0.99999999.......

[A18]

Only if one assumes that 9.99999... refers to a real number, no?

[A18]

Question for the math and number geeks out there.

Could 1 - .9999(repeating) = *, be regarded as true?
* defined as an infinitesimal

As discussed in this thread no. Basically "infinitesimally small" does not exist, it equals zero.

There are no non-zero infinitesimal real numbers. There are non-zero infinitesimal hyperreal numbers, however.

[A18]

Question for the math and number geeks out there.

Could 1 - .9999(repeating) = *, be regarded as true?
* defined as an infinitesimal

I assume you mean positive infinitesimal. I'm not a mathematician, even of the amateur variety, but here's how I would approach your question:

On the real number line, zero is the only infinitesimal number. Thus, we cannot define 0.999... to be such that 1 - 0.999... is positive infinitesimal, for the simple reason that positive infinitesimal does not exist.

The neglected hyperreal number line does recognize non-zero infinitesimals, both positive and negative. But here, we would seem to encounter the opposite problem: "1 - 0.999... is positive infinitesimal" is ambiguous, because there are an infinite number of positive infinitesimals. Thus, 0.999... would refer not to a hyperreal number, but to a range of hyperreal numbers—all hyperreals smaller than 1 for which the standard part is 1.

Anyway, that's my layman's take. Accept it or reject it.

It can't be 1.999...8 because you can't have a digit after an infinite amount.

As intuitive as that rule is, it strikes me as ultimately an arbitrary one.

[A18]

So zero is not considered "infinitesimal" for purposes of the real number line? I stand corrected.

But how does your reasoning transfer over to the hyperreal number line? Granted, the series 9/10, 90/100, 900/1,000, 9,000/10,000, ... converges to 1, but isn't the entire question whether we should define the series in terms of the real number it converges to?

BRTD,

It seems to me that there can be a "last" 9 in 0.999..., so long as an infinite number of 9's precedes it.

[A18]

Looks like we only disagree on my reply to BRTD, then. As I see it, speaking of a "last" 9 does not imply an end to the sequence if it is preceded by an infinite succession of 9's. It is figurative rather than literal. (Of course, for consistency it would be necessary to write 0.99, with the bar over the first 9. Multiplication of that value by two could then be "intuitively" paired with 0.98, with the bar over the first nine.)