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Proof of 1 Not equal 0.999...

Name: FrozenVoid 2007-03-01 8:54 ID:fPuNCwa3

0*infinity=(0+0+0+0...)=0 (x+0=x x=0)
1=0.999... (agreed) 1-0.999...=0
(type 1) 1-0.999...=(1/10)*(1/10)*(1/10)... (from division by 10, 1-0.9 etc)
ab=0 if a or b=0; 1/10=0 ;10*1/10=10*0 ;1=0 ;x*1=x*0 ;x=0
(type 2) (1/10)*(1/10)*(1/10) ...=(1/10)^infinity=1/infinity=0
1/infinity=0 then 0*infinity=1 bu 0*infinity=0 ;1=0 ;x*1=x*0; x=0
Every number equals zero .(by 1=0.999...)

Name: Anonymous 2007-03-04 20:27 ID:yhRxjWMt

I'm not >>40

>>37
if we take .999... to mean a decimal point followed by an "infinite" amount (perhaps it's better to say, uncountable amount) of nines afterwards then the answer to the question:
"what number is between .999... and 1?" is none. Or rather there exists no distinct number between .999... and 1 because they are the same number. Just like there are no numbers between 0 and .000... because they become the same number.

>>38
that number would actually be less than both 1 and .999... since both .999...0 and .999... have the "same" amount of digits (infinitly so for both). If one was subtract the former from the latter, one would get .000...9 which is greater than than the subtraction of 1.000...0 and .999...9 which yields .000... and therefore .999...0 is farther from one than .999... is. You would somehow have to create a number that has more digits than infinity in order to create a number between 1 and .999...

Name: Anonymous 2007-03-04 21:02 ID:McstwcYw

>>41
"(perhaps it's better to say, uncountable amount)"
No, it's not better, because uncountable means something different.

Name: Anonymous 2007-03-05 20:43 ID:Heaven

>>41
>If one was subtract the former from the latter, one would get .000...9

.000...9 is not a number. That expression makes no fucking sense. You can't write something *after* an infinite number of digits, there is no after, THERE IS NO END.

If one was to subtract the former from the latter, one would get ZERO, ZIP, NADA, SAGE THIS FUCKING THREAD DIE DIE DIE

Name: Anonymous 2007-03-05 21:14 ID:KqA9OQxe

In my very first class, on my very first day in university, the professor had everyone in the entire lecture hall chant in unison, "Infinity is not a number."

Best thing he ever taught us.


Also, I just noticed the Wikipedia page "Proof that 0.999... equals 1" has been merged into the 0.999... page. That's kindof boring. In any case, what with FrozenVoid liking Wikipedia so much, how has no one linked this yet?

http://en.wikipedia.org/wiki/0.999...
>In mathematics, the recurring decimal 0.999… , which is also written as 0.\bar{9} , 0.\dot{9} or \ 0.(9), denotes a real number. Notably, this number is equal to 1. In other words, "0.999…" represents the same number as the symbol "1".

Name: GMontag 2007-03-07 2:43 ID:tGho8tNT

>>42
>No, it's not better, because uncountable means something different.

Would 0.999... with aleph_null 9s be a different number than 0.999... with aleph_one 9s?

Name: Anonymous 2007-03-07 4:23 ID:ObWbNNpv

>>47
.999... with aleph-one 9s doesn't really make sense. Aleph-zero is the cardinality of N; it seems obvious that the number of digits in .999... is of similar size.

Name: Anonymous 2007-03-07 8:04 ID:Heaven

>>1
Please die.

Name: Anonymous 2007-03-07 18:16 ID:Heaven

>>49

In a fire.

Name: Anonymous 2007-03-08 13:03 ID:4jaD3hKR

>>47
Would 0.999... with aleph_null 9s be a different number than 0.999... with aleph_one 9s?

As >>48 said, there aren't aleph_one 9s. They're countable, look:

0.9 9 9 9 9 9 9 ...
---1 2 3 4 5 6 7 ...

There isn't a different number with aleph_one 9s or a way of writing it with aleph_one 9s; the decimal expansion is by definition just a sequence of numbers, hence any decimal expansion has countably many digits.

Name: Anonymous 2007-03-09 11:52 ID:4Rr10EF4

Let x=.99999..
Then 10x=9.999..
So 10x-x=(9.999..)-(.999..)
Then 9x=9
Thus, x=1.

GG.

Name: Anonymous 2007-03-11 0:43 ID:eKtBbXdh

1/1 = 1.

1/.999... = 1.000...
10/9.999... = 1.000...

Therefore, 1 = .999...

There is no .000...0001 at the end of .999... because the nines go on infinitely.

Name: Anonymous 2007-03-14 4:17 ID:MsSYv7X+

you assume at the top that .999 = 1 its the only way your formula will work

Name: Anonymous 2007-03-14 8:59 ID:Wsn9dNli

>>52
thread over.

Name: Anonymous 2007-03-14 10:36 ID:o05goJni

If x is equal to .9999 as in step one then it is .9999

Name: Anonymous 2007-03-14 16:18 ID:PIQtEIZN

>>56
Yes, and 0.999... = 1.

Name: Anonymous 2007-03-14 21:36 ID:o05goJni

Chart .999... on a chart and youll never get to point 1. End.

Name: Anonymous 2007-03-15 2:05 ID:Heaven

>>58
How do you chart a single number? Because that's what 0.999... is. It isn't a fucking sequence you useless shithead.

Name: Anonymous 2007-03-15 22:28 ID:WPqjQ+gn

.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 to infinity will eventually (lol) reach the number 1.

didn't read any of the previous posts.

Name: Anonymous 2007-03-16 1:20 ID:Xot0Z6mX

>>62
Really? Ok, graph the sequence 3.01 please.

Name: Anonymous 2007-03-16 5:28 ID:wEakQ7kM

It's the same sequence as .999... (decimal expansion), but with a0 = 3, a2 = 1, and all other a's as zero. The limit of that sequence is 3.01000..., which is what you're looking for.

http://en.wikipedia.org/wiki/Decimal_expansion
First formula there.

Name: Anonymous 2007-03-16 7:21 ID:pkt+anID

Dealing with trolls kills braincells.

Save your braincells, and just move on.

Name: Anonymous 2007-03-16 12:37 ID:Heaven

>>64
Tell me, do the sequences {1/2, 3/4, 7/8, 15/16, ... } and {2/3, 8/9, 26/27, ... } approach the same value? If so, why do you think that the sequences {0.9, 0.99, 0.999, ... } and {1.0, 1.00, 1.000, ...} do not? The article you linked specifically states that a number is equal to an infinite sum of the form Sum[i=0,+inf](a_i/10^i). And, ironically, it specifically links the article about proving 0.999... = 1!

Name: Anonymous 2007-03-16 14:19 ID:Heaven

>>67
I don't. I know that. I have no idea why you thought I didn't.

Name: Anonymous 2007-03-16 17:09 ID:Heaven

>>68
>>69
K, so I reply to >>64 and get two responses from obviously different people. Please try to determine which of you morons I'm addressing before rushing to hit the reply button next time.

Name: Anonymous 2007-03-16 21:42 ID:wEakQ7kM

>>64
>>68
>>71
Same person.

>>70
Only >>69 is guilty of failing to understand whom >>67 was talking to, as well as of ultimate fail in the ways of math.

Name: Anonymous 2007-03-16 22:06 ID:Heaven

>>71
Rereading >>64 I can see you weren't really disagreeing with me, sorry. Nevertheless, you mostly missed my point. Although you can take the sequence of successive one-digit approximations to a number and graph it as a sequence of course, I was objecting to people saying "graph 0.999... and see that it approaches 1 but never equals it", primarily because this kind of statement is just a fancy way of hiding behind the "omg numbers are like, processes, dude!" that so many deniers use. Also because it's silly to say "graph a number as a sequence" when there are infinitely many sequences which converge to any number.

Name: Anonymous 2007-03-17 1:34 ID:Heaven

>>73
Sorry, you must be confused. In math, you have to actually have an argument, not just say "lol no way dude". GTFO.

Name: Anonymous 2007-03-17 12:17 ID:Heaven

>>75
Again, argument? You don't get to just declare yourself correct.

Name: Anonymous 2007-03-17 14:35 ID:Heaven

>>77
Present an argument or GTFO.

Name: Anonymous 2007-03-17 20:11 ID:Heaven

>>79
Ah good, thanks for clearing that up. Now I'll just use my psychic powers to figure out which argument is yours.

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