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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 02:13:12 +0100
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"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:

> On 6/25/2023 4:42 PM, Ben Bacarisse wrote:
>> "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[n] just denotes the iterates of:
>
> i[n] = (floor(1/9 * 10^n) / 10^n) * 9
>
> the infinite iterates are .(9)

So you are not simply saying that lim_{n->oo} i[n] = 1?  OK.  I wonder
what all these other words mean if not the simple fact 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.
>
> i[n] means use n in the formula,

No.  It denotes a (functional) term in an equation (or other formula).
This, for example is correct:

  i[n] < 1

but this

  i[n] = 1

is not.  (Once defined, you don't have to keep repeating the complex
definition you've come up with for i[n].)

> so lets break it down for the first couple
> of iterates:
>
> 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
> ...

All correct.  i[n] = .(9) = 1 is not.  I am not sure what the problem is
other than you can only think in a sort of process/programming mode.

-- 
Ben.