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 methods with puzzles and 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. The Calculators embedded in the page provide helpful resources for
your number searches.
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Contents of this page
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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.
[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" which, from his description of the Pharoah of that time
dates it to
2000 BC or earlier. 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 and for sharing bread and beer (given as wages)
amongst several workers.
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?
Suppose you and 7 other friends go for a pizza. You all like the same kind but you don't want a whole one each.
The 8 of you decide to buy 5 identical pizzas. How do you divide them up between you? Decimals don't help and that you each get 5/8 only tells you
what the problem is (split 5 things into 8 parts) not the solution! It is easier with beer or sacks of grain.
This was an everyday problem in ancient Egypt
when barley loaves may have to be divided amongst workers.
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 today and clear the irrigation channels.
Pause for a minute and decide how YOU would solve this problem before reading on.....
HINT: What if there were only 4 loaves not 5 to be split amongst 8 people?
First Fatima sees that they all get at least half a loaf, so she uses 4 of the loaves
to give all
8 of them half a loaf each. She has 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}
The Egyptian solution has the added benefits of
fewer crumbs/slices than dividing each loaf into 8 and giving 5 slices to each person.
it is easy to see everyone has an identical number of pieces of the same sizes.
each portion received is a fraction 1/n of a loaf. All the fractions are different and unit fractions
Try the following using the Egyptian style of thinking:
You do the Maths...
Suppose Fatima had 3 loaves to share between 4 people.
How would she do it?
3/4 = 1/2 + 1/4
...and what if it was 2 loaves amongst 5 people?
2/5 = 1/3 + 1/15 is the simplest solution
...or 4 loaves between 5 people?
4/5 = 1/2 + 1/4 + 1/20 or
4/5 = 1/2 + 1/5 + 1/10
What about 13 loaves to share among 12 people?
We could give them one loaf each and
divide the 13^{th} into 12 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?
It turns out that Egyptian fractions are not only a very pratical solution to some everyday problems today
but are interesting in their own right.
They had practical uses in the ancient Egyptian
method of multiplying and dividing, and every fraction ^{t}/_{b} can always be written as an Egyptian fraction,
which we will show further down on this page.
There are also many unsolved problems concerning them, which are still a puzzle to mathematicians today.
A Calculator to convert a Fraction to and from an Egyptian Fraction
An Egyptian Fraction for ^{t}/_{b} is a sum of unit fractions, all different.
Enter your
fraction in the boxes below and then click on the Convert to an Egyptian fraction button
and the denominators of an equivalent
Egyptian fraction will be put Denominators box and displayed in the RESULTS window.
Alternatively, type in a list of the unit fraction's denominators of an EF
(they need not all be difference but they should all be bigger than 1) and then press the
button to fill in the fraction boxes with its sum.
Use the Shortest EF Calculator or the
Fixed Length EFs Calculator
to find all the 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. The
disadvantage of the "greedy" method is that sometimes
it will fail to fully convert a fraction if a denominator gets too large for the Calculator.
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 set of distinct unit fractions which sum to
any given fraction ^{t}/_{b}≤1 by actually showing how to find
such a sum.
Fibonacci's Greedy Algorithm for finding Egyptian Fractions
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.
It is the method used in the Fraction ↔ EF CALCULATOR above.
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}:
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},
^{171}/_{1050},
^{7}/_{1050}
The numerator of the remainder is getting smaller each time. If it keeps decreasing then it must
eventually reach 1 and the process stops. In this example, we find the numerator becomes a factor of the denominator and so
the final remainder is a unit fraction and we can stop.
Practice with these examples and then we'll have a look at finding short Egyptian fractions.
What does the greedy method give for ^{5}/_{21}?
What if you started with ^{1}/_{6} (what is the remainder)?
5/12 = 1/5 + 1/27 + 1/945 by the Greedy method but
5/21 = 1/6 + 1/14
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.]
9/20 = 1/3 + 1/9 + 1/180
9/20 = 1/4 + 1/5
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.]
Also, we are choosing the largest u_{1} at each stage, hence the name
of "the greedy algorithm".
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}
To find the largest denominator, evaluate ^{b}/_{t} and
use that value if it is an integer, or else round it up if not.
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 from 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 numeratort.
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:
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
KEY:
–
means the fraction ^{t}/_{b} is not in its lowest form
.
means the fraction ^{t}/_{b} is bigger than 1
The minimum number of unit fractions
that are needed for ^{t}/_{b}
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}
seems to be the one with the smallest numbers (the maximum denominator is the least among all the solutions)
according to David Epstein in
this MathForum post of March 2000.
He also states that he found 2771 separate 7-term EFs 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
=
1
+
1
2i + 1
i + 1
(i + 1)(2i + 1)
We can check it by simplifying the fraction on the right and the algebra will show us that the formula is always true.
Ben Thurston (May 2017) emailed me two other simple formulae:
1
=
1
+
1
a b
a (a + b)
b (a + b)
1
=
1
+
1
+
1
a b c
a (ab + bc + ca)
b (ab + bc + ca)
c (ab + bc + ca)
For example, the first formula applies to ^{1}/_{15} = ^{1}/_{3×5} = ^{1}/_{3×8} + ^{1}/_{5×8} = ^{1}/_{24} + ^{1}/_{40}
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.
In 1954 Breusch proved that every fraction with an odd denominator has an EF consisting of only odd denominators.
Here is a table of one example for each fraction less than 1 with an odd denominator up to 11, giving the EF as a list of the denominators.
All these have the smallest maximum EF denominator:
1/3 → {5, 9, 45}
1/5 → {9, 15, 45}
1/7 → {15, 21, 35}
1/9 → {15, 35, 63}
1/11 → {21, 33, 77}
2/3 → {3,5,9,45}
2/5 → {3,15}
2/7 → {7,15,21,35}
2/9 → {5, 45}
2/11 → {9, 33, 45, 55}
3/5 → {3,5,15}
3/7 → {3,15,35}
1/3
3/11 → {5, 15, 165}
4/5 → {3,5,7,15,21,105}
4/7 → {3,7,15,35}
4/9 → {3, 9}
4/11 → {3, 33}
5/7 → {3,5,9,21,45}
5/9 → {3, 5, 45}
5/11 → {3, 11, 33}
6/7 → {3,5,7,9,21,45}
2/3
6/11 → {3, 9, 11, 99}
7/9 → {3, 5, 9, 15, 35, 45, 63}
7/11 → {3, 5, 15, 33, 165}
8/9 → {3, 5, 7, 9, 21, 35, 63, 105}
8/11 → {3, 5, 9, 21, 45, 77}
9/11 → {3, 5, 9, 11, 21, 45, 77}
10/11 → {3, 5, 7, 11, 15, 21, 55, 105}
4512 Solution:A Special Case of Egyptian Fractions R Breusch Amer Math Monthly (March 1954) pages 200-201
The exercises below introduce a powerful and simple proof technique which uses just the even-ness and odd-ness of numbers, called their parity
to prove some observations about the list above.
We return to this kind of reasoning when we consider EFs for 1 with odd denominatorlater on this page.
You do the Maths...
Looking at the list above, the lengths of the EFs vary. However, a pattern
emerges when we look at whether a list is of even or odd length.
Where do the even-length EFs occur?
Can you make a conjecture?
If the numerator is even, the length of the EF is even.
If the numerator is odd, the length of the EF is odd. The parity of the numerator in a fraction with an odd denominator is the same as the parity
of the length of an EF with only odd denominator unit fractions
Consider an EF of just 2 odd denominators. Let's say 1/(2n+1) + 1/(2m+1).
What can you say about the sum in terms of oddness and evenness (this is called parity)?
Suppose we have 3 odd denominators in an EF. What about the parity of the sum of them?
1/(2n+1) + 1/(2m+1) = (2n +2m + 2)/(2mn+2m+2n+1) = even/odd
For three terms we have the above for two plus 1/odd: even/odd_{1} + 1/odd_{2} = (even×odd_{2} + odd_{1})/(odd_{1}×odd_{2}) = odd/odd
Generalise the above for an all-odd-denominator EF for p/q if q is odd.
The total of any number of odd denominator unit fractions must have an odd denominator.
A total of even/odd can have only even length EFs if all the denominators are odd
A total of odd/odd can have only odd length EFs if all the denominators are odd.
Egyptian Fractions for 1
Earlier we used an Egyptian Fraction for 1 in a proof that every EF can be expanded into an infinite number
of alternative sums for the same fraction. We used:
1 = ^{1}/_{2} + ^{1}/_{3} +
^{1}/_{6}
This is both the the shortest way (3 fractions) and the one with the smallest maximum denominator (6).
Before you read on, you might like to try and find 4 different unit fractions that sum to 1.
Hint:
6 = 3 + 2 + 1 and since it is a sum of different divisors of 6, we can divide by 6 to find our 3-term EF for 1 shown above.
Can you find another number which is also a sum of some of its divisors?
What about 12 = 6 + 4 + 2?
Dividing by 12 gives us 1 = ^{1}/_{2} + ^{1}/_{3} +
^{1}/_{6} which we already know.
EFs for 1 by Length
Here is a fixed length EF for 1 that uses 4 unit fractions:
We do not know any more terms in this series as the only approach so far has been
to perform a computer search and the number of solutions is getting very large very quickly!
A table of the solutions for smaller lengths is given in A073546
but the Calculator above will find these for you.
EFs for 1 by denominator sum
One was to get both short EFs first as well as having smaller denominators is to organize any list
of EFs by the sum of the denominators in each EF.
Here is a list of EFs for 1 arranged in order of denominator sum up to 50:
Denoms of EF for 1
Denom sum
1
1
2, 3, 6
11
2, 4, 6, 12
24
2, 3, 10, 15
30
2, 4, 5, 20
31
2, 3, 9, 18
32
2, 3, 8, 24
37
3, 4, 5, 6, 20
38
2, 4, 10, 12, 15
43
2, 4, 9, 12, 18 2, 5, 6, 12, 20
45
2, 4, 8, 12, 24 3, 4, 6, 10, 12, 15
50
EFs for 1 that begin 1/n
It seems there is always an EF for 1 which begins with 1/n for any n we like!
One such expression for 1 as a sum of distinct unit fractions of which the largest is 1/n is always guaranteed to exist
for any and every n>1 as was shown by Botts in 1967:
A Chain Reaction Process in Number Theory, Truman Botts,
Mathematics Magazine, Vol. 40, No. 2 (1967), pages 55-65
Starting from 1=1/2+1/3+1/6, he uses the Expansion Equation:
1
=
1
+
1
n
n + 1
n ( n + 1)
to replace the smallest 1/n by the right-hand side of this equation. If any duplicates arise, he uses it again on one of the duplicates.
He proves that
by repeating this process we can eventually remove all duplicates with any value 1/n as the largest unit fraction.
However the resulting EFs become quite long quite quickly:
These are not the shortest: 3, 4, 5, 6, 20 is shorter starting with 3
but maybe they have the smallest (maximum) denominators? Can you find any shorter or with smaller numbers?
Prof Greg Martin of the University of British Columbia has found a remarkable EF for 1
with 454 denominators all less than 1000:
There seems to be a pattern here:
The last denominator 1/(2D) can be replaced by two new ones, the smallest of which is 1/(2D+2)
and both the old and the two new denominators are even.
A little algebra shows why:
1
=
1
+
1
..... Let's call this the Expand Even Rule
2D
2D + 2
2D(D + 1)
The expansion preserves the evenness of the denominators.
We can continue this expansion and so show that
There is always an expansion of 1 as an EF with distinct even denominators of any length greater than 3.
You do the Maths...
Above we expanded the final even denominator.
This section helps you to explote what happens if we expanded the others using the Expand Even Rule?
If we start from 2,2 as the denominators of our initial EF for 1,
what does the Rule give if we expand one of the 2's?
2, 4, 4
What about expanding 4?
6,12
If we start with the EF denominators 2, 4, 4 and expanded the final 4,
which EF for 1 do we get?
2, 4, 6, 12
Starting from 2, 4, 6, 12 expand the 6 to get a new EF for 1.
2, 4, 8, 12, 24
In your previous answer, expand the 8 or one of the others. Which EFs do you get?
When it comes to all denominators being odd, the problem is quite different!
There are none of lengths 2, 3, 4, 5, 6, 7 or 8!
The first has 9 unit fractions and there are 5 of them with denominators as
follows:
A computer search reveals that there are none of length 10.
There do not seem to be any EFs for 1 with an even number of odd denominators. Earlier we encountered such a pattern when considering an EF
fraction with solely odd denominators.
Some thought about evenness and oddness (parity) reveals why there never
will be: Can you find a reason why the length must be odd?
The proof uses a parity argument that is, it uses just the oddness and evenness of numbers.
Any odd number can be written as 2n + 1 and every even number is of the form 2n
a SUM of two odd numbers is even:
2n + 1 + 2m + 1 = 2n + 2m + 2 = 2(n + m + 1)
a sum of an even number of odds is always even
a PRODUCT of two odd numbers is odd:
(2n + 1)(2m + 1) = 4nm + 2n + 2m + 1 = 2(2nm + n + m) + 1
a product of any number of odd numbers is always odd.
Therefore an odd number can never have an even factor. Only even numbers have even factors.
Consider just two unit fractions with odd denominator that sum to 1 and add them to get a single fraction as follows
1 =
1
+
1
=
odd1 + odd2
odd1
odd2
odd1×odd2
The numerator is a sum of two odds and so is even.
The denominator is a product of 2 odds and so is odd.
So the numerator and denominator cannot be identical and the fraction cannot be 1.
The same thing happens if we have a sum of 4 unit fractions with odd denominators:
The numerator is a sum of an even number of odds and so must be even
the collected denominator is a product of odd numbers and is therefore odd.
So the fraction can never be reduced to 1 when there are an even number of unit fractions summing to 1.
Thus an EF for 1 which has only odd denominators can never be of even length.
There are EFs for 1 with odd denominators and of odd length from 11 as shown here:
so it seems likely these exist for every odd length from 9 upwards....
...and we can prove this if, as with proving that there are an infinite number of EFs derivable from any EF
in the section above, we can find an expansion formula for 1/odd that uses 3 odd denominators.
Try to find one yourself or else
For instance a bit of algebraic manipulation will show that:
1
=
1
+
1
+
1
2n + 1
2n + 3
(n + 1) (2 n + 3) – 1
(n + 2) (2n + 1) (2 n + 3)
If n itself is odd and only then will each of the 3 denominators on the right be odd too.
So now we know:
There will always be odd length EF for 1 consisting of only odd denominators for all odd lengths from 9 onwards.
What we still don't know is how many there are of odd length n for n from 11 onwards.
Nor indeed do we know the ones with the smallest numbers, that is, the largest denominator in the EFs
for 1 of a given (odd) length which is the smallest possible.
93.20 Egyptian Fraction Representations of 1 with Odd Denominators,
Peter Shiu, The Mathematical Gazette Vol. 93, No. 527 (2009), pages 271-276
EFs for 1 in various arrangements
Here is a list of the EF's for 1 that have the smallest denominators for a given length. We allow any number as denominator
but still maintain the EF condition that all the denominators are different:
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.
Here we list EFs of 1 arranged by maximum denominator irrespective of length and
ordered by the sum of the denominators:
Denominators
Max Den
1
1
2, 3, 6
6
2, 4, 6, 12
12
2, 3, 10, 15 2, 4, 10, 12, 15
15
2, 3, 9, 18 2, 4, 9, 12, 18
18
2, 4, 5, 20 2, 5, 6, 12, 20 3, 4, 5, 6, 20
20
2, 3, 8, 24 2, 4, 8, 12, 24
24
2, 3, 12, 21, 28 2, 4, 6, 21, 28 2, 4, 7, 14, 28
28
Den Sum
Denominators
1
1
11
2, 3, 6
24
2, 4, 6, 12
30
2, 3, 10, 15
31
2, 4, 5, 20
32
2, 3, 9, 18
37
2, 3, 8, 24
38
3, 4, 5, 6, 20
43
2, 4, 10, 12, 15
45
2, 4, 9, 12, 18 2, 5, 6, 12, 20,
50
2, 4, 8, 12, 24 3, 4, 6, 10, 12, 15,
52
3, 4, 6, 9, 12, 18
The series ordered by maximum denominator is 1, 6, 12, 15, 18, 20, 24, 28, 30, 33, 35, 36, 40, 42, 45, 48, ...A092671
The series of sums is
1, 11, 24, 30, 31, 32, 37, 38, 43, 45, 50, 52, 53, 54, 55, 57, 59, 60, 61, 62, 64, 65, 66, 67, 69, 71, 73, 74, 75, 76, 78, ...A052428, the strict Egyptian numbers.
Graham showed that every integer from 78 onwards is in the list! The missing numbers up to 77 are
2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 25, 26, 27, 28, 29, 33, 34, 35, 36, 39, 40, 41, 42, 44, 46, 47, 48, 49, 51, 56, 58, 63, 68, 70, 72, 77
A051882. He uses two more "splitting" expansions one of which is :
1 =
1
+ ... +
1
=
1
+
1
+ ... +
1
d_{1}
d_{k}
2_{ }
2 d_{1}
2 d_{k}
and the second is the same but replacing the 1/2 by
1
=
1
+
1
+
1
+
1
2
3
7
78
91
The first splitting formula shows that there are EFs for 1 consisting of only even denominators and how to derive one from any EF for 1.
The counts of the number of EFs for 1 with sum 1,2,3,... gives the series
1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, ...A051907
but all values after the 77th are non-zero.
The number of solutions is small for sums up to 100 but the larger counts in that range are
6 solutions with a sum of 71,
11 solutions for sums 95 and 99 but
12 solutions for sum 92.
You can verify these counts and see the EFs using the Calculator in the next section.
A theorem on partitions R. L. Graham in Journal of the
Australian Mathematical Society Vol 3 (1963), pages 435-441
If we want those sums that have just 1 EF for 1 we get the list
1, 11, 24, 30, 31, 32, 37, 38, 43, 52, 53, 54, 55, 59, 60, 61, 65, 73, 75, 80, 91A051909
but after this point there may be no more, all the other sums having at least 2 EFs for 1 with that sum.
A Calculator for finding EFs of 1
This Calculator will count the EFs for 1 for a given sum of denominators.
"Find" will also show the EFs.
You can limit the search by giving an 'up to' number of solutions to be found.
Generally a single solution can be found up to a sum of 1000 within a minute or
so, depending on the speed of your computer. There is always at least one solution for any sum bigger than 77.
There is now a total of 14 ways to write 1 as a sum of 4 unit fractions including
those solutions we found above. What are they?
Check your answers with A002966
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.
The Inheritance Puzzle
This is an old puzzle mentioned in many puzzle books in various guises:
A man who had 12 horses and 3 children wrote his will to leave 1/2 of his horses to Pat, 1/3 to Chris and 1/12 to Sam.
However, just after he died one of his horses died too.
How were the children to divide the 11 remaining horses so as to fulfil the terms of the will?
The answer is that a friend calls by and offers to add his own horse to the 11 others.
Now they can split the horses with
1/2 = 6 horses going to Pat
1/3 = 4 horses going to Chris
1/12 = 1 horse going to Sam
a total of 11 horses exactly as specified in the will but
leaving one horse over - the horse the friend brought. The friend could leave taking his horse with him
and the children too go home happy. What happened here?
The answer lies in the peculiar conditions of the will since 1/2 + 1/3 + 1/12 = 11/12 and not 12/12!
It works because the denominators 2, 3 and 12 are all factors of 12.
Here is another variation on the same inheritance puzzle:
A cactus collector had 11 rare cactii and in her will she left 1/2 of the collection to one other collector,
1/4 to another collector and 1/6 to a third.
When she died how could the plants be divided as specified without splitting any?
The same solution applies, with the loan of one extra plant which is then returned on successful division of the 12
and the collectors receive 12/2 = 6, 12/4 = 3 and 12/6 = 2 and of course 6 + 3 + 2 = 11.
This kind of problem is designed from solutions to EFs for 1 where the denominators are all factors of their sum plus 1.
For instance in the cactii puzzle 11 = 6 + 3 + 2 and 6, 3, 2
are all factors of 11 + 1 = 12.
If we can divide these factors into 12 and also include 1/12 then we have an EF for 1 as follows: 1 = 12/12 = 12/6 + 12/3 + 12/2 + 1/12 = 1/2 + 1/4 + 1/6 + 1/12
but note that not all EFs for 1 have this form: 24, 8, 3, 2
are all divisors of 24 and dividing each into 24
gives 1/24 + 1/8 + 1/3 + 1/2 = 1 but 8 + 3 + 2 is only 13.
Numbers that are the sum of a subset of their divisors are called pseudo perfect numbers,
named by Sierpinski in 1965.
For example, the smaller ones are:
n
Divisors with sum n
Divisors/n
6
1, 2, 3
1/6 + 2/6 + 3/6 = 1
12
2, 4, 6
2/12 + 4/12 + 6/12 = 1
12
1, 2, 3, 6
1/12 + 2/12 + 3/12 + 6/12 = 1
18
3, 6, 9
3/12 + 6/18 + 9/18 = 1
18
1, 2, 6, 9
1/18 + 2/18 + 6/18 + 9/18 = 1
20
1, 4, 5, 10
1/20 + 4/20 + 5/20 + 10/20 = 1
The list of pseudo perfect (or semiperfect) numbers starts: 6, 12, 18, 20, 24, 28, 30, 36, 40, 42, 48, 54, 56, 60, 66,... A005835
If n = a + b + ... + c is in the list where the right hand side consists only of divisors of
n then 2n = 2a + 2b + ... + 2c and 3n = 3a + 3b + ... + 3c
and so on must also be in the list.
The pseudo perfect numbers that are not a multiple of another pseudo perfect number are called primitive pseudo perfect numbers: 6, 20, 28, 88, 104, 272, 304, 350, 368, 464, 490, 496, ... A006036
You do the Maths...
There are a total of 7 possible puzzles like the horses and the cactus collection above if there are 3 inheritors.
What are they?
number to share
Proportions
Shares with 1 loaned
7
1/2, 1/4, 1/8
4, 2, 1
11
1/2, 1/3, 1/12
6, 4, 1
11
1/2, 1/4, 1/6
6, 3, 2
17
1/2, 1/3, 1/9
9, 6, 2
19
1/2, 1/4, 1/5
10, 5, 4
23
1/2, 1/3, 1/8
12 ,8, 3
41
1/2, 1/3, 1/7
21, 14, 6
Suppose there are 23 horses and any number of inheritors. What combination of fractions gives a similar puzzle
now?
One collector of rare pottery had 8 people to share the collection between. But one pot got broken.
What size of collection and what proportions give a similar puzzle?
60 pots, one broken, divided in proportions 1/60, 1/30, 1/20, 1/12, 1/10, 1/6, 1/5 and 1/3.
60 has 34 subsets of its divisors that sum to 60. It is pseudoperfect in 34 different ways.
What are the divisors of 60?
1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
Which subsets of divisors sum to 60?
For example, the smallest subset is {10,20,30}.
In how many ways is 120 pseudoperfect?
278 subsets of its divisors sum to 120.
By contrast, 20 and 28 are pseudoperfect in only one way. What are those subsets?
The divisors of 20 are 1, 2, 4, 5, 10, 20
Its unique subset totalling
20 is 1, 4, 5, 10
The divisors of 28 are 1, 2, 4, 7, 14, 28
What is special about the pseudoperfect numbers 6 and 28?
Both are pseudoperfect in only one way but for both these numbers
the subset of divisors includes every divisor except the number itself.
Such numbers are called Perfect numbers. The list of these is sparse and grows
rapidly. It begins: 6, 28, 496, 8128, 33550336, ...A000396
All in this list are even and no one knows if there is an odd perfect number.
536 Puzzles and Curious Problems H E Dudeney (1970 paperback edition of the original of 1967)
Puzzle 172 deals with the first Inheritor problem in this section and expands on it. The puzzles in this
book are quite unique, amusing and very comprehensive. A large proportion of the book is given over to solutions
and the maths behind them. A real treasure.
EFs for other integers
There are EFs for 2 and for 3:
Denominators of an EF for 2: 2, 3, 4, 5, 6, 8, 9, 10, 15, 18, 20, 24
Denominators of an EF for 3: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 30, 34, 100, 11934, 14536368
Can you find an EF for 4? for 5?
Two more unsolved problems
Here are two problems about numbers which can be written using just 3 different unit fractions.
Egyptian fractions for 4/n and the Erdös-Straus Conjecture
Every fraction of the form 3/n where n is not a multiple of 3 and odd can be written as 1/a + 1/b + 1/c for distinct odd a, b and c. For a proof see
A Proof of a Conjecture on Egyptian Fractions T. R. Hagedorn
The American Mathematical Monthly, Vol. 107, (2000), pages 62-63
What about 3/n when n is even?
Can 4/n and 5/n be written as as a sum of just three unit fractions also?
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.
They have no restriction on the three unit fractions being unique, so for this problem, repetitions of unit
fractions are allowed.
It has been verified that 3 unit fractions can found for all values of n up to 51,000,000-th prime
1003162753
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. And what if we insisted that the three fractions be distinct?
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?
4
=
2
=
1
+
1
+
1
2n
n
n
n + 1
n(n + 1)
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
=
2
=
1
+
1
+
1
2n
n
n
2n
2n
This formula duplicates a fraction. Gary Detlef's version has the advantage that all the fractions are different:
4
=
2
=
1
+
1
+
1
2n
n
n
n + 1
n( n + 1)
Here are some of Mordell's cases that take care of many values of n:
4
=
1
+
1
+
1
3n
3n
2n
2n
4
=
1
+
1
+
1
4n−1
2n
2n
n(4n−1)
4
=
1
+
1
+
1
4n−2
n
n(4n−2)
n(4n−2)
4
=
1
+
1
+
1
8n−3
2n
n(2n−3)
n(2n−3)
What about multiples of 3?
4
=
1
+
1
+
1
3n−1
n
3n−1
n(3n−1)
4
=
1
+
1
+
1
=
1
+
1
+
1
3n
2n
2n
3n
n
4n
12n
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. For example:
4
=
1
+
1
+
1
5
2
4
20
If we divide both sides by n and we have a simple formula for three unit fractions for 4/(5n):
4
=
1
+
1
+
1
5n
2n
4n
20n
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).
If you want to check the cases Mordell says are solved, these two equations from Ben Thurston (October 2016) will help fill
in some of the more awkward cases:
4
=
1
+
1
+
1
12n + 7
3(n + 1)
3(n + 1)(3n + 2)
(12n + 7)(3n + 2)
4
=
1
+
1
+
1
24n + 13
2(3n + 2)
2(n + 1)(24n + 13)
2(n + 1)(24n + 13)(3n + 2)
Diophantine Equations Louis J Mordell, Academic Press (1967), pp. 287-290.
You do the Maths...
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.
Let's check some smaller values of n: 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 for n=2. It has been verified for all n up to 922321,
Sur les décompositiones de nombres rationelles en fractions primaries W Sierpinski,
Mathesis 65 (1956), pages 16-32
You do the Maths...
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
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
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 most 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?
Productive Egyptian Fractions
Some fractions have a special form of EF where each denominator is a factor of the next. For example:
4
=
1
+
1
+
1
5
2
2×2
2×2×5
In fact, such representations are useful for a system of measurements such as the pounds-shillings-pence (£ s d) monetary system or
the stones-pounds-ounces system of weights.
There were 12 pence (denoted d) in every shilling (s), 20 shillings in every pound (£) so £2 3s 11d could be represented as
£ 2 + 3/20 + 11/(20×12).
Similarly there are 16 ounces in 1 pound weight and 14 pounds in one stone. 2 stone 9 pounds and 3 ounces
is therefore 2 + 9/14 + 3/(14×16) ounces.
In the Rhind Table above 26 of the 49 Egyptian fractions have this format for 2/n, that is all but two
of those with two unit fractions. The two exceptions are
which applies to a total of 8 two-unit fractions in the Rhind table.
Every fraction has a EF as accumulated Products
Cohen only focuses on a "greedy" method of generating them and restricts his attention to those
where
each new factor is never smaller than the one before it when we order the denominators in increasing size. With this
restriction, he proves a productive Egyptian fraction always exists and that it is unique.
3
=
1
+
1
=
1
+
1
8
3
3×8
4
4×2
factors 3,8 increasing
factors 4,2 decreasing
Cohen called these Egyptian fraction expansions but earlier (1923) Gibbs had called them
Productive Fractions. We will call them ordered productive EFs if the each new factor is
no smaller than the previous one (as in Cohen's paper) and
and productive EFs if there is no restriction on the size of the next new factor.
A good shorthand is to list the new factors of a productive EF, for example
649. Productive Fractions R W M Gibbs, Math. Gaz. (1923) pages 233-234
He also deals with recurring sets of new factors and subtracting unit fractions as well as adding them
Lengths of shortest Productive EFs
Every unit fraction has a productive Egyptian Fraction :
^{1}/_{2}
= ^{1}/_{3}
+ ^{1}/_{6}
^{1}/_{3}
= ^{1}/_{4}
+ ^{1}/_{12}
^{1}/_{4}
= ^{1}/_{5}
+ ^{1}/_{20}
^{1}/_{5}
= ^{1}/_{6}
+ ^{1}/_{30}
^{1}/_{6}
= ^{1}/_{7}
+ ^{1}/_{42}
^{1}/_{7}
= ^{1}/_{8}
+ ^{1}/_{56}
which is summarised by the algebraic identity
1
=
1
+
1
Productive expansion equation
n
n + 1
n ( n + 1)
A similar identity holds for all fractions 2/odd:
2
=
1
+
1
2n + 1
n + 1
(n + 1) ( 2n + 1)
Here is a table of the lengths of the shortest productive EFs for fractions less and one and with a denominator up
to 30, similar to the one above for all EFs.
This table includes all fractions whether in their lowest form or not:
Length of the Shortest Productive EF for Fraction ^{T}/_{B}
KEY:
.
means the fraction ^{t}/_{b} is
not less than 1.
12
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 an infinite number of productive EFs for every fraction?
Since every EF has a productive EF, we can try the method we used above
to show that every productive EF can be expanded.
If the final factor was n (the ratio between the final two unit fractions in a productive EF),
then we can replace it using the Expansion Equation
that we used earlier, as follows:
t
= ...
1
+
1
= ...
1
+
1
+
1
b
a
n a
a
(n+1) a
n (n + 1) a
This is easier when we look at the list of successive factors so that, for example
The final factor n of any productive EF can be replaced by the two factors
n + 1, n:
t/b
productive EF
successive factors
expanded factors
expanded productive EF
13/20
1/2 + 1/8 + 1/40
2, 4, 5
2, 4, 6, 5
1/2 + 1/8 + 1/48 + 1/240
1/2 + 1/8 + 1/48 + 1/240
2, 4, 6, 5
2, 4, 6, 6, 5
1/2 + 1/8 + 1/48 + 1/288 + 1/1440
So every fraction has a productive EF and we have just shown that
we can extend each one by one more factored term as often as we like.
The Harmonic Numbers
Summing the reciprocals from 1 up to n and the series of such sums have long intrigued mathematicians. The sums of the reciprocals
of 1 up to n are called the Harmonic Numbers H(n) and the series
H(1), H(2), H(3), ...
is called the Harmonic Series:
H(n) =
1
+
1
+
1
+
1
+ ... +
1
1
2
3
4
n
The series is called Harmonic because if we stretch a string tightly and twang it, we hear a certain note. Stopping the
string at the half way point makes a sound an octave above the first note. If instead we take just one third of the length we
get another note that seems harmonious to the ear in relation to the whole string. We also get
harmonious sounds if we take one quarter and one fifth and so on for some time.
Pythagoras
first noted this connection with harmonious sound and the lengths of plucked strings.
When a string is plucked as on a violin for instance, there is not only the "pure" note but also other quieter notes produced by
these "harmonic" notes called overtones.
The second table is the same fractions for H(n) but without simplifying the sums.
The numerators give the total number of cycles in all permutations of n+1 and
the fractions give the ratio (probability) of a random permutation
on n+1 letters having exactly two cycles!
What can we say about H(n)? Is any Harmonic number ever exactly an integer for instance?
Subtracting one Harmonic number from another larger one H(n) – H(k)
gives us the sum of a consecutive set of the unit fractions 1/(k+1) + ... + 1/n. Are any of these
ever integers?
Unfortunately, the answers are no; no finite sum of the series of unit fraction starting at 1 or at another
unit fraction will ever sum to a whole number.
The sum of all unit fractions
The series of reciprocals of all the natural numbers from 2 onwards is called the Harmonic Series
and the question of whether or not its sum
has a finite value or not is now a classic maths problem. There is a very easy proof to show
that the harmonic series series "diverges", that is, summed for ever, it gets infinitely large.
In case you want to think about it yourself,
the answer is revealed in an optional section:
Let's look at the first term: 1/2,
The next 2 terms: 1/3 + 1/4 but 1/3 > 1/4 so
1
+
1
>
1
+
1
=
1
3
4
4
4
2
Now look at
the next 4 terms:
1
+
1
+
1
+
1
>
1
+
1
+
1
+
1
=
1
5
6
7
8
8
8
8
8
2
similarly the sum of
the next 8 terms will exceed 1/2.
Since the Harmonic series sum goes on for ever, then we can always find another batch of 2^{n}
terms whose sum adds more than 1/2 to the total,
so the total is always larger than any given number, it never settles down to a fixed value, it grows for ever or "diverges".
This result has been known since at least 1650 (Pietro Mengoli).
75.11 The Noninteger Property of Sums of Reciprocals of Successive Integers Duane W. Detemple
The Mathematical Gazette, Vol. 75, No. 472 (Jun., 1991), pages 193-194
Here is a proof that
no consectuive reciprocals sum to an integer, simpler than that of the original of G.Polyà and G.Szegö of 1976.
The Overhanging books puzzle
There is a surprising application of the divergence of the Harmonic Series that makes a good Science Fair demonstration.
Suppose we have a shelf of identical books (or bricks or dominoes etc).
If we lay them down one on top of another
what is the maximum overhang we can achieve?
Can the top book ever completely overhang the bottom book?
For two books we can get an overhang of 1/2 a book length;
The two books now have a centre of gravity in the centre of their overlap, so the
bottom one can overhang and extra 1/4 of a book length.
The centre of gravity of the three books means they can have an overhang of 1/6,
and so on for more books.
The extra overhang for four and more books is 1/8, 1/10, 1/12, ... .
The total overhang for n books is therefore H(n-1)/2.
How many books will it take before the overhang exceeds the length of one book?
Answer: H(4)/2 = 25/24 which is bigger than 1 so:
5 stacked books are sufficient for the top
one to totally overhang the bottom one!
Can we get a bigger overhang? Yes! Since the Harmonic series diverges, we can get an overhang as large as we like, but it may take a large number of books and
very delicate balancing! Since it is very difficult to find lots of identically sized books, try playing cards.
Theoretically for an overhang of 2 book lengths, we need to find the value of n for which
H(n) > 4
H(30) = 3.99498713092
H(31) = 4.02724519544
so we need 32 books to get an overhang of 2 complete books!
For 3 book lengths, the first n with H(n) > 6 is n = 227
so that stack would be 228 books tall and n = 1674 before H(n) exceeds 8.
A Harmonic Number Calculator
The values of the Harmonic function in this calculator are
computed approximately.
All digits shown are correct and are within the accuracy of the approximation.
If we had 1,000,000 playing cards perfectly arranged to give the maximum overhang of the top card, how many card lengths would it overhang?
H(1 000 000) = 14.392726722865724 so the top card would overhang by more than 7 card lengths.
Concrete Mathematics
(2nd edition, 1994) by Graham, Knuth and Patashnik, Addison-Wesley page 278 expression (6.66):
H(n) ≈ ln(n) + γ +
1
–
1
with error<
1
2n^{ }
12 n^{2}
120 n^{4}
where γ is Euler's constantγ = 0.5772156649015328606....
Problem 52: Overhanging dominoes R T Sharp, Pi Mu Epsilon Journal (1954) 1 (10): 411-412
This seems to be the first appearance of the book stacking problem and its solution.
The Harmonic Series Diverges Again and Again,
S J Kifowit, T A Stamps Twenty proofs, all elementary, of the divergence of the harmonic series.
More Proofs of Divergence of the Harmonic Series S J Kifowit (unpublished)
Proofs 21 to 45 at the level of first year university Calculus.
See also
S J Kifowit publications where there are links to his other
excellent presentations and handouts on the Harmonic series.
EF 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.
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!
Unsolved Problems in Number Theory (3rd edition 2004) by R Guy, Springer
Section D11 (pages 252-262) is all about unsolved problems in Egyptian Fractions
and there is also has an extensive bibliography on the subject.
This is essential reading for the serious researcher!
Problem D11 is about Egyptian Fractions and contains many references and interesting results.
Count Like an Egyptian: A Hands-on Introduction to Ancient Mathematics
David Reimer, Princeton Press (2014)
is a new and detailed book explaining much more about the Rhind papyrus, its methods, the background and history, with a host
of detail and worked examples. An excellent manual for those who want to delve more deeply into the mathematical methods
of the ancient Egyptians. Uses no advanced maths beyond secondary school level. Recommended!
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.