Re: Question


Arturo Magidin wrote:
In article <1146163021.981085.118390@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
zuhair <zaljohar@xxxxxxxxx> wrote:

Arturo Magidin wrote:

Look at the DEFINITION:

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.

still there is a problem, when you say converges to zero,

No, there is no problem when I say that. I am going EXACTLY by the
definition.

You may, perhaps, object to the definition, but then you are objecting
to all of mathematics.

at what
number of terms in that series this will happen,

This question is nonsense. I did NOT say that the sequence is
"eventually equal to zero". That is something else. I said the
sequence CONVERGES to zero. That is something else, and was EXPLICITLY
and CLEARLY defined for you above.

is it at Omega of c_i
or Omega+1 or 2^Omega.

None of them. There is no omega here. Sequences are indexed by NATURAL
numbers. Natural numbers are, BY DEFINITION, finite. Omega is NOT a
natural number. "The sequence converges to zero" is NOT equivalent to
"the sequence is eventually equal to zero".


Your Cauchy priniciple didn't mention that.

Yes, it did. It gave ALL the definitions. You just seem unable to
parse them.

A sequence of rationals was defined to be a function from the NATURALS
to the rationals. We write it as (a_0, a_1, a_2, ..., a_n, ...), where
"a_n" is the value of the function at the natural number n. The only
terms that are defined are the terms indexed by NATURAL numbers, which
are FINITE. There is no "omega", there is no "omega + 1", there is no
"2^omega". There are only finite numbers.

The definition of "cauchy sequence" and "convergence" relates ONLY to
sequences; thus, it relates ONLY to finite indices. The definition of
convergence and of cauchy sequence is given entirely in terms of
FINITE indices and conditions on them.

Period.

It doesn't differentiate
between
0.9999........
which contains Omega of nines in it after the decimal point.


and 0.9999....... 9 which contains Omega+1 of nines in it after the
decimal point.

That expression does NOT represent a sequence, so it is not under
consideration. It is merely a symbolic string of symbols which does
not represent a real number.

In reality I tend to think that it is at what Cantor called once as
"The Absolute Infinity" number of terms in those series that c_i will
be zero

In reality I tend to think that you are not bothering to
think. Certainly, you seem either unwilling or incapable of reading
simple mathematical definitions.

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx

I know that converges into zero is something else other than equal
zero.

The definition states that if the difference converges into zero then
the two Cauchy sequences are equivalent, I disagree with that.

They should be equivalent only when the difference REACHES zero.

and it should specify at which number of terms that would happen.

It is obvious that for any finite number of terms n , c_n is 1/10^n >
0

But if the number of terms in the sequence is infinite then there might
be a possibility of c_i equalling zero, but that would be imaginable at
absolute infinity only.

I agree with you in that you are working according to the definition.
But the definition itself is not convincing.

So I am questioning the definition itself.

Now I want you to answer that question please.

Define: 0.9999......... as 9/10 + 9/10^2 + 9/10^3 + ............. +
9/10^Omega

( 0.9999....... is defined to contain Omega of nines after the decimal
point )

would that number be eqivalent to 1.0000........

My point is that you now that for example the number 0.9999.......nth9
were n is finite , is never eqivalent to 1 , it is always lower than
one by 1/10^n.

Now for the number above which has Omega of nines in it after its
decimal point, would that number
be equivalent to 1. or you think that the difference 1/10^Omega is
larger than zero.

That was my question, can you answer it.

And suppose there is another number 0.9999....... 2^Omega th 9 , ie a
number which contains
2^Omega of nines after the decimal point.

Is that number different from the first number with Omega repeations of
9. or is the same

Is it eqivalent to number 1.0000........ or is lower than it.

Best,

Zuhair

.

Relevant Pages

  • Re: Question
    ... I did NOT say that the sequence is ... "eventually equal to zero". ... A sequence of rationals was defined to be a function from the NATURALS ...
    (sci.math)
  • Re: Question
    ... I did NOT say that the sequence is ... "eventually equal to zero". ... A sequence of rationals was defined to be a function from the NATURALS ...
    (sci.math)
  • Re: Question
    ... this mean that the biggest number of terms in a sequence ... Omega is an ordinal. ... Any how if these are valid both right and left division operators ... cannot be zero because ...
    (sci.math)
  • The Origin of a Sequence
    ... THE ORIGIN OF A SEQUENCE ... implications of the first part for the understanding of natural number as ... In mathematics, so far as I can tell, a sequence is always ... But let us go back to our idea of the Naturals ...
    (sci.math)
  • Re: Question
    ... this mean that the biggest number of terms in a sequence ... Omega is an ordinal. ... And it does not matter how many times you repeat it, ... you repeat it, it is still nonsense. ...
    (sci.math)