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.
Contents of this page
The icon means
there is a Things to do investigation at the end of the section.
indicates an on-line
interactive calculator is provided for the section.
Some of the oldest writing in the world is on a form of paper made from papyrus reeds
that grew all along the Nile river in Egypt.
[The image is a link to
David Joyce's site
on the History of Maths at Clarke University.]
The reeds were squashed and pressed into
long sheets like a roll of wall-paper and left to dry in the sun. When dry, these scrolls
could be rolled up and easily carried or stored.
Some of the papyrus scrolls date back to about 2000 BC,
around the time of the construction of the larger Egyptian pyramids. Because there are deserts
on either side of the Nile, papyrus scrolls
have been well preserved in the dry conditions.
So what was on them 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).
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.
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:
The first reason is a practical one.
Suppose you have 5 sacks of grain to share
between 8 people, so each would receive ^{5}/_{8} of a sack of grain
in terms of present-day fractions.
How are you going to do it simply, without using a calculator?
You could try pouring the 5 sacks of grain into 8 heaps and, by carefully comparing them,
perhaps by weighing them against each other, balance them so they are all
the same! But is there a better way? We will see that using unit 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 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! 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 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):
and
^{5}/_{8} = ^{1}/_{2} + ^{1}/_{8}
Things to do
Suppose Fatima had 3 loaves to share between 4 people.
How would she do it?
...and what if it was 2 loaves amongst 5 people?
...or 4 loaves between 5 people?
What about 13 loaves to share among 12 people?
We could give them one loaf each and
divide the 13^{th} 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?
20^{th}s 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
Which is larger: ^{4}/_{7} or ^{5}/_{8}?
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
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} less than one
(so that T, the number on top, is smaller than B, the bottom number).
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:
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
^{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} 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.)
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
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.]
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
Also, we are choosing the largest u_{1} 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 (u_{1}-1) and expanding the brackets,
then adding T and subtracting B to both sides we have:
B > T (u_{1} - 1)
B > T u_{1} - T
T > T u_{1} - B
Now T*u_{1} - 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)
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
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:
^{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 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
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:
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}
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 fraction needs 8 unit fractions
and gives this example:
A097049 has the numerators and
A097048 the denominators of these "smallest" fractions which need at least 2,3,4,5,... terms in any
Egyptian Fraction.
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:
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 :
i
^{2}/_{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:
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:
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
Things to do
1 = ^{1}/_{2} + ^{1}/_{3} + ^{1}/_{6}
shows 3 different unit fractions with a sum of 1 whereas
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?
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?
Difficult!
Here are the EFs for 1 with the smallest numbers (that is, the largest denominator is smallest) of various lengths:
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}:
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.
Investigate shortest Egyptian fractions for ^{3}/_{n}:
Find a fraction of the form ^{3}/_{n} that is not a sum of two
unit fractions.
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.
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 10^{14}
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:
n
x
y
z
6
3
4
12
8
4
5
20
10
5
6
30
12
6
7
42
...
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.
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:
Here is
a simple formula for even denominators because: ^{4}/_{2n} = ^{2}/_{n} = ^{1}/_{n} + ^{1}/_{2n} + ^{1}/_{2n}
This formula duplicates a fraction. Gary Detlefs version has the advantage that all the fractions are different: ^{4}/_{2n} = ^{1}/_{n} + ^{1}/_{n+1} + ^{1}/_{n(n+1)}
So we only need investigate fractions of the form ^{4}/_{3n+1}
Furthermore, if we have a solution for ^{4}/_{n}
then we automatically have solutions
for ^{4}/_{n k} for all k
by multiplying each of the original denominators
by k. This means that we need only examine prime 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, 11^{2}, 13^{2}, 17^{2},
19^{2} or 23^{2}
when divided by 840 (see A139665).
Things to do
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
Find the single set of 3 unit fractions with a sum of ^{5}/_{6}.
Find the three sets of 3 for ^{5}/_{7}.
What formulae can you find for special cases of ^{5}/_{n} as a sum of 3 unit fractions?
Can you find a fraction that cannot be written using less than 6 unit fractions for its Egyptian fraction?
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.
The shortest for 8/11 is
8/11 = 1/2 + 1/6 + 1/22 + 1/66
and 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 44 and there are two
such EFs both containing 5 unit fractions (out of the 667 of length 5):
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 is a list of the EF's for 1 of various lengths with smallest denominators:
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
n
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
√
8 120 9 45 12 20
17
12
51
68
×
9 153
19
12
76
114
×
10 190
21
14
42
√
11 231 12 84 15 35
23
12
276
√
25
15
75
√
13 325
27
18
54
√
15 135 24 378
29
24
58
174
232
×
15 435
31
20
124
155
×
16 496
33
22
66
√
21 77 18 198 17 561
35
30
42
√
21 105 20 140 18 630
37
24
111
296
×
19 703
39
26
78
√
24 104 21 273 20 780
41
24
246
328
×
21 861
43
42
86
129
301
×
22 946
45
30
90
√
36 60 35 63 27 135 25 225 24 360 23 1035
n
a
b
c
d
shortest?
47
30
141
470
√
24 1128
49
28
196
√
25 1225
51
34
102
√
30 170 27 459 26 1326
53
30
318
795
×
27 1431
55
30
330
√
40 88 33 165 28 1540
57
38
114
√
33 209 30 570 29 1653
59
36
236
531
×
30 1770
61
40
244
488
610
×
31 1891
63
42
126
√
56 72 45 105 36 252 35 315 33 693 32 2016
65
39
195
√
45 117 35 455 33 2145
67
40
335
536
×
34 2278
69
46
138
√
39 299 36 828 35 2415
71
40
568
710
×
36 2556
73
60
219
292
365
×
37 2701
75
50
150
√
60 100 45 225 42 350 40 600 39 975 38 2850
n
a
b
c
d
shortest?
77
44
308
√
63 99 42 462 39 3003
79
60
237
316
790
×
40 3160
81
54
162
√
45 405 42 1134 41 3321
83
60
332
415
498
×
42 3486 also 166 249 498
85
51
255
√
55 187 45 765 43 3655
87
58
174
√
48 464 45 1305 44 3828
89
60
356
534
890
×
45 4005
184 of length 3
91
70
130
√
52 364 49 637 46 4186
93
62
186
√
51 527 48 1488 47 4371
95
60
380
570
×
60 228 57 285 50 950 48 4560
97
56
679
776
×
49 4753
99
66
198
√
90 110 63 231 55 495 54 594 51 1683 50 4950
101
202
303
606
×
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
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:
Is there a pattern common to all the ^{2}/_{5n} forms in the papyrus table?
Is there a pattern common to all the ^{2}/_{7n} forms in the papyrus table?
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 all sorts of information on
Egyptian Fractions and a comprehensive guide to the different algorithms that can be used
to write your own Egyptian Fraction computer programs, although his are described using many different techniques available in
the Mathematica package and there are references to C and C++ sources too.
Dr Scott William's page on
The Rhind 2/n Table
has a list of the fractions ^{2}/_{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. He also includes a discussion and analysis of the fractions chosen and suggestions of the methods the Egyptians
might have used. He has some interesting pages on African mathematics and mathematicians from ancient times to today.
Fibonacci on Egyptian Fractions M. Dunton and R. E. Grimm, Fibonacci Quarterly vol 4 (1966),
pages 339-353. Here
Grimm and Dunton give an English translation and explanation using modern notation of the section in chapter 7 of Fibonacci's Liber Abaci
which gives methods of expressing a fraction as a sum of unit fractions. Fibonacci deals with several special cases
called distinctions before giving the "greedy" algorithm above as the seventh and general method.
Download this paper in PDF
The Rhind Mathematical Papyrus
G Robins, C Shute, British Museum Press, 1987, (88 pages, paperback)
is highly recommended for its explanations of the arithmetic
methods that
may have been used in the 2/n table and the other tables and problems in the papyrus. It has
excellent colour photographs of the papyrus and many illustrations.
Buy it from the Amazon.co.uk site [use the link above] as it is much cheaper than the
Amazon.com site!
The following two books are recommended if you want to read more about
the extraordinary Hungarian mathematician Paul Erdös
The Man Who Loved Only Numbers
The Story of Paul Erdos and the Search for Mathematical Truth by P Hoffmann, Fourth Estate (1999) paperback
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 the Rhind Papyrus, it contents and methods.
Recommended!
The Exact Sciences in Antiquity
by Otto Neugebauer, Dover, second edition 1969, is another great book covering not only Egyptian arithmetic but also
the Babylonian, Sumerian and Greek contributions to both number notation and arithmetic as well as astronomy.
It is about the history of the mathematics more than the maths itself and is now, rightly, a classic on this subject.