Schinazi R.B. From Calculus to Analysis

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Chapter 8

Decimal Representation of Numbers

So far we have freely used the decimal representation of numbers without even

mentioning it. However, the decimal representation of real numbers has deep and

interesting consequences. In particular, it shows that any real number can be repre-

sented by an inﬁnite series. Some of the proofs in this chapter are long and a little

technical. The reader should concentrate on the results and the examples.

We start by dealing with natural numbers.

Representation of naturals

For any natural number n ≥ 1, there exists an integer s ≥ 0 and integers

,...,a

in {0, 1, 2,...,9} with a

> 0 such that

n =a

10 +a

+···+a

s−1

...a

is the decimal representation of n. It is unique.

When we say that the number n is 121, we really mean that the decimal repre-

sentation of n is 121. That is,

n =1 +2 ×10 +1 ×10

The usual convention is to identify numbers with their decimal representation. How-

ever numbers may be represented in different bases. For instance, in base 2, n is

1111001. That is,

n =1 ×2

+1 ×2

+0 ×2

+0 ×2 +1.

We now prove that every natural number has an unique decimal representation.

We will need two steps: existence and uniqueness.

Existence Fix a natural number n.Ifn ≤ 9, then we set s = 0, a

= n, and we

have

n =a

This is the decimal representation of n.

R.B. Schinazi, From Calculus to Analysis,

DOI 10.1007/978-0-8176-8289-7_8, © Springer Science+Business Media, LLC 2012

221

222 8 Decimal Representation of Numbers

Assume now that n ≥10. Let

A =



k ∈N :10



Since 10

>nfor all naturals, A is a nonempty set of naturals (n belongs to A).

Therefore, it has a minimum element t such that 10

>n. Note that t>1 since we

are assuming that n ≥10. Observe that t −1 cannot be in A (otherwise t would not

the minimum). There are two ways for t −1 not to be in A: either it is not in N,or

t−1

≤n. Since t is in N and is at least 2, t −1isalsoinN. Thus, we must have

t−1

≤n. Therefore,

t−1

≤n<10

Set s = t − 1. We now need to perform Euclidean divisions. We ﬁrst recall the

properties.

Long division in the naturals

Given two natural numbers a and b, there exist positive integers q and r such

that

a =bq +r

where 0 ≤ r<b. Moreover, q and r are unique. If r = 0, a is said to be

divisible by b.

This was proved in Sect. 1.1.

We perform the Euclidean division of n by 10

. There are unique positive integers

and r

such that

n =a

where 0 ≤r

< 10

Note that

=10

s+1

>n≥a

and so a

< 10. Observe also that if a

=0, then n = r

< 10

, and this is a contra-

diction since n ≥ 10

. Thus, a

is in {1, 2,...,9}. We now perform the Euclidean

division of r

by 10

s−1

. There are positive integers a

s−1

and r

s−1

such that

s−1

where 0 ≤r

s−1

< 10

s−1

Since r

< 10

,wemusthave

s−1

≤r

< 10

Hence, a

s−1

< 10. Note that

n =a

s−1

We then divide r

s−1

by 10

s−2

, and so on. We denote the quotients of the successive

divisions by a

, a

s−1

,...,a

. The last division we perform is r

by 10. The last re-

mainder r

is therefore strictly less than 10, and we set a

. Using the successive

divisions, we get

n =a

s−1

+···+a

10 +a

8 Decimal Representation of Numbers 223

where each a

(for 1 ≤ i ≤ s)isin{0, 1, 2,...,9}, and a

> 0. That is, we have

proved the existence of a decimal representation for each natural n.

Uniqueness Assume that n has two decimal representations:

n =a

s−1

+···+a

10 +a

n =b

u−1

+···+b

10 +b

with a

> 0 and b

> 0. First note that

n =a

s−1

+···+a

10 +a

≥10

since a

≥0for0≤i ≤s −1 and a

≥1. Using that all a

≤9 and a geometric sum,

we get

n =a

s−1

+···+a

10 +a

≤9 ×10

+9 ×10

s−1

+···+9 ×10 +9



s+1

−1

10 −1



=10

s+1

−1 < 10

s+1

Thus,

≤n<10

s+1

The exact same arguments show that

≤n<10

u+1

This implies that u =s (why?). Therefore,

s−1

+···+a

10 +a

s−1

+···+b

10 +b

In particular,

−b

s−1

+···+b

10 −



s−1

+···+a



On the left-hand side we have a number whose absolute value is less than 9 (since a

and b

are both positive and less than 9). On the right-hand side we have a multiple

of 10. The equality may hold only if both sides are 0. Thus, a

.Wenowhave

s−1

+···+a

10 =b

s−1

+···+b

10.

Dividing both sides by 10 gives

s−1

s−2

+···+a

s−1

s−2

+···+b

We are now back to the initial situation with a

and b

playing the roles of a

and b

, respectively. Using the same argument as above, we get a

. Repeating

the procedure s + 1 times yields a

for 0 ≤i ≤ s. The decimal representation

is unique. That concludes our proof.

224 8 Decimal Representation of Numbers

Example 8.1 The proof of existence of a decimal representation is interesting in that

it provides a method to ﬁnd representations of a number. Moreover, the method may

be applied to ﬁnding a representation in any base. The representation of n in base r

(where r is any natural number) is given by

n =a

r +···+a

where a

for 0 ≤i ≤s is in {0, 1,...,r −1}, and a

≥1.

We illustrate this point by applying the method to ﬁnd a representation in base 3

for the number whose decimal representation is 121. First note that

≤121 < 3

That is, s =4, and the number of digits in base 3 is 5 (s +1 in the notation above).

We perform the division of 121 by 3

to get

121 =3

+40.

Thus, we get a

=1.Wenowdivide40by3

40 =3

+13.

That is, a

=1. Then

13 =3

+4,

and a

=1. Similarly, a

=1 and a

=1. That is, the base 3 representation of 121

is 11111.

We now turn to the decimal representation of real numbers. We start by deﬁning

Integral part of a real

For any real number x, there exists an integer denoted by [x] such that

[x]≤x<[x]+1.

In particular, [x] is the largest integer smaller than x. We call [x] the integral

part of x.

Assume ﬁrst that x ≥0. If x<1, then let [x]=0. We have

0 ≤x<0 +1

for x in [0, 1). Thus, for x in [0,1), [x]=0. We now turn to x ≥1. Deﬁne

B ={n ≥1 :n>x}.

By the Archimedean property, B is nonempty. Since it is a subset of natural num-

bers, it has a minimum n

.Fromx ≥1wehaven

> 1. This implies that n

−1 ≥1.

But n

−1 is not in B. Thus, n

−1 ≤x<n

. Set [x]=n

−1. With this deﬁnition,

[x] is an integer and is such that

[x]≤x<[x]+1.

8 Decimal Representation of Numbers 225

For x<0, we have −x>0, and the existence of an integral part for x is an easy

consequence of the existence of an integral part for −x. This is left as an exercise.

Note that for any real x,wehave

x =[x]+



x −[x]



The integral part may be represented using the decimal representation for naturals.

We now need to represent (x −[x]) which is in [0, 1). Putting together the decimal

representation for [x] and the decimal representation for x −[x], we get a decimal

representation for x.

Decimal representation of a real number

Let x be a real number in [0, 1). Let the sequences a

and d

be deﬁned by





for i ≥1,

−10a

i−1

for all i ≥2.

Then the d

are all in {0, 1,...,9}, and d

< 9 for inﬁnitely many i.The

sequence d

gives the decimal representation

x =

∞



i=1

Moreover, this decimal representation is unique.

Fix x in [0, 1). We will need ﬁve steps to show that x has a unique decimal

representation.

Step 1. We show that all d

are in {0, 1,...,9}.

First note that the a

and therefore the d

are integers. By the deﬁnition of the

integral part, we have

≤10

x<a

+1 for all i ≥1.

Our ﬁrst step is to show that all d

are in {0, 1,...,9}. First note that since x<1,

we have

0 ≤d

=[10x]≤10x<10.

Thus, d

is an integer in {0, 1,...,9}. We now deal with d

for i ≥2. We use

≤10

x<a

and

i−1

≤10

i−1

x<a

i−1

+1.

First note that

x −10 =10



i−1

x −1



< 10a

i−1

226 8 Decimal Representation of Numbers

Thus,

−10a

i−1

< 10

x −



x −10



=10.

Since d

is an integer strictly less than 10, it must be less than or equal to 9. We now

show that it is also positive. We have

−10a

i−1

> 10

x −1 −10



i−1



=−1.

Since d

is an integer strictly larger than −1, it must be larger than or equal to 0. We

have achieved our ﬁrst goal: all d

are in {0, 1,...,9}.

Step 2. We prove that

+10d

i−1

+···+10

i−1

for all i ≥1. (8.1)

By deﬁnition we have

and (8.1) holds for i =1. Assume that (8.1) holds for i. Then we have

i+1

−10a

i+1

−10



+10d

i−1

+···+10

i−1



by the induction hypothesis. Thus,

i+1

+10



+10d

i−1

+···+10

i−1



i+1

+10d

+···+10

and (8.1) holds for i +1. The formula is proved by induction.

Step 3. We prove that

x =

∞



i=1

Deﬁne the sequence s

=10

−i

for i ≥1.

Note that

=10

−i

≤10

−i

x =x

and that

=10

−i

> 10

−i



x −1



=x −10

−i

Thus,

x −10

−i

≤x for all i ≥1. (8.2)

By the squeezing principle the sequence s

converges to x:

lim

i→∞

=x.

Note that

=10

−i

=10

−i



+10d

i−1

+···+10

i−1





j=1

8 Decimal Representation of Numbers 227

Letting i go to inﬁnity, we get

x =

∞



j=1

That is, every real number in [0, 1) has a decimal expansion.

Step 4. We show that inﬁnitely many of the d

must be strictly less than 9.

By contradiction, assume that only ﬁnitely many d

are strictly less than 9. Then

there exists a natural n such that for all i ≥n,wehaved

=9. Therefore,

x −s

∞



j=n+1

∞



j=n+1

n+1

1 −1/10

On the other hand, by (8.2),

0 ≤x −s

< 10

−n

That is, x − s

and x − s

< 10

−n

. We reach a contradiction. Therefore,

inﬁnitely many of the d

must be strictly less than 9.

Step 5. The decimal expansion is unique.

We have just shown that for any x in [0, 1), there are integers d

in {0, 1 ...,9}

such that inﬁnitely many of the d

are not 9 and such that

x =

∞



i=1

Moreover, the sequence d

may be computed using





for i ≥1

and

−10a

i−1

for all i ≥2.

Assume now that there is another decimal representation for x:

x =

∞



i=1



where the d



are integers in {0, 1 ...,9} such that inﬁnitely many of the d



are not 9.

We are now going to prove that for each i ≥1, we have d



For any natural n,wehave

x =

∞



i=1

n−i





i=1

n−i



∞



i=n+1

n−i



Since there are inﬁnitely many d



that are not 9, there are inﬁnitely many i such that

i>nand d



< 9. Thus,

∞



i=n+1

n−i



∞



i=n+1

n−i

×9 =1.

228 8 Decimal Representation of Numbers

Therefore,



i=1

n−i



≤10



i=1

n−i



+1.

Note also that

k =



i=1

n−i



=10

n−1



+10

n−2



+···+d



is an integer. That is, we have

k ≤10

x<k+1,

where k is an integer. By the deﬁnition of the integral part of a real number, we have





=k =



i=1

n−i



. (8.3)

We now are ready to prove by induction that for all i ≥ 1, we have d



.Let

n =1in(8.3) to get

[10x]=d



Thus, d



. Assume now that for i ≤ j , d



.Letn =j +1in(8.3):



j+1



j+1



i=1

j+1−i





i=1

j+1−i

+10

j+1−j −1



j+1

=10



i=1

j−i



j+1

However, (8.1) states that

+10d

j−1

+···+10

j−1



i=1

j−i

for all j ≥1.

Therefore,



j+1



=10a



j+1

That is,



j+1

−10a

But d

j+1

is, by deﬁnition, a

j+1

−10a

. Hence,



j+1

This concludes the proof that decimal expansions are unique.

8 Decimal Representation of Numbers 229

Remark Putting together the decimal representation of the naturals and the decimal

representation of real numbers in [0,1), we see that all positive real numbers can be

represented by

10 +b

+···+b

∞



i=1

where s ≥0 is an integer, the b

and d

are integers in {0, 1, 2,...,9}, and inﬁnitely

many of the d

are not 9. It is a remarkable fact that ALL real numbers can be written

in this simple form.

Example 8.2 Find d

for i =1, 2,...,5forx =

√

We ﬁrst compute a

=[10x]. Note that

(10x)

=50.

Thus,

7 < 10x<8,

and a

= 7. This is also the ﬁrst decimal d

= 7. We now compute a

=[10

x].

Using (10

=5000, we get that

70 < 10

x<71.

So a

=70 and d

−10a

=0.

We have that (10

=500000 and

707 < 10

x<708.

So a

=707 and d

−10a

=707 −700 =7. And so on. We get

x =

√

=0.70710....

Application 8.1 The elementary way to obtain decimal expansions for rational

numbers uses repeated Euclidean divisions (Note that this method cannot be de-

ﬁned for irrationals!). For instance, by dividing 1 by 3 one gets 1/3 = 0.3333....

We are now going to check that this algorithm gives the same result as the general

method described above.

Assume that the rational number x in (0, 1) can be represented by a/b where

a<bare natural numbers. We start by performing the division of 10a by b to get

10a =q

b +r

where 0 ≤r

<b. We then divide 10r

by b to get

10r

b +r

where 0 ≤ r

<b. We deﬁne inductively the sequences q

and r

of positive or

zero integers by performing successive divisions. More precisely, let r

=a, and for

i ≥1, let

10r

i−1

b +r