Re: Question

In article <1146156962.005648.209040@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
zuhair <zaljohar@xxxxxxxxx> wrote:

[.snip.]

Alternatively, one can define a "distance" of rational numbers by the
usual means: the distance between a/b and c/d is |ad-bc|/bd, where
|ad-bc| is the absolute value of ad-bc.

A sequence of rationals is function from the natural numbers to the
rationals.

We say a sequence (a_0, a_1,...) (usually denoted {a_i}) converges to
the rational number Q if and only for every N>0 there exists M>0 such
that if n>M, then |a_n-Q|<1/N.

We say a sequence (a_0, a_1, ...) is a "Cauchy sequence" if and only
if for every N>0 there exists M>0 such that if n,m>M, then
|a_n-a_m|<1/N.

It is easy to verify that if a sequence converges to some rational,
then it is Cauchy, though the converse does not hold.

We can define an equivalence relation among sequences by saying that
the sequence {a_i} and the sequence {b_i} are "equivalent" if and only
if the sequence {a_i-b_i} is a Cauchy sequence.

Note the correction: this should read "the sequence {a_i-b_i}
converges to 0".

It is this latter definition that gives rise to the numerical
representation. A decimal expansion

N.d1d2d3....

with N an integer, di an integer between 0 and 9, is short hand for
the sequence

(N, N+d1/10, N+(d1/10)+(d2/100), ..., N + (d1/10) + (d2/100) + ... + (dn/10^n),...)

which can easily be verified is a Cauchy sequence; so the decimal
expansion represents the EQUIVALENCE CLASS of cauchy sequences
corresponding to this sequence. It is again a trivial exercise to
show, for example, that the Cauchy sequence represented by
1.000000... (which is the constant sequence (1,1,1,1,...)) and the
Cauchy sequence represented by 0.9999.... (which is the sequence
(9/10, 99/100, 999/1000, ....)) are equivalent Cauchy
sequence. Therefore, a fortiori, they represent the same "real
number".

Tell me what is that trivial exercise.

Can you be bothered to try anything that contradicts your
preconceptions or challenges your ignorance, or must everything be
done for you?

The first sequence is {a_i} with a_i = 1 for all i. The second
sequence is {b_j} with b_j = (10^j-1)/10^j for each j.

By definition, {a_i} is equivalent ot {b_j} if and only if
{a_i - b_i} converges to 0. First, let c_i = a_i - b_i. Then

c_i = 1 - [(10^j-1)/10^j] = [10^j - 10^j + 1]/10^j = 1/10^j.

Does the sequence {1/10^j} converge to 0? According to the DEFINITION,
the sequence {c_i} converges to 0 if and only if for every N>0 there
exists M>0 such that, for all n>M, |c_n - 0| < 1/N.

So, let N>0. Then there exists M such that 10^M > N. Therefore, for
all n>M, we have

|c_n-0| = |c_n| = c_n = 1/10^n < 1/10^M < 1/N.

Thus, for every N>0 there exists M>0 such that for all n>M,
|c_n-0|<1/N. This proves, BY DEFINITIION, that the sequence
{c_i}={a_i-b_i} converges to 0. BY DEFINITION, this means that the
sequence {a_i} and the sequence {b_i} are equivalent. This means, BY
DEFINITION, that the real number corresponding to the equivalence
class of the sequence {a_i} (which was the real number represented by
the decimal expansion 1.0000....) and the real number corresponding to
the equivalence class of the sequence {b_i} (which was the real number
represented by the decimal expansion 0.9999....) are the same real
number, since the two equivalence classes are the same.

--
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"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
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Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx

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