Re: .999... = 1

On Mon, 29 Aug 2005 23:48:43 +0100, Robert Low <mtx014@xxxxxxxxxxxxxx>
wrote:

>David C. Ullrich wrote:
>> On Mon, 29 Aug 2005 16:02:59 +0100, Robert Low <mtx014@xxxxxxxxxxxxxx>
>>>That's pretty much it: you can think of a real number as
>>>precisely being an equivalence class of Cauchy sequences
>>>of rationals.
>>>>However, to me, .99999... represents a sequence that tends to 1,
>>>>rather than 1.
>>>OK, so it's a sequence of rationals that tends to 1.
>> No, 0.999... is most certainly not a sequence of rationals.
>> It is the limit of a certain sequence of rationals.
>
>Eventually it is.

"Eventually"? It is or it isn't. (It is.)

>But surely 0.999... means 'the real number
>defined by the Cauchy sequence 0.9, 0.99, 0.999 ...', since we
>have to make sense of the infinite sum. And reals are defined
>as equivalence classes of Cauchy sequences of rationals, so
>we have a Cauchy sequence 0.9, 0.99, 0.999 etc which lies
>in the same equivalence class as the Cauchy sequence 1,1,1 etc,
>so they represent the same real.

By definition this means that you have to prove that

(*) 0.9, 1, 0.99, 1, 0.999, 1, ...

is a Cauchy sequence. Proving that that is a Cauchy sequence
is _exactly_ equivalent to proving that the sequence 0.9, 0.99, etc
converges to 1. The details are the same, except that of course
the notation will if anything be more complicated.

Hence introducing equivalence classes of Cauchy sequences
does not simplify anything here. (That's an actual qed -
it's easy to show that one can convert _any_ proof that
(*) is a Cauchy sequence into a proof that the sequence
0.9, etc converges to 1.)

>Unless, of course, you think of the reals as the unique
>complete ordered field...


************************

David C. Ullrich
.

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