representation of real numbers

representation of real numbers

0.1

Introduction

It is well-known that there are several methods to introduce the

real numbers

. We shall follow an inductive method which is instructive as well as

elementary

. Apart from that such treatment is modern, interesting and is obtained through two

theorems

and a lemma, which are relatively easy to understand. So that our aim will be to prove the above

propositions

. Our starting point is the following theorem.

Theorem 1

.

Let

{

a

i

}

be a

sequence

of

positive integers

such that

a

i

>

1

, for all

i

≥

1

. Then any real number

ρ

is uniquely expressible by

ρ

=

b

0

+

∑

i

=

1

∞

b

i

∏

j

=

1

i

a

j

,

(1)

where the

b

i

are

integers

satisfying the

inequalities

0

≤

b

i

≤

a

i

-

1

for all

i

≥

1

, and

b

i

<

a

i

-

1

for infinitely many

i

.

Proof.

Let

{

b

i

}

0

∞

be a sequence of integers and

{

ρ

i

}

1

∞

a sequence of real numbers defined by the equations

b

0

=

[

ρ

]

,

ρ

1

=

ρ

-

b

0

,

(2)

and for all

i

≥

1

b

i

=

[

a

i

⁢

ρ

i

]

,

ρ

i

+

1

=

a

i

⁢

ρ

i

-

b

i

,

(3)

denoting

[

⋅

]

the integral part

function

. Clearly

ρ

i

+

1

is the

fractional part

of

a

i

⁢

ρ

i

, therefore for all

i

≥

1

we have,

0

≤

ρ

i

<

1

.

(4)

Next we multiply (4) by

a

i

whence

0

≤

a

i

⁢

ρ

i

<

a

i

, but

b

i

=

[

a

i

⁢

ρ

i

]

, so that

0

≤

b

i

≤

a

i

-

1

,

an inequality required by the theorem.

Now, by (3) and (4), and applying

induction

on

ρ

i

, we can establish that

ρ

=

b

0

+

ρ

1

=

b

0

+

b

1

a

1

+

ρ

2

a

1

=

b

0

+

b

1

a

1

+

b

2

a

1

⁢

a

2

+

ρ

3

a

1

⁢

a

2

=

⋯

=

b

0

+

∑

i

=

1

n

b

i

∏

j

=

1

i

a

j

+

ρ

n

+

1

∏

j

=

1

n

a

j

.

(5)

Now we define

d

n

=

b

0

+

∑

i

=

1

n

b

i

∏

j

=

1

i

a

j

,

thus from (4), (5) and by the

hypothesis

a

i

≥

2

, we arrive to

0

≤

ρ

-

d

n

=

ρ

n

+

1

∏

j

=

1

n

a

j

<

1

2

n

,

because the fractional part

ρ

n

+

1

<

1

. Then if we let

n

grows beyond of any bound,

ρ

-

d

n

will be so close to zero as we want and such a observation implies the representation (1).

We still need to prove the another inequality of the theorem, i.e.

b

i

<

a

i

-

1

for infinitely many

i

, but also the uniqueness of representation (1). To do that we need to make use of the

identity

∑

i

=

1

∞

a

n

+

i

-

1

∏

j

=

1

i

a

n

+

j

=

1

.

(6)

(It is legitimate to consider this identity as a lemma, as we need it to prove this theorem as well as the next one).

We shall prove later this identity. Let us prove the inequality

b

i

<

a

i

-

1

by

tertio excluso

; thus suppose that there is a fixed integer

n

such that

b

i

=

a

i

-

1

for all

i

>

n

. From (1) and (6) we get

ρ

=

b

0

+

∑

i

=

1

n

b

i

∏

j

=

1

i

a

j

+

∑

i

=

n

+

1

∞

a

i

-

1

∏

j

=

1

i

a

j

=

b

0

+

∑

i

=

1

n

b

i

∏

j

=

1

i

a

j

+

1

∏

j

=

1

n

a

j

⁢

∑

i

=

1

∞

a

n

+

i

-

1

∏

j

=

1

i

a

n

+

j

=

b

0

+

∑

i

=

1

n

b

i

∏

j

=

1

i

a

j

+

1

∏

j

=

1

n

a

j

,

and comparing this with (5) one realizes that

ρ

n

+

1

=

1

, contradicting (4).

Finally we must prove the uniqueness of the representation (1). So that we suppose

ρ

=

c

0

+

∑

i

=

1

∞

c

i

∏

j

=

1

i

a

j

.

Here the integers

c

i

satisfy

the same conditions as do the

b

i

. It is necessary (and

sufficient

!) to show that

c

i

=

b

i

for every

i

. The condition

c

i

<

a

i

-

1

, for infinitely many

i

, altogether with the identity (6) imply that

∑

i

=

1

∞

c

i

∏

j

=

1

i

a

j

<

1

,

so we see that

c

0

is the integral part of

ρ

, i.e.

c

0

=

[

ρ

]

, but also from (2)

b

0

=

[

ρ

]

, therefore

c

0

=

b

0

. Next we shall again use

tertio excluso

. On this way let us suppose that for some

n

≥

1

the pair

b

n

and

c

n

are unequal. There is no loss of generality in assuming that

n

is the smallest integer with this property (which is justified by a simple inductive

argument

), and that

b

n

>

c

n

, so that

b

n

-

c

n

≥

1

. Thus we have

∑

i

=

n

∞

b

i

∏

j

=

1

i

a

j

=

∑

i

=

n

∞

c

i

∏

j

=

1

i

a

j

.

(There is no

contradiction

at all in this equality, as it is easily seen in the next below equation).

It is obvious that these series are

absolutely convergent

, so we may rearrange terms to obtain

∑

i

=

n

+

1

∞

c

i

-

b

i

∏

j

=

1

i

a

j

=

b

n

-

c

n

∏

j

=

1

n

a

j

≥

1

∏

j

=

1

n

a

j

.

But then recalling that

c

i

<

a

i

-

1

, we see tat

c

i

-

b

i

<

a

i

-

1

. From this fact and (6), we can write

∑

i

=

n

+

1

∞

c

i

-

b

i

∏

j

=

1

i

a

j

<

∑

i

=

n

+

1

∞

a

i

-

1

∏

j

=

1

i

a

j

=

1

∏

j

=

1

n

a

j

⁢

∑

i

=

1

∞

a

n

+

i

-

1

∏

j

=

1

i

a

n

+

j

=

1

∏

j

=

1

n

a

j

.

This is a contradiction with respect to the inequality found before, thus the proof of this theorem is

complete

.
∎

0.2

Some implications

First we remarked that Theorem 1 is a

generalization

of the standard

decimal expansion

for a real number

ρ

. This may be seen by taking all the integers

a

i

=

10

. Thus, if

ρ

>

0

, (1) gives the decimal representation

ρ

=

b

0

+

∑

i

=

1

∞

b

i

10

i

=

b

0

.

b

1

⁢

b

2

⁢

⋯

.

(7)

Second, if

ρ

<

0

, we must write its decimal representation on the form (7)and then changing all signs.
Third, any real

ρ

could have an ambiguous decimal representation, e.g.

ρ

=

1.5699

⁢

⋯

, having an

infinite

successive sequence of

9

’s, which also involves a

geometric series

in

10

-

i

in turns implying (at the limit) that also

ρ

=

1.57

. For that reason, (7) represents that number with an infinite succession of

0

,s, that is,

ρ

=

1.57

=

1.5700

⁢

⋯

. The reason for this resides in that an infinite succession of

9

’s is ruled out by the condition of the theorem that

b

i

<

a

i

-

1

for infinitely many

i

, a condition that in the present example takes the form

b

i

<

9

for infinitely many

i

.

Now we prove (6).

Lemma 1

.

For the

integers sequence

{

a

i

}

, where

a

i

>

1

for every

i

≥

1

, we have

∑

i

=

1

∞

a

n

+

i

-

1

∏

j

=

1

i

a

n

+

j

=

1

.

Proof.

Let us take the

partial sum

S

m

=

∑

i

=

1

m

a

n

+

i

-

1

∏

j

=

1

i

a

n

+

j

=

∑

i

=

1

m

a

n

+

i

∏

j

=

1

i

a

n

+

j

-

∑

i

=

1

m

1

∏

j

=

1

i

a

n

+

j

=

a

n

+

1

a

n

+

1

+

∑

i

=

2

m

a

n

+

i

∏

j

=

1

i

a

n

+

j

-

∑

i

=

1

m

1

∏

j

=

1

i

a

n

+

j

=

1

+

∑

i

=

2

m

a

n

+

i

∏

j

=

1

i

a

n

+

j

-

∑

i

=

2

m

+

1

1

∏

j

=

1

i

a

n

+

j

-

1

=

1

+

∑

i

=

2

m

a

n

+

i

∏

j

=

1

i

a

n

+

j

-

∑

i

=

2

m

+

1

a

n

+

i

a

n

+

i

⁢

∏

j

=

1

i

a

n

+

j

-

1

=

1

+

∑

i

=

2

m

a

n

+

i

∏

j

=

1

i

a

n

+

j

-

∑

i

=

2

m

a

n

+

i

∏

j

=

1

i

a

n

+

j

-

n

+

m

+

1

(

n

+

m

+

1

)

⁢

∏

j

=

1

m

+

1

a

n

+

j

-

1

=

1

+

∑

i

=

2

m

a

n

+

i

∏

j

=

1

i

a

n

+

j

-

∑

i

=

2

m

a

n

+

i

∏

j

=

1

i

a

n

+

j

-

1

∏

j

=

1

m

a

n

+

j

,

since when in the

∏

operator

a

n

+

j

-

1

↦

a

n

+

j

, then the index value of

j

, at its

upper limit

,

m

+

1

↦

m

, but its

lower limit

does not. Thus, central sums cancel and the last term

vanishes

because, by hypothesis, we have

lim

m

→

∞

⁡

S

m

=

∑

i

=

1

∞

a

n

+

i

-

1

∏

j

=

1

i

a

n

+

j

=

1

-

lim

m

→

∞

⁡

1

∏

j

=

1

m

a

n

+

j

,

and the lemma is proved.
∎

Theorem 1 also represents an

irrational

number whenever we add a couple of additional conditions. Thus we have the following important theorem.

Theorem 2

.

Let us consider the same integers sequence

{

a

i

}

described in the preceding theorem, and that the integers

b

i

satisfying the inequalities of that result. In

addition

, let us assume that infinite integers

b

i

are

positive

, and that each

prime number

divides infinitely many

a

i

. Then

ρ

is irrational.

Proof.

We contradict the thesis by supposing

ρ

=

p

/

q

is

rational

(

p

,

q

,

coprime

). By the last hypothesis in the preceding theorem, we can choose an integer

n

sufficiently large in order to

q

be a

divisor

of

∏

j

=

1

n

a

j

. Now we may use (1) replacing the LHS by our

rational number

assumption

, next multiplying both side by the latter

product

, and rearranging terms we get (we do the

partition

sum

i

=

1

,

…

,

n

;

n

+

1

,

…

⁢

∞

)

∏

j

=

1

n

a

j

⁢

(

p

-

b

0

⁢

q

)

q

-

∑

i

=

1

n

b

0

⁢

∏

j

=

1

n

a

j

∏

j

=

1

i

a

j

=

∑

i

=

1

∞

b

n

+

i

∏

j

=

1

i

a

n

+

j

.

(8)

By hypothesis, the LHS of (8) is obviously an integer. However, we have proved already that the last inequality of theorem 1 requires that

b

n

+

i

<

a

n

+

i

-

1

for infinitely many

i

, so that, from the RHS of (8) and the lemma we see that

∑

i

=

1

∞

b

n

+

i

∏

j

=

1

i

a

n

+

j

<

∑

i

=

1

∞

a

n

+

i

-

1

∏

j

=

1

i

a

n

+

j

=

1

,

a clear contradiction, proving the theorem.
∎

0.3

Example

e

is

irrational

.

All we know there are different ways to prove the irrationality of

e

. In particular, it results illustrative if we adapt our theorems and its hypotheses (all of which are true in this case) to this problem. From Taylor’s expansion

e

=

∑

i

=

0

∞

1

i

!

.

Let us use (1), by setting

ρ

=

e

,

b

0

=

2

,

b

i

-

1

=

1

for all

i

≥

2

, and

a

j

=

j

+

1

,

0

≤

j

≤

i

-

1

. Thus,

e

=

2

+

∑

i

=

1

∞

1

∏

j

=

1

i

j

.

So far our discussion on real numbers. An interesting approach on

transcendental numbers

as well as and extensive bibliography on real numbers are given, for

instance

, in

[

1

]

.

References

1

I. Niven,

IRRATIONAL NUMBERS

, Ch. VII, pp. 83-88; also p. 157, The

Mathematical Association of America

, 2005.

Title

representation of real numbers

Canonical name

RepresentationOfRealNumbers

Date of creation

2013-03-22 19:10:36

Last modified on

2013-03-22 19:10:36

Owner

perucho (2192)

Last modified by

perucho (2192)

Numerical id

10

Author

perucho (2192)

Entry type

Topic

Classification

msc 11A63

Related topic

UniquenessOfDigitalRepresentation

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