Re: .99999....=1

On Oct 26, 11:58 pm, Proginoskes <CCHeck...@xxxxxxxxx> wrote:
On Oct 26, 3:34 pm, lwal...@xxxxxxxxx wrote:
0.999... = 1 - 10^N.
I think you mean 1 - (0.1)^N here.

Of course I did. 1 - 10^N makes no sense here. Naturally,
you are right here.

Since (0.1)^N is an infinitesimal,
and due to the Transfer Principle, this equation implies that

0.999... = 1.

I decided to check out some old threads, both here in this
newsgroup and in other nonmathematical boards. (In fact, I
once found a physics board where the majority of the posters
believed that 0.999... < 1, and it was a small minority of
mathematicians who argued that 0.999... = 1!)

As far as the hyperreals are concerned, there are some who
believe that a decimal such as 0.abcd... represents the
following hyperreal

{0.a, 0.ab, 0.abc, 0.abcd, ...}

in the sense that _every_ sequence of reals -- regardless of
whether the sequence even converges -- corresponds to a
unique hyperreal. In the case where the sequence does
converge, as in this case, the standard part of the hyperreal
is in fact that limit. (This is mentioned in the Keisler link
you provided above.) But in this example, the infinitesimal
part is never zero unless only finitely many of the digits
are nonzero. By this approach, 0.999... is 1 - (0.1)^N, just
as 0.333... is exactly one-third of this, (1 - (0.1)^N) / 3.

The other approach is to consider that 0.999.. already has
a meaning as a standard real, namely 1, since the Cauchy
sequence converges to unity. And since the hyperreals are
an extension of the reals, every real, including 0.999..., must
be a hyperreal and must equal the same real. So by this
approach, 0.999... = 1, and any number with a decimal
expansion must equal its own standard part. Hyperreals
such as 1 - (0.1)^N cannot be represented as a decimal.

The Wikipedia article to which I linked in another post also
mentions Lightstone's extended decimal representation of
the hyperreals:

"In his formalism, there are two natural extensions of 0.333...,
neither of which falls short of 1/3 by an infinitesimal:

0.333...;...000... does not exist, while
0.333...;...333... = 1/3 exactly."

As we see here, Lightstone is allowing the decimal places
to be indexed, not by the standard naturals (as with the
standard reals), but by the hypernaturals (the analogue of
the naturals in the hyperreals). We see that if every digit,
whether in a standard natural position or in a hypernatural
position, is three, then the number is 1/3 (and thus the
number 0.999...;...999... = 1 exactly).

One may wonder why 0.333...;...000... does not exist. This
reminds me of Internal Set Theory, the version of set theory
which allows for nonstandard hyperreals. In IST, N actually
_is_ the set of hyperreals. The set of standard naturals
can not exist in IST -- it is an example of what is known as
"illegal set formation." So with 0.333...;...000..., the set of
all hypernaturals whose digit is three would be the set of
all standard natural numbers -- which is an instance of
illegal set formation. So 0.333...;...000... does not exist.

Of course, if 1/3 = 0.333...;...333... in Lightstone's notation,
what about 1/11 = 0.090909...? Or even worse 1/7, which
would be 0.142857142857...? The answer is that it would
depend on the choice of ultrafilter. And of course, trying
to write sqrt(2) or pi in Lightstone's notation would be an
impossible task. (Such a notation would still _exist_, but
one would not be able to calculate it.)

Notice that 0.999...;...999... notation is reminiscent of some
of Tony Orlow's notation for his infinitesimals. (See the
Cantor threads for more info on T-riffic numbers.) This also
comes up in the Archimedes Plutonium threads, where it
is sometimes asked whether the digits of the AP-adic
numbers (a generalization of p-adics) are themselves
indexed by AP-adics. (Once again, please refer to the
relevant AP threads for more info.)

.

Relevant Pages

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