The Old 0.99...=1 - Politics Forum.org | PoFo

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By Vivisekt
#656209
I've seen it - in elementary school. What about it?
By Kytro
#656215
Well the number of people that claim it is false is rather high...
User avatar
By Vivisekt
#656219
I see it as true.

First, keep in mind that X={.99...} is an infinite set.

In order for X to be less than one, it would have to terminate before 1; there would be a number between the terminus of X and 1. Infinity, however, does not ever terminate - and thus it cannot be less than 1. The infinite number encompasses all possible divisions of the set inclusive (.9, 1], and thereby runs directly into 1.

Consider: 1/9 ={.11...}, 2/9={.22...}, but 9/9=1.

It only looks misleading because the decimal system is a whore. That's all. Point nine repeating is a decimal representation of 9/9, eg, 1.
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By Truth-a-naut
#656310
Oh g-d. Another forum I frequent has this come up a lot, the threads explode into 40 page monsters.
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By Potemkin
#656351
Oh g-d. Another forum I frequent has this come up a lot, the threads explode into 40 page monsters.

I don't see why. Vivisect has given a very good demonstration of why 0.99999 recurring equals 1. Another way of thinking of it is to say that if 0.9999 recurring is not equal to 1, then there must be some finite quantity by which it is less than 1. In other words, 1-x=0.9999 recurring. What is x? Clearly, as you add '9's to the end of 0.9999 recurring, x becomes infinitesimally small, and as you approach the limit, it becomes zero. So x=0, and 1=0.99999 recurring. QED.
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By Yeddi
#656423
infinitesimally small, and as you approach the limit, it becomes zero.


If it is indeed infinite then it will never reach zero, it's what they call an asymptote... from memory.

Of course I'm shit at maths, but my understanding is that .999etc. doesn't equal 1 just like 99 doesn't equal 100
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By Potemkin
#656465
If it is indeed infinite then it will never reach zero, it's what they call an asymptote... from memory.

But you can take the limit. It may never 'reach' the limit (whatever that means), but we know how it behaves as it approaches the limit. If we couldn't 'take the limit', then most of maths just wouldn't work.

Of course I'm shit at maths, but my understanding is that .999etc. doesn't equal 1 just like 99 doesn't equal 100

False. (100-99)/100 =/= 1-0.9999etc.

When you write 0.9999etc, what are you doing? You are taking the limit as you add an infinite number of 9s to the end of that. Until the limit is reached, you don't have 0.9999etc, you have something less than it. And in that limit, x in the equation 1-x=0.9999etc is zero.

But my question remains, if 1-x=0.9999etc, then what is x? Answer me that and I'll be willing to consider your argument that 0.99999etc =/= 1.
User avatar
By Vivisekt
#656469
Yeddi - the repeating nine in our example, though infinite, remains a symptom of division and this has a limit.

Say you want to go down to the corner store and buy a Pepsi. You think to yourself: since the store is 10 miles away, 5 miles is the halfway point between your body and that refreshing filthwater of capitalism. You conclude that you will invariably cross that halfway mark before you reach the store. Seems logical enough on the surface.

So, you start driving down the street.

Sure enough, you pass the 5 mile halfway point on the way to the store. You're 5 miles into the trip, and you think to yourself that the halfway point between you and the store is now 2.5 miles. If your earlier logic holds, you posit, then you must also cross this halfway point before you get there. As you continue to drive down the street (you haven't stopped moving, remember), you eventually travel another 2.5 miles and are now 2.5 miles from the store. You realize, at this point, that the new halfway point between you and the store is 1.25 miles. As you continue traveling, another 1.25 miles goes by. The halfway point between you and the store is now .625 miles. Continuing on, once you've traveled an additional .625 miles, the new halfway point between you and the store is .3125 miles...

By your logic, applied to our {.99...}, you would never get to the store. You'd get really close, but for all of eternity you'd just fall short of it because you've always got to cross that halfway point between you and the store before you get there.

This, as we know from the fact that we can move at all, is false. You'll get there.

9/9 = {.99...} = 1

What you're thinking of is the philosophical concept of an infinite number - which, really, isn't math. Calculus doesn't work that way.

;)
User avatar
By Yeddi
#656482
What you're thinking of is the philosophical concept of an infinite number - which, really, isn't math. Calculus doesn't work that way.


Like I said Viv, i'm not a maths guy. Right Lads, I'll take your words for it and inform all my friends n such.
But before I go I must clear this up
False. (100-99)/100 =/= 1-0.9999etc.


I was not suggesting that they were equal, i was suggesting that just as 99 cannot be equal to 100, .99~ also would not equal 1.
But as Viv assures me, Calculus doesn't work that way.

...is the halfway point between your body and that refreshing filthwater of capitalism. You conclude that you will invariably cross that halfway mark before you reach the store...


Xeno's Paradox, how I've missed thee, but you have a new refreshing twist that i like and respect :up:
By Saf
#656504
x=.999
10x=9.99

10(.999...) - .999... = 9
9(.999...) = 9
8.999999999999... = 9

This shit is bananas.
User avatar
By Yeddi
#656508
Okay, i've look a bit more around the interweb and apparently it's because decimals are only ever approx. But of course I could be wrong here too, the internet is a big scary place! But it would certainly make sense.
User avatar
By Vivisekt
#656513
Saf The Pope wrote:This shit is bananas.

{.9999999...} - 1 = {.0000000...}

;)
By Russkie
#656542
{9(.99) = 8.91} not {9}
This proof is retarded...
I already proved it's wrong...
User avatar
By Todd D.
#656633
Of course I'm shit at maths, but my understanding is that .999etc. doesn't equal 1 just like 99 doesn't equal 100

Think about it:
1/3 = .33333...
2/3 = .66666...

1/3 + 2/3 = 3/3 = 1

Therefore .3333 + .6666 = .99999 = 1
User avatar
By Vivisekt
#656643
Russkie wrote:{9(.99) = 8.91} not {9}
This proof is retarded...
I already proved it's wrong...

Uh. No.

How is it that you don't even know what a repeating decimal is? Have you never taken a basic arithmetic class in your life, or something? I mean, I can totally understand how a person might have no idea how the logic articulates in Calculus, but jesus christ...

You're also horribly misusing set notation in your little 'proof' there.



edits: wtf! typos!
Last edited by Vivisekt on 07 Jun 2005 19:21, edited 2 times in total.
By Saf
#656673
I think that's the best way to put it, Todd. Quite clear.
User avatar
By enLight
#656718
I'm no mathematician and I understand how 0.999... (i.e. 9/9) equals 1 but I have a question:

1 - 0.1111111... = ?

Because look, that would equal 0.999... which is 1. So you could say that:

1 - 0.111... = 1

Except 0.1111... is the same as 1/9 and then we run into this problem:

9/9 - 1/9 = 8/9

8/9 is .8888... and that surely does not equal 1.

Yeah...it's fucked up. :hmm:
User avatar
By Vivisekt
#656720
enLight wrote:I understand how 0.999... (i.e. 9/9) equals 1 but I have a question:

1 - 0.1111111... = ?

Because look, that would equal 0.999... which is 1. So you could say that:

1 - 0.111... = 1

Not quite. When you subtract the repeating point one from 1, you're just taking {0.11...} from {0.99...} to arrive at X.

Ergo X={0.88...} = 8/9
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