Re: how to list all of the real numbers

On Aug 23, 9:15 pm, Jonathan Hoyle <jonho...@xxxxxxx> wrote:
On Aug 23, 7:04 pm, "Ross A. Finlayson" <r...@xxxxxxxxxxxxxxx> wrote:
The continuum of real numbers, where any numbers between zero and one
are real numbers else they wouldn't be between zero and one, including
the nilpotent infinitesimal iota and iota-sums and iota-multiples,
such that infinitely many iota's in multiple equal exactly one, leads
to reformulation of the infinitesimal analysis as infinitesimal
analysis, where that is nonstandard and not necessarily Non-Standard.


<snip>

This run-on sentence is ill-defined and not even well thought out.
You need to break this into steps. By doing so, the errors will
become more apparent.

As soon as RF mentioned "nilpotent infinitesimal," I immediately
consider the Dual Numbers:

http://en.wikipedia.org/wiki/Dual_number

We consider the field of real numbers, R, and the corresponding
polynomial ring R[iota]. Now we mod out the ideal generated by
the polynomial iota^2. The resulting ring is the ring of dual
numbers, and it obviously has the property that iota^2 = 0, so
iota is nilpotent. This iota acts like an infinitesimal, since
one can prove that for any polynomial P, P evaluated at x+iota
has the value P(x) + iota P'(x).

Then, where there is the consideration that 1/x for
x = 0, 1/0 = oo, 1/oo = 0, yet that e^oo would be considered a
discontinuity, in the asymptotic analysis reveals that 2^x < 3^x < ...
and log x < x < e^x etcetera, preserving trichotomy.

Now it sounds as if RF is discussing the hyperreals, where
every positive hyperreal x, whether finite or infinite, has the
property that log x < x < 2^x < e^x < 3^x.

But unfortunately, this is incompatible with the dual numbers
mentioned earlier. The dual numbers, being nilpotent, have no
inverses, yet the hyperreal infinitesimals have infinite
hyperreals as inverses. The hyperreals, in fact, form a field
while the dual numbers do not.

.

Relevant Pages

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  • Re: infinity
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  • Re: how to list all of the real numbers
    ... As soon as RF mentioned "nilpotent infinitesimal," I immediately ... iota is nilpotent. ... every positive hyperreal x, whether finite or infinite, has the ... Also I think the hyperreals are the reals, hyperintegers integers, ...
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