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Decimal Expansion of
Rational Numbers
By
Arogya Singh
Prashant Rajbhandari
Seema KC
SUMMARY
This project is based on the decimal expansion of rational numbers. Rational number is
any number that can be expressed as the ratio of two integers p/q, where q  0. The
resulting decimal expansion of the rational number will either terminate or repeat. The
first problem in this project has pretty much dealing with the conditions of the decimal
expansion of p/q to terminate or repeat. Our assumption in this problem is to satisfy the
equation of the product of prime factors that is (Q = P1n1*P2n2………..…Pknk). The result is
that a fraction terminates in its decimal form, if the prime factors of the denominator are
only 2’s and 5’s or a product of prime factors, 2’s and 5’s. Otherwise it repeats. The
Problem 2 has represented the repeating decimal in the rational number form p/q. Our
result for this problem is based on our assumption of five rational numbers that has
different numbers of repeating numbers after the decimal. The pattern of the problem 3 is
also similar to problem 2 but here our emphasis was on the given numbers.
Equal contribution has been made on this project by our team members. The group
members that have been assigned for this project is Arogya Singh, Prashant Rajbhandari
and Seema KC. Arogya Singh has illustrated his idea on how a repeating decimal can be
represented in the rational number form P/Q. His main goal for this project was to
elaborate as many examples as he could. Prashant Rajbhandari has illustrated his thought
on “what conditions the decimal expansion of p/q will terminate or repeat?” Seema KC
has come up with the idea that a fraction terminates in its decimal form, if the prime
factors of the denominator are only 2’s and 5’s or a product of prime factors, 2’s and 5’s.
Otherwise it repeats. So this project is a combined effort of all the participants.
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Problem 1
The decimal from of a rational number p/q can be obtained simply by dividing the
denominator into the numerator. The result will be either a terminating decimal or a
repeating decimal.
Under what conditions will the decimal expansion of p/q terminate? Repeat?
a. The decimal expansion of a rational number can either terminates or repeats. A
rational number is any number that can be expressed as the ratio of two integers p/q,
where q  0. Therefore rational numbers include the integers. Some examples of
rational numbers are
2 1 1  50 22
,
, ,
,
, 0,
1 3 4
2
7
25 , and 1.2. Rational numbers that
can be expressed in a decimal form either terminates or repeats. For Example;
3
= 0.6 (Terminate)
5
27
= 1.35 (Terminate)
20
2
= 0. 6 (Repeat)
3
Note that the overbar indicates that 0. 6 = 0.66666666…..
b. A fraction (in simplest form/lowest terms) terminates in its decimal form, if the prime
factors of the denominator are only 2’s and 5’s or a product of prime factors, 2’s and
5’s. Otherwise it repeats. The product of prime factors can also be expressed as
(Q = P1n1*P2n2………..……………Pknk). In the examples listed below, factor the
denominators into the product of prime factors and see what we get!
2
Terminate
Repeat
1
, where 2 in the denominator is the prime
2
1
Here, 3 in the denominator is not the
3
factor of (2*1).
4
, where 5 in the denominator is the prime
5
factor of (5*1).
3
, where 10 in the denominator is the
10
prime factor of (2*5).
4
,where 25 in the denominator is the
25
prime factor of (5*5).
3
, where 16 in the denominator is the
16
prime factor of (2*2*2*2).
7
, where 20 in the denominator is the
20
prime factor of (2*2*5).
prime factor of 2’s & 5’s.
1
Where, 15 in the denominator is not
15
the prime factor of 2’s & 5’s.
7
Where, 30 in the denominator is not
30
the prime factor of 2’s & 5’s.
1
Where, 66 in the denominator is not
66
the prime factor of 2’s & 5’s.
7
Where, 21 in the denominator is not
21
the prime factor of 2’s & 5’s.
4
Where, 29 in the denominator is not
29
the prime factor of 2’s & 5s.
c. The decimal expansion of an irrational number neither terminates nor repeats. Since
every fraction has an equivalent decimal form, real numbers include rational
numbers. However, some real numbers cannot be expressed by fractions. These
numbers are called irrational numbers. Some examples of irrational numbers are:
2 = 1.414213562
 = 3.141592653
r = 0.10110111011110
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What is your conjecture?
Since p/q = p(1/q), it is sufficient to investigate the decimal expansions of 1/q. Our
conjecture or assumption is that the decimal expansion of 1/q for enough positive
integers can either be terminates or repeats. As it has already been describe above that
a fraction (in simplest form/lowest terms) terminates in its decimal form if the prime
factors of the denominator are only 2’s and 5’s or a product of primes factors, 2’s and
5’s. Otherwise it repeats. The product of prime factors can also be expressed in the
equation of;
Q = P1n1*P2n2………..……………Pknk, Therefore
1/2 = 0.5
Terminate
1/3 = 0. 3
Repeat
1/4 = 0.25
Terminate
1/5 = 0.20
Terminate
1/6 = 0.1 6
Repeat
1/7 = 0. 142857
Repeat
1/8 = 0.125
Terminate
1/9 = 0. 1
Repeat
1/10 = 0.1
Terminate
If we take 100 as a denominator it terminates because 100 is the product of prime
factors of 2*2*5*5. Similarly, if we take 66 it repeats because 66 is the product of
prime factors of 2*3*11. Therefore any number that is in the denominator and is the
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combination of 2’s and 5’s is terminated. The denominators of the first few unit
fractions having repeating decimals are 3, 6, 7, 9, 11, 12, 13, 14, 15, 17, 18, 19, 21,
22, 23, 24, 26, 27, 28, and 29.
Problem 2
In a decimal expansion of a rational number, it is easy to represent terminating
decimal in the form p/q. For example, 3.74 = 374/100 = 187/50 and 1.2516 =
12,516/10,000 = 3129/2500. Similarly, to represent a repeating decimal in the rational
number form p/q, we have taken few examples.
(a) Example 1
To represent the repeating decimal in the form of p/q, first we have taken 3. 7 . Note
that the overbar indicates that the 3. 7 is equal to 3.777777………
Set,
r = 3.777777………….
Then
10r = 37.77777……
And
10r – r = 34
So
9r = 34
Therefore r =
34
Ans.
9
(b) Example 2
In this example we have taken two different repeating numbers after the decimal.
Therefore we have chosen 5. 16 this is equal to 5.1616161616…….
Set,
r = 5.161616…………….
Then
100r = 516.1616…………….
And
100r – r = 511
5
So
99r = 511
Therefore r =
511
Ans.
99
(c) Example 3
In this example we have taken three different repeating numbers after the decimal.
We have selected 4. 312 this is equal to 4.312312312………………
Set,
r = 4.312312312………………….
Then
1000r = 4312.312312………………………
And
1000r – r = 4308
So
999r = 4308
Therefore r =
4308 1436
=
Ans.
999
333
(d) Example 4
In this example we have taken four different repeating numbers after the decimal.
Therefore, we have selected 4. 3761 and this is equal to 4.37613761………….
Set,
r = 4.37613761………………..
Then
10000r = 43761.37613761…………….
And
10000r – r = 43757
So
9999r = 43757
Therefore r =
43757
Ans.
9999
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(e) Example 5
In this example we have taken five different repeating numbers after the decimal.
Therefore, we have selected 3. 15213 and this is equal to 3.152131521315213………
Set,
r = 3.152131521315213…………….
Then
100000r = 315213.1521315213…………….
And
100000r – r = 315210
So
99999r = 315210
Therefore r =
315210
Ans.
99999
Problem 3
We have expressed each of the given repeating decimals in the rational number form
p/q.
(a) 13. 201
The overbar can also be expressed as 13.201201201……
Set
r = 13.201201201…………….
Then
1000r = 13201.201201………….
And
1000r – r = 13188
So
999r = 13188
Therefore r =
13188
Ans.
999
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(b) 0. 27
The above bar can also be expressed as 0.2727272727……..
Set
r = 0.272727………….
Then
100r = 27.2727………..
And
100r – r = 27
So
99r = 27
Therefore r =
27 3
=
Ans.
99 11
(c) 0. 23
The above bar can also be expressed as 0.2323232323……
Let
r = 0.23232323………….
Then
100r = 23.2323…………..
And
100r – r = 23
So
99r = 23
Therefore r =
23
Ans.
99
(d) 4.16 3
The above bar can also be expressed as 4.163333333…….
Let
r = 4.163333……………..
Then
1000r = 4163.33………
And
1000r – r = 4159.17
So
999r = 4159.17
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Again
99900r = 415917
415917
Ans.
99900
Therefore r =
Showing that the repeating decimal 0.999… = 0. 9 represents the number 1 and also
that 1=1.0, it follows that a rational number may have more than one decimal
representation. To follow this criterion, we have assumed few other rational numbers
that have more than one decimal representation. For example we have taken
2.99999… and 9.9999…..
a) 2.9999…..= 2. 9
Let
r = 2.9999…
Then
10r = 29.9999….
And
10r – r = 27
So
9r = 27
Therefore, r =
27
= 3 Ans.
9
b) 9.9999…… = 9. 9
Let
r = 9.9999….
Then
10r = 99.9999….
And
10r – r = 90
So
9r = 90
Therefore r =
90
= 10 Ans.
9
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Conclusion
The first new thing we gained from this project was from the solution of
problem1. “A fraction terminates in its decimal form, if the prime factors of the
denominators are only 2s and 5s or a product of the prime factors, 2s and 5s.” This
statement alone gave fractions a new look. From now on we can easily state if a
fraction terminates in its decimal form just by looking at it.
Apart from the above gain, we also learned that “a rational number may have
more than one decimal representation.” We got this from the last part of problem 3.
0. 9 = 10, 2. 9 = 3. All three of us, Seema KC, Prashant Rajbhandari, and Arogya
Singh, are business majors, so even 0.1 makes a lot of difference in our fields.
Therefore this very problem and solution amazed us in the field of Mathematics.
Signature,
Seema KC
Prashant Rajbhandari
Arogya Singh
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