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The question: Is 0.999... (infinite number of 9s) equal to 1?

The answer is, of course, yes. There are numerous proofs showing this. Here are a couple of simpler proofs I remember seeing:

1/3 = 0.333... and 2/3 = 0.666... and 0.333... + 0.666... = 0.999... and 1/3 + 2/3 = 3/3 = 1

Then there's:

10x = 9.999...

x = 0.999...

10x - x = 9x

9.999... - 0.999... = 9

9x = 9, so 0.999... is the same as 1

And ultimately there's the proof which says that 0.999... can be expressed as the sum of a geometric series and the sum of a geometric series is,

**by definition**, equal to its limit.

Which is where I have a bit of a problem. All other proofs seem to be referring to this one bit:

**by definition**0.999... as the sum of a geometric series is equal to its limit, 1.

I did some reading up and, remembering some advanced maths courses I took at the uni, looked to hyperreal numbers and non-standard analysis.

The difference seems to be that in non-standard analysis, the sum of a geometric series that would give us 0.999... is not equal to its limit. The sum would actually be different from the limit by an

*infinitesimal*number.

In non-standard analysis, by definition, 0.999... is not equal to 1. Which gives us funky things like infinitesimals and their reciprocals, infinitely large numbers.

Some mathematicians seem to not like non-standard analysis all that much. The main criticism seems to be that it allows for things that are not Lebesgue measurable. Then again, Zermelo-Fraenkel set theory with the Axiom of Choice added allows for non-measurable sets, giving rise to things like the Banach-Tarski paradox, yet these results seem to be accepted for, well, practical purposes.

Basically, my theory is that whether 0.999... is equal to 1 or not depends on if you allow for non-standard analysis or not. If you allow for infinitesimals, then 0.999... is not equal to 1, by definition. If you use real analysis, then 0.999... is equal to 1.

And, basically, both answers seem to be equally valid.

What do you think? Discuss!