Oh, please show proof one way or another.

[ 06-11-2004: Message edited by: TheKiller ]

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1 divided by 3 is .3333 repeating
one third times 3 is 1
.3333 repeating times 3 is .9999

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Also, 1 - 0.9999(...) = 0.0000(...)1 = 0

However, this is very creative math, since normal math doesn't really accept infinite numbers in any form.
IIRC, the "correct" way to look at this is that the fact that 1/3*3 appears to not be 1 is actually just a rounding error, not some proof that 1=0.9999(...)

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If x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1

quote:

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RaveyBoy said:
Also, 1 - 0.9999(...) = 0.0000(...)1 = 0

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I like the 0.0000(...)1 part. The one left over after the result of the subtract is after infinity zeros.

Did anyone see the April Fools jokes that Blizzard had posted?

[ 06-11-2004: Message edited by: TheKiller ]

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One such proof starts with the definition that for a number to be smaller than another number there needs to be a number that you can add to the supposed smaller number to make a sum equal to the supposed larger number. There is no number that can be added to .99999... to make a sum of 1, therefore .99999... is equal to 1.

You can also use series to prove it. The number .99999... can be writen as the infinite geometric series:

(sum from n=1 to infinity) of the general term (.9)*(.1)^(n-1)

By the definition of an infinite geometric series, the sum of this series is:

(.9)/(1-(.1))

Now we simplify the expression:

(.9)/(.9) = 1

Therefore, .99999... is equal to 1.

Another proof is the one Tea Bagger posted.

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It's the same and it is different. I believe that because of this paradox of numbers, there is no surface understanding of it, but a deeper clarity that we have of numbers to allow us to define and use them no matter how abstract they are.

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1 = 1 + 0

1 = 1 + ( - 1 + 1)

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

1 = 1 + ( - 1 + 1) + ( - 1 + 1)

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

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

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

1 = 0

yeah... uh... infinite series are something to be really careful about.

        n
lim     S(9/(10^j)
n->¥   j=1

[ 06-12-2004: Message edited by: WillyTrombone ]

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[ 06-12-2004: Message edited by: FS ]

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Originally posted by FS:
Wow, I can't believe I'm agreeing with FS on this one

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quote:

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

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

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You can't do that!
There are n*"-1", and (n+1)*"+1"

I'm aware that many (most?) claim that all infinite sets have the same size, but I'm going to have to call bullsquid on that one.

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The infinite series WillyTrombone posted is a good example. That is a conditionally convergent alternating series. It has been proven that you can rearrange the terms of a conditionally covergent series to add up to any number you want. It's some crazy stuff.

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I'm aware that many (most?) claim that all infinite sets have the same size, but I'm going to have to call bullsquid on that one.

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I certainly wouldn't say most. I haven't taken any classes on set theory but I recall each of my calculus professors specifically mentioning that there are different infinities. (for instance, if you have the set of all positive integers, that would be less infinite than the set of all integers. And if you had an infinite set of the set of all integers, that's like a whole different degree of infinitude.)

as for the original question in this thread, in terms of logic and practice, .99999999... = 3/3 = 1. Rigorously speaking, it is actualy infintessimally less than one and therefore not equal to one. The distinction is one that is not really useful except to mathemeticians.

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Originally posted by WillyTrombone:
Rigorously speaking, it is actualy infintessimally less than one and therefore not equal to one. The distinction is one that is not really useful except to mathemeticians.

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There again is our good friend infinity. "Infintessimally less than" means they are equal.

By definition, an infinitesimally small amount is one that approaches 0 as a limit. Therefore the difference is 0. Therefore .99999... = 1

You cannot have an infinitesimally small amount as a difference because that is not a number. If your function cannot be evaluated at the point (infinity) then you must take the limit as the function approaches that point.

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By definition, an infinitesimally small amount is one that approaches 0 as a limit

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Originally posted by Snag:
And it never reaches 0, therefore it is NOT equal to 0 for anything more than convenience and lack of significance.

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True, and .99999... is still equal to 1.

Try arguing your points to a math professor if you believe you are correct.

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Originally posted by The Jakeman:
Conventional reasoning goes out the window when you are dealing with any flavor of infinity, so don't try.

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0! = 1

The fact that .99999... equals 1 is not for convenience... it is a proven fact.

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I understand "how" it equals 1...just being an ass...as usual I suppose

1/9=.111_
2/9=.222_
3/9=.333_
4/9=.444_
5/9=.555_
6/9=.666_
7/9=.777_
8/9=.888_
9/9=1

Not the hardest of proof but it shows the series...

ed. for typo

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By definition, an infinitesimally small amount is one that approaches 0 as a limit. Therefore the difference is 0. Therefore .99999... = 1

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it's not equal. it approaches 1 as a limit at infinity. I think snag said it well. It may as well be regarded as one since any rounding will go up since the next digit will always be nine. That is why we need the limit. it is not to find out the value AT infinity, because such reasoning is not possible within the constraints of our universe. Therefore, we look at the limit as it approaches infinity, rather than the value at an infinite coordinate. Limits are not necessarily evaluations. Since there is no way to find the limit as x->oo+, there is no way to test for continuity, and no way to say that the limit is the value.

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Originally posted by WillyTrombone:
it's not equal...

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Tell that to a Math professor.

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Originally posted by WillyTrombone:
...it approaches 1 as a limit at infinity. I think snag said it well. It may as well be regarded as one since any rounding will go up since the next digit will always be nine. That is why we need the limit. it is not to find out the value AT infinity, because such reasoning is not possible within the constraints of our universe. Therefore, we look at the limit as it approaches infinity, rather than the value at an infinite coordinate...

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That is exactly what I said in the rest of my post that you excluded in your quote of me.

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Originally posted by WillyTrombone:
Limits are not necessarily evaluations.

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True. But in this case it is an evaluation. Look at the problem in the context of infinite series. You use a limit to evaluate the sum of the infinite series. It is a real evaluation to get a real sum.

quote:

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Originally posted by WillyTrombone:
Since there is no way to find the limit as x->oo+, there is no way to test for continuity, and no way to say that the limit is the value.

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I'm not sure what continuity has to do with this, but you are correct when you say you cannot take the right hand limit at infinity. Infinity is not a number to begin with so there is no right or left handedness to it.

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Originally posted by Cacophonous:
Snag no offense but I take it that math is not your best subject. =)

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If you say it is equal to 1, the math professor will laugh at you. Just as Snag said: "only for convenience and lack of significance" it is equal to 1. I remember the hell of working with Limits and it taught me one thing, just because a number is so damned close to a counterpart, don't assume they are the same thing. Of course you can for convenience, but also if you feel like looking like an idiot.

It's funny though because in all of my advanced math classes, I was told to not expect .99999~ = 1 because in laymens terms:

Oh yeah and .999999~ = 1

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1 = Sqrt(1)=sqrt(-1 * -1)=sqrt(-1) * sqrt(-1) = -1
=)

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