DISCOVERING UNIT FRACTIONS
Lukáš Lednický
Department of Mathematics, Faculty of Natural Sciences, Constantine the Philosopher University in Nitra,
Trieda A. Hlinku 1, 949 74 Nitra, Slovakia
Corresponding author: [email protected]
Abstract
We introduce in this contribution an inquiry approach to solve the problem of writing a fraction as a sum of a
certain number of different unit fractions.We use some knowledge from the history of mathematics and from
physics.We use the parallel connection of resistors in the electric circuit to introduce the problem in example
1.We focus only on the solving of a mathematical problem in the further examples.
Key words: inquiry-based learning, unit fraction.
1 Introduction
The main goal of inquiry-based learning is to bring the work of scientists to students
through investigation and problem solving. Students gain not only new knowledge, but also
they develop problem-solving skills. We can show the students by using various problems,
which mathematicians had to solve, that mathematics is not only algorithmic counting of
exercises.
2 Material and Methods
Unit fractions are fractions, which numerator is equal to 1. We will consider in this
contribution only those unit fractions, which denominator is an integer greater than 1. The
history of unit fractions dates back to ancient Egypt. Egyptians used them to represent other
fractions. They wrote every fraction, that is not a unit fraction, as a sum of different unit
fractions.They created tables, which contained decomposition of some fractions to a sum of
different unit fractions. The Rhind Papyrus (1600 – 1800 BC) involved such a table. It
contains decomposition of fractions of the form 2 for odd value of n from 3 to 101. There
n
was only one decomposition for a certain value of n in the table and it was not always the
shortest one. Often there appearsa decomposition according to a pattern, that occurs in
example 3. The denominators of the unit fractions were often even numbers, but always
smaller then 1000. However there are some mysteries about the table. For example, why the
table ends at the number 101, or why they used
2
1
1
1



95 60 380 570 ,
instead of the much simpler
2
1
1
.


95 60 228
3 Results and discussion
We want to show a solution of the problem to represent a fraction as a sum of distinct unit
fractions through an inquiry approach. We will use a situation from middle school physics,
which is familiar to students. The given solutions are not the only one. It is possible to find
different solutions or procedures of the solution.
Example 1: We want to connect two resistors whit different values of resistance to a
parallel electric circuit. How should we choose the resistances of the resistors, if the total
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resistance has to be 10 Ώ? Find a pattern for these resistances for an arbitrary value of the
total resistanceR. The value of total resistance and the values of resistance of both resistors
have to be an integer.
Solution: We know from physics, that the total resistance of parallel connection of two
resistors is given by the formula
1 1 1
  .
R R1 R2
After substituting R  10 , we obtain an equation with two unknown quantities R1 and R2 ,
1 1 1
  .
10 R1 R2
We can proceed now in two ways.
1. We will subtract the fraction
1
from the both sides of the equationand so we obtain an
R1
equation
1 1 1
  .
10 R1 R2
The remainder on the left side has to be positive, therefore is
substitute for
R1
R1  10 .
Step by step we shall
integers greater than 10. Those integers, for which the remainder
1 1
 is
10 R1
an unit fraction, are solutions. So we discover following solutions:
R1  11 , R2  110 ,
R1  12 , R2  60 ,
R1  14 , R2  35 ,
R1  15 , R2  30 .
For R1  20 is R2  20 . This solution fails to satisfy the condition, that values of R1 and R2
have to be different. We do not need to test integers greater then 20, because we would
discover the same solutions.
2. We shall express one unknown quantity from the equation by equivalent changes. For
example we shall express the unknown quantity R1 :
R1 
10 R2
.
R2  10
Now we get two conditions, because R1 has to be an integer: R2  10 and 10R2 has to be
divisible by R2  10 . We find the following solutions by substituting integers from 11 to 20
for R2 .
R1  110 , R2  11 ,
R1  60 , R2  12 ,
R1  35 , R2  14 ,
R1  30 , R2  15 .
These solutions differ from those in procedure 1 only in order.
In the search for the solution for the arbitrary R, we shall proceed analog as in the solution
2. We shall start from the equation
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1 1 1
  ,
R R1 R2
and we shall express the unknown quantity R1 :
R1 
RR2
.
R2  R
We get two conditions: R2  R and RR2 has to be divisible by R2  R  . From the second
condition we see the solution R2  R  1 and R1  RR  1 , because R2  R  R  1  R  1 and
1 divides RR2 . So we found the following pattern
1
1
1


.
R R  1 RR  1
Example 2: Find a decomposition of a fraction 1 , n  N , to a sum of three, four, five, …
n
different unit fractions.
Solution: An analogic procedure from example 1 is hard to use in this example. We shall
choose a different procedure. If n  1 , then we search for three numbers x, y, z  N , for which
is true
1 1 1
1   .
x y z
We can proceed as follows. We will substitute 1 by the greatest possible unit fraction, 1
x
2
. Then modify the equation to the form
1 1 1
  .
2 y z
We can solve this equation using the procedure from example 1. Also we can proceed by
substituting
1
by the greatest possible unit fraction, 1 . The equation can be then modified to
y
3
the form
1 1
 .
6 z
A unit fraction occurs on the left side of the last equation, therefor we can write
1
1 1 1
  .
2 3 6
If there is not a unit fraction on the left side of the equation, then it is necessary to choose
for
1
or possibly for 1 , different unit fractions. This procedure is called the greedy
y
x
algorithm.
If n  1 , we can use the greedy algorithm or the already found decomposition for number
1. When we divide the equality
1
1 1 1
 
2 3 6
by number n, we obtain the equality
1
1
1
1
.



n 2n 3n 6n
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We have discovered a decomposition of the fraction 1 , n  N , to a sum of three different
n
unit fractions.
We can use the greedy algorithm to find the decomposition to a sum of four, five and more
distinct unit fractions, but sometimes we cannot choose the largest unit fraction. It is possible,
that the decomposition would not contain the given number of unit fractions. We shall show it
on an example. We want to find the decomposition of the fraction 25 to a sum of three
28
different unit fractions. We are searching for a solution of the equation
25 1 1 1
   .
28 x y z
The fraction 25 is greater than 1 , therefor we choose a substitution 1  1 . Now let’s
28
2
x
2
count,
1 1 25 1 11
    .
y z 28 2 28
1 1
 , because 11  1 , and we obtain,
Now we choose the substitution
y 3
28 3
1 11 1 5
.

 
z 28 3 84
1
We can see, that 1 is not a unit fraction. We need to choose a different unit fraction for .
y
z
We can try the next possible unit fraction 1 and we obtain,
4
1 11 1
4
1

 
 .
z 28 4 28 7
The algorithm ends, because 1 is a unit fraction and we can write
z
25 1 1 1
   .
28 2 4 7
We can find out the correct solution by using already found results. We know that
1
1
1
1
and



n 2n 3n 6n
1
1
1


.
n n  1 nn  1
If we apply the second equality on the last unit fraction in the first one, we obtain
1 1 1
1
1
  

.
n 2n 3n 6n  1 6n6n  1
Specifically if n  3 , we obtain the following decomposition
1 1 1 1
1
.
  

3 6 9 19 342
We have discovered decomposition to a sum of four different unit fractions. We can find a
decomposition of the fraction 1 to a sum of any number of different unit fractions by
n
applying both equalities.
Example 3: Decompose the fraction 2 , n  3 , to sum of two different unit fractions.
n
Solution: We will divide the example into two cases.
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If n is even, the fraction can be reduced by number 2. The whole task simplifies, because
we are looking for a decomposition of the fraction 1 , k  2 . We have solved this case already
k
in example 1.
If n is odd, we try to find the decomposition of some fractions. By using the greedy
algorithm we find out, that
2 1 1
  ,
3 2 6
2 1 1
  ,
5 3 15
2 1 1
 
,
7 4 28
2 1 1
 
, etc.
9 5 45
We can notice, that the denominator of the first fraction in the decompostition is a half of
number n increased by 1. If we mark the denominator as x, then
x
n 1
.
2
The number x is an integer, because the number n  1 is always even. We can notice also,
that the denominator of the second fraction of the decomposition is always a product of
numbers n and x. If we mark it as y, then
y  nx 
nn  1
.
2
We will check the created assumptions. Count
1 1
1
1
n 1
2n  1 2
 



 .




n

1
n
n

1
n
n

1
x y
nn  1 n
2
2
2
Our assumptions were correct. If the odd number n is replaced by 2k  1, k  N , then we
can write the decomposition in the form
2
1
1


.
2k  1 k  1 k  12k  1
The decomposition of fractions to a sum of different unit fractions is not used often.
However there exist situations, in which it is useful to use it. Consider a situation, that we
have to divide 5 breads between 8 people. Every person should receive 5 of a bread. We
8
could cut all the breads to eighths and every person would get 5 eighths. If we express the
fraction 5 as a sum of unit fractions, we see, that 5  4  1  1  1 . From this we see, that
8
8
8
8
2
8
every person receives a half and an eighth of a bread. It is sufficient to divide four breads to
halves and one to eighths.
Unit fractions are useful sometimes for comparing of fractions. Compare the fractions 5
6
6
5 1 1
6 1 1
1
and . We express both fractions as sum of unit fractions,   and   
. It
7
6 2 3
7 2 3 42
is obvious from these decompositions, that 5  6 .
6 7
The physical application from example 1 is not the only opportunity to introduce the
problem of unit fractions. May, f be the focal length of a lens, xbe the distance of an object
from the lens and y be the distance of the image from the lens. Then the focal length and the
distances are related by the formula
1 1 1
  .
f x y
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Another opportunity is the common work task. If n workerswork at the same time on one
task and the i-th worker would finish the task in time Ti , then they can finish the task in time
T, which can be calculated from the equation
1 1 1
1
   .
T T1 T2
Tn
4 Conclusion
We present some inquiry examples, that deal with the problem of unit fractions, in this
contribution. The solutions are based on the knowledge about fractions. We aim also on the
use of discovered results in mathematics or in common practice. According to our opinion
implementing such examples to mathematical lessons, we can increase the interest of students
in mathematics and better develop their problem-solving skills.
This task could be used as an extension of fractions in mathematical lessons. There are
implemented some inquiry-based learning processes in the solutions, for example simplifying
and structuring, controlling variables, hypothesizing.
5 References
Kolman, A. 1969.Dejiny matematiky vestarověku. Praha: Academia, 1969. 224 s.
Znám, Š. 1977. Teória čísel. Bratislava: Alfa, 1977. 205 s.
http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fractions/egyptian.html#ef
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