[electricpanda]

But I thought that was because it was so close to one, you could essentially just call it that because the difference would be indistinguishable (and also hard to fit on a calculator)?
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Dubbed Pants by AsA, via the way of the electric pants.

[Tamir]

No, it really is one. =/

Here's a basic proof that you might not like due to infinities:

x = 0.9999999999....
10x = 9.999999999.....
10x - x = 9
9x = 9
x = 1

But it's true, honest. I can find you other proofs if you want. (I know another one off the top of my head, but it's a bit strange for people who haven't heard that sort of proof before.)

[electricpanda]

No, you've frightened me enough for one day. I'll just give you the benefit of the doubt XD No point arguing what you're ill-equipped to argue with.
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Dubbed Pants by AsA, via the way of the electric pants.

[Nem]

See that looks just as wrong as the one the other person posted to me. You've moved everything up by a decimal point, in other words you've added a zero to the far end, but the sequence has no end to add that to. It's like trying to add something to itself.
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We stroll along the roof of hell
Gazing at flowers.
- Issa

[Tamir]

The other one was problematic because if you want the following to be true,

(1+1+1+1+.....) - (1+1+1+1+.....) = 0

then the number of ones in each infinity must be equal. This is ridiculous. If you're claiming that the number of ones is equal in each one, than there's a finite amount of them, and you can number them, and it's all good. You can't equate infinities like that.

In my proof, no infinities got equated. I assumed the nines go on infinitely, and therefore even if I move the decimal number a bit, I still have an infinitude past the decimal point.

[Nem]

[Tamir]

[theBSDude]

Well, .999... is 1 minus an infinitely small number; basically .000...0001 or 1 divided by infinity, which is zero.
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TheBSDude, dubbed 'BS' by Asa and nicknamed 'Thebes.'
Keeper of the Tristam, The Amazing Non-Emo Vampire!(tm) and a Massive List of Webcomics.

"I'm like a ninja with no hopes or dreams." --Wally of Dilbert

[Tyris]

Here's a proof that isn't even slightly mathematically rigorous, but sounds good:

The difference between 0.9 recurring and 1 is 0.(0 recurring)1. But because the tail of 0s is infinitely long, you never actually reach the 1... the 0s go on forever, and it's simply 0. And if the difference between two numbers is 0, then they must be equal.
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Keeper of Nanashi's ghostly blue visage and Felucca's hypothetical ejecting nose-cone.

"It's a sad thing when a man loses his cookie." - Nem

[theBSDude]

That's what I said.

My Calc3 teacher (the one who told me this) used this "proof," which, in my opinion, is really just a mind-screw:

1/9 = .111 repeating
2/9 = .222 repeating
3/9 = .333 repeating
...and so on...
8/9 = .888 repeating
so 9/9 should = .999 repeating
_________________
...In accordance with the prophesy.

TheBSDude, dubbed 'BS' by Asa and nicknamed 'Thebes.'
Keeper of the Tristam, The Amazing Non-Emo Vampire!(tm) and a Massive List of Webcomics.

"I'm like a ninja with no hopes or dreams." --Wally of Dilbert

[Tamir]

Sure, but for that you need to accept that 1/9 is 0.111....

*shrug*

Analysis style proof time! First, an axiom: there exists no positive number x such that for every natural number n, x < 1/n. Translation: there are no infinitesimal numbers.

Proof: For any natural number n, we wish to show that that (1 - 0.999...) < 1/n.
If that holds, then the difference between 1 and 0.999... must be zero, since no infinitesimal numbers exist, and therefore 0.999... = 1.
How do we show that (1 - 0.999...) < 1/n? Well, let's say we choose n=10000, so 1/n = 0.00001. Now, we know that (1 - 0.99999) = 0.00001, but 0.999.... > 0.99999 of course, so naturally (1 - 0.999...) < 0.00001. The same goes for any n you can choose.

QED?

[Nem]

Isn’t the whole 0.9... thing extending to an infinitesimally small number?

[Tamir]

[Nem]

[Tamir]