Re: .999... = 1
- From: David C. Ullrich <ullrich@xxxxxxxxxxxxxxxx>
- Date: Tue, 30 Aug 2005 06:08:23 -0500
On 30 Aug 2005 00:13:03 -0700, "MoeBlee" <jazzmobe@xxxxxxxxxxx> wrote:
>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?
Right. Now exactly how do you prove that those two sequences
are equivalent? By definition you have to show that
(*) 0.9, 1, 0.99, 1, 0.999, 1, ...
is a Cauchy sequence. The details involved in showing that
that's a Cauchy sequence are exactly the same as the details
in just showing that the first sequence converges to 1.
>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.
Look at the definitions. Showing that (*) is a Cauchy
sequence is exactly the same thing as showing that
the sequence 0.9, 0.99, etc, converges to 1. Any
proof you give for the first fact _includes_ a
proof for the second fact.
>> 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?
Yes, that's one of various things I'm 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?
Yes. I never thought about this question before, realized a
minute after I sent my post that it was utterly trivial.
There is some subtlety in showing the reals are archimedean;
after all non-archimedean ordered fields do exist, and we
need to make sure that one didn't sneak in when we constructed
the reals. But showing that the rationals are archimedean
is literally trivial:
"Theorem" If e > 0 is rational then there exists a
positive integer n with 1/n < e.
Proof: Suppose e = p/q where p and q are positive integers.
Let n = 2q. QED.
(Now to finish the proof that 0.999... = 1, show by
induction that 10^n > n.)
>Thanks,
>
>MoeBlee
************************
David C. Ullrich
.
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