Re: Question

In article <1146165177.094108.274750@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"zuhair" <zaljohar@xxxxxxxxx> wrote:

Arturo Magidin wrote:

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

One cannot "disagree" with a definition in mathematics. One either
accepts it or rejects it.

They should be equivalent only when the difference REACHES zero.

Using that as your equivalence relation, what you would get is only
another copy of the rational numbers, not the usual real numbers at all.


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.

Such objects are not any part of the real numbers at all, as 10^omega is
itself not a real number

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.

Since it requires the existence of a real number which cannot be a real
number, the answer is that no such thing exists as a real number.
.

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