Aspiring Novice
2005-10-15 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
2005-10-15 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=sqrt-1 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 1-P= 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
2005-10-15 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
2005-10-15 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
2005-10-18 06:23:14 |
Re: 0.9999.... = 1?
thanks guys. i always wondered about that. :) |
tanx
2005-10-25 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
2005-11-25 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
2005-11-29 22:51:45 |
Tanx
Tanx: your first and last equations show that 0.999...=1, even though the value of X rests unknown! |
JayDeeKay
2006-06-09 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
2006-06-12 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
2006-07-27 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. |
