CLASSROOM
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In this section of Resonance, we invite readers to pose questions likely to be raised in a
classroom situation. We may suggest strategies for dealing with them, or invite responses,
or both. “Classroom” is equally a forum for raising broader issues and sharing personal
experiences and viewpoints on matters related to teaching and learning science.
Syed Abbas
School of Basic Sciences
Sheep Distribution Problem Through Egyptian Fractions
Indian Institute of Technology
Mandi 175001
Himachal Pradesh
Email: [email protected]
Keywords
Greedy algorithm of Fibonacci,
unit fractions, Egyptian fractions.
840
We investigate and further explore the problem
of dividing x = n + m (m, n are coprime) sheep in
1
, m1 ratio.
n
Let us begin with an ancient story. An old person had
5 sheep and in his last wish he asked his two children to
distribute his sheep among them in the ratio of 12 and
1
. Now the problem was that the sheep could not be
3
divided in half and one third. So they went to a wise
man asking for his help. The wise man added one extra
sheep to these 5 sheep, which made the total six. Then
he gave half of these six which is 3 to one brother and
one third which is 2 to the other brother. After distribution the wise man was still left with 1 sheep which he
included in his herd. We see that the total sum of the
distribution is 5 and hence the brothers were satisfied
with this distribution. Now a natural question to ask is,
whether this is true for any number of sheep? Can we
always find a number such that if we add this number
and then distribute the sheep into half and one third,
we are left with this extra added number? This is easy
if the number of sheep is a multiple of 5. Suppose one
has 5x sheep, then we can always add x such that the
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CLASSROOM
total number of sheep becomes 6x. Now half of 6x is 3x
and one third of 6x is 2x, which make a total of 5x and
we still have the added extra sheep with us.
Now suppose, one wants to divide y = n + m (m, n are
coprime, m, n > 1; this is possible if y > 6) number of
sheep in a ratio of n1 and m1 . Notice here that y is not
divisible by n and m. The question here is, whether one
can always find a number l such that if we add it to y
and divide this new number y + l by n and m, we get the
natural numbers whose sum is y. Writing this statement
y+l
y+l
in the form
we get n + m = y, which
of an equation,
Suppose one
wants to divide y =
n +m (m,n are
coprime, m, n > 1;
this is possible if
y > 6) number of
sheep in a ratio of
1/n and 1/m.
Notice here that y
is not divisible by
n and m.
1
= 0. As n1 + m1 is
gives y n1 + m1 − 1 + l n1 + m
not equal to one, the above equation can be written
as y(n + m − nm) + l(n + m) = 0. Thus we obtain
l = y(nm−n−m)
= nm − n − m and hence we can always
(n+m)
choose an integer l = nm − n − m. It is interesting
to note that if the number of sheep is some multiple of
n+m, say a(n+m), then one needs to add a(nm−m−n)
sheep. Hence it is always possible to add a number of
sheep in the herd so that after distributing it in m1 and
1
ratio, we are still left with the added number of sheep.
n
One more similar story is as follows, A farmer dies leaving his 17 sheep to be divided between his 3 sons in the
following proportions, one half to the eldest, one third
to the middle son, and one ninth to the youngest. This
was a difficult calculation for the sons and so they asked
an intelligent man for help. The man added one sheep
to make the total number 18. The calculation is easy
now. One half of 18 is 9, one third of 18 is 6, and one
ninth of 18 is 2, giving a total of 17. The man can now
take his one sheep away leaving the three sons happy.
The above problems can be generalized to any number,
say, y = n + m + k of sheep and dividing these into
1
, m1 , k1 parts. It is easy to see that the number y is
n
not divisible by any of these numbers. One can prove
by using similar arguments as above that there exists a
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841
CLASSROOM
It is well known
that every rational
number admits a
proper Egyptian
fraction
representation,
and every positive
fraction has
infinitely many
expressions as
sums of unit
fractions.
natural number l such that if we divide y + l the number
by n, m and k, we get the total y. Interestingly this
problem can be further generalized to any finite number
of divisions.
The answer of the above problems lies in expressing any
number in terms of an Egyptian fraction. In particular, in the first two cases one wishes to write any rational number of the form n−1
as a sum of Egyptian fracn
tions. Egyptian fractions were used by ancient Egyptian
mathematicians to represent rational numbers in terms
of unit fractions. A unit fraction is a fraction of the form
1
, where n ≥ 2. An Egyptian fraction is an expression
n
of the form
m
1
1
1
+
+ ··· + ,
=
n
n1 n2
nk
where 0 <
m
< 1, k ≥ 1.
n
It is well known that every rational number admits a
proper Egyptian fraction representation, and every positive fraction has infinitely many expressions as sums of
unit fractions. The representation is said to be proper
if all ni , i = 1, · · · k are distinct. Europeans have known
how to compute Egyptian numbers since the 12th century. The proof is given by Fibonacci in his book Liber
Abaci written in 1202. The proof involves the application
of greedy algorithm of Fibonacci [1], which transforms
rational numbers into Egyptian fractions. Let us first
explore the process by picking the number 58 . This number lies between 12 and 1 because 58 = 113 . Subtracting 12
5
from 58 , gives 58 − 12 = 18 and hence 58 = 12 + 18 , which is
the Egyptian fraction representation of 58 . Let us explore
4
this process a little bit more – now pick the number 13
.
4
As 13
= 311 , it clearly lies between 14 and 13 . Subtracting
1
4
4
4
3
from this number, we obtain 13
− 14 = 52
. Now we re3
peat this process for the number 52 to get unit fractions.
3
1
1
Clearly 52
= 171 1 , and thus it lies between 18
and 17
. By
subtracting
842
1
,
18
3
we get
5
52
−
1
18
=
1
.
468
Combining both
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equalities, we get
1
3
1
1
1
4
= +
= +
+
.
13
4 52
4 18 468
From the above calculation, it is clear how to proceed
with any number, say, m
. If m/n is not of the form 1/k,
n
one can always choose a largest natural number N such
that N1 < m
< N 1−1 . Let us subtract N1 from m
, where
n
n
m is not 1, which gives
If m/n is not of the
form 1/k, one can
always choose a
largest natural
number N such
that 1/N < m/n
< 1/(N–1).
1
mN − n
m
−
=
.
n
N
nN
Using the above inequality, we get
mN − n < 0 and
mN − m < n or mN − n < m,
which implies that 0 < mN − n < m. Hence our numerator is less than m and positive, which means that
−n
our new number mN
is smaller than the original numnN
m
ber n and this is what we want. Now, if mN − n is
equal to one, then we do not proceed further; otherwise
−n
we repeat the same process for the number mN
until
nN
we get one as the numerator. The algorithm is called
greedy because we are taking off the maximum possible
fraction at every step, which is a greedy thing to do.
One can observe that for fractions of the form n2 , the
greedy algorithm uses at most two terms and for the
fraction n3 , this algorithm uses at most three terms. We
now generalize the above method. Suppose m
< 1 is
n
written in its lowest terms; then there is an Egyptian
fraction with at most m terms whose sum is m
. Denote
n
m
1
r
=
− n ,
s
n
[ m]
which gives m
= [ n1 ] + rs , where [·] denotes the greatest
n
m
integer function. If r = 1, we do not proceed further;
otherwise, repeat the process. Further, repeated appli1
1
cations of the equality n1 = (n+1)
+ n(n+1)
guarantees
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The algorithm is
called greedy
because we are
taking off the
maximum possible
fraction at every
step, which is a
greedy thing to do.
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CLASSROOM
Egyptian fractions
also give the best
way to divide, say,
m pieces of bread
among n workers.
that each unit fraction has infinitely many representations. Consider, for example, the number 17
. Then,
18
r
17
1
4
18
r
= 18 − 2 = 9 , as [ 17 ] = 2. Because s is not equal to
s
one, we repeat the process again. Let rs11 = 49 − 13 = 19 ,
hence 17
= 12 + 13 + 19 , which is exactly what we dis18
cussed above in the case of 17 sheep. Notice that in this
case there can be infinitely many ways of distributing 17
sheep to 3, 4, 5 or a maximum of 17 people. Moreover,
1
1
one can further write 19 = 10
+ 90
, which gives another
1
1
representation. Also becasue 2 = 13 + 16 and 13 = 14 + 12
1
which gives another representation 17
= 13 + 14 + 16 + 19 + 12
.
18
But this representation is not useful in the case of distributing 17 sheep, as 18 is not divisible by 12; it is
only applicable if 17 + 1 is divisible by all denominators. Hence, in this case, only numbers allowed in the
denominator are 2, 3, 6, 9 or 18.
Egyptian fractions are useful to check whether a given
number is larger than another given number. Suppose
we want to know whether 34 is larger than 45 or not.
Representing both numbers by Egyptian fractions gives
3
1
= 12 + 14 and 45 = 12 + 14 + 20
, which clearly tells us that
4
4
3
is larger than 4 .
5
Egyptian fractions also give the best way to divide, say,
m pieces of bread among n workers. For example, let
us say we have 5 pieces of bread and want to equally
distribute these among 8 workers. It is not difficult to
see that each one should get 58 part. Now if we represent
this number in terms of Egyptian fractions, we get 58 =
1
+ 18 . It clearly shows that it is better to divide the
2
first 4 pieces each into 2 portions and the last piece into
8 portions, and then give each one of them a 12 and 18
portion (Figure 1).
Figure 1. Distribution of 5
pieces of bread among 8
workers.
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CLASSROOM
The sum of the unit fractions that adds upto one is of
special interest. It is an open problem to find
kthe 1smallest possible value of nk , say p(k), such that i=1 ni = 1,
where n1 < n2 < n3 < · · · . For some values one can
check that this is possible, for example, p(3) = 6, p(4) =
12, but, in general, it is not known.
One very famous
conjecture called
Erdös–Straus
conjecture on
Egyptian fractions
states that for all
integers n ≥ 2, the
rational number 4/n
can be expressed as
the sum of three unit
fractions, that is,
We conclude the article by stating one very famous conjecture called ‘Erdös-Straus conjecture’ on Egyptian fractions given by Paul Erdös and Ernst G Straus in 1948.
It states that for all integers n ≥ 2, the rational number
4
can be expressed as the sum of three unit fractions,
n
that is,
4
1
1
1
+
+ .
=
n
n1 n2 n3
For example if n = 5, then we can write the following
Egyptian fraction
4 1 1 1 .
= + +
n n1 n2 n3
1 1
1
1 1
1
4
= + +
= + + .
5
2 4 20
2 5 10
In this case there are two solutions. The main difficulty here is to prove the conjecture for positive triplets
(n1 , n2, n3 ). The conjecture has been verified up to n ≤
1014 [2] using computers, but it is still an open problem
for all n. The above equation falls into the category of
Diophantine equations. In general, a Diophantine equation is a polynomial equation that allows only integer
solutions. The question regarding the existence of an
algorithm which tells whether an arbitrary Diophantine
equation has a solution or not is the famous ‘Hilbert’s
10th problem [3].
Suggested Reading
[1]
Wikipedia,en.wikipedi.
orgwiki/Greedy-algorithm-forEgyptian-fractions
[2]
Allan Swett, The Erdös–
Straus Conjecture, math.
uindy.edu/swett/esc.htm
[3]
M Davis, Hilbert's Tenth
Prolem is Unsolvable,
Amer. Math. Monthly,
Vol. 80, pp.233–269, 1973.
Acknowledgement
The author is thankful to the anonymous reviewer for
his/her constructive comments and suggestions, which
helped to improve the article considerably.
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