. The number of solutions to 4/n as a sum of 3 unit fractions is:
    The first value is 4/3:
    1 solution for 4/3 = 1/1+1/4+1/12
    1 solution for 4/4 = 1/2+1/3+1/6
    2 ways for 4/5
    5 ways for 4/6 = 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 series are 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 same applied 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,... which is A075248 in Neil Sloane's Online Encyclopedia of Integer Sequences. If the conjecture is true, then there are no zeroes in this series apart from the starting value.

/ Things to do /

    1. Find the single set of 3 unit fractions with a sum of 5/6.
    2. Find the three sets of 3 for 5/7.
    3. What formulae can you find for special cases of 5/n as a sum of 3 unit fractions?
  1. From the table of lengths of the shortest Egyptian fractions above, find a fraction that needs 5 unit fractions for its Egyptian fraction.
  2. Can you find a fraction that cannot be written using less than 6 unit fractions for its Egyptian fraction?
  3. Investigate Egyptian fractions which have only odd denominators.
    Is it possible to find a sum of odd Egyptian fractions for every fraction a/b?

The above will give you some ideas for your own experiments and the References below point to more information and ideas.
Happy calculating!

Smallest Denominators

Apart from the shortest Egyptian fractions (those with the fewest unit fractions), 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 Egyptian fractions. Here is a list of the EF's for 1 of various lengths with smallest denominators:
Length Denominators with sum of 1
32 3 6
42 4 6 12
52 4 10 12 15
63 4 6 10 12 15
73 4 9 10 12 15 18
83 5 9 10 12 15 18 20
4 5 6 9 10 15 18 20
9 4 5 8 9 10 15 18 20 24
4 6 8 9 10 12 15 18 24
10 5 6 8 9 10 12 15 18 20 24
11 5 6 8 9 10 15 18 20 21 24 28
5 7 8 9 10 14 15 18 20 24 28
6 7 8 9 10 12 14 15 18 24 28
12 4 8 9 10 12 15 18 20 21 24 28 30
6 7 8 9 10 14 15 18 20 24 28 30
13 4 8 9 11 12 18 20 21 22 24 28 30 33
4 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
5 8 9 10 11 12 18 21 22 24 28 30 33
5 8 9 10 11 15 18 20 21 22 24 28 33
6 7 8 9 11 14 18 20 22 24 28 30 33
6 7 8 10 11 14 15 20 22 24 28 30 33
6 7 9 10 11 14 15 18 20 22 28 30 33
6 8 9 10 11 12 15 18 20 22 24 30 33
6 8 9 10 11 12 15 18 21 22 24 28 33
7 8 9 10 11 12 14 15 18 22 24 28 33
146 8 9 10 11 15 18 20 21 22 24 28 30 33
7 8 9 10 11 14 15 18 20 22 24 28 30 33
156 8 9 11 14 15 18 20 21 22 24 28 30 33 35
166 8 11 12 14 15 18 20 21 22 24 28 30 33 35 36
176 10 11 12 14 15 18 20 21 22 24 28 30 33 35 36 40
187 10 11 12 14 15 18 20 21 22 24 28 30 33 35 36 40 42
So of all the EFs for 1 with 3 fractions, the smallest has all denominators no bigger than 6.
Of those EFs for 1 with 4 fractions, the smallest has no denominator bigger than 12. and for 5 fractions, the smallest has no denominator bigger than 15.
The series of these smallest maximum denominators (the minimax solution) in the 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 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 a longer decomposition. Only one sum of unit fractions is given when several are possible. The scribe tends to favour unit fractions with even denominators, since this makes their use in multiplication and division easier. The Egyptian multiplication method was based on doubling and adding, in exactly the same way 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 was decimal 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. All the shortest forms and alternative shortest forms are given here in an extra column.
2 = 1 + 1 + 1 + 1
nabcd

nabcdshortest?
5315   
7428   
9618  5 45
11666   
13852104 ×7 91
151030  8 120
9 45
12 20
17125168 ×9 153
191276114 ×10 190
211442  11 231
12 84
15 35
2312276   
251575  13 325
271854  15 135
24 378
292458174232×15 435
3120124155 ×16 496
332266  21 77
18 198
17 561
353042  21 105
20 140
18 630
3724111296 ×19 703
392678  24 104
21 273
20 780
4124246328 ×21 861
434286129301×22 946
453090  36 60
35 63
27 135
25 225
24 360
23 1035
nabcdshortest?
4730141470 24 1128
4928196  25 1225
5134102  30 170
27 459
26 1326
5330318795 ×27 1431
5530330  40 88
33 165
28 1540
5738114  33 209
30 570
29 1653
5936236531 ×30 1770
6140244488610×31 1891
6342126  56 72
45 105
36 252
35 315
33 693
32 2016
6539195  45 117
35 455
33 2145
6740335536 ×34 2278
6946138  39 299
36 828
35 2415
7140568710 ×36 2556
7360219292365×37 2701
7550150  60 100
45 225
42 350
40 600
39 975
38 2850
nabcdshortest?
7744308  63 99
42 462
39 3003
7960237316790×40 3160
8154162  45 405
42 1134
41 3321
8360332415498×42 3486
also
166 249 498
8551255  55 187
45 765
43 3655
8758174  48 464
45 1305
44 3828
8960356534890×45 4005
184 of length 3
9170130  52 364
49 637
46 4186
9362186  51 527
48 1488
47 4371
9560380570 ×60 228 
57 285
50 950
48 4560
9756679776 ×49 4753
9966198  90 110
63 231
55 495
54 594
51 1683
50 4950
101202303606 ×51 5151
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 + 1/618
2/103 = 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 the 2/5n forms in the papyrus table?
  2. Is there a pattern common to all the 2/7n forms in the papyrus table?
  3. Which fractions in the table could be found by the Fibonacci method?

Links and References

The following two books are recommended if you want to read more about the extraordinary Hungarian mathematician Paul Erdös On the History of Egyptian mathematics, I recommend:
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