Aspiring Novice
20051015 09:10:14 
0.9999.... = 1?
if 1/3 = 0.3333~
and 1/3 + 1/3 + 1/3 = 3/3 = 1
and 0.3~ + 0.3~ + 0.3~ = 0.9~
does 0.9~ = 1 ???
note: 0.n~ represents infintly repeating (no rounding). 
Percy
20051015 14:23:14 
Are... the Percy number!
I used to think that maybe there was this number which was infinitely small, numerically I called it 0.000`...1, thats to say it has an infinite number of noughts between the dp and the one. Lets call this number the Percy number or P for short. Now better mathematicians than me assured me that in conventional maths this number does not exist! These ppl will tell you that 0.999`=1. I'm not denying the logic of conventional wisdom, but all maths is made up, i=sqrt1 is a good example but even the concept of oneness only exists in the human mind (and twoness etc). So with my new number we can say stuff like 1P= 0.999` and maybe even x/infinity=P.
Can anyone think of more uses for/equations with P?
(I have to go now and give my dad a map of Tryfan before he gets lost.) 
Charlie
20051015 15:27:08 
Re: 0.9999.... = 1?
x = .9999...(to infinity)...
10 x = 9.9999...(to infinity)...
Subtracting the top from the bottom equation:
9 x = 9
Dividing each side by 9:
x = 1 
Gamer
20051015 16:08:19 
Re: 0.9999.... = 1?
This problem is already at http://perplexus.info/show.php?pid=86 so posting on it would just repeat the comments said there. 
Aspiring Novice
20051018 06:23:14 
Re: 0.9999.... = 1?
thanks guys. i always wondered about that. :) 
tanx
20051025 16:21:48 
Re: 0.9999.... = 1?
LOL  the percy number
x = 0.999... + P
10(x) = 9.999... + 10(P)
9(x) = 9 + 9(P)
9(x) = 9(1 + P)
x = 1 + P
so are you saying percy that 0.999... > 1 ???

Mindy Rodriguez
20051125 21:13:37 
Re: 0.9999.... = 1?
No, he is saying that P = 10P = 0.
So, 0.99999~ is equal to 1.
(plus or minus P) 
Federico Kereki
20051129 22:51:45 
Tanx
Tanx: your first and last equations show that 0.999...=1, even though the value of X rests unknown! 
JayDeeKay
20060609 16:20:49 
Re: 0.9999.... = 1?
Maybe I'm missing something profound here, but the proof that 0.999.. = 1 does not require any multiplication or addition of infinite decimals (shades of Euler!). It follows rigorously from the *definition* of that infinite decimal fraction:
0.9999...9 = 9/10 + 9/100 + 9/1000 +...+ 9/10^n [there are n 9s]
= 9/10 * (1  (1/10)^n) / (1  1/10) [sum of a geometric series]
= 1 * (1  (1/10)^n)
> 1 as n > infinity.
Voila! 
Gamer
20060612 19:47:46 
Re: 0.9999.... = 1?
Yep. It approaches 1, the same way 1/n approaches infinity as n goes to 0. Note that it never gets there though  that is the problem. 
Waldo Pepper
20060727 15:02:29 
Re: 0.9999.... = 1?
This is how I learned it in calculus:
.999 down to infinity can be written as
9/10 + 9/100 + 9/1000 + ...+ 9/(10^n) now we set this equal to S. So
9/10 + 9/100 + 9/1000 + ...+ 9/(10^n) = S Note that every term is 1/10 the previous one. So we multiply the above by 1/10 and get 9/100 + 9/1000 + 9/10000 + ...
+ 9/(10^(n+1)) = 1/10 S. Then you subtract the two equations and get
9/10  9/(10^(n+1)) = S  1/10 S = 9/10 S. A 9/10 cancels from both sides of the equation and we get 1  1/(10^n) = S. Now as n goes to infinity, 1/(10^n) is going to go to zero and we get 1 = S. 