Re: .99999... still=/= 1

From: Tapio (hurmecom_at_dlc.fi)
Date: 11/15/04 

Date: Mon, 15 Nov 2004 20:15:19 GMT

"Josh Purinton" <usenet-noreply.a.jp@xoxy.net> wrote in message
news:8sXld.338823$wV.181078@attbi_s54...
> In article <9Oald.579$Ol.210@read3.inet.fi>, Tapio <hurmecom@dlc.fi>
> wrote:
>>There are two important point to consider:
>>1) The string between the start and the end is always infinite, let's mark
>>it [].
>>2) The point of reference means how we define the counting results
>>compared
>>the point of reference.
>
> Let's call your infinitely-long strings of decimal digits "supernatural
> numbers" (a term due to George Greene, I believe.)
>
> A fundamental property of the integers is that if k is an integer, then
> so is k+1.
>
> If your supernatural numbers are integers, then 999...999 is an integer.
> But then what is 999...999 + 1? Such a number doesn't seem to exist.

First of all 999...999 exists. This number is defined as a sum ( 0 --> oo)
9*10^n. (Do not worry about limit, I will explain it few rows down below)
Second: You can always add 1. (Euclides). Thus You and Euclides are right.
:-)

I told earlier that definition: omega (w) is the number that is greater than
any integer, i.e. 999...999 + 1. This definition, which You can find also in
some textbooks, defines indirectly that there is an upper bound or limit for
the string length and the members in the placeholders of the string.

If You seemingly prefer "supernatural numbers" instead of so called finite
integers, then we can define: omega (w) is the number that is greater than
any supernatural integer.
Note that supernatural numbers are that I have called earlier infinite
integers. The critical point is to observe - as I consider the strings - if
there are infinite many reals in the set [0,1] with the start - first member
0 - and the end - the last member 1, then as a consequence it is at least
possible and trivially evident that there can be infinite strings with the
start and the end. This means in practise that "supernatural integers" are
finite, with the start and with the end, although they are infinite. Paradox
? No!
Thus there is no difference between finite or infinite. (This may sound at
the first glance paradoxical) .Consider as on example ...01 with infinite
string with meaningless zeros. This integer is familiar finite integer 1,
but infinite string.

We can mark the omega with the symbol 1 that is not the same as integer 1 as
omega 1 is behind our standard infinity in the next infinity.
The important point is that now we have defined the end of infinite string.
Note now that you asked exactly just like I asked couple of years ago in
this newsgroup, see "Infinite integers".

My answer was: the limit of 999...999 equals to omega (w). The infinite
string itself, namely 999...999 < omega 1, because as you correctly asked
999...999 + 1. So you have to add 1 because you can always add 1 according
to Euclides, but all the placeholders in the infinite string were occupied
with the maximal member or digit of base system, namely with digit 9 (in
this case the base is - of course - 10).
As a consequence you have "carry overflow" into the next infinity or into
the next infinite string, which cannot be on or within the standard finite
or infinite integer area, because it's on the next overflow area, i.e. in
the omega area.
But - note this! - your point of reference is still on the right hand side
(=rhs) of the infinite string. You are talking about the standard point of
reference. :-)

If we change the point of the reference into the left hand side (=lhs) of
the string, i.e. into the omega point of view. Then you should immediately
see that 999...999 (the point of reference is on rhs) is exactly 0.999...
(the point of reference is on lhs). Now you can find the familiar limit of
0.999... =1, where 1 equals to omega 1. But, as we are taking about
supernatural integers you easy know and observe that you have to add
standard integer 1 into 0.999... (the point of reference is on lhs, NOT on
rhs), because it is actually 999...999, before you can say that the sum
equals to omega 1. Exacly in the way you asked correctly!

Now let's go back to the title of the thread: Therefore you have to add an
infinitesimal or epsilon into 0.999... to reach equality to the standard
integer 1. Of course, we can agree that the limit of 0.999... equals to 1
(=standard integer or ...01). But how do I explain the epsilon environment
so that I have to add epsilon into 0.999 so that the result equals to 1?.
There are many ways. :-)
I have referred earlier delta - epsilon theory: "for every delta there
exists epsilon infinitely" inspite of limit! Therefore you have to add
epsilon. But perhabs this delta-epsilon theory does not satisfy your
requirements?

But I have simpler example. Just shift the point of reference into rhs in
the case odf standard decimal real 0.999... Then the string behaves like
your 999..999 and you will see all the reals between [0,1) like standard
integers (actually they are not integers, but they behave now like standard
integers, because we have shifted the point of reference.They are still
decimal numbers). You have already answered yourself by asking :

> "But then what is 999...999 + 1? Such a number doesn't seem to exist".

Of course it exists. It's the standard integer 1 that what you and
mathematicians are claiming whole the time!!! I do accept that limit. The
limit of 0.999... equals to 1, but the string itself is - of course
0.999...<1, because you had to add 1, i.e the smallest or least element.

Thus the limit is a different thing than the sum of the string, which you
have proofed yourself by asking my explanation. (I explained that couple of
years ago in this newsgroups)

But as a lemma
1) it is now proofed that the reals are well-ordered, because there exists
the least element, namely ...01 as and only as the point of reference is on
rhs. :-)
2) there exists bijection between supernatural integers and decimal reals
between [0,1).
3) Cantors proof is still valid as he did not consider standard integers as
supernatural integers, that can be finite and infinite- too.

Any questions?

with best regards

Tapio.

> Therefore your "supernatural numbers" are not integers.

> --
> Josh Purinton

Relevant Pages

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