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2004-12-17 23:54:20
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0.999... = 1

?

Yes

Why?

See below.




x = 0.999....

10x = 9.999...

10x - x = 9

10x - x = 9x

9 = 9x

1 = x

1 = 0.999...




Allow me to give an extra proof.

ormula:

I) 1+x+x^2+x^3.... = 1/(1-x)

if the absolute value of x is smaller than 1.

We can use this also as:

II) x+x^2+x^3.... = 1/(1-x) - 1


in our case:

0.999.... = 9/10 + 9/10^2 + 9/10^3 + .... =
= 9*(1/10 + 1/10^2 + 1/10^3 + ...) =
= 9*((1/10) + (1/10)^2 + (1/10)^3 + ...) =
(according to the formula II)
= 9*{1/[1-(1/10)] - 1} =
= 9*{1/[9/10]-1} =
= 9*{10/9 - 1} =
9*{1/9} = 1




As you probably know I didn't created this one myself. :p
Copy/past all the way.

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2004-12-16 [Aradon Templar]: The problem lies in the second step. Assuming that .999... went on forever, when you multiply it by ten, there's a zero after 'forever'. 9.999...0 . so 10x doesn't quite equal 9.999..., and thus 10x-x doesn't quite equal 9...

2004-12-16 [Kayne]: That is the "correction" I get a lot. Z zero at the end of a 0.9999... number - doesn't count. But that doesn't matter anyway cause there can't be a number after forever.

2004-12-16 [Aradon Templar]: thus you can't really multiply it by ten... :\

2004-12-16 [Aradon Templar]: And yes, I know all abouts how it probably doesn't apply, its just the explanation for why there is a valid arguement against .9... equalling 1.

2004-12-16 [Kayne]: Why can't you mulitiply by ten? Cause it is an infinte number?

2004-12-16 [Aradon Templar]: cause you can't add a zero at the end :P If you could multiply things like that, I could prove that 1=.9... in other fashions... like, 1/3 = .333..., then mulitply by three, 1=.999.... Doesn't work...

2004-12-16 [Kayne]: Actually - you gave another proof for the fact that 1 = 0.9999... :p. I don't like that one really but it is correct too. You don't need to add a zero so there is no need to add one. :/

2004-12-16 [Aradon Templar]: no, actually, what it does, kayne, is that it states that 1/3 = .333... is incorrect. It is an approximation. Just as 10x (where x is .999...) = 9.999... is incorrect, and also an approximation.

2004-12-16 [Kayne]: No, actually 1/3 = 0.3333... is correct. You just need an infinite number of 3's and that is a bit hard to wright.

2004-12-16 [Aradon Templar]: No, becuase of the above proof. An infinite number of three's still doesn't quite reach a third.

2004-12-16 [Kayne]: Actually it does. But if you think it doesn't then you have a different opinion and therefore wrong.

2004-12-16 [Aradon Templar]: No, it obviously doesn't, because 1 (or 3 x 1/3) doesn't equal .999... (or 3 x .333...) Simply because there is obviously a number difference. 1 = 1 and ONLY 1, or a fraction in the occasion that the numerator = denomenator.

2004-12-16 [Kayne]: 0.999... = 1 Believe it then or not. I'm not having this discussion anymore today. I'm at the moment having a 400 and counting comments discussion about it. Not today anymore:p

2004-12-16 [Aradon Templar]: Alright. You know I'm right :P

2004-12-16 [Kayne]: No - Templar you are wrong. This is the second time in the history of elftown that I know that I'm right and you are wrong.

2004-12-16 [windowframe]: This is as bullshit as the 'all numbers are equal to 0' theory - it looks nice but it's fundamentally flawed, and bad maths. Deal. Or sulk - you seem to be doing that quite well, and I don't really see a reason to stop you.

2004-12-16 [windowframe]: Your maths is flawed kayne, Templar is absolutly right, 1/3 does not equal exactly o.33333 recurring. That's why good mathematicians prefer using 1/3 - because it's entirely more accurate than 0.333333 can ever be.

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