Re: .999... = 1

David C. Ullrich wrote:
> On 28 Aug 2005 23:39:46 -0700, "MoeBlee" <jazzmobe@xxxxxxxxxxx> wrote:
> Talking about reals being equivalence classes of rationals
> has nothing to do with your question. That's one way to say
> exactly what the phrase "real number" means, but it doesn't
> have any effect on the proof that the limit of the sequence
> 0.9, 0.99, etc is 1.

Let x = 0, .9, .99, .999, ..., which is to represent a particular
Cauchy sequence of rationals.

Let y = 1, 1, 1, 1, ..., which is to represent a particular Cauchy
sequence of rationals.

If x is in the same equivalence class as y, then we've proven that
..999... = 1, right?

Perhaps there's a dead end trying to prove this way, but I don't see
how we could know that before we even try.

> In fact one doesn't even need to _talk_ about real
> numebers at all in order to answer your question.
> Suppose that it's 2000 years ago and the only numbers
> in existence are the rationals - instead of saying
> that sqrt(2) is irrational we say that 2 does not
> have a square root. In that context it is _still_
> true that 0.999... = 1, and the proof is exactly
> the same as before

I think I see. The identity is more basic than anything having to do
with real numbers if we can show the 9's sequence converges to 1,
without even mentioning real numbers. Is that what you're saying? But,
as you mention below, you did invoke the completeness property of reals
to finish the proof.

> (Except that the technical detail
> "if e > 0 then there exists n such that
> 10^(-n) < e" needs to be given a different
> justification - this morning I essentially
> invoked completeness of the reals to show
> that the reals, and hence the rationals, are
> archimedean. The rationals are still archimedean
> whether the reals exist or not.)

Then there's a proof that doesn't use the completeness property of the
reals?

Thanks,

MoeBlee

.

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