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From: Ben Bacarisse <ben.usenet@bsb.me.uk>
Newsgroups: sci.math
Subject: Re: .999 repeating is almost an infinitely large sequence
Date: Mon, 26 Jun 2023 00:42:14 +0100
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"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:

> On 6/25/2023 4:19 PM, Chris M. Thomasson wrote:
>> On 6/25/2023 4:14 PM, Ben Bacarisse wrote:
>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>
>>>> Check this out:
>>>>
>>>> i[n] = (floor(1/9 * 10^n) / 10^n) * 9 = .999....
>>>>
>>>> Right?
>>>
>>> No.  There is no n (in ℕ) that makes that equation hold.
>>>
>>>> Lets check:
>>>
>>> OK...
>>>
>>>> i[0] = (floor(1/9 * 10^0) / 10^0) * 9 = 0
>>>> i[1] = (floor(1/9 * 10^1) / 10^1) * 9 = .9
>>>> i[2] = (floor(1/9 * 10^2) / 10^2) * 9 = .99
>>>> i[3] = (floor(1/9 * 10^3) / 10^3) * 9 = .999
>>>> ...
>>>>
>>>> See?
>>>
>>> Do you?  None of i[n] = 0.999...
>>>
>> When n is taken to infinity, it equals .(9), right?
>
> n is without limit. Therefore the iterates go on forever and always get
> closer and closer to one. .999... = 1

Are you trying to say that lim_{n->oo} i[n] = 1?

> I am not exactly sure why:
>
> i[n] = (floor(1/9 * 10^n) / 10^n) * 9
>
> would not equal .(9) when n is taken to infinity.

lim_{n->oo} i[n] = 1.  Is that what you mean?

My objection was to this:

  i[n] = (floor(1/9 * 10^n) / 10^n) * 9 = .999....

It looks like an equation, but there is no n that satisfies it.

-- 
Ben.