# The annoying old 0.999 vs 1, hopefully with a twist

1. Jun 22, 2006

### miscellanea

This question is not new by any definition of the word. However, it's been bugging me for quite a while now and I've done some reading up on the issue.

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!

2. Jun 22, 2006

### StatusX

But 0.999... is not a hyperreal number, it is a real number.

3. Jun 22, 2006

### miscellanea

Hyperreal numbers are a proper extensions of the reals. You get reals for free and if you add non-standard analysis, you also get infinitesimals and infinitely large numbers.

4. Jun 22, 2006

### StatusX

Let me rephrase: 0.999... is a real number (and yes, technically also a hyperreal number - I was thinking of "hyperreal" like "irrational", ie, those that are hyperreal but not real, but you're correct), so it cannot differ from 1 by an infinitessimal, since the difference of any two real numbers is another real number, and there are no real infinitessimals besides 0.

Last edited: Jun 22, 2006
5. Jun 22, 2006

### AlphaNumeric

Hyperreals have been mentioned before on these forums in relation to this matter of often heated contention.

Though I don't remember it's name, there is a theorem which states any result true in normal analysis is true for non-standard analysis. If 0.9r = 1 in standard analysis then 0.9r* = 1* in non-standard analysis where 0.9r* and 1* are the elements in hyper reals. While 1 = 1*, 0.9r* is different in construction to 0.9r. Do a search to get a better explaination from someone who actually knows this stuff.

6. Jun 22, 2006

### miscellanea

Yes, I've heard of this theorem as well, though I don't know its name. That theorem is one of the reasons I posted here in the first place -- this difference shouldn't actually exist, but everything I've seen suggests that it does.

Which is really strange.

7. Jun 22, 2006

### HallsofIvy

No, there is no difference between 0.9999... and 1. 0.9999... is a regular real number, 1. It has nothing to do with "hyper-reals". The "definition" y0u are talking about is simply the definition of the base 10 number system.

8. Jun 22, 2006

### Hurkyl

Staff Emeritus
In the reals, we have:

$$\sum_{i = 1}^{+\infty} 9 \cdot 10^{-i} = 1$$

and in the hyperreals, we also have

$${}^{\star}\sum_{i = {}^{\star}1}^{+\infty} {}^\star 9 \cdot (^\star 10)^{-i} = {}^\star 1$$

(Bleh, I'm gonna stop putting stars on things)

One fact that may be being overlooked as that these two sums are summing over a different set of terms. The first one is the (real) sum of $9 \cdot 10^{-i}$ for every positive natural number i. The latter is the (hyperreal) sum of $9 \cdot 10^{-i}$ for every hypernatural number i.

The interesting fact that demonstrates a reason why might want to study nonstandard analysis is that we can consider the hyperfinite hyperreal sum:

$$\sum_{i = 1}^H 9 \cdot 10^{-i}$$

where H is some transfinite hypernatural number. (But still, $H < +\infty$, by definition of $+\infty$). This sum is the one that differs from 1 by a positive infinitessimal.

Another angle is to look at the hyperdecimal numbers. If we have a transfinite, but hyperfinite number of 9's (that is, the number of 9's is a transfinite hypernatural number) in 0.999.....9, then this does, in fact, denote a number infinitessimally close, but unequal, to 1. However, this is a terminating hyperdecimal. The nonterminating hyperdecimal 0.999... is, in fact, equal to 1.

Last edited: Jun 22, 2006
9. Jul 7, 2006

### Hurkyl

Staff Emeritus
I've moved the posts not dealing with nonstandard analysis into a new thread.

10. Jul 7, 2006

### HallsofIvy

Why? I would have thought it should be the responses that DO deal with non-standard analysis that should be moved. The original post had nothing to do with non-standard analysis!