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40 At Right Angles | Vol. 3, No. 1, March 2014 in the classroom

I

n the article on ‘Recurring Decimals’ published in the November

2013 issue of At Right Angles, many questions remained

unanswered in the end. They had emerged as empirical

observations during the course of the exploration. We study these

observations closely here, examine their validity and explain them

using simple principles of divisibility.

(A) Is it true that all fractions give rise to either terminating

decimals or to recurring decimals of some periodicity?

Yes! Let the fraction be m/n where m, n are positive integers with

no common factors. Let the decimal form of m/n be computed.

If the decimal terminates, well and good; so we examine what

happens in the case of non-termination. When the digits of m (the

dividend) ‘run out’, what do we do? — we simply tack on zeros to

the units ‘end’ of the number and continue the division. At each

stage we get some non-zero remainder which is one of the num- bers 1,2,3, . . ., n−1. So, after at most n divisions, we necessarily get

a remainder that we had already got earlier.

Exploration of

Recurring Decimals:

Some Explanations

Shailesh Shirali

Padmapriya Shirali

Keywords: Recurring decimal, terminating decimal, repetend, prime factorization

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At Right Angles | Vol. 3, No. 1, March 2014 41

From this point onwards, the string of digits we

obtain from the division will be identical to what

had been obtained earlier; in other words, the

decimal will recur from this point onwards.

An example will make this clear. Consider the

fraction 10/13. The long division working (of

10 ÷ 13) yields the display shown in Table 1.

Observe that in the last row, the remainder is the

same as in the �irst row. �t follows that the row

following this one will be identical to the second

row, the row after that will be identical to the

third row, and so on. Note the sequence of

quotients: 0, 7, 6, 9, 2, 3, 0, .... Therefore,

10/13 = 0.076923 076923 ... = 0.076923.

(B) Is it true that the decimal expansion of 1111/1111

terminates precisely when 1111 1 nnnnaaaa × 5555bbbb where aaaa

and bbbb are non-negative integers? (The values of

nnnn for which the decimal expansion of 1/nnnn

terminates were found to be the following: 2, 4, 5,

8, 10, 16, 20, 25, 32, 40, 50, 64, 80, 100, .... All

these numbers have the stated form.)

Yes, again! For, suppose that the decimal

expansion of 1/nnnn terminates; say

1/nnnn = 0.nnnn�nnnn� ... nnnn� where nnnn�, nnnn�, ... , nnnn� are decimal

digits. Let AAAA be the kkkk-digit number nnnn�nnnn� ... nnnn�.

Then, clearly,

1

nnnn = AAAA

10� , ∴ AAAA = 10�

nnnn . (1)

Relation (eq:1) tells us that nnnn is a divisor of 10�.

Since the prime factors of 10 are 2 and 5, we

deduce that nnnn cannot have any prime factors other

than 2 and 5. Hence nnnn = 2� × 5� for some

non-negative integers nnnn and bbbb.

A numerical instance will make this clear. Take

nnnn = 32; we get: 1/32 = 0.03125, so AAAA = 3125 and

kkkk = 5. Relation (eq:1) reduces to:

1

32 = 3125

100000, ∴ 3125 =

100000

32 .

The converse statement too is true: if nnnn = 2� × 5�

for some non-negative integers nnnn and bbbb, then the

decimal expansion of 1/nnnn terminates. For, if nnnn aa bbbb

then we have:

1

2� × 5� = 5���

2� × 5� = 5���

10� .

This yields a terminating decimal with nnnn digits

after the decimal point. Example: Consider

nnnn = 2� × 5� = 80:

1

80 = 1

2� × 5� = 5�

2� × 5� = 125

10000 = 0.0125.

Similarly, if bbbb aa aaaa we have:

1

2� × 5� = 2���

2� × 5� = 2���

10� .

Now we get a terminating decimal with bbbb digits

after the decimal point. Example: Consider

nnnn = 2� × 5� = 250:

1

250 = 1

2� × 5� = 2�

2� × 5� = 4

1000 = 0.004.

(C) Is it true that the values of 1111 for which the

repetend of 1111/1111 is a single-digit number are

given by the formulas 1111 1 nnnn × nnnnaaaa × 5555bbbb and

1111 1 nnnnnnnn × nnnnaaaa × 5555bbbb? (The values of nnnn we found for

which the repetend of 1/nnnn is a single-digit number

were: 3, 6, 9, 12, 15, 18, 24, 30, 36, 45, 48, 60, 72,

75, 90, 96, .... All these �it the given formula.) We

shall show that the answer again is Yes. What

does it mean for the repetend to have just one

digit? Suppose that the repetend is the single digit

dddd. Then the decimal expansion of 1/nnnn must be of

the form

0.nnnn�nnnn�nnnn� ... nnnn�dddddddddddddddd ...

where nnnn�, nnnn�, nnnn�, ... , nnnn� are digits. Let AAAA be the

kkkk-digit number nnnn�nnnn�nnnn� ... nnnn�. From the relation

1

nnnn

= 0.nnnn�nnnn�nnnn� ... nnnn�dddddddddddddddd ... ,

we get the following two relations, by

multiplication by 10� and then by 10 (these

multiplications shift the decimal point, �irst by kkkk

steps and then by 1 more step):

���

� � ������ ... ��.���� ... � �.���� ... , (2)

TABLE 1. Steps in the division 10÷13

Quotient Remainder Updated

dividend

0 10 100

7 9 90

6 12 120

9 3 30

2 4 40

3 1 10

0 10 100

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42 At Right Angles | Vol. 3, No. 1, March 2014

�����

� � ������ ... ���.���� ... � (��� � �AAAAAAAAA

(3)

Here the term 10AAAA AA AAAA is the number aaaa�aaaa�aaaa� ... aaaa�AAAA

which equals aaaa�aaaa�aaaa� ... aaaa�0 AA AAAA, i.e.,

10 × aaaa�aaaa�aaaa� ... aaaa� AA AAAA dd 10AAAA AA AAAA.

If we do “(eq:3) minus (eq:2)”, the portion after the

decimal point (.AAAAAAAAAAAAAAAA A) gets wiped out by the

subtraction, and we get:

10��� − 10�

nnnn dd 10AAAA AA AAAA − AAAA dd AAAAAA AA AAAA. (4)

From this relation we deduce that nnnn is a divisor of

the number 10��� − 10� dd 10� × AA. Since the prime

divisors of 10� × AA are 2, 5, 3, it follows that nnnn is the

product of a divisor of a power of 10 and a divisor of

AA (which can only be 3 or AA; if the divisor were 1,

then the decimal would terminate). Hence

nnnn dd ddd� × 5� or nnnn dd d� × 2� × 5�. This

conclusion matches the observed �inding.

Example: Working with actual numbers will make

the argument clear. Take nnnn dd dnn. Then we have

1/nnnn dd dddddddd, so AAAA dd 5 (this is the repetend).

Also, kkkk dd d (there is one digit before the repeating

portion). So we multiply both sides �irst by 10� dd 10

and then by 10� dd 100. We now get:

10

1nn dd 0.55555 ... ,

100

1nn dd 5.55555 ... .

Subtracting, we get:

100

1nn − 10

1nn dd 5, i.e., AA0

1nn dd 5.

So 1nn is a divisor of AA0 dd 10 × AA. Note that

1nn nn 2 × AA. This will illustrate what we mean when

we say that “nnnn is the product of a divisor of a power

of 10 and a divisor of AA”. (Here the divisor of 10 is 2,

and the divisor of AA is AA itself.)

(D) For which values of nnnn does the decimal

expansion of 1111/nnnn have a two-digit repetend?

To answer this we argue exactly the same way as

we did earlier. We see that the decimal expansion

must be of the form

0.aaaa�aaaa�aaaa� ... aaaa�AAAA�AAAA�AAAA�AAAA� ... (5)

where aaaa�, aaaa�, aaaa�, ... , aaaa�, AAAA�, AAAA� are digits, and the

repetend is the two-digit number DDDD dd AAAA�AAAA�. Let AAAA

be the kkkk-digit number aaaa�aaaa�aaaa� ... aaaa�. From the

relation

1

nnnn dd 0.aaaa�aaaa�aaaa� ... aaaa�AAAA�AAAA�AAAA�AAAA� ... ,

we get the following two relations, by

multiplication by 10� and then by 10�:

10�

nnnn dd AAAA.AAAA�AAAA�AAAA�AAAA� ... , (6)

10���

nnnn dd (100AAAA AA DDDDAA.AAAA�AAAA�AAAA�AAAA� ... . (7)

Subtraction, (eq:7) minus (eq:6), now yields:

10��� − 10�

nnnn dd 100AAAA AA DDDD D AAAA dd AAAAAAAA AA DDDDD (8)

From this we deduce that nnnn is a divisor of the

number 10��� − 10� dd 10� × AAAA. Hence nnnn is the

product of a divisor of a power of 10 and a divisor

of AAAA. The latter divisor (of AAAA) cannot be a

divisor of AA, because in that case we would have a

repetend consisting of only one digit. So the

divisor of AAAA must be one of the following: 11, 33,

AAAA. These numbers when multiplied by divisors of

powers of 10 yield the nnnn's we want. So nnnn must be

one of the following: 11, 22, 33, 44, 55, 66, nnnn, AAAA,

110, 132, .... This conclusion matches the

observed �inding exactly.

(E) For which values of nnnn does the decimal

expansion of 1111/nnnn have a three-digit repetend?

As the strategy by now will be familiar, we skip

the initial steps. The conclusion now is: the

decimal expansion of 1/nnnn will have a three-digit

repetend precisely when nnnn is a divisor of the

number 10� × AAAAAA (for some kkkk) but not a divisor

or either 10� × AAAA or 10� × AA. Since the prime

factorization of AAAAAA is AAAAAA AA 3� × 37, the divisor

of AAAAAA contained in nnnn must be one of the

following: 27, 37, 111, 333, AAAAAA. These numbers

when multiplied by divisors of powers of 10 yield

the nnnn's we want. So nnnn must be one of the

following: 27, 37, 54, 74, 10nn, 14nn, 135, 175, ....

Yet again, this conclusion matches the observed

�inding exactly.

(F) For which values of nnnn does the decimal

expansion of 1111/nnnn have a four-digit repetend?

While answering this, we must explain the

observed fact that there is no nnnn in the range

1 ≤ nnnn ≤ 100 for which the repetend is a four-digit

number.

Following the same steps, we see that the decimal

expansion of 1/nnnn will have a four-digit repetend