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.999999999999... = 1 ?


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#81 Overlord

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Posted 31 August 2004 - 05:11 PM

it's about what huhn said, .5*((.99999...)+(.99999..)) = .5 * (2 times the same number) so this is equal to the number itself (.999...)

and in orwell's argument, he says "there is allways a third one C between A and B, strictly different from them"

.9999... (the one from huhn's sum) is the same as .9999... (the number that he tries to prove that it's not equal to 1), so it's not "strictly different"
so the " You cracked their "logic" explanation" is not true

#82 Orwell

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Posted 31 August 2004 - 05:23 PM

and no, its not enough untill I here an explenation that makes sense.

<{POST_SNAPBACK}>

We are the only ones here bringing many arguments and demonstrations to show that 0.999...=1, and you, the only arguments you have are "it is clear that...", "it's very simple", "it is not logic" etc, but you never found something really consistant to convince us that 0.999!=1 :o

#83 huhn_m

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Posted 31 August 2004 - 06:56 PM

so

0.899999... is also equal to 0.9
because their is no number in between ... this is what you are saying

In fact what you say is that there is no number with an inifinite number
of nines afterwards because thei are always "equal" to the larger one.

0.199999.. = 0.2
0.299999.. = 0.3

and so on.

And you still have not proven the thing with the limit.
After your terms

lim x->inf. of 1/sqrt(x)

can also become 0 because it
approaches 0 indefinitely. Tell this your math teacher and he'll kill you.

So we have your "C between A and B" thesis
agains my "the limit does not mean equal" thesis.

So what ... they seem to exculde each other ...

#84 2072

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Posted 31 August 2004 - 07:18 PM

lets say x= .99999...
then

.5*(x + x) = .5*(2*x) = .5*2*(x) = .5*2*(.99999...)

note: I've also learned that .9999...=1 in math class at university

#85 huhn_m

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Posted 31 August 2004 - 07:26 PM

yeah ... you're right ... I noticed that this was senseless :)
But it still does not say why a limit should be equal to "equal"

#86 Orwell

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Posted 31 August 2004 - 07:40 PM

I have to admit that i don't really see the problem you are pointing out with the limits.. :unsure:

lim (x->inf) 1/sqrt(x) is actually exactly equal to 0, don't you agree with that? There is no problem to say a number is equal to a limit, because every number A is the limit for x-> inf of f( x ) = A for example...

#87 huhn_m

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Posted 31 August 2004 - 08:00 PM

yeah but the equation is never zero.

you can say:

lim x->inf of 1/sqrt(x) is 0 and
lim 0.999.. -> inf is 1

but never

1/sqrt(x) = 0 since this will never happen
0.9999... = 1 since this will never happen.

Don't you see there is a difference between a limit and an equation?
My teacher'd kill me if I forgot the limit sign!

#88 Overlord

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Posted 31 August 2004 - 08:05 PM

for me, 1/sqrt(x) = 0 has no real solution, but if you include +inf in the subset of acceptable solutions, you can say that 1/sqrt(+inf) = 0.
at the same time, you can say (either you include +inf or not) that lim 1/sqrt(x) | x->+inf = 0

there is a little thing i don't understand, you can always write 1/sqrt(x)=0 or 1/x=0 or 0x=48971, these are valid equations, only with no real answers :huh:

what do you mean by "the equation is never zero" ? an equation is never equal to zero, it's not a number :huh:

#89 huhn_m

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Posted 01 September 2004 - 05:47 AM

1) I ment the left side never equals zero.
2) Of course every equation has a solution as soon as
you assume sqr(i)=-1 ... but this is no longer a real number.
3) Inf. actually is no number. YOu can't calc with infinity.
So calculate infinity-(infinity/2)+5 what is the result?
Inifnity/2+5? You can't calculate with this and even in infinity
the aclculaton gets NEVER 1 since there are only more 9s appended
... unless you loose exactness.

#90 CrimsonCasio

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Posted 01 September 2004 - 05:59 AM

again, my thoughts exactly. :D
keep it up huhn ;)

#91 Orwell

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Posted 01 September 2004 - 11:21 AM

2) Of course every equation has a solution as soon as
you assume sqr(i)=-1 ... but this is no longer a real number.

<{POST_SNAPBACK}>

My calc doesn't want to find the solution for the equation "1+2=0", even in complex mode... :o Can you help me? :lol:

what you have written

lim 0.999.. -> inf is 1

is uncorrect. 0.999.. cannot tend to inf, 'cause it is not a real variable like x.

but you can write lim (x->inf) (1-10^(-x)) = 1 , and of course i agree with that ;)
but still, the sequence (1-10^(-x)) can be written { 0, 0.9, 0.99, 0.999, ... } and thus, for x->inf, the limit is 0.999... .
But we already know that lim (x->inf) (1-10^(-x)) = 1 ... So? :rolleyes:

#92 huhn_m

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Posted 01 September 2004 - 12:19 PM

My calc doesn't want to find the solution for the equation "1+2=0", even in complex mode...  :o Can you help me?  :lol:


Well ... don't know ... try dropping your calc to the floor at least 10 times.
Maybe it works then B) :thumbsup:

No honestly ... it was my mistake :)

is uncorrect. 0.999.. cannot tend to inf, 'cause it is not a real variable like x.

but you can write lim (x->inf) (1-10^(-x)) = 1 , and of course i agree with that  ;)
but still, the sequence (1-10^(-x)) can be written { 0, 0.9, 0.99, 0.999, ... } and thus, for x->inf, the limit is 0.999... .
But we already know that lim (x->inf) (1-10^(-x)) = 1 ... So?  :rolleyes:

<{POST_SNAPBACK}>


?!?!? So what have you prooven ... I think you just prooved that I'm, right haven't you
... I must confess that I didn't really understand you :)

#93 Orwell

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Posted 01 September 2004 - 06:44 PM

we know that

lim (x->inf) (1-10^(-x)) = 1 - 10^(-inf) = 1

However, we see that

lim (x->inf) (1-10^(-x)) = 0 + 0.9 + 0.09 + 0.009 + ... = 0.999...

And, because "if a limit exists then it is unique", we have directly 1 = 0.999... :ph34r:

#94 R00KIE

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Posted 02 September 2004 - 12:34 AM

@Orwell

when you say that you can?t find a number between 1 and 0.999(9) just because 0.999(9) has an infinite number of 9 then you are wrong, you just need to add another number at the end of it that's the beauty of infinity it never ends :crazy:

#95 Daruosh

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Posted 02 September 2004 - 12:39 AM

As you can see in the Poll Results, there two groups (like math scientists).
If you can prove that 0.99999... is equal to 1, someone else can prove that .9999... is not equal.
But it's very valuable to discuss about it. This will imrove our knowledge ;)

#96 Overlord

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Posted 02 September 2004 - 11:25 AM

"you just need to add another number at the end of it that's the beauty of infinity it never ends"

if you "add" a number < 9, your number becomes < 0.9999... but it must be 0.9999... < x < 1
if you "add" only number = 9, your number = 0.9999... but it must be 0.9999... < x < 1

#97 Webness

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Posted 03 September 2004 - 10:46 AM

"equal to 0" + "equal to 1" = 1 :thumbsup:
I solved it.. *lol*

#98 huhn_m

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Posted 04 September 2004 - 12:10 PM

And, because "if a limit exists then it is unique", we have directly 1 = 0.999...  :ph34r:

<{POST_SNAPBACK}>


Huh?? I don't really understand the sentence

if a limit exists then it is unique

It is not ... there are many equations that have the limit 1 .... or other limits :ph34r:

#99 Overlord

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Posted 04 September 2004 - 01:32 PM

i don't understand, can you give an example ?

#100 Orwell

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Posted 04 September 2004 - 07:53 PM

if a limit exists then it is unique

This is part of the definition of a "limit".
A certain sequence can't have two different limit at the same point. Take 1/x for example:
lim (x->0 pos) 1/x = inf
lim (x->0 neg) 1/x = - inf
so we say that lim (x->0) 1/x doesn't exist :)

In the case of lim (x->inf) (1-10^(-x)), we already know it is equal to 1, so this limit exists, but we can see that it is also equal to 0.999... , so the condtion for the existence of this limit is that 1 = 0.9999... ;)

#101 huhn_m

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Posted 05 September 2004 - 11:34 AM

yeah ... I misunderstood the statement ... for
me it sounded like no other equation can have this limit
what would be senseless.

But in this case you're right.

Actually your arguments are in the mathematical sense quite
good but still, from the point of a logic thinking man I will NEVER
say the a number equals another if they are not really the same

(e.g. String 1 = String 2).

#102 Orwell

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Posted 05 September 2004 - 11:42 AM

So , 1/3 is not equal to 0.3333... ?

#103 huhn_m

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Posted 05 September 2004 - 05:57 PM

yeah. got me there but I think you know what I mean don't you ...

#104 2072

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Posted 05 September 2004 - 06:36 PM

let's say 1/1=0.9999.... :P

#105 huhn_m

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Posted 06 September 2004 - 05:52 AM

now this was a good one ... 2072 is right :) .. accordign to your theory a
equation can be turned upside down (don'T take my by the word :) )

#106 genesis

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Posted 06 September 2004 - 11:45 AM

This reminds me of an arguement I had with a friend that:

1 / infinity =/= 0 (similar to the 0.000...1 =/= 0 statement)
The reason can be explained using chance, if there are infinite paths to take and one of them is correct, there IS A CHANCE that that path will be chosen, so the probability is NOT 0. Maybe the same arguement can be applied to your question.

As for the mathematical proofs, skeptics should respect the 'clarity' (is that the word?) of their logic. The same method is used to convert recurring decimals into fractions. But then again, there is always uncertainty to everything...

NOTE: These proofs are CORRECT, not like the (although ingenius) proofs that 1=2 (using the trivial division by 0) and 1=-1 (using the incorrect formula sqrt X / Y = sqrt X / sqrt Y). There is no error involved. Using gut instincts that 0.999... does not look like 1 is not comparable to respectability of calculus.

#107 Orwell

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Posted 06 September 2004 - 12:00 PM

1 / infinity =/= 0 (similar to the 0.000...1 =/= 0 statement)
The reason can be explained using chance, if there are infinite paths to take and one of them is correct, there IS A CHANCE that that path will be chosen, so the probability is NOT 0.

<{POST_SNAPBACK}>

"infinite paths"? "correct path"? I don't understand it, can you explain? :huh:

#108 genesis

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Posted 06 September 2004 - 12:03 PM

sorry, i didnt want to be too specific... for example, an infinite-sided die and the number 1 on one of the sides being what you want.

#109 Orwell

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Posted 06 September 2004 - 12:04 PM

Okay, but you assume there is a the number 1 on one of the sides... How can you se that?

#110 genesis

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Posted 06 September 2004 - 02:17 PM

They are numbered one to infinite, it really doesn't matter! What matters is the chance of picking a side and getting it is NOT zero. The reason I don't like to use the dice analogy is the prospect of an infinite-sided shape. I prefer, say, infinite pebbles.

#111 Orwell

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Posted 06 September 2004 - 02:52 PM

OK again, but what if there is no side with the '1' value :)

#112 huhn_m

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Posted 07 September 2004 - 01:44 PM

huh? There is one. It's the same as with the
0.0...1 there is a one at the end
And with the labyrinth with endless ways ... there IS one that is valid.

Your question does not fit orwell. If genesis says "they are numbered 1 to infinite" you
can't say "but what if there is no side with a one".

This is as if I'd say. 0.9999... = 1 : but what if there is no chiffre 9?

#113 genesis

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Posted 07 September 2004 - 05:20 PM

Good idea with the labyrinth :) but now I've contradicted myself!

0.999... = 1 - 0.000...1 and 0.000...1 =/= 0, so 0.999... =/= 1. However, I do believe 0.999... = 1!

Although, you could argue that 0.000...1 + 0.999... =/= 1 because if there are infinite digits, you can't be certain that the numbers have the right place values for addition. This way, both theories can be true.

#114 huhn_m

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Posted 07 September 2004 - 07:07 PM

no but this can also applied for the other theorie since
there are endless ways and all but one (the missing 1 at the end to get one)
are the wrong ones. So you must pick the one so there is still a chance you pick it.

1 Chances, 0.0...1 right ways, 0,999.. wrong ways.
^ insert as many 0s as 9s

#115 Orwell

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Posted 07 September 2004 - 11:14 PM

Er... Actually for me 0.000...1 is equal to 0... :)

Anyway, if we are speaking about limits here, don't forget a limit can't have 2 different valors... So if there exist at least one "correct way", and a least one "wrong way" (to re-use genesis' words), then the limit doesn't exist, it is as simple as that ;)

#116 bcgsr

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Posted 08 September 2004 - 11:00 AM

wow, i like this post... just came across it :)

when you happen to have a result you all can agree on i thought you might discuss if

1,999999...=2


;) Have fun!

#117 Clive_88

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Posted 20 September 2004 - 08:51 PM

It's very simple. The fraction 1/3 is equal to 0.333333333333 (0.3 recurring). Everyone knows that three thirds (3/3) is equal to 1. Therefore:
3 x 0.33333333333333 = 0.99999999999999 = 1

:plol:

#118 CrimsonCasio

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Posted 20 September 2004 - 09:20 PM

yeah, i asked my math teacher about this (he's going on his second phd in math...), at first he thought you were wrong but then he tried to prove it and came up with the exact oppisite. i'll get him to write down what he did, it was diffrent than everything else posted.

#119 Marco

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Posted 01 October 2004 - 05:11 PM

Why so difficult? 0.999... with unlimited digits "9" definitively is equal to one.

Let?s turn it around and subtract both from 1:

I: 1-1 = 0 (true)
II: 1-0.999... = 0 (true or not??)

II is true, because:

a = 0.9; 1-0.9 = 0.1 = b
a = 0.99; 1-0.99 = 0.01 = b

and so on. Thus, 1-a = b = 10^-n, where a has n digits "9".

And what does this mean for the digits of b? that it has n digits "0", and the last digit would be an "1" (example: 1-0.99 = b = 0.01 with a has n = 2 digits "9" and b has n = 2 digits "0" before its digit "1").

But if a has an unlimited number n of digits "9", b also will have an UNLIMITED NUMBER OF DIGITS "0". If b has an unlimited number of digits, it can?t have a LAST digit, thus it can?t contain "1" anymore, as the count of digits would not be infinitiv then.

That?s why 1-0.999... = 0!

From I and II follows: 1 = 0.999...

#120 Bija

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Posted 01 October 2004 - 06:04 PM

i agree with marco

and i really think that 0.99999...=1




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